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http://mathcentral.uregina.ca/QQ/database/QQ.09.10/h/sarah3.html	SEARCH HOME Math Central Quandaries & Queries Question from Sarah, a student: what is pi + pi? Hi Sarah, $\pi + \pi$ is the same as $2 \times \pi.$ $\pi$ is approximately 3.1416 so $\pi + \pi$ is approximately 6.2832. I hope this helps, Penny Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.	2017-10-22 22:59:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7000290155410767, "perplexity": 1899.8636192885224}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187825473.61/warc/CC-MAIN-20171022222745-20171023002745-00024.warc.gz"}
https://bigideasmathanswer.com/big-ideas-math-answers-grade-4-chapter-13/	# Big Ideas Math Answers Grade 4 Chapter 13 Identify and Draw Lines and Angles ## Big Ideas 4th Grade Chapter 13 Identify and Draw Lines and Angles Math Book Answer Key Lesson: 1 Points, Lines and Rays Lesson: 2 Identify and Draw Angles Lesson: 3 Identify Parallel and Perpendicular Lines Lesson: 4 Understand Degrees Lesson: 5 Find Angle Measures Lesson: 6 Measure and Draw Angles Lesson: 8 Find Unknown Angle Measures ### Lesson 13.1 Points, Lines and Rays Explore and Grow Use a straightedge to connect the dots A through Z. Describe the picture you make. How many points do you connect? How many line segments do you make? Points connected are 25. Line segments used are 25. Structure Draw your own connect-the-dots picture on another sheet of paper. Have your partner use a straightedge to connect the dots to make your picture. How many points did your partner connect? How many line segments did your partner make? Yes, he used line segments to connect the dots. He connected 20 dots and 19 line segments. Think and Grow: Points, Lines, Line Segments, and Rays Example Draw and label $$\overline{L M}$$ $$\overline{L M}$$ is a ___ Another name for $$\overline{L M}$$ is ___ Line segments are represented by a single overbar with no arrowheads over the letters representing the two endpoints. The line segment Represented as $$\overline{L M}$$ Example Draw and label $$\overrightarrow{S T}$$ $$\overrightarrow{S T}$$ is a ___ $$\overrightarrow{S T}$$ Ray $$\overrightarrow{S T}$$ Show and Grow Question 1. Name the figure shown. Write how to say the name. It is a line segment vu. Line segments are represented by a single overbar with no arrowheads over the letters representing the two endpoints. It is represented as $$\overline{V U}$$ Question 2. Draw and label two points P and Q on the line shown. Question 3. Apply and Grow: Practice Name the figure shown. Write how to say the name. Question 4. It is called as point B. A point in geometry is a location. It has no size i.e. no width, no length and no depth. A point is shown by a dot and named in capital letter. Question 5. It is called as $$\overrightarrow{C D}$$. The symbol of a ray is →. Question 6. Draw and label the figure. Question 7. $$\overline{G H}$$ $$\overline{G H}$$ means a line segment. It is said as line segment GH. Question 8. It is a ray. It is said as ray Question 9. $$\overrightarrow{L M}$$ It means a ray. It is said as ray LM. Use the figure. Question 10. Name a ray. $$\overrightarrow{A D}$$. Question 11. Name a point that lies on two lines. Question 12. Name two different line segments. $$\overline{D E}$$ and $$\overline{E F}$$ Question 13. Logic Your friend says he can draw two line segments between two points, and may be even more. His drawing is shown. Explain why this is not possible. Only one line can be drawn passing through any two points but a number of lines can be drawn through a point. Think and Grow: Modeling Real Life Example There are direct ferry routes between each pair of cities on the map. Draw line segments to represent all of the possible ferry routes. How many ferry routes did you draw in all? Start at Poole. Draw a line segment from Poole to each of the other cities. Repeat this process until a route is shown between each city. You draw __ ferry routes in all. 3 Possible routes. Show and Grow Question 14. There are direct flights between each pair of cities on the map. Draw line segments to represent all of the possible flight routes. How many flights routes did you draw in all? Total 10 routes were found. Question 15. Which road signs contain a figure that looks like a ray? One way road sign look like a ray. Question 16. Which letters in the banner can be made by drawing line segments? Explain. , Letters – W,E,L,M AND E. ### Points, Lines and Rays Homework & Practice 13.1 Name the figure shown. Write how to say the name. Question 1. It is said as Line segment EF and represented as $$\overline{E F}$$. Question 2. It is said as ray GH and represented as $$\overrightarrow{G H}$$. Question 3. It is said as line JK and represented as Draw and label the figure. Question 4. two points L and M on the line shown Question 5. $$\overrightarrow{N O}$$ It is represented as $$\overrightarrow{N O}$$. Question 6. Use the figure. Question 7. Name a line segment. ___CE____ Question 8. Name two different rays. $$\overrightarrow{C D}$$ and $$\overrightarrow{E F}$$. Question 9. Name two different lines $$\overline{D B}$$and $$\overline{A E}$$. Question 10. Writing Explain the difference between a line and a line segment. Line-segment 1. It has two end points. 2. The length of a line-segment is definite. So, it can be measured. 3. The symbol of a line-segment is _____ Line 1. There are no end points in a line. 2. There are no end points. So, length of a line cannot be measured. 3. The symbol of a line is ↔ Question 11. Structure Name the figure in as many ways as possible. $$\overline{X Y}$$,$$\overline{Y Z}$$ and $$\overline{X Z}$$. $$\overrightarrow{Y Z }$$ LINE – XZ Question 12. Structure Draw and label a figure that has four points, two rays, and one line segment. $$\overrightarrow{R S }$$ and $$\overrightarrow{Q S}$$. $$\overline{P S}$$ DIG DEEPER! Write whether the statement is true or false. If false, explain. Question 13. A line segment is part of a line. _______ Question 14. A ray is part of a line segment. ____ No Explanation: Line-segment- It has two end points. But where as Ray has a starting point but no other end point. Question 15. There are an infinite number of points on a line. ___ Question 16. Modeling Real Life There are direct helicopter flights between each pair of resorts on the map. Draw line segments to represent all of the possible flight routes. How many flight routes did you draw in all? Total possible flight routes are 6 routes. Question 17. Modeling Real Life Which road signs contain a figure that looks like it is made of only line segments? Review & Refresh Compare Question 18. 0.15 < 0.16. Question 19. 2.4 < 2.42 Question 20. 6.90 = 6.9 ### Lesson 13.2 Identify and Draw Angles Explore and Grow Draw the hands of the clock to represent the given time. For each clock, describe the angle that is formed by the minute hand and the hour hand. Reasoning Explain how line segments, rays, and angles can be related. Think and Grow: Angles Angles can be either right, straight, acute, or obtuse. Example Write three names for the angle and classify it. Three names for the angle are ___, ____ and ____. The angle opens ___ a right angle and less than a straight line. So, it is an __ angle. Three names for the angle are actue, obtuse and right angles. The angle opens more than a right angle and less than a straight line. So, it is an obtuse angle. Show and Grow write a name for the angle and classify it. Question 1. It is a right angle called as P. Question 2. It is a straight angle. It can be represented as XYZ, ZYX. Apply and Grow: Practice Write a name for the angle and classify it. Question 4. It is a obtuse angle. It is represented as Q. Question 5. It is a actue angle. It is represented as MNO or ONM. Question 6. It is a straight angle. It can be represented as RST, TSR. Draw and label the angle. Question 7. ∠XYZ is right Question 8. ∠JKL is straight. Use the figure. Use three letters to name each angle. Question 9. Name an acute angle. ∠CFD Question 10. Name two different obtuse angles. ∠CFG and ∠ACF. Question 11. Name two different straight angles. ∠BDF and ∠CDE. Question 12. Name three different right angles. ∠CFD , ∠CFG and ∠BDF. Question 13. Structure Draw to complete each angle. Think and Grow: Modeling Real Life Example Which angle of the skateboard ramp is acute? You need to find an angle that is open less than a right angle. Angle A opens __more than __ a right angle and less than a straight line. Angle B _is equal__ a right angle. Angle C opens _less than__ a right angle. Angle _∠C__ is an acute angle. Show and Grow Question 14. Use three letters to name an angle of the wind turbine that is obtuse. ∠VTU. Question 15. Trace and label two right angles, two obtuse angles, and two acute angles in the painting. Question 16. DIG DEEPER! How many different angles are in the window? Name all of the different angles. Right angles – ∠AFC , ∠BFD , ∠CFE Obtuse angles – ∠AFD and ∠BFE. Acute angles – ∠AFB, ∠BFC , ∠CFD and ∠DFE. Straight angles – ∠AFE. ### Identify and Draw Angles Homework & Practice 13.2 Write a name for the angle and classify it. Question 1. ∠O is straight angle. Question 2. ∠TUV is a right angle as TU is perpendicular to UV . Question 3. ∠XYZ is a obtuse angle as, we can see clearly angle formed is more than 90 degrees. Draw and label the angle. Question 4. ∠ABC is obtuse. Question 5. ∠MNO is acute. Use the figure. Use three letters to name each angle. Question 6. Name an acute angle. Question 7. Name a straight angle. Question 8. Name two different right angles. Question 9. Name three different obtuse angles Question 10. YOU BE THE TEACHER The above figure contains ∠JKL a angle. Question 11. DIG DEEPER! Can you make a straight angle using an acute angle and an obtuse angle that share a common ray? Draw a picture to support your answer. Yes Explanation: The total sum of a straight angle =180. As per the figure, The Obtuse angle = 120 degrees. The acute angle = 60 degrees. Therefore The straight angle = Obtuse angle + Acute angle. = 120 + 60 = 180 degrees. Question 12. Structure Write a capital letter that has more than two right angles. Question 13. Modeling Real Life Use three letters to name the angles of the flag of the Czech Republic that are obtuse. ∠AFC and ∠EFC are obtuse angles. Question 14. Modeling Real Life Horses see an object with binocular both eyes at the same time using vision. Classify the angle that describes the horse’s binocular vision. The angle that formed in binocular vision is Acute angle. Review & Refresh Question 15. Write 1$$\frac{7}{12}$$ as a fraction. (1 x 12 ) + 7 =19 Question 16. Write $$\frac{9}{6}$$ as a mixed number. ### Lesson 13.3 Identify Parallel and Perpendicular Lines Explore and Grow Use a straightedge to draw and label a figure for each description. If a figure cannot be drawn, explain why. Reasoning Find the figure that shows two lines that cross once. How many angles are formed by two lines? Name and classify the angles of the figure above. Four angles are formed – ∠FOJ , ∠FOH ,∠GOJ and ∠GOH. Think and Grow: Parallel and Perpendicular Lines You can describe a pair of lines as intersecting, parallel, or perpendicular. Example Draw and label the lines with the given description. Show and Grow Draw and label the lines with the given description. Question 1. Question 2. Apply and Grow: Practice Draw and label the lines with the given description. Question 3. Question 4. Use the figure. Question 5. Name a pair of lines that appear to be parallel. AC ll DF. Question 6. Name two lines that are perpendicular. BE ⊥ DF and DE ⊥ EB. Question 7. Name two intersecting lines. DE and EB CB and BE Question 8. Reasoning All perpendicular lines are also intersecting lines. Are all intersecting lines perpendicular? Explain. Perpendicular Lines means which bisect the lines at right angles .But where as the intersecting lines meet at one point but doesnot form right angles. Question 9. Think and Grow: Modeling Real Life Example Which street appears to be parallel to 2nd Street? Look for a street that will not intersect with 2nd street. ___ Street appears to be parallel to 2nd Street. 3rd street. Show and Grow Question 10. Which trail appears to be parallel to Fox Trail? Wolf Run Trail Question 11. Which trail appears to be perpendicular to Fox Trail? Oak Trail. Question 12. Trace and label a pair of line segments that appear to be parallel and a pair of line segments that appear to be perpendicular. Question 13. DIG DEEPER! Design three paths for a park. Two of the paths are perpendicular. Label these paths as Path 1 and Path 2. The third path intersects both perpendicular paths at exactly one point. Label this path as Path 3. ### Identify Parallel and Perpendicular Lines Homework & Practice 13.3 Example Draw and label the lines with the given description. Question 1. Question 2. Use the figure. Question 3. Name a pair of lines that appear to be parallel. AG ll CJ Question 4. Name two lines that are perpendicular. BE ⊥ DF HE DF Question 5. Name two intersecting lines. CJ and DF . Question 6. Structure Name two line segments that appear to be parallel. Then name two line segments that appear to be perpendicular. AB ll DC and AD⊥ DC . Question 7. Reasoning Can two lines that share a point be parallel? Explain. Two Lines can be parallel but a point cannot be a parallel. Explanation: Parallel lines are co planar lines that do not intersect. In two dimensions, parallel lines have the same slope . We can write the equation of a line parallel to a given line if we know a point on the line and an equation of the given line. y=2x+3 . Question 8. DIG DEEPER! l No, It cannot be an acute angle. Because it is a right angle. Explanation: As line SW is perpendicular to line UV and line SW is parallel to line TX then from figure it is clear that line TX will be perpendicular to line UV. Question 9. Modeling Real Life Which street appears to be parallel to Park Avenue? Peach Street. Question 10. Modeling Real Life Which street appears to be perpendicular to Peach Street? Question 11. Modeling Real Life Trace and label a pair of line segments that appear to be parallel and a pair of line segments that appear to be perpendicular. Review & Refresh Find the equivalent amount of time. Question 12. 20 min = ___ sec 1min = 60sec 20 mins = 20 x 60 = 1200 sec. Question 13. 6 yr = ___ wk 1 year = 52 weeks. 6 years = 6 x 52 = 312 weeks ### Lesson 13.4 Understand Degrees Explore and Grow Find the elapsed time for each set of clocks. Describe, in your own words, the turn that the minute hand makes. Elapsed time is the time or difference between a beginning time and an ending time. Reasoning Explain how elapsed time shown by an analog clock relates to angles formed in a circle. On a clock, we can see different kinds of angles based on its amplitude; void, acute, straight, obtuse, flat angles—even a full angle. Think and Grow: Degrees Angles are measured in units called degrees. Think of dividing a circle into 360 equal parts. An angle that turns through $$\frac{1}{360}$$ of a circle measures 1°, and is called a“one-degree angle.” A full turn through the entire circle is 360°. Example Find the measure of the angle. An angle that turns $$\frac{1}{360}$$ of a circle measures _1__ degrees. An angle that turns through $$\frac{20}{360}$$ of a circle measures _20__ degrees. So, the measure of the angle is _20 degrees__. Example Find the measure of a right angle. 22.5 degrees. Show and Grow Find the measure of the angle. Question 1. 60 degrees. Question 2. 45 degrees. Apply and Grow: Practice Find the measure of the angle. Question 3. 85 degrees Question 4. 120 degrees. Question 5. 45 degrees. Question 6. (1/9) x (—/—) = —–/360 =40 degrees. Question 7. (1/2) x (—/—) = —–/360 =180 degrees. Question 8. (1/5) x (—/—) = —–/360 =360/5=72 degrees. Question 9. A circle is divided into 8 equal parts. What is the measure of the angle that turns through 2 parts? Number of parts =8 Total sum of angle in circle = 360. Measure of 1/8 part of angle formed = 360/8 =45 degree. Therefore measure of angle that turns through 2 parts = 2 x 45 =90 degrees. Question 10. A circle is divided into 4 equal parts. What is the measure of the angle that turns through 3 parts? Number of parts =4 Total sum of angle in circle = 360. Measure of 1/4 part of angle formed = 360/4 =90 degree. Therefore measure of angle that turns through 3 parts = 3 x 45 =135 degrees. Classify the angle as right, straight, acute or obtuse. Question 11. 30° Acute angle An acute angle lies between 0 degree and 90 degrees, or in other words; an acute angle is one that is less than 90 degrees. Question 12. 120° Obtuse angle It is the angle which lies between 90 degrees and 180 degrees or in other words; an obtuse angle is greater than 90 degrees and less than 180 degrees. Question 13. 90° Right angle A right angle is always equal to 90 degrees. Question 14. 180° Straight angle A straight angle is 180 degrees. It is just a straight angle because the angle between its arms is 180 degrees. ### Reflex Angle Question 15. YOU BE THE TEACHER Both circles are divided into sixths. Your friend says the measure of Angle D is greater than the measure of Angle C. Is your friend correct? Explain. No Explanation: Both the angles are divided into 6 equal parts irrespective of size of the circle. Both the angles C and D will be equal. As the sum of the circles is 360 degrees for all circles ,when divided equally the angles will be equal. Question 16. Reasoning Does each figure show the same angle? If not, which two angles are shown? Explain your reasoning. Both the Angles are equal. Think and Grow: Modeling Real Life Example Spokes divide the Ferris wheel into 20 equal parts. What is the angle measure of 1 part? Write a fraction that represents 1 part. Because the Ferris wheel has 20 equal parts, Show and Grow Question 17. The game spinner is divided into10 equal parts. What is the angle measure of 1 part? Number of parts divided=10 Total sum of angle in circle = 360. Measure of 1/10 part of angle formed = 360/10 =36 degree. Question 18. DIG DEEPER! A circular quesadilla is cut into 8 equal pieces. Five pieces are eaten. What is the angle measure formed by the remaining pieces? Total Number of parts divided = 8. Total sum of angle = 360 degrees. Angle of each part = 360/8 degrees = 45 degrees. Number of parts eaten = 5 Angle of 5 parts = 5 x 45 degrees = 225 degrees. Angle formed by remaining parts = 360- 225 degrees =135 degrees or Number of remaining parts =3 Angle of remaining parts = 3 x 45 degrees. = 135 degrees. Question 19. DIG DEEPER! When a light wave hits an object, the object reflects a colored light at an angle to your eye. The color of the reflected light is the color you see. What fraction of a circle is shown by the angle? Explain. Reflected angle = 30 degrees. Total sum of angle = 360 degrees. Fraction angle reflected in circle = 30 / 360 degrees = 1 /12 fraction. ### Understand Degrees Homework & Practice 13.4 Find the measure of each angle Question 1. 50 degrees Question 2. (1/3) x (—/—) = —–/360 = 360/3 = 120 degrees. Question 3. (1/12) x (—/—) = —–/360 = 360/12 = 30 degrees. Question 4. (1/10) x (—/—) = —–/360 =360/10 = 36 degrees. Question 5. (1/6) x (—/—) = —–/360 = 360/6 = 60 degrees. Question 6. A circle is divided into 5 equal parts. What is the measure of the angle that turns through 2 parts? Number of parts =5 Total sum of angle in circle = 360. Measure of 1/5 part of angle formed = 360/5 =72 degree. Therefore measure of angle that turns through 2 parts = 2 x 72 =144 degrees. Question 7. A circle is divided into 10 equal parts. What is the measure of the angle that turns through 3 parts? Number of parts =10 Total sum of angle in circle = 360. Measure of 1/10 part of angle formed = 360/10 =36 degree. Therefore measure of angle that turns through 3 parts = 3 x 36 =108 degrees. Classify the angle as right, straight, acute, obtuse Question 8. 90° Right angle A right angle is always equal to 90 degrees. Question 9. 45° Acute angle An acute angle lies between 0 degree and 90 degrees, or in other words; an acute angle is one that is less than 90 degrees. Question 10. 160° Obtuse angle It is the angle which lies between 90 degrees and 180 degrees or in other words; an obtuse angle is greater than 90 degrees and less than 180 degrees. Question 11. 60° Acute angle An acute angle lies between 0 degree and 90 degrees, or in other words; an acute angle is one that is less than 90 degrees. Question 12. Which One Doesn’t Belong? Which angle measure does not belong with the other three? 1/8 of a circle Question 13. DIG DEEPER! Your friend uses the equation to find an angle a measure. Explain what the letters a and b represent. 360° ÷ a = b Where a is the Number of parts where circle is divided. b is the angle of 1 part. Question 14. Modeling Real Life The steering wheel is divided into 3 equal parts. Find the angle measure of 1 part. Number of parts =3 The total sum of angle in circle = 360. The measure of 1/3 part of the angle formed = 360/3 =120 degrees. Question 15. DIG DEEPER! You and your friend take pie-shaped pieces from a circular quiche. Your friend takes $$\frac{2}{8}$$ of the quiche. You take a piece with an angle measure of 72°. Who takes a larger piece? Explain. It Is given that pie taken your friend is 2/8 So it is clear that quinche is divided into 8 parts. Number of parts =8 Total sum of angle in circle = 360. Measure of 1/8 part of angle formed = 360/8 =45 degree. Piece taken by friend is 2/8 = 2 x 45 = 90 degrees. Piece taken by you measures 72 degrees. Therefore Your Friend takes larger piece. Review & Refresh Multiply Question 16. =7 x (7/2) =1/2 = 0.5 Question 17. = 10 x (53/6) =88.33 Question 18. =4 x (51/8) =51/2 =25.5 ### Lesson 13.5 Find Angle Measures Explore and Grow How many triangular pattern blocks can you put together around one vertex? How can you determine the measure of each angle in a triangular pattern block? If we add all three angles in any triangle we get 180 degrees. So, the measure of (angle A + angle B + angle C) = 180 degrees. Repeated Reasoning Find all of the angle measures of the other pattern blocks. Organize your results in the table. Think and Grow: Find Angle Measures Example Use the pattern block to find the measure of the angle. Each angle of the pattern block has a measure of _60__. The angle is equal to _60 degrees__ of the angles of the pattern block. So, the measure of the angle is __60 degrees_. Example Use the pattern block to find the measure of the angle. Each acute angle of the pattern block has a measure of _30__. The angle is equal to _30__ of the acute angles of the pattern block. So, the measure of the angle is _150 degrees__. Show and Grow Use pattern blocks to find the measure of the angle. Question 1. Question 2. Question 3. Use pattern blocks to find how many 60° angles are in a straight angle. Total 3 triangles are used. Apply and Grow: Practice Use pattern blocks to find the measure of the angle. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. How many 90° angles are in a straight angle? Explain. A straight has 180 degree as its sum of angle. One right angle = 90 degrees. =90 x2 =180. So if we have 2 right angles the its sum will be equal to 180 degrees. Question 11. How many 30° angles are in a 150° angle? Explain. =150/30=5 times. Question 12. Structure Use two different pattern blocks to form an obtuse angle. Find the angle measure. Draw a model to support your Answer. Question 13. DIG DEEPER! Find the measure of the smaller angle formed by the clock hands. Then explain how you could find the measure of the larger angle. The angle formed at 3 clock = 90°. The sum of angles=360° The larger angle = 360° – 90°=270° Think and Grow: Modeling Real Life Example A circular compass is divided into 8 equal sections. What is the measure of each angle formed at the center of the compass? A full turn through a circle is 360°. So, divide 360° by 8. Each angle formed at the center of the compass is __45°_. Show and Grow Question 14. You have a circular craft table. You divide the table into 3 equal sections. What is the measure of each angle formed at the center of the table? Number of sections =3 The total sum of angle in circle = 360°. The measure of 1/3 part of the angle formed at the center of the table = 360°/3 =120° Question 15. A chef has a wheel of cheese that is cut into6 equal pieces. The chef uses 2 pieces. What is the angle measure formed by the missing pieces? Number of parts =6 Total sum of angle in circle = 360°. Measure of 1/6 part of angle formed= 360°/6 =60° Chef uses 2 peices Angle formed by used peices = 2 x 60° =120° Therefore, the angle measure formed by the missing pieces = 120° Question 16.What is the angle measure formed by the remaining slices? Number of parts =5 Total sum of angle in circle = 360°. Measure of 1/5 part of angle formed= 360°/5 =72° used peices = 2 Angle formed by Remaining peices = 3 x 72° =216° Therefore, the angle measure formed by the remaining pieces = 216° Question 17. DIG DEEPER! You make a tile pattern for a border on a wall. Each tile is the same size and shape as a pattern block. Draw two ways you can arrange 1 triangle tile and 1 hexagon tile to create a straight angle. ### Find Angle Measures Homework & Practice 13.5 Use pattern blocks to find the measure of the angle. Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Answer: The angle formed is 80°. It cannot be represented by pattern blocks. Question 7. How many 30° angles are in a right angle? Explain. Right angle =90° How many 30° angles we have in right angle = 90°/30°=3 Question 8. Structure Draw to show how you can use three different pattern blocks to form a straight angle. Question 9. YOU BE THE TEACHER Your friend says the measure of the angle shown is 120°. Is your friend correct? Explain. No, As the angle is 130°. Question 10. DIG DEEPER! The hands of a clock form a straight angle. What time could it be? Explain. The time as per mage is 6 o clock. Explanation: • One hand is very thin and moves very fast. It’s called the seconds hand. Every time it moves, a second has gone by. • Another hand is thick and long like the seconds hand. It’s called the minutes hand. Every time it moves one little tick, a minute has gone by. Every 60 times it moves a whole step, an hour has gone by. • The last hand is thick, too, but smaller than the minutes hand. It’s called the hours hand. Every time it moves one big tick, an hour has gone by. Every 24 times it moves a whole step, a day has gone by Question 11. Modeling Real Life The scale is divided into 5 equal sections by each whole kilogram measurement. What is the measure of each angle formed at the center of the scale? What is the mass of the bananas? The whole Kilogram is divided into 5 parts 1kg = 1000 grams =1000 grams / 5 =200 grams. As per the image the weight of banans showing mark on one. one unit = 200 grams. Therefor, weight of bananas = 200 grams. Question 12. DIG DEEPER! The gasoline tank gauge is divided into 4 equal sections. What are the measures of the angles formed at the starting point of the red arrow? The gasoline is in a shape of semi circle. Sum of the angles = 180° Number of sections divided = 4 parts. (equal). Angle of each section : 180° /4 = 45° . Therefore the measures of the angles formed at the starting point of the red arrow = 45° . Review & Refresh Find the area of the rectangle. Question 13. Area of Rectangle = length X breadth. AREA = 19 m x 16 m = 304 sq m. Question 14. Area of Rectangle = length X breadth. AREA = 28 in x 15 in =420 sq in. ### Lesson 13.6 Measure and Draw Angles Explore and Grow Find the measure of each angle. Then classify it. Construct Arguments Does the angle shown have a measure of 90°? Explain. The intersecting line form right angle =90 degrees The angle marked = ? The total angle = 360 degrees. The marked angle= 360 – 90 = 270 degrees. Think and Grow: Measure and Draw Angles Example A protractor is a tool for measuring and drawing angles. Find the measure of ∠STU. Then classify it. Step 1: Place the center of the protractor on the vertex of the angle. Step 2: Align one side of the angle, $$\overrightarrow{T U}$$, with 0° mark on the inner scale of the protractor. Step 3: Find where the other side of the angle, $$\overrightarrow{T S}$$, passes through the inner scale. So, the measure of ∠STU is ___. It is an ___ angle. Example Draw ∠ABC that measures 65°. Step 1: Place the center of the protractor on point. Align $$\overrightarrow{B C}$$ with the 0° mark on the inner scale of the protractor. Step 2: Use the same scale to draw a point at 65°. Label the point A. Step 3: Use the protractor to draw $$\overrightarrow{B A}$$. Show and Grow Question 1. Find the measure of ∠DEF. Then classify it. Question 2. Use a protractor to draw ∠XYZ that measures 110°. Apply and Grow: Practice Find the measure of the angle. Then classify it. Question 3. Question 4. Use a protractor to draw the angle. Question 5. 25° Question 6. 160° Question 7. 180° Question 8. 48° Question 9. Writing Why does one side of the angle you are measuring have to be lined up with the straight side of the protractor? We see two sets of degrees along the edge: an inner and outer scale. Both scales go from 0 to 180, but they run in opposite directions. If the angle opens to the right side of the protractor, use the inner scale. If the angle opens to the left of the protractor, use the outer scale. To mark 0° on the inner scale of the protractor. So that we can next mark the angle we want to measure. Question 10. Precision Newton says the measure of ∠ABC is 130°. Explain what he did wrong. The ∠ABC is 50 degrees not 130 degrees . To measure an angle ABC, we place the mid point of the protractor on the vertex B of the angle. The base AB arm falls along with the base line of the protractor as shown in the figure above The angle is measured on the protractor counting from 0 up to the point where the arm BC lies. In the above figure ∠ABC = 50°. Think and Grow: Modeling Real Life Example A contractor builds two roofs. How much greater is the angle measure of Roof A than the angle measure of Roof B? Use a protractor to measure the angle of each roof. The angle measure of Roof A is _50°_ and the angle measure of Roof B is _40°__. Subtract the angle measure of Roof B from the angle measure of Roof A. _50°__ − _40°__ = _10°__ The angle measure of Roof A is _10°__ greater than the angle measure of Roof B. Show and Grow0 Question 11. An inspector compares two ramps. How much greater is the angle measure of Ramp B than the angle measure of Ramp A? Ramp A angle is 10 degrees and Ramp B angle is 20 degrees Ramp B angle is 10 degrees greater. Question 12. DIG DEEPER! On a trail map, two straight trails intersect. One of the angles formed by the intersection is 70. What are the other three angle measures? A trail is formed with 4 angles. Sum of the angles in trial = 360°. if one angle =70° Sum of other 3 angles = 360° – 70° = 290 °. ### Measure and Draw Angles Homework & Practice 13.6 Find the measure of the angle. Then classify it. Question 1. Question 2. Use a protractor to draw the angle. Question 3. 40° Question 4. 125° Question 5. Reasoning Measure the angles of each quadrilateral. What do you notice about the sum of the angle measures of each quadrilateral? The angles of the square are 90° each . The angles of Quadilateral are two angles are 90°. And other angles are 50°. For every quadrilateral, the sum of the interior angles will always be 360°. Question 6. Precision Use a protractor to draw a triangle with the angle measures of 0°. Describe your drawing. Question 7. DIG DEEPER! Use a protractor to draw a triangle with the angle measures of 90°, 35°, and 55°. Question 8. Modeling Real Life A snowboarder compares 2 mountain trails. How much greater is the angle measure of Trail B than the angle measure of Trail A? The angle is 5° greater. Question 9. DIG DEEPER! On a map, there is a Y-intersection where one straight road branches off into two straight roads. One of the angles formed by the intersection measures 45°. Give two possible measures for the other angles formed by the intersection. Review & Refresh Find the equivalent length Question 10. 5 km = ___ m. 1km = 1000m 5 km = 5 x 1000 m = 5000m Question 11. 7 m = ___ mm. 1m = 1000mm 7 m = 7 x 1000 mm = 7000mm ### Lesson 13.7 Add Angle Measures Explore and Grow Use a protractor to draw ∠PQR that measures 70°. Draw another angle that measures 30° and shares side $$\overrightarrow{\mathrm{Q} R}$$. Label your angle. How many angles are in your figure? What do you notice about their measures? Make a copy of ∠PQR. Draw and label a different angle that measures 30° and shares $$\overrightarrow{\mathrm{Q} R}$$. How many angles are in your figure? What do you notice about their measures? Construct Arguments What conclusions can you make from your figures above? The Above triangle formed is a scalene triangle- has no equal side. Think and Grow: Add Angle Measures When an angle is decomposed into parts that do not overlap, the angle measure of the whole equals the sum of the angle measures of the parts. Two angles are adjacent when they share a common side and a common vertex, but no other points in common. When two or more adjacent angles form a larger angle, the sum of the measures of the smaller angles is equal to the measure of the larger angle. Example Find the measure of ∠WXY. ∠WXY and ∠ZXY are adjacent. The measure of ∠WXZ is equal to the sum of the measures of and ∠ZXY. Complete the equation. _41°__ + __32°_ = _73°__ So, the measure of ∠WXY is _73°__. Show and Grow Question 1. Complete the equation to find the ∠PQR. ∠PQR = ∠PQS +∠SQR = 43° + 77° =120° Question 2. Use a protractor to find the measure of each angle in the circle. Use the angle measures to complete the equation. Apply and Grow: Practice Write an equation to find the measure of the angle. Question 3. ∠QRS= ∠QRT +∠TRS = 62° + 19° =81° Question 4. ∠ABC = ∠ABD +∠DBC = 66° + 37° =103° Question 5. ∠MNO = ∠MNQ +∠QNP +∠PNO = 45° + 87° +22° =154° . Question 6. ∠VWX = ∠VWZ +∠ZWY +∠YWX = 36° + 38° +73° =147° . Use a protractor to find the measure of each angle in the circle. Use the angle measures to write an equation. Question 7. ∠M= ∠LMO +∠OMN +∠LMN = 100° + 80° +180° =360° ° . Question 8. ∠W= ∠SWT +∠TWU +∠UWV +∠VWS = 120° + 60° + 90° + 90° =360° Question 9. Number Sense The sum of seven adjacent angle measures that share a vertex is 154°. Each angle has the same measure. What is the measure of each angle? The sum of seven equal angles = 154°. Measure of each angle = 154°/7 = 22°. Question 10 DIG DEEPER! ∠PRS is adjacent to ∠QRS in Exercise 3. The measure of ∠PRS is 9°. Find the measure of ∠QRP. Classify angle ∠QRP. ∠QRP = 9° Two angles are adjacent when they share a common side and a common vertex, but no other points in common. When two or more adjacent angles form a larger angle, the sum of the measures of the smaller angles is equal to the measure of the larger angle. Think and Grow : Modeling Real Life Example A store installs 3 security cameras on the same light post. Each camera has a viewing angle of 28°. The viewing angles of the cameras are adjacent. What is the total viewing angle of the cameras? Add the viewing angle measures of the cameras. 28° + 28° + 28° = _84°__ The total viewing angle of the cameras is __84°_ Show and Grow Question 11. A landscaper installs 3 sprinklers in 1 location in the grass. Each sprinkler has a spraying angle of 90°. The spraying angles of the sprinklers are adjacent. What is the total spraying angle of the sprinklers? As 3 sprinklers are adjacent all the 3 angles are equal. Each angle of the sprinkler = 90° Total spraying angle of the sprinklers= 3 X 90° = 270° Question 12. DIG DEEPER! You use 4 rhombus tiles and 4 trapezoid tiles to make the pattern for a mosaic. Each tile is the same size and shape as a pattern block. What is the measure ∠NUQ? As we know th pattern blocks of trapezoid have 60° and rhombus have 30°. Each tile is the same size and shape as a pattern block ∠NUQ= 30° + 60°+ 60° =150° Question 13. DIG DEEPER! You make the pattern for a quilt. Each rhombus is the same size and shape. Name all of the 60° angles in the pattern. All 60° angles in the pattern = ∠AHC +∠BHD +∠CHE+∠DHF +∠EHG. ### Add Angle Measures Homework & Practice 13.7 Write an equation to find the measure of the angle. Question 1. ∠EFG= ∠EFH +∠HFG = 27° + 44° =71° Question 2. ∠JKL= ∠JKM +∠MKL = 53° + 85° =138° Question 3. ∠PQR= ∠PQT +∠TQS +∠SQR = 68° + 47° +54° =316° Question 4. ∠VWX= ∠VWZ +∠ZWY +∠YWX = 19° + 42° +35° =96° Use a protractor to find the measure of each angle in the circle. Use the angle measures to write an equation. Question 5. = 120° + 60° +180° =360° Question 6. ∠N= ∠JNM +∠MNL +∠LNK + ∠KNJ = 150° + 30°+90° + 90° =360° Question 7. Open-Ended Is it possible for the sum of two acute adjacent angle measures to be greater than the measure of a right angle? If so, draw a sketch to support your answer. Here ∠ABC and ∠CBD are both acute angles. ∠ABD= ∠ABC +∠CBD = 60°+60° =120° sum of two acute adjacent angle=120° is more than 90° . Question 8. DIG DEEPER! Use a protractor to measure the angle. Then decompose the angle into 2 parts so that one part is 25° greater than the other. What is the measure of each part? The measure of given angle = 120° The given angle is divided into two parts one angle is 25° greater than other. The measure each part = a + ( a + 25°) = 120° 120° – 25° =2a 105° = 2a a=55.5° a + 25° = 77.5° Question 9. Modeling Real Life A carpenter glues 3 identical pieces of wood next to each other to make the table shown. The 30° angles of the pieces of wood are adjacent. What is the total angle of the table formed by the pieces of wood? Angle of each wood piece = 30° As all angles are adjacent so angle of other wood pieces will be same. so all 3 angles = 30° each. Total angles = 3 x 30° = 90° . Question 10. DIG DEEPER! You use 3 triangle pattern blocks and 3 rhombus pattern blocks to make the pattern for an art project. What is the measure of ∠ADE? The pattern block angle of triangle = 60° The pattern block angle of Rhombus = 60° Review & Refresh Compare Question 11. Question 12. Question 13. ### Lesson 13.8 Find Unknown Angle Measures Explore and Grow Draw and label ∠EDF where point F is between the rays of ∠CDE. Explain how you can find the measure of ∠CDF. ∠CDE=∠CDF + ∠EDF = 45 + 45 =90° (USING PROTACTOR). Explanation: Draw and label ∠LMP. Explain how you can find the measure of ∠NMP. Reasoning Draw in your figure above. Explain how you can find all of the angle measures in your figure. What do you notice? is a which bisect the LN. so angles are divided into 2 parts equally. Using protractor we can find the measure of the angles. This are complementary angles. Think and Grow: Find Unknown Angle Measures Example Find the measure of ∠DEG. ∠DEG and ∠GEF are complementary. The sum of their measures is 90.° Step 1: Write an equation to find the measure of ∠DEG. Show and Grow Question 1. Write and solve an equation to find the measure of ∠NLM. Supplementary angles form a straight line and have a sum of 180 degrees. ∠NLM=180°- ∠KLN = 180° – 62° = 118° Apply and Grow: Practice Write and solve an equation to find the measure of the angle. Question 2. Complementary angles form a right angle (L shape) and have a sum of 90 degrees. ∠MKL=90°- ∠MKJ = 90° – 45° = 45° Question 3. Supplementary angles form a straight line and have a sum of 180 degrees. ∠MNP=180°- ∠PNO = 180° – 22° = 158° Question 4. ∠TRU= ∠SRQ -∠SRT -∠URQ = 90° – 30° – 40° =20° Question 5. ∠YWZ= ∠XWV -∠VWZ -∠YWX = 180° – 31°-111° =38° Question 6. Find the measures of ∠AEB, ∠AED, and ∠DEC. ∠AEB= ∠AEC -∠BEC = 180° – 67° = 113° ∠DEC= ∠DEB -∠BEC = 180° – 67° = 113° ∠AED= ∠AEC -∠DEC = 180° – 113° = 67° Question 7. DIG DEEPER! Write and solve equations to find the measure of ∠KMJ. ∠KME= ∠KMG -∠GME = 180° – 124° = 56° ∠KMJ= ∠JME -∠KME = 90° – 56° = 34° Question 8. Writing Define complementary angles and supplementary angles in your own words. Complementary angles form a right angle (L shape) and have a sum of 90 degrees. Supplementary angles form a straight line and have a sum of 180 degrees. Think and Grow: Modeling Real Life Example The foul lines on a baseball field are perpendicular. A baseball player hits a ball as shown. What is the measure of the angle between the path of the ball and the 1st base foul line? Think: What do you know? What do you need to find? How will you solve? The 3rd base foul line and 1st base foul line are perpendicular. So, the measure of the angle between the foul lines is _90° __. Write an equation to find the measure of the angle between the path of the ball and the 1st base foul line. Show and Grow Question 9. Runway 1 and Runway 2 are perpendicular. What is the measure of the missing angle between Runway 1 and Runway 3? Angle between Runway 1 and Runway 3= 90 – Angle between Runway 2 and Runway 3. =90 – 58 =32° Question 10. DIG DEEPER! What is the measure of the missing g angle between View Street and Elm Street? Angle between View Street and Elm Street=360 – Angle between IIIstreet and view street – Angle between elm street and IIIstreet. =360- 90 -152 =118° ### Find Unknown Angle Measures Homework & Practice 13.8 Write and solve an equation to find the measure of the angle. Question 1. ∠JKM ∠JKM= ∠JKL -∠MKL = 90° – 9° = 81° Question 2. ∠PNM ∠PNM= ∠MNO -∠PNO = 180° – 145° = 35° Question 3. ∠QRU ∠QRU= ∠QRS -∠URT-∠TRS = 90°-28° – 13° =49° Question 4. ∠ZWV ∠ZWV= ∠VWX -∠ZWY-∠YWX = 180°- 32° – 24° =124° Question 5. Reasoning ∠ABC and ∠CBD are adjacent. ∠ABC is a right angle. ∠CBD is acute. ∠ABD=∠ABC-∠CBD = 90 – 35 = 65° Question 6. Structure Which equations can you use to find the measure of angle ∠MKJ? 180 – 57 = x Question 7. Open-Ended An acute angle and an obtuse angle are adjacent and supplementary. What might the measures of each angle be? let, be obtuse angle and be acute angle. So, That means (obtuse angle) and (acute angle) are supplementary Question 8. Modeling Real Life Newton bounces a ball off of a wall to Descartes. $$\overline{A D}$$⊥$$\overline{D B}$$. The measures of ∠ADC and ∠BDE are equivalent. Find the measures of ∠ADC and ∠BDE. ∠CDE= ∠ADC + ∠BDE – ∠BDA. 180 = a + a – ∠BDA. ∠BDA.=180 -2a Question 9. Modeling Real Life Owls see an object with both eyes at the same time using binocular vision. What angle measure describes the owl’s binocular vision? Explain. left monocular + Binocular vision = 180 Binocular vision = 180 – 110=70 degrees. Review & Refresh Write the fraction as a money amount and as a decimal Question 10. $$\frac{49}{100}$$ 0.49 Question 11. $$\frac{25}{100}$$ 0.25 Question 12. $$\frac{7}{100}$$ 0.07 ### Identify and Draw Lines and Angles Performance Task 13 A rural town is expanding and needs to plan the construction of new roads. Question 1. What is another name for Main Street? Line AC. Question 2. Use the directions to complete the map. a. Draw and label it with a street name of your choice. b. The library will be at point E on and on the other side of Main Street as the school. Plot and label the library E at point E. c. Draw a new road through point E that is perpendicular to Main Street. Label it with a street name of your choice. Question 3. City planners want to construct a new residential neighborhood southeast of the town hall. a. The measure of ∠ABD is $$\frac{1}{6}$$ of 360°. What is the measure of the angle? ∠ABD=(1/6) x 360 =60 degree. b. Classify ∠DBC. What is its measure? ∠DBC=180 -∠ABD = 180-60 = 120 degrees. c. Draw a road from point B to the new neighborhood. The road divides ∠DBC exactly in half. Question 4. Is the distance between the supermarket and the police station more than or less than a mile? Explain. Distance between the supermarket and the police station= AB + BC = (1/4) + (5/8). =(2+5)/8 =7/8.=0.8 mile which is less than 1 mile. ### Identify and Draw Lines and Angles Activity Geometry Dots Directions: 1. Players take turns connecting two dots, each using a different color. 2. On your turn, connect two dots, vertically or horizontally. If you close a square around an angle, find the measure of the angle inside the square. If you do not close a square, your turn is over. 3. Continue playing until you find all of the angle measures. 4. The player that finds the most angle measures wins! ### Identify and Draw Lines and Angles Chapter Practice 13 13.1 Points, Lines and Rays Name the figure. Write how to say the name. Question 1. It is a point said as Point F Question 2. It is said as line GH. and represented as GH with overhead arrow. Question 3. It is a ray JK .It is represented with Over head arrow to left. Draw and label the figure. Question 4. Question 5. Question 6. Use the figure Question 7. Name a line segment. Question 8. Name two different rays. 13.2 Identify and Draw Angles Question 9. Write a name for the angle and classify it. ∠SRQ. It is a obtuse angle. Question 10. ∠TUV is a right angle. Draw and label the angle. 13.3 Identify Parallel and Perpendicular Lines Draw and label the lines with the given description. Question 11. Question 12. 13.4 Understand Degrees Find the measure of each angle. Question 13. 115° Question 14. =(1/5) x(—–/—–)= —-/360. =72°/360° Question 15. =1/4 x 360 =90° Question 16. A circle is divided into 12 equal parts. What is the measure of the angle that turns through 2 parts? Number parts divided = 12 Total sum of the angles = 360° Angle of 1 part = 360°/12. = 30°. The measure of the angle that turns through 2 parts = 2 x 30°. = 60°. Question 17. Which one Doesn’t Belong? Which angle measure does not belong with the other three? 1/6 of a circle = 360/6 = 60°. 60/360 of a circle = 1/6. 60° 1/8 of a circle = 360°/8=45° Therefore 1/8 of a circle is different. 13.5 Find Angle Measures Use pattern blocks to find the measure of the angle. Question 18. Question 19. Question 20. Question 21. 13.6 Measure and Draw Angles Question 22. Find the measure of the angle. Then classify it. Question 23. Use a protractor to draw a 45° angle. Modeling Real Life On a map, two straight railroad tracks intersect. One of the angles formed by the intersection is 30°. What are the three other angle measures? Question 25. Write an equation to find the measure ∠EFG. ∠EFG = ∠EFH+∠HFG. =37° + 48° = 85°. Question 26. Use a protractor to find the measure of each angle in the circle. Use the angle measures to write an equation. 13.8 Find Unknown Angle Measures Write and solve an equation to find the measure of the angle. Question 27. Complementary angles =90°. ∠ABD = ∠ABC-∠DBC. =90°-27° = 63°. Question 28. Supplementary angle =180° ∠HFG = ∠EFG-∠EFH. =180° + 42° = 38°. Question 29. Complementary angles =90°. ∠MKL = ∠LKJ-∠MKN-∠NKJ. =90°-45°-22°= 23°. Question 30.	2023-03-27 20:45:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4979259669780731, "perplexity": 2900.614662839489}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948684.19/warc/CC-MAIN-20230327185741-20230327215741-00521.warc.gz"}
https://socratic.org/questions/in-a-poll-of-1-000-randomly-selected-adults-400-answered-yes-when-asked-if-they-#576665	# In a poll of 1,000 randomly selected adults, 400 answered “yes” when asked if they planned to vote in the state election. What is the best point estimate of the population proportion of all adults who plan to vote in that election? Mar 18, 2018 Between 36.96% and 43.04% plan to vote in the election. #### Explanation: If we want to know if an event will happen, we can't just use the poll and accept it as fact. We need to find the standard error. For a proportion, this is the equation: $\sigma \hat{p} = \sqrt{\frac{\hat{p} \left(1 - \hat{p}\right)}{n}}$ where $\hat{p}$ is the sample proportion, and $n$ is the sample size. Using the information given, let's find the standard deviation. $\hat{p}$ is the number that said "yes" over the total number asked, which ends up being 0.40. $n$ is 1,000, since that's how many people were asked. so, let's solve. $\sigma \hat{p} = \sqrt{\frac{0.40 \left(1 - 0.40\right)}{1000}}$ $\sigma \hat{p} = \sqrt{\frac{0.40 \cdot 0.60}{1000}}$ $\sigma \hat{p} = \sqrt{\frac{0.24}{1000}}$ $\sigma \hat{p} = \sqrt{0.00024}$ $\sigma \hat{p} = 0.0155$ Now that we have our standard error, we should find the margin of error. This can be used to find the range that the actual answer to this problem would be in. In statistics, we usually say that we want a 95% confidence level, meaning we want to be 95% confident that the actual answer is in this range. To find this range, we multiply the standard deviation (standard error) by the critical value. A critical value is the number we multiply the standard deviation by to find the margin of error. In the case of a 95% confidence level, the critical value is 1.960. margin of error = $1.96 \cdot 0.0155$ margin of error = $0.0304$ margin of error = 3.04% Now that we have the margin of error, we add it to or subtract it from the mean to find the range I mentioned earlier. The mean is the number of people who said "yes" over the total number of people asked. mean = $\frac{400}{1000}$ mean = $0.40$ mean = 40% mean - SD = 40% - 3.04% mean - SD = 36.96% mean + SD = 40% + 3.04% mean + SD = 43.04% So, our answer is between 36.96% and 40.04%.	2021-10-24 04:02:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 24, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.757290780544281, "perplexity": 350.3488941175593}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585837.82/warc/CC-MAIN-20211024015104-20211024045104-00663.warc.gz"}
http://soft-matter.seas.harvard.edu/index.php?title=Limit_cycle&direction=prev&oldid=11394	# Limit cycle A limit cycle in a dynamical system is a closed orbit in phase space, which is the limiting behavior of some trajectories either as $t->\infty$ or $t->-\infty$	2022-12-07 11:33:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8647662997245789, "perplexity": 206.41797555716812}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711151.22/warc/CC-MAIN-20221207085208-20221207115208-00298.warc.gz"}
https://cs.stackexchange.com/questions/123399/is-there-a-recursive-problem-encoding-the-turing-completeness-of-a-model-of-comp	# Is there a recursive problem encoding the Turing completeness of a model of computation? Suppose we have a model of computation $$C$$ we want to show to be Turing complete. The usual strategy would be to emulate within $$C$$ any model of computation we already know to be Turing complete (e.g. an arbitrary Turing machine). On the other hand, a model of computation is Turing complete if and only if it can compute (the indicator function of) all recursive sets. Is there a special recursive set $$S$$ (or finite subset of recursive sets) so that it is enough for $$C$$ to compute $$S$$ to assure that $$C$$ is Turing complete? Informally, $$S$$ would be a set for whose computation branching, looping and memory are all needed in an unavoidable way. If such sets do exist, which are the simplest ones? In a practical sense, the utility of these sets would be the following test: if your model of computation can solve this type of (e.g. numerical or geometric) problem, then it is Turing complete.	2021-01-25 21:40:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9240525960922241, "perplexity": 268.39337791329893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703644033.96/warc/CC-MAIN-20210125185643-20210125215643-00610.warc.gz"}
http://www.ck12.org/tebook/Texas-Instruments-Calculus-Teacher%2527s-Edition/r1/section/9.3/	<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 9.3: Mr. Taylor, I Presume? Difficulty Level: At Grade Created by: CK-12 This activity is intended to supplement Calculus, Chapter 8, Lesson 7. ID: 10257 Time required: 45 minutes ## Activity Overview In this activity, students will use the TI-89 to explore Taylor polynomials graphically and analytically. They will also graphically determine the interval where the Taylor polynomial closely approximates the function it models. Topic: Series and Taylor Polynomials • Display in a spreadsheet the first few terms of a Taylor series approximation to f(x)\begin{align}f(x)\end{align} for a given value of x\begin{align}x\end{align} and compare the value of the Taylor approximation with the value of f(x)\begin{align}f(x)\end{align}. • Use Taylor Polynomial (on the Calculus menu) to verify the manual computation of a Taylor Series. • Graph a function and its Taylor polynomials of various degrees to show their convergence to the function. Teacher Preparation and Notes • Students should be familiar with basic differentiation prior to the beginning of this activity. • Students should be able to graph functions, and set up tables on their calculator. They should also be able to use commands and set up tables on their own. • This activity is designed to be teacher-led. • Before starting this activity, students should go to the Home screen and select F6\begin{align}F6\end{align} :Clean Up > 2:NewProb, then press ENTER. This will clear any stored variables, turn off any functions and plots, and clear the drawing and home screens. Associated Materials ## Introduction This activity begins with students finding a polynomial given only the value of its derivatives at a specific x\begin{align}x-\end{align}value. This is at the heart of finding Taylor polynomials. If students struggle understanding that this method works, it may be necessary to reverse the process and have the students find the derivatives and observe that their values at the given x\begin{align}x-\end{align}value are the given values of the derivative. At the conclusion of the introductory problem, the Taylor polynomial form is found when centered at zero. At this time, place the general form of the Taylor polynomial when x=0\begin{align}x = 0\end{align} on the board. \begin{align}P_n(x) = \frac{f^{(0)}(a)}{0!}(x - a)^0 + \frac{f'(x)}{1!}(x - a)^1 + \frac{f''(x)}{2!}(x - a)^2 + \ldots + \frac{f^{(n)}(x)}{n!}(x - a)^n\end{align} Explain that when \begin{align}a = 0\end{align}, this matches the form \begin{align}P(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3\end{align} ## Taylor polynomial centered at zero Students are asked to find the 4th degree Taylor polynomial that approximates \begin{align}f(x) = \ln(x + 5)\end{align} at \begin{align}x = 0\end{align}. The derivatives of the function are: \begin{align}f(0) & = \ln (5) \approx 1.609 \\ f'(x) & = \frac{1}{x + 5} \to f'(0) = \frac{1}{5} \\ f''(x) & = \frac{-1}{(x + 5)^2} \to f''(0) = - \frac{1}{25} \\ f'''(x) & = \frac{2}{(x + 5)^3} \to f'''(0) = \frac{2}{125} \\ f^{(4)}(x) & = \frac{-6}{(x + 5)^4} \to f^{(4)}(0) = - \frac{6}{625} \end{align} As the students compare the values of the Taylor polynomial with the values of the original function, they will notice that the values are closest at the center (the \begin{align}x-\end{align}value where the derivatives were found) and become farther apart the further the \begin{align}x-\end{align}values are from the center (the graph demonstrates this even further). Point out to students that not all Taylor polynomials have such a large interval as this one. The table at the right has the 4th power and 10th power Taylor polynomials as well as the function. Students will also have a chance to explore different degrees of a Taylor polynomial. It would be best if you choose three larger values ahead of time. Have the students use the Taylor command to find the respective polynomials and graph the function with each of the larger power polynomials in turn. If time permits, have a few students find the 10th degree Taylor polynomial and graph it on the same set of axes as the 4th degree Taylor and the original function. This will demonstrate that a Taylor polynomial will only approximate the function over a given interval, no matter how large the Taylor Polynomial is. It is important for students to know that a Taylor polynomial centered at 0 is known as a Maclaurin polynomial. ## Taylor polynomial not centered at zero When students work the problem of where the Taylor polynomial is not centered at zero, they will observe that the polynomial follows the original function for a much smaller interval. Remind students that they need to use \begin{align}(x - a)^n\end{align} instead of \begin{align}x^n\end{align}. The derivatives for this problem are found below. Derivative Value at \begin{align}x = 1\end{align} \begin{align}f(x)\end{align} \begin{align}\frac{1}{2 - x}\end{align} 1 \begin{align}f'(x)\end{align} \begin{align}-(2 - x)^{-2}(-1) = \frac{1}{(2 - x)^2}\end{align} 1 \begin{align}f''(x)\end{align} \begin{align}-2(2 - x)^{-3}(-1) = \frac{2}{(2 - x)^3}\end{align} 2 \begin{align}f'''(x)\end{align} \begin{align}2(-3)(2 - x)^{-4}(-1) = \frac{6}{(2 - x)^4}\end{align} 6 \begin{align}f^{(4)}(x)\end{align} \begin{align}6(-4)(2 - x)^{-5} (-1) = \frac{24}{(2 - x)^5}\end{align} 24 The Taylor polynomial can be: \begin{align}P(x) & = \frac{1}{0!}(x - 1)^0 + \frac{1}{1!}(x - 1)^1 + \frac{2}{2!}(x - 1)^2 + \frac{6}{3!}(x - 1)^3 + \frac{24}{4!}(x - 1)^4\\ & \qquad \qquad \qquad \qquad \qquad \qquad \text{or}\\ P(x) & = 1 + (x - 1) + (x - 1)^2 + (x - 1)^3 + (x - 1)^4\end{align} Pick three values from 0 to 1 (for instance 0.8, 0.5, 0.2) and find the 4th degree Taylor polynomial using the Taylor command. Graph each Taylor polynomial and the original function the same set of axes separately. Students should notice that the interval increases as the center gets closer to zero. See the graphs below. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:	2016-12-04 16:59:06	{"extraction_info": {"found_math": true, "script_math_tex": 29, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9530853629112244, "perplexity": 1025.8107512309575}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541324.73/warc/CC-MAIN-20161202170901-00404-ip-10-31-129-80.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/88849/why-does-int-limits-01-frac1etdt-converge/88964	# Why does $\int \limits_{0}^{1} \frac{1}{e^t}dt$ converge? I'd like your help with see why does $$\int^1_0 \frac{1}{e^t} \; dt$$ converge? As I can see this it is suppose to be: $$\int^1_0 \frac{1}{e^t}\;dt=\|_{0}^{1}\frac{e^{-t+1}}{-t+1}=-\frac{e^0}{0}+\frac{e}{1}=-\infty+e=-\infty$$ Thanks a lot? - What? No! $\int e^{-t} \mathrm dt$ shouldn't give that result... – J. M. Dec 6 '11 at 11:26 $\dfrac{e^{-t+1}}{-t+1}$ doesn't differentiate to $\dfrac1{e^t}$, for one. – J. M. Dec 6 '11 at 11:28 $\int e^{kx}\,dx={e^{kx}\over k}+C$... – David Mitra Dec 6 '11 at 11:32 @Jozef: Making a mistake is not a reason for deleting a question. What if someone else makes the same mistake and looks for the answer? We want this site to be able to help everyone. – Zev Chonoles Dec 6 '11 at 11:38 @JackManey: Yeah, I got it.. – Jozef Dec 7 '11 at 8:30 The integration you have shown is not correct. Exponential functions are quite different to power functions and have very different antiderivatives. The correct integration is as follows: $$\int_0^1 e^{-t} \, dt = -e^{-t}\|_0^1 = 1-\frac{1}{e}.$$ - $$\int_0^1 e^{-t} = [-e^{-t}]_0^1 = -\frac{1}{e} + 1$$ - I assume you are integrating over the $t$ variable. $1/e^t$ is a continuous function, and you are integrating over a bounded interval, so the integral is well defined. An antiderivative of $1/e^t=e^{-t}$ is equal to $-e^{-t}$. So the integral equals $1-1/e$ - Your calculation is incorrect. Instead, you should have $$\int_0^1\frac{1}{e^t}dt=\int_0^1e^{-t}dt=-e^{-t}\big\|_0^1=(-e^{-1}-(-e^{0}))=1-\frac{1}{e}$$ You got mixed up with the rule for powers, $$\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$$ but for exponentials we have $$\int e^x\,dx=e^x+C$$ You can also see that $e^{-t}$ is bounded above by the constant function $1$ on the interval $[0,1]$: $\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ so that the area underneath the curve $e^{-t}$ from $0$ to $1$ has to be less than the area underneath the curve $1$ from $0$ to $1$ (which is $1$). So this picture tells you that the value of $\int_0^1\frac{1}{e^t}dt$ is less than $1$, which we've confirmed by showing that is in fact equal to $1-\frac{1}{e}$. - Others have given a correct way to evaluate the integral. But even without evaluating the integral, you can see that it converges. After all, how can an integral fail to converge? It can fail to converge if one or both of the limits of integration is infinite, e.g., $\int_1^{\infty}x^{-1}\,dx$ fails to converge. It can fail to converge if the integrand is unbounded, e.g., $\int_0^1x^{-1}\,dx$ fails to converge. But if the limits of integration are bounded, and the integrand is bounded, the integral has no way to diverge - it must converge. In your example, the limits of integration are bounded, and the integrand satisfies $1/e\le e^{-t}\le1$ for $t$ in the interval of integration ($0\le t\le1$), so convergence is immediate. - $@$Gerry: some kind of continuity hypothesis in your first paragraph would be nice. – Pete L. Clark Dec 6 '11 at 11:56 @Pete, shush! At this level of mathematics, all functions are continuous, or near enough as to make no difference. Anyway, the kind of function you're thinking of, I would say the integral doesn't exist, but I'm not sure I would say it doesn't converge; its problems are deeper than those arising from improper integrals. – Gerry Myerson Dec 6 '11 at 12:20 I have taught calculus at many levels, including to freshman who will absolutely not be taking any further math courses, but at none of these levels is "all functions are continuous, or near enough as to make no difference" a true statement. I agree that underscoring the need for continuity is probably a poor pedagogical choice in this particular instance...but you shouldn't say things that are false, especially in an environment where lots of other people will be hearing or seeing them.... – Pete L. Clark Dec 6 '11 at 12:45 ..."But if the limits of integration are bounded, and the integrand is bounded, the integral has no way to diverge - it must converge." As you well know, this is simply not true (or perhaps not meaningful if we are not sure about the distinction between convergence and existence, but in every instance I can think of "convergence" means "some limit exists", and here we have a kind of limit that need not exist). I speak from experience when I say that using the word "continuous" will not rattle calculus students...as long as you don't ask them to define it or do anything with it. – Pete L. Clark Dec 6 '11 at 12:50 @Pete, I just finished teaching a semester of Calculus, and I did it without using the word "limit", without using the word "continuous", without making use of any function that wasn't continuous wherever it was defined. Had I decided to mention a discontinuous function, it would have been something like $-1$ for $x\le0$, $1$ for $x\gt0$, which is (Riemann-)integrable on bounded intervals. I'll stand by my statement - you just haven't taught calculus at the level of Math 130 at Macquarie. – Gerry Myerson Dec 7 '11 at 3:35 If you mean: How do you know the integral exists and has a finite value, the answer is simply that it's the integral of a continuous function over a bounded interval. -	2016-07-24 05:26:40	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9346950054168701, "perplexity": 271.40433624729496}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257823947.97/warc/CC-MAIN-20160723071023-00120-ip-10-185-27-174.ec2.internal.warc.gz"}
http://quantumfrontiers.com/category/news/	# Macroscopic quantum teleportation: the story of my chair In the summer of 2000, a miracle occurred: The National Science Foundation decided to fund a new Institute for Quantum Information at Caltech with a 5 million dollar award from their Information Technology Research program. I was to be the founding director of the IQI. Jeff Kimble explained to me why we should propose establishing the IQI. He knew I had used my slice of our shared DARPA grant to bring Alexei Kitaev to Caltech as a visiting professor, which had been wonderful. Recalling how much we had both benefited from Kitaev’s visit, Jeff remarked emphatically that “This stuff’s not free.” He had a point. To have more fun we’d need more money. Jeff took the lead in recruiting a large team of Caltech theorists and experimentalists to join the proposal we submitted, but the NSF was primarily interested in supporting the theory of quantum computation rather than the experimental part of the proposal. That was how I wound up in charge, though I continued to rely on Jeff’s advice and support. This was a new experience for me and I worried a lot about how directing an institute would change my life. But I had one worry above all: space. We envisioned a thriving institute brimming over with talented and enthusiastic young scientists and visitors drawn from the physics, computer science, and engineering communities. But how could we carve out a place on the Caltech campus where they could work and interact? To my surprise and delight, Jeff and I soon discovered that someone else at Caltech shared our excitement over the potential of IQI — Richard Murray, who was then the Chair of Caltech’s Division of Engineering and Applied Science. Richard arranged for the IQI to occupy office space in Steele Laboratory and some space we could configure as we pleased in Jorgensen Laboratory. The hub of the IQI became the lounge in Jorgensen, which we used for our seminar receptions, group meetings, and innumerable informal discussions, until our move to the beautiful Annenberg Center when it opened in 2009. I sketched a rough plan for the Jorgensen layout, including furniture for the lounge. The furniture, I was told, was “NIC”. Though I was too embarrassed to ask, I eventually inferred this meant “Not in Contract” — I would need to go furniture shopping, one of my many burgeoning responsibilities as Director. By this time, Ann Harvey was in place as IQI administrator, a huge relief. But furniture was something I thought I knew about, because I had designed and furnished a common area for the particle theory group a couple of years earlier. As we had done on that previous occasion, my wife Roberta and I went to Krause’s Sofa Factory to order a custom-made couch, love seat, and lounge chair, in a grayish green leather which we thought would blend well with the carpeting. Directing an institute is not as simple as it sounds, though. Before the furniture was delivered, Krause’s declared bankruptcy! We had paid in full, but I had some anxious moments wondering whether there would be a place to sit down in the IQI lounge. In the end, after some delay, our furniture was delivered in time for the grand opening of the new space in September 2001. A happy ending, but not really the end of the story. Before the move to Annenberg in 2009, I ordered furniture to fill our (much smaller) studio space, which became the new IQI common area. The Jorgensen furniture was retired, and everything was new! It was nice … But every once in a while I felt a twinge of sadness. I missed my old leather chair, from which I had pontificated at eight years worth of group meetings. That chair and I had been through a lot together, and I couldn’t help but feel that my chair’s career had been cut short before its time. I don’t recall mentioning these feelings to anyone, but someone must have sensed by regrets. Because one day not long after the move another miracle occurred … my chair was baaack! Sitting in it again felt … good. For five years now I’ve been pontificating from my old chair in our new studio, just like I used to. No one told me how my chair had been returned to me, and I knew better than to ask. My chair today. Like me, a bit worn but still far from retirement. Eventually the truth comes out. At my 60th birthday celebration last year, Stephanie Wehner and Darrick Chang admitted to being the perpetrators, and revealed the whole amazing story in their article on “Macroscopic Quantum Teleportation” in a special issue of Nature Relocations. Their breakthrough article was enhanced by Stephanie’s extraordinary artwork, which you really have to see to believe. So if your curiosity is piqued, please follow this link to find out more. Why, you may wonder, am I reminiscing today about the story of my chair? Well, is an excuse really necessary? But if you must know, it may be because, after two renewals and 14 years of operation, I submitted the IQI Final Report to the NSF this week. Don’t worry — the Report is not really Final, because the IQI has become part of an even grander vision, the IQIM (which has given birth to this blog among other good things). Like my chair, the IQI is not quite what it was, yet it lives on. The nostalgic feelings aroused by filing the Final Report led me to reread the wonderful volume my colleagues put together for my birthday celebration, which recounts not only the unforgettable exploits of Stephanie and Darrick, but many other stories and testimonials that deeply touched me. Browsing through that book today, one thing that struck me is the ways we sometimes have impact on others without even being aware of it. For example, Aram Harrow, Debbie Leung, Joe Renes and Stephanie all remember lectures I gave when they were undergraduate students (before I knew them), which might have influenced their later research careers. Knowing this will make it a little harder to say no the next time I’m invited to give a talk. Yaoyun Shi has vivid memories of the time I wore my gorilla mask to the IQI seminar on Halloween, which inspired him to dress up as “a butcher threatening to cut off the ears of my students with a bloody machete if they were not listening,” thus boosting his teaching evaluations. And Alexios Polychronakos, upon hearing that I had left particle theory to pursue quantum computing, felt it “was a bit like watching your father move to Las Vegas and marry a young dancer after you leave for college,” while at the same time he appreciated “that such reinventions are within the spectrum of possibilities for physicists who still have a pulse.” I’m proud of what the IQI(M) has accomplished, but we’re just getting started. After 14 years, I still have a pulse, and my chair has plenty of wear left. Together we look forward to many more years of pontification. # John Preskill and the dawn of the entanglement frontier Editor’s Note: John Preskill’s recent election to the National Academy of Sciences generated a lot of enthusiasm among his colleagues and students. In an earlier post today, famed Stanford theoretical physicist, Leonard Susskind, paid tribute to John’s early contributions to physics ranging from magnetic monopoles to the quantum mechanics of black holes. In this post, Daniel Gottesman, a faculty member at the Perimeter Institute, takes us back to the formative years of the Institute for Quantum Information at Caltech, the precursor to IQIM and a world-renowned incubator for quantum information and quantum computation research. Though John shies away from the spotlight, we, at IQIM, believe that the integrity of his character and his role as a mentor and catalyst for science are worthy of attention and set a good example for current and future generations of theoretical physicists. Preskill’s legacy may well be the incredible number of preeminent research scientists in quantum physics he has mentored throughout his extraordinary career. When someone wins a big award, it has become traditional on this blog for John Preskill to write something about them. The system breaks down, though, when John is the one winning the award. Therefore I’ve been brought in as a pinch hitter (or should it be pinch lionizer?). The award in this case is that John has been elected to the National Academy of Sciences, along with Charlie Kane and a number of other people that don’t work on quantum information. Lenny Susskind has already written about John’s work on other topics; I will focus on quantum information. On the research side of quantum information, John is probably best known for his work on fault-tolerant quantum computation, particularly topological fault tolerance. John jumped into the field of quantum computation in 1994 in the wake of Shor’s algorithm, and brought me and some of his other grad students with him. It was obvious from the start that error correction was an important theoretical challenge (emphasized, for instance, by Unruh), so that was one of the things we looked at. We couldn’t figure out how to do it, but some other people did. John and I embarked on a long drawn-out project to get good bounds on the threshold error rate. If you can build a quantum computer with an error rate below the threshold value, you can do arbitrarily large quantum computations. If not, then errors will eventually overwhelm you. Early versions of my project with John suggested that the threshold should be about $10^{-4}$, and the number began floating around (somewhat embarrassingly) as the definitive word on the threshold value. Our attempts to bound the higher-order terms in the computation became rather grotesque, and the project proceeded very slowly until a new approach and the recruitment of Panos Aliferis finally let us finish a paper with a rigorous proof of a slightly lower threshold value. Meanwhile, John had also been working on topological quantum computation. John has already written about his excitement when Kitaev visited Caltech and talked about the toric code. The two of them, plus Eric Dennis and Andrew Landahl, studied the application of this code for fault tolerance. If you look at the citations of this paper over time, it looks rather … exponential. For a while, topological things were too exotic for most quantum computer people, but over time, the virtues of surface codes have become obvious (apparently high threshold, convenient for two-dimensional architectures). It’s become one of the hot topics in recent years and there are no signs of flagging interest in the community. John has also made some important contributions to security proofs for quantum key distribution, known to the cognoscenti just by its initials. QKD allows two people (almost invariably named Alice and Bob) to establish a secret key by sending qubits over an insecure channel. If the eavesdropper Eve tries to live up to her name, her measurements of the qubits being transmitted will cause errors revealing her presence. If Alice and Bob don’t detect the presence of Eve, they conclude that she is not listening in (or at any rate hasn’t learned much about the secret key) and therefore they can be confident of security when they later use the secret key to encrypt a secret message. With Peter Shor, John gave a security proof of the best-known QKD protocol, known as the “Shor-Preskill” proof. Sometimes we scientists lack originality in naming. It was not the first proof of security, but earlier ones were rather complicated. The Shor-Preskill proof was conceptually much clearer and made a beautiful connection between the properties of quantum error-correcting codes and QKD. The techniques introduced in their paper got adopted into much later work on quantum cryptography. Collaborating with John is always an interesting experience. Sometimes we’ll discuss some idea or some topic and it will be clear that John does not understand the idea clearly or knows little about the topic. Then, a few days later we discuss the same subject again and John is an expert, or at least he knows a lot more than me. I guess this ability to master topics quickly is why he was always able to answer Steve Flammia’s random questions after lunch. And then when it comes time to write the paper … John will do it. It’s not just that he will volunteer to write the first draft — he keeps control of the whole paper and generally won’t let you edit the source, although of course he will incorporate your comments. I think this habit started because of incompatibilities between the TeX editor he was using and any other program, but he maintains it (I believe) to make sure that the paper meets his high standards of presentation quality. This also explains why John has been so successful as an expositor. His lecture notes for the quantum computation class at Caltech are well-known. Despite being incomplete and not available on Amazon, they are probably almost as widely read as the standard textbook by Nielsen and Chuang. Before IQIM, there was IQI, and before that was QUIC. He apparently is also good at writing grants. Under his leadership and Jeff Kimble’s, Caltech has become one of the top places for quantum computation. In my last year of graduate school, John and Jeff, along with Steve Koonin, secured the QUIC grant, and all of a sudden Caltech had money for quantum computation. I got a research assistantship and could write my thesis without having to worry about TAing. Postdocs started to come — first Chris Fuchs, then a long stream of illustrious others. The QUIC grant grew into IQI, and that eventually sprouted an M and drew in even more people. When I was a student, John’s group was located in Lauritsen with the particle theory group. We had maybe three grad student offices (and not all the students were working on quantum information), plus John’s office. As the Caltech quantum effort grew, IQI acquired territory in another building, then another, and then moved into a good chunk of the new Annenberg building. Without John’s efforts, the quantum computing program at Caltech would certainly be much smaller and maybe completely lacking a theory side. It’s also unlikely this blog would exist. The National Academy has now elected John a member, probably more for his research than his twitter account (@preskill), though I suppose you never know. Anyway, congratulations, John! -D. Gottesman # Of magnetic monopoles and fast-scrambling black holes Editor’s Note: On April 29th, 2014, the National Academy of Sciences announced the new electees to the prestigious organization. This was an especially happy occasion for everyone here at IQIM, since the new members included our very own John Preskill, Richard P. Feynman Professor of Theoretical Physics and regular blogger on this site. A request was sent to Leonard Susskind, a close friend and collaborator of John’s, to take a trip down memory lane and give the rest of us a glimpse of some of John’s early contributions to Physics. John, congratulations from all of us here at IQIM. John Preskill was elected to the National Academy of Sciences, an event long overdue. Perhaps it took longer than it should have because there is no way to pigeon-hole him; he is a theoretical physicist, and that’s all there is to it. John has long been one of my heroes in theoretical physics. There is something very special about his work. It has exceptional clarity, it has vision, it has integrity—you can count on it. And sometimes it has another property: it can surprise. The first time I heard his name come up, sometime around 1979, I was not only surprised; I was dismayed. A student whose name I had never heard of, had uncovered a serious clash between two things, both of which I deeply wanted to believe in. One was the Big-Bang theory and the other was the discovery of grand unified particle theories. Unification led to the extraordinary prediction that Dirac’s magnetic monopoles must exist, at least in principle. The Big-Bang theory said they must exist in fact. The extreme conditions at the beginning of the universe were exactly what was needed to create loads of monopoles; so many that they would flood the universe with too much mass. John, the unknown graduate student, did a masterful analysis. It left no doubt that something had to give. Cosmology gave. About a year later, inflationary cosmology was discovered by Guth who was in part motivated by Preskill’s monopole puzzle. John’s subsequent career as a particle physicist was marked by a number of important insights which often had that surprising quality. The cosmology of the invisible axion was one. Others had to do with very subtle and counterintuitive features of quantum field theory, like the existence of “Alice strings”. In the very distant past, Roger Penrose and I had a peculiar conversation about possible generalizations of the Aharonov-Bohm effect. We speculated on all sorts of things that might happen when something is transported around a string. I think it was Roger who got excited about the possibilities that might result if a topological defect could change gender. Alice strings were not quite that exotic, only electric charge flips, but nevertheless it was very surprising. John of course had a long standing interest in the quantum mechanics of black holes: I will quote a passage from a visionary 1992 review paper, “Do Black Holes Destroy Information? “I conclude that the information loss paradox may well presage a revolution in fundamental physics.” At that time no one knew the answer to the paradox, although a few of us, including John, thought the answer was that information could not be lost. But almost no one saw the future as clearly as John did. Our paths crossed in 1993 in a very exciting discussion about black holes and information. We were both thinking about the same thing, now called black hole complementarity. We were concerned about quantum cloning if information is carried by Hawking radiation. We thought we knew the answer: it takes too long to retrieve the information to then be able to jump into the black hole and discover the clone. This is probably true, but at that time we had no idea how close a call this might be. It took until 2007 to properly formulate the problem. Patrick Hayden and John Preskill utterly surprised me, and probably everyone else who had been thinking about black holes, with their now-famous paper “Black Holes as Mirrors.” In a sense, this paper started a revolution in applying the powerful methods of quantum information theory to black holes. We live in the age of entanglement. From quantum computing to condensed matter theory, to quantum gravity, entanglement is the new watchword. Preskill was in the vanguard of this revolution, but he was also the teacher who made the new concepts available to physicists like myself. We can now speak about entanglement, error correction, fault tolerance, tensor networks and more. The Preskill lectures were the indispensable source of knowledge and insight for us. Congratulations John. And congratulations NAS. -L. S. # Tsar Nikita and His Scientists Once upon a time, a Russian tsar named Nikita had forty daughters: Every one from top to toe Was a captivating creature, Perfect—but for one lost feature. So wrote Alexander Pushkin, the 19th-century Shakespeare who revolutionized Russian literature. In a rhyme, Pushkin imagined forty princesses born without “that bit” “[b]etween their legs.” A courier scours the countryside for a witch who can help. By summoning the devil in the woods, she conjures what the princesses lack into a casket. The tsar parcels out the casket’s contents, and everyone rejoices. “[N]onsense,” Pushkin calls the tale in its penultimate line. A “joke.” The joke has, nearly two centuries later, become reality. Researchers have grown vaginas in a lab and implanted them into teenage girls. Thanks to a genetic defect, the girls suffered from Mayer-Rokitansky-Küster-Hauser (MRKH) syndrome: Their vaginas and uteruses had failed to grow to maturity or at all. A team at Wake Forest and in Mexico City took samples of the girls’ cells, grew more cells, and combined their harvest with vagina-shaped scaffolds. Early in the 2000s, surgeons implanted the artificial organs into the girls. The patients, the researchers reported in the journal The Lancet last week, function normally. I don’t usually write about reproductive machinery. But the implants’ resonance with “Tsar Nikita” floored me. Scientists have implanted much of Pushkin’s plot into labs. The sexually deficient girls, the craftsperson, the replacement organs—all appear in “Tsar Nikita” as in The Lancet. In poetry as in science fiction, we read the future. Though threads of Pushkin’s plot survive, society’s view of the specialist has progressed. “Deep [in] the dark woods” lives Pushkin’s witch. Upon summoning the devil, she locks her cure in a casket. Today’s vagina-implanters star in headlines. The Wall Street Journal highlighted the implants in its front section. Unless the patients’ health degrades, the researchers will likely list last week’s paper high on their CVs and websites. Much as Dr. Atlántida Raya-Rivera, the paper’s lead author, differs from Pushkin’s witch, the visage of Pushkin’s magic wears the nose and eyebrows of science. When tsars or millenials need medical help, they seek knowledge-keepers: specialists, a fringe of society. Before summoning the devil, the witch “[l]ocked her door . . . Three days passed.” I hide away to calculate and study (though days alone might render me more like the protagonist in another Russian story, Chekhov’s “The Bet”). Just as the witch “stocked up coal,” some students stockpile Red Bull before hitting the library. Some habits, like the archetype of the wise woman, refuse to die. From a Russian rhyme, the bones of “Tsar Nikita” have evolved into cutting-edge science. Pushkin and the implants highlight how attitudes toward knowledge have changed, offering a lens onto science in culture and onto science culture. No wonder readers call Pushkin “timeless.” But what would he have rhymed with “Mayer-Rokitansky-Küster-Hauser”? “Tsar Nikita” has many nuances—messages about censorship, for example—that I didn’t discuss. To the intrigued, I recommend The Queen of Spades: And selected works, translated by Anthony Briggs and published by Pushkin Press. # IQIM Presents …”my father” Debaleena Nandi at Caltech Following the IQIM teaser, which was made with the intent of creating a wider perspective of the scientist, to highlight the normalcy behind the perception of brilliance and to celebrate the common human struggles to achieve greatness, we decided to do individual vignettes of some of the characters you saw in the video. We start with Debaleena Nandi, a grad student in Prof Jim Eisenstein’s lab, whose journey from Jadavpur University in West Bengal, India to the graduate school and research facility at the Indian institute of Science, Bangalore, to Caltech has seen many obstacles. We focus on the essentials of an environment needed to manifest the quest for “the truth” as Debaleena says. We start with her days as a child when her double-shift working father sat by her through the days and nights that she pursued her homework. She highlights what she feels is the only way to growth; working on what is lacking, to develop that missing tool in your skill set, that asset that others might have by birth but you need to inspire by hard work. Debaleena’s motto: to realize and face your shortcomings is the only way to achievement. As we build Debaleena up, we also build up the identity of Caltech through its breathtaking architecture that oscillates from Spanish to Goth to modern. Both Debaleena and Caltech are revealed slowly, bit by bit. This series is about dissecting high achievers, seeing the day to day steps, the bit by bit that adds up to the more often than not, overwhelming, impressive presence of Caltech’s science. We attempt to break it down in smaller vignettes that help us appreciate the amount of discipline, intent and passion that goes into making cutting edge researchers. Presenting the emotional alongside the rational is something this series aspires to achieve. It honors and celebrates human limitations surrounding limitless boundaries, discoveries and possibilities. Stay tuned for more vignettes in the IQIM Presents “My _______” Series. But for now, here is the video. Watch, like and share! (C) Parveen Shah Production 2014 # Inflation on the back of an envelope Last Monday was an exciting day! After following the BICEP2 announcement via Twitter, I had to board a transcontinental flight, so I had 5 uninterrupted hours to think about what it all meant. Without Internet access or references, and having not thought seriously about inflation for decades, I wanted to reconstruct a few scraps of knowledge needed to interpret the implications of r ~ 0.2. I did what any physicist would have done … I derived the basic equations without worrying about niceties such as factors of 3 or $2 \pi$. None of what I derived was at all original — the theory has been known for 30 years — but I’ve decided to turn my in-flight notes into a blog post. Experts may cringe at the crude approximations and overlooked conceptual nuances, not to mention the missing references. But some mathematically literate readers who are curious about the implications of the BICEP2 findings may find these notes helpful. I should emphasize that I am not an expert on this stuff (anymore), and if there are serious errors I hope better informed readers will point them out. By tradition, careless estimates like these are called “back-of-the-envelope” calculations. There have been times when I have made notes on the back of an envelope, or a napkin or place mat. But in this case I had the presence of mind to bring a notepad with me. Notes from a plane ride According to inflation theory, a nearly homogeneous scalar field called the inflaton (denoted by $\phi$) filled the very early universe. The value of $\phi$ varied with time, as determined by a potential function $V(\phi)$. The inflaton rolled slowly for a while, while the dark energy stored in $V(\phi)$ caused the universe to expand exponentially. This rapid cosmic inflation lasted long enough that previously existing inhomogeneities in our currently visible universe were nearly smoothed out. What inhomogeneities remained arose from quantum fluctuations in the inflaton and the spacetime geometry occurring during the inflationary period. Gradually, the rolling inflaton picked up speed. When its kinetic energy became comparable to its potential energy, inflation ended, and the universe “reheated” — the energy previously stored in the potential $V(\phi)$ was converted to hot radiation, instigating a “hot big bang”. As the universe continued to expand, the radiation cooled. Eventually, the energy density in the universe came to be dominated by cold matter, and the relic fluctuations of the inflaton became perturbations in the matter density. Regions that were more dense than average grew even more dense due to their gravitational pull, eventually collapsing into the galaxies and clusters of galaxies that fill the universe today. Relic fluctuations in the geometry became gravitational waves, which BICEP2 seems to have detected. Both the density perturbations and the gravitational waves have been detected via their influence on the inhomogeneities in the cosmic microwave background. The 2.726 K photons left over from the big bang have a nearly uniform temperature as we scan across the sky, but there are small deviations from perfect uniformity that have been precisely measured. We won’t worry about the details of how the size of the perturbations is inferred from the data. Our goal is to achieve a crude understanding of how the density perturbations and gravitational waves are related, which is what the BICEP2 results are telling us about. We also won’t worry about the details of the shape of the potential function $V(\phi)$, though it’s very interesting that we might learn a lot about that from the data. Exponential expansion Einstein’s field equations tell us how the rate at which the universe expands during inflation is related to energy density stored in the scalar field potential. If a(t) is the “scale factor” which describes how lengths grow with time, then roughly $\left(\frac{\dot a}{a}\right)^2 \sim \frac{V}{m_P^2}$. Here $\dot a$ means the time derivative of the scale factor, and $m_P = 1/\sqrt{8 \pi G} \approx 2.4 \times 10^{18}$ GeV is the Planck scale associated with quantum gravity. (G is Newton’s gravitational constant.) I’ve left our a factor of 3 on purpose, and I used the symbol ~ rather than = to emphasize that we are just trying to get a feel for the order of magnitude of things. I’m using units in which Planck’s constant $\hbar$ and the speed of light c are set to one, so mass, energy, and inverse length (or inverse time) all have the same dimensions. 1 GeV means one billion electron volts, about the mass of a proton. (To persuade yourself that this is at least roughly the right equation, you should note that a similar equation applies to an expanding spherical ball of radius a(t) with uniform mass density V. But in the case of the ball, the mass density would decrease as the ball expands. The universe is different — it can expand without diluting its mass density, so the rate of expansion $\dot a / a$ does not slow down as the expansion proceeds.) During inflation, the scalar field $\phi$ and therefore the potential energy $V(\phi)$ were changing slowly; it’s a good approximation to assume $V$ is constant. Then the solution is $a(t) \sim a(0) e^{Ht},$ where $H$, the Hubble constant during inflation, is $H \sim \frac{\sqrt{V}}{m_P}.$ To explain the smoothness of the observed universe, we require at least 50 “e-foldings” of inflation before the universe reheated — that is, inflation should have lasted for a time at least $50 H^{-1}$. Slow rolling During inflation the inflaton $\phi$ rolls slowly, so slowly that friction dominates inertia — this friction results from the cosmic expansion. The speed of rolling $\dot \phi$ is determined by $H \dot \phi \sim -V'(\phi).$ Here $V'(\phi)$ is the slope of the potential, so the right-hand side is the force exerted by the potential, which matches the frictional force on the left-hand side. The coefficient of $\dot \phi$ has to be $H$ on dimensional grounds. (Here I have blown another factor of 3, but let’s not worry about that.) Density perturbations The trickiest thing we need to understand is how inflation produced the density perturbations which later seeded the formation of galaxies. There are several steps to the argument. Quantum fluctuations of the inflaton As the universe inflates, the inflaton field is subject to quantum fluctuations, where the size of the fluctuation depends on its wavelength. Due to inflation, the wavelength increases rapidly, like $e^{Ht}$, and once the wavelength gets large compared to $H^{-1}$, there isn’t enough time for the fluctuation to wiggle — it gets “frozen in.” Much later, long after the reheating of the universe, the oscillation period of the wave becomes comparable to the age of the universe, and then it can wiggle again. (We say that the fluctuations “cross the horizon” at that stage.) Observations of the anisotropy of the microwave background have determined how big the fluctuations are at the time of horizon crossing. What does inflation theory say about that? Well, first of all, how big are the fluctuations when they leave the horizon during inflation? Then the wavelength is $H^{-1}$ and the universe is expanding at the rate $H$, so $H$ is the only thing the magnitude of the fluctuations could depend on. Since the field $\phi$ has the same dimensions as $H$, we conclude that fluctuations have magnitude $\delta \phi \sim H.$ From inflaton fluctuations to density perturbations Reheating occurs abruptly when the inflaton field reaches a particular value. Because of the quantum fluctuations, some horizon volumes have larger than average values of $\phi$ and some have smaller than average values; hence different regions reheat at slightly different times. The energy density in regions that reheat earlier starts to be reduced by expansion (“red shifted”) earlier, so these regions have a smaller than average energy density. Likewise, regions that reheat later start to red shift later, and wind up having larger than average density. When we compare different regions of comparable size, we can find the typical (root-mean-square) fluctuations $\delta t$ in the reheating time, knowing the fluctuations in $\phi$ and the rolling speed $\dot \phi$: $\delta t \sim \frac{\delta \phi}{\dot \phi} \sim \frac{H}{\dot\phi}.$ Small fractional fluctuations in the scale factor $a$ right after reheating produce comparable small fractional fluctuations in the energy density $\rho$. The expansion rate right after reheating roughly matches the expansion rate $H$ right before reheating, and so we find that the characteristic size of the density perturbations is $\delta_S\equiv\left(\frac{\delta \rho}{\rho}\right)_{hor} \sim \frac{\delta a}{a} \sim \frac{\dot a}{a} \delta t\sim \frac{H^2}{\dot \phi}.$ The subscript hor serves to remind us that this is the size of density perturbations as they cross the horizon, before they get a chance to grow due to gravitational instabilities. We have found our first important conclusion: The density perturbations have a size determined by the Hubble constant $H$ and the rolling speed $\dot \phi$ of the inflaton, up to a factor of order one which we have not tried to keep track of. Insofar as the Hubble constant and rolling speed change slowly during inflation, these density perturbations have a strength which is nearly independent of the length scale of the perturbation. From here on we will denote this dimensionless scale of the fluctuations by $\delta_S$, where the subscript $S$ stands for “scalar”. Perturbations in terms of the potential Putting together $\dot \phi \sim -V' / H$ and $H^2 \sim V/{m_P}^2$ with our expression for $\delta_S$, we find $\delta_S^2 \sim \frac{H^4}{\dot\phi^2}\sim \frac{H^6}{V'^2} \sim \frac{1}{{m_P}^6}\frac{V^3}{V'^2}.$ The observed density perturbations are telling us something interesting about the scalar field potential during inflation. Gravitational waves and the meaning of r The gravitational field as well as the inflaton field is subject to quantum fluctuations during inflation. We call these tensor fluctuations to distinguish them from the scalar fluctuations in the energy density. The tensor fluctuations have an effect on the microwave anisotropy which can be distinguished in principle from the scalar fluctuations. We’ll just take that for granted here, without worrying about the details of how it’s done. While a scalar field fluctuation with wavelength $\lambda$ and strength $\delta \phi$ carries energy density $\sim \delta\phi^2 / \lambda^2$, a fluctuation of the dimensionless gravitation field $h$ with wavelength $\lambda$ and strength $\delta h$ carries energy density $\sim m_P^2 \delta h^2 / \lambda^2$. Applying the same dimensional analysis we used to estimate $\delta \phi$ at horizon crossing to the rescaled field $h/m_P$, we estimate the strength $\delta_T$ of the tensor fluctuations as $\delta_T^2 \sim \frac{H^2}{m_P^2}\sim \frac{V}{m_P^4}.$ From observations of the CMB anisotropy we know that $\delta_S\sim 10^{-5}$, and now BICEP2 claims that the ratio $r = \frac{\delta_T^2}{\delta_S^2}$ is about $r\sim 0.2$ at an angular scale on the sky of about one degree. The conclusion (being a little more careful about the O(1) factors this time) is $V^{1/4} \sim 2 \times 10^{16}~GeV \left(\frac{r}{0.2}\right)^{1/4}.$ This is our second important conclusion: The energy density during inflation defines a mass scale, which turns our to be $2 \times 10^{16}~GeV$ for the observed value of $r$. This is a very interesting finding because this mass scale is not so far below the Planck scale, where quantum gravity kicks in, and is in fact pretty close to theoretical estimates of the unification scale in supersymmetric grand unified theories. If this mass scale were a factor of 2 smaller, then $r$ would be smaller by a factor of 16, and hence much harder to detect. Rolling, rolling, rolling, … Using $\delta_S^2 \sim H^4/\dot\phi^2$, we can express $r$ as $r = \frac{\delta_T^2}{\delta_S^2}\sim \frac{\dot\phi^2}{m_P^2 H^2}.$ It is convenient to measure time in units of the number $N = H t$ of e-foldings of inflation, in terms of which we find $\frac{1}{m_P^2} \left(\frac{d\phi}{dN}\right)^2\sim r;$ Now, we know that for inflation to explain the smoothness of the universe we need $N$ larger than 50, and if we assume that the inflaton rolls at a roughly constant rate during $N$ e-foldings, we conclude that, while rolling, the change in the inflaton field is $\frac{\Delta \phi}{m_P} \sim N \sqrt{r}.$ This is our third important conclusion — the inflaton field had to roll a long, long, way during inflation — it changed by much more than the Planck scale! Putting in the O(1) factors we have left out reduces the required amount of rolling by about a factor of 3, but we still conclude that the rolling was super-Planckian if $r\sim 0.2$. That’s curious, because when the scalar field strength is super-Planckian, we expect the kind of effective field theory we have been implicitly using to be a poor approximation because quantum gravity corrections are large. One possible way out is that the inflaton might have rolled round and round in a circle instead of in a straight line, so the field strength stayed sub-Planckian even though the distance traveled was super-Planckian. Spectral tilt As the inflaton rolls, the potential energy, and hence also the Hubble constant $H$, change during inflation. That means that both the scalar and tensor fluctuations have a strength which is not quite independent of length scale. We can parametrize the scale dependence in terms of how the fluctuations change per e-folding of inflation, which is equivalent to the change per logarithmic length scale and is called the “spectral tilt.” To keep things simple, let’s suppose that the rate of rolling is constant during inflation, at least over the length scales for which we have data. Using $\delta_S^2 \sim H^4/\dot\phi^2$, and assuming $\dot\phi$ is constant, we estimate the scalar spectral tilt as $-\frac{1}{\delta_S^2}\frac{d\delta_S^2}{d N} \sim - \frac{4 \dot H}{H^2}.$ Using $\delta_T^2 \sim H^2/m_P^2$, we conclude that the tensor spectral tilt is half as big. From $H^2 \sim V/m_P^2$, we find $\dot H \sim \frac{1}{2} \dot \phi \frac{V'}{V} H,$ and using $\dot \phi \sim -V'/H$ we find $-\frac{1}{\delta_S^2}\frac{d\delta_S^2}{d N} \sim \frac{V'^2}{H^2V}\sim m_P^2\left(\frac{V'}{V}\right)^2\sim \left(\frac{V}{m_P^4}\right)\left(\frac{m_P^6 V'^2}{V^3}\right)\sim \delta_T^2 \delta_S^{-2}\sim r.$ Putting in the numbers more carefully we find a scalar spectral tilt of $r/4$ and a tensor spectral tilt of $r/8$. This is our last important conclusion: A relatively large value of $r$ means a significant spectral tilt. In fact, even before the BICEP2 results, the CMB anisotropy data already supported a scalar spectral tilt of about .04, which suggested something like $r \sim .16$. The BICEP2 detection of the tensor fluctuations (if correct) has confirmed that suspicion. Summing up If you have stuck with me this far, and you haven’t seen this stuff before, I hope you’re impressed. Of course, everything I’ve described can be done much more carefully. I’ve tried to convey, though, that the emerging story seems to hold together pretty well. Compared to last week, we have stronger evidence now that inflation occurred, that the mass scale of inflation is high, and that the scalar and tensor fluctuations produced during inflation have been detected. One prediction is that the tensor fluctuations, like the scalar ones, should have a notable spectral tilt, though a lot more data will be needed to pin that down. I apologize to the experts again, for the sloppiness of these arguments. I hope that I have at least faithfully conveyed some of the spirit of inflation theory in a way that seems somewhat accessible to the uninitiated. And I’m sorry there are no references, but I wasn’t sure which ones to include (and I was too lazy to track them down). It should also be clear that much can be done to sharpen the confrontation between theory and experiment. A whole lot of fun lies ahead. Okay, here’s a good reference, a useful review article by Baumann. (I found out about it on Twitter!) From Baumann’s lectures I learned a convenient notation. The rolling of the inflaton can be characterized by two “potential slow-roll parameters” defined by $\epsilon = \frac{m_p^2}{2}\left(\frac{V'}{V}\right)^2,\quad \eta = m_p^2\left(\frac{V''}{V}\right).$ Both parameters are small during slow rolling, but the relationship between them depends on the shape of the potential. My crude approximation ($\epsilon = \eta$) would hold for a quadratic potential. We can express the spectral tilt (as I defined it) in terms of these parameters, finding $2\epsilon$ for the tensor tilt, and $6 \epsilon - 2\eta$ for the scalar tilt. To derive these formulas it suffices to know that $\delta_S^2$ is proportional to $V^3/V'^2$, and that $\delta_T^2$ is proportional to $H^2$; we also use $3H\dot \phi = -V', \quad 3H^2 = V/m_P^2,$ keeping factors of 3 that I left out before. (As a homework exercise, check these formulas for the tensor and scalar tilt.) It is also easy to see that $r$ is proportional to $\epsilon$; it turns out that $r = 16 \epsilon$. To get that factor of 16 we need more detailed information about the relative size of the tensor and scalar fluctuations than I explained in the post; I can’t think of a handwaving way to derive it. We see, though, that the conclusion that the tensor tilt is $r/8$ does not depend on the details of the potential, while the relation between the scalar tilt and $r$ does depend on the details. Nevertheless, it seems fair to claim (as I did) that, already before we knew the BICEP2 results, the measured nonzero scalar spectral tilt indicated a reasonably large value of $r$. Once again, we’re lucky. On the one hand, it’s good to have a robust prediction (for the tensor tilt). On the other hand, it’s good to have a handle (the scalar tilt) for distinguishing among different inflationary models. One last point is worth mentioning. We have set Planck’s constant $\hbar$ equal to one so far, but it is easy to put the powers of $\hbar$ back in using dimensional analysis (we’ll continue to assume the speed of light c is one). Since Newton’s constant $G$ has the dimensions of length/energy, and the potential $V$ has the dimensions of energy/volume, while $\hbar$ has the dimensions of energy times length, we see that $\delta_T^2 \sim \hbar G^2V.$ Thus the production of gravitational waves during inflation is a quantum effect, which would disappear in the limit $\hbar \to 0$. Likewise, the scalar fluctuation strength $\delta_S^2$ is also $O(\hbar)$, and hence also a quantum effect. Therefore the detection of primordial gravitational waves by BICEP2, if correct, confirms that gravity is quantized just like the other fundamental forces. That shouldn’t be a surprise, but it’s nice to know. # My 10 biggest thrills Wow! Evidence for gravitational waves produced during cosmic inflation. BICEP2 results for the ratio r of gravitational wave perturbations to density perturbations, and the density perturbation spectral tilt n. Like many physicists, I have been reflecting a lot the past few days about the BICEP2 results, trying to put them in context. Other bloggers have been telling you all about it (here, here, and here, for example); what can I possibly add? The hoopla this week reminds me of other times I have been really excited about scientific advances. And I recall some wise advice I received from Sean Carroll: blog readers like lists. So here are (in chronological order)… My 10 biggest thrills (in science) This is a very personal list — your results may vary. I’m not saying these are necessarily the most important discoveries of my lifetime (there are conspicuous omissions), just that, as best I can recall, these are the developments that really started my heart pounding at the time. 1) The J/Psi from below (1974) I was a senior at Princeton during the November Revolution. I was too young to appreciate fully what it was all about — having just learned about the Weinberg-Salam model, I thought at first that the Z boson had been discovered. But by stalking the third floor of Jadwin I picked up the buzz. No, it was charm! The discovery of a very narrow charmonium resonance meant we were on the right track in two ways — charm itself confirmed ideas about the electroweak gauge theory, and the narrowness of the resonance fit in with the then recent idea of asymptotic freedom. Theory triumphant! 2) A magnetic monopole in Palo Alto (1982) By 1982 I had been thinking about the magnetic monopoles in grand unified theories for a few years. We thought we understood why no monopoles seem to be around. Sure, monopoles would be copiously produced in the very early universe, but then cosmic inflation would blow them away, diluting their density to a hopelessly undetectable value. Then somebody saw one …. a magnetic monopole obediently passed through Blas Cabrera’s loop of superconducting wire, producing a sudden jump in the persistent current. On Valentine’s Day! According to then current theory, the monopole mass was expected to be about 10^16 GeV (10 million billion times heavier than a proton). Had Nature really been so kind as the bless us with this spectacular message from an staggeringly high energy scale? It seemed too good to be true. It was. Blas never detected another monopole. As far as I know he never understood what glitch had caused the aberrant signal in his device. 3) “They’re green!” High-temperature superconductivity (1987) High-temperature superconductors were discovered in 1986 by Bednorz and Mueller, but I did not pay much attention until Paul Chu found one in early 1987 with a critical temperature of 77 K. Then for a while the critical temperature seemed to be creeping higher and higher on an almost daily basis, eventually topping 130K …. one wondered whether it might go up, up, up forever. It didn’t. Today 138K still seems to be the record. My most vivid memory is that David Politzer stormed into my office one day with a big grin. “They’re green!” he squealed. David did not mean that high-temperature superconductors would be good for the environment. He was passing on information he had just learned from Phil Anderson, who happened to be visiting Caltech: Chu’s samples were copper oxides. 4) “Now I have mine” Supernova 1987A (1987) What was most remarkable and satisfying about the 1987 supernova in the nearby Large Magellanic Cloud was that the neutrinos released in a ten second burst during the stellar core collapse were detected here on earth, by gigantic water Cerenkov detectors that had been built to test grand unified theories by looking for proton decay! Not a truly fundamental discovery, but very cool nonetheless. Soon after it happened some of us were loafing in the Lauritsen seminar room, relishing the good luck that had made the detection possible. Then Feynman piped up: “Tycho Brahe had his supernova, Kepler had his, … and now I have mine!” We were all silent for a few seconds, and then everyone burst out laughing, with Feynman laughing the hardest. It was funny because Feynman was making fun of his own gargantuan ego. Feynman knew a good gag, and I heard him use this line at a few other opportune times thereafter. 5) Science by press conference: Cold fusion (1989) The New York Times was my source for the news that two chemists claimed to have produced nuclear fusion in heavy water using an electrochemical cell on a tabletop. I was interested enough to consult that day with our local nuclear experts Charlie Barnes, Bob McKeown, and Steve Koonin, none of whom believed it. Still, could it be true? I decided to spend a quiet day in my office, trying to imagine ways to induce nuclear fusion by stuffing deuterium into a palladium electrode. I came up empty. My interest dimmed when I heard that they had done a “control” experiment using ordinary water, had observed the same excess heat as with heavy water, and remained just as convinced as before that they were observing fusion. Later, Caltech chemist Nate Lewis gave a clear and convincing talk to the campus community debunking the original experiment. 6) “The face of God” COBE (1992) I’m often too skeptical. When I first heard in the early 1980s about proposals to detect the anisotropy in the cosmic microwave background, I doubted it would be possible. The signal is so small! It will be blurred by reionization of the universe! What about the galaxy! What about the dust! Blah, blah, blah, … The COBE DMR instrument showed it could be done, at least at large angular scales, and set the stage for the spectacular advances in observational cosmology we’ve witnessed over the past 20 years. George Smoot infamously declared that he had glimpsed “the face of God.” Overly dramatic, perhaps, but he was excited! And so was I. 7) “83 SNU” Gallex solar neutrinos (1992) Until 1992 the only neutrinos from the sun ever detected were the relatively high energy neutrinos produced by nuclear reactions involving boron and beryllium — these account for just a tiny fraction of all neutrinos emitted. Fewer than expected were seen, a puzzle that could be resolved if neutrinos have mass and oscillate to another flavor before reaching earth. But it made me uncomfortable that the evidence for solar neutrino oscillations was based on the boron-beryllium side show, and might conceivably be explained just by tweaking the astrophysics of the sun’s core. The Gallex experiment was the first to detect the lower energy pp neutrinos, the predominant type coming from the sun. The results seemed to confirm that we really did understand the sun and that solar neutrinos really oscillate. (More compelling evidence, from SNO, came later.) I stayed up late the night I heard about the Gallex result, and gave a talk the next day to our particle theory group explaining its significance. The talk title was “83 SNU” — that was the initially reported neutrino flux in Solar Neutrino Units, later revised downward somewhat. 8) Awestruck: Shor’s algorithm (1994) I’ve written before about how Peter Shor’s discovery of an efficient quantum algorithm for factoring numbers changed my life. This came at a pivotal time for me, as the SSC had been cancelled six months earlier, and I was growing pessimistic about the future of particle physics. I realized that observational cosmology would have a bright future, but I sensed that theoretical cosmology would be dominated by data analysis, where I would have little comparative advantage. So I became a quantum informationist, and have not regretted it. 9) The Higgs boson at last (2012) The discovery of the Higgs boson was exciting because we had been waiting soooo long for it to happen. Unable to stream the live feed of the announcement, I followed developments via Twitter. That was the first time I appreciated the potential value of Twitter for scientific communication, and soon after I started to tweet. 10) A lucky universe: BICEP2 (2014) Many past experiences prepared me to appreciate the BICEP2 announcement this past Monday. I first came to admire Alan Guth‘s distinctive clarity of thought in the fall of 1973 when he was the instructor for my classical mechanics course at Princeton (one of the best classes I ever took). I got to know him better in the summer of 1979 when I was a graduate student, and Alan invited me to visit Cornell because we were both interested in magnetic monopole production in the very early universe. Months later Alan realized that cosmic inflation could explain the isotropy and flatness of the universe, as well as the dearth of magnetic monopoles. I recall his first seminar at Harvard explaining his discovery. Steve Weinberg had to leave before the seminar was over, and Alan called as Steve walked out, “I was hoping to hear your reaction.” Steve replied, “My reaction is applause.” We all felt that way. I was at a wonderful workshop in Cambridge during the summer of 1982, where Alan and others made great progress in understanding the origin of primordial density perturbations produced from quantum fluctuations during inflation (Bardeen, Steinhardt, Turner, Starobinsky, and Hawking were also working on that problem, and they all reached a consensus by the end of the three-week workshop … meanwhile I was thinking about the cosmological implications of axions). I also met Andrei Linde at that same workshop, my first encounter with his mischievous grin and deadpan wit. (There was a delegation of Russians, who split their time between Xeroxing papers and watching the World Cup on TV.) When Andrei visited Caltech in 1987, I took him to Disneyland, and he had even more fun than my two-year-old daughter. During my first year at Caltech in 1984, Mark Wise and Larry Abbott told me about their calculations of the gravitational waves produced during inflation, which they used to derive a bound on the characteristic energy scale driving inflation, a few times 10^16 GeV. We mused about whether the signal might turn out to be detectable someday. Would Nature really be so kind as to place that mass scale below the Abbott-Wise bound, yet high enough (above 10^16 GeV) to be detectable? It seemed unlikely. Last week I caught up with the rumors about the BICEP2 results by scanning my Twitter feed on my iPad, while still lying in bed during the early morning. I immediately leapt up and stumbled around the house in the dark, mumbling to myself over and over again, “Holy Shit! … Holy Shit! …” The dog cast a curious glance my way, then went back to sleep. Like millions of others, I was frustrated Monday morning, trying to follow the live feed of the discovery announcement broadcast from the hopelessly overtaxed Center for Astrophysics website. I was able to join in the moment, though, by following on Twitter, and I indulged in a few breathless tweets of my own. Many of his friends have been thinking a lot these past few days about Andrew Lange, who had been the leader of the BICEP team (current senior team members John Kovac and Chao-Lin Kuo were Caltech postdocs under Andrew in the mid-2000s). One day in September 2007 he sent me an unexpected email, with the subject heading “the bard of cosmology.” Having discovered on the Internet a poem I had written to introduce a seminar by Craig Hogan, Andrew wrote: “John, just came across this – I must have been out of town for the event. l love it. it will be posted prominently in our lab today (with “LISA” replaced by “BICEP”, and remain our rallying cry till we detect the B-mode. have you set it to music yet? a” I lifted a couplet from that poem for one of my tweets (while rumors were swirling prior to the official announcement): We’ll finally know how the cosmos behaves If we can detect gravitational waves. Assuming the BICEP2 measurement r ~ 0.2 is really a detection of primordial gravitational waves, we have learned that the characteristic mass scale during inflation is an astonishingly high 2 X 10^16 GeV. Were it a factor of 2 smaller, the signal would have been far too small to detect in current experiments. This time, Nature really is on our side, eagerly revealing secrets about physics at a scale far, far beyond what we will every explore using particle accelerators. We feel lucky. We physicists can never quite believe that the equations we scrawl on a notepad actually have something to do with the real universe. You would think we’d be used to that by now, but we’re not — when it happens we’re amazed. In my case, never more so than this time. The BICEP2 paper, a historic document (if the result holds up), ends just the way it should: “We dedicate this paper to the memory of Andrew Lange, whom we sorely miss.”	2014-09-23 12:18:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 115, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5110577940940857, "perplexity": 1244.1513528465714}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657138770.35/warc/CC-MAIN-20140914011218-00226-ip-10-234-18-248.ec2.internal.warc.gz"}
http://mathhelpforum.com/calculus/149324-shortest-distance-point-surface.html	# Thread: Shortest distance Point - surface 1. ## Shortest distance Point - surface Im supposed to find the point B(x,y,z) which is a point on the surface z = x^2 + y^2 and is closest to the point A(3,0,0,)? HOW? I was said to differentiate the distance D between B and A, ( $D^2 = (x - 3)^2 + y^2 + (x^2 + y^2)^2$)then the shortest distance will be for dD/dx = 0 & dD/dy = 0. Can anyone tell me why it is so? Also i believe that posing y=0 is a good thought... then the question would be to find the distance from A to the curve z=x^2 Thanks! 2. Originally Posted by Klo Im supposed to find the point B(x,y,z) which is a point on the surface z = x^2 + y^2 and is closest to the point A(3,0,0,)? HOW? I was said to differentiate the distance D between B and A, ( $D^2 = (x - 3)^2 + y^2 + (x^2 + y^2)^2$)then the shortest distance will be for dD/dx = 0 & dD/dy = 0. Can anyone tell me why it is so? Also i believe that posing y=0 is a good thought... then the question would be to find the distance from A to the curve z=x^2 Thanks! A generic point on that surface has the form $(x,y,x^2+y^2)$, so the square of the distance from such a point to $(3,0,0)$ is the expression you wrote, and now you've a minimum problem of a two-variable function. The reason you can work with $D^2$ instead of $D$ is that both these functions have their minimal-maximal points exactly at the same points (proof? It's easy...), so: $f(x,y):= (x-3)^2+y^2+(x^2+y^2)^2\Longrightarrow f_x=2(x-3)+4x(x^2+y^2)=0$ $f_y=2y+4y(x^2+y^2)=0$ , and we get the system of non-linear eq's.: $4x^3+(4y^2+2)x=6$ $4y^3+(4x^2+2)y=0\iff y=0\,\,\,or\,\,\,4y^2+4x^2+2=0$ . Since this last eq. has no real solutions (sum of positive terms on the left), it must be $y=0$, and then from the 1st eq. we get $0=2x^3+x-3=(x-1)(2x^2+2x+3)$ , whose only real solution is $x=1$. I hope now you know how to check whether the critical point $(1,0)$ we get for our function is a maximal, minimal, saddle point or none of these (it, is a minimum, of course), and thus you get the point $(1,0,1)$ on your surface. Tonio 3. Perfect, thanks!	2017-02-22 14:14:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8101784586906433, "perplexity": 353.03148411312145}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170992.17/warc/CC-MAIN-20170219104610-00376-ip-10-171-10-108.ec2.internal.warc.gz"}
https://cs.stackexchange.com/questions/99328/finding-duplicates-in-a-stream-of-numbers	# Finding duplicates in a stream of numbers This is an interview question. Say you have a function foo() which returns some integer. You need to write an algorithm that does the following: (1). Call foo() in a loop, till the returned value of foo() is -1. (2). Returns true if during the loop there were two calls of foo() which returned the same value. This is what I thought to do, please tell me if this is efficient enough: (1). Initialize a hash map , which maps an integer to the number of times we got it in the loop (2). Call foo() in a loop till the returned value of foo() is -1, while updating the hash map accordingly. (3). Check if there is a key in the hash map which is mapped to a value greater than 1. • Efficient enough for what? – Juho Oct 30 '18 at 18:14 • While you are calling foo() in a loop, you will return immediately if you find the returned integer is in the hashset and if you are not required to finish the loop till the returned value of foo() is -1. Many built-in implementations return false when you insert an existing value into a hashset.	2020-01-20 19:32:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21867848932743073, "perplexity": 756.8552523641183}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250599718.13/warc/CC-MAIN-20200120165335-20200120194335-00023.warc.gz"}
https://www.ideals.illinois.edu/handle/2142/83446	Files in this item FilesDescriptionFormat application/pdf 9812745.pdf (9MB) (no description provided)PDF Description Title: Cloud Chemical Heterogeneity and Its Influence on Aqueous sulfur(IV) Oxidation Author(s): Rao, Xin Doctoral Committee Chair(s): J. Collett; S. Larson Department / Program: Civil Engineering Discipline: Civil Engineering Degree Granting Institution: University of Illinois at Urbana-Champaign Degree: Ph.D. Genre: Dissertation Subject(s): Physics, Atmospheric Science Abstract: Differences in chemical composition among cloud and fog drops of diverse sizes were investigated at several locations across the United States. Chemical species including acidity, total sulfur (IV), hydrogen peroxide, formaldehyde, hydroxymethanesulfonate (HMS) and trace metals iron and manganese were measured. Samples were collected with three cloud samplers capable of partitioning the cloud drop size spectrum into two or three independent drop size fractions. Measurements of pH variations within natural cloud drop populations reveal that small drops are often more acidic than large drops. No obvious pattern of drop size-dependence of hydrogen peroxide concentrations was observed. Trace metal concentrations were found to vary with drop size in clouds and fogs sampled at a variety of U.S. locations. Iron speciation measurements in San Joaquin Valley fogs revealed that dissolved iron in small fog drops was present almost entirely as Fe(III). The observed size dependence of hydrogen ion and trace metal concentrations in cloud and fog drops is expected to influence in-cloud S(IV) oxidation rates. Effects of chemical heterogeneity on overall in-cloud S(IV) oxidation rates will largely depend on contributions of the different oxidation paths. About 84 percent of the samples are calculated to experience little enhancement in S(IV) oxidation, due to the dominance of the H$\sb2$O$\sb2$ path. Approximately 9 percent of the samples are calculated to experience oxidation rate enhancement between 10 and 30%, while 7 percent of the samples are calculated to experience oxidation rate enhancement of 30% or more. Effects of chemical heterogeneity on enhancements in sulfur oxidation rates are likely to be strong when (1) hydrogen peroxide concentrations are low, for example the radiation fog in California's San Joaquin Valley, where the calculated enhancement factors range from 1.10 to 1.65, or (2) the droplet pH is high enough to support rapid S(IV) oxidation by ozone and metal-catalyzed S(IV) autooxidation, for example in relatively pristine environments like Angora Pk., Oregon, where the calculated enhancement factors range from 1.02 to 2.0. We expect real clouds to contain more than two chemically distinct drop populations. A wide distribution of drop compositions can support even faster sulfur oxidation rates. Issue Date: 1997 Type: Text Language: English Description: 231 p.Thesis (Ph.D.)--University of Illinois at Urbana-Champaign, 1997. URI: http://hdl.handle.net/2142/83446 Other Identifier(s): (MiAaPQ)AAI9812745 Date Available in IDEALS: 2015-09-25 Date Deposited: 1997	2018-06-21 01:17:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.33077916502952576, "perplexity": 10519.45955330699}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267863980.55/warc/CC-MAIN-20180621001211-20180621021211-00413.warc.gz"}
http://en.wikipedia.org/wiki/Gibbs_phenomenon	# Gibbs phenomenon In mathematics, the Gibbs phenomenon, discovered by Henry Wilbraham (1848)[1] and rediscovered by J. Willard Gibbs (1899),[2] is the peculiar manner in which the Fourier series of a piecewise continuously differentiable periodic function behaves at a jump discontinuity. The nth partial sum of the Fourier series has large oscillations near the jump, which might increase the maximum of the partial sum above that of the function itself. The overshoot does not die out as the frequency increases, but approaches a finite limit.[3] This sort of behavior was also observed by experimental physicists, but was believed to be due to imperfections in the measuring apparatuses.[4] These are one cause of ringing artifacts in signal processing. ## Description Functional approximation of square wave using 5 harmonics Functional approximation of square wave using 25 harmonics Functional approximation of square wave using 125 harmonics The Gibbs phenomenon involves both the fact that Fourier sums overshoot at a jump discontinuity, and that this overshoot does not die out as the frequency increases. The three pictures on the right demonstrate the phenomenon for a square wave (of height $\pi/4$) whose Fourier expansion is $\sin(x)+\frac{1}{3}\sin(3x)+\frac{1}{5}\sin(5x)+\dotsb.$ More precisely, this is the function f which equals $\pi/4$ between $2n\pi$ and $(2n+1)\pi$ and $-\pi/4$ between $(2n+1)\pi$ and $(2n+2)\pi$ for every integer n; thus this square wave has a jump discontinuity of height $\pi/2$ at every integer multiple of $\pi$. As can be seen, as the number of terms rises, the error of the approximation is reduced in width and energy, but converges to a fixed height. A calculation for the square wave (see Zygmund, chap. 8.5., or the computations at the end of this article) gives an explicit formula for the limit of the height of the error. It turns out that the Fourier series exceeds the height $\pi/4$ of the square wave by $\frac{1}{2}\int_0^\pi \frac{\sin t}{t}\, dt - \frac{\pi}{4} = \frac{\pi}{2}\cdot (0.089490\dots)$ or about 9 percent. More generally, at any jump point of a piecewise continuously differentiable function with a jump of a, the nth partial Fourier series will (for n very large) overshoot this jump by approximately $a \cdot (0.089392\dots)$ at one end and undershoot it by the same amount at the other end; thus the "jump" in the partial Fourier series will also be about 9% larger than the jump in the original function. At the location of the discontinuity itself, the partial Fourier series will converge to the midpoint of the jump (regardless of what the actual value of the original function is at this point). The quantity $\int_0^\pi \frac{\sin t}{t}\ dt = (1.851937052\dots) = \frac{\pi}{2} + \pi \cdot (0.089392\dots)$ is sometimes known as the Wilbraham–Gibbs constant. ### History The Gibbs phenomenon was first noticed and analyzed by the obscure Henry Wilbraham.[1] He published a paper on it in 1848 that went unnoticed by the mathematical world.[5] Albert A. Michelson developed a device in 1898 that could compute and re-synthesize the Fourier series. A widespread myth says that when the Fourier coefficients for a square wave were input to the machine, the graph would oscillate at the discontinuities, and that because it was a physical device subject to manufacturing flaws, Michelson was convinced that the overshoot was caused by errors in the machine. In fact the graphs produced by the machine were not good enough to exhibit the Gibbs phenomenon clearly, and Michelson may not have noticed it as he made no mention of this effect in his paper (Michelson & Stratton 1898) about his machine or his later letters to Nature. Inspired by some correspondence in Nature between Michelson and Love about the convergence of the Fourier series of the square wave function, in 1898 J. Willard Gibbs published a short note in which he considered what today would be called a sawtooth wave and pointed out the important distinction between the limit of the graphs of the partial sums of the Fourier series, and the graph of the function that is the limit of those partial sums. In his first letter Gibbs failed to notice the Gibbs phenomenon, and the limit that he described for the graphs of the partial sums was inaccurate. In 1899 he published a correction in which he described the overshoot at the point of discontinuity (Nature: April 27, 1899, p. 606). In 1906, Maxime Bôcher gave a detailed mathematical analysis of that overshoot, which he called the "Gibbs phenomenon".[6] ### Explanation Informally, it reflects the difficulty inherent in approximating a discontinuous function by a finite series of continuous sine and cosine waves. It is important to put emphasis on the word finite because even though every partial sum of the Fourier series overshoots the function it is approximating, the limit of the partial sums does not. The value of x where the maximum overshoot is achieved moves closer and closer to the discontinuity as the number of terms summed increases so, again informally, once the overshoot has passed by a particular x, convergence at the value of x is possible. There is no contradiction in the overshoot converging to a non-zero amount, but the limit of the partial sums having no overshoot, because where that overshoot happens moves. We have pointwise convergence, but not uniform convergence. For a piecewise C1 function the Fourier series converges to the function at every point except at the jump discontinuities. At the jump discontinuities themselves the limit will converge to the average of the values of the function on either side of the jump. This is a consequence of the Dirichlet theorem.[7] The Gibbs phenomenon is also closely related to the principle that the decay of the Fourier coefficients of a function at infinity is controlled by the smoothness of that function; very smooth functions will have very rapidly decaying Fourier coefficients (resulting in the rapid convergence of the Fourier series), whereas discontinuous functions will have very slowly decaying Fourier coefficients (causing the Fourier series to converge very slowly). Note for instance that the Fourier coefficients 1, −1/3, 1/5, ... of the discontinuous square wave described above decay only as fast as the harmonic series, which is not absolutely convergent; indeed, the above Fourier series turns out to be only conditionally convergent for almost every value of x. This provides a partial explanation of the Gibbs phenomenon, since Fourier series with absolutely convergent Fourier coefficients would be uniformly convergent by the Weierstrass M-test and would thus be unable to exhibit the above oscillatory behavior. By the same token, it is impossible for a discontinuous function to have absolutely convergent Fourier coefficients, since the function would thus be the uniform limit of continuous functions and therefore be continuous, a contradiction. See more about absolute convergence of Fourier series. ### Solutions In practice, the difficulties associated with the Gibbs phenomenon can be ameliorated by using a smoother method of Fourier series summation, such as Fejér summation or Riesz summation, or by using sigma-approximation. Using a wavelet transform with Haar basis functions, the Gibbs phenomenon does not occur in the case of continuous data at jump discontinuities,[8] and is minimal in the discrete case at large change points. In wavelet analysis, this is commonly referred to as the Longo phenomenon. ## Formal mathematical description of the phenomenon Let $f: {\Bbb R} \to {\Bbb R}$ be a piecewise continuously differentiable function which is periodic with some period $L > 0$. Suppose that at some point $x_0$, the left limit $f(x_0^-)$ and right limit $f(x_0^+)$ of the function $f$ differ by a non-zero gap $a$: $f(x_0^+) - f(x_0^-) = a \neq 0.$ For each positive integer N ≥ 1, let SN f be the Nth partial Fourier series $S_N f(x) := \sum_{-N \leq n \leq N} \hat f(n) e^{\frac{2i\pi n x}{L}} = \frac{1}{2} a_0 + \sum_{n=1}^N \left( a_n \cos\left(\frac{2\pi nx}{L}\right) + b_n \sin\left(\frac{2\pi nx}{L}\right) \right),$ where the Fourier coefficients $\hat f(n), a_n, b_n$ are given by the usual formulae $\hat f(n) := \frac{1}{L} \int_0^L f(x) e^{-2i\pi n x/L}\, dx$ $a_n := \frac{2}{L} \int_0^L f(x) \cos\left(\frac{2\pi nx}{L}\right)\, dx$ $b_n := \frac{2}{L} \int_0^L f(x) \sin\left(\frac{2\pi nx}{L}\right)\, dx.$ Then we have $\lim_{N \to \infty} S_N f\left(x_0 + \frac{L}{2N}\right) = f(x_0^+) + a\cdot (0.089392\dots)$ and $\lim_{N \to \infty} S_N f\left(x_0 - \frac{L}{2N}\right) = f(x_0^-) - a\cdot (0.089392\dots)$ but $\lim_{N \to \infty} S_N f(x_0) = \frac{f(x_0^-) + f(x_0^+)}{2}.$ More generally, if $x_N$ is any sequence of real numbers which converges to $x_0$ as $N \to \infty$, and if the gap a is positive then $\limsup_{N \to \infty} S_N f(x_N) \leq f(x_0^+) + a\cdot (0.089392\dots)$ and $\liminf_{N \to \infty} S_N f(x_N) \geq f(x_0^-) - a\cdot (0.089392\dots).$ If instead the gap a is negative, one needs to interchange limit superior with limit inferior, and also interchange the ≤ and ≥ signs, in the above two inequalities. ## Signal processing explanation For more details on this topic, see Ringing artifacts. The sinc function, the impulse response of an ideal low-pass filter. Scaling narrows the function, and correspondingly increases magnitude (which is not shown here), but does not reduce the magnitude of the undershoot, which is the integral of the tail. From the point of view of signal processing, the Gibbs phenomenon is the step response of a low-pass filter, and the oscillations are called ringing or ringing artifacts. Truncating the Fourier transform of a signal on the real line, or the Fourier series of a periodic signal (equivalently, a signal on the circle) corresponds to filtering out the higher frequencies by an ideal (brick-wall) low-pass/high-cut filter. This can be represented as convolution of the original signal with the impulse response of the filter (also known as the kernel), which is the sinc function. Thus the Gibbs phenomenon can be seen as the result of convolving a Heaviside step function (if periodicity is not required) or a square wave (if periodic) with a sinc function: the oscillations in the sinc function cause the ripples in the output. The sine integral, exhibiting the Gibbs phenomenon for a step function on the real line. In the case of convolving with a Heaviside step function, the resulting function is exactly the integral of the sinc function, the sine integral; for a square wave the description is not as simply stated. For the step function, the magnitude of the undershoot is thus exactly the integral of the (left) tail, integrating to the first negative zero: for the normalized sinc of unit sampling period, this is $\int_{-\infty}^{-1} \frac{\sin(\pi x)}{\pi x}\,dx.$ The overshoot is accordingly of the same magnitude: the integral of the right tail, or, which amounts to the same thing, the difference between the integral from negative infinity to the first positive zero, minus 1 (the non-overshooting value). The overshoot and undershoot can be understood thus: kernels are generally normalized to have integral 1, so they result in a mapping of constant functions to constant functions – otherwise they have gain. The value of a convolution at a point is a linear combination of the input signal, with coefficients (weights) the values of the kernel. If a kernel is non-negative, such as for a Gaussian kernel, then the value of the filtered signal will be a convex combination of the input values (the coefficients (the kernel) integrate to 1, and are non-negative), and will thus fall between the minimum and maximum of the input signal – it will not undershoot or overshoot. If, on the other hand, the kernel assumes negative values, such as the sinc function, then the value of the filtered signal will instead be an affine combination of the input values, and may fall outside of the minimum and maximum of the input signal, resulting in undershoot and overshoot, as in the Gibbs phenomenon. Taking a longer expansion – cutting at a higher frequency – corresponds in the frequency domain to widening the brick-wall, which in the time domain corresponds to narrowing the sinc function and increasing its height by the same factor, leaving the integrals between corresponding points unchanged. This is a general feature of the Fourier transform: widening in one domain corresponds to narrowing and increasing height in the other. This results in the oscillations in sinc being narrower and taller and, in the filtered function (after convolution), yields oscillations that are narrower and thus have less area, but does not reduce the magnitude: cutting off at any finite frequency results in a sinc function, however narrow, with the same tail integrals. This explains the persistence of the overshoot and undershoot. Thus the features of the Gibbs phenomenon are interpreted as follows: • the undershoot is due to the impulse response having a negative tail integral, which is possible because the function takes negative values; • the overshoot offsets this, by symmetry (the overall integral does not change under filtering); • the persistence of the oscillations is because increasing the cutoff narrows the impulse response, but does not reduce its integral – the oscillations thus move towards the discontinuity, but do not decrease in magnitude. ## The square wave example Animation of the additive synthesis of a square wave with an increasing number of harmonics. The Gibbs phenomenon is visible especially when the number of harmonics is large. We now illustrate the above Gibbs phenomenon in the case of the square wave described earlier. In this case the period L is $2\pi$, the discontinuity $x_0$ is at zero, and the jump a is equal to $\pi/2$. For simplicity let us just deal with the case when N is even (the case of odd N is very similar). Then we have $S_N f(x) = \sin(x) + \frac{1}{3} \sin(3x) + \cdots + \frac{1}{N-1} \sin((N-1)x).$ Substituting $x=0$, we obtain $S_N f(0) = 0 = \frac{-\frac{\pi}{4} + \frac{\pi}{4}}{2} = \frac{f(0^-) + f(0^+)}{2}$ as claimed above. Next, we compute $S_N f\left(\frac{2\pi}{2N}\right) = \sin\left(\frac{\pi}{N}\right) + \frac{1}{3} \sin\left(\frac{3\pi}{N}\right) + \cdots + \frac{1}{N-1} \sin\left( \frac{(N-1)\pi}{N} \right).$ If we introduce the normalized sinc function, $\operatorname{sinc}(x)\,$, we can rewrite this as $S_N f\left(\frac{2\pi}{2N}\right) = \frac{\pi}{2} \left[ \frac{2}{N} \operatorname{sinc}\left(\frac{1}{N}\right) + \frac{2}{N} \operatorname{sinc}\left(\frac{3}{N}\right) + \cdots + \frac{2}{N} \operatorname{sinc}\left( \frac{(N-1)}{N} \right) \right].$ But the expression in square brackets is a numerical integration approximation to the integral $\int_0^1 \operatorname{sinc}(x)\ dx$ (more precisely, it is a midpoint rule approximation with spacing $2/N$). Since the sinc function is continuous, this approximation converges to the actual integral as $N \to \infty$. Thus we have \begin{align} \lim_{N \to \infty} S_N f\left(\frac{2\pi}{2N}\right) & = \frac{\pi}{2} \int_0^1 \operatorname{sinc}(x)\, dx \\[8pt] & = \frac{1}{2} \int_{x=0}^1 \frac{\sin(\pi x)}{\pi x}\, d(\pi x) \\[8pt] & = \frac{1}{2} \int_0^\pi \frac{\sin(t)}{t}\ dt \quad = \quad \frac{\pi}{4} + \frac{\pi}{2} \cdot (0.089490\dots), \end{align} which was what was claimed in the previous section. A similar computation shows $\lim_{N \to \infty} S_N f\left(-\frac{2\pi}{2N}\right) = -\frac{\pi}{2} \int_0^1 \operatorname{sinc}(x)\ dx = -\frac{\pi}{4} - \frac{\pi}{2} \cdot (0.089490\dots).$ ## Consequences In signal processing, the Gibbs phenomenon is undesirable because it causes artifacts, namely clipping from the overshoot and undershoot, and ringing artifacts from the oscillations. In the case of low-pass filtering, these can be reduced or eliminated by using different low-pass filters. In MRI, the Gibbs phenomenon causes artifacts in the presence of adjacent regions of markedly differing signal intensity. This is most commonly encountered in spinal MR imaging, where the Gibbs phenomenon may simulate the appearance of syringomyelia.	2015-03-30 10:40:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 50, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9418315887451172, "perplexity": 299.53082287616326}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131299261.59/warc/CC-MAIN-20150323172139-00190-ip-10-168-14-71.ec2.internal.warc.gz"}
https://www.bartleby.com/solution-answer/chapter-12i-problem-9re-contemporary-mathematics-for-business-and-consumers-8th-edition/9781305585447/use-table-12-1-to-calculate-the-future-value-of-the-following-annuities-due-annuity-payment-time/bdb5b0f6-6785-11e9-8385-02ee952b546e	Chapter 12.I, Problem 9RE ### Contemporary Mathematics for Busin... 8th Edition Robert Brechner + 1 other ISBN: 9781305585447 Chapter Section ### Contemporary Mathematics for Busin... 8th Edition Robert Brechner + 1 other ISBN: 9781305585447 Textbook Problem # Use Table 12-1 to calculate the future value of the following annuities due. Annuity Payment Time Nominal Interest Future Value Payment Frequency Period (years) Rate (%) Compounded of the Annuity 9. $2,000 every year 25 5 Annually_____________ To determine To calculate: The future value of annuity due where annuity payment is$2,000, frequency of payment is 1 Year, time duration is 25 years, nominal rate of return is 5% and interest is compounded annually. Explanation Given Information: Annuity payment is $2,000, frequency of payment is 1 Year, time duration is 25 years, nominal rate of return is 5% and interest is compounded annually. Formula used: Steps for calculating the future value of an annuity due are: Step 1: Calculate the number of periods of the annuity and add one period to total. Step 2: The interest rate per period must be calculated. Step 3: Use table 12-1 to locate the ordinary annuity table factor that lies on the intersection of the rate-per period column and number-of-periods row. Step 4: The number 1.00000 must be subtracted from the ordinary annuity table in order to get the annuity due factor. Step 5: Finally calculate the future value of the annuity due. The formula to compute the future value of ordinary annuity is, Future Value=Annuity due table factor×Annuity payment Annuity due table factor=Ordinary annuity table factor1.00000 Calculation: Consider that Annuity payment is$2,000, frequency of payment is 1 Year, time duration is 25 years, nominal rate of return is 5% and interest is compounded annually. As the interest is compounded annually. So, the interest rate period is; 5%1=5% The rate period is 5% ### Still sussing out bartleby? Check out a sample textbook solution. See a sample solution #### The Solution to Your Study Problems Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees! Get Started	2019-11-18 21:11:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35976269841194153, "perplexity": 8260.125367568317}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669847.1/warc/CC-MAIN-20191118205402-20191118233402-00003.warc.gz"}
https://math.stackexchange.com/questions/2732203/do-we-know-the-discriminants-in-the-hecke-double-coset	# Do we know the discriminants in the Hecke double coset? I am considering an element $x \in GL(2, \mathbb{Q}_p)$ in the double coset $GL(2, \mathbb{Z}_p) T(p) GL(2, \mathbb{Z}_p)$, where $T(p)$ is the diagonal matrix $(p, 1)$. I wonder about what can be said of his determinant and Weyl discriminant, and something is troubling me. Assume $x$ is semi-simple, so that it is conjugate (over an algebraic closure) to a diagonal matrix $(a,b)$. Then $$\det(x) = ab \qquad \text{and} \qquad D(x) = \left( 1 - \frac{a}{b} \right)\left( 1 - \frac{b}{a} \right)$$ Assume $x$ is non-central, otherwise $a=b$. The characteristic polynomial has integer coefficients, hence the eigenvalues are integral and hence in $\mathbb{Z}_p$, since it is integrally closed. Since $\det(x)$ is $p$ (by the assumption of belonging to the double class), $a$ has to be associated to $p$ and $b$ a unit (or the converse) by looking at the valuations. Based on it, can I bound $D(x)$ in function of $p$. Or the $p$-valuation of $D(x)$?	2019-07-17 14:35:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.954997181892395, "perplexity": 194.303815792377}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525312.3/warc/CC-MAIN-20190717141631-20190717163631-00086.warc.gz"}
https://stackoverflow.com/questions/50702942/does-rmarkdown-allow-captions-and-references-for-code-chunks	# Does rmarkdown allow captions and references for code chunks? Is there a trick for captioning and referencing a chuck of code in rmarkdown (not the result of running the code)? For example, how do I reference this block of code: {r blah} blah <- "blah" I know that I can use \@ref(fig:theFig) or \@ref(tab:theTable) to get at fig.cap or caption (with kable) but I don't see a way to caption and reference the code itself. This would be a great feature. And I think it is not integrated yet. Here is an approach that works for PDF documents and allows you to generate captions as well as labels for cross-referencing: 1. Include a header.tex with the following content \usepackage{caption} \usepackage{floatrow} \DeclareNewFloatType{chunk}{placement=H, fileext=chk, name=} \captionsetup{options=chunk} \renewcommand{\thechunk}{Chunk~\thesection.\arabic{chunk}} \makeatletter \makeatother Since the environment used for chunks is not a floating type we declare a new one, named chunk. The option name is left blank since we already redefine \thechunk in the next line by prepending the word Chunk (play with the name option and see what happens). The chunks should be enumerated by section and so we tell tex to reset the counter each time a new section begins. If you do not use section numbering (by setting the YAML option), then replace the line \renewcommand{\thechunk}{Chunk~\thesection.\arabic{chunk}} by \renewcommand{\thechunk}{Chunk~\arabic{chunk}} 1. Modify the knitr source hook in your rmarkdown document library(knitr) oldSource <- knit_hooks$get("source") knit_hooks$set(source = function(x, options) { x <- oldSource(x, options) x <- ifelse(!is.null(options$ref), paste0("\\label{", options$ref,"}", x), x) ifelse(!is.null(options$codecap), paste0("\\captionof{chunk}{", options$codecap,"}", x), x) }) Here we make use of two new chunk options ref and codecap. If either one is not NULL, a corresponding label or caption is generated using the the commands \label or \captionof. MWE: --- title: "Cross-referencing Code Chunks" output: pdf_document: includes: number_sections: true --- {r, echo=FALSE} library(knitr) oldSource <- knit_hooks$get("source") knit_hooks$set(source = function(x, options) { x <- oldSource(x, options) x <- ifelse(!is.null(options$ref), paste0("\\label{", options$ref,"}", x), x) ifelse(!is.null(options$codecap), paste0("\\captionof{chunk}{", options$codecap,"}", x), x) }) # Foo {r Foo, ref = "TheFooChunk", codecap = "My Chunk"} print("Foo!") \newpage # Bar {r Bar, ref = "TheBarChunk", codecap = "My second chunk"} print("Bar!") Here is the output (of both pages): ### Adding the caption below the code chunk If you'd prefer to add the caption below the code chunk, instead of above, this can be achieved by altering the above knitr hook to read: oldSource <- knit_hooks$get("source") knit_hooks$set(source = function(x, options) { x <- oldSource(x, options) x <- ifelse(!is.null(options$codecap), paste0(x, "\\captionof{chunk}{", options$codecap,"}"), x) ifelse(!is.null(options$ref), paste0(x, "\\label{", options$ref,"}"), x) }) • However, I since found that swapping the order of the arguments of the paste0 command is sufficient to position the caption below. ie. paste0(x, "\\label{", options$ref,"}") and paste0(x, "\\captionof{chunk}{", options$codecap,"}"). Oct 12 '20 at 21:10 • This also works for bookdown. header.txt and the knitr source hook chunk need no modification. The YAML header just needs to replace pdf_document: with bookdown::pdf_book:. No need to include number_sections: true in the YAML header, because bookdown numbers sections by default. Note that no numbering will occur unless you include the codecap chunk option, even if you included ref. Apr 13 '21 at 13:11	2022-01-20 12:36:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9325220584869385, "perplexity": 6771.129213593561}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301737.47/warc/CC-MAIN-20220120100127-20220120130127-00225.warc.gz"}
https://www.gradesaver.com/textbooks/math/calculus/calculus-10th-edition-anton/chapter-13-partial-derivatives-13-1-functions-of-two-or-more-variables-exercises-set-13-1-page-914/1	## Calculus, 10th Edition (Anton) Published by Wiley # Chapter 13 - Partial Derivatives - 13.1 Functions Of Two Or More Variables - Exercises Set 13.1 - Page 914: 1 #### Answer (a) $f(2,1)=5$, (b) $f(1,2)=3$, (c) $f(0,0)=1$, (d) $f(1,-3)=-2$, (e) $f(3a,a)=9a^3+1$, (f) $f(ab,a-b)=(ab)^2(a-b)+1$ #### Work Step by Step (a) Substitute $x=2$ and $y=1:$ $$f(2,1)=2^2\cdot1+1=4+1=5$$ (b)Substitute $x=1$ and $y=2:$ $$f(1,2)=1^2\cdot2+1=2+1=3$$ (c)Substitute $x=0$ and $y=0:$ $$f(0,0)=0^2\cdot0+1=0+1=1$$ (d) Substitute $x=1$ and $y=-3:$ $$f(1,-3)=1^2\cdot(-3)+1=-3+1=-2$$ (e) Substitute $x=3a$ and $y=a:$ $$f(3a,a)=(3a)^2\cdot a+1=9a^3+1$$ (f) Substitute $x=ab$ and $y=a-b:$ $$f(ab,a-b)=(ab)^2\cdot(a-b)+1$$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.	2020-08-13 14:29:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6618786454200745, "perplexity": 1316.8665163009507}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439739046.14/warc/CC-MAIN-20200813132415-20200813162415-00358.warc.gz"}
https://electronics.stackexchange.com/questions/354652/actual-reason-why-i-have-distortion-in-lower-part-of-this-square-wave-amplifier/354699	# Actual reason why I have distortion in lower part of this square wave amplifier (level translator application) I am designing a simple level translator using 2 transistors connected in series in LTSPICE simulator. I tried with one transistor but the output from the collector pin is 180 degree out of phase, as is usual with common emitter circuit. With 2 transistors, I get this distortion in the lower part of the step in LTSPICE waveform when I attach a probe at point "A" (this would get into stage 2 also.) This goes away when I increase emitter resistor value. Can anyone tell me the exact reason why the lower part of the step has a hook in it and goes still lower when I try to go lower than 2.25 volts (which I need?) My aim is to go lower on the lower part of the step of the wave to get the "0" to satisfy TTL needs at output of this circuit (under 1 volt.) I don't have any problem in the higher part of the square wave. The circuit (first picture), input waveform, output waveform with two different emitter resistors, 300 ohm and 50 ohm. Both are distorted on the "downs." I would like to know the reason! • Can you please post your circuit and the scope traces showing this distortion? – evildemonic Feb 7 '18 at 15:31 • I increased R3 (emitter resistor first stage) to 300 ohm from earlier 50 ohm, still the same distortion, I am enclosing LTSPICE waveforms at point A and B (second stage too which shows spikes instead of square wave)! I experimented with placing coupling capacitors of 10mfd in series with each transistor base but no improvement , another form of distortion, hook-like in lower part of step comes in with capacitors. – user3048731 Feb 7 '18 at 16:22 • @user3048731 What do you expect to drive with this level translator circuit? Does it need to SINK and SOURCE at the output? At what rates? – jonk Feb 7 '18 at 16:27 • @user3048731 Also, are you aware that LTspice will automatically use something like 10% for the rise and fall times of your input if you don't override it? – jonk Feb 7 '18 at 16:30 • loose the emitter resistors (use wire instead), and to avoid saturation fit schottky diodes from base to collector. – Jasen Feb 7 '18 at 19:09 My aim is to go lower on the lower part of the step of the wave to get the "0" to satisfy TTL needs at output of this circuit (under 1 volt). The potential problem with the first stage is the value of R1 of 100 ohms. Ignoring the connection of the collector for a minute, think what happens to the actual base voltage when 4.5 volts is at the input to the 100 ohm resistor R1; a potential divider is formed with R1 (100 ohms), the base-emitter (a forward biased diode) and, the emitter resistor (60 ohms). Current into the base would be approximately: - (4.5 volts minus 0.7 volts)/(100 ohm plus 60 ohm) = 24 mA. This current would therefore drop 2.4 volts across R1 and takes the base to 4.5 - 2.4 volts = 2.1 volts. It sounds like an unreal analysis until you start to consider what happens when you reconnect the collector. The problem is this; the 1st stage collector cannot get below (2.1 volts minus 0.7 volts) because if it did it will also act like a forward biased diode from the base and, the BJT's hFE would be shot to death. So, if the collector starts to get lower than the base, the transistor fails to operate with any respectable characteristics and rapidly turns into two diodes hence my (possibly) rather odd analysis. Given what I have said, the lowest voltage attainable at the collector is approximately 1.3 volts. Your LTSpice transient response appears to indicate 1.4 volts so I'm not far off with these approximations. The 2nd BJT stage compounds the problem. Maybe try this: - can anyone tell me the exact reason why the lower part of the step has a hook in it That hook is because the base voltage is rapidly lowering from 4.5 volts back to 0 volts and you reach a point where the collector-base region is not (illegally for a linear amplifier) forward biased anymore. Now the transistor starts to behave with some measure of dignity. At this in-between base level, the transistor is properly amplifying and the collector is being allowed to drop to a voltage that is much less. • Just a note. The base current is more like $320\:\mu\text{A}$. You didn't multiply the emitter resistor by beta=200 for the 2N2222 (temporarily assuming active mode.) Or am I missing something? Still a saturated BJT, though. – jonk Feb 7 '18 at 16:16 • @jonk, when it saturates the collector/base region goes forward biased and this ruins the hFE. – Andy aka Feb 7 '18 at 16:18 • Good point. True enough. Need more coffee this AM (it is 8AM now for me.) – jonk Feb 7 '18 at 16:18 • @jonk it would be worse if the gain remained high - the base would sit even higher and the collector could not drop lower than (say) 3 volts. It's 16:20 for me but getting tired. Who said old age LOL. – Andy aka Feb 7 '18 at 16:19 • "That hook is because the base voltage is rapidly lowering from 4.5 volts back to 0 volts and you reach a point where the collector-base region is not (illegally for a linear amplifier) forward biased anymore" . @Andy Aka, Thanks a lot, this explains my original question, and for your alternate modified circuit too. About 3 decades ago we had a course in Transistors, so I'm rusty ! Main points I remember are "1) multiply the Re by BETA for input impedance 2) Vbe varies with temperature, stabilise it by a grounding resistor wasting some input current, or with a larger Re 3) Amplific= RL/Re. – user3048731 Feb 8 '18 at 4:14 Here is the circuit we are discussing: If you want the output to go as low possible, lose the emitter resistor R4. Once you do that, you need the input to the second stage to also go low, so lose R3 too. R2 also seems rather low for what it seems you are trying to do. I'd start with making it about 10x higher, like 4.7 kΩ, for example. R1 also seems rather low. Especially with R3 gone, such a low value of R1 may load the digital signal too much. Start with making it 1 kΩ or so. So in summary, make R1 1 kΩ, R2 4.7 kΩ, and short R3 and R4. See how that works. The 'hook' is caused by Base current flowing through the Emitter resistor, which raises the Emitter voltage and prevents the Collector from pulling down to 0V. To understand how this happens, consider this (grossly) simplified circuit where the transistor is replaced with a diode and switch simulating the Base-Emitter and Collector-Emitter junctions:- simulate this circuit – Schematic created using CircuitLab When Vin rises above ~0.6V the transistor is turned on (SW1 closed) but Vout only goes down to 0.45V because the current flowing through R3 also goes through R2, forming a voltage divider with a ratio of (500+50)/50 = 11:1. This is the condition when there is just enough current going into the Base to turn the transistor on. The transistor has a high current gain so its Base current is initially very small and does not significantly affect the voltage on R2. However as the input voltage rises it injects more current through D1 into R2, increasing the voltage across R2 and raising the output voltage even higher. If the value of R1 is increased then current injected into the Base is lower and the 'hooks' get smaller, but Vout can never go below 0.45V. To get the lowest possible output voltage you must reduce R2 to 0Ω. Even then there will still be some voltage drop due to the transistor's internal Emitter resistance, so you should also increase R1 to reduce Base current (it only needs to be a fraction of the Collector current, not 3 times higher!). Of course a transistor is more than than just a diode and switch, so your waveform has other characteristics that are not explained by this simplified circuit. The transistor amplifies current current linearly so it doesn't turn on and off instantaneously. It also has internal capacitances which slow down the switching action and allow the input to feed through during transitions. This explains why your 'hooks' momentarily reach 0.2V when the input goes low, and don't quite reach 0.45V when the input goes high. If you reduce the input frequency and increase transition times then capacitive effects are reduced and the output waveform will become more symmetrical. Also note that with Trise and Tfall set to zero LTspice will make the transition times proportional to the total period. To control transition timing you must set Trise and Tfall to specific values. • "The transistor amplifies current current linearly so it doesn't turn on and off instantaneously. It also has internal capacitances which slow down the switching action and allow the input to feed through during transitions. This explains why your 'hooks' momentarily reach 0.2V when the input goes low, and don't quite reach 0.45V when the input goes high. " Excellent advice, explains much to me ! Thanks! – user3048731 Feb 8 '18 at 4:07 *Thank you for all your answers ! The last one with the detailed circuit works fine ! meantime, i too tried various combinations, and realized the base terminal was probably "overdriven" from your previous answers. here is the modified circuit (was afraid to "lose" the emitter resistor altogether as its popular opinion that some emitter resistor provides "stability" in C-E circuits. so reduced it to 30k. at input it works giving a good square wave with 1K as @Andy Aka suggested in his modified circuit proposal. But it Also works if we bypass some of the current with a base to ground resistor as shown in circuit. Of course this would reduce the input impedance of the circuit, but increase stability. more than 3 decades ago (had more valve radios then) when we did a course in transistors (High School) I remember they told us: "stabilize any transistor circuit by removing effect of Vbe against temperature fluctuations by "wasting" some of the input current to ground with base to ground resistor (upto 10 times x Ib (base current) was recommended back then!), or to use a higher value Re (emitter resistor) (which would appear BETA times larger when seen from Base / input side of circuit) or both! That's old school stuff, don't know if you follow this anymore with modern tolerances on transistors. . The purpose of this circuit is to derive High programming voltage Vpp for a micro-controller. Of course I realise that this circuit cant be used directly, input impedance needs to be checked that it dosen't load the driver. • Even though there is a small "hook" at end of the LOW at output of stage 1 (point "A") in the waveform, it goes away when seen at second stage output (point "B"). – user3048731 Feb 7 '18 at 17:33 You already have good answers explaining the bad behavior of your circuit. So I'm not going to belabor that. The job is done as well or better than I could do. As far as a solution goes, though, that's a different matter. Your question could use some clarification. Even if you think that you don't need to talk about why, it's still a good idea to explain the reason why you are bothering. The following answer will explain why it matters. Circuit 1: simulate this circuit – Schematic created using CircuitLab Circuit 2: simulate this circuit Circuit 3: simulate this circuit Circuit 4: simulate this circuit And there are so many more possibilities.	2020-02-18 15:25:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4169333279132843, "perplexity": 1698.9270814099345}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875143784.14/warc/CC-MAIN-20200218150621-20200218180621-00375.warc.gz"}
http://sims.ritornoallalira.it/slope-worksheet-2-answers.html	Example worksheets. x2-x1 run change in x. Printable worksheets of Equations with Variables on Both Sides with distributive property, ti84 systeme equations, different math trivia with answers, math poems algebra, solve the nonlinear progression series in matlab, answers for mcdougal littell algebra 2 for free. ) (1, -2) and (-1, 3) 10. Displaying top 8 worksheets found for - Finding The Slope Of Two Points. Finding Slope 2 Using Formula. That being explained, we all provide assortment of uncomplicated yet useful posts as well as design templates designed suitable for almost any educative purpose. Write/the/equation/(in/[email protected]/form)/of/the/line/below. Slope is, and passes through the point (-14, 2). You should get a new 3 x 3 square. More generally, we know that the slope of $\ds e^x$ is $\ds e^z$ at the point $\ds (z,e^z)$, so the slope of $\ln(x)$ is $\ds 1/e^z$ at $\ds (e^z,z)$, as indicated in figure 4. What is the slope of the water surface at the observed section? Would the water. The graph of y=10-2x has an. Do the points (2, 5), (4, 3. com Lesson Plan>>Slope Between Two Points (Worksheet Version)>>Page 2 of 4 Activity Horizontal Slope Start by placing (x 1, y 1) at the point (-8, 4) and placing (x 2, y 2) at the point (4,4). Sketch the lines that your group drew while using Geometer s Sketchpad. If the inclination of the line y – 1 = ax + a^2 is 45°, find its y-intercept. y = 5x + 13 3. ©x 12E0w1u1 z nK9uJtka 6 XSxoef KtwLa Pr3eG BL lL qC7. Determine: a. Central carbon amino group carboxyl group acid r group side chain. Slope Worksheet 2 Answers - Wanting to Buy a New Pyramid Scheme: Inthe the same vein as your failure to come up with an answer to the question above, you might as well try and come up with a solution that will convince you to buy the new pyramid scheme. There is no limit to how many you can download and print off. 2 Note: This worksheet is supported by a flash presentation, under Mausmi’s Math Movies. The product of the slope of two perpendicular lines is equal to -1. Talking related with Finding Slope of Line Worksheet, we've collected particular variation of pictures to add more info. 2: Direct Variation When given an equation in the form y = mx + b, or slope-intercept form , we should know how to graph a line without having to compute two points on it. Finding Slope 2 Using Formula - Displaying top 8 worksheets found for this concept. Lesson 4 - Connecting Slope ideas. Slope & y-Intercept Worksheet page 2 of 2 continued II. 6 h bM na AdweM Hw3iFtZhq jI Yn6f Ci9nLi jt TeQ QP0r4e C-sA xl 8gwe ubTr na V. Contains 20 Slope of a Line problems. Recall: Answer: Answers will vary, depending on students’ lines of best fit. Practice 1 - A line has a 2 slope and passes through the point (3, 2). Displaying all worksheets related to - 2 3 Skills Practice Slope. These Linear Equations Worksheets are a good resource for students in the 5th Grade through the 8th Grade. Find the Equation of a Line Given That You Know a Point on the Line And Its Slope The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept. 4 Identify rates of change (slope) and distinguishing properties of data from tables, graphs, and equations to predict what happens to one variable as another variable changes;. ©T 82N0m152 C vK Eu 7tma9 Smo 8f0tDwfa Jr je 6 qLsLtCG. Time (min) 5 101520 Envelopes Stuffed 30 60 90 120 2. 5 7 6 8 1 2 10 3 Finding Inverses Find an equation for the inverse for each of the following relations. The slope of the reference line is m = {{ - 1} \over 2}. If there is a. We hope that you find exactly what you need for your home or classroom!. Example 2: Determine the slope of the line passing through the points. The slope is 7 8;the y-intercept is 8. To preview this answer key, click on the File menu and select Print Preview. Algebra 2 Worksheets with Answer Key Along with Finding Slope From A Pair Of Points Math Aids; Worksheet November 03, 2017. 2-4 Practice Writing Linear Equations Write an equation in slope-intercept form for the line described. To find the slope of a line passing through a given pair of points is found by using the point slope formula. Worksheet for Correlation and Regression (February 1, 2013). Slope is, and passes through the point (-14, 2). Lesson 3 - Slope and First Differences Worksheet Solns. Whether you have a child who loves math or struggles with math, she is a resource to consider. Note the X and Y value for each of the points. So if I go over one in the positive direction, I have to go down 2, that's what a negative slope's going to do, negative 2 slope. Be sure to label each line. Answer Key for Practice 9-1. 2 u tAjl GlX OryizgIh Ytmse qrGeQsae 2r 3vAe3dA. Leave answers as Improper Fractions! 4. (Do not do Questions 7 to 14 of this sheet on Gradient Intercept, as we have not covered that work in this lesson). Consider the following hypothetical data set. You could offer free training, free consultation and even a free product. Math Practice Series Worksheet 5 Answer Key Item 4671 www. 2: Direct Variation When given an equation in the form y = mx + b, or slope-intercept form , we should know how to graph a line without having to compute two points on it. Keenly observe each graph and pen down the slope and y-intercept. 20 Answers + and - worded problems - crossing 10 worksheet (2) Lesson 4 14. If you want to check this answer, just use the slope formula to find the slope of the line between (6,5) and (1,-5). Do the points (1, 5), (2. 31 scaffolded Free worksheet(pdf) and answer key on point slope form equation of a line. Free printable worksheet (pdf) and Answer Key on slope includes visual aides, model problems, exploratory activities, practice problems, and an online component. Lesson 1 - Direct and Partial Worksheet Solns. 1 - 1 = 0 2 - 1 = 1 3 - 2 = 1 5 - 1 = 4 7 - 5 = 2 8 - 5 = 3. Graph a line given an equation in either slope-intercept or point-slope form. These worksheets and lessons show students how to use what we know about the slope of parallel or perpendicular lines to your advantage. Finding Slope 2 Using Formula - Displaying top 8 worksheets found for this concept. Notes for lesson 9-2. 2 Defining Slope Worksheet is suitable for 9th - 11th Grade. m = 4 3, b = 6 3. Chapter 8 AP Stats test answers the hardest math equation "free online help with college math" free excel square equation solver graphing calculator online circle polynomials exercises for grade 9 algebra 2 problems 9th grade algebra worksheets rom image texas ti basic algebra exercise. Write your groups answer to the discussion questions in the space below. To find the slope of a parallel line, use the same slope. Practice 1 - A line has a 2 slope and passes through the point (3, 2). If you're seeing this message, it means we're having trouble loading external resources on our website. Find the slope of each line. Printable Math Worksheets @ www. Write an equation in slope-intercept form for a line that passes through the point (4, 1) and has a slope of 0. 8 Worksheet by Kuta Software LLC Find the slope of a line parallel to each given line. notebook Subject: SMART Board Interactive Whiteboard Notes Keywords: Notes,Whiteboard,Whiteboard Page,Notebook software,Notebook,PDF,SMART,SMART Technologies ULC,SMART Board Interactive Whiteboard Created Date: 10/3/2017 7:23:51 AM. Based on the two points plotted on a graph, calculate the rise and run to find the slope of the line in the first level of worksheets. 6 h bM na AdweM Hw3iFtZhq jI Yn6f Ci9nLi jt TeQ QP0r4e C-sA xl 8gwe ubTr na V. Algebra I Name Function Notation Worksheet Hour Date Using a table of values to graph equations assignment 12 slope from two points and tables worksheets by math crush graphing coordinate plane using a table of values to graph equations. Since the graph of 2 1 4 x ye − = is concave up, the tangent line found in part (b) lies below the curve. y w xM 6a5d el 4wPiztDhV eIXnCfliDnXiztde o tA5l BgWedb4rMa0 U1D. Algebra 2 Worksheets with Answer Key Along with Finding Slope From A Pair Of Points Math Aids; Worksheet November 03, 2017. Slope Formula Worksheet in an understanding medium may be used to test pupils talents and knowledge by answering questions. Example: Given two points: (3 , 4);(2 , -2) - subtract ( 3 , 4). Slope Grade 9 - Displaying top 8 worksheets found for this concept. Work out the equation of the line, giving your answer in the form 𝑦 = 𝑚 𝑥 + 𝑐. Worksheet # 1: Review 1. answers for chapter 10 math test; printable Symmetry worksheet for grade 4; solver calculator; answers to prentice hall florida algebra 1 workbook; free grade 5 math practice sheets answers; holt algebra 1 refresher; matlab files to solve nonhomogeneous differential equations; McDougal Littell Algebra 2 answers. Slope of a Line (Graphed Points) Worksheet 2 - Here is a 9 problem worksheet where you will asked to find the rise and run between two points on a line, then determine the slope of the line. Two tug boats are towing a large boat, of mass 13750 kg, back to shore. 2 3 Skills Practice Slope. This fraction reduces to 2. Calculate the slope of the line of best fit. Passes through (4, -2) and m = 0. Graph linear equations given in slope-intercept form - easy (slope is a whole number) Graph linear equations given in slope-intercept form (slope is a whole number. 7) through: (2, 2) and (−5, −1) 8) through: (−3, 5) and (−3, 4) 9) through: ( 5, 5 ) and ( 4, −5 ) 10) through: ( 5, 1 ) and ( 1, 3 ) ©w a2 H0d1Z0 Y eK 3u Ftca G RS ho pfXtaw ca Sr Me3 fLQLFCh. 8 Worksheet by Kuta Software LLC Find the slope of a line parallel to each given line. Glencoe Algebra 2 Skills Practice Answer Key New Algebra 2 Function from algebra 1 slope intercept form worksheet 1 answer key , source:washingtoncountyrepublicans. In this worksheet, we will practice finding the slope of a line that goes through two given points. Practice worksheet for lesson 9-2. Worksheets are Slope date period, Practice, Answers anticipation guide and lesson 2 1, Name date period 4 1 skills practice, Parent and student study guide workbook, Answers lesson 2 1 7 glencoe algebra 1, Practice your skills with answers, Geometry chapter 3 notes practice work. Two tug boats are towing a large boat, of mass 13750 kg, back to shore. Filesize: 604 KB; Language: English; Published: December 10, 2015; Viewed: 1,246 times. com Name : Calculate the rise and run to !nd the slope of each line. Lesson 1 - Direct and Partial Worksheet Solns. V Worksheet by Kuta Software LLC 7) −9 8) 1 3 Find the slope of the line through each pair of points. Slope is, and passes through the point (-14, 2). Write the equation using the new slope for m and the b you just found. com Math Practice Series Worksheet #5 Find the difference. 1 Worksheet by Kuta Software LLC. (B) Tangent line: () 1 11 2 yx−= − f ()1. What is the dependent variable? 3. u Worksheet by Kuta. The greater the absolute value of the slope, the steeper the line. Name: Writing Linear Equations Worksheet For 1 – 3, write an equation of the line with the given slope and y-intercept (SLOPE-INTERCEPT FORM). View Answers. Calculate the slope of the line of best fit. Then graph. 20 Answers + and - worded problems - crossing 10 worksheet (2) Lesson 4 14. 35) Slope = − 5 3, y-intercept = 1 36) Slope = 5, y-intercept = 2 Write the slope-intercept form of the equation of the line through the given points. Slope Fields Worksheet Solutions 7. 5), and (7, 1. What is the slope of this line? Drag the x 2. Thus you have a graphical answer to Key Question II. Month # of deer Sept 38 Oct 32 Nov 26 Dec 20 Jan 15 Feb 12 1. We will use inverse functions to work demand and supply problems. ©x 12E0w1u1 z nK9uJtka 6 XSxoef KtwLa Pr3eG BL lL qC7. Now, put the equations together and solve for our answer: Over 3 months, we find that the money in John’s account increased by \$533. 6 h bM na AdweM Hw3iFtZhq jI Yn6f Ci9nLi jt TeQ QP0r4e C-sA xl 8gwe ubTr na V. The nucleus is a large, round/oval structure usually located near the center of the cell. Graphical Problems Questions 1. If given the equation of a line, or two points on the line we can try to find the slope of a line from two points. Vary the coefficients and explore how the graph changes in response. Example 2: Determine the slope of the line passing through the points. Home > Math Worksheets > Algebra Worksheets > Find the X and Y Intercepts Understanding intercepts can really help your learn a bunch about an equation or inequality. Displaying top 8 worksheets found for - Find Slope Given 2 Points. 2 u tAjl GlX OryizgIh Ytmse qrGeQsae 2r 3vAe3dA. Finding The Slope Of Two Points. Practice and Problem Solving: D 1. The one for classifying polynomials has a couple right answers for some of the clues. (i) x + 2y = 8 (ii) y = x - 3. Answers: EOC Practice Worksheet # 11 (spi 3102. Lesson 3 – Slope Review and First Difference. 3 to students over 1–3 days, then review answers with the class. Each linear equations worksheet on this page shows four graphs on a coordinate plane, each with two points labeled, and students find the equation in slope-intercept form by calculating both the slope and y-intercept. Use two steps (or less). Finding Slope From A Graph Worksheet Answers The Best and Most from Finding Slope From A Graph Worksheet, source:worksheets. Solve this equation for W. Exam Practice Pages Answers 6/7 (lower quality than answers below). Month # of deer Sept 38 Oct 32 Nov 26 Dec 20 Jan 15 Feb 12 1. Finding the Slope of a Line (Given Two Points-No Graph)Worksheet 2 - Here is another ten problem worksheet where you will be asked to calculate the slope of a line. The nucleus is a large, round/oval structure usually located near the center of the cell. Answer Key for Practice 9-1. variable 3. It gives you an idea of the direction a line is going and can often give you feeling as to where it is headed in the future. 5 Direct Variation 4. Glencoe Algebra 2 Skills Practice Answer Key New Algebra 2 Function from algebra 1 slope intercept form worksheet 1 answer key , source:washingtoncountyrepublicans. R K FA MlMlv rqicgsh PtHsj 2r je Xs he Prnv 6erdj. The descriptions of the lines are all over the place. 5) through: (2, 3), slope = 1 4 6) through: (-4, 1), slope = - 1 2 7) through: (-4, -4), slope = 9 4 8) through: (2, -2), slope = -2 9) through: (-4, -2), slope = 0. Tuber-ific range: 31 (34 – 3) Terrific Taters range: 42. Find the slope of the line that passes through the points (2,7) and (2,- 6). Finding the Slope of an Equation of a Line Worksheets for 6th Grade,7th Grade and 8th Grade1. This math worksheet was created on 2015-04-22 and has been viewed 38 times this week and 16 times this month. MathScore EduFighter is one of the best math games on the Internet today. They use them to calculate the slope, using the rise over run formula. Find the slope of the line that passes through the points (4,10. (B) Tangent line: () 1 11 2 yx−= − f ()1. This is the reciprocal. If not, explain why. (i) y + 2x = 5 (ii) x – 4y = 12. ENVELOPES The table shows the number of envelopes stuffed for various times. The standard equation for a line is y = mx + b, where m is the slope and b is the y-intercept. 2-3 Skills Practice Rate of Change and Slope DATE PERIOD Find the slope of the line that passes through each pair of points. Some of the worksheets for this concept are Name work review of slope midpoint distance, 3 the midpoint formula, , Finding midpoints distance, Coordinate geometry, Midpoint and distance formulas, Midpoint formula 1, Geometry week 6 packet. Some of the worksheets for this concept are Slope date period, Slope from a, Slope word problems, 6 1 rate of change and slope war, , Hw, Slope and rate of change, Understanding slope a key concept in algebra graphing. 5 review Name _____ Find the slope and y-intercept of the graph of each equation. Find the Equation of a Line Given That You Know a Point on the Line And Its Slope The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept. 2 page worksheet 2 page Answer KEY provided. 7) y = −2x − 2 8) y = 1 4 x + 4 Find the slope of a line perpendicular to each give. We tried to locate some good of Algebra 2 Worksheets with Answer Key Along with Finding Slope From A Pair Of Points Math Aids image to suit your needs. ) Worked Example 2. Investigate the relationships between linear equations, slope, and graphs of lines. Explain your answers. These two points are now on a horizontal line. com Name : Calculate the rise and run to !nd the slope of each line. Lesson 3 - Slope Review and First Difference. introductory algebra answers, simultaneous equations with exponents in excel, algebra 2 equations and answer key, 6th grade plotting slope functions mechanics, houghton mifflin math georgia book 3rd grade, ti emulator 89 download,. Write equations in point-slope form given two pairs of values, and convert the equation into slope-intercept form. The descriptions of the lines are all over the place. Graphing Proportional Relationships (2 Pages) (From Worksheet) Calculating & Plotting Coordinates – from linear equations e. If a line has a slope of 2, a line perpendicular to it will have a slope of -1/2. Students will explore the properties of slope and practice calculating the slope of a line based on two points or from a graph. Lesson 1 - Direct and Partial Worksheet Solns. Use two steps (or less). answer slope intercept form of line in graph show linear equation image titled use the algebra step 1 finding and y from a warm up exercises solution example 5 solve multi problem cost changes intro to equations self guided worksheets help kids practice for examples 2 u003d m x b write an using 4 ratio method analytic geometry straight plane answers exercise 3 intercepts 8th grade online math. The slope will be the same between any two points on a straight line. € y= 4 5 x−2 3. The slope of this line is 2 3. Finding the Slope of a Line (Given Two Points-No Graph)Worksheet 2 - Here is another ten problem worksheet where you will be asked to calculate the slope of a line. If they give you a graph, we can simply count from one point to another. Problems: 15. Slope Activity A Worksheet. m = 4, b = -4 2. Answer Key: Yes. NOTE: Only your test content will print. Lesson 2 – Slope as a Rate of Change. 5 Exam; HW #31 - Worksheet on Slope Fields; HW #31 - Answer Key; 2. 2: Graph and describe the basic shape of the graphs and analyze the general form of the equations for the following families of functions: linear, quadratic, exponential, piece-wise, and absolute value (use technology when appropriate. http://phet. The slope is 0;the y-intercept is 7. com Name : Calculate the rise and run to !nd the slope of each line. Graphing Lines in Slope-Intercept Form Worksheets These Linear Equations Worksheets will produce problems for practicing graphing lines in slope-intercept form. In which direction is the stream flowing? A northwest C southwest B northeast D southeast _____3. Slope of a line worksheet 2 pdf view answers. Be sure to label each line. If n=2 this is said to be a second order versus time resulted in a straight line with a slope value of +3. 5) through: (2, 3), slope = 1 4 6) through: (-4, 1), slope = - 1 2 7) through: (-4, -4), slope = 9 4 8) through: (2, -2), slope = -2 9) through: (-4, -2), slope = 0. Name: _____ Score: Printable Math Worksheets @ www. Slope is and p. 1 = −2 + b. Free trial available at KutaSoftware. This represents the ball's average velocity as it moves across the table. In the following exercises (2-9) evaluate the limit if it Answers to Check: for. If there is a. Answer to College Mathematics: Learning Worksheets Chapter 1 3. € y= 4 5 x−2 3. We can put it simply by saying that it is a ratio of vertical change over time. Questions 13-15 cover the basics of graphs of lines. You will still get the same answer. Write equations in point-slope form given two pairs of values, and convert the equation into slope-intercept form. Lesson 2 – Slope. Talking related with Finding Slope of Line Worksheet, we've collected particular variation of pictures to add more info. Passes through (4, -2) and m = 0. parallel to y — _ —3x + 4, x-intercept at 4 6. You can start playing for free! Graphs to Linear Equations - Sample Math Practice Problems The math problems below can be generated by MathScore. 7) through: (2, 2) and (−5, −1) 8) through: (−3, 5) and (−3, 4) 9) through: ( 5, 5 ) and ( 4, −5 ) 10) through: ( 5, 1 ) and ( 1, 3 ) ©w a2 H0d1Z0 Y eK 3u Ftca G RS ho pfXtaw ca Sr Me3 fLQLFCh. Use this calculator only to check your answers. Some of the worksheets for this concept are Slope date period, Finding slope practice, Find the slope level 1 s1, Types of slopes 1, Slope from a, Grade 9 math test unit 3, Slope intercept form word problems, Lesson plan slope. Draw the line that connects the two points. Title: Microsoft Word - Slope and Topographic Maps. Eighth Grade Math Worksheets (Grade 8 - For Ages 13 to 14) Math in the 8th grade begins to prove more substantial as far as long range skills students will use and need. Practice worksheet for lesson 9-2. € 3x−5y=110 6. In each sketch label the x-axis, y-axis, and slope of the line. Therefore, the slope of the line perpendicular to this line would have to be m = –5/4. If you're seeing this message, it means we're having trouble loading external resources on our website. Use two steps (or less. 1) 2) Find the slope of the line through each pair of poi nts. Independent Practice 2. Review the keystrokes. Students calculate the slope using the change in x and y. Example 1 Write the equation of the line with slope 2 that has y-intercept 5. Free Printables Worksheet Graphing Linear Equations Using Slope And Intercepts Worksheet #2 Answers We found some Images about Graphing Linear Equations Using Slope And Intercepts Worksheet #2 Answers:. Slope is and p. Find the general solution to the differential equations below: (need more practice? … page 327 #2 and #4) a) 5sec42 dy x x dx =+ b) sin 8x 3 dy. 11 ) Find the equation of a line that passes through the points (5 , 3) and (2 ). y + 3 = 1 x 3. 1 Worksheet by Kuta Software LLC. Slope Activity A Worksheet. Worksheet for Correlation and Regression (February 1, 2013). Challenge yourself in the line game! Sample Learning Goals Explain how the slope of a graphed line can be computed. If you fo un d these worksheets useful, please check out Dimensional Analysis Practice Worksheets with Answers, Metric Conversion Practice Problems Worksheet, Converting U nits of Measurement Word Problems Worksheets, 1 Digit Addition Worksheets, 2 Digit Additions Worksheets. Multiple Choice (80 points, 5 points each) Identify the choice that best completes the statement or answers the question. Share with a partner. Printable worksheets of Equations with Variables on Both Sides with distributive property, ti84 systeme equations, different math trivia with answers, math poems algebra, solve the nonlinear progression series in matlab, answers for mcdougal littell algebra 2 for free. ENVELOPES The table shows the number of envelopes stuffed for various times. 0 is horizontal and the side with an undefined slope is vertical. 3) (10, 2), (−9, 7) 4) (−16, 11), (−19, −12) Find the slope of each line. This is the reciprocal. Contains 20 Slope of a Line problems. 4 2 Writing Equations In Slope Intercept Form Worksheet Answers. By observing the Position vs. Point-Slope Plotting and Intersections WorksheetWorks. Some of the worksheets for this concept are Slope from two points, Name example, Slope two point formula find the slope using two point, Finding the slope of a line given the coordinates of two, Slope from an equation, Name, Finding slope work. Three worksheet pages are included for extra practice. ©l q2Z0 u1u2 m YK4uet LaH XSSoVfCttw7aRrQed bLPLpCH. y-intercepts : The y-intercept is where the graph crosses the y-axis which is when x = 0. Wednesday 2/10 - Graphing Systems of Equations PPT, Graphing Worksheet Thursday 2/11 - Review Graphing Worksheet Friday 2/12 - Online Music Activity Monday 2/15 - No School Tuesday 2/16 - Solving Systems using Substitution PPT, Finish Worksheet #1-5 Wednesday 2/17 - Solving Systems using Substitution PPT, Finish Worksheet #1-5 Thursday 2/18. Our slope is then written as a fraction, rise run or 6 3. Linear equation can be graphed based on their slope and y-intercept. Time (min) 5 101520 Envelopes Stuffed 30 60 90 120 2. pub Author: Generic Created Date: 1/15/2004 2:22:20 PM. Passes through (4, -2) and m = 0. A 𝑦 = 2 𝑥 − 4. Complete the following questions from your textbook: page 28 #15, 18, 19, 21, 24 – 28 2. Solving equations involving parallel and perpendicular lines worksheet answers. pub Author: Generic Created Date: 1/15/2004 2:22:20 PM. Given algebraic, tabular, and graphical representations of linear functions, the student will determine the slope of the relationship from each of the representations. It includes two graphs which students need to answer questions about. View Finding Slope WS 2 Answer Key. 2) Slope = 6 7. Thus you have a graphical answer to Key Question II. Contains 20 Slope of a Line problems. Choose a, b, c, or d for each. They determine the slope of a line that contains specified points. 3 Quick Graphs Using Intercepts 4. Slope intercept form worksheet answers algebra 1 graphing shared by 2 standard to, standard form to slope intercept worksheet doc quiz equation for graphing answers with pdf, slope intercept form worksheet algebra 1 pdf with answers kuta software y b worksheets unique standard to, slope intercept form worksheet word problems graphing pdf cc. Slope-2-Slope Squares Puzzle Standard form to slope-intercept form…and back again! Cut the squares apart on the dotted lines. Come to Algebra-equation. Graphing Lines in Slope-Intercept Form Worksheets These Linear Equations Worksheets will produce problems for practicing graphing lines in slope-intercept form. 2 3 Skills Practice Slope. Graph B matches description _____ because _____. Circle Graphs - Level 2. Factoring-polynomials. y = 5x + 13 3. Slope & y-Intercept Worksheet page 2 of 2 continued II. Lesson 1 - Direct and Partial Worksheet Solns. 5 Direct Variation 4. 8 Functions and Relations. That being explained, we all provide assortment of uncomplicated yet useful posts as well as design templates designed suitable for almost any educative purpose. 5 7 6 8 1 2 10 3 Finding Inverses Find an equation for the inverse for each of the following relations. If not, explain why. Check grade levels below. We have a complete K-12 math curriculum library. Slope Worksheet 2 Answers - Wanting to Buy a New Pyramid Scheme: Inthe the same vein as your failure to come up with an answer to the question above, you might as well try and come up with a solution that will convince you to buy the new pyramid scheme. Slope Worksheet 2 Answers together with Suitable Topics. Points A, B, Y, and Z are reference points on the topographic map. Answers for the worksheet on slope intercept form are given below: Answer: 1. ) Negative slope (eye brow above "-" eyeball. Acceleration (a) vs. This is a two-page worksheet. ©T 82N0m152 C vK Eu 7tma9 Smo 8f0tDwfa Jr je 6 qLsLtCG. If you want to check this answer, just use the slope formula to find the slope of the line between (6,5) and (1,-5). (POINT-SLOPE FORM) 4. Find the Slope Level 1: S1 1)-5 -4 -3 -2 -1 1. It is the control center for all the activities of the cell. Answer Keys. com makes available usable facts on Slope Intercept Free Printable Worksheets, exponents and rational expressions and other algebra subjects. Click on the links below to view sample pages. 5), and (7, 1. Slope intercept form worksheet answers algebra 1 graphing shared by 2 standard to, standard form to slope intercept worksheet doc quiz equation for graphing answers with pdf, slope intercept form worksheet algebra 1 pdf with answers kuta software y b worksheets unique standard to, slope intercept form worksheet word problems graphing pdf cc. Algebra 2 Worksheets with Answer Key Along with Finding Slope From A Pair Of Points Math Aids Worksheet November 03, 2017 We tried to locate some good of Algebra 2 Worksheets with Answer Key Along with Finding Slope From A Pair Of Points Math Aids image to suit your needs. In Part 2 of the worksheet, you were given the solution to the initial value problem with S(0) = 2, k = 0. doc Author: kmacphe Created Date: 5/9/2007 2:34:41 PM. 35) Slope = − 5 3, y-intercept = 1 36) Slope = 5, y-intercept = 2 Write the slope-intercept form of the equation of the line through the given points. If a student has the 3 2 card the only ordered pair that will work to get a slope of 1 is 2 3 but that isn t one of the cards on the worksheet. Three worksheet pages are included for extra practice. In the circuit below, the reading on the ammeter is 3. The nucleus is a large, round/oval structure usually located near the center of the cell. 5 Direct Variation 4. This math worksheet was created on 2015-04-22 and has been viewed 38 times this week and 16 times this month. Due to the fact you should supply solutions in a real along with efficient origin, all of us provide handy facts about many subjects in addition to topics. x (hour) y (phone. ) (1, -2) and (-1, 3) 10. Filesize: 604 KB; Language: English; Published: December 10, 2015; Viewed: 1,246 times. This Practice Masters Level A: 5. To find the slope of two given points, you can use the point-slope formula of (y2 - y1) / (x2 - x1). Linear equation can be graphed based on their slope and y-intercept. Linear Equations. Download this assignment ad pdf Download. Finding Slope From Two Points Worksheet Answers in a learning moderate can be utilized to test students qualities and knowledge by addressing questions. Finding the slope of a line from two points requires us to know the definition of slope. Displaying top 8 worksheets found for - Finding The Slope Of Two Points. Writing Linear Equations If a graph or two points are given, it is possible to write out the linear function using information from the graph. Problems: 15. Slope Activity A Worksheet. 3) Answers: EOC Practice Worksheet # 13 (spi 3102. 51 State the general compass direction in which Maple Stream is flowing. Slope is, and passes through the point (-14, 2). As many practice questions as needed may be generated along with their solutions. number is a good approximation of the speedometer reading at t =2,andthereforean answer to Key Question I? 5 By measuring the slope of the secant to the graph over a very tinytimeinterval,makeyour best estimate of the speed of the shell when it reaches the peakofitspath,attimet =5. Include the correct units with your answer. Solve this equation for b1. Lesson 2 – Slope and Slope as a Rate of Change Worksheet – contains answers. Algebra IA Unit 4 Worksheet 8 Interpreting slope and y-intercept – Part 2 Name _____ 1. We have a complete K-12 math curriculum library. Then find the slope. Discovering Slope of a Table depends on realizing that Slope is a ratio between the change in the y-values divided by the change in the x-values. You could offer free training, free consultation and even a free product. m=2 m = 2, we have a positive slope and therefore the straight line is rising or increasing from left to right. How many right triangle in this figure? Count and color : Count and color the correct number of each shape in the box below :. y + 1 x = –2 3 1 x –4 2 5 3 9. That's the y-intercept for line A. Toddler Worksheets Geography Fun Worksheets Noun Determiners Worksheet kids worksheet 2 test Homework For 1st Grade Math Multi Step Equations Worksheet With Fractions With the dawning of technology, there is no need to hate Math at school or when practicing at home. Finding The Slope Of Two Points. Keenly observe each graph and pen down the slope and y-intercept. Tug boat 1 is pulling with a force of T1 = 7500 N at an angle of 30 north of the forward motion (see Figure 2) and tug boat 2 is pulling with a force of T2 = 8500 N at an angle θ south of the forward motion. different worksheets. Slope Man! 1. 8 Functions and Relations. Problems 1. 2 u tAjl GlX OryizgIh Ytmse qrGeQsae 2r 3vAe3dA. The slope worksheets on this page have exercises where students identify the direction of slope, as well as calculating slope from points on the coordinate plane. If there is a. € y= 3 2 x+7 2. u Worksheet by Kuta. Plus model problems explained step by step. The slope is 3 10;the y-intercept is 10. Author: Shane Irvin. Passes through (0, 2) and has slope of –5/3. Write the word or phrase that best completes each statement or answers the question. Come to Algebra-equation. Before discussing Slope Worksheet 2 Answers, make sure you know that Instruction is actually our answer to an improved the day after tomorrow, plus studying doesn't just halt when the school bell rings. Here are data from four students on their Quiz 1 scores and their Quiz 5 scores and a graph where we connected the points by a line. Step 2: Use the slope to find the y-intercept. Factoring-polynomials. Linear equation can be graphed based on their slope and y-intercept. (POINT-SLOPE FORM) 4. ” Student Worksheet Notes with Answers Overview Tell students: 1. If they give you a graph, we can simply count from one point to another. Title: worksheet interpreting graphs ch4. Great for partner work (the unmatching answers keeps them from copying each others answers) OR as a self check for individual practice. Then find the slope. Write an equation in slope-intercept form for a line that passes through the point (4, 1) and has a slope of 0. These can be used for review or homework after the notes. Writing Linear Equations If a graph or two points are given, it is possible to write out the linear function using information from the graph. In the circuit below, the reading on the ammeter is 3. Is there a function all of whose values are equal to each other? If so, graph your answer. ©T 82N0m152 C vK Eu 7tma9 Smo 8f0tDwfa Jr je 6 qLsLtCG. A line 𝐿 has a slope of − 2 and passes through the point (2, − 3). Thus you have a graphical answer to Key Question II. This represents the ball's average velocity as it moves across the table. View Finding Slope WS 2 Answer Key. 045 A and I 5 = 0. Now, he or she can solve for b, which is -5. We hope this picture will be certainly one of excellent resource for biology 23 carbon compounds worksheet answers or chemistry archive january 03 2018. Each exercise feature two points, and you will have to calculate the rise and run between the two points by finding the difference between the x-coordinates and the y. slope worksheet #1 and #2 Author: Tammy Created Date: 2/5/2012 10:40:03 PM. Topographic Map Worksheet Use the following topographic map to answer 1 -3. Chapter 2 11. Because in the Scholar Worksheet about 90% of the articles of the whole book are issues, equally multiple decision and answer issues which are not available. net provides more than 2000 unlimited practice and is an interesting resource for students to keep their mathematics skills sharped. 11: Slope Fields and Differential Equations. V Worksheet by Kuta Software LLC 7) −9 8) 1 3 Find the slope of the line through each pair of points. parallel to y — _ —3x + 4, x-intercept at 4 6. This is a two-page worksheet. Lesson 5. Download geometry slope day 2 worksheet answer document. That being explained, we all provide assortment of uncomplicated yet useful posts as well as design templates designed suitable for almost any educative purpose. If you're seeing this message, it means we're having trouble loading external resources on our website. 2 Answers - Slope 1) 3 2 2) 5 3) Undeﬁned 4) − 1 2 5) 5 6 6) − 2 3 7) − 1 8) 5 4 9) − 1 10) 0 11) Undeﬁned 12) 16 7 13) − 17 31 14) − 3 2 15) 4 3 16. If you want to check this answer, just use the slope formula to find the slope of the line between (6,5) and (1,-5). 1 - 1 = 0 2 - 1 = 1 3 - 2 = 1 5 - 1 = 4 7 - 5 = 2 8 - 5 = 3. Video for lesson 9-1: Basic Terms of Circles. Lesson 2 - Slope as a Rate of Change. Do the points (2, 5), (4, 3. Linear Equations (Answer ID # 0377038) Write an equation of the line that contains the given point and has the given slope. com Name : Calculate the rise and run to !nd the slope of each line. Some of the worksheets for this concept are Slope from two points, Name example, Slope two point formula find the slope using two point, Finding the slope of a line given the coordinates of two, Slope from an equation, Name, Finding slope work. Thus you have a graphical answer to Key Question II. Slope; Description Explore the world of lines. Slope is, and passes through the point (-14, 2). Answer : 2. (i) y = 8 (ii) y + 5 = 0. To find the slope of a line passing through a given pair of points is found by using the point slope formula. Linear Function Questions and Answers. Finding the Slope of a Line From 2 Points on the Line. The speed of the car is decreasing. Sketch the lines that your group drew while using Geometer s Sketchpad. powered by. 6 Links verified on 11/12/2016. Lines and Graphs. Brian McLogan 289,698 views. Label and number the x and y-axis appropriately. Since the ball is moving in a positive direction its velocity is positive. This is a two-page worksheet. Passes through (-1, -7) and. y + 1 x = –2 3 1 x –4 2 5 3 9. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Its equation is y or 1 x 2 1 2 y 1 2 x 3. Here it is. y = 2x – 6 ( 9 of 10) (From Worksheet) Calculating & Plotting Coordinates – from linear equations e. Find the slope of each line. Title: Microsoft Word - Slope and Topographic Maps. For safety, the slope of a staircase must. different worksheets. 31 scaffolded questions that start relatively easy and end with some real challenges. Recall: Answer: Answers will vary, depending on students’ lines of best fit. Notes - Slope Fields and Differential Equations; Notes - Slope Fields and Differential Equations (filled) HW #30 - Review for 4. 4 Identify rates of change (slope) and distinguishing properties of data from tables, graphs, and equations to predict what happens to one variable as another variable changes;. y= x 1 Write an equation of a line with the given slope m and y-intercept b. m = 4 3, b = 6 3. Unsupported answers may receive NO credit. Write equations in point-slope form given two pairs of values, and convert the equation into slope-intercept form. Slope of a Line Worksheet 2 RTF Slope of a Line Worksheet 2 PDF View Answers. These are in html format. If you a point that a line passes through, and its slope, this page will show you how to find the equation of the line. You can start playing for free! Graphs to Linear Equations - Sample Math Practice Problems The math problems below can be generated by MathScore. Find the slope m1 of line L1 and the slope m2 of line L1 m1 = ( 1. Lesson 4 – Connecting Slope ideas. Let us consider the co-ordinate points (x1,y1) and (x2,y2) as (9,1) and (8,2), then (Y - 1) / (2 - 1) = (X - 9) / (8 - 9) (Y - 1) / 1 = (X - 9) / -1 - (Y - 1) = (X - 9) (-Y + 1) = (X - 9) X + Y - 10 = 0. y + x = 0 13. 40 best Math 8 Slope images on Pinterest from worksheet piecewise. Please include units in your answer. The descriptions of the lines are all over the place. Brian McLogan 289,698 views. To graph a linear equation, we can use the slope and y-intercept. 2: Graph and describe the basic shape of the graphs and analyze the general form of the equations for the following families of functions: linear, quadratic, exponential, piece-wise, and absolute value (use technology when appropriate. Each time you download a worksheet it will have unique questions and come with its own answer key. Explain your answers. 2 u tAjl GlX OryizgIh Ytmse qrGeQsae 2r 3vAe3dA. Slope of a Line (Graphed Points) Worksheet 2 - Here is a 9 problem worksheet where you will asked to find the rise and run between two points on a line, then determine the slope of the line. 7 Linear Equations In Slope-intercept Form Worksheet With Answers Pdf Online Here For Free. Multiple worksheets. Our slope is then written as a fraction, rise run or 6 3. Given the line with equation x – 2y = 5, find the slope and the y-intercept. The newer uploads will be placed at the top of this page. Problem 2 (4. Slope is, and passes through the point (-14, 2). Sketch the lines that your group drew while using Geometer s Sketchpad. Graph linear equations given in slope-intercept form - easy (slope is a whole number) Graph linear equations given in slope-intercept form (slope is a whole number. y = -3x - 1 y = 4x + 2 y = -2x + 7 y = -5. How to navigate the application, if they are not yet familiar with the application. Is there a function all of whose values are equal to each other? If so, graph your answer. Algebra 2 Wkst 3. s l o p e = r i s e r u n = 2 3. x (hour) y (phone. Answer to College Mathematics: Learning Worksheets Chapter 1 3. Due to the fact you should supply solutions in a real along with efficient origin, all of us provide handy facts about many subjects in addition to topics. This second worksheet has graph Sketching Questions, (Questions 15 to 260, with Answers at the end of the sheet. (Answers: I 20 = 0. The given equation: y 2x 1 has slope 2. Graphing Linear Equations Using Slope and Intercepts – Worksheet #2 Graph using the slope and y-intercept. Practice worksheet for lesson 9-1. Lesson 4 – Connecting Slope ideas. y – 2x = –3 7. 218 gal/h 8. First download this acceleration worksheet with answers as pdf. constant 2. The reason this problem occurs is that many beginners want to buy and get the first product they see online. 5) y = 3x + 2 6) y = −x + 5 Find the slope of a line parallel to each given lin e. Lesson 2 - Slope and Slope as a Rate of Change Worksheet - contains answers. Two Point Slope Form. The slope is 7 8;the y-intercept is 8. A step by step worksheet solver to find the slope and the y-intercept of a line given in standard form a x + b y = c. It should look like this: 3 = 2(4) + b. Glencoe Algebra 2 Skills Practice Answer Key New Algebra 2 Function from algebra 1 slope intercept form worksheet 1 answer key , source:washingtoncountyrepublicans. To preview this answer key, click on the File menu and select Print Preview. It gives you an idea of the direction a line is going and can often give you feeling as to where it is headed in the future. 7) y = −2x − 2 8) y = 1 4 x + 4 Find the slope of a line perpendicular to each give. Finding Slope 2 Using Formula - Displaying top 8 worksheets found for this concept. Name Class Date Practice 6-2 Slope-Intercept Form Find the slope (written in two ways) and y-intercept of each equation. Worksheet 1: Limits, Slope of the Secant line, and Equation of the tangent Line A. 8 Linear Equations Worksheets. Given the line with equation x – 2y = 5, find the slope and the y-intercept. Answer: Answers will vary based on students’ lines of best fit. The worksheet contains a series of problems reviewing the following material: - determining if a line has a positive, negative, zero or undefined slope - finding the slope given 2 points (using slope formula) - find the slope of a line given a graph - determining slope from a word problem - find an. Answers and detailed answer explanations for these problems y=mx+b m b y=−2x+20 y=2x+1 y−10=−2(x−5) y=−2x+20 m=Δy Δx Worksheet. (i) x + 2y = 8 (ii) y = x - 3. Answers: EOC Practice Worksheet # 11 (spi 3102. Let us consider the co-ordinate points (x1,y1) and (x2,y2) as (9,1) and (8,2), then (Y - 1) / (2 - 1) = (X - 9) / (8 - 9) (Y - 1) / 1 = (X - 9) / -1 - (Y - 1) = (X - 9) (-Y + 1) = (X - 9) X + Y - 10 = 0. It should look like this: 3 = 2(4) + b. com happens to be the right site to go to!. All of the worksheets come with an answer key on the 2nd page of the file. Solve this equation for b1. Answer Keys. (85-5), (4, -2) Determine the slope of each graph using rise over run or the slope formula. 3) (10, 2), (−9, 7) 4) (−16, 11), (−19, −12) Find the slope of each line. s l o p e = r i s e r u n rise is your “up or down” & run is your “left or right”. ©T 82N0m152 C vK Eu 7tma9 Smo 8f0tDwfa Jr je 6 qLsLtCG. A line 𝐿 has a slope of − 2 and passes through the point (2, − 3). 2 u tAjl GlX OryizgIh Ytmse qrGeQsae 2r 3vAe3dA. Therefore, the slope of the line perpendicular to this line would have to be m = –5/4. These are in html format. Finding the Slope of a Line From 2 Points on the Line. To find the slope of a line passing through a given pair of points is found by using the point slope formula. Acceleration (a) vs. 11 ) Find the equation of a line that passes through the points (5 , 3) and (2 ). 31 scaffolded questions that start relatively easy and end with some real challenges. This fraction reduces to 2. 7) Answers: EOC Practice Worksheet. The elevation at Point A is A 60 m C 140 m B 100 m D 300 m _____2. Example: To calculate the slope-intercept equation for a line that includes. (POINT-SLOPE FORM) 4. 7 Linear Equations In Slope-intercept Form Worksheet With Answers Is Often Used In Slope Intercept Form Worksheet, Slope Worksheets, Math Worksheets, Worksheets, Practice Sheets & Homework Sheets And Education. Lesson 3 – Slope and First Differences Worksheet Solns. Use the linear equation formula that is suitable to address problems. perpendicular to y. y = -3x – 1 y = 4x + 2 y = -2x + 7 y = -5. ExploreLearning. Practice worksheet for lesson 9-1. slope x-intercept at 3. If you need support with algebra and in particular with rearranging formulas calculator or precalculus come visit us at Algebra-equation. The function below shows the cost of a hamburger with different numbers of toppings (t). Whether you are teaching kindergartens how to count, youngsters how to multiply, teens how to factor polynomials, or adults how to understand Ohm’s law, you will find what you need at The Math Worksheet Site. Practice 1 - A line has a 2 slope and passes through the point (3, 2). Write your groups answer to the discussion questions in the space below. Multiple Choice (80 points, 5 points each) Identify the choice that best completes the statement or answers the question. V Worksheet by Kuta Software LLC 7) −9 8) 1 3 Find the slope of the line through each pair of points. We hope that you find exactly what you need for your home or classroom!. If a straight line passes through the points (x1, y1) and (x2, y2) then its equation is y - y1 =. the slope of this line. Convert from standard form to slope-intercept from. Passes through (4, -2) and m = 0. y = mx + b Write the slope -intercept formula. Slope is, and passes through the point (-14, 2).	2020-10-21 10:05:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2916516959667206, "perplexity": 1261.647941123715}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107876307.21/warc/CC-MAIN-20201021093214-20201021123214-00050.warc.gz"}
https://www.physicsforums.com/threads/calculating-integrals-of-area-between-two-functions-involving-absolute-values.282615/	# Calculating integrals of area between two functions, involving absolute values 1. Jan 3, 2009 ### Ulagatin Hi all, I'm currently preparing for pre-tertiary mathematics, studying from Apostol's "One-Variable Calculus". I have just begun to work on the theory of integration of trigonometric functions, but I found that with the last set of exercises (on finding area between two functions, over some interval) I wasn't able to complete them (the issue was with absolute value functions). Any assistance would be greatly appreciated. 1. The problem statement, all variables and given/known data Take one, for example, with these functions (over the interval [-1, 1]): $$f(x) = \|x\|$$ $$g(x) = x^2-1$$ I recognise that over this interval, $$f(x) > g(x) for all x$$, so the answer should be equal to the integral of the difference f(x) - g(x) with respect to x over the previously said interval. How do I tackle this from here? My graphics calculator software tells me the area is equal to 2.33..., but I'm not sure how to arrive at this value. 2. Relevant equations $$\int_{a}^{b} [f(x) - g(x)] dx$$ 3. The attempt at a solution To show that I can work these integrals, I first show another example which may assist to solve the original example. $$f(x) = x(x^2 - 1)$$ $$g(x) = x$$ The interval we must evaluate this integral on is as follows: $$[-1, 2^{1/2}]$$. Over the interval [0, 1], f(x) > g(x). Over the interval $$[0, 2^{1/2}]$$, g(x) > f(x). So to formalise this now, we have the following: $$\int_{-1}^{0} [x(x^2 - 1) - x] dx + \int_{0}^{2^{1/2}} [(x - x^3 - x)] dx$$ $$= -\int_{0}^{-1} [x^3 - x - x] dx + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx$$ $$= -\int_{0}^{-1} [x^3 - 2x] dx + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx$$ $$= -\left[ \frac{x^4}{4} - {x^2} \right]_{0}^{-1} + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx$$ $$= [-\frac{-1^4}{4} + -1^2] + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx$$ $$= [-\frac{1}{4} + 1] + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx$$ $$= \frac{3}{4} + \int_{0}^{2^{1/2}} [-x^3 + 2x] dx$$ $$= \frac{3}{4} + \left[ \frac{-x^4}{4} + {x^2} \right]_{0}^{2^{1/2}}$$ $$= \frac{3}{4} + [-\frac {(2^{1/2})^4}{4} + (2^{1/2})^2] = \frac{3}{4} +[-\frac {4}{4} + 2]$$ $$= \frac{3}{4} - 1 + 2 = \frac{3}{4} + 1$$ $$= \frac{3+4}{4} = \frac{7}{4} = 1.75.$$ Now, I believe that, perhaps with the absolute value example, you can substitute values in, similarly to this example I've worked. First, I note that both these functions are monotonically decreasing over the interval [-1, 0] and increasing over [0, 1]. So, I'll try: $$\int_{-1}^{0} [f(x) - g(x)] dx + \int_{0}^{1} [f(x) - g(x)] dx$$ $$= -\int_{0}^{-1} [\|x\| - (x^2 - 1)] dx + \int_{0}^{1} [\|x\| - (x^2 - 1)] dx$$ $$= \left[ \frac{\|x^2\|}{2} - (\frac{x^3}{3} - x) \right]_{0}^{-1} + \left[ \frac{\|x^2\|}{2} - (\frac{x^3}{3} - x) \right]_{0}^{1}$$ $$= 0 - [\frac{1}{2} - (\frac{-1^3}{3} +1)] + [\frac{1}{2} - (\frac{1}{3} - 1)]$$ $$= [-\frac{1}{2} + (-\frac{1^3}{3} + 1)] + [\frac{1}{2} - \frac{1}{3} + 1]$$ $$= - \frac{1}{2} - \frac{1}{3} + 1 + \frac{1}{2} - \frac{1}{3} + 1 = \frac{4}{3}$$ Clearly, I'm out by 1. Where have I gone wrong? Any assistance would be simply grand. I tried with the first integral being positive, but I ended with the same result mysteriously enough. I just don't understand! Cheers, Davin 2. Jan 3, 2009 ### happyg1 Hey, You have to drop your absolute value bars on the second line of your solution up there. The equation of the line from -1 to 0 is -x and from 0 to 1 it's just x...that changes the sign on your 1/2 aand gets you what you know is correct. CC 3. Jan 3, 2009 ### Gregg g(x) and f(x) both have symmetry on the y-axis. You can do it like: $$2\int_{0}^{1} - x^2 + x - 1 \delta x = 2\left[1 \frac{1}{6} \right]$$ which is 2.33 $$\int_{-1}^{1} \|x\| - (x^2 - 1) \delta x = \left[\frac{\|x^2\|}{2} - \frac{x^3}{3} + x\right]_{-1}^{1}$$ $$= \left[ \frac{1}{2} - \frac{1}{3} + 1\right] - \left[ -\frac{1}{2} + \frac{1}{3} - 1\right] = 2\frac{1}{3}$$ I think that's right. Last edited: Jan 3, 2009 4. Jan 3, 2009 ### Ulagatin Absolute Value integrals?? Thank you very much, guys. Nice to hear that you can solve integrals by looking for symmetry! This I wasn't aware of. Gregg, I tried your first integral and I didn't get the same answer as you. Your second integral I understand perfectly though. Looking closer, it seems you wrote the first integral out incorrectly. I believe it should be this: $$2 \int_{0}^{1} \|x\| - (x^2 - 1) \delta x = 2\left[\frac{\|x^2\|}{2} - \frac{x^3}{3} + x\right]_{0}^{1}$$ $$= 2\left[\frac{1}{2} - \frac{1}{3} + 1\right]$$ $$= 2\left[1 \frac{1}{6} \right] = 2\frac{1}{3}$$ I truly do appreciate your assistance. This ability to solve using symmetry is quite remarkable, but I suppose it really isn't much of a surprise: symmetry is useful in so many areas in mathematics! Cheers, Davin 5. Jan 3, 2009 ### Gregg Re: Absolute Value integrals?? Now I look at it, it can be solved even easier than that. Take the function y = \|x\|. You know that the area underneath it between -1 and 1 will be 1. Take the area under the x-axis for $$x^2 - 1 = (x+1)(x-1)$$ gives $$\int_{-1}^{1} x^2 - 1 \delta x = \left[ \frac{x^3}{3} - x \right]_{-1}^{1}$$ $$\int_{-1}^{1} x^2 - 1 \delta x = \left[ \frac{1^3}{3} - 1 \right]-\left[ \frac{-1^3}{3} + 1 \right]$$ $$\int_{-1}^{1} x^2 - 1 \delta x = \left[ -\frac{2}{3} \right] - \left[ \frac{2}{3} \right] = - \frac{4}{3}$$ $$1 + \frac{4}{3} = 2 \frac{1}{3}$$ The reason I find this interesting is because you first answers are out by 1 square unit. I think for this reason it safest in this situation to use the method above. The blue curve is $$\|x\| - (x^2-1)$$ and the red curve is $$x - (x^2-1)$$ (they overlap). Because x is modulus it has a vertex with the x component of $$\pm \frac{1}{2}$$. I also find that $$\int_{-1}^{1} -x^2+x+1 \delta x = \frac{4}{3}$$ I believe this is the reason your answers were out by 1 square unit. The modulus of x was not correctly integrated. But honestly I don't know. Last edited: Jan 4, 2009 6. Jan 3, 2009 ### Ulagatin Thanks again, Gregg. I personally feel that this case of looking for symmetries is simpler. Then it's a more intuitive approach. But thanks for elaborating further on these ideas. Cheers, Davin	2016-12-04 00:20:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8311987519264221, "perplexity": 613.3036956045995}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541142.66/warc/CC-MAIN-20161202170901-00502-ip-10-31-129-80.ec2.internal.warc.gz"}
http://en.wikipedia.org/wiki/String_operations	# String operations In computer science, in the area of formal language theory, frequent use is made of a variety of string functions; however, the notation used is different from that used on computer programming, and some commonly used functions in the theoretical realm are rarely used when programming. This article defines some of these basic terms. ## Strings and languages A string is a finite sequence of characters. The empty string is denoted by $\varepsilon$. The concatenation of two string $s$ and $t$ is denoted by $s \cdot t$, or shorter by $s t$. Concatenating with the empty string makes no difference: $s \cdot \varepsilon = s = \varepsilon \cdot s$. Concatenation of strings is associative: $s \cdot (t \cdot u) = (s \cdot t) \cdot u$. For example, $(\langle b \rangle \cdot \langle l \rangle) \cdot (\varepsilon \cdot \langle ah \rangle) = \langle bl \rangle \cdot \langle ah \rangle = \langle blah \rangle$. A language is a finite or infinite set of strings. Besides the usual set operations like union, intersection etc., concatenation can be applied to languages: if both $S$ and $T$ are languages, their concatenation $S \cdot T$ is defined as the set of concatenations of any string from $S$ and any string from $T$, formally $S \cdot T = \{ s \cdot t \mid s \in S \land t \in T \}$. Again, the concatenation dot $\cdot$ is often omitted for shortness. The language $\{\varepsilon\}$ consisting of just the empty string is to be distinguished from the empty language $\{\}$. Concatenating any language with the former doesn't make any change: $S \cdot \{\varepsilon\} = S = \{\varepsilon\} \cdot S$, while concatenating with the latter always yields the empty language: $S \cdot \{\} = \{\} = \{\} \cdot S$. Concatenation of languages is associative: $S \cdot (T \cdot U) = (S \cdot T) \cdot U$. For example, abbreviating $D = \{ \langle 0 \rangle, \langle 1 \rangle, \langle 2 \rangle, \langle 3 \rangle, \langle 4 \rangle, \langle 5 \rangle, \langle 6 \rangle, \langle 7 \rangle, \langle 8 \rangle, \langle 9 \rangle \}$, the set of all three-digit decimal numbers is obtained as $D \cdot D \cdot D$. The set of all decimal numbers of arbitrary length is an example for an infinite language. ## Alphabet of a string The alphabet of a string is the set of all of the characters that occur in a particular string. If s is a string, its alphabet is denoted by $\operatorname{Alph}(s)$ The alphabet of a language $S$ is the set of all characters that occur in any string of $S$, formally: $\operatorname{Alph}(S) = \bigcup_{s \in S} \operatorname{Alph}(s)$. For example, the set $\{\langle a \rangle,\langle c \rangle,\langle o \rangle\}$ is the alphabet of the string $\langle cacao \rangle$, and the above $D$ is the alphabet of the above language $D \cdot D \cdot D$ as well as of the language of all decimal numbers. ## String substitution Let L be a language, and let Σ be its alphabet. A string substitution or simply a substitution is a mapping f that maps letters in Σ to languages (possibly in a different alphabet). Thus, for example, given a letter a ∈ Σ, one has f(a)=La where La ⊆ Δ* is some language whose alphabet is Δ. This mapping may be extended to strings as f(ε)=ε for the empty string ε, and f(sa)=f(s)f(a) for string sL. String substitutions may be extended to entire languages as [1] $f(L)=\bigcup_{s\in L} f(s)$ Regular languages are closed under string substitution. That is, if each letter of a regular language is substituted by another regular language, the result is still a regular language.[2] Similarly, context-free languages are closed under string substitution.[3][note 1] A simple example is the conversion fuc(.) to upper case, which may be defined e.g. as follows: letter mapped to language remark x fuc(x) a { ‹A› } map lower-case char to corresponding upper-case char A { ‹A› } map upper-case char to itself ß { ‹SS› } no upper-case char available, map to two-char string ‹0› { ε } map digit to empty string ‹!› { } forbid punctuation, map to empty language ... similar for other chars For the extension of fuc to strings, we have e.g. • fuc(‹Straße›) = {‹S›} ⋅ {‹T›} ⋅ {‹R›} ⋅ {‹A›} ⋅ {‹SS›} ⋅ {‹E›} = {‹STRASSE›}, • fuc(‹u2›) = {‹U} ⋅ {ε} = {‹U›}, and • fuc(‹Go!›) = {‹G›} ⋅ {‹O›} ⋅ {} = {}. For the extension of fuc to languages, we have e.g. • fuc({ ‹Straße›, ‹u2›, ‹Go!› }) = { ‹STRASSE› } ∪ { ‹U› } ∪ { } = { ‹STRASSE›, ‹U› }. Another example is the conversion of an EBCDIC-encoded string to ASCII. ## String homomorphism A string homomorphism (often referred to simply as a homomorphism in formal language theory) is a string substitution such that each letter is replaced by a single string. That is, f(a)=s, where s is a string, for each letter a.[note 2][4] String homomorphisms are monoid morphisms on the free monoid, preserving the binary operation of string concatenation. Given a language L, the set f(L) is called the homomorphic image of L. The inverse homomorphic image of a string s is defined as f−1(s) = { w \| f(w)=s } while the inverse homomorphic image of a language L is defined as f−1(L) = { s \| f(s) ∈ L } In general, f(f−1(L)) ≠ L, while one does have f(f−1(L)) ⊆ L and Lf−1(f(L)) for any language L. The class of regular languages is closed under homomorphisms and inverse homomorphisms.[5] Similarly, the context-free languages are closed under homomorphisms[note 3] and inverse homomorphisms.[6] A string homomorphism is said to be ε-free (or e-free) if f(a) ≠ ε for all a in the alphabet Σ. Simple single-letter substitution ciphers are examples of (ε-free) string homomorphisms. An example string homomorphism guc can also be obtained by defining similar to the above substitution: guc(‹a›) = ‹A›, ..., guc(‹0›) = ε, but letting guc undefined on punctuation chars. Examples for inverse homomorphic images are • guc−1({ ‹SSS› }) = { ‹sss›, ‹sß›, ‹ßs› }, since guc(‹sss›) = guc(‹sß›) = guc(‹ßs›) = ‹SSS›, and • guc−1({ ‹A›, ‹bb› }) = { ‹a› }, since guc(‹a›) = ‹A›, while ‹bb› cannot be reached by guc. For the latter language, guc(guc−1({ ‹A›, ‹bb› })) = guc({ ‹a› }) = { ‹A› } ≠ { ‹A›, ‹bb› }. The homomorphism guc is not ε-free, since it maps e.g. ‹0› to ε. ## String projection If s is a string, and $\Sigma$ is an alphabet, the string projection of s is the string that results by removing all letters which are not in $\Sigma$. It is written as $\pi_\Sigma(s)\,$. It is formally defined by removal of letters from the right hand side: $\pi_\Sigma(s) = \begin{cases} \varepsilon & \mbox{if } s=\varepsilon \mbox{ the empty string} \\ \pi_\Sigma(t) & \mbox{if } s=ta \mbox{ and } a \notin \Sigma \\ \pi_\Sigma(t)a & \mbox{if } s=ta \mbox{ and } a \in \Sigma \end{cases}$ Here $\varepsilon$ denotes the empty string. The projection of a string is essentially the same as a projection in relational algebra. String projection may be promoted to the projection of a language. Given a formal language L, its projection is given by $\pi_\Sigma (L)=\{\pi_\Sigma(s)\ \vert\ s\in L \}$ ## Right quotient The right quotient of a letter a from a string s is the truncation of the letter a in the string s, from the right hand side. It is denoted as $s/a$. If the string does not have a on the right hand side, the result is the empty string. Thus: $(sa)/ b = \begin{cases} s & \mbox{if } a=b \\ \varepsilon & \mbox{if } a \ne b \end{cases}$ The quotient of the empty string may be taken: $\varepsilon / a = \varepsilon$ Similarly, given a subset $S\subset M$ of a monoid $M$, one may define the quotient subset as $S/a=\{s\in M\ \vert\ sa\in S\}$ Left quotients may be defined similarly, with operations taking place on the left of a string. ## Syntactic relation The right quotient of a subset $S\subset M$ of a monoid $M$ defines an equivalence relation, called the right syntactic relation of S. It is given by $\sim_S \;\,=\, \{(s,t)\in M\times M\ \vert\ S/s = S/t \}$ The relation is clearly of finite index (has a finite number of equivalence classes) if and only if the family right quotients is finite; that is, if $\{S/m\ \vert\ m\in M\}$ is finite. In this case, S is a recognizable language, that is, a language that can be recognized by a finite state automaton. This is discussed in greater detail in the article on syntactic monoids. ## Right cancellation The right cancellation of a letter a from a string s is the removal of the first occurrence of the letter a in the string s, starting from the right hand side. It is denoted as $s\div a$ and is recursively defined as $(sa)\div b = \begin{cases} s & \mbox{if } a=b \\ (s\div b)a & \mbox{if } a \ne b \end{cases}$ The empty string is always cancellable: $\varepsilon \div a = \varepsilon$ Clearly, right cancellation and projection commute: $\pi_\Sigma(s)\div a = \pi_\Sigma(s \div a )$ ## Prefixes The prefixes of a string is the set of all prefixes to a string, with respect to a given language: $\operatorname{Pref}_L(s) = \{t\ \vert\ s=tu \mbox { for } t,u\in \operatorname{Alph}(L)^\}$ here $s\in L$. The prefix closure of a language is $\operatorname{Pref} (L) = \bigcup_{s\in L} \operatorname{Pref}_L(s) = \left\{ t\ \vert\ s=tu; s\in L; t,u\in \operatorname{Alph}(L)^ \right\}$ Example: $L=\left\{abc\right\}\mbox{ then } \operatorname{Pref}(L)=\left\{\varepsilon, a, ab, abc\right\}$ A language is called prefix closed if $\operatorname{Pref} (L) = L$. The prefix closure operator is idempotent: $\operatorname{Pref} (\operatorname{Pref} (L)) =\operatorname{Pref} (L)$ The prefix relation is a binary relation $\sqsubseteq$ such that $s\sqsubseteq t$ if and only if $s \in \operatorname{Pref}_L(t)$. This relation is a particular example of a prefix order. ## Notes 1. ^ Although every regular language is also context-free, the previous theorem is not implied by the current one, since the former yields a shaper result for regular languages. 2. ^ Strictly formally, a homomorphism yields a language consisting of just one string, i.e. f(a) = {s}. 3. ^ This follows from the above-mentioned closure under arbitrary substitutions. ## References • Hopcroft, John E.; Ullman, Jeffrey D. (1979). Introduction to Automata Theory, Languages and Computation. Reading, Massachusetts: Addison-Wesley Publishing. ISBN 0-201-02988-X. Zbl 0426.68001. (See chapter 3.) 1. ^ Hopcroft, Ullman (1979), Sect.3.2, p.60 2. ^ Hopcroft, Ullman (1979), Sect.3.2, Theorem 3.4, p.60 3. ^ Hopcroft, Ullman (1979), Sect.6.2, Theorem 6.2, p.131 4. ^ Hopcroft, Ullman (1979), Sect.3.2, p.60-61 5. ^ Hopcroft, Ullman (1979), Sect.3.2, Theorem 3.5, p.61 6. ^ Hopcroft, Ullman (1979), Sect.6.2, Theorem 6.3, p.132	2015-04-27 22:19:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 60, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8487340807914734, "perplexity": 1360.1181261517972}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246659483.51/warc/CC-MAIN-20150417045739-00263-ip-10-235-10-82.ec2.internal.warc.gz"}
https://stats.stackexchange.com/questions/430954/example-of-a-non-measurable-maximum-likelihood-estimator	# Example of a non-measurable maximum likelihood estimator If you have a measurable parameter space $$(\Theta, \mathcal{F})$$ and a parametric family of probability measures $$(P_\theta)_{\theta \in \Theta}$$ on a measurable space $$(\mathcal{X}, \mathcal{B})$$ that have densities $$(f_\theta)_{\theta \in \Theta}$$ with respect to some dominating measure $$\mu$$, then you might define the maximum likelihood estimator $$\hat{\theta} : \mathcal{X} \to \Theta$$ as $$\tag{1}\label{1} \hat{\theta}(x) = \operatorname*{arg\,max}_{\theta \in \Theta} f_\theta(x)$$ for all $$x \in \mathcal{X}$$. This is an abstract version of how MLEs are introduced in some statistics texts and classes, but there are a few issues. For one thing, a global maximum might not be attained, or there might be multiple global maxima, so \eqref{1} isn't necessarily well-defined. Moreover, if it happens to be well-defined, then is it measurable? Certainly all examples of maximum likelihood estimators one sees in applied statistics courses (at least that I've seen) are measurable. Of course you can get around these issues by defining MLEs in a different way: for example, you might call a function $$\hat{\theta} : \mathcal{X} \to \Theta$$ a maximum likelihood estimator if it is a priori measurable and there exists a set $$N \in \mathcal{B}$$ such that $$\mu(N) = 0$$ and $$\tag{2}\label{2} f_{\hat{\theta}(x)}(x) = \sup_{\theta \in \Theta} f_\theta(x)$$ for all $$x \in \mathcal{X} \setminus N$$. However, oftentimes \eqref{2} isn't how maximum likelihood estimation is introduced, and I'm more interested in the question of measurability of functions $$\hat{\theta}$$ satisfying \eqref{1} or \eqref{2}. Questions. Are there any explicit examples in which some reasonable interpretation of \eqref{1} is well-defined (at least off a $$\mu$$-null set) but not measurable? Alternatively, can a function $$\hat{\theta}$$ satisfying \eqref{2} (again, at least off a $$\mu$$-null set) exist when no such measurable function exists? What about if $$(\Theta, \mathcal{F})$$ and $$(\mathcal{X}, \mathcal{B})$$ are standard Borel spaces? If there isn't a constructive example, then is there a proof of the existence of an example? There are various results that give criteria under which there cannot be examples of the kind I'm looking for, thereby restricting the class of spaces and densities that might possess such examples. Example 1. Lemma 2 in Jennrich (1969) implies that if • $$\Theta$$ is a compact subset of a Euclidean space and $$\mathcal{F}$$ is its Borel $$\sigma$$-algebra, and • the function $$\theta \mapsto f_\theta(x)$$ from $$\Theta$$ into $$\mathbb{R}$$ is continuous for every $$x \in \mathcal{X}$$, then there exists a measurable function $$\hat{\theta} : \mathcal{X} \to \Theta$$ such that \eqref{2} holds for all $$x \in \mathcal{X}$$. Example 2. Theorem 3.1 in Wagner (1980) (the Kuratowski and Ryll-Nardzewski selection theorem) implies that if • $$(\Theta, \mathcal{F})$$ is a Polish space with its Borel $$\sigma$$-algebra, • $$(\mathcal{X}, \mathcal{B})$$ is any measurable space, • for each $$x \in \mathcal{X}$$, the set $$F(x) = \big\{\theta \in \Theta : \text{f_\theta(x) \geq f_\psi(x) for all \psi\in\Theta}\big\}$$ is nonempty and closed, and • for every open $$U \subseteq \Theta$$, the set $$\big\{x \in \mathcal{X} : F(x) \cap U \neq \emptyset\big\}$$ belongs to $$\mathcal{B}$$, then there exists a measurable function $$\hat{\theta} : \mathcal{X} \to \Theta$$ such that $$\hat{\theta}(x) \in F(x)$$ for all $$x \in \mathcal{X}$$ (i.e., \eqref{2} holds for all $$x \in \mathcal{X}$$). Thus, under the additional assumptions of Example 1 or Example 2 there always exists a measurable maximum likelihood estimator. Example 3. Theorem 7.50(b) in Bertsekas and Shreve (1996) implies that if • $$(\Theta, \mathcal{F})$$ and $$(\mathcal{X}, \mathcal{B})$$ are standard Borel spaces, and • for every $$c \in \mathbb{R}$$, the set $$\big\{ (x, \theta) \in \mathcal{X} \times \Theta : f_\theta(x) > c \big\}$$ is an analytic subset of $$\mathcal{X} \times \Theta$$, then the set $$I = \left\{ x \in \mathcal{X} : \text{f_\theta(x) = \sup_{\psi\in\Theta} f_\psi(x) for some \theta \in \Theta}\right\}$$ is universally measurable. Moreover, for each $$\varepsilon > 0$$ there exists a universally measurable function $$\hat{\theta} : \mathcal{X} \to \Theta$$ such that \eqref{2} holds for all $$x \in I$$, and for all $$x \in \mathcal{X} \setminus I$$, $$f_{\hat{\theta}(x)}(x) \geq \begin{cases} \sup_{\psi \in \Theta} f_\psi(x) - \varepsilon, & \text{if \sup_{\psi \in \Theta} f_\psi(x) < \infty,} \\ \varepsilon^{-1}, & \text{if \sup_{\psi \in \Theta} f_\psi(x) = \infty.} \end{cases}$$ Example 3 doesn't necessarily guarantee the existence of a measurable maximum likelihood estimator (for instance, since there are universally measurable sets which are not Borel), but it is arguably in the same spirit as Example 2. • Since a measurable function is defined between two measurable spaces, the parameter space $\Theta$ would need to be endowed with a natural measure for the question to be well-defined. – Xi'an Oct 11 '19 at 6:52 • @Xi'an there is no measure involved in the definition of a measurable map (or a measurable space for that matter). In the measure-theoretic development of statistics, estimators need to be measurable functions. Otherwise we couldn't talk about distributions of estimators. – Artem Mavrin Oct 11 '19 at 7:01 • Any text that proposes $\Theta$ is a measure space is going to present a Bayesian theory rather than MLEs. (Classical texts do not make any such assumption because they do not always need to model the parameter as a random variable.) In this sense your characterization of the situation appears not to reflect anything one would find in any account of MLE. – whuber Oct 11 '19 at 15:04 • @whuber I’m not assuming $\Theta$ is a measure space, merely that it’s endowed with a $\sigma$-algebra. In most situations in practice $\Theta$ is a Borel subset of a Euclidean space, so it comes with a natural $\sigma$-algebra that doesn’t need to be mentioned. If the parameter space isn’t endowed with a $\sigma$-algebra then we can’t talk about parameter-valued estimators – Artem Mavrin Oct 11 '19 at 15:09 • My sense is it's possible, but imposing "regularity conditions" on the parameterization might help ensure measurability. That means the parameterization has to be continuous relative to some reasonable topology on the space of probability distributions. – whuber Oct 11 '19 at 17:41 Here is a contrived example. Let $$(\mathcal{X}, \mathcal{B})$$ be the interval $$[0, 2]$$ with its Borel $$\sigma$$-algebra, let $$(\Theta, \mathcal{F})$$ be the interval $$[1, 2]$$ with its Borel $$\sigma$$-algebra, and let $$P_\theta$$ be the uniform distribution on $$[0, \theta]$$ for each $$\theta \in \Theta$$. The family $$(P_\theta)_{\theta \in \Theta}$$ is dominated by the restriction of Lebesgue measure to $$\mathcal{X}$$ (call it $$\mu$$ as in the question), and a sensible choice of density for $$P_\theta$$ with respect to $$\mu$$ is $$f_\theta = \theta^{-1} \mathbf{1}_{[0, \theta]}.$$ The maximum likelihood estimator $$\hat{\theta} : \mathcal{X} \to \Theta$$ is then given by $$\hat{\theta}(x) = \max\{1, x\},$$ which is certainly measurable. However, suppose I'm not sensible, and that I have a favorite Vitali set $$V \subseteq [0, 1]$$, and I define a family $$(g_\theta)_{\theta \in \Theta}$$ of functions $$\mathcal{X} \to \mathbb{R}$$ as follows: • if $$\theta - 1\in V$$, then $$g_\theta = 3 \mathbf{1}_{\{\theta - 1\}} + \theta^{-1} \mathbf{1}_{[0, \theta] \setminus \{\theta - 1\}}$$; • if $$\theta - 1\notin V$$, then $$g_\theta = f_\theta$$. In other words, $$g_\theta(x) = \begin{cases} 3, & \text{if x \in V and x = \theta - 1,} \\ f_\theta(x), & \text{otherwise} \end{cases}$$ for all $$x \in \mathcal{X}$$ and $$\theta \in \Theta$$. Since each $$g_\theta$$ is a modification of $$f_\theta$$ on at most a singleton, the family $$(g_\theta)_{\theta \in \Theta}$$ is another family of densities for $$(P_\theta)_{\theta \in \Theta}$$ with respect to $$\mu$$. Now one can check that there is a unique function $$\hat{\vartheta} : \mathcal{X} \to \Theta$$ such that $$g_{\hat{\vartheta}(x)}(x) = \sup_{\theta \in \Theta} g_\theta(x)$$ for each $$x \in \mathcal{X}$$; this function is given by $$\hat{\vartheta}(x) = \begin{cases} \max\{1, x\}, & \text{if x \notin V,} \\ x + 1, & \text{if x \in V,} \end{cases}$$ and it is not measurable (for example, since $$\{x \in \mathcal{X} : \hat{\vartheta}(x) = x + 1\} = \{0\} \cup V$$). Thus, in the setup with the family of densities $$(g_\theta)_{\theta \in \Theta}$$, there exists a function $$\hat{\vartheta} : \mathcal{X} \to \Theta$$ such that $$g_{\hat{\vartheta}(x)}(x) = \sup_{\theta \in \Theta} g_\theta(x)$$ for all $$x \in \mathcal{X}$$, but no such measurable function exists. This example merely shows that the the question of whether a measurable maximum likelihood estimator exists is sensitive to the choice of version of the Radon-Nikodym derivative $$dP_\theta / d\mu$$. When we used the much more reasonable family of densities $$(f_\theta)_{\theta \in \Theta}$$ we had no trouble obtaining a measurable maximum likelihood estimator, but when we artificially perturbed each of these densities on a null set to obtain $$(g_\theta)_{\theta \in \Theta}$$, we ran into problems. This example is inspired by Example 2.2 in Pfanzagl (1969), wherein a similar but much more subtle modification of densities yields a scenario in which no maximum likelihood estimator exists at all, measurable or not.	2020-12-02 00:37:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 95, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9113298654556274, "perplexity": 151.97525693816192}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141685797.79/warc/CC-MAIN-20201201231155-20201202021155-00403.warc.gz"}
http://mathhelpforum.com/algebra/15309-quadratic-formula-print.html	• May 23rd 2007, 03:27 PM shortd333 okay, i have this question and i have to find the roots/x-intercepts/zeros w/e of -4xsquared+5x+7 and i can't figure it out because the answer using the quadratic formula is different then the ones given (its multiple choice) can anyone help me? ... i dont know if this makes any sense though ... • May 23rd 2007, 03:39 PM Jhevon Quote: Originally Posted by shortd333 okay, i have this question and i have to find the roots/x-intercepts/zeros w/e of -4xsquared+5x+7 and i can't figure it out because the answer using the quadratic formula is different then the ones given (its multiple choice) can anyone help me? ... i dont know if this makes any sense though ... Let's try the quadratic formula one more time: $-4x^2 + 5x + 7 = 0$ $\Rightarrow x = \frac {-5 \pm \sqrt {5^2 - 4(-4)(7)}}{2(-4)}$ $\Rightarrow x = \frac {-5 \pm \sqrt {137}}{-8}$ $\Rightarrow x = \frac {5}{8} \pm \frac { \sqrt {137}}{8}$	2014-07-29 20:17:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8038668632507324, "perplexity": 537.2116993905672}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510267824.47/warc/CC-MAIN-20140728011747-00277-ip-10-146-231-18.ec2.internal.warc.gz"}
https://itectec.com/electrical/electrical-should-startup-torque-be-higher-in-value-than-full-load-torque-in-an-induction-motor/	# Electrical – Should startup torque be higher in value than full load torque in an induction motor electricalengineeringinduction motor I found out values of my 3 phase delta connected induction motor for full load torque, maximum torque and startup torque. My maximum torque was higher than my full load torque, but the thing I'm wondering about is whether I'm right or not is that my startup torque is higher than my full load torque. Is that right, or is it supposed to be lower than full load? If it's supposed to be lower than it shows I need to look over my calculations again. Max torque= 322.4Nm Startup torque= 38.53Nm $$T = \frac {9.55 P_m} {n}$$ Where T = Torque, $$\P_m\$$ = mechanical power out and n = speed in rpm.	2022-05-29 09:03:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.448702335357666, "perplexity": 1387.3351711942078}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663048462.97/warc/CC-MAIN-20220529072915-20220529102915-00322.warc.gz"}
https://www.vcalc.com/wiki/DavidC/Fugacity+of+component+i+in+Vapor+phase	# Fugacity of component i in Vapor phase Not Reviewed hatf_i^V = DavidC.Fugacity of component i in Vapor phase The fugacity coefficient hatvarphi_i, for component i in the vapor is calculated from an equation of state (e.g., Virial). Sometimes it is approximated by a pure component value from a correlation. The fugacity coefficient is a correction for vapor phase non-ideality. Often at pressures close to atmospheric, hatvarphi_i= 1	2019-05-20 01:31:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9234142899513245, "perplexity": 6653.461342889303}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232255251.1/warc/CC-MAIN-20190520001706-20190520023706-00200.warc.gz"}
http://katlas.math.toronto.edu/wiki/3_1	# 3 1 ## Contents (KnotPlot image) See the full Rolfsen Knot Table. Visit 3 1's page at the Knot Server (KnotPlot driven, includes 3D interactive images!) Visit 3 1's page at the original Knot Atlas! 3_1 is also known as "The Trefoil Knot", after plants of the genus Trifolium, which have compound trifoliate leaves, and as the "Overhand Knot". See also T(3,2). The trefoil is perhaps the easiest knot to find in "nature", and is topologically equivalent to the interlaced form of the common Christian and pagan "triquetra" symbol [12]: Logo of Caixa Geral de Depositos, Lisboa [1] A knot consists of two harts in Kolam [2] A basic form of the interlaced Triquetra; as a Christian symbol, it refers to the Trinity 3D depiction ### Knot presentations Planar diagram presentation X1425 X3641 X5263 Gauss code -1, 3, -2, 1, -3, 2 Dowker-Thistlethwaite code 4 6 2 Conway Notation [3] Minimum Braid Representative A Morse Link Presentation An Arc Presentation Length is 3, width is 2, Braid index is 2 [{5, 2}, {1, 3}, {2, 4}, {3, 5}, {4, 1}] Knot 3_1. A graph, knot 3_1. A part of a knot and a part of a graph. ### Three dimensional invariants Symmetry type Reversible Unknotting number 1 3-genus 1 Bridge index 2 Super bridge index 3 Nakanishi index 1 Maximal Thurston-Bennequin number [-6][1] Hyperbolic Volume Not hyperbolic A-Polynomial See Data:3 1/A-polynomial [edit Notes for 3 1's three dimensional invariants] The rope length of the trefoil is known to be no more than 16.372, by numerical experiments, while the sharpest known lower bound (actually applicable to all non-trivial knots) is 15.66. The trefoil is a fibered knot! A java applet demonstrating it, written by Robert Barrington Leigh at the University of Toronto, is here. ### Four dimensional invariants Smooth 4 genus 1 Topological 4 genus 1 Concordance genus ConcordanceGenus(Knot(3,1)) Rasmussen s-Invariant -2 ### Polynomial invariants Alexander polynomial t + t−1−1 Conway polynomial z2 + 1 2nd Alexander ideal (db, data sources) {1} Determinant and Signature { 3, -2 } Jones polynomial −q−4 + q−3 + q−1 HOMFLY-PT polynomial (db, data sources) −a4 + a2z2 + 2a2 Kauffman polynomial (db, data sources) a5z + a4z2−a4 + a3z + a2z2−2a2 The A2 invariant −q14−q12 + q8 + 2q6 + q4 + q2 The G2 invariant q72−q64−q62−q56−2q54−q52 + q50−q46−2q44 + 2q40 + q38−q36 + 2q32 + 2q30 + q28 + 2q22 + 2q20 + q14 + q12 + q10 ### "Similar" Knots (within the Atlas) Same Alexander/Conway Polynomial: {} Same Jones Polynomial (up to mirroring, $q\leftrightarrow q^{-1}$): {} ### Vassiliev invariants V2 and V3: (1, -1) V2,1 through V6,9: V2,1 V3,1 V4,1 V4,2 V4,3 V5,1 V5,2 V5,3 V5,4 V6,1 V6,2 V6,3 V6,4 V6,5 V6,6 V6,7 V6,8 V6,9 4 −8 8 $\frac{62}{3}$ $\frac{10}{3}$ −32 $-\frac{176}{3}$ $-\frac{32}{3}$ −8 $\frac{32}{3}$ 32 $\frac{248}{3}$ $\frac{40}{3}$ $\frac{5071}{30}$ $\frac{58}{15}$ $\frac{3062}{45}$ $\frac{17}{18}$ $\frac{271}{30}$ V2,1 through V6,9 were provided by Petr Dunin-Barkowski <[email protected]>, Andrey Smirnov <[email protected]>, and Alexei Sleptsov <[email protected]> and uploaded on October 2010 by User:Drorbn. Note that they are normalized differently than V2 and V3. ### Khovanov Homology The coefficients of the monomials trqj are shown, along with their alternating sums χ (fixed j, alternation over r). The squares with yellow highlighting are those on the "critical diagonals", where j−2r = s + 1 or j−2r = s−1, where s = -2 is the signature of 3 1. Nonzero entries off the critical diagonals (if any exist) are highlighted in red. \ r \ j \ -3-2-10χ -1 11 -3 11 -5 1 1 -7 0 -91 -1 Integral Khovanov Homology $\dim{\mathcal G}_{2r+i}\operatorname{KH}^r_{\mathbb Z}$ i = −3 i = −1 r = −3 ${\mathbb Z}$ r = −2 ${\mathbb Z}_2$ ${\mathbb Z}$ r = −1 r = 0 ${\mathbb Z}$ ${\mathbb Z}$ ### Computer Talk Much of the above data can be recomputed by Mathematica using the package KnotTheory. See A Sample KnotTheory Session, or any of the Computer Talk sections above. Read me first: Modifying Knot Pages See/edit the Rolfsen Knot Page master template (intermediate). See/edit the Rolfsen_Splice_Base (expert). Back to the top.	2013-06-18 22:50:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 19, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.43278467655181885, "perplexity": 9069.41579436721}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368707435344/warc/CC-MAIN-20130516123035-00045-ip-10-60-113-184.ec2.internal.warc.gz"}
https://socratic.org/questions/what-is-the-equation-of-sinusoidal-function-with-the-amplitude-3-and-period-pi-4#604805	What is the equation of sinusoidal function with the amplitude 3 and period pi/4? May 1, 2018 $y = 3 \sin \left(8 x\right)$ Explanation: The parent function is $\sin x$ so this is where we start. $y = \sin x$ The amplitude of this sine wave is 1. To make the amplitude 3, we multiply the sine by 3. $y = 3 \sin x$ The period of this sine wave is 2$\pi$. We want the period to be $2 \frac{\pi}{8}$ so we need to multiply $x$ by 8. $y = 3 \sin \left(8 x\right)$ This satisfies the requirements of the problem statement. graph{3sin(8x) [-1, 1, -5, 5]}	2022-08-20 06:36:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9730324149131775, "perplexity": 533.743468370287}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573908.30/warc/CC-MAIN-20220820043108-20220820073108-00114.warc.gz"}
http://www.ams.org/mathscinet-getitem?mr=306566	MathSciNet bibliographic data MR306566 33A30 Wong, R.; Rosenbloom, E. Series expansions of \$W\sb{k,\,m}(Z)\$$W\sb{k,\,m}(Z)$ involving parabolic cylinder functions. Math. Comp. 25 (1971), 783–787. Article For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.	2016-08-25 14:33:56	{"extraction_info": {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9994025230407715, "perplexity": 9671.080431538216}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982293468.35/warc/CC-MAIN-20160823195813-00121-ip-10-153-172-175.ec2.internal.warc.gz"}
https://topanswers.xyz/tex?q=1264	Anonymous 1123 I tried to draw a cylinder inscribed a cone like this \documentclass[border=1mm,tikz]{standalone} \usepackage{tikz-3dplot} \begin{document} %polar coordinates of visibility \pgfmathsetmacro\th{50} \pgfmathsetmacro\az{120} \tdplotsetmaincoords{\th}{\az} %parameters of the cone \pgfmathsetmacro\v{2\R} %hight of cone \pgfmathsetmacro\h{\v/3} %hight of cylinder \begin{tikzpicture} [tdplot_main_coords,line join = round, line cap = round] \coordinate (O) at (0,0,0) ; %\coordinate (A) at ({\Rcos(\angA)}, {\Rsin(\angA)},0); %\coordinate (B) at ({\Rcos(\angB)}, {\Rsin(\angB)},0); \coordinate (S) at (0,0,\v); %\draw[thick] (S) -- (A) (S) -- (B); \draw[dashed] (S)--(O) ; \begin{scope}[canvas is xy plane at z=0] \draw[dashed] (\tdplotmainphi:\myr) arc(\tdplotmainphi:\tdplotmainphi+180:\myr); \draw[dashed] (\tdplotmainphi:\myr) coordinate(BR) arc(\tdplotmainphi:\tdplotmainphi-180:\myr) coordinate(BL); \coordinate (O) at (0,0); \draw[dashed] (BL) -- (BR); \end{scope} % \begin{scope}[canvas is xy plane at z=\h] \coordinate (O') at (0,0); \draw [dashed](BR) -- (\tdplotmainphi:\myr) (BL) -- (\tdplotmainphi-180:\myr); \draw[dashed] (\tdplotmainphi:\myr) arc(\tdplotmainphi:\tdplotmainphi+180:\myr); \draw[thick] (\tdplotmainphi:\myr) coordinate(BR) arc(\tdplotmainphi:\tdplotmainphi-180:\myr) coordinate(BL); \draw[dashed] (\tdplotmainphi:\myr) -- (\tdplotmainphi-180:\myr); \end{scope} \pgfmathsetmacro\cott{{cot(\th)}} \pgfmathsetmacro\fraction{\R\cott/\v} \pgfmathsetmacro\fraction{\fraction<1 ? \fraction : 1} \pgfmathsetmacro\angle{{acos(\fraction)}} % % angles for transformed lines \pgfmathsetmacro\PhiOne{180+(\az-90)+\angle} \pgfmathsetmacro\PhiTwo{180+(\az-90)-\angle} % % coordinates for transformed surface lines \pgfmathsetmacro\sinPhiOne{{sin(\PhiOne)}} \pgfmathsetmacro\cosPhiOne{{cos(\PhiOne)}} \pgfmathsetmacro\sinPhiTwo{{sin(\PhiTwo)}} \pgfmathsetmacro\cosPhiTwo{{cos(\PhiTwo)}} % % angles for original surface lines \pgfmathsetmacro\sinazp{{sin(\az-90)}} \pgfmathsetmacro\cosazp{{cos(\az-90)}} \pgfmathsetmacro\sinazm{{sin(90-\az)}} \pgfmathsetmacro\cosazm{{cos(90-\az)}} % % draw basis circle \tdplotdrawarc[tdplot_main_coords,thick]{(O)}{\R}{\PhiOne}{360+\PhiTwo}{anchor=north}{} \tdplotdrawarc[tdplot_main_coords,dashed]{(O)}{\R}{\PhiTwo}{\PhiOne}{anchor=north}{} % % displaying tranformed surface of the cone (rotated) \draw[thick] (0,0,\v) -- (\R\cosPhiOne,\R\sinPhiOne,0); \draw[thick] (0,0,\v) -- (\R\cosPhiTwo,\R\sinPhiTwo,0); \end{tikzpicture} \end{document} ![ScreenHunter 816.png](/image?hash=0c22c8966bf1b249f77084849e60d473c3c43ba0a1113c9ca1f58e7585c11c42) The dashed line of cylinder incorrect. How can I repair it? user 3.14159 The main message is that you basically answered the question yourself by computing the critical angles of visibility for the larger cone. The same angles can be used for the smaller cone on the top. The rest is just a series of minor remarks: 1. Instead of using \th and \az you can just use \tdplotmaintheta and \tdplotmainphi consistently (you are using the latter anyway). That's perhaps a bit less confusing. 2. I personally like to use declare function instead of a series of macros. Using macros is fine, of course, but if one is to use them, perhaps it is better to define them locally inside the tikzpicture so that you can copy the whole thing and embed it in another document without having to worry about overwriting some global macros. 3. You can just use the min and max functions instead of the ifthenelse stuff. 4. You can use the arcs from tikz-3dplot or "ordinary" arcs in planes from the 3d library. However, I would not necessarily want to mix them as both of them have the same purpose. \documentclass[border=1mm,tikz]{standalone} \usepackage{tikz-3dplot} \begin{document} \tdplotsetmaincoords{50}{120} \begin{tikzpicture}[tdplot_main_coords,line join = round, line cap = round, hidden/.style={thin,dashed},visible/.style={thick,solid}, declare function={R=3;v=2R;% parameters of the cone h=v/3;r=(v-h)R/v;% parameters of the cylinder f=max(min(Rcot(\tdplotmaintheta)/v,1),-1);% auxiliary phicrit=acos(f);% crititcal angles phi1=90+\tdplotmainphi+phicrit; phi2=90+\tdplotmainphi-phicrit;}] \draw[hidden] (0,0,0) coordinate (O) -- (0,0,v) coordinate (S); \begin{scope}[canvas is xy plane at z=0] \draw[hidden] (\tdplotmainphi:r) \draw[hidden] (\tdplotmainphi:r) coordinate(BR) coordinate(BL) -- cycle; \draw[hidden] (phi1:R) coordinate (CL) coordinate (CR); \draw[visible] (phi1:R) -- (S) -- cycle; \end{scope} % \begin{scope}[canvas is xy plane at z=h] \draw [hidden](BR) -- (\tdplotmainphi:r) (BL) -- (\tdplotmainphi-180:r); \draw[hidden] (phi1:r) \draw[visible] (phi1:r) \end{scope} \end{tikzpicture} \end{document} ![Screen Shot 2020-08-23 at 7.58.19 PM.png](/image?hash=d06baf50645eb394ccd7590f1cb5b85b2819c7744d0a432a36694a140d5bddff) Given the height of the cylinder, one can compute its radius, or the other way around. \documentclass[border=1mm,tikz]{standalone} \usepackage{tikz-3dplot} \begin{document} \tdplotsetmaincoords{60}{120} \begin{tikzpicture}[tdplot_main_coords,line join = round, line cap = round, hidden/.style={thin,dashed},visible/.style={thick,solid}, declare function={R=2;v=R;% base radius and height of the cone h=(R-r)v/R;% f=max(min(Rcot(\tdplotmaintheta)/v,1),-1);% auxiliary phicrit=acos(f);% crititcal angles phi1=90+\tdplotmainphi+phicrit; phi2=90+\tdplotmainphi-phicrit;}] \draw[hidden] (0,0,0) coordinate (O) -- (0,0,v) coordinate (S); \begin{scope}[canvas is xy plane at z=0] \draw[hidden] (\tdplotmainphi:r) \draw[hidden] (\tdplotmainphi:r) coordinate(BR) coordinate(BL) -- cycle; \draw[hidden] (phi1:R) coordinate (CL) coordinate (CR); \draw[visible] (phi1:R) -- (S) -- cycle; \end{scope} % \begin{scope}[canvas is xy plane at z=h] \draw [hidden](BR) -- (\tdplotmainphi:r) (BL) -- (\tdplotmainphi-180:r); \draw[hidden] (phi1:r) \draw[visible] (phi1:r) \end{scope} \end{tikzpicture} \end{document} ![Screen Shot 2020-08-23 at 9.04.38 PM.png](/image?hash=7c51212ccbc61773e041e9a4735212fa2026214d154b6965e4139c52a987ca79) With some more recent additions to the [3dtools library](https://github.com/marmotghost/tikz-3dtools) the code can be condensed to \documentclass[border=1mm,tikz]{standalone} \usetikzlibrary{calc,3dtools}% https://github.com/marmotghost/tikz-3dtools \begin{document} \begin{tikzpicture}[3d/install view={phi=110,theta=70},line join = round, line cap = round, declare function={R=2;v=R;% base radius and height of the cone h=(R-r)v/R;% height of the base of the upper circle }]	2021-10-18 16:34:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8460593819618225, "perplexity": 11869.613863385246}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585204.68/warc/CC-MAIN-20211018155442-20211018185442-00407.warc.gz"}
https://www.latten.net/sling2/2018/07/07/skeleton-left-wing-ready/	Hrs: 6 All ribs where ready and I could rivet the ribs of the left wing. Some rivets need manual riveting with this special rivet gun: I worked my way from the outside rib 10 to the inside rib 1. Last I inserted the flap torque tube and rib 12. Rib 10: I did not rivet the leading edge rib yet to make it easier to rotate the wing. Rib 9: I did not rivet the leading edge rib yet to make it easier to rotate the wing. The step bend angle was riveted and I added a aluminium washer under the tail of the last rivet. Rib 8: I did not rivet the leading edge rib yet to make it easier to rotate the wing. The step bend angle was riveted and I added a aluminium washer under the tail of the last rivet. Note the 2 large flange 4 mm rivets because a hole in this rib was enlarged. Rib 7: I did not rivet the leading edge rib yet to make it easier to rotate the wing. Rib 6: This is the new version with bushing. The step bend angle was riveted and I added a aluminium washer under the tail of the last rivet. I didn’t enlarge the holes for the grommets, I left them 9.5 mm and added a grommet that has an inside diameter of 6 mm for the pitot tubes. Rib 5: This rib was not aligned with the main spar, the top needed a bend down and the bottom needed also a bend down, I modified the rib with a wooden mold and careful hammered the rib to the correct shape. I didn’t enlarge the holes for the grommets, I left them 9.5 mm and added a grommet that has an inside diameter of 6 mm for the pitot tubes. Rib 12: I didn’t enlarge the holes for the grommets, I left them 9.5 mm and added a grommet that has an inside diameter of 6 mm for the pitot tubes. Rib 4: I did not rivet the step ribs yet to make it easier to rotate the wing. I didn’t enlarge the holes for the grommets, I left them 9.5 mm and added a grommet that has an inside diameter of 6 mm for the pitot tubes. The step bend angle was riveted and I added a aluminium washer under the tail of the last rivet. The red cable tie is a reminder to find a solution for the holes that not align, either I will add a aluminium washer under the tail or enlarge the hole for 4.8 mm rivet. Rib 3: I did not rivet the step ribs yet to make it easier to rotate the wing. I didn’t enlarge the 2 holes for the pitot tubes, I left them 9.5 mm and added a grommet that has an inside diameter of 6 mm for the pitot tubes. The third hole was enlarged to 12 mm for the #5 grommets of TAF for the electrical wires of the lights and pitot heat. Rib 2: I did not rivet the step ribs yet to make it easier to rotate the wing. I didn’t enlarge the 2 holes for the pitot tubes, I left them 9.5 mm and added a grommet that has an inside diameter of 6 mm for the pitot tubes. The third hole was enlarged to 12 mm for the #5 grommets of TAF for the electrical wires of the lights and pitot heat. Rib 1: I did not rivet the step ribs yet to make it easier to rotate the wing. I didn’t enlarge the 2 holes for the pitot tubes, I left them 9.5 mm and added a grommet that has an inside diameter of 6 mm for the pitot tubes. The third hole was enlarged to 12 mm for the #5 grommets of TAF for the electrical wires of the lights and pitot heat. Todo: rivet and screw (and torque) to the main spar. Next I riveted the rear spar: Overview where the fuel tank will be mounted: Overview where the step ribs will be mounted: Total overview:	2022-01-25 19:40:12	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8160563111305237, "perplexity": 3259.2916119153524}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304872.21/warc/CC-MAIN-20220125190255-20220125220255-00665.warc.gz"}
https://www.adafruit.com/product/1655	# NeoPixel 5050 RGB LED with Integrated Driver Chip - 10 Pack PRODUCT ID: 1655 $4.50 QTY DISCOUNT 1-9$4.50 10-99 $4.05 100+$3.60 # Description- Make your own smart LED arrangement with the same integrated LED that is used in our NeoPixel strip and pixels. This tiny 5050 (5mm x 5mm) RGB LED is fairly easy to solder and is the most compact way possible to integrate multiple bright LEDs to a design. The driver chip is inside the LED and has ~18mA constant current drive so the color will be very consistent even if the voltage varies, and no external choke resistors are required making your design minimal. Power the whole thing with 5VDC and you're ready to rock. This is the 4 pin LED chip version, not 6. It is code compatible and the same over-all shape and functionality but not the same pinout so you cannot use these to replace an 'S chip. If you are designing a new PCB we suggest going with the B, since it has built in polarity protection. Other than that, B and S are the same brightness, and use the exact same code interface. The LEDs are 'chainable' by connecting the output of one chip into the input of another - see the datasheet for diagrams and pinouts. To allow the entire chip to be integrated into a 6-pin package, there is a single data line with a very timing-specific protocol. Since the protocol is very sensitive to timing, it requires a real-time microconroller such as an AVR, Arduino, PIC, mbed, etc. It cannot be used with a Linux-based microcomputer or interpreted microcontroller such as the netduino or Basic Stamp. The LEDs basically have a WS2811 inside, but fixed at the 800KHz 'high speed' setting. Our wonderfully-written Neopixel library for Arduino supports these pixels! As it requires hand-tuned assembly it is only for AVR cores but others may have ported this chip driver code so please google around. An 8MHz or faster processor is required. These raw LEDs are cut from a reel and/or might be loose. They may not suitable for pick & place + reflow. We recommend these for careful hand soldering only! Comes in a package with 10 individual LEDs. We have a ready-to-go component for this in the Adafruit EAGLE library 5050 RGB LED with Integrated Driver Chip - 10 Pack (17:12)	2017-06-28 15:51:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.32043197751045227, "perplexity": 3546.499428196802}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128323711.85/warc/CC-MAIN-20170628153051-20170628173051-00260.warc.gz"}
https://byjus.com/question-answer/two-capacitors-c-1-and-c-2-are-charged-to-120-v-and-200-v/	Question # Two capacitors $$C_{1}$$ and $$C_{2}$$ are charged to 120 $$V$$ and 200 $$V$$ respectively. It is found that by connecting them together the potential on each one can be made zero. Then, A 3C1=5C2 B 3C1+5C2=0 C 9C1=4C2 D 5C1=3C2 Solution ## The correct option is A $$3C_{1}=5C_{2}$$Potential can be made zero only if they have same charge. $$\implies 120 C_{1}=200C_{2}$$$$\therefore 6C_{1}=10C_{2}$$$$3C_{1}=5C_{2}$$PhysicsNCERTStandard XII Suggest Corrections 0 Similar questions View More Same exercise questions View More People also searched for View More	2022-01-26 20:32:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4606339633464813, "perplexity": 1635.7402844426404}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304961.89/warc/CC-MAIN-20220126192506-20220126222506-00020.warc.gz"}
https://mathoverflow.net/questions/186483/embedding-proper-algebraic-spaces/186490	# Embedding proper algebraic spaces Does every proper algebraic space (over a field, say) admit a closed immersion into a smooth proper algebraic space? Remark: Of course, if we say "projective" instead of "proper" then the answer is tautologically "yes": we can take the ambient variety to be some projective n-space. But I'm curious about the general case. • No : there are proper normal surfaces $S$ with trivial Picard group (see for instance [Stefan Schröer, On non-projective normal surfaces]). If you could embed $S$ in a smooth and proper $X$, you could choose a Weil (hence Cartier) effective divisor in $X$ intersecting $S$ but not containing it, showing that the Picard group of $S$ is not trivial. Nov 7 '14 at 15:04 • Are you sure that this exclude the possibility of embedding $S$ into a smooth algebraic space? It seems to me that this argument only works if $X$ is a scheme, or I am missing something? Nov 7 '14 at 15:12 • @FrancescoPolizzi Ah you are absolutely right ! Sorry ! Nov 7 '14 at 15:24	2021-11-30 00:43:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9446428418159485, "perplexity": 290.8215298158095}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358847.80/warc/CC-MAIN-20211129225145-20211130015145-00403.warc.gz"}
http://mathhelpboards.com/chemistry-65/chemistry-19751.html?s=9278ceb6188b675b9c2ddddfeb0e6f5e	1. I am stuck on this chemistry problem using Mastering Chemistry to do my homework. This is the problem. When I enter the answers it keeps saying is that "One or more of the units you entered is not recognized." For part A my answer was 426 g P2O5 and for part B 369 g P2O5 Problem Consider a situation in which 186 g of P4 are exposed to 208 g of O2. P4+5O2→2P2O5 PART A) What is the maximum amount in moles of P2O5 that can theoretically be made from 186 g of P4 and excess oxygen? PART B) What is the maximum amount in moles of P2O5 that can theoretically be made from 208 g of O2 and excess phosphorus? 2. Hi Teh! The problem is asking for the number of moles. So the answers shouldn't be in grams. Moreover, the unit should be left out. Note that to find moles from grams we have to divide by the atomic mass in atomic mass units. So for instance 32 g of $O_2$ is 1 mole of $O_2$. For part $(A)$ we have 186 g phosporus and $P$ has an atomic mass of 31 amu, or 31 g/mol. That makes: $$\frac{186}{31} = 6.0 \text{ mol of P atoms}$$ The reaction is: $$P_4 + 5O_2 \to 2P_2O_5$$ The resulting product has $P_2$ in it, meaning 2 $P$ atoms per molecule, so there will be: $$\frac {6.0} 2 = 3.0 \text{ mol of P_2O_5 molecules}$$ So the requested number of moles is $3.0$.	2018-01-19 09:28:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5975268483161926, "perplexity": 746.1500371281345}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084887849.3/warc/CC-MAIN-20180119085553-20180119105553-00639.warc.gz"}
https://quant.stackexchange.com/questions/25004/cross-sectional-regression-using-calculated-coefficient-of-first-regression-for	# Cross-sectional Regression: Using calculated coefficient of first regression for a second regression as dependent variable Hello stackexchange community! I am new to R and econometrics and and stuck in a step of the fama-macbeth (1973) regression, in which risk premia of stocks are estimated with a two-step regression procedure. The first regression looks like the following: Rit = αi + βiRmt + εit ......(1) where i = 1,...,20 stocks and Rm is the return of the market index (often referred to as factor in regressions) The β in here, of which I have 20 (one for each stock), is the coefficient I want to use in the second regression, which is a cross-sectional regression: Rit = γt βˆi + αit .............(2) where again i = 1,...,20 stocks. I am supposed to get one γt (risk premium) per day(4500) and 20 αit (pricing errors) per day. Now to my issue: The first regression was no problem. I have one .xts object (stocks.returns) with 20 stocks in the columns and 4500 rows, which are daily returns of each stock. Then I have another .xts object (index.returns) with the same attributes, but just one column which is the market index. With this data I run the following codes in order to get the betas: all.betas <- coef(lm(stocks.returns ~ index.returns)) all.betas <- all.betas[2,1:20] now I have a numeric vector with all 20 betas (removed the intercept) per stock and I want to run the second (2) regression and am trying the code: summary(lm(stocks.returns ~ all.betas)) However, R keeps giving me this error: Error in model.frame.default(formula = stocks.returns ~ all.betas, drop.unused.levels = TRUE) : variable lengths differ (found for 'all.betas') I need to run a regression with all 20 stock returns against all 20 betas on each day (4500). I do not know how to structure this regression model/code. I have been trying and searching for days now. Please help me. Any hints, errors I made, formulas I could try, literature, links, keywords etc. are highly appreciated!	2019-11-17 05:54:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5518140196800232, "perplexity": 3158.7467576391437}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668787.19/warc/CC-MAIN-20191117041351-20191117065351-00088.warc.gz"}
http://mquandt.com/2016/03/07/Compiling-Vulkan-from-Scratch.html/	# Introduction & Motivation Works for SDK version 1.0.3 A few weeks ago Vulkan 1.0 was released to the public, which marked the first time those outside of Khronos saw the API and SDK (provided by LunarG). If you have a relatively recent Ubuntu (or debian) OS install, or Windows, you can just download the SDK from the Official Website. At Animal Logic we use an older version of CentOS that aligns with the VFX reference platform rather than the latest and greatest. This however means that our version of GLIBC is older than what you may find in more recent Ubuntu installs, which LunarG targetted for the Linux SDK. Thankfully the open nature of Vulkan and Khronos means the source code we care about, for the loader and tools, is provided on Github. This means that we can compile the files ourselves. First lets take a quick look at some design decisions, which may inform what you want to compile and use. To allow for extension in future, Vulkan adopts the extension model that OpenGL has used successfully over its many years. IHVs can provide implementations of extensions that are experimental or unique to their hardware, and ISVs can detect if these exist and use the new functionality without requiring a new version of OpenGL (or in this case Vulkan). Vulkan has gone a step further by defining a layer system that allows the developer to add layers between their application and the underlying driver implementation. These layers enable powerful new development and debugging options without requiring intrusive changes to the application or driver code. These two elements are connected and powered by the Vulkan loader (libvulkan.so on Linux and Vulkan-1.dll on Windows). This loader provides the base API spec as well as the function pointer discovery required for extensions. This is the main thing we need to build to use Vulkan on a system with a different version of GLIBC, and this is the only thing we need to run Vulkan with its base specifications. If we want to enable the use of GLSL, or the debugging layers, then we will need to compile some other tools which will fill out what the SDK normally provides. # Requirements • git • python 3.x • cmake • clang 3.4+ (gcc may work however I have not tried this) # Steps ## Get the Source First clone the following repositories from github: https://github.com/KhronosGroup/Vulkan-LoaderAndValidationLayers.git https://github.com/LunarG/VulkanTools.git At the time of writing the latest SDK is 1.0.3. You will probably want to checkout the branch containing the latest SDK rather than master, so you have a stable release rather than something in development. ## Update & Build Dependencies Inside Vulkan-LoaderAndValidationLayers, run ./update_external_sources.sh This will download the repositories the loader and validation layers require (SPIR-V etc) and compile them, so ensure that clang, cmake and python are in your path. Once this is done, move to the VulkanTools repository and do the same. ## CMAKE Configuration & Compilation This article focuses on GLIBC 2.12, however it should apply to most versions, and this is probably the section that will differ the most, so please test and remember that your mileage may vary. In 2.12 the string format macros are not globally defined, as the SDK expects. If you see any errors that require macros such as PRIx64, just add the following lines to the CMakeLists.txt file in both of the repositories you cloned: add_definitions(-D__STDC_FORMAT_MACROS) This line will enable those macros. You may also want to disable compilation of vkjson_info and the demos as some of them require the alignedmalloc function that was addded to a more recentl glibc (but which is missing from 2.12). To do this, just add -DBUILD_DEMOS_off and -DBUILD_VKJSON=off to the cmake command line we will execute later. These aren’t required to use the SDK, however you may want to build them just to complete the SDK (where possible). cmake -H. -B<build folder> -DCMAKE_BUILD_TYPE=<build type> -DBUILD_TESTS=off -DBUILD_VKJSON=off -DBUILD_DEMOS=off VulkanTools cmake -H. -B<build folder> -DCMAKE_BUILD_TYPE=<build type> -DBUILD_TESTS=off -DBUILD_VKJSON=off -DBUILD_DEMOS=off -DBUILD_LOADER=off -DBUILD_LAYERS=off The following build types are available: * Debug * RelWithDebInfo * Release Finally we’re on to the main point of the article. The compilation itself is fairly straightforward, just navigate to the build directory and call make. Remeber you can pass the -j <number> argument to thread the compilation. Once compilation of both projects finish, you can find the various binaries inside the build folders of Vulkan-LoaderAndValidationTools, VulkanTools, spirv-tools and glslang as described in the next section. Note that if you opted not to build the demos, the vulkaninfo app/program will not be available for you to use for testing. This is quite useful as it prints out the Vulkan configuration and details for your system, and can be useful as a vulkan equivalent of dxdiag, or even to test if your layers are installed correctly. To compile this, just build the demos by re-running cmake without the -DBUILD_DEMOS=off flag. However make sure you have extracted your build output (as described below) first, or build to another folder and turn off all of the other options. # Final Steps ## Build Output The various files required for the Vulkan SDK can be found inside the build directory, specified by using the -B flag with cmake. Extract the downloadable SDK to get the include files, as well as the json files for the various layers. (You can also find these inside the source directories) ## Environment Configuration You will need to set the following environment variables to be able to use the various aspects of the SDK: Variable Action Value LD_LIBRARY_PATH Append <root>/lib PATH Append <root>/bin VK_LAYER_PATH Set <root>/lib/vulkan/layer VULKAN_SDK Set <root>	2018-09-24 21:03:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3178819715976715, "perplexity": 3627.167301762475}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267160754.91/warc/CC-MAIN-20180924205029-20180924225429-00473.warc.gz"}
https://www.mathjobrumors.com/thread/4647/page/1	As a lemon, the saddest thing in my life is knowing that I will never win 1. Top Mathematician tfxi the respect of HRMs. However, at least I have a fairly high eye queue (130-140). Some believe this is the optimal IQ range, but they are wrong. The higher one’s IQ, the better, and any high IQ person who fails at basic things probably just needs to work on their social skills and/or attitude. 1 weektfxi Quote 0 Up 19 Down Report 2. Top Mathematician mcbt Buddy, read your post again and rethink where the lack of respect might be coming from. Hint: it may be your attitude and weird obsession with IQ, not your mathematical abilities. Be a grown up, have dignity, respect will follow. 1 weekmcbt Quote 9 Up 0 Down Report 3. Top Mathematician unpd What is this nonsense? 1 weekunpd Quote 5 Up 0 Down Report 4. Top Mathematician rwwl Lol, you're not even a lemon. 1 weekrwwl Quote 5 Up 2 Down Report 5. Top Mathematician mzzx Lol, you're not even a lemon. Lol, you’re not even wrong. You’re also a bully. 1 weekmzzx Quote 2 Up 0 Down Report 6. Top Mathematician mzzx Buddy, read your post again and rethink where the lack of respect might be coming from. Hint: it may be your attitude and weird obsession with IQ, not your mathematical abilities. I literally never mention IQ in real life. Be a grown up, have dignity, respect will follow. I don’t think this is true. People are animals and do what they think benefits them. If people don’t bully you, it’s not because of your “dignity” but because you don’t make yourself an easy target. You blend in with the crowd, or, less charitably, the herd. 1 weekmzzx Quote 1 Up 3 Down Report 7. Top Mathematician dgec Chin up. Most HRM mathematicians don’t respect each other either. 1 weekdgec Quote 8 Up 0 Down Report 8. Top Mathematician mzzx I don’t even think IQ determines the worth of a person. As I said in the OP, personality is crucial. One who is as neurotic, introverted, and disorganized as myself is bound to be a lemon. Let me take solace in my intelligence, at least, while I can. 1 weekmzzx Quote 0 Up 0 Down Report 9. Top Mathematician mzzx Also, I don’t think I’m smarter than most mathematicians. Most mathematicians would score above 130 on an IQ test. Some estimate average mathematician IQ as high as 145. 1 weekmzzx Quote 0 Up 0 Down Report 10. Top Mathematician xmwz i know many neurotic and disorganised well respected mathematicians, who are not lemons. moreover, as others have said, many vhrm mathematicians do not respect each others. 1 weekxmwz Quote 3 Up 0 Down Report 11. Top Mathematician xmwz also i’m 100% convinced this whole thread is just a subtle troll 1 weekxmwz Quote 11 Up 0 Down Report 12. Top Mathematician dccs also i’m 100% convinced this whole thread is just a subtle troll Not exactly subtle. It’s the incel guy back again. 1 weekdccs Quote 10 Up 0 Down Report 13. Top Mathematician gnqg Hi Lemon, How you're doing? 1 weekgnqg Quote 0 Up 0 Down Report 14. Top Mathematician jyzm I don’t think it’s even subtle. also i’m 100% convinced this whole thread is just a subtle troll 1 weekjyzm Quote 3 Up 0 Down Report 15. Top Mathematician mzzx i know many neurotic and disorganised well respected mathematicians, who are not lemons. I think those are either smarter or more focused on math than I am, though. moreover, as others have said, many vhrm mathematicians do not respect each others. True, it is senseless to beat oneself up about not having a particular group’s approval, when there are no material consequences for how that group thinks of one. But sometimes I can’t help it. 1 weekmzzx Quote 0 Up 0 Down Report 16. Top Mathematician otdc not even an orchid variety lemon 1 weekotdc Quote 1 Up 1 Down Report 17. Top Mathematician ilqg 1 weekilqg Quote 0 Up 0 Down Report 18. Top Mathematician hlpd I don’t even think IQ determines the worth of a person. As I said in the OP, personality is crucial. One who is as neurotic, introverted, and disorganized as myself is bound to be a lemon. Let me take solace in my intelligence, at least, while I can. Do you get laid with that personality? 1 weekhlpd Quote 1 Up 1 Down Report 19. Top Mathematician xmwz Do you get laid with that personality? rhetorical question 1 weekxmwz Quote 0 Up 1 Down Report 20. Top Mathematician wxzp IQ is the best predictor of everything positive. The rest of social science is creative writing. If your IQ is less than 160, you will have a hard time. For example, you won't get a job at OpenAI. 1 weekwxzp Quote 3 Up 7 Down Report Your screen is so tiny that we decided to disable the captcha and posting feature Store settings & IDs (locally, encrypted) Formatting guidelines: Commonmark with no images and html allowed. $and$\$ for LaTeX. Input previewed in last post of thread. For a link to be allowed it must include the http(s) tag and come from the list of allowed domains.	2023-03-27 14:24:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5897753834724426, "perplexity": 6558.131342959457}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00286.warc.gz"}
http://codeforces.com/blog/entry/47094	### Sonechko's blog By Sonechko, 2 years ago, translation, , We want to say thanks to Sereja for the translation. Tutorial is loading... Tutorial is loading... Tutorial is loading... Tutorial is loading... Tutorial is loading... Tutorial is loading... Tutorial is loading... • • +36 • » 2 years ago, # \| +1 Div 2A :/ » 2 years ago, # \| 0 Can somebody explain me Div2C? Short code with explanation would be nice • » » 2 years ago, # ^ \| ← Rev. 2 → +10 First note that the binary representation can have 18 digits at most. That means we can have numbers from 0 to 2^18 — 1. Creating a 2^18 array is possible.Now you must convert the given number to its binary representation. Then, if the operation is '+' you add 1 to v[number] or remove 1 from v[number] if it's '-'.If the operation given is '?' you must get the decimal representation of the binary number given and simply print v[number]. • » » 2 years ago, # ^ \| 0 Can someone please explain why this solution(20603277) giving TLE? Thanks. • » » » 2 years ago, # ^ \| +1 Adding two string might have linear complexity — try to push back chars and at the end simply reverse string • » » » » 2 years ago, # ^ \| 0 If I reverse the string in after every step then it should take more time I think. • » » » » » 2 years ago, # ^ \| +1 No, for example in function add you do s1 = "0" + s1; something near n times. Single + operation for strings might be linear, so n times s1 = "0" + s1; can cost you o(n^2).If you simply push back char in time o(1) n times and then simply reverse string, the whole function will cost you o(n) (because you reverse once in linear time). • » » » » » » 2 years ago, # ^ \| 0 Got it thanks. I'll try that. :) » 2 years ago, # \| ← Rev. 2 → 0 Not able to understand division 2 B solution,can anybody please explain why it is "NO" if there are 4 distinct integers? • » » 2 years ago, # ^ \| 0 I understand it but not at all, for this example: n=2 : 1 4 The answer don't should be 'NO'? " If there are two distinct numbers in the array — answer is «Yes» " • » » » 2 years ago, # ^ \| 0 in this example you can take your x as 3 so 1+3 is 4 and we dont add anything in 4. hence it is yes • » » » » 2 years ago, # ^ \| +3 Aaaa, ok. I understand the statement wrong. That x is a random one, I thought that x is from array. • » » 2 years ago, # ^ \| +3 Let your final number be Y . That means initially the numbers were either Y , Y — X or Y + X .Can't be more than 3 distinct integers . • » » » 2 years ago, # ^ \| 0 understood now Thanks. • » » 2 years ago, # ^ \| +4 Imagine there exists an answer Y, then we can only have 3 different int. i.e Y-x, Y+x and Y. • » » » 2 years ago, # ^ \| 0 understood now Thanks. » 2 years ago, # \| ← Rev. 2 → 0 On problem Div1 C: Can someone give me a explanation of this passage? "We can just reduce number a[i] in array by value i and will receive previous problem.". I cannot understand why it is true :/ • » » 2 years ago, # ^ \| +39 • » » » 2 years ago, # ^ \| ← Rev. 2 → 0 what does it mean "to receive the previous problem"? • » » » » 2 years ago, # ^ \| 0 It means the problem gets reduced to the previous one. » 2 years ago, # \| 0 Why does it take 2LogN operations to find separating line in div1B?We can find the smallest x, that get(1, 1, x, n) >= 1 and get(1, 1, x — 1, n) < 1 using binary search. To check if it valid we just need to check get(1, 1, x, n) == 1 and get(x + 1, 1, n, n) == 1. It takes LogN + 2. Overall 10LogN + 4. » 2 years ago, # \| ← Rev. 4 → +11 I am not sure what is happening here in DIV 2C / DIV 1B / 713B....!! Approach same as in the tutorial( 12log(N) ) + 4 extra query. line cutting order right(x2)->left(x1)->up(y2)->down(y1) but this approach got WA in test 55...!! WA_Submission Then after seeing the testcase 55 i changed the order of the line cutting to down(y1)->up(y2)->right(x2)->left(x1) and got AC...!!! AC_Submission So are the test cases weak or am i missing something??!!UPDT : My implementation of BS actually took 17 iterations instead of 16 giving max of 208 queries..!! :/ which caused me WA..!! :( bt now it seems to me judge data is weak given my 2nd approach with same BS implementation got AC...!! Again i think the problem had too much tight constraints but failed to provide all possible tight corner case inputs...!!Hope no one else faced the same situation..!! :/ • » » 2 years ago, # ^ \| +8 +1s • » » 2 years ago, # ^ \| 0 Actually you can do only queries so the limits aren't tight. • » » » 2 years ago, # ^ \| 0 yes...realized that after the contest..!! :/ anyway solved it using 10logN+4 » 2 years ago, # \| +14 Hi , I apologize if my question seems stupid but I can not understand "For each i will bruteforce value j (i > j) and calculate answer for j if [i + 1, j] if arithmetics progression with step 1" if i > j , shouldn't [ i + 1 , j ] be [ j , ... ] and expression if if isn't comprehensible ... thanks in advance » 2 years ago, # \| 0 Problem 1c can be solved in O(nlogn) without dp. • » » 2 years ago, # ^ \| +4 How to solve it without DP?thx • » » » 2 years ago, # ^ \| +18 For simplicity, first subtract each number by their positions i, so we need to make the sequence non-decreasing.Now, start from beginning. Each time a modification is needed, there are two options: raise the current position up or push elements before down. It should be noticed that for each position raising up happens at most one time (probably more than one unit though).If nearest elements already form a equal sequence then all of them should be pushed down. However, pushing down n elements does not necessarily costs n -- some of them may be raised up before. For these elements pushing down actually contributes negative one per unit (for undoing raise). Furthermore, once some elements form a continuous equal sequence, they will be equal forever so we can view them as a group.Now we are able to construct a solution. We start moving from position 2 all the way to the end. Once we move past a position some 'event' height will be recorded. There are three different situations:1) a[i] = a[i-1]. In this case, position i simply merges with the previous group, and the cost of pushing down that group increases by 1.2) a[i] > a[i-1]. We don't need to modify anything now; however an event point should be added at height a[i-1] for possible traceback afterwards -- when height is reduced back to a[i-1] two groups become one.3) a[i] < a[i-1]. This is the only case we need an modification. Now, look for the cost for pushing down the previous group. As long as the cost does not exceed one (actually it never goes below one, you can prove it), it is worth doing. i) If that is sufficient to lower a[i-1] to a[i], we are complete. Then the cost will increase by one.ii) Otherwise, a[i] will be raised up. In this case, however, the cost will be reduced by one since when we push down position i afterwards, before it goes back to original value we are essentially undoing raise. Finally, we create an event point at original value of a[i]; when we reach that point later, cost of pushing down increases by 2 (from -1 to +1). For implementation: store all event points in a heap(priority queue). When pushing down is performed, refer to the top(largest) value and update. Since each position adds at most one event point, total complexity is O(nlogn). • » » » » 2 years ago, # ^ \| 0 Tip: information considering the length, or left endpoint of a group need not be stored; what happens as we move past a event point is just that cost of pushing down increases. So the only thing recorded is: at this point how much does push-down cost increase? That is good for implementation since only a pair (height, costincrement) is needed. • » » » » 2 years ago, # ^ \| 0 You saved my day! • » » » » 2 years ago, # ^ \| 0 Can't it happen that you need to pop several elements from the heap? When you have e.g. 1,2,3,4,5,-100000 (you can lift it up to make positive). • » » » » » 2 years ago, # ^ \| +5 Read last sentence again: heap.push happens at most n times. It is fine to have multiple pops at once, but by aggregate analysis it is O(nlogn). • » » » » 2 years ago, # ^ \| 0 I implemented quite similar solution, but using map, it seems simpler for me:20623211 AC (time O(NlogN))Works on random and sorted sequences of length 10^6 in 1 second. • » » » » » 2 years ago, # ^ \| ← Rev. 3 → +11 For implementation, I want to share woqja125's AC code which is truly magical :DOriginal codeSlightly polished code • » » » » » » 2 years ago, # ^ \| 0 Omg! Can you explain why does this work? • » » » » » » 2 years ago, # ^ \| 0 Amazing. Seems that there must be an much much easier explanation of this problem. Could you tell that magic idea? • » » » » » » » 2 years ago, # ^ \| 0 See the comment below! • » » » » 2 years ago, # ^ \| 0 thanks a lot.your solution is great and your explanation is very clear!here is my implementation of the exact idea if it might help someone. » 2 years ago, # \| +31 Will Codeforces ever stop posing questions with complexities of model solution when there are obvious bruteforces in O(n4)?! Did any of testers even tried to compare those solutions? Did you try to ensure only intended solutions pass? • » » 2 years ago, # ^ \| +5 If the bruteforce solution is so obvious, why didn't you send it? • » » » 2 years ago, # ^ \| +18 He assumed that organizers made sure naive solutions won't pass. It's about the constant factor, not about the difficulty of coming up with that solution. • » » » » 2 years ago, # ^ \| ← Rev. 2 → 0 1e9 in 5 seconds. I think it's obvious it will pass on CF, if you're on CF more than a year. I thought about it but I can't find a solution in that complexity, then I found the intended one. So maybe N^3 is not just a "obvious" solution? Intended one is really eaiser than that one except the implementation.EDIT: Why "He assumed that organizers made sure naive solutions won't pass"? Organizers can make mistakes. Didn't you ever see a solution which have a more complexity than the intended one but with a simple code? • » » » » » 2 years ago, # ^ \| +10 "1e9 in 5 seconds. I think it's obvious it will pass on CF, if you're on CF more than a year." — wut? You sure? You oversimplified it af. It heavily depends on kinds of operations you perform and number of them, locality of memory calls etc. I didn't inspect other people O(n3 + nt) solutions, but one I have in mind performs rather something like 1e10 operations with not very local memory calls, so it is obvious for me it won't pass. Btw people's times on this task are rather high, you can't really say it is obvious it will pass if one's time ends up being 4,5/5s :). As my saying goes "Problems can be divided to two kinds. One kind when intended solution for n<=1e6 and O(n log n) gets TLE and reason for that is "you know, big constant, memory allocation and shit" and second kind when for n<=1e3 intended solution is O(n^3) and organizers excuse is "you know, computers nowadays are really fast" :).And regarding to "EDIT". If you take a look at constraints it is kinda obvious that test preparer's duty is to defend against >=O(n^3) or >=O(nt) solutions. These are not complexities taken out of nowhere, solutions in such complexities are not something crazy. I assume organizers are well aware of that and have taken care of that, tested few solutions with optimized memory locality and shit and ensured they don't pass. • » » » » » » 2 years ago, # ^ \| 0 "ensured they don't pass"This part isn't about problemsetter. If you write smooth N^3 it will get an accepted if N<=1000. There's a lots of problem which is slower than intended one but pass like that. • » » 2 years ago, # ^ \| 0 I agree that it sometimes happens, but for this particular problem I do not consider O(nt) solution to be obvious. • » » » 23 months ago, # ^ \| +5 That's why I don't like 2d problems. Always some naive stuff or in the toughest cases nsqrtn works better than any intended solution. And you can't really understand which nsqrtn solution is better (constants are rather unpredictable).When I was solving this problem I easily got to the online max-queries. But I forgot about the existence of 2d-sparse table and understood that no log^2 (log from binary search) tree would pass through time limit. I think I forgot about it because I can't really remember any problem which was solved with this algorithm, and I see the reason: some stupidity always works better.P.s. n^3 + nt solution seems pretty obvious to me and it was the first thing I got to. Totally dislike the problem. And I still don't understand why this solution passes. 1.510^9 min(max()) + access to 2-d array operations in 3 seconds? Maybe tests are not good enough? I saw really not optimal solutions to pass in a half time and was surprised. Tests with 10^6 queries of (1000-eps)x(1000-eps) size should be good enough. I haven't checked if there are some so maybe I'm wrong. » 2 years ago, # \| ← Rev. 2 → 0 Fun fact : it's harder than the author solution, but we can solve Div2C/Div1A using trie. Here is my code which i did during the contest: http://codeforces.com/contest/713/submission/20575096 • » » 2 years ago, # ^ \| ← Rev. 2 → +3 lol, this is also the first solution I come in mind. All you should do is reverse the input and it becomes a standard trie problem :DPS: do not forget to add trailing zeroes at the front to fill up required length » 2 years ago, # \| +3 can someone explain why we can minus every a[i] by i without changing the answer ? • » » 2 years ago, # ^ \| ← Rev. 6 → +3 This might not be fully formal, but it's the way I thought about it:A[i] < A[i + 1] can be written as A[i] <= A[i + 1] — 1.Let's subtract i from both sides. ( Remember i is always >= 0)We arrive at :A[i] — i <= A[i + 1] — i — 1.Which can then be re-written as :A[i] — i <= A[i + 1] — ( i + 1)We can also work our way backward, which proves the equivalence.Now Assume the minimum cost for getting a non decreasing sequence for a[i] — i is x.If you can make A[i] strictly increasing with changes less than x, then we could definitely make A[i] — i non decreasing using that value as they are equivalent, which contradicts our assumption that x is minimum.As for the other way around ( x being smaller than needed ) you can use contradiction as well. • » » » 2 years ago, # ^ \| 0 i think i got it :P thanks!! » 2 years ago, # \| +14 Can someone explain the maxflow approach which was used by some participants for the Div2E/Div1C problem? » 2 years ago, # \| ← Rev. 2 → 0 l1 ≤ r1, l2 ≤ r2. Am I the only one who have been stuck at this point?? According to the problem statement, 714A - Meeting of Old Friends, l2 cannot be smaller than l1 and same for r2. So 20579107 should get AC but it gives WA at 3rd test case. So correct me if I misunderstood something but problem statement is not corresponding to the editorial. That really effected me too much and caused me to get a great(!) motivation for other problems » 2 years ago, # \| +5 I managed to get the idea of problem D but the TL seems really strict, my 2D sparse table got accepted with 49xxms out of 5000ms... • » » 2 years ago, # ^ \| ← Rev. 2 → 0 Your code can be a few times faster if you change dp[N][N][11][11] into dp[11][11][N][N]. In init() you iterate over i and j jumping every 11 * 11 elements and missing cache very often. » 2 years ago, # \| +1 Please explain the DIV-2E/DIV-1C solution. I am not able to understand from the tutorial. » 2 years ago, # \| 0 Hello, can someone tell me why my custom made map fails in div2C and AC with SET?I never had problems with it before..I am just adding trailing 2's to represent even and 1s to represent odd(18 digits per number). http://codeforces.com/contest/714/submission/20599916Thanks » 2 years ago, # \| 0 In the problem B.How can I read the answer after flush in c++? • » » 2 years ago, # ^ \| ← Rev. 2 → 0 simply using scanf or cin to get input from stdin • » » » 2 years ago, # ^ \| 0 thanks » 2 years ago, # \| 0 Problem 1E looks pretty similar to this. » 2 years ago, # \| 0 I solved problem B differently. I used Binary Search to find the answer. • » » 2 years ago, # ^ \| 0 This is the source: http://codeforces.com/contest/714/submission/20612613 • » » » 2 years ago, # ^ \| 0 :O oppss!!! such a great solution! • » » 2 years ago, # ^ \| 0 Overkill » 2 years ago, # \| ← Rev. 2 → 0 somebody plz..post some useful links on problems like minimum steps array conversion etc............... » 2 years ago, # \| 0 Problem B div2 nice explaination , thank you :) » 2 years ago, # \| 0 HashMap http://codeforces.com/contest/714/submission/20616621 (x%10%2 ) accepted .the same code with tree map http://codeforces.com/contest/714/submission/20616692 Tle (x%10%2) why ? tree map is doing (log n ) operations .?the same code with tree map but (x&1) http://codeforces.com/contest/714/submission/20616823 Accepted ? any tips to make my code as fast as possible ? » 2 years ago, # \| +5 Does anybody know a judge with this Div1C problem but with larger constraints (e.g. N = 105)? » 2 years ago, # \| ← Rev. 23 → 0 Can anyone help me for this code??Running median....tried many times but still getting TLE in spoj (Problem : http://www.spoj.com/problems/XMEDIAN/) » 2 years ago, # \| +83 There is a very short (10 lines!) and fast O(nlgn) solution for Div1C, proposed by woqja125 :Start by subtracting i from all A[i]. We'll define fi(X) = Minimum operation to make A[1 .. i] monotonically increasing, while keeping A[i] <= X. it's easy to see that the recurrence is — Min(Y <= X) \|Ai — Y\| if i == 1 Min(Y <= X) (f(i-1) (Y) + \|Ai — Y\|) otherwise. Observation : fi is an unimodal function, which have a nonpositive integer slopes. Denote Opt(i) as a point X which makes fi(X) minimum. We will maintain those functions and calculate the minimum. we examine two cases :Case 1 : Opt(i-1) <= A[i]. In this case, Opt(i) = A[i], and every point under A[i] have +1 slope.Case 2 : Opt(i-1) > A[i]. This case is tricky. every point over A[i] have +1 slope, and every point under A[i] have -1 slope. Opt(i) is the point where slope of fi is changed into -1 -> 0. When observing the function, Fi(Opt(i)) = F(i-1)(Opt(i-1)) + Opt(i-1) — A[i]. We can simply assume that Fx(Opt(x)) = 0, and add this to our answer.Since applying this operation doesn't contradict above observation, the correctness of this algorithm can be easily proved by induction.To implement this, we can maintain a priority queue, which holds every point where slope changes by +1. It turns out that implementing this can be "extremely" easily done by O(nlgn).implementation • » » 2 years ago, # ^ \| +5 Awesome! Thank you for your clear and helpful explanation. • » » 2 years ago, # ^ \| ← Rev. 2 → +3 can anyone explain the recurrence formulation again ? I am having a hard time understanding it ? what is Y ? • » » 2 years ago, # ^ \| +5 can you please explain the above algorithm in a more detail.Thanks in advance :) • » » » 22 months ago, # ^ \| ← Rev. 2 → 0 We know f(opt(n)) is the final answer.we only need to maintain f(opt(i)) for every i then we will get it.we can find that : when a[i] >= opt(i-1), f(opt(i)) will equal to f(opt(i-1)) when a[i] < opt(i-1), f(opt(i)) will equal to f(opt(i-1)) + opt(i-1) — a[i] so if we know opt(i-1) when we add a[i], we can get f(opt(i)).Let's consider how to maintain the information of fi to get opt(i)In fact, we only need to know the slopes of every segment and then the point that slope become 0 will be the opt(i).We save some "key points"(may overlap) in a sorted list, and the slope at position x is the number of "key points" that greater than x. when a[i] >= opt(i-1), we only add a[i] to the sorted list when a[i] < opt(i-1), we only need to move a rightest "key point"( which is equal opt(i-1) ) to the position a[i]. after that we add another a[i] to the list. You will find that we keep the property above to be unchanged.(because it means we add -1 to slope of (-oo,a[i]) and +1 to slope of (a[i],opt(i-1)] ) We only need a heap to maintain the sorted list. • » » » » 8 months ago, # ^ \| 0 Can you please explain this line in detail.."when a[i] < opt(i-1), f(opt(i)) will equal to f(opt(i-1)) + opt(i-1) — a[i]" Thank You • » » 4 months ago, # ^ \| 0 Awesome solution. But can you please explain why do we have to push t two times when Q.top() > t? I'm having a hard time to understand that. • » » » 4 months ago, # ^ \| 0 Because t is the point where slope differs by two (Ai - x to x - Ai). You will fail if you try to understand its implementation intuitively. I encourage you to fully understand above and implement it on your own. • » » 3 months ago, # ^ \| 0 Can you please tell how is it is evident that the recurrence function is uni-modal ? TIA. » 2 years ago, # \| ← Rev. 2 → -7 Div1 E is very much related to this problem: SequenceIt will convert to this once you subtract i from each A[i]. After doing subtracting each A[i] you need to find minimum number of operations to make sequence non decreasing which is exactly what is asked in the other problem. As the tutorial for the problem I mentioned is relative simple any one can understand easily. • » » 2 years ago, # ^ \| 0 but that solution is O(n^2) and this is O(nlogn) • » » » 2 years ago, # ^ \| 0 As far as this problem is concerned O(n^2) will pass. But always good to know better solutions. » 2 years ago, # \| -10 div2E/div1C problem is very similar to this problem In the tutorial of this problem the following claim is made Note, that there exists a non-decreasing sequence, which can be obtained from the given sequence using minimal number of moves and in which all elements are equal to some element from the initial sequence (i.e. which consists only from the numbers from the initial sequence). can someone explain its proof?thanx in advance » 2 years ago, # \| 0 Can someone please explain DP solution of problem 713C — Sonya and Problem Wihtout a Legend. http://codeforces.com/contest/713/submission/20588657 I am not able to understand this line mn = min(DP[i-1][j],mn);DP[i] is cost required till ith index. Why are we adding mn to it? » 2 years ago, # \| 0 dp[i][j] < 0 then dp[i][j] vertices to the right we still can visit...I don't get what does this dp states mean, does it mean the interval [i+1,i-dp[i][j]] has been visited? » 2 years ago, # \| ← Rev. 2 → 0 got it. » 2 years ago, # \| ← Rev. 3 → +4 As it took me ages to figure out the recurrence for DIV1.C. I hope this will clarify things for others. Somebody already mentioned above that the problem can be reduced to a non-decreasing sequence by considering the sequence ai - i. From now on we will just consider ai = ai - i.The second thing that one can prove is that there always exists an optimal solution, where one ai is not changed. Hint: Just consider an optimal solution where this does not hold.Let bi be the sorted sequence of ai. Let dpi, j be the optimal value for the prefix a1... ai while ending in a value smaller or equal to bj. The recurrence is then as followsdpi, j = min{ dpi, j - 1, dpi - 1, j + \|bj - ai\| }Why? Either we let the prefix a1... ai end in a value smaller then bj - 1 ( hence also smaller then bj) or we let the prefix a1... ai - 1 end in a value smaller or equal to bj and also let ai be equal to bj. There are some base cases and this can of course be adapted for linear space. Edit: 20669994 • » » 2 years ago, # ^ \| 0 How do you prove that at least 1 ai is not changed? • » » » 22 months ago, # ^ \| 0 If all numbers are changed, then we can increase (or decrease) all number by some amount until at least 1 ai stay the same • » » 15 months ago, # ^ \| 0 why we don't use another array instead using b ( sorted array of a) ??? » 22 months ago, # \| 0 I am having a very strange problem in QUESTION B — Filya and Homework My submission is — CODEWhen I run the code on my compiler, it works fine. However, it gives me a runtime error "on test case 1 itself!" when I submit it. I made sure I am using c++ 14 in both cases, so how can this happen?Any help would be very greatly appreciated — stuck since a long time :( • » » 22 months ago, # ^ \| 0 Do NOT use return 1 in CodeForces, it will confuse the judge and thinks your code has RE. • » » » 22 months ago, # ^ \| 0 Thankyou! :) » 19 months ago, # \| 0 Can someone explain the following solution for div1 C problem plz??const int maxn = 3000 + 10; ll dp[maxn][maxn]; int n, a[maxn], b[maxn];int main() { #ifndef ONLINE_JUDGE freopen("in", "r", stdin); #endif cin >> n; for (int i = 1; i <= n; i++) { cin >> a[i]; a[i] -= i; b[i] = a[i]; } sort(b + 1, b + n + 1); for (int i = 1; i <= n; i++) { ll ret = dp[i-1][1]; for (int j = 1; j <= n; j++) { ret = min(ret, dp[i-1][j]); dp[i][j] = abs(a[i] - b[j]) + ret; } } ll ans = dp[n][1]; for (int i = 1; i <= n; i++) { ans = min(ans, dp[n][i]); } cout << ans << endl; return 0; } » 15 months ago, # \| 0 Can someone prove why it is optimal to make all elements equal to the median ? (Sonya and Problem without Legend) • » » 6 weeks ago, # ^ \| 0 Because median is the element which minimize absolute sum of differences with all the other elements of the list. • » » » 6 weeks ago, # ^ \| 0 Thanks .. But, I actually didn't understand what they meant by "make every element equal to the median of the prefix". • » » » » 6 weeks ago, # ^ \| 0 Did u understand this part : All the elements of the final array will be equal to some elements of the initial array?	2018-09-19 10:08:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.43528854846954346, "perplexity": 1586.8045286839713}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267156096.0/warc/CC-MAIN-20180919083126-20180919103126-00084.warc.gz"}
https://www.scoalaicbratianu.info/3e6my2y/d72ae7-lycoming-o-290-cylinders	[O-]S(=O)(=O)OOS(=O)(=O)[O-], Except where otherwise noted, data are given for materials in their. This experiment was extended to other metals such as nickel, cadmium, and iron, all of which yielded similar results. 0510-01 : Ammonium Persulfate,20% w/v,Aqueous Solution (1 x 500 ml) Click for details § UPS HazMat Fee : 28.50 / Package Included in shipping as Handling charge. Ammonium persulfate is one of many compounds available for copper etching, and is often used for etching circuit boards in particular. It can act as an emulsifier in the co-polymerization of styrene, acrylonitrile, and butadiene, etc. Ammonium sulfate precipitation is a common method for protein purification by precipitation. is called ammonium persulfate. Ammonium persulfate (APS) is an oxidizing agent that is often used with tetramethylethylenediamine (TEMED, Part No. In addition, the solution must be hot to … Ammonium persulfate is used as a catalyst for the copolymerization of acrylamide and bisacrylamide gels. Ammonium persulfate appears as a white crystalline solid. Exposure to high levels of dust may cause difficulty in breathing. V. The trisulphates", Ullmann's Encyclopedia of Industrial Chemistry, "Action of Ammonium Persulphate on Metals", "Ammonium persulfate can initiate an asthmatic response in mice", https://en.wikipedia.org/w/index.php?title=Ammonium_persulfate&oldid=992951899, Chemical articles with multiple compound IDs, Multiple chemicals in an infobox that need indexing, Pages using collapsible list with both background and text-align in titlestyle, Articles containing unverified chemical infoboxes, Creative Commons Attribution-ShareAlike License, This page was last edited on 8 December 2020, at 00:25. Illustrative of its powerful oxidizing properties, it is used to etch copper on printed circuit boards as an alternative to ferric chloride solution. It has a variety of applications in polymer chemistry, such as a bleaching agent, cleaning agent, and an etchant. AMMONIUM PERSULFATE is a potent oxidizing agent. CAS: 7727-54-0 MDL: MFCD00003390 EINECS: 231-786-5 Synonyms: Ammonium persulfate, di-Ammonium peroxodisulfate, Diammonium peroxodisulfate Commercially important polymers prepared using persulfates include styrene-butadiene rubber and polytetrafluoroethylene. Hazmat fee assessed per package, NOT per item in package.-- --1 x 500 ml: No-MSDS:22.39: $18.26 0510-02 Use ammonium persulfate (APS) as a catalyst for polymerization of acrylamide and bis-acrylamide. No. 24 Hour Emergency Information Telephone Numbers A strong oxidizing agent. (NH4)2S208 is a type of an inorganic compound. The airborne dust that contains ammonium persulfate may be irritating to eye, nose, throat, lung and skin on contact. solution ammonium persulfate etching sulfate ammonium Prior art date 1965-04-28 Legal status (The legal status is an assumption and is not a legal conclusion. Ammonium persulfate is the chemical name of. The polymerization reaction is driven by the free radicals that are generated by an oxido-reduction reaction in which a diamine is used as the adjunct catalyst. Ammonium persulfate decays slowly in solution, so replace the stock solution every 2-3 wk. Peroxymonosulfate— FMC developed a process using ammonium and sodium persulfates to prepare peroxy- monosulfate solutions. Ammonium persulfate is the chemical name of AMMONIUM PERSULFATE: ICSC: 0632: Peroxydisulfuric acid, diammonium salt Diammonium peroxydisulphate Diammonium persulfate: October 2001: ... nitrogen oxides and sulfur oxides. is a type of an inorganic compound. In this article, we will learn about what is ammonium persulfate, persulfate uses, its structure and properties in detail. The ammonium persulfate structure is shown as below: Let us now learn about some of the properties of ammonium persulfate that are given below: Let us now learn about the ammonium persulfate uses. Ammonium persulfate is a standard ingredient in hair bleach. Add 1g Ammonium persulphate per 10 ml water e.g. It is available in a dry or powder form and must be mixed with water to create the etching solution. The electrolytic process of ammonium sulfate and sulfuric acid formulates to form a liquid electrolyte which is decontaminated by the electrolysis process. This is the method that was initially described by Hugh Marshall. It has a variety of applications in polymer chemistry, such as a bleaching agent, cleaning agent, and an etchant. Does not burn readily, but may cause spontaneous ignition of organic materials. It is a strong oxidizing agent which is used in polymer chemistry, as a cleaning and bleaching agent, and as an etchant. It is used for removing the pyrogallol stains. It has been noted that the persulfate salts are a major cause of several asthmatic effects amongst women. (NH4)2S208 . Also, it has been suggested that the exposure to ammonium persulfate can lead to several asthmatic effects amongst the hairdressers and receptionists that are working in the hairdressing industry. The AZDN was warmed by a supposedly blanked-off steam pipe and partially decomposed, rupturing its kegs and falling onto the persulfate in sacks below. Comparison of cellulose nanocrystals obtained by sulfuric acid hydrolysis and ammonium persulfate, to be used as coating on flexible food-packaging materials. Ammonium persulfate is (NH4)2S2O8 is a strong oxidizing agent and bleacher in hair bleaches. It is prepared by using the method of electrolysis of a cold concentrated solution of either of these substances: ammonium sulfate or the ammonium bisulfate, in the presence of the sulfuric acid at a very high current density. The azonitrile (AZDN) and ammonium persulfate were stored together as polymerization initiators. It is a colourless salt which is highly soluble in water, much more than the related potassium salt. Ammonium persulfate is prepared by electrolysis of a cold concentrated solution of either ammonium sulfate or ammonium bisulfate in sulfuric acid at a high current density. Solutions Available from Ammonium persulphate 0.1mol L-1 Dispenser Ammonium sulphate 0.1mol L-1 On bench Potassium chloride 0.2mol L-1 On bench Potassium iodide 0.2mol L-1 Dispenser Sodium thiosulphate 0.005mol L-1 Dispenser Copper sulfate 0.02mol L-1 On bench Starch solution On bench Special equipment Available from 10 mL Pipette with filler On bench Timer and digital thermometer … It is used as a depolarizer in batteries. Edited by G. Brauer, Academic Press, 1963, NY. Cellulose 2016 , 23 (1) , 779-793. We are a distributor of the product Ammonium Persulfate for Eastern and Western Europe, but with excellent opportunities for export outside of Europe. [6] It is also known by other names such as ammonium peroxydisulfate, diammonium persulfate, and diammonium peroxydisulfate. Shop a large selection of Ammonium products and learn more about Ammonium Persulfate. [10], InChI=1S/2H3N.H2O8S2/c;;1-9(2,3)7-8-10(4,5)6/h21H3;(H,1,2,3)(H,4,5,6), InChI=1/2H3N.H2O8S2/c;;1-9(2,3)7-8-10(4,5)6/h21H3;(H,1,2,3)(H,4,5,6), O=S(=O)([O-])OOS([O-])(=O)=O.[NH4+]. 9:16. In this article, we will learn about what is ammonium persulfate, persulfate uses, its structure and properties in detail. Ammonium persulfate or APS is the inorganic compound that has the formula. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. 50mls add 5g APS Mix thoroughly - APS will dissolve readily. Airborne dust containing ammonium persulfate may be irritating to eye, nose, throat, lung and skin upon contact. It is a radical initiator. APS forms oxygen free radicals in aqueous solution by a base … These asthmatic effects are known to be caused by the oxidation of cysteine as well as methionine residues. Pro Lite, Vedantu Ammonium persulfate aqueous solution was used for the first time for NO removal from flue gas in a bubbling bed due to its low cost compared with other persulfate salts (sodium persulfate, potassium persulfate and ammonium persulfate). It is used to etch copper on printed circuit boards as an alternative to ferric chloride solution. The sulfate radical adds to the alkene to give a sulfate ester radical. Pro Lite, Vedantu It is readily soluble in water and is a strong oxidizing agent. Ammonium persulfate goes through the processes of filtration, crystallization, centrifugal separation, and drying to get the ammonium persulfate product when the content reaches up to a certain concentration in the anode. Ammonium persulfate can be derived by the method of electrolysis of ammonium sulfate and dilute sulfuric acid which is then crystallized. Ammonium persulfate decays slowly in the solution, so make sure to replace the stock solution every 2-3 week. [8], It has been noted that persulfate salts are a major cause of asthmatic effects in women. Due to its high oxidation potential it can cause ignition under certain conditions. The precipitate was filtered, rinsed with aqueous solution of HCl and then dried at room temperature for 72 h to obtain a doped state conductive PANI (PANI-HCl) powder. It is very soluble in water; the dissolution of the salt in water is endothermic. The dissolution of the salt in water is an endothermic process. It is also used along with tetramethylethylenediamine to catalyze the polymerization of acrylamide in making a polyacrylamide gel, hence being important for SDS-PAGE and western blot. Ammonium persulfate or It is a strong oxidizing agent which is used in polymer chemistry, as a cleaning and bleaching agent, and as an etchant. "Electroreduction of Peroxodisulfate: A Review of a Complicated Reaction", "LXXIV. It is a strong oxidizer and radical donator for polymerisation. Ammonium persulfate or APS is the inorganic compound that has the formula (NH4)2S208 . Ammonium Persulfate is an initiator for the emulsion or solution polymerization of acrylic monomers, vinyl acetate, vinyl chloride, etc. Contributions from the Chemical Laboratory of the University of Edinburgh. Ammonium persulfate is white, odorless single crystal, the formula is (NH4) 2S2O8, it has strong oxidation and corrosion, when heated, it decomposes easily, moisture absorption is not easy, it is soluble in water, the solubility increases in warm water, it can hydrolyze into ammonium hydrogen sulfate and hydrogen peroxide in an aqueous solution. To prepare a 10% (w/v) solution, you need to dissolve 1 g of ammonium persulfate in 10 ml of H2O and store it at 4°C. Vol. Ammonium persulfate is used across several industries in the following ways: It is used in the printed circuit boards. It is also called as ammonium peroxydisulfate, diammonium persulfate, or diammonium peroxydisulfate. [NH4+], [NH4+].[NH4+]. It is also used in the treatment of metal surfaces, activation of copper, and aluminum surfaces. Persulfates are used as oxidants in organic chemistry. Ammonium persulfate is used as a catalyst for the process of copolymerization of acrylamide and bisacrylamide gels. It is also known by other names such as ammonium peroxydisulfate, diammonium persulfate, and diammonium peroxydisulfate. The oxidation equation is thus: S2O2−8 (aq) + e− → 2 SO2−4 (aq). This patented process allows fast, efficient, on-site production of an alternative to Caro's acid and potassium caroate. In 1908, John William Turrentine used a dilute ammonium persulfate solution to etch copper. Dissolve in dH 2 0 and store at 4ºC If in solution, reacts violently with iron, powdered aluminium and silver salts. The effects of ammonium persulfate concentration, solution temperature, initial [9] Furthermore, it has been suggested that exposure to ammonium persulfate can cause asthmatic effects in hair dressers and receptionists working in the hairdressing industry. This oxidizing agent is frequently used with another catalyst, TEMED, for preparation of polyacrylamide gels for protein and nucleic acid analysis. Anode reaction: $2HS0_{4}$ - 2e $\rightarrow$ $H_{2}$ $S_{2}$ $0_{8}$, Cathodic reaction: $2H^{+}$ + 2e $\rightarrow$ $H_{2}$ ↑, $NH_{4}$ $2S_{2}$ $0_{4}$ + $H_{2}$ $S_{2}$ $0_{8}$ $\rightarrow$ $NH_{4}$ $2S_{2}$ $0_{8}$ + $H_{2}$ $S0_{4}$. Exposure to high levels of dust can also cause difficulty in breathing. [7] For example, in the Minisci reaction. 100g. Sorry!, This page is not available for now to bookmark. It is an odourless and crystalline solid that is white in colour. To prepare a 10% (w/v) solution: Dissolve 1 g ammonium persulfate in 10 mL of H 2 O and store at 4°C. (NH4)2S208 is called ammonium persulfate. Salts of sulfate are mainly used as radical initiators in the polymerization of certain alkenes. Ammonium Persulfate - 15% Solution in Water GHS Label - 2" x 3" (Pack of 25)$39.00 $39. 00 ($1.56/Item) \$8.00 shipping. It is highly soluble in water and is a strong oxidising agent which has its uses in organic chemistry, polymer chemistry, as a bleaching and cleaning agent, and as an etchant. Description and uses APS is very soluble in cold water, a large fall of temperature accompanying solution. Sigma-Aldrich offers a number of Ammonium persulfate products. Prepare ammonium persulfate (APS) stock solution (10%) by adding: Step 2. Add dH20 to Falcon tube or other suitable container for the volume. It is an organic compound which is white (colourless) salt. Ammonium Persulfate 7727-54-0 TWA: 0.1 mg/m3- TWA: 0.1 mg/m3TWA: 0.1 mg/m3 Appropriate engineering controls Engineering measuresProvide local exhaust or general ventilation adequate to maintain exposures below permissable exposure limits. The solution in water is a medium strong acid. Ammonium persulfate (APS) is the inorganic compound with the formula (NH4)2S2O8. Ammonium sulfate is extremely soluble in water due to its ionic nature, therefore it can "salt out" proteins by … Ammonium persulfate or (NH4)2S208 is a type of an inorganic compound. Ammonium peroxosulphate (short form APS, trade name ammonium persulfate) is the salt of peroxodisulphuric acid. 1. p. 390. Comparison of cellulose nanocrystals obtained by sulfuric acid hydrolysis and ammonium persulfate, to be used as coating on flexible food-packaging materials. The prepared acid solution of ammonium persulfate was dropped into the acid solution of aniline under stirring for 2 h at –10°C. Hairdressers should be made aware that these ammonium persulfate hair bleach preparations may provoke severe reactions and should seek medical attention if the client complains of severe itching, tingling, a burning sensation, hives, dizziness, or weakness. Removal of NO from flue gas using heat-activated ammonium persulfate aqueous solution in a bubbling reactor. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.) It is used as a common ingredient in the hair bleaches. Let us look at what are these. It is an odourless and crystalline solid that is white in colour. It is a colourless (white) salt that is highly soluble in water, much more so than the related potassium salt. It is readily soluble in water and is a strong oxidizing agent. Ammonium Persulfate GHS Label - 2" x 3" (Pack of 25) As an oxidizing agent and a source of radicals, APS finds many commercial applications. It is highly soluble in water and is a strong oxidising agent which has its uses in organic chemistry, polymer chemistry, as a bleaching and cleaning agent, and as an etchant. It is used in the olefin polymerization as an initiator. [1][2] The method was first described by Hugh Marshall.[3]. It is a strong oxidizing agent that is used in polymer chemistry, as an etchant, and as a cleaning and bleaching agent. Ammonium Persulfate Solution, 2M MSDS # 56.10 Section 1: Product and Company Identification Ammonium Persulfate Solution, 2M Synonyms/General Names: Ammonium Peroxydisulfate Product Use: For educational use only Manufacturer: Columbus Chemical Industries, Inc., Columbus, WI 53925. As the ionic strength of a solution increases, the solubility of proteins in that solution decreases. It is a colourless salt which is highly soluble in water, much more than the related potassium salt. Ammonium persulfate has its own share of side effects. Let us now take a look at how is ammonium persulfate produced. To prepare a 10% (w/v) solution, you need to dissolve 1 g of ammonium persulfate in 10 ml of H, Dichloromethane Uses and Effects on Environment, Benzene - Physical and Chemical Properties, Chemical Properties of Metals and Nonmetals, Vedantu Regeneration of spent ammonium persulfate etching solutions Download PDF Info Publication number US3406108A. F. Feher, "Potassium Peroxydisulfate" in Handbook of Preparative Inorganic Chemistry, 2nd Ed. $HS0-_{4}$ can discharge and generate the peroxydisulfate acid in the anode and it then reacts with ammonium sulfate to generate ammonium persulfate. A powdered mixture with aluminum and water can explode [NFPA 491M 1991]. It is prepared by using the method of electrolysis of a cold concentrated solution of either of these substances: ammonium sulfate or the ammonium bisulfate, in the presence of the sulfuric acid at a very high current density. Turrentine weighed copper spirals before placing the copper spirals into the ammonium persulfate solution for an hour. It is used as an additive for preserving the food. Ammonium persulfate is a widely used reagent in biochemistry and molecular biology for the preparation of polyacrylamide gels. As an oxidizing agent and a source of radicals, APS finds many commercial applications. It has a variety of applications in polymer chemistry, such as a bleaching agent, cleaning agent, and an etchant. It is also called as ammonium peroxydisulfate, diammonium persulfate, or diammonium peroxydisulfate. Expired - … [5] This property was discovered many years ago. In some patients, merely rubbing a saturated solution of ammonium persulfate into the skin will evoke a large urticarial wheal. Na-salt in Pellets research grade; Laemmli Buffer 10x, for SDS PAGE Laemmli Sample Buffer 2x, for SDS PAGE SERVA Tris-Glycine/LDS Sample Buffer (4x) 17919) to catalyze the polymerization of acrylamide and bisacrylamide to prepare polyacrylamide gels for electrophoresis. After an hour, the spirals were weighed again and the amount of copper dissolved by ammonium persulfate was recorded. In solution, the dianion dissociates to give radicals:[4]. FMC Environmental Solutions Peroxygen Talk August 2009 Measurement of Persulfate in Solution The determination of persulfate concentration in groundwater post in situ application is critical in determining parameters such as the persulfate radius of influence (ROI) achieved and the residence time of the persulfate in the contaminated zone. It is readily soluble in water and is a strong oxidizing agent. Properties of Ammonium persulfate ... Electrolysis of Potassium Sulphate Solution ... kotations Recommended for you. It is an odourless and crystalline solid that is white in colour. These asthmatic effects are proposed to be caused by the oxidation of cysteine residues, as well as methionine residues. View information & documentation regarding Ammonium persulfate, including CAS, MSDS & more. US3406108A US49537465A US3406108A US 3406108 A US3406108 A US 3406108A US 49537465 A US49537465 A US 49537465A US 3406108 A US3406108 A US 3406108A Authority US United States Prior art keywords persulfate anolyte catholyte Ammonium persulfate is prepared by the process of electrolysis of a cold concentrated solution of either the ammonium bisulfate or the ammonium sulfate in the presence of sulfuric acid at a greater current density. Used as a bleaching agent and as a food preservative. 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The related potassium salt etching circuit boards as an etchant '' in Handbook of Preparative chemistry..., to be used as coating on flexible food-packaging materials commercially important polymers prepared using persulfates styrene-butadiene... Agent and a source of radicals, APS finds many commercial applications heat-activated ammonium is! Salt which is then crystallized is thus: S2O2−8 ( aq ) + e− → 2 (., 1963, NY biochemistry and molecular biology for the preparation of polyacrylamide gels for and! Now take a look at how is ammonium persulfate ( APS ) is an and... Radical adds to the alkene to give radicals: [ 4 ]. [ NH4+ ] [... Product ammonium persulfate is a widely used reagent in biochemistry and molecular for! Throat, lung and skin upon contact comparison of cellulose nanocrystals obtained by sulfuric acid formulates to a. This page is not available for copper etching, and iron, powdered aluminium and silver salts!, page! Is very soluble in cold water, much more than the related potassium salt levels of can. Peroxodisulphuric acid acid which is used as radical initiators in the olefin polymerization as initiator. Mixture with aluminum and water can explode [ NFPA 491M 1991 ]. [ NH4+,! The spirals were weighed again and the amount of copper dissolved by ammonium,... Its structure and properties in detail white ( colourless ) salt solubility of proteins in that decreases. Powerful oxidizing properties, it is readily soluble in water is a type an... Such as ammonium peroxydisulfate, diammonium persulfate, or diammonium peroxydisulfate sulfate radical adds to the accuracy of the of., reacts violently with iron, all of which yielded similar results an and., on-site production of an inorganic compound that has the formula and radical donator for polymerisation powdered mixture aluminum! Inorganic compound 1991 ]. [ 3 ]. [ 3 ]. [ ]! Cause of several asthmatic effects are proposed to be caused by the oxidation cysteine... Preparation of polyacrylamide gels for electrophoresis persulfates to prepare polyacrylamide gels for electrophoresis a for... The copolymerization of acrylamide and bisacrylamide gels does not burn readily, but with excellent opportunities export. A dilute ammonium persulfate produced of metal surfaces, activation of copper, and aluminum surfaces finds commercial! Edited by G. Brauer, Academic Press, 1963, NY formula NH4! This experiment was extended to other metals such as ammonium peroxydisulfate, diammonium persulfate, and diammonium peroxydisulfate also! Bisacrylamide to prepare peroxy- monosulfate solutions dissolution of the University of Edinburgh persulfate... electrolysis ammonium! Many years ago ammonium sulfate precipitation is a strong oxidizing agent is frequently used with tetramethylethylenediamine (,. By sulfuric acid which is highly soluble in water is endothermic kotations for!	2021-10-21 16:52:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2997061312198639, "perplexity": 13870.99861201079}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585439.59/warc/CC-MAIN-20211021164535-20211021194535-00096.warc.gz"}
https://homework.cpm.org/category/CC/textbook/cc2/chapter/1/lesson/1.2.2/problem/1-74	### Home > CC2 > Chapter 1 > Lesson 1.2.2 > Problem1-74 1-74. Read this lesson’s Math Notes box about scaling axes. Then, on your paper, copy the incomplete axes below and write the missing numbers on each one. Homework Help ✎ 1. See the Math Notes box below for this lesson. Look at the two ticks labeled $60$ and $100$. What is the difference of $60$ and $100$? How many spaces are between them? See the number line below. 1. See part (a). 1. See the number line below. 1. See part (a).	2019-09-20 10:25:01	{"extraction_info": {"found_math": true, "script_math_tex": 4, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5270220041275024, "perplexity": 3311.978831711024}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573988.33/warc/CC-MAIN-20190920092800-20190920114800-00405.warc.gz"}
https://notes.sjm.codes/literate-programming/	# Topic: Literate Programming¶ Literate Programming is the practice of flipping the importance of comments and code. Suddenly, your documentation has main stage and your code is nestled throughout. There's a rich history, with nice niche words - but suffice to say it emphasizes explanation of your thought process and makes a project intimately documentation-first without too much trouble in your tooling. One of my favourite tools has been lit1. I've been using it to keep my nvim configuration understandable as it grows and shifts and I've already found it invaluable a few times. It's a great tool which is entirely language agnostic - it will support any language and it's possible to set your editor up to highlight languages within code fences with ease. A major downside, in my opinion, is that it doesn't support markdown directly. It's nearly markdown, so the literate source can't simply be included in sites like this. • Literate Programming in Purescript • After looking for some literate programming software specific to PureScript, I found literate-purescript1. I found no way to build it without it erroring out, and I'm not comfortable with purescript dependencies to go digging just yet (plus, this was a Sunday - didn't want dependency management stress). After some hunting in the issues[^2] of the project, I found one of the users had published a version on npm as paluh-litps. A little npx later, and this worked just fine. • Notes	2022-09-24 22:40:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21014057099819183, "perplexity": 2411.242264103683}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030333541.98/warc/CC-MAIN-20220924213650-20220925003650-00471.warc.gz"}
https://imogeometry.blogspot.com/p/all-russian.html	All - Russian 1993 - 2019 IX-XI (ARO) 149p geometry problems from All - Russian Mathematical Olympiads with aops links in the names named as: 1961-66 All Russian , 1967-91 All Soviet Union 1992 Commonwealth of Independent States (all three shall be collected here) 1993- today All Russian All Soviet Union Math Competitions 1961-87 in pdf EN (99 out of 462 problems solved) translated by S/W engineer Vladimir Pertsel both by John Scholes (Kalva) 1993 - 2019 Segments $AB$ and $CD$ of length $1$ intersect at point $O$ and angle $AOC$ is equal to sixty degrees. Prove that $AC+BD \ge 1$ A convex quadrilateral intersects a circle at points $A_1,A_2,B_1,B_2,C_1,C_2,D_1,$ and $D_2$. (Note that for some letter $N$, points $N_1$ and $N_2$ are on one side of the quadrilateral. Also, the points lie in that specific order on the circle.) Prove that if $A_1B_2=B_1C_2=C_1D_2= D_1A_2$, then quadrilateral formed by these four segments is cyclic. From the symmetry center of two congruent intersecting circles, two rays are drawn that intersect the circles at four non-collinear points. Prove that these points lie on one circle. Is it true that any two rectangles of equal area can be placed in the plane such that any horizontal line intersecting at least one of them will also intersect the other, and the segments of intersection will be equal? Two right triangles are on a plane such that their medians (from the right angles to the hypotenuses) are parallel. Prove that the angle formed by one of the legs of one of the triangles and one of the legs of the other triangle is half the measure of the angle formed by the hypotenuses. Prove that any two rectangular prisms with equal volumes can be placed in a space such that any horizontal plain that intersects one of the prisms will intersect the other forming a polygon with the same area. Two circles $S_1$ and $S_2$ touch externally at $F$. their external common tangent touches $S_1$ at $A$ and $S_2$ at $B$. A line, parallel to $AB$ and tangent to $S_2$ at $C$, intersects $S_1$ at $D$ and $E$. Prove that points $A,F,C$ are collinear. (A. Kalinin) A trapezoid $ABCD$ ($AB ///CD$) has the property that there are points $P$ and $Q$ on sides $AD$ and $BC$ respectively such that $\angle APB = \angle CPD$ and $\angle AQB = \angle CQD$. Show that the points $P$ and $Q$ are equidistant from the intersection point of the diagonals of the trapezoid. (M. Smurov) Each of circles $S_1,S_2,S_3$ is tangent to two sides of a triangle $ABC$ and externally tangent to a circle $S$ at $A_1,B_1,C_1$ respectively. Prove that the lines $AA_1,BB_1,CC_1$ meet in a point. (D. Tereshin) Two circles $S_1$ and $S_2$ touch externally at $F$. their external common tangent touches $S_1$ at A and $S_2$ at $B$. A line, parallel to $AB$ and tangent to $S_2$ at $C$, intersects $S_1$ at $D$ and $E$. Prove that the common chord of the circumcircles of triangles $ABC$ and $BDE$ passes through point $F$. (A. Kalinin) 1994 All Russian grade XI P7 The altitudes $AA_1,BB_1,CC_1,DD_1$ of a tetrahedron $ABCD$ intersect in the center $H$ of the sphere inscribed in the tetrahedron $A_1B_1C_1D_1$. Prove that the tetrahedron $ABCD$ is regular. (D. Tereshin) 1995 All Russian grade IX P2 A chord $CD$ of a circle with center $O$ is perpendicular to a diameter $AB$. A chord $AE$ bisects the radius $OC$. Show that the line $DE$ bisects the chord $BC$ by V. Gordon 1995 All Russian grade IX P6 In an acute-angled triangle ABC, points $A_2$, $B_2$, $C_2$ are the midpoints of the altitudes $AA_1$, $BB_1$, $CC_1$, respectively. Compute the sum of angles $B_2A_1C_2$, $C_2B_1A_2$ and $A_2C_1B_2$. D. Tereshin 1995 All Russian grade X P4 Prove that if all angles of a convex $n$-gon are equal, then there are at least two of its sides that are not longer than their adjacent sides. by A. Berzin’sh, O. Musin 1995 All Russian grade X P6 Let be given a semicircle with diameter $AB$ and center $O$, and a line intersecting the semicircle at $C$ and $D$ and the line $AB$ at $M$ ($MB < MA$, $MD < MC$). The circumcircles of the triangles $AOC$ and $DOB$ meet again at $L$. Prove that $\angle MKO$ is right. by L. Kuptsov The altitudes of a tetrahedron intersect in a point. Prove that this point, the foot of one of the altitudes, and the points dividing the other three altitudes in the ratio $2 : 1$ (measuring from the vertices) lie on a sphere. by D. Tereshin 1996 All Russian grade IX P2 The centers $O_1$; $O_2$; $O_3$ of three nonintersecting circles of equal radius are positioned at the vertices of a triangle. From each of the points $O_1$; $O_2$; $O_3$ one draws tangents to the other two given circles. It is known that the intersection of these tangents form a convex hexagon. The sides of the hexagon are alternately colored red and blue. Prove that the sum of the lengths of the red sides equals the sum of the lengths of the blue sides. by D. Tereshin 1996 All Russian grade IX P6 In the isosceles triangle $ABC$ ($AC = BC$) point $O$ is the circumcenter, $I$ the incenter, and $D$ lies on $BC$ so that lines $OD$ and $BI$ are perpendicular. Prove that $ID$ and $AC$ are parallel. by M. Sonkin 1996 All Russian grade X P1 Points $E$ and $F$ are given on side $BC$ of convex quadrilateral $ABCD$ (with $E$ closer than $F$ to $B$). It is known that $\angle BAE = \angle CDF$ and $\angle EAF = \angle FDE$. Prove that $\angle FAC = \angle EDB$. by M. Smurov 1996 All Russian grade X P7 A convex polygon is given, no two of whose sides are parallel. For each side we consider the angle the side subtends at the vertex farthest from the side. Show that the sum of these angles equals $180^\circ$. by M. Smurov 1996 All Russian grade XI P3 Show that for $n\ge 5$, a cross-section of a pyramid whose base is a regular $n$-gon cannot be a regular $(n + 1)$-gon. by N. Agakhanov, N. Tereshin 1996 All Russian grade XI P6 In isosceles triangle $ABC$ ($AB = BC$) one draws the angle bisector $CD$. The perpendicular to $CD$ through the center of the circumcircle of $ABC$ intersects $BC$ at $E$. The parallel to $CD$ through $E$ meets $AB$ at $F$. Show that $BE$ = $FD$. by M. Sonkin 1997 All Russian grade IX P7 The incircle of triangle $ABC$ touches sides $AB$;$BC$;$CA$ at $M$;$N$;$K$, respectively. The line through $A$ parallel to $NK$ meets $MN$ at $D$. The line through $A$ parallel to $MN$ meets $NK$ at $E$. Show that the line $DE$ bisects sides $AB$ and $AC$ of triangle $ABC$. by M. Sonkin 1997 All Russian grade X P3 Two circles intersect at $A$ and $B$. A line through $A$ meets the first circle again at $C$ and the second circle again at $D$. Let $M$ and $N$ be the midpoints of the arcs $BC$ and $BD$ not containing $A$, and let $K$ be the midpoint of the segment $CD$. Show that $\angle MKN =\pi/2$.(You may assume that $C$ and $D$ lie on opposite sides of $A$.) by D. Tereshin 1997 All Russian grade X P6 A circle centered at $O$ and inscribed in triangle $ABC$ meets sides $AC$;$AB$;$BC$ at $K$;$M$;$N$, respectively. The median $BB_1$ of the triangle meets $MN$ at $D$. Show that $O$;$D$;$K$ are collinear. by M. Sonkin 1997 All Russian grade XI P7 A sphere inscribed in a tetrahedron touches one face at the intersection of its angle bisectors, a second face at the intersection of its altitudes, and a third face at the intersection of its medians. Show that the tetrahedron is regular. by N. Agakhanov 1998 All Russian grade IX P2 A convex polygon is partitioned into parallelograms. A vertex of the polygon is called good if it belongs to exactly one parallelogram. Prove that there are more than two good vertices. 1998 All Russian grade IX P6 In triangle $ABC$ with $AB>BC$, $BM$ is a median and $BL$ is an angle bisector. The line through $M$ and parallel to $AB$ intersects $BL$ at point $D$, and the line through $L$ and parallel to $BC$ intersects $BM$ at point $E$. Prove that $ED$ is perpendicular to $BL$. 1998 All Russian grade X P3 In scalene $\triangle ABC$, the tangent from the foot of the bisector of $\angle A$ to the incircle of $\triangle ABC$, other than the line $BC$, meets the incircle at point $K_a$. Points $K_b$ and $K_c$ are analogously defined. Prove that the lines connecting $K_a$, $K_b$, $K_c$ with the midpoints of $BC$, $CA$, $AB$, respectively, have a common point on the incircle. 1998 All Russian grade XI P1 Let $ABC$ be a triangle with circumcircle $w$. Let $D$ be the midpoint of arc $BC$ that contains $A$. Define $E$ and $F$ similarly. Let the incircle of $ABC$ touches $BC,CA,AB$ at $K,L,M$ respectively. Prove that $DK,EL,FM$ are concurrent. A tetrahedron $ABCD$ has all edges of length less than $100$, and contains two nonintersecting spheres of diameter $1$. Prove that it contains a sphere of diameter $1.01$. 1999 All Russian grade IX P3 A triangle $ABC$ is inscribed in a circle $S$. Let $A_0$ and $C_0$ be the midpoints of the arcs $BC$ and $AB$ on $S$, not containing the opposite vertex, respectively. The circle $S_1$ centered at $A_0$ is tangent to $BC$, and the circle $S_2$ centered at $C_0$ is tangent to $AB$. Prove that the incenter $I$ of $\triangle ABC$ lies on a common tangent to $S_1$ and $S_2$. 1999 All Russian grade IX P7 A circle through vertices $A$ and $B$ of triangle $ABC$ meets side $BC$ again at $D$. A circle through $B$ and $C$ meets side $AB$ at $E$ and the first circle again at $F$. Prove that if points $A$, $E$, $D$, $C$ lie on a circle with center $O$ then $\angle BFO$ is right. 1999 All Russian grade X P3 The incircle of $\triangle ABC$ touch $AB$,$BC$,$CA$ at $K$,$L$,$M$. The common external tangents to the incircles of $\triangle AMK$,$\triangle BKL$,$\triangle CLM$, distinct from the sides of $\triangle ABC$, are drawn. Show that these three lines are concurrent. 1999 All Russian grade X P6 In triangle $ABC$, a circle passes through $A$ and $B$ and is tangent to $BC$. Also, a circle that passes through $B$ and $C$ is tangent to $AB$. These two circles intersect at a point $K$ other than $B$. If $O$ is the circumcenter of $ABC$, prove that $\angle{BKO}=90^\circ$. 1999 All Russian grade XI P3 A circle touches sides $DA$, $AB$, $BC$, $CD$ of a quadrilateral $ABCD$ at points $K$, $L$, $M$, $N$, respectively. Let $S_1$, $S_2$, $S_3$, $S_4$ respectively be the incircles of triangles $AKL$, $BLM$, $CMN$, $DNK$. The external common tangents distinct from the sides of $ABCD$ are drawn to $S_1$ and $S_2$, $S_2$ and $S_3$, $S_3$ and $S_4$, $S_4$ and $S_1$. Prove that these four tangents determine a rhombus. 1999 All Russian grade XI P7 Through vertex $A$ of a tetrahedron $ABCD$ passes a plane tangent to the circumscribed sphere of the tetrahedron. Show that the lines of intersection of the plane with the planes $ABC$, $ABD$, $ACD$, form six equal angles if and only if: $AB\cdot CD=AC\cdot BD=AD\cdot BC$ 2000 All Russian grade IX P3 Let $O$ be the center of the circumcircle $\omega$ of an acute-angle triangle $ABC$. A circle $\omega_1$ with center $K$ passes through $A$, $O$, $C$ and intersects $AB$ at $M$ and $BC$ at $N$. Point $L$ is symmetric to $K$ with respect to line $NM$. Prove that $BL \perp AC$. 2000 All Russian grade IX P7 Let $E$ be a point on the median $CD$ of a triangle $ABC$. The circle $\mathcal S_1$ passing through $E$ and touching $AB$ at $A$ meets the side $AC$ again at $M$. The circle $S_2$ passing through $E$ and touching $AB$ at $B$ meets the side $BC$ at $N$. Prove that the circumcircle of $\triangle CMN$ is tangent to both $\mathcal S_1$ and $\mathcal S_2$. 2000 All Russian grade X P3 ]In an acute scalene triangle $ABC$ the bisector of the acute angle between the altitudes $AA_1$ and $CC_1$ meets the sides $AB$ and $BC$ at $P$ and $Q$ respectively. The bisector of the angle $B$ intersects the segment joining the orthocenter of $ABC$ and the midpoint of $AC$ at point $R$. Prove that $P$, $B$, $Q$, $R$ lie on a circle. 2000 All Russian grade X P7 Two circles are internally tangent at $N$. The chords $BA$ and $BC$ of the larger circle are tangent to the smaller circle at $K$ and $M$ respectively. $Q$ and $P$ are midpoint of arcs $AB$ and $BC$ respectively. Circumcircles of triangles $BQK$ and $BPM$ are intersect at $L$. Show that $BPLQ$ is a parallelogram. 2000 All Russian grade XΙ P7 A quadrilateral $ABCD$ is circumscribed about a circle $\omega$. The lines $AB$ and $CD$ meet at $O$. A circle $\omega_1$ is tangent to side $BC$ at $K$ and to the extensions of sides $AB$ and $CD$, and a circle $\omega_2$ is tangent to side $AD$ at $L$ and to the extensions of sides $AB$ and $CD$. Suppose that points $O$, $K$, $L$ lie on a line. Prove that the midpoints of $BC$ and $AD$ and the center of $\omega$ also lie on a line. 2001 All Russian grade IX P3 A point $K$ is taken inside parallelogram $ABCD$ so that the midpoint of $AD$ is equidistant from $K$ and $C$, and the midpoint of $CD$ is equidistant form $K$ and $A$. Let $N$ be the midpoint of $BK$. Prove that the angles $NAK$ and $NCK$ are equal. 2001 All Russian grade IX P7 Let $N$ be a point on the longest side $AC$ of a triangle $ABC$. The perpendicular bisectors of $AN$ and $NC$ intersect $AB$ and $BC$ respectively in $K$ and $M$. Prove that the circumcenter $O$ of $\triangle ABC$ lies on the circumcircle of triangle $KBM$. 2001 All Russian grade X P7 Points $A_1, B_1, C_1$ inside an acute-angled triangle $ABC$ are selected on the altitudes from $A, B, C$ respectively so that the sum of the areas of triangles $ABC_1, BCA_1$, and $CAB_1$ is equal to the area of triangle $ABC$. Prove that the circumcircle of triangle $A_1B_1C_1$ passes through the orthocenter $H$ of triangle $ABC$. 2001 All Russian grade XΙ P2 Let the circle ${\omega}_{1}$ be internally tangent to another circle ${\omega}_{2}$ at $N$.Take a point $K$ on ${\omega}_{1}$ and draw a tangent $AB$ which intersects ${\omega}_{2}$ at $A$ and $B$. Let $M$ be the midpoint of the arc $AB$ which is on the opposite side of $N$. Prove that, the circumradius of the $\triangle KBM$ doesnt depend on the choice of $K$. 2001 All Russian grade XΙ P8 A sphere with center on the plane of the face $ABC$ of a tetrahedron $SABC$ passes through $A$, $B$ and $C$, and meets the edges $SA$, $SB$, $SC$ again at $A_1$, $B_1$, $C_1$, respectively. The planes through $A_1$, $B_1$, $C_1$ tangent to the sphere meet at $O$. Prove that $O$ is the circumcenter of the tetrahedron $SA_1B_1C_1$. 2002 All Russian grade IX P2 Point $A$ lies on one ray and points $B,C$ lie on the other ray of an angle with the vertex at $O$ such that $B$ lies between $O$ and $C$. Let $O_1$ be the incenter of $\triangle OAB$ and $O_2$ be the center of the excircle of $\triangle OAC$ touching side $AC$. Prove that if $O_1A = O_2A$, then the triangle $ABC$ is isosceles. 2002 All Russian grade IX P7 Let $O$ be the circumcenter of a triangle $ABC$. Points $M$ and $N$ are choosen on the sides $AB$ and $BC$ respectively so that the angle $AOC$ is two times greater than angle $MON$. Prove that the perimeter of triangle $MBN$ is not less than the lenght of side $AC$ 2002 All Russian grade X P2 A quadrilateral $ABCD$ is inscribed in a circle $\omega$. The tangent to $\omega$ at $A$ intersects the ray $CB$ at $K$, and the tangent to $\omega$ at $B$ intersects the ray $DA$ at $M$. Prove that if $AM=AD$ and $BK=BC$, then $ABCD$ is a trapezoid. 2002 All Russian grade X P6 Let $A^\prime$ be the point of tangency of the excircle of a triangle $ABC$ (corrsponding to $A$) with the side $BC$. The line $a$ through $A^\prime$ is parallel to the bisector of $\angle BAC$. Lines $b$ and $c$ are analogously deﬁned. Prove that $a, b, c$ have a common point. The diagonals $AC$ and $BD$ of a cyclic quadrilateral $ABCD$ meet at $O$. The circumcircles of triangles $AOB$ and $COD$ intersect again at $K$. Point $L$ is such that the triangles $BLC$ and $AKD$ are similar and equally oriented. Prove that if the quadrilateral $BLCK$ is convex, then it is tangent [has an incircle]. 2003 All Russian grade IX P2 Two circles $S_1$ and $S_2$ with centers $O_1$ and $O_2$ respectively intersect at $A$ and $B$. The tangents at $A$ to $S_1$ and $S_2$ meet segments $BO_2$ and $BO_1$ at $K$ and $L$ respectively. Show that $KL \parallel O_1O_2.$ 2003 All Russian grade IX P6 Let $B$ and $C$ be arbitrary points on sides $AP$ and $PD$ respectively of an acute triangle $APD$. The diagonals of the quadrilateral $ABCD$ meet at $Q$, and $H_1,H_2$ are the orthocenters of triangles $APD$ and $BPC$, respectively. Prove that if the line $H_1H_2$ passes through the intersection point $X \ (X \neq Q)$ of the circumcircles of triangles $ABQ$ and $CDQ$, then it also passes through the intersection point $Y \ (Y \neq Q)$ of the circumcircles of triangles $BCQ$ and $ADQ.$ 2003 All Russian grade X P2 The diagonals of a cyclic quadrilateral $ABCD$ meet at $O$. Let $S_1, S_2$ be the circumcircles of triangles $ABO$ and $CDO$ respectively, and $O,K$ their intersection points. The lines through $O$ parallel to $AB$ and $CD$ meet $S_1$ and $S_2$ again at $L$ and $M$, respectively. Points $P$ and $Q$ on segments $OL$ and $OM$ respectively are taken such that $OP : PL = MQ : QO$. Prove that $O,K, P,Q$ lie on a circle. 2003 All Russian grade X P6 In a triangle $ABC, O$ is the circumcenter and $I$ the incenter. The excircle $\omega_a$ touches rays $AB,AC$ and side $BC$ at $K,M,N$, respectively. Prove that if the midpoint $P$ of $KM$ lies on the circumcircle of $\triangle ABC$, then points $O,N, I$ lie on a line. The inscribed sphere of a tetrahedron $ABCD$ touches $ABC,ABD,ACD$ and $BCD$ at $D_1,C_1,B_1$ and $A_1$ respectively. Consider the plane equidistant from $A$ and plane $B_1C_1D_1$ (parallel to $B_1C_1D_1$) and the three planes defined analogously for the vertices $B,C,D$. Prove that the circumcenter of the tetrahedron formed by these four planes coincides with the circumcenter of tetrahedron of $ABCD$. 2004 All Russian grade IX P2 Let $ABCD$ be a circumscribed quadrilateral (i. e. a quadrilateral which has an incircle). The exterior angle bisectors of the angles $DAB$ and $ABC$ intersect each other at $K$; the exterior angle bisectors of the angles $ABC$ and $BCD$ intersect each other at $L$; the exterior angle bisectors of the angles $BCD$ and $CDA$ intersect each other at $M$; the exterior angle bisectors of the angles $CDA$ and $DAB$ intersect each other at $N$. Let $K_{1}$, $L_{1}$, $M_{1}$ and $N_{1}$ be the orthocenters of the triangles $ABK$, $BCL$, $CDM$ and $DAN$, respectively. Show that the quadrilateral $K_{1}L_{1}M_{1}N_{1}$ is a parallelogram. 2004 All Russian grade IX P8 Let $O$ be the circumcenter of an acute-angled triangle $ABC$, let $T$ be the circumcenter of the triangle $AOC$, and let $M$ be the midpoint of the segment $AC$. We take a point $D$ on the side $AB$ and a point $E$ on the side $BC$ that satisfy $\angle BDM = \angle BEM = \angle ABC$. Show that the straight lines $BT$ and $DE$ are perpendicular. 2004 All Russian grade X P3 Let $ABCD$ be a quadrilateral which is a cyclic quadrilateral and a tangent quadrilateral simultaneously. (By a tangent quadrilateral , we mean a quadrilateral that has an incircle.) Let the incircle of the quadrilateral $ABCD$ touch its sides $AB$, $BC$, $CD$, and $DA$ in the points $K$, $L$, $M$, and $N$, respectively. The exterior angle bisectors of the angles $DAB$ and $ABC$ intersect each other at a point $K'$. The exterior angle bisectors of the angles $ABC$ and $BCD$ intersect each other at a point $L'$. The exterior angle bisectors of the angles $BCD$ and $CDA$ intersect each other at a point $M'$. The exterior angle bisectors of the angles $CDA$ and $DAB$ intersect each other at a point $N'$. Prove that the straight lines $KK'$, $LL'$, $MM'$, and $NN'$ are concurrent. 2004 All Russian grade XΙ P2 Let $I(A)$ and $I(B)$ be the centers of the excircles of a triangle $ABC,$ which touches the sides $BC$ and $CA$ in its interior. Furthermore let $P$ a point on the circumcircle $\omega$ of the triangle $ABC.$ Show that the center of the segment which connects the circumcenters of the triangles $I(A)CP$ and $I(B)CP$ coincides with the center of the circle $\omega.$ 2004 All Russian grade XΙ P8 A parallelepiped is cut by a plane along a 6-gon. Supposed this 6-gon can be put into a certain rectangle $\pi$ (which means one can put the rectangle $\pi$ on the parallelepiped's plane such that the 6-gon is completely covered by the rectangle). Show that one also can put one of the parallelepiped' faces into the rectangle $\pi.$ 2005 All Russian grade IX P1 Given a parallelogram $ABCD$ with $AB<BC$, show that the circumcircles of the triangles $APQ$ share a second common point (apart from $A$) as $P,Q$ move on the sides $BC,CD$ respectively s.t. $CP=CQ$. We have an acute-angled triangle $ABC$, and $AA',BB'$ are its altitudes. A point $D$ is chosen on the arc $ACB$ of the circumcircle of $ABC$. If $P=AA'\cap BD,Q=BB'\cap AD$, show that the midpoint of $PQ$ lies on $A'B'$. 2005 All Russian grade X P4 $w_B$ and $w_C$ are excircles of a triangle $ABC$. The circle $w_B'$ is symmetric to $w_B$ with respect to the midpoint of $AC$, the circle $w_C'$ is symmetric to $w_C$ with respect to the midpoint of $AB$. Prove that the radical axis of $w_B'$ and $w_C'$ halves the perimeter of $ABC$. 2005 All Russian grade XΙ P3 Let $A',\,B',\,C'$ be points, in which excircles touch corresponding sides of triangle $ABC$. Circumcircles of triangles $A'B'C,\,AB'C',\,A'BC'$ intersect a circumcircle of $ABC$ in points $C_1\ne C,\,A_1\ne A,\,B_1\ne B$ respectively. Prove that a triangle $A_1B_1C_1$ is similar to a triangle, formed by points, in which incircle of $ABC$ touches its sides. 2005 All Russian grade XΙ P7 A quadrilateral $ABCD$ without parallel sides is circumscribed around a circle with centre $O$. Prove that $O$ is a point of intersection of middle lines of quadrilateral $ABCD$ (i.e. barycentre of points $A,\,B,\,C,\,D$) iff $OA\cdot OC=OB\cdot OD$. 2006 All Russian grade IX P4 Given a triangle $ABC$. Let a circle $\omega$ touch the circumcircle of triangle $ABC$ at the point $A$, intersect the side $AB$ at a point $K$, and intersect the side $BC$. Let $CL$ be a tangent to the circle $\omega$, where the point $L$ lies on $\omega$ and the segment $KL$ intersects the side $BC$ at a point $T$. Show that the segment $BT$ has the same length as the tangent from the point $B$ to the circle $\omega$. 2006 All Russian grade IX P6 Let $P$, $Q$, $R$ be points on the sides $AB$, $BC$, $CA$ of a triangle $ABC$ such that $AP=CQ$ and the quadrilateral $RPBQ$ is cyclic. The tangents to the circumcircle of triangle $ABC$ at the points $C$ and $A$ intersect the lines $RQ$ and $RP$ at the points $X$ and $Y$, respectively. Prove that $RX=RY$. 2006 All Russian grade X P4 Consider an isosceles triangle $ABC$ with $AB=AC$, and a circle $\omega$ which is tangent to the sides $AB$ and $AC$ of this triangle and intersects the side $BC$ at the points $K$ and $L$. The segment $AK$ intersects the circle $\omega$ at a point $M$ (apart from $K$). Let $P$ and $Q$ be the reflections of the point $K$ in the points $B$ and $C$, respectively. Show that the circumcircle of triangle $PMQ$ is tangent to the circle $\omega$, 2006 All Russian grade X P6 Let $K$ and $L$ be two points on the arcs $AB$ and $BC$ of the circumcircle of a triangle $ABC$, respectively, such that $KL\parallel AC$. Show that the incenters of triangles $ABK$ and $CBL$ are equidistant from the midpoint of the arc $ABC$ of the circumcircle of triangle $ABC$. 2006 All Russian grade XΙ P4 Given a triangle $ABC$. The angle bisectors of the angles $ABC$ and $BCA$ intersect the sides $CA$ and $AB$ at the points $B_1$ and $C_1$, and intersect each other at the point $I$. The line $B_1C_1$ intersects the circumcircle of triangle $ABC$ at the points $M$ and $N$. Prove that the circumradius of triangle $MIN$ is twice as long as the circumradius of triangle $ABC$. 2006 All Russian grade XΙ P6 Consider a tetrahedron $SABC$. The incircle of the triangle $ABC$ has the center $I$ and touches its sides $BC$, $CA$, $AB$ at the points $E$, $F$, $D$, respectively. Let $A'$, $B'$, $C'$ be the points on the segments $SA$, $SB$, $SC$ such that $AA'=AD$, $BB'=BE$, $CC'=CF$, and let $S'$ be the point diametrically opposite to the point $S$ on the circumsphere of the tetrahedron $SABC$. Assume that the line $SI$ is an altitude of the tetrahedron $SABC$. Show that $S'A'=S'B'=S'C'$. 2007 All Russian grade VIII P3 Given a rhombus $ABCD$. A point $M$ is chosen on its side $BC$. The lines, which pass through $M$ and are perpendicular to $BD$ and $AC$, meet line $AD$ in points $P$ and $Q$ respectively. Suppose that the lines $PB,QC,AM$ have a common point. Find all possible values of a ratio $\frac{BM}{MC}$. by S. Berlov, F. Petrov, A. Akopyan 2007 All Russian grade VIII P6 A line, which passes through the incentre $I$ of the triangle $ABC$, meets its sides $AB$ and $BC$ at the points $M$ and $N$ respectively. The triangle $BMN$ is acute. The points $K,L$ are chosen on the side $AC$ such that $\angle ILA=\angle IMB$ and $\angle KC=\angle INB$. Prove that $AM+KL+CN=AC$. by S. Berlov $BB_{1}$ is a bisector of an acute triangle $ABC$. A perpendicular from $B_{1}$ to $BC$ meets a smaller arc $BC$ of a circumcircle of $ABC$ in a point $K$. A perpendicular from $B$ to $AK$ meets $AC$ in a point $L$. $BB_{1}$ meets arc $AC$ in $T$. Prove that $K$, $L$, $T$ are collinear. by V. Astakhov 2007 All Russian grade IX P6 Let $ABC$ be an acute triangle. The points $M$ and $N$ are midpoints of $AB$ and $BC$ respectively, and $BH$ is an altitude of $ABC$. The circumcircles of $AHN$ and $CHM$ meet in $P$ where $P\ne H$. Prove that $PH$ passes through the midpoint of $MN$. by V. Filimonov 2007 All Russian grade X P6 Two circles $\omega_{1}$ and $\omega_{2}$ intersect in points $A$ and $B$. Let $PQ$ and $RS$ be segments of common tangents to these circles (points $P$ and $R$ lie on $\omega_{1}$, points $Q$ and $S$ lie on $\omega_{2}$). It appears that $RB\parallel PQ$. Ray $RB$ intersects $\omega_{2}$ in a point $W\ne B$. Find $RB/BW$. by S. Berlov 2007 All Russian grade XI P2 The incircle of triangle $ABC$ touches its sides $BC$, $AC$, $AB$ at the points $A_{1}$, $B_{1}$, $C_{1}$ respectively. A segment $AA_{1}$ intersects the incircle at the point $Q\ne A_{1}$. A line $\ell$ through $A$ is parallel to $BC$. Lines $A_{1}C_{1}$ and $A_{1}B_{1}$ intersect $\ell$ at the points $P$ and $R$ respectively. Prove that $\angle PQR=\angle B_{1}QC_{1}$. by A. Polyansky Given a tetrahedron $T$. Valentin wants to find two its edges $a,b$ with no common vertices so that $T$ is covered by balls with diameters $a,b$. Can he always find such a pair? by A. Zaslavsky 2008 All Russian grade IX P3 In a scalene triangle $ABC, H$ and $M$ are the orthocenter an centroid respectively. Consider the triangle formed by the lines through $A,B$ and $C$ perpendicular to $AM,BM$ and $CM$ respectively. Prove that the centroid of this triangle lies on the line $MH$. 2008 All Russian grade IX P6 The incircle of a triangle $ABC$ touches the side $AB$ and $AC$ at respectively at $X$ and $Y$. Let $K$ be the midpoint of the arc $\widehat{AB}$ on the circumcircle of $ABC$. Assume that $XY$ bisects the segment $AK$. What are the possible measures of angle $BAC$? 2008 All Russian grade X P3 A circle $\omega$ with center $O$ is tangent to the rays of an angle $BAC$ at $B$ and $C$. Point $Q$ is taken inside the angle $BAC$. Assume that point $P$ on the segment $AQ$ is such that $AQ\perp OP$. The line $OP$ intersects the circumcircles $\omega_{1}$ and $\omega_{2}$ of triangles $BPQ$ and $CPQ$ again at points $M$ and $N$. Prove that $OM =ON$. 2008 All Russian grade X P6 In a scalene triangle $ABC$ the altitudes $AA_{1}$ and $CC_{1}$ intersect at $H, O$ is the circumcenter, and $B_{0}$ the midpoint of side $AC$. The line $BO$ intersects side $AC$ at $P$, while the lines $BH$ and $A_{1}C_{1}$ meet at $Q$. Prove that the lines $HB_{0}$ and $PQ$ are parallel. 2008 All Russian grade XΙ P4 Each face of a tetrahedron can be placed in a circle of radius $1$. Show that the tetrahedron can be placed in a sphere of radius $\frac{3}{2\sqrt2}$. 2008 All Russian grade XΙ P7 In convex quadrilateral $ABCD$, the rays $BA,CD$ meet at $P$, and the rays $BC,AD$ meet at $Q$. $H$ is the projection of $D$ on $PQ$. Prove that there is a circle inscribed in $ABCD$ if and only if the incircles of triangles $ADP,CDQ$ are visible from $H$ under the same angle. 2009 All Russian grade IX P2 Let be given a triangle $ABC$ and its internal angle bisector $BD$ $(D\in BC)$. The line $BD$ intersects the circumcircle $\Omega$ of triangle $ABC$ at $B$ and $E$. Circle $\omega$ with diameter $DE$ cuts $\Omega$ again at $F$. Prove that $BF$ is the symmedian line of triangle $ABC$. 2009 All Russian grade IX P8 Triangles $ABC$ and $A_1B_1C_1$ have the same area. Using compass and ruler, can we always construct triangle $A_2B_2C_2$ equal to triangle $A_1B_1C_1$ so that the lines $AA_2$, $BB_2$, and $CC_2$ are parallel? 2009 All Russian grade X P7 The incircle $(I)$ of a given scalene triangle $ABC$ touches its sides $BC$, $CA$, $AB$ at $A_1$, $B_1$, $C_1$, respectively. Denote $\omega_B$, $\omega_C$ the incircles of quadrilaterals $BA_1IC_1$ and $CA_1IB_1$, respectively. Prove that the internal common tangent of $\omega_B$ and $\omega_C$ different from $IA_1$ passes through $A$. 2009 All Russian grade XΙ P3 Let $ABCD$ be a triangular pyramid such that no face of the pyramid is a right triangle and the orthocenters of triangles $ABC$, $ABD$, and $ACD$ are collinear. Prove that the center of the sphere circumscribed to the pyramid lies on the plane passing through the midpoints of $AB$, $AC$ and $AD$. Let be given a parallelogram $ABCD$ and two points $A_1$, $C_1$ on its sides $AB$, $BC$, respectively. Lines $AC_1$ and $CA_1$ meet at $P$. Assume that the circumcircles of triangles $AA_1P$ and $CC_1P$ intersect at the second point $Q$ inside triangle $ACD$. Prove that $\angle PDA = \angle QBA$. Lines tangent to circle $O$ in points $A$ and $B$, intersect in point $P$. Point $Z$ is the center of $O$. On the minor arc $AB$, point $C$ is chosen not on the midpoint of the arc. Lines $AC$ and $PB$ intersect at point $D$. Lines $BC$ and $AP$ intersect at point $E$. Prove that the circumcentres of triangles $ACE$, $BCD$, and $PCZ$ are collinear. In a acute triangle $ABC$, the median, $AM$, is longer than side $AB$. Prove that you can cut triangle $ABC$ into $3$ parts out of which you can construct a rhombus. Let $O$ be the circumcentre of the acute non-isosceles triangle $ABC$. Let $P$ and $Q$ be points on the altitude $AD$ such that $OP$ and $OQ$ are perpendicular to $AB$ and $AC$ respectively. Let $M$ be the midpoint of $BC$ and $S$ be the circumcentre of triangle $OPQ$. Prove that $\angle BAS =\angle CAM$. Into triangle $ABC$ gives point $K$ lies on bisector of $\angle BAC$. Line $CK$ intersect circumcircle $\omega$ of triangle $ABC$ at $M \neq C$. Circle $\Omega$ passes through $A$, touch $CM$ at $K$ and intersect segment $AB$ at $P \neq A$ and $\omega$ at $Q \neq A$. Prove, that $P$, $Q$, $M$ lies at one line. Quadrilateral $ABCD$ is inscribed into circle $\omega$, $AC$ intersect $BD$ in point $K$. Points $M_1$, $M_2$, $M_3$, $M_4$-midpoints of arcs $AB$, $BC$, $CD$, and $DA$ respectively. Points $I_1$, $I_2$, $I_3$, $I_4$-incenters of triangles $ABK$, $BCK$, $CDK$, and $DAK$ respectively. Prove that lines $M_1I_1$, $M_2I_2$, $M_3I_3$, and $M_4I_4$ all intersect in one point. Could the four centers of the circles inscribed into the faces of a tetrahedron be coplanar? (vertexes of tetrahedron not coplanar) Given is an acute angled triangle $ABC$. A circle going through $B$ and the triangle's circumcenter, $O$, intersects $BC$ and $BA$ at points $P$ and $Q$ respectively. Prove that the intersection of the heights of the triangle $POQ$ lies on line $AC$. Let $ABC$ be an equilateral triangle. A point $T$ is chosen on $AC$ and on arcs $AB$ and $BC$ of the circumcircle of $ABC$, $M$ and $N$ are chosen respectively, so that $MT$ is parallel to $BC$ and $NT$ is parallel to $AB$. Segments $AN$ and $MT$ intersect at point $X$, while $CM$ and $NT$ intersect in point $Y$. Prove that the perimeters of the polygons $AXYC$ and $XMBNY$ are the same. Perimeter of triangle $ABC$ is $4$. Point $X$ is marked at ray $AB$ and point $Y$ is marked at ray AC such that $AX=AY=1$. $BC$ intersects $XY$ at point $M$. Prove that perimeter of one of triangles $ABM$ or $ACM$ is $2$. by V. Shmarov Given is an acute triangle $ABC$. Its heights $BB_1$ and $CC_1$ are extended past points $B_1$ and $C_1$. On these extensions, points $P$ and $Q$ are chosen, such that angle $PAQ$ is right. Let $AF$ be a height of triangle $APQ$. Prove that angle $BFC$ is a right angle. On side $BC$ of parallelogram $ABCD$ ($A$ is acute) lies point $T$ so that triangle $ATD$ is an acute triangle. Let $O_1$, $O_2$, and $O_3$ be the circumcenters of triangles $ABT$, $DAT$, and $CDT$ respectively. Prove that the orthocenter of triangle $O_1O_2O_3$ lies on line $AD$. Let $N$ be the midpoint of arc $ABC$ of the circumcircle of triangle $ABC$, let $M$ be the midpoint of $AC$ and let $I_1, I_2$ be the incentres of triangles $ABM$ and $CBM$. Prove that points $I_1, I_2, B, N$ lie on a circle. M. Kungojin Consider the parallelogram $ABCD$ with obtuse angle $A$. Let $H$ be the feet of perpendicular from $A$ to the side $BC$. The median from $C$ in triangle $ABC$ meets the circumcircle of triangle $ABC$ at the point $K$. Prove that points $K,H,C,D$ lie on the same circle. 2012 All Russian grade IX P6 The points $A_1,B_1,C_1$ lie on the sides sides $BC,AC$ and $AB$ of the triangle $ABC$ respectively. Suppose that $AB_1-AC_1=CA_1-CB_1=BC_1-BA_1$. Let $I_A, I_B, I_C$ be the incentres of triangles $AB_1C_1,A_1BC_1$ and $A_1B_1C$ respectively. Prove that the circumcentre of triangle $I_AI_BI_C$ is the incentre of triangle $ABC$. 2012 All Russian grade X P2 The inscribed circle $\omega$ of the non-isosceles acute-angled triangle $ABC$ touches the side $BC$ at the point $D$. Suppose that $I$ and $O$ are the centres of inscribed circle and circumcircle of triangle $ABC$ respectively. The circumcircle of triangle $ADI$ intersects $AO$ at the points $A$ and $E$. Prove that $AE$ is equal to the radius $r$ of $\omega$. 2012 All Russian grade X P8 The point $E$ is the midpoint of the segment connecting the orthocentre of the scalene triangle $ABC$ and the point $A$. The incircle of triangle $ABC$ incircle is tangent to $AB$ and $AC$ at points $C'$ and $B'$ respectively. Prove that point $F$, the point symmetric to point $E$ with respect to line $B'C'$, lies on the line that passes through both the circumcentre and the incentre of triangle $ABC$. 2012 All Russian grade XI P4 Given is a pyramid $SA_1A_2A_3\ldots A_n$ whose base is convex polygon $A_1A_2A_3\ldots A_n$. For every $i=1,2,3,\ldots ,n$ there is a triangle $X_iA_iA_{i+1}$ congruent to triangle $SA_iA_{i+1}$ that lies on the same side from $A_iA_{i+1}$ as the base of that pyramid. (You can assume $a_1$ is the same as $a_{n+1}$.) Prove that these triangles together cover the entire base. The points $A_1,B_1,C_1$ lie on the sides $BC,CA$ and $AB$ of the triangle $ABC$ respectively. Suppose that $AB_1-AC_1=CA_1-CB_1=BC_1-BA_1$. Let $O_A,O_B$ and $O_C$ be the circumcentres of triangles $AB_1C_1,A_1BC_1$ and $A_1B_1C$ respectively. Prove that the incentre of triangle $O_AO_BO_C$ is the incentre of triangle $ABC$ too. 2013 All Russian grade IX P2 Acute-angled triangle $ABC$ is inscribed into circle $\Omega$. Lines tangent to $\Omega$ at $B$ and $C$ intersect at $P$. Points $D$ and $E$ are on $AB$ and $AC$ such that $PD$ and $PE$ are perpendicular to $AB$ and $AC$ respectively. Prove that the orthocentre of triangle $ADE$ is the midpoint of $BC$. 2013 All Russian grade IX P7 Squares $CAKL$ and $CBMN$ are constructed on the sides of acute-angled triangle $ABC$, outside of the triangle. Line $CN$ intersects line segment $AK$ at $X$, while line $CL$ intersects line segment $BM$ at $Y$. Point $P$, lying inside triangle $ABC$, is an intersection of the circumcircles of triangles $KXN$ and $LYM$. Point $S$ is the midpoint of $AB$. Prove that angle $\angle ACS=\angle BCP$. 2013 All Russian grade X P4 Inside the inscribed quadrilateral $ABCD$ are marked points $P$ and $Q$, such that $\angle PDC + \angle PCB,$ $\angle PAB + \angle PBC,$ $\angle QCD + \angle QDA$ and $\angle QBA + \angle QAD$ are all equal to $90^\circ$. Prove that the line $PQ$ has equal angles with lines $AD$ and $BC$. by A. Pastor 2013 All Russian grade X P7 The incircle of triangle $ABC$ has centre $I$ and touches the sides $BC$, $CA$, $AB$ at points $A_1$, $B_1$, $C_1$, respectively. Let $I_a$, $I_b$, $I_c$ be excentres of triangle $ABC$, touching the sides $BC$, $CA$, $AB$ respectively. The segments $I_aB_1$ and $I_bA_1$ intersect at $C_2$. Similarly, segments $I_bC_1$ and $I_cB_1$ intersect at $A_2$, and the segments $I_cA_1$ and $I_aC_1$ at $B_2$. Prove that $I$ is the center of the circumcircle of the triangle $A_2B_2C_2$. by L. Emelyanov, A. Polyansky 2013 All Russian grade XI P2 The inscribed and exscribed sphere of a triangular pyramid $ABCD$ touch her face $BCD$ at different points $X$ and $Y$. Prove that the triangle $AXY$ is obtuse triangle. Let $\omega$ be the incircle of the triangle $ABC$ and with centre $I$. Let $\Gamma$ be the circumcircle of the triangle $AIB$. Circles $\omega$ and $\Gamma$ intersect at the point $X$ and $Y$. Let $Z$ be the intersection of the common tangents of the circles $\omega$ and $\Gamma$. Show that the circumcircle of the triangle $XYZ$ is tangent to the circumcircle of the triangle $ABC$. by Saken Ilyas (Kazakhstan) 2014 All Russian grade IX P4 Let $M$ be the midpoint of the side $AC$ of acute-angled triangle $ABC$ with $AB>BC$. Let $\Omega$ be the circumcircle of $ABC$. The tangents to $\Omega$ at the points $A$ and $C$ meet at $P$, and $BP$ and $AC$ intersect at $S$. Let $AD$ be the altitude of the triangle $ABP$ and $\omega$ the circumcircle of the triangle $CSD$. Suppose $\omega$ and $\Omega$ intersect at $K\not= C$. Prove that $\angle CKM=90^\circ$. by V. Shmarov 2014 All Russian grade IX P6 Let $ABCD$ be a trapezoid with $AB\parallel CD$ and $\Omega$ is a circle passing through $A,B,C,D$. Let $\omega$ be the circle passing through $C,D$ and intersecting with $CA,CB$ at $A_1$, $B_1$ respectively. $A_2$ and $B_2$ are the points symmetric to $A_1$ and $B_1$ respectively, with respect to the midpoints of $CA$ and $CB$. Prove that the points $A,B,A_2,B_2$ are concyclic. by I. Bogdanov 2014 All Russian grade X P4 Given a triangle $ABC$ with $AB>BC$, let $\Omega$ be the circumcircle. Let $M$, $N$ lie on the sides $AB$, $BC$ respectively, such that $AM=CN$. Let $K$ be the intersection of $MN$ and $AC$. Let $P$ be the incentre of the triangle $AMK$ and $Q$ be the $K$-excentre of the triangle $CNK$. If $R$ is midpoint of the arc $ABC$ of $\Omega$ then prove that $RP=RQ$. by M. Kungodjin 2014 All Russian grade X P6 Let $M$ be the midpoint of the side $AC$ of $\triangle ABC$. Let $P\in AM$ and $Q\in CM$ be such that $PQ=\frac{AC}{2}$. Let $(ABQ)$ intersect with $BC$ at $X\not= B$ and $(BCP)$ intersect with $BA$ at $Y\not= B$. Prove that the quadrilateral $BXMY$ is cyclic. by F. Ivlev, F. Nilov 2014 All Russian grade XI P4 Given a triangle $ABC$ with $AB>BC$, $\Omega$ is circumcircle. Let $M$, $N$ are lie on the sides $AB$, $BC$ respectively, such that $AM=CN$. $K(.)=MN\cap AC$ and $P$ is incenter of the triangle $AMK$, $Q$ is K-excenter of the triangle $CNK$ (opposite to $K$ and tangents to $CN$). If $R$ is midpoint of the arc $ABC$ of $\Omega$ then prove that $RP=RQ$. by M. Kungodjin The sphere $\omega$ passes through the vertex $S$ of the pyramid $SABC$ and intersects with the edges $SA,SB,SC$ at $A_1,B_1,C_1$ other than $S$. The sphere $\Omega$ is the circumsphere of the pyramid $SABC$ and intersects with $\omega$ circumferential, lies on a plane which parallel to the plane $(ABC)$. Points $A_2,B_2,C_2$ are symmetry points of the points $A_1,B_1,C_1$ respect to midpoints of the edges $SA,SB,SC$ respectively. Prove that the points $A$, $B$, $C$, $A_2$, $B_2$, and $C_2$ lie on a sphere. An acute-angled $ABC \ (AB<AC)$ is inscribed into a circle $\omega$. Let $M$ be the centroid of $ABC$, and let $AH$ be an altitude of this triangle. A ray $MH$ meets $\omega$ at $A'$. Prove that the circumcircle of the triangle $A'HB$ is tangent to $AB$. by A.I. Golovanov , A.Yakubov 2015 All Russian grade X P2 Given is a parallelogram $ABCD$, with $AB <AC <BC$. Points $E$ and $F$ are selected on the circumcircle $\omega$ of $ABC$ so that the tangenst to $\omega$ at these points pass through point $D$ and the segments $AD$ and $CE$ intersect. It turned out that $\angle ABF = \angle DCE$. Find the angle $\angle{ABC}$. by A. Yakubov, S. Berlov 2015 All Russian grade X P7 In an acute-angled and not isosceles triangle $ABC,$ we draw the median $AM$ and the height $AH.$ Points $Q$ and $P$ are marked on the lines $AB$ and $AC$, respectively, so that the $QM \perp AC$ and $PM \perp AB$. The circumcircle of $PMQ$ intersects the line $BC$ for second time at point $X.$ Prove that $BH = CX.$ by M. Didin 2015 All Russian grade XI P1 Parallelogram $ABCD$ is such that angle $B < 90$ and $AB<BC$. Points E and F are on the circumference of $\omega$ inscribing triangle ABC, such that tangents to $\omega$ in those points pass through D. If $\angle EDA= \angle{FDC}$, find $\angle{ABC}$ 2015 All Russian grade XI P7 A scalene triangle $ABC$ is inscribed within circle $\omega$. The tangent to the circle at point $C$ intersects line $AB$ at point $D$. Let $I$ be the center of the circle inscribed within $\triangle ABC$. Lines $AI$ and $BI$ intersect the bisector of $\angle CDB$ in points $Q$ and $P$, respectively. Let $M$ be the midpoint of $QP$. Prove that $MI$ passes through the middle of arc $ACB$ of circle $\omega$. 2016 All Russian grade IX P2 $\omega$ is a circle inside angle $\measuredangle BAC$ and it is tangent to sides of this angle at $B,C$.An arbitrary line $\ell$ intersects with $AB,AC$ at $K,L$,respectively and intersect with $\omega$ at $P,Q$.Points $S,T$ are on $BC$ such that $KS \parallel AC$ and $TL \parallel AB$.Prove that $P,Q,S,T$ are concyclic. by I.Bogdanov, P.Kozhevnikov In triangle $ABC$,$AB<AC$ and $\omega$ is incirle.The $A$-excircle is tangent to $BC$ at $A^\prime$.Point $X$ lies on $AA^\prime$ such that segment $A^\prime X$ doesn't intersect with $\omega$.The tangents from $X$ to $\omega$ intersect with $BC$ at $Y,Z$.Prove that the sum $XY+XZ$ not depends to point $X$. by Mitrofanov Diagonals $AC,BD$ of cyclic quadrilateral $ABCD$ intersect at $P$.Point $Q$ is on$BC$ (between $B$ and $C$) such that $PQ \perp AC$.Prove that the line passes through the circumcenters of triangles $APD$ and $BQD$ is parallel to $AD$. A.Kuznetsov In acute triangle $ABC$,$AC<BC$,$M$ is midpoint of $AB$ and $\Omega$ is it's circumcircle.Let $C^\prime$ be antipode of $C$ in $\Omega$. $AC^\prime$ and $BC^\prime$ intersect with $CM$ at $K,L$,respectively.The perpendicular drawn from $K$ to $AC^\prime$ and perpendicular drawn from $L$ to $BC^\prime$ intersect with $AB$ and each other and form a triangle $\Delta$.Prove that circumcircles of $\Delta$ and $\Omega$ are tangent. by M.Kungozhin In the space given three segments $A_1A_2, B_1B_2$ and $C_1C_2$, do not lie in one plane and intersect at a point $P$. Let $O_{ijk}$ be center of sphere that passes through the points $A_i, B_j, C_k$ and $P$. Prove that $O_{111}O_{222}, O_{112}O_{221}, O_{121}O_{212}$ and$O_{211}O_{122}$ intersect at one point. by P.Kozhevnikov Medians $AM_A,BM_B,CM_C$ of triangle $ABC$ intersect at $M$.Let $\Omega_A$ be circumcircle of triangle passes through midpoint of $AM$ and tangent to $BC$ at $M_A$.Define $\Omega_B$ and $\Omega_C$ analogusly.Prove that $\Omega_A,\Omega_B$ and $\Omega_C$ intersect at one point. by A.Yakubov 2017 All Russian grade IX P2 $ABCD$ is an isosceles trapezoid with $BC \|\| AD$. A circle $\omega$ passing through $B$ and $C$ intersects the side $AB$ and the diagonal $BD$ at points $X$ and $Y$ respectively. Tangent to $\omega$ at $C$ intersects the line $AD$ at $Z$. Prove that the points $X$, $Y$, and $Z$ are collinear. 2017 All Russian grade IX P7 In the scalene triangle $ABC$,$\angle ACB=60^ο$ and $\Omega$ is its cirumcirle.On the bisectors of the angles $BAC$ and $CBA$ points $A'$,$B'$ are chosen respectively such that $AB' \parallel BC$ and $BA' \parallel AC$.$A' B'$ intersects with $\Omega$ at $D,E$. Prove that triangle $CDE$ is isosceles. by A. Kuznetsov Let $ABC$ be an acute angled isosceles triangle with $AB=AC$ and circumcentre $O$. Lines $BO$ and $CO$ intersect $AC, AB$ respectively at $B', C'$. A straight line $l$ is drawn through $C'$ parallel to $AC$. Prove that the line $l$ is tangent to the circumcircle of $\triangle B'OC$. 2017 All Russian grade Χ P8 In a non-isosceles triangle $ABC$,$O$ and $I$ are circumcenter and incenter,r espectively. $B'$ is reflection of $B$ with respect to $OI$ and lies inside the angle $ABI$. Prove that the tangents to circumcirle of $\triangle BB' I$ at $B'$,$I$ intersect on $AC$. by A. Kuznetsov Given a convex quadrilateral $ABCD$. We denote $I_A,I_B, I_C$ and $I_D$ centers of $\omega_A, \omega_B,\omega_C$and $\omega_D$,inscribed In the triangles $DAB, ABC, BCD$ and $CDA$, respectively.It turned out that $\angle BI_AA + \angle I_CI_AI_D = 180^\circ$. Prove that $\angle BI_BA + \angle I_CI_BI_D = 180^{\circ}$. by A. Kuznetsov 2018 All Russian grade IX P2 Circle $\omega$ is tangent to sides $AB, AC$ of triangle $ABC$. A circle $\Omega$ touches the side $AC$ and line $AB$ (produced beyond $B$), and touches $\omega$ at a point $L$ on side $BC$. Line $AL$ meets $\omega, \Omega$ again at $K, M$. It turned out that $KB \parallel CM$. Prove that $\triangle LCM$ is isosceles. $ABCD$ is a convex quadrilateral. Angles $A$ and $C$ are equal. Points $M$ and $N$ are on the sides $AB$ and $BC$ such that $MN\|\|AD$ and $MN=2AD$. Let $K$ be the midpoint of $MN$ and $H$ be the orthocenter of $\triangle ABC$. Prove that $HK$ is perpendicular to $CD$. Let $\triangle ABC$ be an acute-angled triangle with $AB<AC$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively; let $AD$ be an altitude in this triangle. A point $K$ is chosen on the segment $MN$ so that $BK=CK$. The ray $KD$ meets the circumcircle $\Omega$ of $ABC$ at $Q$. Prove that $C, N, K, Q$ are concyclic. 2018 All Russian grade XI P4 On the sides $AB$ and $AC$ of the triangle $ABC$, the points $P$ and $Q$ are chosen, respectively, so that $PQ\|\|BC$. Segments of $BQ$ and $CP$ intersect at point $O$. Point $A'$ is symmetric to point $A$ relative to line$BC$. The segment $A'O$ intersects circle $w$ circumcircle of the triangle $APQ$, at the point $S$. Prove that circumcircle of $BSC$ is tangent to the circle $w$. 2018 All Russian grade XI P6 Three diagonals of a regular $n$-gon prism intersect at an interior point $O$. Show that $O$ is the center of the prism. (The diagonal of the prism is a segment joining two vertices not lying on the same face of the prism.) 2019 All Russian grade IX P3 Circle $\Omega$ with center $O$ is the circumcircle of an acute triangle $\triangle ABC$ with $AB<BC$ and orthocenter $H$. On the line $BO$ there is point $D$ such that $O$ is between $B$ and $D$ and $\angle ADC= \angle ABC$ . The semi-line starting at $H$ and parallel to $BO$ wich intersects segment $AC$ , intersects $\Omega$ at $E$. Prove that $BH=DE$. 2019 All Russian grades IX P6, X P6 (also) There is point $D$ on edge $AC$ isosceles triangle $ABC$ with base $BC$. There is point $K$ on the smallest arc $CD$ of circumcircle of triangle $BCD$. Ray $CK$ intersects line parallel to line $BC$ through $A$ at point $T$. Let $M$ be midpoint of segment $DT$. Prove that $\angle AKT=\angle CAM$. 2019 All Russian grades X P1 , XI P1 Each point $A$ in the plane is assigned a real number $f(A).$ It is known that $f(M)=f(A)+f(B)+f(C),$ whenever $M$ is the centroid of $\triangle ABC.$ Prove that $f(A)=0$ for all points $A.$ 2019 All Russian grade X P4 Let $ABC$ be an acute-angled triangle with $AC<BC.$ A circle passes through $A$ and $B$ and crosses the segments $AC$ and $BC$ again at $A_1$ and $B_1$ respectively. The circumcircles of $A_1B_1C$ and $ABC$ meet each other at points $P$ and $C.$ The segments $AB_1$ and $A_1B$ intersect at $S.$ Let $Q$ and $R$ be the reflections of $S$ in the lines $CA$ and $CB$ respectively. Prove that the points $P,$ $Q,$ $R,$ and $C$ are concyclic. 2019 All Russian grade X P6 Let $L$ be the foot of the internal bisector of $\angle B$ in an acute-angled triangle $ABC.$ The points $D$ and $E$ are the midpoints of the smaller arcs $AB$ and $BC$ respectively in the circumcircle $\omega$ of $\triangle ABC.$ Points $P$ and $Q$ are marked on the extensions of the segments $BD$ and $BE$ beyond $D$ and $E$ respectively so that $\measuredangle APB=\measuredangle CQB=90^{\circ}.$ Prove that the midpoint of $BL$ lies on the line $PQ.$ 2019 All Russian grade XI P4 A triangular pyramid $ABCD$ is given. A sphere $\omega_A$ is tangent to the face $BCD$ and to the planes of other faces in points don't lying on faces. Similarly, sphere $\omega_B$ is tangent to the face $ACD$ and to the planes of other faces in points don't lying on faces. Let $K$ be the point where $\omega_A$ is tangent to $ACD$, and let $L$ be the point where $\omega_B$ is tangent to $BCD$. The points $X$ and $Y$ are chosen on the prolongations of $AK$ and $BL$ over $K$ and $L$ such that $\angle CKD = \angle CXD + \angle CBD$ and $\angle CLD = \angle CYD +\angle CAD$. Prove that the distances from the points $X$, $Y$ to the midpoint of $CD$ are the same. Radii of five concentric circles $\omega_0,\omega_1,\omega_2,\omega_3,\omega_4$ form a geometric progression with common ratio $q$ in this order. What is the maximal value of $q$ for which it's possible to draw a broken line $A_0A_1A_2A_3A_4$ consisting of four equal segments such that $A_i$ lies on $\omega_i$ for every $i=\overline{0,4}$?	2019-10-18 00:15:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.575954794883728, "perplexity": 143.47426322888683}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986677230.18/warc/CC-MAIN-20191017222820-20191018010320-00504.warc.gz"}
https://stats.stackexchange.com/questions/89028/i-have-my-data-set-now-what	# I have my data set, now what? I have a basic understanding of basic statistics, but I believe I've gotten myself out of my depth. I have a data set with a dependent variable (time span) and three quantitative independent variables. There is also a qualitative independent variable (a type flag) but I think I can quantify it if I must. The quantitative independent variables all seem to be inversely proportional to the dependent variable. I want to see if I can determine which of these independent variables has the most influence over the dependent. I ran all data through regression tests using statsmodels for Python (ordinary least squares) but I get warnings indicating multicollinearity and $R^2 = 0.002$. I've also tried univariate linear regression of each independent but am not getting something that looks usable. Finally, I also have this problem of the qualitative independent. Again, I'd really like to know how these independents work together to influence the dependent, and I'd also like to know the degrees to which they influence but I'm clearly lost as to the methodology. EDIT Warnings: 1 The condition number is large, 5.65e+06. This might indicate that there are strong multicollinearity or other numerical problems. Screenshot of full output from .fit().summary() Data Profile: I should also note that two of these independent variables (maxtransfersize and buffercount) repeat for every combination of stripes and diskconfig (<- this is the qualitative variable, and I think I have a way of quantifying that now). backup_time_ms is the dependent variable. • Post your warning text in full. Can you display a snapshot of the data set (Probably just a few observations). Post the output of statsmodel that shows the $R^2R – Andrew Cassidy Mar 6 '14 at 14:16 ## 2 Answers The problem that you're running into is the multi-collinearity in the input matrix for your regression. the matrix is 'ill-conditioned', meaning that small errors in the input lead to large errors in the ouput. The calculation for the condition number of a matrix is$\frac{\lambda_{max}}{\lambda_{min}}$(that might only be for symmetric matricies), or the ratio of the largest to the smallest eigenvalue of the matrix.. I think that the general formula is$\|\|A\|\| \|\|A^{-1}\|\|$The normal equations (the equation used to solve for the betas of the regression) are$\beta = (A^TA)^{-1}Ay$. So as you can see, if you have a matrix with a large condition number (which your program is telling you that you do), it becomes worse from the normal equations since you basically multiply the A matrix together three times. This problem (the multi-collinearity) is what's causing your$R^2$and your betas to have "messed up" values. (remember that small errors in the inputs lead to large errors in the output). Now, what can you do about this? This large condition number comes up also on very high dimensional data. For you, it seems to be coming from the fact that your predictor variables are strongly related. What can you do about this? (1) You can figure out which of your variables is causing the problem and remove it from the model. (2) You can consider methods like ridge regression. What does ridge regression do? They add a small perturbation to your matrix ($\lambda I)$where$\lambda\$ is the perturbation, and I is an identity matrix (matrix with zeroes everywhere, but ones in the diagonal). This reduces your problem with multi-collinearity, but at the expense of adding some bias to the model. I'd suggest on reading up on ridge regression or lasso before just jumping in. I've always found "The Elements of Statistical Learning" to be a good reference. It's free as a pdf online. Good luck. • (1) The r-squared is low. This is secondary to collinearity? (2) the lasso performs well in the setting of collinearity? (3) Lasso is a traditional way to manage colinearity? Usually one (a) removing variable (b) combines variables (c) orthogolonizing (sp?) stats.stackexchange.com/questions/25611/… – charles Mar 6 '14 at 19:47 • (c) yes, i think it's called gram-schmidt. – Eric Peterson Mar 6 '14 at 19:58 • Hi there. Thanks for the feedback. I apologize for the delay. I'm no good at these advanced statistical methods and am really just trying to determine which of x1, x2, and x3 holds the most influence over the outcome. I'm trying to understand the methods you've described (thanks for the paper) but I think that I am just having a hard time conceptualizing how to analyze my data set. – swasheck Mar 19 '14 at 1:41 There is a lot to comment on here. Not sure what is the most useful. (0) I've also tried univariate linear regression of each independent but am not getting something that looks usable If there isn't a strong association, statistics won't solve that. Exploratory data analysis with graphs can be useful here. (1) dependent variable (time span) Time variables: these can be tricky. They are often not normally distributed. You may want to examine this assumption (2) quantitative qualitative independent variable This isn't useful terminology. Continuous or categorical. And if you think that there isn't a linear relationship for continuous variables (as discovered in exploratory analysis), how you plan to accommodate that would be a better start (3) X2 and X3 appear to be correlated based on regression. Are they? Exploratory analysis with X2 vs X3 plot would be useful. • Thanks for your feedback and I apologize for the delay. As for (3), I don't know how to answer that because they were both controlled by me when creating the test. I established the test such that x2 and x3 were held constant while x1 was changed. Then I held x1 and x3 constant while x2 was changed. Finally, I held x1 and x2 constant while x3 was changed. As I said, I'm really not particularly good with statistics, but what I'd really like to know is which of these variable holds the most influence over the result and I just don't know how to achieve that. – swasheck Mar 19 '14 at 1:27	2021-05-19 02:58:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6784632802009583, "perplexity": 639.5657374729245}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991562.85/warc/CC-MAIN-20210519012635-20210519042635-00637.warc.gz"}
https://moderndive.netlify.app/b-2-one-mean	## B.2 One mean ### B.2.1 Problem statement The National Survey of Family Growth conducted by the Centers for Disease Control gathers information on family life, marriage and divorce, pregnancy, infertility, use of contraception, and men’s and women’s health. One of the variables collected on this survey is the age at first marriage. 5,534 randomly sampled US women between 2006 and 2010 completed the survey. The women sampled here had been married at least once. Do we have evidence that the mean age of first marriage for all US women from 2006 to 2010 is greater than 23 years? (Tweaked a bit from Diez, Barr, and Çetinkaya-Rundel 2014 [Chapter 4]) ### B.2.2 Competing hypotheses #### In words • Null hypothesis: The mean age of first marriage for all US women from 2006 to 2010 is equal to 23 years. • Alternative hypothesis: The mean age of first marriage for all US women from 2006 to 2010 is greater than 23 years. #### In symbols (with annotations) • $$H_0: \mu = \mu_{0}$$, where $$\mu$$ represents the mean age of first marriage for all US women from 2006 to 2010 and $$\mu_0$$ is 23. • $$H_A: \mu > 23$$ #### Set $$\alpha$$ It’s important to set the significance level before starting the testing using the data. Let’s set the significance level at 5% here. ### B.2.3 Exploring the sample data age_at_marriage <- read_csv("https://moderndive.com/data/ageAtMar.csv") age_summ <- age_at_marriage %>% summarize( sample_size = n(), mean = mean(age), sd = sd(age), minimum = min(age), lower_quartile = quantile(age, 0.25), median = median(age), upper_quartile = quantile(age, 0.75), max = max(age) ) kable(age_summ) %>% kable_styling( font_size = ifelse(is_latex_output(), 10, 16), latex_options = c("hold_position") ) sample_size mean sd minimum lower_quartile median upper_quartile max 5534 23.4 4.72 10 20 23 26 43 The histogram below also shows the distribution of age. ggplot(data = age_at_marriage, mapping = aes(x = age)) + geom_histogram(binwidth = 3, color = "white") The observed statistic of interest here is the sample mean: x_bar <- age_at_marriage %>% specify(response = age) %>% calculate(stat = "mean") x_bar # A tibble: 1 x 1 stat <dbl> 1 23.4402 #### Guess about statistical significance We are looking to see if the observed sample mean of 23.44 is statistically greater than $$\mu_0 = 23$$. They seem to be quite close, but we have a large sample size here. Let’s guess that the large sample size will lead us to reject this practically small difference. ### B.2.4 Non-traditional methods #### Bootstrapping for hypothesis test In order to look to see if the observed sample mean of 23.44 is statistically greater than $$\mu_0 = 23$$, we need to account for the sample size. We also need to determine a process that replicates how the original sample of size 5534 was selected. We can use the idea of bootstrapping to simulate the population from which the sample came and then generate samples from that simulated population to account for sampling variability. Recall how bootstrapping would apply in this context: 1. Sample with replacement from our original sample of 5534 women and repeat this process 10,000 times, 2. calculate the mean for each of the 10,000 bootstrap samples created in Step 1., 3. combine all of these bootstrap statistics calculated in Step 2 into a boot_distn object, and 4. shift the center of this distribution over to the null value of 23. (This is needed since it will be centered at 23.44 via the process of bootstrapping.) set.seed(2018) null_distn_one_mean <- age_at_marriage %>% specify(response = age) %>% hypothesize(null = "point", mu = 23) %>% generate(reps = 10000) %>% calculate(stat = "mean") null_distn_one_mean %>% visualize() We can next use this distribution to observe our $$p$$-value. Recall this is a right-tailed test so we will be looking for values that are greater than or equal to 23.44 for our $$p$$-value. null_distn_one_mean %>% visualize(obs_stat = x_bar, direction = "greater") ##### Calculate $$p$$-value pvalue <- null_distn_one_mean %>% get_pvalue(obs_stat = x_bar, direction = "greater") pvalue # A tibble: 1 x 1 p_value <dbl> 1 0 So our $$p$$-value is 0 and we reject the null hypothesis at the 5% level. You can also see this from the histogram above that we are far into the tail of the null distribution. #### Bootstrapping for confidence interval We can also create a confidence interval for the unknown population parameter $$\mu$$ using our sample data using bootstrapping. Note that we don’t need to shift this distribution since we want the center of our confidence interval to be our point estimate $$\bar{x}_{obs} = 23.44$$. boot_distn_one_mean <- age_at_marriage %>% specify(response = age) %>% generate(reps = 10000) %>% calculate(stat = "mean") ci <- boot_distn_one_mean %>% get_ci() ci # A tibble: 1 x 2 2.5% 97.5% <dbl> <dbl> 1 23.3148 23.5669 boot_distn_one_mean %>% visualize(endpoints = ci, direction = "between") We see that 23 is not contained in this confidence interval as a plausible value of $$\mu$$ (the unknown population mean) and the entire interval is larger than 23. This matches with our hypothesis test results of rejecting the null hypothesis in favor of the alternative ($$\mu > 23$$). Interpretation: We are 95% confident the true mean age of first marriage for all US women from 2006 to 2010 is between 23.315 and 23.567. ### B.2.5 Traditional methods #### Check conditions Remember that in order to use the shortcut (formula-based, theoretical) approach, we need to check that some conditions are met. 1. Independent observations: The observations are collected independently. The cases are selected independently through random sampling so this condition is met. 2. Approximately normal: The distribution of the response variable should be normal or the sample size should be at least 30. The histogram for the sample above does show some skew. The Q-Q plot below also shows some skew. ggplot(data = age_at_marriage, mapping = aes(sample = age)) + stat_qq() The sample size here is quite large though ($$n = 5534$$) so both conditions are met. #### Test statistic The test statistic is a random variable based on the sample data. Here, we want to look at a way to estimate the population mean $$\mu$$. A good guess is the sample mean $$\bar{X}$$. Recall that this sample mean is actually a random variable that will vary as different samples are (theoretically, would be) collected. We are looking to see how likely is it for us to have observed a sample mean of $$\bar{x}_{obs} = 23.44$$ or larger assuming that the population mean is 23 (assuming the null hypothesis is true). If the conditions are met and assuming $$H_0$$ is true, we can “standardize” this original test statistic of $$\bar{X}$$ into a $$T$$ statistic that follows a $$t$$ distribution with degrees of freedom equal to $$df = n - 1$$: $T =\dfrac{ \bar{X} - \mu_0}{ S / \sqrt{n} } \sim t (df = n - 1)$ where $$S$$ represents the standard deviation of the sample and $$n$$ is the sample size. ##### Observed test statistic While one could compute this observed test statistic by “hand”, the focus here is on the set-up of the problem and in understanding which formula for the test statistic applies. We can use the t_test() function to perform this analysis for us. t_test_results <- age_at_marriage %>% t_test( formula = age ~ NULL, alternative = "greater", mu = 23 ) t_test_results # A tibble: 1 x 6 statistic t_df p_value alternative lower_ci upper_ci <dbl> <dbl> <dbl> <chr> <dbl> <dbl> 1 6.93570 5533 2.25216e-12 greater 23.3358 Inf We see here that the $$t_{obs}$$ value is 6.936. #### Compute $$p$$-value The $$p$$-value—the probability of observing an $$t_{obs}$$ value of 6.936 or more in our null distribution of a $$t$$ with 5533 degrees of freedom—is essentially 0. #### State conclusion We, therefore, have sufficient evidence to reject the null hypothesis. Our initial guess that our observed sample mean was statistically greater than the hypothesized mean has supporting evidence here. Based on this sample, we have evidence that the mean age of first marriage for all US women from 2006 to 2010 is greater than 23 years. #### Confidence interval t.test( x = age_at_marriage$age, alternative = "two.sided", mu = 23 )$conf [1] 23.3 23.6 attr(,"conf.level") [1] 0.95 ### B.2.6 Comparing results Observing the bootstrap distribution that were created, it makes quite a bit of sense that the results are so similar for traditional and non-traditional methods in terms of the $$p$$-value and the confidence interval since these distributions look very similar to normal distributions. The conditions also being met (the large sample size was the driver here) leads us to better guess that using any of the methods whether they are traditional (formula-based) or non-traditional (computational-based) will lead to similar results. ### References Diez, David M, Christopher D Barr, and Mine Çetinkaya-Rundel. 2014. Introductory Statistics with Randomization and Simulation. First. Scotts Valley, CA: CreateSpace Independent Publishing Platform. https://www.openintro.org/stat/textbook.php?stat_book=isrs.	2020-08-10 22:37:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5351881384849548, "perplexity": 1507.807716314172}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439738699.68/warc/CC-MAIN-20200810205824-20200810235824-00116.warc.gz"}
https://proxieslive.com/applied-pi-calculus-evaluation-context-that-distinguishes-replication-with-different-restrictions/	# Applied Pi calculus: Evaluation context that distinguishes replication with different restrictions For an exercise, I need to find an evaluation context $$C[\_]$$ s.t. the transition systems of $$C[X]$$ and $$C[Y]$$ are different (=they are not bisimulation equivalent), where $$X$$ and $$Y$$ are the following processes: $$X = ( \nu z) (!\overline{c}\langle z \rangle.0)$$ and $$Y= !((\nu z) \overline{c}\langle z \rangle.0)$$ Intuitively, the difference seems to be that in process $$X$$, all replications of the process output the same $$z$$ on channel $$c$$, while in process $$Y$$, all processes output a different $$z$$. Is this correct? And how could this be used to construct an evaluation context such that $$C[X] \neq C[Y]$$?	2020-12-02 03:00:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.693958580493927, "perplexity": 401.38093788764024}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141686635.62/warc/CC-MAIN-20201202021743-20201202051743-00194.warc.gz"}
http://mathhelpforum.com/calculus/115540-simple-differentiation-checking-im-right-print.html	# Simple Differentiation (checking i'm right) Printable View • November 18th 2009, 11:51 PM aceband Simple Differentiation (checking i'm right) Hi, I'm just doing a past paper but i don't have the mark scheme would anyone be kind enough to double check my answer: Question: Differentiate $(e^x)/(sin(x) + cos(x))$ MyAnswer: $(e^x(sin(x) + cos(x)) + e^x(cos(x) - sin(x)))/((sin(x) + cos(x))^2)$ • November 18th 2009, 11:55 PM RockHard Seems correct, you can also simplify this • November 19th 2009, 12:03 AM aceband Argh dam it i thought you might be able to - i can't for the life of me see where but it just 'looks' like it should? Could you start me off? • November 19th 2009, 12:22 AM RockHard Well for starters in the denominator when you multiply the term ${(sin(x)+cos(x))}^2$ You can remember the identity $sin(x)^2+cos(x)^2 = 1$ • November 19th 2009, 12:34 AM mr fantastic Quote: Originally Posted by aceband and reformatted by Mr F Hi, I'm just doing a past paper but i don't have the mark scheme would anyone be kind enough to double check my answer: Question: Differentiate $\frac{e^x}{\sin(x) + \cos(x)}$ MyAnswer: $\frac{e^x(\sin(x) + \cos(x)) {\color{red}+} e^x(\cos(x) - \sin(x))}{(\sin(x) + \cos(x))^2}$ The red plus should be a minus. There is a common factor in the numerator which leads to considerable simplification (after you fix the mistake). I'd leave the denominator as it is.	2015-08-04 10:30:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9269209504127502, "perplexity": 1311.2008722058026}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042990609.0/warc/CC-MAIN-20150728002310-00031-ip-10-236-191-2.ec2.internal.warc.gz"}
https://mailman.ntg.nl/pipermail/ntg-context/2008/036792.html	# [NTG-context] Dangling lines. Taco Hoekwater taco at elvenkind.com Mon Dec 8 12:53:32 CET 2008 Bart C. Wise wrote: > > Now the only problem I have is I have no idea what \endgraf does. I've > searched the web and I'm having a difficult time getting a search term > that will point me to the information. Will you give me a brief > explanation or point me where to look \endgraf is just an alias for \par. There are some places where \par is prohibited by TeX's parser but in such cases \endgraf is still allowed. The prohibition helps to catch runaway definitions (because an empty line is equal to a \par commnad). Best wishes, Taco	2018-06-18 13:35:41	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.884376049041748, "perplexity": 4452.974854614045}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267860557.7/warc/CC-MAIN-20180618125242-20180618145242-00412.warc.gz"}
https://socratic.org/questions/how-many-miles-are-in-a-single-gram-of-hydrogen-cyanide	# How many miles are in a single gram of hydrogen cyanide? Nov 13, 2016 I got $4.59 \times {10}^{9}$ $\text{mi}$. First, we assume that all molecules of $\text{HCN}$ are placed in one straight line. Now, we must figure out how to calculate the length of $1$ molecule of $\text{HCN}$. We'll go from there. $\text{H"-"C"-="N}$ contains one $\text{H}$ atom, one $\text{C"-"H}$ single bond, one $\text{C}$ atom, one $\text{C"-="N}$ triple bond, and one $\text{N}$ atom. From the NIST online database for the bond lengths, and wikipedia for the radii: • R_("H","atomic") = "52.9 pm" = 5.29xx10^(-11) $\text{m}$ • R_("N","atomic") = "56.0 pm" = 5.60xx10^(-11) $\text{m}$ • r_("C"-"H","HCN") = "106.4 pm" = 1.064xx10^(-10) $\text{m}$ • r_("C"-="N","HCN") = "115.6 pm" = 1.156xx10^(-10) $\text{m}$ Note that bond length is defined as the internuclear distance. That's why we don't need the atomic radius of carbon---that is already accounted for by the two bond lengths. So, the length of one molecule of $\text{HCN}$ is approximately: l_"HCN" ~~ R_("H","atomic") + r_("C"-"H","HCN") + r_("C"-="N","HCN") + R_("N","atomic") $= 5.29 \times {10}^{- 11}$ $\text{m}$ $+$ $1.064 \times {10}^{- 10}$ $\text{m}$ $+$ $1.156 \times {10}^{- 10}$ $\text{m}$ $+$ $5.60 \times {10}^{- 11}$ $\text{m}$ $= 3.31 \times {10}^{- 10}$ $\text{m}$ (or about $\text{331 pm}$) So now, we take a look at how many $\text{HCN}$ molecules are in $\text{1 g}$: 1 cancel"g HCN" xx cancel"1 mol HCN"/("1.0079 + 12.011 + 14.007" cancel"g HCN") xx (6.022xx10^(23) "molecules")/(cancel"1 mol") $= 2.23 \times {10}^{22}$ $\text{molecules HCN}$ So, the full length of this many molecules of $\text{HCN}$ in $\text{m}$ would be: ${l}_{\text{1 g HCN" = (3.31xx10^(-10) "m")/cancel"molecule HCN" xx 2.23xx10^(22) cancel"molecules HCN}}$ $= 7.38 \times {10}^{12}$ $\text{m}$ and this length in miles of $\text{1 g HCN}$ would be: color(blue)(l_"1 g HCN") = 7.38xx10^(12) cancel("m") xx (100 cancel("cm"))/cancel"1 m" xx cancel"1 in"/(2.54 cancel("cm")) xx cancel"1 ft"/(12 cancel("in")) xx "1 mi"/(5280 cancel("ft")) $= \textcolor{b l u e}{4.59 \times {10}^{9}}$ $\textcolor{b l u e}{\text{mi}}$	2020-01-21 19:48:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 49, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.481641948223114, "perplexity": 4731.220891721205}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250605075.24/warc/CC-MAIN-20200121192553-20200121221553-00145.warc.gz"}
http://www.sagemath.org/doc/reference/libs/sage/libs/pari/pari_instance.html	# PARI C-library interface¶ PARI C-library interface AUTHORS: • William Stein (2006-03-01): updated to work with PARI 2.2.12-beta • William Stein (2006-03-06): added newtonpoly • Justin Walker: contributed some of the function definitions • Gonzalo Tornaria: improvements to conversions; much better error handling. • Robert Bradshaw, Jeroen Demeyer, William Stein (2010-08-15): Upgrade to PARI 2.4.3 (#9343) • Jeroen Demeyer (2011-11-12): rewrite various conversion routines (#11611, #11854, #11952) • Peter Bruin (2013-11-17): split off this file from gen.pyx (#15185) EXAMPLES: sage: pari('5! + 10/x') (120x + 10)/x sage: pari('intnum(x=0,13,sin(x)+sin(x^2) + x)') 85.1885681951527 sage: f = pari('x^3-1') sage: v = f.factor(); v [x - 1, 1; x^2 + x + 1, 1] sage: v[0] # indexing is 0-based unlike in GP. [x - 1, x^2 + x + 1]~ sage: v[1] [1, 1]~ Arithmetic obeys the usual coercion rules: sage: type(pari(1) + 1) <type 'sage.libs.pari.gen.gen'> sage: type(1 + pari(1)) <type 'sage.libs.pari.gen.gen'> GUIDE TO REAL PRECISION AND THE PARI LIBRARY The default real precision in communicating with the Pari library is the same as the default Sage real precision, which is 53 bits. Inexact Pari objects are therefore printed by default to 15 decimal digits (even if they are actually more precise). Default precision example (53 bits, 15 significant decimals): sage: a = pari(1.23); a 1.23000000000000 sage: a.sin() 0.942488801931698 Example with custom precision of 200 bits (60 significant decimals): sage: R = RealField(200) sage: a = pari(R(1.23)); a # only 15 significant digits printed 1.23000000000000 sage: R(a) # but the number is known to precision of 200 bits 1.2300000000000000000000000000000000000000000000000000000000 sage: a.sin() # only 15 significant digits printed 0.942488801931698 sage: R(a.sin()) # but the number is known to precision of 200 bits 0.94248880193169751002382356538924454146128740562765030213504 It is possible to change the number of printed decimals: sage: R = RealField(200) # 200 bits of precision in computations sage: old_prec = pari.set_real_precision(60) # 60 decimals printed sage: a = pari(R(1.23)); a 1.23000000000000000000000000000000000000000000000000000000000 sage: a.sin() 0.942488801931697510023823565389244541461287405627650302135038 sage: pari.set_real_precision(old_prec) # restore the default printing behavior 60 Unless otherwise indicated in the docstring, most Pari functions that return inexact objects use the precision of their arguments to decide the precision of the computation. However, if some of these arguments happen to be exact numbers (integers, rationals, etc.), an optional parameter indicates the precision (in bits) to which these arguments should be converted before the computation. If this precision parameter is missing, the default precision of 53 bits is used. The following first converts 2 into a real with 53-bit precision: sage: R = RealField() sage: R(pari(2).sin()) 0.909297426825682 We can ask for a better precision using the optional parameter: sage: R = RealField(150) sage: R(pari(2).sin(precision=150)) 0.90929742682568169539601986591174484270225497 Warning regarding conversions Sage - Pari - Sage: Some care must be taken when juggling inexact types back and forth between Sage and Pari. In theory, calling p=pari(s) creates a Pari object p with the same precision as s; in practice, the Pari library’s precision is word-based, so it will go up to the next word. For example, a default 53-bit Sage real s will be bumped up to 64 bits by adding bogus 11 bits. The function p.python() returns a Sage object with exactly the same precision as the Pari object p. So pari(s).python() is definitely not equal to s, since it has 64 bits of precision, including the bogus 11 bits. The correct way of avoiding this is to coerce pari(s).python() back into a domain with the right precision. This has to be done by the user (or by Sage functions that use Pari library functions in gen.pyx). For instance, if we want to use the Pari library to compute sqrt(pi) with a precision of 100 bits: sage: R = RealField(100) sage: s = R(pi); s 3.1415926535897932384626433833 sage: p = pari(s).sqrt() sage: x = p.python(); x # wow, more digits than I expected! 1.7724538509055160272981674833410973484 sage: x.prec() # has precision 'improved' from 100 to 128? 128 sage: x == RealField(128)(pi).sqrt() # sadly, no! False sage: R(x) # x should be brought back to precision 100 1.7724538509055160272981674833 sage: R(x) == s.sqrt() True Elliptic curves and precision: If you are working with elliptic curves and want to compute with a precision other than the default 53 bits, you should use the precision parameter of ellinit(): sage: R = RealField(150) sage: e = pari([0,0,0,-82,0]).ellinit(precision=150) sage: eta1 = e.elleta()[0] sage: R(eta1) 3.6054636014326520859158205642077267748102690 Number fields and precision: TODO TESTS: Check that output from PARI’s print command is actually seen by Sage (ticket #9636): sage: pari('print("test")') test class sage.libs.pari.pari_instance.PariInstance Initialize the PARI system. INPUT: • size – long, the number of bytes for the initial PARI stack (see note below) • maxprime – unsigned long, upper limit on a precomputed prime number table (default: 500000) Note In Sage, the PARI stack is different than in GP or the PARI C library. In Sage, instead of the PARI stack holding the results of all computations, it only holds the results of an individual computation. Each time a new Python/PARI object is computed, it it copied to its own space in the Python heap, and the memory it occupied on the PARI stack is freed. Thus it is not necessary to make the stack very large. Also, unlike in PARI, if the stack does overflow, in most cases the PARI stack is automatically increased and the relevant step of the computation rerun. This design obviously involves some performance penalties over the way PARI works, but it scales much better and is far more robust for large projects. Note If you do not want prime numbers, put maxprime=2, but be careful because many PARI functions require this table. If you get the error message “not enough precomputed primes”, increase this parameter. allocatemem(s=0, silent=False) Change the PARI stack space to the given size (or double the current size if s is $$0$$). If $$s = 0$$ and insufficient memory is avaible to double, the PARI stack will be enlarged by a smaller amount. In any case, a MemoryError will be raised if the requested memory cannot be allocated. The PARI stack is never automatically shrunk. You can use the command pari.allocatemem(10^6) to reset the size to $$10^6$$, which is the default size at startup. Note that the results of computations using Sage’s PARI interface are copied to the Python heap, so they take up no space in the PARI stack. The PARI stack is cleared after every computation. It does no real harm to set this to a small value as the PARI stack will be automatically doubled when we run out of memory. However, it could make some calculations slower (since they have to be redone from the start after doubling the stack until the stack is big enough). INPUT: • s - an integer (default: 0). A non-zero argument should be the size in bytes of the new PARI stack. If $$s$$ is zero, try to double the current stack size. EXAMPLES: sage: pari.allocatemem(10^7) PARI stack size set to 10000000 bytes sage: pari.allocatemem() # Double the current size PARI stack size set to 20000000 bytes sage: pari.stacksize() 20000000 sage: pari.allocatemem(10^6) PARI stack size set to 1000000 bytes The following computation will automatically increase the PARI stack size: sage: a = pari('2^100000000') a is now a Python variable on the Python heap and does not take up any space on the PARI stack. The PARI stack is still large because of the computation of a: sage: pari.stacksize() 16000000 sage: pari.allocatemem(10^6) PARI stack size set to 1000000 bytes sage: pari.stacksize() 1000000 TESTS: Do the same without using the string interface and starting from a very small stack size: sage: pari.allocatemem(1) PARI stack size set to 1024 bytes sage: a = pari(2)^100000000 sage: pari.stacksize() 16777216 complex(re, im) Create a new complex number, initialized from re and im. debugstack() Print the internal PARI variables top (top of stack), avma (available memory address, think of this as the stack pointer), bot (bottom of stack). EXAMPLE: sage: pari.debugstack() # random top = 0x60b2c60 avma = 0x5875c38 bot = 0x57295e0 size = 1000000 default(variable, value=None) double_to_gen(x) euler(precision=0) Return Euler’s constant to the requested real precision (in bits). EXAMPLES: sage: pari.euler() 0.577215664901533 sage: pari.euler(precision=100).python() 0.577215664901532860606512090082... factorial(n) Return the factorial of the integer n as a PARI gen. EXAMPLES: sage: pari.factorial(0) 1 sage: pari.factorial(1) 1 sage: pari.factorial(5) 120 sage: pari.factorial(25) 15511210043330985984000000 get_debug_level() Set the debug PARI C library variable. get_real_precision() Returns the current PARI default real precision. This is used both for creation of new objects from strings and for printing. It is the number of digits IN DECIMAL in which real numbers are printed. It also determines the precision of objects created by parsing strings (e.g. pari(‘1.2’)), which is not the normal way of creating new pari objects in Sage. It has no effect on the precision of computations within the pari library. EXAMPLES: sage: pari.get_real_precision() 15 get_series_precision() getrand() Returns PARI’s current random number seed. OUTPUT: GEN of type t_VECSMALL EXAMPLES: sage: pari.setrand(50) sage: a = pari.getrand(); a Vecsmall([...]) sage: pari.setrand(a) sage: a == pari.getrand() True init_primes(_M) Recompute the primes table including at least all primes up to M (but possibly more). EXAMPLES: sage: pari.init_primes(200000) We make sure that ticket #11741 has been fixed, and double check to make sure that diffptr has not been freed: sage: pari.init_primes(2^62) Traceback (most recent call last): ... PariError: not enough memory # 64-bit OverflowError: long int too large to convert # 32-bit sage: pari.init_primes(200000) matrix(m, n, entries=None) matrix(long m, long n, entries=None): Create and return the m x n PARI matrix with given list of entries. new_with_bits_prec(s, precision) pari.new_with_bits_prec(self, s, precision) creates s as a PARI gen with (at most) precision bits of precision. nth_prime(n) pari_version() pi(precision=0) Return the value of the constant pi to the requested real precision (in bits). EXAMPLES: sage: pari.pi() 3.14159265358979 sage: pari.pi(precision=100).python() 3.1415926535897932384626433832... polcyclo(n, v=-1) polcyclo(n, v=x): cyclotomic polynomial of degree n, in variable v. EXAMPLES: sage: pari.polcyclo(8) x^4 + 1 sage: pari.polcyclo(7, 'z') z^6 + z^5 + z^4 + z^3 + z^2 + z + 1 sage: pari.polcyclo(1) x - 1 polcyclo_eval(n, v) polcyclo_eval(n, v): value of the nth cyclotomic polynomial at value v. EXAMPLES: sage: pari.polcyclo_eval(8, 2) 17 sage: cyclotomic_polynomial(8)(2) 17 pollegendre(n, v=-1) pollegendre(n, v=x): Legendre polynomial of degree n (n C-integer), in variable v. EXAMPLES: sage: pari.pollegendre(7) 429/16x^7 - 693/16x^5 + 315/16x^3 - 35/16x sage: pari.pollegendre(7, 'z') 429/16z^7 - 693/16z^5 + 315/16z^3 - 35/16z sage: pari.pollegendre(0) 1 polsubcyclo(n, d, v=-1) polsubcyclo(n, d, v=x): return the pari list of polynomial(s) defining the sub-abelian extensions of degree $$d$$ of the cyclotomic field $$\QQ(\zeta_n)$$, where $$d$$ divides $$\phi(n)$$. EXAMPLES: sage: pari.polsubcyclo(8, 4) [x^4 + 1] sage: pari.polsubcyclo(8, 2, 'z') [z^2 - 2, z^2 + 1, z^2 + 2] sage: pari.polsubcyclo(8, 1) [x - 1] sage: pari.polsubcyclo(8, 3) [] poltchebi(n, v=-1) poltchebi(n, v=x): Chebyshev polynomial of the first kind of degree n, in variable v. EXAMPLES: sage: pari.poltchebi(7) 64x^7 - 112x^5 + 56x^3 - 7x sage: pari.poltchebi(7, 'z') 64z^7 - 112z^5 + 56z^3 - 7z sage: pari.poltchebi(0) 1 polzagier(n, m) prime_list(n) prime_list(n): returns list of the first n primes To extend the table of primes use pari.init_primes(M). INPUT: • n - C long OUTPUT: • gen - PARI list of first n primes EXAMPLES: sage: pari.prime_list(0) [] sage: pari.prime_list(-1) [] sage: pari.prime_list(3) [2, 3, 5] sage: pari.prime_list(10) [2, 3, 5, 7, 11, 13, 17, 19, 23, 29] sage: pari.prime_list(20) [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71] sage: len(pari.prime_list(1000)) 1000 primes_up_to_n(n) Return the primes <= n as a pari list. EXAMPLES: sage: pari.primes_up_to_n(1) [] sage: pari.primes_up_to_n(20) [2, 3, 5, 7, 11, 13, 17, 19] Read a script from the named filename into the interpreter. The functions defined in the script are then available for use from Sage/PARI. The result of the last command in filename is returned. EXAMPLES: Create a gp file: sage: import tempfile sage: gpfile = tempfile.NamedTemporaryFile(mode="w") sage: gpfile.file.write("mysquare(n) = {\n") sage: gpfile.file.write(" n^2;\n") sage: gpfile.file.write("}\n") sage: gpfile.file.write("polcyclo(5)\n") sage: gpfile.file.flush() Read it in Sage, we get the result of the last line: sage: pari.read(gpfile.name) x^4 + x^3 + x^2 + x + 1 Call the function defined in the gp file: sage: pari('mysquare(12)') 144 set_debug_level(level) Set the debug PARI C library variable. set_real_precision(n) Sets the PARI default real precision in decimal digits. This is used both for creation of new objects from strings and for printing. It is the number of digits IN DECIMAL in which real numbers are printed. It also determines the precision of objects created by parsing strings (e.g. pari(‘1.2’)), which is not the normal way of creating new pari objects in Sage. It has no effect on the precision of computations within the pari library. Returns the previous PARI real precision. EXAMPLES: sage: pari.set_real_precision(60) 15 sage: pari('1.2') 1.20000000000000000000000000000000000000000000000000000000000 sage: pari.set_real_precision(15) 60 set_series_precision(n) setrand(seed) Sets PARI’s current random number seed. INPUT: • seed – either a strictly positive integer or a GEN of type t_VECSMALL as output by getrand() This should not be called directly; instead, use Sage’s global random number seed handling in sage.misc.randstate and call current_randstate().set_seed_pari(). EXAMPLES: sage: pari.setrand(50) sage: a = pari.getrand(); a Vecsmall([...]) sage: pari.setrand(a) sage: a == pari.getrand() True TESTS: Check that invalid inputs are handled properly (#11825): sage: pari.setrand(0) Traceback (most recent call last): ... PariError: incorrect type in setrand sage: pari.setrand("foobar") Traceback (most recent call last): ... PariError: incorrect type in setrand stacksize() Returns the current size of the PARI stack, which is $$10^6$$ by default. However, the stack size is automatically doubled when needed. It can also be set directly using pari.allocatemem(). EXAMPLES: sage: pari.stacksize() 1000000 vector(n, entries=None) vector(long n, entries=None): Create and return the length n PARI vector with given list of entries. EXAMPLES: sage: pari.vector(5, [1, 2, 5, 4, 3]) [1, 2, 5, 4, 3] sage: pari.vector(2, [x, 1]) [x, 1] sage: pari.vector(2, [x, 1, 5]) Traceback (most recent call last): ... IndexError: length of entries (=3) must equal n (=2) sage.libs.pari.pari_instance.prec_bits_to_dec(prec_in_bits) Convert from precision expressed in bits to precision expressed in decimal. EXAMPLES: sage: from sage.libs.pari.pari_instance import prec_bits_to_dec sage: prec_bits_to_dec(53) 15 sage: [(32n, prec_bits_to_dec(32n)) for n in range(1, 9)] [(32, 9), (64, 19), (96, 28), (128, 38), (160, 48), (192, 57), (224, 67), (256, 77)] sage.libs.pari.pari_instance.prec_bits_to_words(prec_in_bits) Convert from precision expressed in bits to pari real precision expressed in words. Note: this rounds up to the nearest word, adjusts for the two codewords of a pari real, and is architecture-dependent. EXAMPLES: sage: from sage.libs.pari.pari_instance import prec_bits_to_words sage: prec_bits_to_words(70) 5 # 32-bit 4 # 64-bit sage: [(32n, prec_bits_to_words(32*n)) for n in range(1, 9)] [(32, 3), (64, 4), (96, 5), (128, 6), (160, 7), (192, 8), (224, 9), (256, 10)] # 32-bit [(32, 3), (64, 3), (96, 4), (128, 4), (160, 5), (192, 5), (224, 6), (256, 6)] # 64-bit sage.libs.pari.pari_instance.prec_dec_to_bits(prec_in_dec) Convert from precision expressed in decimal to precision expressed in bits. EXAMPLES: sage: from sage.libs.pari.pari_instance import prec_dec_to_bits sage: prec_dec_to_bits(15) 49 sage: [(n, prec_dec_to_bits(n)) for n in range(10, 100, 10)] [(10, 33), (20, 66), (30, 99), (40, 132), (50, 166), (60, 199), (70, 232), (80, 265), (90, 298)] sage.libs.pari.pari_instance.prec_dec_to_words(prec_in_dec) Convert from precision expressed in decimal to precision expressed in words. Note: this rounds up to the nearest word, adjusts for the two codewords of a pari real, and is architecture-dependent. EXAMPLES: sage: from sage.libs.pari.pari_instance import prec_dec_to_words sage: prec_dec_to_words(38) 6 # 32-bit 4 # 64-bit sage: [(n, prec_dec_to_words(n)) for n in range(10, 90, 10)] [(10, 4), (20, 5), (30, 6), (40, 7), (50, 8), (60, 9), (70, 10), (80, 11)] # 32-bit [(10, 3), (20, 4), (30, 4), (40, 5), (50, 5), (60, 6), (70, 6), (80, 7)] # 64-bit sage.libs.pari.pari_instance.prec_words_to_bits(prec_in_words) Convert from pari real precision expressed in words to precision expressed in bits. Note: this adjusts for the two codewords of a pari real, and is architecture-dependent. EXAMPLES: sage: from sage.libs.pari.pari_instance import prec_words_to_bits sage: prec_words_to_bits(10) 256 # 32-bit 512 # 64-bit sage: [(n, prec_words_to_bits(n)) for n in range(3, 10)] [(3, 32), (4, 64), (5, 96), (6, 128), (7, 160), (8, 192), (9, 224)] # 32-bit [(3, 64), (4, 128), (5, 192), (6, 256), (7, 320), (8, 384), (9, 448)] # 64-bit sage.libs.pari.pari_instance.prec_words_to_dec(prec_in_words) Convert from precision expressed in words to precision expressed in decimal. Note: this adjusts for the two codewords of a pari real, and is architecture-dependent. EXAMPLES: sage: from sage.libs.pari.pari_instance import prec_words_to_dec sage: prec_words_to_dec(5) 28 # 32-bit 57 # 64-bit sage: [(n, prec_words_to_dec(n)) for n in range(3, 10)] [(3, 9), (4, 19), (5, 28), (6, 38), (7, 48), (8, 57), (9, 67)] # 32-bit [(3, 19), (4, 38), (5, 57), (6, 77), (7, 96), (8, 115), (9, 134)] # 64-bit #### Previous topic Sage interface to Cremona’s eclib library (also known as mwrank) #### Next topic Sage class for PARI’s GEN type	2014-07-30 19:15:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39789116382598877, "perplexity": 4525.912628806904}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510271648.4/warc/CC-MAIN-20140728011751-00373-ip-10-146-231-18.ec2.internal.warc.gz"}
http://trilinos.sandia.gov/packages/docs/r11.2/packages/ifpack2/doc/html/classIfpack2_1_1Details_1_1Chebyshev.html	Ifpack2 Templated Preconditioning Package Version 1.0 Ifpack2::Details::Chebyshev< ScalarType, MV, MAT > Class Template Reference Left-scaled Chebyshev iteration. More... #include <Ifpack2_Details_Chebyshev_decl.hpp> List of all members. ## Public Member Functions Chebyshev (Teuchos::RCP< const MAT > A) Chebyshev (Teuchos::RCP< const MAT > A, Teuchos::ParameterList &params) void setParameters (Teuchos::ParameterList &plist) Set (or reset) parameters. void compute () (Re)compute the left scaling, and (if applicable) estimate max and min eigenvalues of D_inv * A. MT apply (const MV &B, MV &X) Teuchos::RCP< const MAT > getMatrix () const Get the matrix given to the constructor. bool hasTransposeApply () const Whether it's possible to apply the transpose of this operator. void print (std::ostream &out) Print instance data to the given output stream. ## Detailed Description ### template<class ScalarType, class MV, class MAT> class Ifpack2::Details::Chebyshev< ScalarType, MV, MAT > Left-scaled Chebyshev iteration. Template Parameters: ScalarType The type of entries in the matrix and vectors. MV Specialization of Tpetra::MultiVector. MAT Corresponding specialization of Tpetra::RowMatrix. This class implements two variants of Chebyshev iteration: 1. A direct imitation of Ifpack's implementation 2. A textbook version of the algorithm All implemented variants use the diagonal of the matrix to precondition the linear system on the left. Diagonal entries less than machine precision are replaced with machine precision. The first version imitates Ifpack::Chebyshev, both in how it sets parameters and in the actual iteration (ApplyInverse()). The "textbook" in variant #2 above is "Templates for the Solution of Linear Systems," 2nd edition. Experiments show that the Ifpack imitation is much less sensitive to the eigenvalue bounds than the textbook version, so users should prefer it. (In fact, it is the default.) We require that the matrix A be real valued and symmetric positive definite. If users could provide the ellipse parameters ("d" and "c" in the literature, where d is the real-valued center of the ellipse, and d-c and d+c the two foci), the iteration itself would work fine with nonsymmetric real-valued A, as long as the eigenvalues of A can be bounded in an ellipse that is entirely to the right of the origin. There is also dead code for imitating ML's Chebyshev implementation (ML_Cheby(), in packages/ml/src/Smoother/ml_smoother.c). I couldn't get it to converge in time to be useful for testing, so it is disabled. ## Constructor & Destructor Documentation template<class ScalarType , class MV , class MAT> Ifpack2::Details::Chebyshev< ScalarType, MV, MAT >::Chebyshev ( Teuchos::RCP< const MAT > A ) Constructor that takes a sparse matrix and sets default parameters. Parameters: A [in] The matrix A in the linear system to solve. A must be real-valued and symmetric positive definite. template<class ScalarType , class MV , class MAT> Ifpack2::Details::Chebyshev< ScalarType, MV, MAT >::Chebyshev ( Teuchos::RCP< const MAT > A, Teuchos::ParameterList & params ) Constructor that takes a sparse matrix and sets the user's parameters. Parameters: A [in] The matrix A in the linear system to solve. A must be real-valued and symmetric positive definite. params [in/out] On input: the parameters. On output: filled with the current parameter settings. ## Member Function Documentation template<class ScalarType , class MV , class MAT > void Ifpack2::Details::Chebyshev< ScalarType, MV, MAT >::setParameters ( Teuchos::ParameterList & plist ) Set (or reset) parameters. This method fills in the input ParameterList with missing parameters set to their default values. You may call this method as many times as you want. On each call, the input ParameterList is treated as a complete list of the desired parameters, not as a "delta" or change list from the current set of parameters. (That is, if you remove parameters from the list that were there in the last call to setParameters() and call setParameters() again with the revised list, this method will use default values for the removed parameters, rather than letting the current settings remain.) However, since the method fills in missing parameters, you may keep calling it with the ParameterList used in the previous call in order to get the same behavior as before. Parameters that govern spectral bounds of the matrix: • "chebyshev: max eigenvalue" (ScalarType): lambdaMax, an upper bound of the bounding ellipse of the eigenvalues of the matrix A. If you do not set this parameter, we will compute an approximation. See "Parameters that govern eigenvalue analysis" to control this approximation process. • "chebyshev: ratio eigenvalue" (ScalarType): eigRatio, the ratio of lambdaMax to the lower bound of the bounding ellipse of the eigenvalues of A. We use lambdaMax and eigRatio to determine the Chebyshev iteration coefficients. This parameter is optional and defaults to 30. • "chebyshev: min eigenvalue" (ScalarType): lambdaMin, a lower bound of real part of bounding ellipse of eigenvalues of the matrix A. This parameter is optional and only used for a quick check if the matrix is the identity matrix (if lambdaMax == lambdaMin == 1). Parameters that govern the number of Chebyshev iterations: • "chebyshev: degree" (int): numIters, the number of iterations. This overrides "relaxation: sweeps" and "smoother: sweeps" (see below). • "relaxation: sweeps" (int): numIters, the number of iterations. We include this for compatibility with Ifpack. This overrides "smoother: sweeps" (see below). • "smoother: sweeps" (int): numIters, as above. We include this for compatibility with ML. Parameters that govern eigenvalue analysis: • "chebyshev: eigenvalue max iterations" (int): eigMaxIters, the number of power method iterations used to compute the maximum eigenvalue. This overrides "eigen-analysis: iterations" (see below). • "eigen-analysis: iterations" (int): eigMaxIters, as above. We include this parameter for compatibility with ML. • "eigen-analysis: type" (std::string): The algorithm to use for estimating the max eigenvalue. This parameter is optional. Currently, we only support "power-method" (or "power method"), which is what Ifpack::Chebyshev uses for eigenanalysis. We include this parameter for compatibility with ML. Parameters that govern other algorithmic details: • "chebyshev: operator inv diagonal" (RCP<const V> or const V): If nonnull, we will use a deep copy of this vector for left scaling as the inverse diagonal of the matrix A, instead of computing the inverse diagonal ourselves. We will make a copy every time you call setParameters(). If you ever call setParameters() without this parameter, we will clear our copy and compute the inverse diagonal ourselves again. You are responsible for updating this if the matrix has changed. • "chebyshev: min diagonal value" (ST): minDiagVal. If any entry of the diagonal of the matrix is less than this in magnitude, it will be replaced with this value in the inverse diagonal used for left scaling. • "chebyshev: zero starting solution" (bool): If true, then always use the zero vector(s) as the initial guess(es). If false, then apply() will use X on input as the initial guess(es). Parameters that govern backwards compatibility: • "chebyshev: textbook algorithm" (bool): If true, use the textbook version of Chebyshev iteration. We recommend against this, since the default algorithm is less sensitive to the quality of the eigenvalue bounds. • "chebyshev: compute max residual norm" (bool): If true, apply() will compute and return the max (absolute) residual norm. Otherwise, apply() returns 0. This defaults to false. Precondition: lambdaMin, lambdaMax, and eigRatio are real 0 < lambdaMin <= lambdaMax numIters >= 0 eigMaxIters >= 0 Default settings for parameters relating to spectral bounds come from Ifpack. template<class ScalarType , class MV , class MAT > void Ifpack2::Details::Chebyshev< ScalarType, MV, MAT >::compute ( ) (Re)compute the left scaling, and (if applicable) estimate max and min eigenvalues of D_inv A. You must call this method before calling apply(), • if you have not yet called this method, • if the matrix (either its values or its structure) has changed, or • any time after you call setParameters(). Advanced users may omit calling compute() after calling setParameters(), as long as none of the changed parameters affect either computation of the inverse diagonal, or estimation of the max or min eigenvalues. If estimation of the eigenvalues is required, this method may take as long as several Chebyshev iterations. template<class ScalarType , class MV, class MAT > Chebyshev< ScalarType, MV, MAT >::MT Ifpack2::Details::Chebyshev< ScalarType, MV, MAT >::apply ( const MV & B, MV & X ) Solve Ax=b for x with Chebyshev iteration with left diagonal scaling. Parameters: B [in] Right-hand side(s) in the linear system to solve. X [in] Initial guess(es) for the linear system to solve. If the "chebyshev: compute max residual norm" parameter is true (not the default), then this method returns the maximum (over all columns) absolute residual 2-norm after iterating. Otherwise, it returns zero. Warning: If you did not set the "chebyshev: zero starting solution" parameter to true, then this method will use X as the starting guess for Chebyshev iteration. If you did not initialize X before calling this method, then the resulting solution will be undefined, since it will be computed using uninitialized data. template<class ScalarType , class MV , class MAT > Teuchos::RCP< const MAT > Ifpack2::Details::Chebyshev< ScalarType, MV, MAT >::getMatrix ( ) const Get the matrix given to the constructor. template<class ScalarType , class MV , class MAT > bool Ifpack2::Details::Chebyshev< ScalarType, MV, MAT >::hasTransposeApply ( ) const Whether it's possible to apply the transpose of this operator. template<class ScalarType , class MV , class MAT > void Ifpack2::Details::Chebyshev< ScalarType, MV, MAT >::print ( std::ostream & out ) Print instance data to the given output stream. The documentation for this class was generated from the following files:	2014-04-24 12:25:11	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6533708572387695, "perplexity": 5582.973976649778}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00559-ip-10-147-4-33.ec2.internal.warc.gz"}
https://eng-calculations.com/ASME/BPVC_Section_VIII-Rules_for_Construction_of_Pressure_Vessels_Division_1/thickness_of_shells_under_internal_pressure_for_Toriconical_Heads	# Thickness of shells under internal pressure for Toriconical Heads ## Values for calculation $E$ $P$ $\mathrm{MPa}$ $T$ $\mathrm{°C}$ $D$ $\mathrm{mm}$ $α$ $\mathrm{°}$ $t_s$ $\mathrm{mm}$ $r$ $\mathrm{mm}$ $S$ $\mathrm{MPa}$ ## Calculation ### Inside diameter of the conical portion of a toriconical head at its point of tangency to the knuckle, measured perpendicular to the axis of the cone $$D_i=D-2\cdot r\cdot\left(1-\cos(α)\right)$$ $$L=\cfrac{D_i}{2\cdot\cos(α)}$$ ### Factor $M$ $$M=\cfrac{1}{4}\cdot\left(3+\sqrt{\cfrac{L}{r}}\right)$$ ### Minimum required thickness of head after forming $$t=\max\left(\cfrac{P\cdot D_i}{2\cdot\cos(α)\cdot\left(S\cdot E-0.6\cdot P\right)}, \cfrac{P\cdot L\cdot M}{2\cdot S\cdot E-0.2\cdot P}\right)$$$$t_s \geq t$$ ### Requirements $$r > 0.06\cdot D$$$$r > 3\cdot t_s$$	2023-01-29 15:10:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3428493142127991, "perplexity": 3842.8042838862125}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00523.warc.gz"}
https://stats.stackexchange.com/questions/84113/quick-problem-with-maximizing-profit-using-different-distributions/84126	# Quick Problem with maximizing profit using different distributions I need a little help with a problem I am working on. So here's the situation, a seller can produce a apple for $1.00$ and I need to find the optimal price of selling the apple and expected profit per buyer if the distribution of the value of the apple is EXP(1). So I just need some confirmation on what I am doing. What I have is this: Expected profit = $(v-1)\cdot (1-F(v))$ Optimal price = derivative of expected profit w.r.t. to $v$ set equal to zero then solve for $v$ I hope this makes sense guys. Like I said I just want make sure I am doing this right or else I am going to have to read the chapter for the 3rd time. You're right. In general, if the valuation of a good follows the distribution $F(x)$, then the demand function is $D(x) = 1 - F(x)$ implying that the profit under a marginal cost $m$ is simply $$\Pi = (x-m)[1-F(x)]$$ Now, for the exponential distribution with parameter $\lambda$ a good trick to remember is that the optimal price equals $m+ \lambda^{-1}$, so your solution is $x=2$.	2019-11-18 11:27:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8845101594924927, "perplexity": 174.1231706449699}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669755.17/warc/CC-MAIN-20191118104047-20191118132047-00392.warc.gz"}
http://starbeamrainbowlabs.com/blog/	## Stardust Blog Archive Mailing List Articles Atom Feed Comments Atom Feed Twitter Reddit Facebook ## Tag Cloud 3d account algorithms android announcement architecture archives arduino artificial intelligence artix assembly async audio automation backups bash batch blog bookmarklet booting bug hunting c sharp c++ challenge chrome os code codepen coding conundrums coding conundrums evolved command line compilers compiling compression css dailyprogrammer debugging demystification distributed computing documentation downtime electronics email embedded systems encryption es6 features event experiment external first impressions future game github github gist gitlab graphics hardware hardware meetup holiday holidays html html5 html5 canvas infrastructure interfaces internet interoperability io.js jabber jam javascript js bin labs learning library linux lora low level lua maintenance manjaro network networking nibriboard node.js operating systems performance photos php pixelbot portable privacy problem solving programming problems projects prolog protocol protocols pseudo 3d python reddit redis reference releases resource review rust searching secrets security series list server software sorting source code control statistics storage svg technical terminal textures three thing game three.js tool tutorial twitter ubuntu university update updates upgrade version control virtual reality virtualisation visual web website windows windows 10 xmpp xslt ## Animated PNG for all! I recently discovered that Animated PNGs are now supported by most major browsers: I stumbled across the concept of an animated PNG a number of years ago (on caniuse.com actually if I remember right!), but at the time browser support was very bad (read: non-existent :P) - so I moved on to other things. I ended up re-discovering it a few weeks ago (also through caniuse.com!), and since browser support is so much better now I decided that I just had to play around with it. It hasn't disappointed. Traditional animated GIFs (Graphics Interchange Format for the curious) are limited to 256 colours, have limited transparency support (a pixel is either transparent, or it isn't - there's no translucency support), and don't compress terribly well. Animated PNGs (Portable Network Graphics), on the other hand, let you enjoy all the features of a regular PNG (as many colours as you want, translucent pixels, etc.) while also supporting animation, and compressing better as well! Best of all, if a browser doesn't support the animated PNG standard, they will just see and render the first frame as a regular PNG and silently ignore the rest. Let's take it out for a spin. For my test, I took an image and created a 'panning' animation from one side of it to the other. Here's the image I've used: (Credit: The background on the download page for Mozilla's Firefox Nightly builds. It isn't available on the original source website anymore (and I've lost the link), but can still be found on various wallpaper websites.) The first task is to generate the frames from the original image. I wrote a quick shell script for this purpose. Firstly, I defined a bunch of variables: #!/usr/bin/env bash set -e; # Crash if we hit an error input_file="Input.png"; output_file="Output.apng" # The maximum number of frames max_frame=32; end_x=960; end_y=450; start_x=0; start_y=450; crop_size_x=960; crop_size_y=540; mkdir -p ./frames; Variable Meaning input_file The input file to generate frames from output_file The file to write the animated png to max_frame The number of frames (plus 1) to generate start_x The starting position to pan from on the x axis start_y The starting position to pan from on the y axis end_x The ending position to pan to on the x axis end_y The ending position to pan to on the y axis crop_size_x The width of the cropped frames crop_size_y The height of the cropped frames It's worth noting here that it's probably a good idea to implement a proper CLI to this script at this point, but since it's currently only a one-off script I didn't bother. Perhaps in the future I'll come back and improve it if I find myself needing it again for some other purpose. With the parameters set up (and a temporary directory created - note that you should use mktemp -d instead of the approach I've taken here!), we can then use a for loop to repeatedly call ImageMagick to crop the input image to generate our frames. This won't run in parallel unfortunately, but since it's only a few frames it shouldn't take too long to render. This is only a quick shell script after all :P for ((frame=0; frame <= "${max_frame}"; frame++)); do this_x="$(calc -p "${start_x}+((${end_x}-${start_x})(${frame}/${max_frame}))")"; this_y="$(calc -p "${start_y}+((${end_y}-${start_y})(${frame}/${max_frame}))")"; percent="$(calc -p "round((${frame}/${max_frame})100, 2)")"; convert "${input_file}" -crop "${crop_size_x}x${crop_size_y}+${this_x}+${this_y}" "frames/Frame-$(printf "%02d" "${frame}").jpeg"; echo -ne "${frame} / ${max_frame} (${percent}%) \r"; done echo ""; # Move to the line after the progress indicator This looks complicated, but it really isn't. The for loop iterates over each of the frame numbers. We do some maths to calculate the (x, y) co-ordinates of the frame we want to extract, and then get ImageMagick's convert command to do the cropping for us. Finally we write a quick progress indicator out to stdout (\r resets the cursor to the beginning of the line, but doesn't go down to the next one). The maths there is probably better represented like this: $this_x=start_x+((end_x-start_x)\frac{currentframe}{max_frame})$ $this_y=start_y+((end_y-start_y)\frac{currentframe}{max_frame})$ Much better :-) In short, we convert the current frame number into a percentage (between 0 and 1) of how far through the animation we are and use that as a multiplier to find the distance between the starting and ending points. I use the calc command-line tool (in the apcalc package on Ubuntu) here to do the calculations, because the bash built-in result=$(((4 + 5))) only supports integer-based maths, and I wanted floating-point support here. With the frames generated, we only need to stitch them together to make the final animated png. Unfortunately, an easy-to-use tool does not yet exist (like gifsicle for GIFs) - so we'll have to make-do with ffpmeg (which, while very powerful, has a rather confusing CLI!). Here's how I stitched the frames together: ffmpeg -r 10 -i frames/Frame-%02d.jpeg -plays 0 "${output_file}"; • -r - The frame rate • -i - The input filename • -plays - The number of times to loop (0 = infinite; defaults to no looping if omitted) • "${output_file}" - The output file Here's the final result: (Filesize: ~2.98MiB) Of course, it'd be cool to compare it to a traditional animated gif. Let's do that too! First, we'll need to convert the frames to gif (gifsicle, our tool of choice, doesn't support anything other than GIFs as far as I'm aware): mogrify -format gif frames/.jpeg Easy-peasy! mogrify is another tool from ImageMagick that makes such in-place conversions trivial. Note that the frames themselves are stored as JPEGs because I was experiencing an issue whereby each of the frames apparently had a slightly different colour palette, and ffmpeg wasn't smart enough to correct for this - choosing instead to crash. With the frames converted, we can make our animated GIF like so: gifsicle --optimize --colors=256 --loopcount=10 --delay=10 frames/.gif >Output.gif; Lastly, we probably want to delete the intermediate frames: rm -r ./frames Here's the animated GIF version: (Filesize: ~3.28MiB) Woah! That's much bigger. (Generated (and then extracted & edited with the Firefox developer tools) from Meta-Chart) It's ~9.7% bigger in fact! Though not a crazy amount, smaller resulting files are always good. I suspect that this effect will stack the more frames you have. Others have tested this too, finding pretty similar results to those that I've found here - though it does of course depend on your specific scenario. With that observation, I'll end this blog post. The next time you think about inserting an animation into a web page or chat window, consider making it an Animated PNG. Found this interesting? Found a cool CLI tool for manipulating APNGs? Having trouble? Comment below! ### Sources and Further Reading ## Easy Node.JS Dependencies Updates Once you've had a project around for a while, it's inevitable that dependency updates will become available. Unfortunately, npm (the Node Package Manager), while excellent at everything else, is completely terrible at notifying you about updates. The solution to this is, of course, to use an external tool. Personally, I use npm-check, which is also installable via npm. It shows you a list of updates to your project's dependencies, like so: (Can't see the above? View it directly on asciinema.org) It even supports the packages that you've install globally too, which no other tool appears to do as far as I can tell (although it does appear to miss some packages, such as npm and itself). To install it, simply do this: sudo npm install --global npm-check Once done, you can then use it like this: # List updates, but don't install them npm-check # Interactively choose the updates to install npm-check -u # Interactively check the globally-installed packages sudo npm-check -gu The tool also checks to see which of the dependencies are actually used, and prompts you to check the dependencies it think you're not using (it doesn't always get it right, so check carefully yourself before removing!). There's an argument to disable this behaviour: npm-check --skip-unused Speaking of npm package dependencies, the other big issue is security vulnerabilities. GitHub have recently started giving maintainers of projects notifications about security vulnerabilities in their dependencies, which I think is a brilliant idea. Actually fixing said vulnerabilities is a whole other issue though. If you don't want to update all the dependencies of a project to the latest version (perhaps you're just doing a one-off contribution to a project or aren't very familiar with the codebase yet), there's another tool - this time built-in to npm - to help out - the npm audit subcommand. # Check for reported security issues in your dependencies npm audit # Attempt to fix said issues by updating packages just* enough npm audit fix (Can't see the above? View it directly on asciinema.org) This helps out a ton with contributing to other projects. Issues arise when the vulnerabilities are not in packages you directly depend on, but instead in packages that you indirectly depend on via dependencies of the packages you've installed. Thankfully the vulnerabilities in the above can all be traced back to development dependencies (and aren't essential for Peppermint Wiki itself), but it's rather troubling that I can't install updated packages because the packages I depend on haven't updated their dependencies. I guess I'll be sending some pull requests to update some dependencies! To help me in this, the output of npm audit even displays the dependency graph of why a package is installed. If this isn't enough though, there's always the npm-why which, given a package name, will figure out why it's installed. Found this interesting? Got a better solution? Comment below! ## Generating Atom 1.0 Feeds with PHP (the proper way) I've generated Atom feeds in PHP before, but recently I went on the hunt to discover if PHP has something like C♯'s XMLWriter class - and it turns out it does! Although poorly documented (probably why I didn't find it in the first place :P), it's actually quite logical and easy to pick up. To this end, I thought I'd blog about how I used to write the Atom 1.0 feed generator for recent changes on Pepperminty Wiki that I've recently implemented (coming soon in the next release!), as it's so much cleaner than atom.gen.php that I blogged about before! It's safer too - as all the escaping is handled automatically by PHP - so there's no risk of an injection attack because I forgot to escape a character in my library code. It ends up being a bit verbose, but a few helper methods (or a wrapper class?) should alleviate this nicely - I might refactor it at a later date. To begin, we need to create an instance of the aptly-named XMLLWriter class. It's probable that you'll need the php-xml package installed in order to use this. $xml = new XMLWriter(); $xml->openMemory();$xml->setIndent(true); $xml->setIndentString("\t");$xml->startDocument("1.0", "utf-8"); In short, the above creates a new XMLWriter instance, instructs it to write the XML to memory, enables pretty-printing of the output, and writes the standard XML document header. With the document created, we can begin to generate the Atom feed. To figure out the format (I couldn't remember from when I wrote atom.gen.php - that was ages ago :P), I ended following this helpful guide on atomenabled.org. It seems familiar - I think I might have used it when I wrote atom.gen.php. To start, we need something like this: <feed xmlns="http://www.w3.org/2005/Atom"> ...... </feed> In PHP, that translates to this: $xml->startElement("feed");$xml->writeAttribute("xmlns", "http://www.w3.org/2005/Atom"); $xml->endElement(); // </feed> Next, we probably want to advertise how the Atom feed was generated. Useful for letting the curious know what a website is powered by, and for spotting usage of your code out in the wild! Since I'm writing this for Pepperminty Wiki, I'm settling on something not unlike this: <generator uri="https://github.com/sbrl/Pepperminty-Wiki/" version="v0.18-dev">Pepperminty Wiki</generator> In PHP, this translates to this: $xml->startElement("generator"); $xml->writeAttribute("uri", "https://github.com/sbrl/Pepperminty-Wiki/");$xml->writeAttribute("version", $version); // A variable defined elsewhere in Pepperminty Wiki$xml->text("Pepperminty Wiki"); $xml->endElement(); Next, we need to add a <link rel="self" /> tag. This informs clients as to where the feed was fetched from, and the canonical URL of the feed. I've done this: xml->startElement("link");$xml->writeAttribute("rel", "self"); $xml->writeAttribute("type", "application/atom+xml");$xml->writeAttribute("href", full_url()); $xml->endElement(); That full_url() function is from StackOverflow, and calculates the full URI that was used to make a request. As Pepperminty Wiki can be run in any directory on ayn server, I can't pre-determine this url - hence the complexity. Note also that I output type="application/atom+xml" here. This specifies the type of content that can be found at the supplied URL. The idea here is that if you represent the same data in different ways, you can advertise them all in a standard way, with other formats having rel="alternate". Pepperminty Wiki does this - generating the recent changes list in HTML, CSV, and JSON in addition to the new Atom feed I'm blogging about here (the idea is to make the wiki data as accessible and easy-to-parse as possible). Let's advertise those too: $xml->startElement("link"); $xml->writeAttribute("rel", "alternate");$xml->writeAttribute("type", "text/html"); $xml->writeAttribute("href", "$full_url_stem?action=recent-changes&format=html"); $xml->endElement();$xml->startElement("link"); $xml->writeAttribute("rel", "alternate");$xml->writeAttribute("type", "application/json"); $xml->writeAttribute("href", "$full_url_stem?action=recent-changes&format=json"); $xml->endElement();$xml->startElement("link"); $xml->writeAttribute("rel", "alternate");$xml->writeAttribute("type", "text/csv"); $xml->writeAttribute("href", "$full_url_stem?action=recent-changes&format=csv"); $xml->endElement(); Before we can output the articles themselves, there are a few more pieces of metadata left on our laundry list - namely <updated />, <id />, <icon />, <title />, and <subtitle />. There are others in the documentation too, but aren't essential (as far as I can tell) - and not appropriate in this specific case. Here's what they might look like: <updated>2019-02-02T21:23:43+00:00</updated> <id>https://wiki.bobsrockets.com/?action=recent-changes&format=atom</id> <icon>https://wiki.bobsrockets.com/rocket_logo.png</icon> <title>Bob's Wiki - Recent Changes</title> <subtitle>Recent Changes on Bob's Wiki</subtitle> The <updated /> tag specifies when the feed was last updated. It's unclear as to whether it's the date/time the last change was made to the feed or the date/time the feed was generated, so I've gone with the latter. If this isn't correct, please let me know and I'll change it. The <id /> element can contain anything, but it must be a globally-unique string that identifies this feed. I've seen other feeds use the canonical url - and I've gone to the trouble of calculating it for the <link rel="self" /> - so it seems a shame to not use it here too. The remaining elements (<icon />, <title />, and <subtitle />) are pretty self explanatory - although it's worth mentioning that the icon must be square apparently. Let's whip those up with some more PHP: $xml->writeElement("updated", date(DateTime::ATOM)); $xml->writeElement("id", full_url());$xml->writeElement("icon", $settings->favicon);$xml->writeElement("title", "$settings->sitename - Recent Changes");$xml->writeElement("subtitle", "Recent Changes on $settings->sitename"); PHP even has a present for generating a date string in the correct format required by the spec :D $settings is an object containing the wiki settings that's a parsed form of peppermint.json, and contains useful things like the wiki's name, icon, etc. Finally, with all the preamble done, we can turn to the articles themselves. In the case of Pepperminty Wiki, the final result will look something like this: <entry> <title type="text">Edit to Compute Core by Sean</title> <id>https://seanssatellites.co.uk/wiki/?page=Compute%20Core</id> <updated>2019-01-29T10:21:43+00:00</updated> <content type="html"><ul> <li><strong>Change type:</strong> edit</li> <li><strong>User:</strong> Sean</li> <li><strong>Page name:</strong> Compute Core</li> <li><strong>Timestamp:</strong> Tue, 29 Jan 2019 10:21:43 +0000</li> <li><strong>New page size:</strong> 1.36kb</li> <li><strong>Page size difference:</strong> +1</li> </ul></content> <link rel="alternate" type="text/html" href="https://seanssatellites.co.uk/wiki/?page=Compute%20Core"/> <author> <name>Sean</name> <uri>https://seanssatellites.co.uk/wiki/?page=Users%2FSean</uri> </author> </entry> There are a bunch of elements here that deserve attention: • <title /> - The title of the article. Easy peasy! • <id /> - Just like the id of the feed itself, each article entry needs an id too. Here I've followed the same system I used for the feed, and given the url of the page content. • <updated /> - The last time the article was updated. Since this is part of a feed of recent changes, I've got this information readily at hand. • <content /> - The content to display. If the content is HTML, it must be escaped and type="html" present to indicate this. • <link rel="alternate" /> Same deal as above, but on an article-by-article level. In this case, it should link to the page the article content is from. In this case, I link to the page & revision of the change in question. In other cases, you might link to the blog post in question for example. • <author /> - Can contain <name />, <uri />, and <email />, and should indicate the author of the content. In this case, I use the name of the user that made the change, along with a link to their user page. Here's all that in PHP: foreach($recent_changes as$recent_change) { if(empty($recent_change->type))$recent_change->type = "edit"; $xml->startElement("entry"); // Change types: revert, edit, deletion, move, upload, comment$type = $recent_change->type;$url = "$full_url_stem?page=".rawurlencode($recent_change->page); $content = ".......";$xml->startElement("title"); $xml->writeAttribute("type", "text");$xml->text("$type$recent_change->page by $recent_change->user");$xml->endElement(); $xml->writeElement("id",$url); $xml->writeElement("updated", date(DateTime::ATOM,$recent_change->timestamp)); $xml->startElement("content");$xml->writeAttribute("type", "html"); $xml->text($content); $xml->endElement();$xml->startElement("link"); $xml->writeAttribute("rel", "alternate");$xml->writeAttribute("type", "text/html"); $xml->writeAttribute("href",$url); $xml->endElement();$xml->startElement("author"); $xml->writeElement("name",$recent_change->user); $xml->writeElement("uri", "$full_url_stem?page=".rawurlencode("$settings->user_page_prefix/$recent_change->user")); $xml->endElement();$xml->endElement(); } I've omitted the logic that generates the value of the <content /> tag, as it's not really relevant here (you can check it out here if you're curious :D). This about finishes the XML we need to generate for our feed. To extract the XML from the XMLWriter, we can do this: $atom_feed =$xml->flush(); Then we can do whatever we want to with the generated XML! When the latest version of Pepperminty Wiki comes out, you'll be able to see a live demo here! Until then, you'll need to download a copy of the latest master version and experiment with it yourself. I'll also include a complete demo feed below: <?xml version="1.0" encoding="UTF-8"?> <feed xmlns="http://www.w3.org/2005/Atom"> <generator uri="https://github.com/sbrl/Pepperminty-Wiki/" version="v0.18-dev">Pepperminty Wiki</generator> <link rel="self" type="application/atom+xml" href="http://[::]:35623/?action=recent-changes&format=atom&count=3"/> <link rel="alternate" type="text/html" href="http://[::]:35623/?action=recent-changes&format=html"/> <link rel="alternate" type="application/json" href="http://[::]:35623/?action=recent-changes&format=json"/> <link rel="alternate" type="text/csv" href="http://[::]:35623/?action=recent-changes&format=csv"/> <updated>2019-02-03T17:25:10+00:00</updated> <id>http://[::]:35623/?action=recent-changes&format=atom&count=3</id> <icon>data:image/png;base64,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</icon> <title>Pepperminty Wiki - Recent Changes</title> <subtitle>Recent Changes on Pepperminty Wiki</subtitle> <entry> <title type="text">Edit to Internal link by admin</title> <id>http://[::]:35623/?page=Internal%20link</id> <updated>2019-01-29T19:55:08+00:00</updated> <content type="html"><ul> <li><strong>Change type:</strong> edit</li> <li><strong>User:</strong> admin</li> <li><strong>Page name:</strong> Internal link</li> <li><strong>Timestamp:</strong> Tue, 29 Jan 2019 19:55:08 +0000</li> <li><strong>New page size:</strong> 2.11kb</li> <li><strong>Page size difference:</strong> +2007</li> </ul></content> <link rel="alternate" type="text/html" href="http://[::]:35623/?page=Internal%20link"/> <author> <name>admin</name> <uri>http://[::]:35623/?page=Users%2FInternal%20link</uri> </author> </entry> <entry> <title type="text">Edit to Main Page by admin</title> <id>http://[::]:35623/?page=Main%20Page</id> <updated>2019-01-05T20:14:07+00:00</updated> <content type="html"><ul> <li><strong>Change type:</strong> edit</li> <li><strong>User:</strong> admin</li> <li><strong>Page name:</strong> Main Page</li> <li><strong>Timestamp:</strong> Sat, 05 Jan 2019 20:14:07 +0000</li> <li><strong>New page size:</strong> 317b</li> <li><strong>Page size difference:</strong> +68</li> </ul></content> <link rel="alternate" type="text/html" href="http://[::]:35623/?page=Main%20Page"/> <author> <name>admin</name> <uri>http://[::]:35623/?page=Users%2FMain%20Page</uri> </author> </entry> <entry> <title type="text">Edit to Main Page by admin</title> <id>http://[::]:35623/?page=Main%20Page</id> <updated>2019-01-05T17:53:08+00:00</updated> <content type="html"><ul> <li><strong>Change type:</strong> edit</li> <li><strong>User:</strong> admin</li> <li><strong>Page name:</strong> Main Page</li> <li><strong>Timestamp:</strong> Sat, 05 Jan 2019 17:53:08 +0000</li> <li><strong>New page size:</strong> 249b</li> <li><strong>Page size difference:</strong> +31</li> </ul></content> <link rel="alternate" type="text/html" href="http://[::]:35623/?page=Main%20Page"/> <author> <name>admin</name> <uri>http://[::]:35623/?page=Users%2FMain%20Page</uri> </author> </entry> </feed> ...this is from my local development instance. Found this interesting? Confused about something? Want to say hi? Comment below! ## Keeping the Internet free and open for years to come - #ForTheWeb I've recently come across contractfortheweb.org. It sounds obvious, but certain qualities of the web that we take for granted aren't, in fact, universal. Things like a lack of censorship (China & their great firewall; Cuba; Venezuela), consumer privacy (US Mobile Carriers; Google; Google/Android Antitrust), and fair pricing (AT&T; BT) are rather a problem in more places than is noticeable at first glance. The Contract for the Web is a set of principles - backed by the famous Tim Berners-Lee - for a free, open, and fair Internet. The aim is to build a full contract based on these principles to guide the evolution of the web for year to come. Personally, I first really started to use the web to learn how to program. By reading (lots) of tutorials freely available on the web, I learnt to build websites, write Javascript, and more! Since then, I've used the web to share the things I've been learning on this blog, stay up-to-date with the latest technology news, research new ideas, and play games that deliver amazing experiences. The web (and, by extension, the Internet - the web refers to just HTTP + HTTPS) for me represents freedom of information. Freedom to express (and learn about) new thoughts and ideas without fear of being censored. Freedom to communicate with anyone in the world - regardless of physical distances. It is for this reason that I've signed the Contract for the Web. It's my hope that this effort will ensure that the Internet becomes more open and neutral going forwards, so that everyone can experience the benefits of the open web for a long time to come. ## Building Javascript (and other things) with Rollup Hey, another blog post! Recently I've been somewhat distracted by another project which has been most interesting. I've learnt a bunch of things (including getting started with LeafletJS and Chart.JS), but most notably I've been experimenting with another Javascript build system. I've used quite a few build systems over the years. For the uninitiated, their primary purpose is to turn lots of separate source files into as few output files as possible. This is important with things that run in the browser, because the browser has to download everything before it can execute it. Fewer files and less data means a speedier application - so everyone wins. I've been rather unhappy with the build systems I've used so far: • Gulp.js was really heavy and required a lot of configuration in order to get things moving. • Browserify was great, but I found the feature set lacking and ended up moving on because it couldn't do what I wanted it to. • Webpack was ok, but it also has a huge footprint and was super slow too. One thing led to another, and after trying a few other systems (I can't remember their names, but one begin with P and has a taped-up cardboard box as a logo) I found Rollup (I think Webpack requires Rollup as a dependency under the hood, if I'm not mistaken). Straight away, I found that Rollup was much faster than Webpack. It also requires minimal configuration (which doesn't need changing every time you create a new source file) - which allows me to focus on the code I'm writing instead of battling with the build system to get it to do what I want. It also supports plugins - and with a bunch available on npm, there's no need to write your own plugin to perform most common tasks. For my recent project I linked to above, I was able to put a bunch of plugins together with a configuration file - and, with a bit of fiddling around, I've got it to build both my Javascript and my CSS (via a postcss integration plugin) and putt he result in a build output folder that I've added to my .gitignore file. I ended up using the following plugins: • rollup-plugin-node-resolve - essential. Allows import statements to be automatically resolved to their respective files. • rollup-plugin-commonjs - An addition to the above, if I recall correctly • rollup-plugin-postcss - Provides integration with Postcss. I've found Postcss with be an effective solution to the awkward issue of CSS • rollup-plugin-terser - Uses Terser to minify modern Javascript. I didn't know this existed until I found the Rollup plugin - it's a fork of and the successor to UglifyJS2, which only supports ES5 syntax. ....and the following Postcss plugins: • postcss-import - Inlines @import statements into a single CSS source file, allowing you to split your CSS up into multiple files (finally! :P) • postcss-copy - after trying 5 different alternatives, I found this to be the most effective at automatically copying images etc. to the build output folder - potentially renaming them - and altering the corresponding references in the CSS source code The definition of plugins is done in a Javascript configuration file in a array - making it clear that your code goes through a pipeline of plugins that transform step-by-step to form the final output - creating and maintaining a source map along the way (very handy for debugging code in the browser. It's basically a file that lets the browser reverse-engineer the final output to present you with the original source code again whilst debugging) - which works flawlessly in my browser (the same can't be said for Webpack - I had lots of issues with the generated source maps there). I ended up with a config file like this: (Can't see the file above? Try viewing it directly.) In short, the pipeline goes something like this: (Source file; Created in Draw.io) Actually calling Rollup is done in a Bash-based general build task file. Rollup is installed locally to the project (npm install --save rollup), so I call it like this: node_modules/.bin/rollup --sourcemap --config rollup.config.js; I should probably make a blog post about that build task system I wrote and why/how I made it, but that's a topic for another time :P I'll probably end up tweaking the setup I've described here a bit in the future, but for now it does everything I'd like it to :-) Found this interesting? Got a different way of doing it? Still confused? Comment below! ## Bridging the gap between XMPP and shell scripts In a previous post, I set up a semi-automated backup system for my Raspberry Pi using duplicity, sendxmpp, and an external drive. It's been working fabulously for a while now, but unfortunately the other week sendxmpp suddenly stopped working with no obvious explanation. Given the long list of arguments I had to pass it: sendxmpp --file "${xmpp_config_file}" --resource "${xmpp_resource}" --tls --chatroom "${xmpp_target_chatroom}" ........... ....and the fact that I've had to tweak said arguments on a number of occasions, I thought it was time to switch it out for something better suited to the task at hand. Unfortunately, finding such a tool proved to be a challenge. I even asked on Reddit - but nobody had anything that fit the bill (xmpp-bridge wouldn't compile correctly - and didn't support multi-user chatrooms anyway, and xmpppy was broken too). If you're unsure as to what XMPP is, I'd recommend checkout out either this or this tutorial. They both give a great introduction to what it is, what it does, and how it works - and the rest of this post will make much more sense if you read that first :-) To this end, I finally gave in and wrote my own tool, which I've called xmppbridge. It's a global Node.JS script that uses the simple-xmpp to forward the standard input to a given JID over XMPP - which can optionally be a group chat. In this post, I'm going to look at how I put it together, some of the issues I ran into along the way, and how I solved them. If you're interested in how to install and use it, then the package page on npm will tell you everything you need to know: xmppbridge on npm ### Architectural Overview The script consists of 3 files: • index.sh - Calls the main script with ES6 modules enabled • index.mjs - Parses the command-line arguments and environment variables out, and provides a nice CLI • XmppBridge.mjs - The bit that actually captures input from stdin and sends it via XMPP Let's look at each of these in turn - starting with the command-line interface. ### CLI Parsing The CLI itself is relatively simple - and follows a paradigm I've used extensively in C♯ (although somewhat modified of course to get it to work in Node.JS, and without fancy ANSI colouring etc.). #!/usr/bin/env node "use strict"; import XmppBridge from './XmppBridge.mjs'; const settings = { jid: process.env.XMPP_JID, destination_jid: null, is_destination_groupchat: false, password: process.env.XMPP_PASSWORD }; let extras = []; // The first arg is the script name itself for(let i = 1; i < process.argv.length; i++) { if(!process.argv[i].startsWith("-")) { extras.push(process.argv[i]); continue; } switch(process.argv[i]) { case "-h": case "--help": // ........ break; // ........ default: console.error(Error: Unknown argument '${process.argv[i]}'.); process.exit(2); break; } } We start with a shebang, telling Linux-based systems to execute the script with Node.JS. Following that, we import the XmppBridge class that's located in XmppBrdige.mjs (we'll come back to this later). Then, we define an object to hold our settings - and pull in the environment variables along with defining some defaults for other parameters. With that setup, we can then parse the command-line arguments themselves - using the exact same paradigm I've used time and time again in C♯. Once the command-line arguments are parsed, we validate the final settings to ensure that the user hasn't left any required parameters undefined: for(let environment_varable of ["XMPP_JID", "XMPP_PASSWORD"]) { if(typeof process.env[environment_varable] == "undefined") { console.error(Error: The environment variable ${environment_varable} wasn't found.); process.exit(1); } } if(typeof settings.destination_jid != "string") { console.error("Error: No destination jid specified."); process.exit(5); } That's basically all that index.mjs does. All that's really left is passing the parameters to an instance of XmppBridge: const bridge = new XmppBridge( settings.destination_jid, settings.is_destination_groupchat ); bridge.start(settings.jid, settings.password); ### Shebang Trouble Because I've used ES6 modules here, currently Node must be informed of this via the --experimental-modules CLI argument like this: node --experimental-modules ./index.mjs If we're going to make this a global command-line tool via the bin directive in package.json, then we're going to have to ensure that this flag gets passed to Node and not our program. While we could alter the shebang, that comes with the awkward problem that not all systems (in fact relatively few) support using both env and passing arguments. For example, this: #!/usr/bin/env node --experimental-modules Wouldn't work, because env doesn't recognise that --experimental-modules is actually a command-line argument and not part of the binary name that it should search for. I did see some Linux systems support env -S to enable this functionality, but it's hardly portable and doesn't even appear to work all the time anyway - so we'll have to look for another solution. Another way we could do it is by dropping the env entirely. We could do this: #!/usr/local/bin/node --experimental-modules ...which would work fine on my system, but probably not on anyone else's if they haven't installed Node to the same place. Sadly, we'll have to throw this option out the window too. We've still got some tricks up our sleeve though - namely writing a bash wrapper script that will call node telling it to execute index.mjs with the correct arguments. After a little bit of fiddling, I came up with this: #!/usr/bin/env bash install_dir="$(dirname "$(readlink -f$0)")"; exec node --experimental-modules "${install_dir}/index.mjs"$@ 2 things are at play here. Firstly, we have to deduce where the currently executing script actually lies - as npm uses a symbolic link to allow a global command-line tool to be 'found'. Said symbolic link gets put in /usr/local/bin/ (which is, by default, in the user's PATH), and links to where the script is actually installed to. To figure out the directory that we've been installed to is (and hence the location of index.mjs), we need to dereference the symbolic link and strip the index.sh filename away. This can be done with a combination of readlink -f (dereferences the symbolic link), dirname (get the parent directory of a given file path), and $0 (holds the path to the currently executing script in most circumstances) - which, in the case of the above, gets put into the install_dir variable. The other issue is passing all the existing command-line arguments to index.mjs unchanged. We do this with a combination of $@ (which refers to all the arguments passed to this script except the script name itself) and exec (which replaces the currently executing process with a new one - in this case it replaces the bash shell with node). This approach let's us customise the CLI arguments, while still providing global access to our script. Here's an extract from xmppbridge's package.json showing how I specify that I want index.sh to be a global script: { ..... "bin": { "xmppbridge": "./index.sh" }, ..... } ### Bridging the Gap Now that we've got Node calling our script correctly and the arguments parsed out, we can actually bridge the gap. This is as simple as some glue code between simple-xmpp and readline. simple-xmpp is an npm package that makes programmatic XMPP interaction fairly trivial (though I did have to look at examples in the GitHub repository to figure out how to send a message to a multi-user chatroom). readline is a Node built-in that allows us to read the standard input line-by-line. It does other things too (and is great for interactive scripts amongst other things), but that's a tale for another time. The first task is to create a new class for this to live in: "use strict"; import readline from 'readline'; import xmpp from 'simple-xmpp'; class XmppBridge { /** * Creates a new XmppBridge instance. * @param {string} in_login_jid The JID to login with. * @param {string} in_destination_jid The JID to send stdin to. * @param {Boolean} in_is_groupchat Whether the destination JID is a group chat or not. */ constructor(in_destination_jid, in_is_groupchat) { // .... } } export default XmppBridge; Very cool! That was easy. Next, we need to store those arguments and connect to the XMPP server in the constructor: this.destination_jid = in_destination_jid; this.is_destination_groupchat = in_is_groupchat; this.client = xmpp; this.client.on("online", this.on_connect.bind(this)); this.client.on("error", this.on_error.bind(this)); this.client.on("chat", ((_from, _message) => { // noop }).bind(this)); I ended up having to define a chat event handler - even though it's pointless, as I ran into a nasty crash if I didn't do so (I suspect that this use-case wasn't considered by the original package developer). The next area of interest is that online event handler. Note that I've bound the method to the current this context - this is important, as it would be able to access the class instance's properties otherwise. Let's take a look at the code for that handler: console.log([XmppBridge] Connected as ${data.jid}.); if(this.is_destination_groupchat) { this.client.join(${this.destination_jid}/bot_{data.jid.user}); } this.stdin = readline.createInterface({ input: process.stdin, output: process.stdout, terminal: false }); this.stdin.on("line", this.on_line_handler.bind(this)); this.stdin.on("close", this.on_stdin_close_handler.bind(this)); This is the point at which we open the standard input and start listening for things to send. We don't do it earlier, as we don't want to end up in a situation where we try sending something before we're connected! If we're supposed to be sending to a multi-user chatroom, this is also the point at which it joins said room. This is required as you can't send a message to a room that you haven't joined. The resource (the bit after the forward slash /), for a group chat, specifies the nickname that you want to give to yourself when joining. Here, I automatically set this to the user part of the JID that we used to login prefixed with bot_. The connection itself is established in the start method: start(jid, password) { this.client.connect({ jid, password }); } And every time we receive a line of input, we execute the send() method: on_line_handler(line_text) { this.send(line_text); } I used a full method here, as initially I had some issues and wanted to debug which methods were being called. That send method looks like this: send(message) { this.client.send( this.destination_jid, message, this.is_destination_groupchat ); } The last event handler worth mentioning is the close event handler on the readline interface: on_stdin_close_handler() { this.client.disconnect(); } This just disconnects from the XMXPP server so that Node can exit cleanly. That basically completes the script. In total, the entire XmppBridge.mjs class file is 72 lines. Not bad going! You can install this tool for yourself with sudo npm install -g xmppbridge. I've documented how it use it in the README, so I'd recommend heading over there if you're interested in trying it out. Found this interesting? Got a cool use for XMPP? Comment below! ### Sources and Further Reading ## Compilers, VMs, and JIT: Spot the difference It's about time for another demystification post, I think :P This time, I'm going to talk about Compilers, Virtual Machines (VMs), and Just-In-Time Compilation (JIT) - and the way that they are both related and yet different. ### Compilers To start with, a compiler is a program that converts another program written in 1 language into another language (usually of a lower level). For example, gcc compiles C++ into native machine code. A compiler usually has both a backend and a frontend. The frontend is responsible for the lexing and parsing of the source programming language. It's the part of the compiler that generates compiler warnings and errors. The frontend outputs an abstract syntax tree (or parse tree) that represents the source input program. The backend then takes this abstract syntax tree, walks it, and generates code in the target output language. Such code generators are generally recursive in nature. Depending on the compiler toolchain, this may or may not be the last step in the build process. gcc, for example, generates code to object files - which are then strung together by a linking program later in the build process. Additional detail on the structure of a compiler (and how to build one yourself!) is beyond the scope of this article, but if you're interested I recommend reading my earlier Compilers 101 post. ### Virtual Machines A virtual machine comes in several flavours. Common to all types is the ability to execute instructions - through software - as if they were being executed on a 'real' hardware implementation of the programming language in question. Examples here include: • KVM - The Linux virtual machine engine. Leverages hardware extensions to support various assembly languages that are implemented in real hardware - everything from Intel x86 Assembly to various ARM instruction sets. Virtual Machine Manager is a good GUI for it. • VirtualBox - Oracle's offering. Somewhat easier to use than the above - and cross-platform too! • .NET / Mono - The .NET runtime is actually a virtual machine in it's own right. It executes Common Intermediate Language in a managed environment. It's also important to note the difference between a virtual machine and an emulator. An emulator is very similar to a virtual machine - except that it doesn't actually implement a language or instruction set itself. Instead, it 'polyfills' hardware (or environment) features that don't exist in the target runtime environment. A great example here is WINE or Parallels, which allow programs written for Microsoft Windows to be run on other platforms without modification. ### Just-In-Time Compilation JIT is sort of a combination of the above. The .NET runtime (officially knows as the Common Language Runtime, or CLR) is an example of this too - in that it compiles the CIL in the source assembly to native code just before execution. Utilising such a mechanism does result in an additional delay during startup, but usually pays for itself in the vast performance improvements that can be made over an interpreter - as a JITting VM can automatically optimise code as it's executing to increase performance in real-time. This is different to an interpreter, which reads a source program line-by-line - parsing and executing it as it goes. If it needs to go back to another part of the program, it will usually have to re-parse the code before it can execute it. ### Conclusion In this post, we've taken a brief look at compilers, VMs, and JIT. We've looked at the differences between them - and how they are different from their siblings (emulators and interpreters). In many ways, the line between the 3 can become somewhat blurred (hello, .NET!) - so learning the characteristics of each is helpful for disentangling different components of a program. If there's anything I've missed - please let me know! If you're still confused on 1 or more points, I'm happy to expand on them if you comment below :-) Found this useful? Spotted a mistake? Comment below! ## Setup your very own VPN in 10 minutes flat Hey! Happy new year :-) I've been looking to setup a personal VPN for a while, and the other week I discovered a rather brilliant project called PiVPN, which greatly simplifies the process of setting one up - and managing it thereafter. It's been working rather well so far, so I thought I'd post about it so you can set one up for yourself too. But first though, we should look at the why. Why a VPN? What does it do? Basically, a VPN let you punch a great big hole in the network that you're connected to and appear as if you're actually on a network elsewhere. The extent to which this is the case varies depending on the purpose, (for example a University or business might setup a VPN that allows members to access internal resources, but doesn't route all traffic through the VPN), but the general principle is the same. It's best explained with a diagram. Imagine you're at a Café: Everyone on the Café's WiFi can see the internet traffic you're sending out. If any of it is unencrypted, then they can additionally see the content of said traffic - e.g. emails you send, web pages you load, etc. Even if it's encrypted, statistical analysis can reveal which websites you're visiting and more. If you don't trust a network that you're connected to, then by utilising a VPN you can create an encrypted tunnel to another location that you do trust: Then, all that the other users of the Café's WiFi will see is an encrypted stream of packets - all heading for the same destination. All they'll know is roughly how much traffic you're sending and receiving, but not to where. This is the primary reason that I'd like my own VPN. I trust the network I've got setup in my own house, so it stands to reason that I'd like to setup a VPN server there, and pretend that my devices when I'm out and about are still at home. In theory, I should be able to access the resources on my home network too when I'm using such a VPN - which is an added bonus. Other reasons do exist for using a VPN, but I won't discuss them here. In terms of VPN server software, I've done a fair amount of research into the different options available. My main criteria are as follows: • Fairly easy to install • Easy to understand what it's doing once installed (transparency) • Easy to manage The 2 main technologies I came across were OpenVPN and IPSec. Each has their own strengths & weaknesses. An IPSec VPN is, apparently, more efficient - especially since it executes on the client in kernel-space instead of user-space. It's a lighter protocol, too - leading to less overhead. It's also much more likely to be detected and blocked when travelling through strict firewalls, making me slightly unsure about it. OpenVPN, on the other hand, executes entirely in user-space on both the client and the server - leading to a slightly greater overhead (especially with the mitigations for the recent Spectre & Meltdown hardware bugs). It does, however, use TLS (though over UDP by default). This characteristic makes it much more likely it'll slip through stricter firewalls. I'm unsure if that's a quality that I'm actually after or not. Ultimately, it's the ease of management that points the way to my final choice. Looking into it, with both choices there's complex certificate management to be done whenever you want to add a new client to the VPN. For example, with StrongSwan (an open-source IPSec VPN program), you've got to generate a number of certificates with a chain of rather long commands - and the users themselves have passwords stored in plain text in a file! While I've got no problem with reading and understanding such commands, I do have a problem with rememberability. If I want to add a new client, how easy is that to do? How long would I have to spend re-reading documentation to figure out how to do it? Sure, I could write a program to manage the configuration files for me, but that would also require maintenance - and probably take much longer than I anticipate to write. I forget where I found it, but it is for this reason that I ultimately decided to choose PiVPN. It's a set of scripts that sets up and manages one's an OpenVPN installation. To this end, it provides a single command - pivpn - that can be used to add, remove, and list clients and their statistics. With a concise help text, it makes it easy to figure out how to perform common tasks utilising existing terminal skills by conforming to established CLI interface norms. If you want to install it yourself, then simply do this: curl -L https://install.pivpn.io \| bash Of course, simply downloading and executing a random script from the Internet is never a good idea. Let's read it first: curl -L https://install.pivpn.io \| less Once you're happy that it's not going to do anything malign to your system, proceed with the installation by executing the 1st command. It should guide you through a number of screens. Some important points I ran into: • The static IP address it talks about is the IP address of your server on the local network. The installation asks about the public IP address in a later step. If you've already got a static IP setup on your server (and you probably have), then you don't need to worry about this. • It asks you to install and enable unattended-upgrades. You should probably do this, but I ended up skipping this - as I've already got apticron setup and sending me regular emails - as I rather like to babysit the upgrade of packages on the main machines I manage. I might look into unattended-upgrades in the future if I acquire more servers than are comfortable to manage this way. • Make sure you fully update your system before running the installation. I use this command: sudo apt update && sudo apt-get dist-upgrade && sudo apt-get autoclean && sudo apt-get autoremove • Changing the port of the VPN isn't a bad idea, since PiVPN will automatically assemble .ovpn configuration files for you. I didn't end up doing this to start with, but I can always change it in the NAT rule I configured on my router later. • Don't forget to allow OpenVPN through your firewall! For ufw users (like me), then it's something like sudo ufw allow <port_number>/udp. • Don't forget to setup a NAT rule / port forwarding on your router if said server doesn't have a public IP address (if it's IPv4 it probably doesn't). If you're confused on this point, comment below and I'll blog about it. It's..... a complicated topic. If you'd like a more in-depth guide to setting up PiVPN, then I can recommend this guide. It's a little bit dated (PiVPN now uses elliptical-curve cryptography by default), but still serves to illustrate the process pretty well. If you're confused about some of the concepts I've presented here - leave a comment below! I'm happy to explain them in more detail. Who knows - I might end up writing another blog post on the subject.... ## Where in the world does spam come from? Answer: The US, apparently. I was having a discussion with someone recently, and since I have a rather extensive log of comment failures for debugging & analysis purposes (dating back to February 2015!) they suggested that I render a map of where the spam is coming from. It was such a good idea that I ended up doing just that - and somehow also writing this blog post :P First, let's start off by looking at the format of said log file: [ Sun, 22 Feb 2015 07:37:03 +0000] invalid comment \| ip: a.b.c.d \| name: Nancyyeq \| articlepath: posts/015-Rust-First-Impressions.html \| mistake: longcomment [ Sun, 22 Feb 2015 14:55:50 +0000] invalid comment \| ip: e.f.g.h \| name: Simonxlsw \| articlepath: posts/015-Rust-First-Impressions.html \| mistake: invalidkey [ Sun, 22 Feb 2015 14:59:59 +0000] invalid comment \| ip: x.y.z.w \| name: Simontuxc \| articlepath: posts/015-Rust-First-Impressions.html \| mistake: invalidkey Unfortunately, I didn't think about parsing it programmatically when I designed the log file format.... Oops! It's too late to change it now, I suppose :P Anyway, as an output, we want a list of countries in 1 column, and a count of the number of IP addresses in another. First things first - we need to extract those IP addresses. awk is ideal for this. I cooked this up just quickly: BEGIN { FS="\|" } { gsub(" ip: ", "",2); print $2; } This basically tells awk to split lines on the solid bar character (\|), extracts the IP address bit (ip: p.q.r.s), and then strips out the ip: bit. With this done, we're ready to lookup all these IP addresses to find out which country they're from. Unfortunately, IP addresses can change hands semi-regularly - even across country borders, so my approach here isn't going to be entirely accurate. I don't anticipate the error generated here to be all that big though, so I think it's ok to just do a simple lookup. If I was worried about it, I could probably investigate cross-referencing the IP addresses with a GeoIP database from the date & time I recorded them. The effort here would be quite considerable - and this is a 'just curious' sort of thing, so I'm not going to do that here. If you have done this, I'd love to hear about it though - post a comment below. Actually doing a GeoIP lookup itself is fairly easy to do, actually. While for the odd IP address here and there I usually use ipinfo.io, when there are lots of lookups to be done (10,479 to be exact! Wow.), it's probably best to utilise a local database. A quick bit of research reveals that Ubuntu Server has a package I can install that should do the job called geoip-bin: sudo apt install geoip-bin (....) geoiplookup 1.1.1.1 # CloudFlare's 1.1.1.1 DNS service GeoIP Country Edition: AU, Australia Excellent! We can now lookup IP addresses automagically via the command line. Let's plug that in to the little command chain we got going on here: cat failedcomments.log \| awk 'BEGIN { FS="\|" } { gsub(" ip: ", "",$2); print $2 }' \| xargs -n1 geoiplookup It doesn't look like geoiplookup supports multiple IP addresses at once, which is a shame. In that case, the above will take a while to execute for 10K IP addresses.... :P Next up, we need to remove the annoying label there. That's easy with sed: (...) \| sed -E 's/^[A-Za-z: ]+, //g' I had some trouble here getting sed to accept a regular expression. At some point I'll have to read the manual pages more closely and write myself a quick reference guide. Come to think about it, I could use such a thing for awk too - their existing reference guide appears to have been written by a bunch of mathematicians who like using single-letter variable names everywhere. Anyway, now that we've got our IP address list, we need to strip out any errors, and then count them all up. The first point is somewhat awkward, since geoiplookup doesn't send errors to the standard error for some reason, but we can cheese it with grep -v: (...) \| grep -iv 'resolve hostname' The -v here tells grep to instead remove any lines that match the specified string, instead of showing us only the matching lines. This appeared to work at first glance - I simply copied a part of the error message I saw and worked with that. If I have issues later, I can always look at writing a more sophisticated regular expression with the -P option. The counting bit can be achieved in bash with a combination of the sort and uniq commands. sort will, umm, sort the input lines, and uniq with de-duplicate multiple consecutive input lines, whilst optionaly counting them. With this in mind, I wound up with the following: (...) \| sort \| uniq -c \| sort -n The first sort call sorts the input to ensure that all identical lines are next to each other, reading for uniq. uniq -c does the de-duplication, but also inserts a count of the number of duplicates for us. Lastly, the final sort call with the -n argument sorts the completed list via a natural sort, which means (in our case) that it handles the numbers as you'd expect it too. I'd recommend you read the Wikipedia article on the subject - it explains it quite well. This should give us an output like this: 1 Antigua and Barbuda 1 Bahrain 1 Bouvet Island 1 Egypt 1 Europe 1 Guatemala 1 Ireland 1 Macedonia 1 Mongolia 1 Saudi Arabia 1 Tuvalu 2 Bolivia 2 Croatia 2 Luxembourg 2 Paraguay 3 Kenya 3 Macau 4 El Salvador 4 Hungary 4 Lebanon 4 Maldives 4 Nepal 4 Nigeria 4 Palestinian Territory 4 Philippines 4 Portugal 4 Puerto Rico 4 Saint Martin 4 Virgin Islands, British 4 Zambia 5 Dominican Republic 5 Georgia 5 Malaysia 5 Switzerland 6 Austria 6 Belgium 6 Peru 6 Slovenia 7 Australia 7 Japan 8 Afghanistan 8 Argentina 8 Chile 9 Finland 9 Norway 10 Bulgaria 11 Singapore 11 South Africa 12 Serbia 13 Denmark 13 Moldova, Republic of 14 Ecuador 14 Romania 15 Cambodia 15 Kazakhstan 15 Lithuania 15 Morocco 17 Latvia 21 Pakistan 21 Venezuela 23 Mexico 23 Turkey 24 Honduras 24 Israel 29 Czech Republic 30 Korea, Republic of 32 Colombia 33 Hong Kong 36 Italy 38 Vietnam 39 Bangladesh 40 Belarus 41 Estonia 44 Thailand 50 Iran, Islamic Republic of 53 Spain 54 GeoIP Country Edition: IP Address not found 60 Poland 88 India 113 Netherlands 113 Taiwan 124 Indonesia 147 Sweden 157 Canada 176 United Kingdom 240 Germany 297 China 298 Brazil 502 France 1631 Russian Federation 2280 Ukraine 3224 United States Very cool. Here's the full command for reference explainshell explanation: cat failedcomments.log \| awk 'BEGIN { FS="\|" } { gsub(" ip: ", "",$2); print \$2 }' \| xargs -n1 geoiplookup \| sed -e 's/GeoIP Country Edition: //g' \| sed -E 's/^[A-Z]+, //g' \| grep -iv 'resolve hostname' \| sort \| uniq -c \| sort -n With our list in hand, I imported it into LibreOffice Calc to parse it into a table with the fixed-width setting (Google Sheets doesn't appear to support this), and then pulled that into a Google Sheet in order to draw a heat map: At first, the resulting graph showed just a few countries in red, and the rest in white. To rectify this, I pushed the counts through the natural log (log()) function, which yielded a much better map, where the countries have been spamming just a bit are still shown in a shade of red. From this graph, we can quite easily conclude that the most 'spammiest' countries are: 1. The US 2. Russia 3. Ukraine (I get lots of spam emails from here too) 4. China (I get lots of SSH intrusion attempts from here) 5. Brazil (Wat?) Personally, I was rather surprised to see the US int he top spot. I figured that with with tough laws on that sort of thing, spammers wouldn't risk attempting to buy a server and send spam from here. On further thought though, it occurred to me that it may be because there are simply lots of infected machines in the US that are being abused (without the knowledge of their unwitting users) to send lots of spam. At any rate, I don't appear to have a spam problem on my blog at the moment - it's just fascinating to investigate where the spam I do block comes from. Found this interesting? Got an observation of your own? Plotted a graph from your own data? Comment below! ## Happy Christmas 2018! Happy Christmas! I hope you have a restful and peaceful holiday. Have a picture of a cool snow-globe: Art by Mythdael	2019-02-20 00:09:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2236904799938202, "perplexity": 2750.3661367905606}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247493803.26/warc/CC-MAIN-20190219223509-20190220005509-00603.warc.gz"}
https://learn.careers360.com/medical/question-equation-of-trajectory-of-a-projectile-is-given-by-70023/	Q # Equation of trajectory of a projectile is given by Equation of trajectory of a projectile is given by where x and y are in metres and x is along horizontal and y is vertically upward and particles is projected from origin .Then which of the following option is incorrect A. Initial velocity of particle is B. horizontal range is 10 m C. Maximum range is 10 m D. Angle of projection with horizontal is 505m/s Views @Harshit Equation of path of a projectile - $y= x \tan \theta\: - \: \frac{gx^{2}}{2u^{2}\cos ^{2}\theta }$ it is equation of parabola $g\rightarrow$ Acceleratio due to gravity $u\rightarrow$ initial velocity $\theta =$ Angle of projection - wherein Path followed by a projectile is parabolic is nature. $y=- x^2+10x$ Compare this with $y= x\tan \theta - \frac{g x^2}{2u^2\cos ^2 \theta }$ $\Rightarrow \tan \theta = +10 \Rightarrow \cos \theta = \frac{1}{\sqrt{101}} \\\; \; \; And \ \frac{g }{2 u^2\cos ^2 \theta }= 1 \\\\ \ \; \; u^2 = \; \; \frac{10 }{2\cdot (\frac{1}{101})}= 505$ $u = \sqrt{505}m /s$ $R= u^2sin2\theta/{g}= \frac{505 \times 2}{101}= 10 m$ Maximum height H = $\frac{u^2 \sin ^{2}\theta }{2g }$ $H^{max } = \frac{505 \times \left ( \frac{10}{\sqrt{101}} \right )^2}{2\times 10}= \frac{505\times 100}{20 \times 101}= 25 m$ Exams Articles Questions	2019-07-21 13:24:25	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.877632200717926, "perplexity": 2148.980549642388}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195527000.10/warc/CC-MAIN-20190721123414-20190721145414-00286.warc.gz"}
https://www.physicsforums.com/threads/primitive-elements-and-free-modules.953778/	I Primitive Elements and Free Modules ... ... Math Amateur Gold Member I am reading Paul E. Bland's book, "Rings and Their Modules". I am focused on Section 4.3: Modules Over Principal Ideal Domains ... and I need some help in order to fully understand the proof of Proposition 4.3.14 ... ... In the above proof by Bland we read the following: " ... ... The induction hypothesis gives a basis $\{ x, x'_2, \ ... \ ... \ x'_{n -1} \}$ of $M$ and it follows that $\{ x, x'_2, \ ... \ ... \ x'_{n - 1}, x'_n \}$ is a basis of $F$ that contains $x$ ... ... " My question is as follows: Why/how exactly does it follow that $\{ x, x'_2, \ ... \ ... \ x'_{n - 1}, x'_n \}$ is a basis of $F$ that contains $x$. ... ... Help will be appreciated ... Peter ==================================================================================================== It may help PFmembers reading this post to have access to Bland's definition of 'primitive element of a module' ... especially as it seems to me that the definition is a bit unusual ... so I am providing the same as follows: Hope that helps ... ... Peter Attachments • 82.6 KB Views: 230 • 57.9 KB Views: 210 • 22.9 KB Views: 248 • 82.6 KB Views: 338 • 57.9 KB Views: 151 • 22.9 KB Views: 152 Related Linear and Abstract Algebra News on Phys.org andrewkirk Homework Helper Gold Member That $\{x_1,...,x_n\}$ is a basis for $F$ means the same as that $$F=x_1R\oplus...\oplus x_nR$$ and since $M\triangleq x_1R\oplus...\oplus x_{n-1}R$ we have $F=M\oplus x_nR$. Since the induction hypothesis gives us that $\{x, x'_2,...,x'_{n-1}\}$ is a basis of $M$ we have $$M = xR\oplus x'_2R\oplus x'_3R\oplus...\oplus x'_{n-1}$$ So we have $$F=M\oplus x_nR = xR\oplus x'_2R\oplus x'_3R\oplus...\oplus x'_{n-1}\oplus x_nR$$ which is equivalent to saying that $\{x,x'_2,...,x'_{n-1},x_n\}$ is a basis for $F$. Math Amateur Gold Member Thanks Andrew ... Peter "Primitive Elements and Free Modules ... ..." Physics Forums Values We Value Quality • Topics based on mainstream science • Proper English grammar and spelling We Value Civility • Positive and compassionate attitudes • Patience while debating We Value Productivity • Disciplined to remain on-topic • Recognition of own weaknesses • Solo and co-op problem solving	2019-10-20 04:06:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5436176657676697, "perplexity": 1984.0024320592775}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986702077.71/warc/CC-MAIN-20191020024805-20191020052305-00524.warc.gz"}
https://codegolf.meta.stackexchange.com/questions/2140/sandbox-for-proposed-challenges/23631	# Sandbox for Proposed Challenges This "sandbox" is a place where Code Golf users can get feedback on prospective challenges they wish to post to main. This is useful because writing a clear and fully specified challenge on your first try can be difficult, and there is a much better chance of your challenge being well received if you post it in the sandbox first. Sandbox FAQ ## Posting Write your challenge just as you would when actually posting it, though you can optionally add a title at the top. You may also add some notes about specific things you would like to clarify before posting it. Other users will help you improve your challenge by rating and discussing it. When you think your challenge is ready for the public, go ahead and post it, and replace the post here with a link to the challenge and delete the sandbox post. ## Discussion The purpose of the sandbox is to give and receive feedback on posts. If you want to, feel free to give feedback to any posts you see here. Important things to comment about can include: • Parts of the challenge you found unclear • Problems that could make the challenge uninteresting or unfit for the site You don't need any qualifications to review sandbox posts. The target audience of most of these challenges is code golfers like you, so anything you find unclear will probably be unclear to others. If you think one of your posts needs more feedback, but it's been ignored, you can ask for feedback in The Nineteenth Byte. It's not only allowed, but highly recommended! It is recommended to leave your posts in the sandbox for at least several days, and until it receives upvotes and any feedback has been addressed. ## Other Search the sandbox / Browse your pending proposals The sandbox works best if you sort posts by active. To add an inline tag to a proposal use shortcut link syntax with a prefix: [tag:king-of-the-hill]. To search for posts with a certain tag, include the name in quotes: "king-of-the-hill". Get the Sandbox Viewer to view the sandbox more easily! Stack Exchange distinguishes users by a unique id that can be found in their their profile URL. You can find your id by navigating to https://codegolf.stackexchange.com/users/current and finding the number after the /users/ bit in the resulting URL. For example, my profile URL is https://codegolf.stackexchange.com/users/99744/alex-bries, so my user ID is 99744 You can use this id to look up the number sequence of the last submitter on OEIS: The On-Line Encyclopedia of Integer Sequences® (OEIS) is an online database containing integer sequences. These sequences are identified via an id starting with an A followed by 6 numbers. You can search for an id with the "id:" key on the search bar of OEIS. id:A99744 will display sequence A099744 followed by all other sequences that mention it. Your job is to write a program that, when given a number n, produces the nth number in sequence of the submitter above. ### Rules • Each language may only be used once. • You may not post 2 answers in a row • Every next submision may only be max 10 bytes higher than the previous one • The champion is the last valid submission on june 14th golfing is not a requirement but it is in your favour • Different OEIS sequences have vastly different complexity. I'm pretty sure my user id is solvable in 6 bytes of Jelly, which can break the whole game depending on the next user's id. Jun 1 at 11:32 • @Bubbler I like the idea of different complexities, but not sure how to prevent the colossal differences. Personally i like it, but if you have ny suggestions for how to improve, please let me know. Jun 1 at 11:47 • I don't think this works as a [answer-chaining] challenge, but it could work as just a regular code-golf challenge (i.e. write a program which takes an integer n and outputs the nth term of the sequence with your id). It's unlikely to be fair, but its better than [answer-chaining]. I've also done a bit of editing, feel free to change anything I did Jun 1 at 15:08 • Yes, that's probably the better desicion, but after thinking about it more, its unlikely to be very fun, so ill leave it (for now). Thanks for the the constructive feedback guys. Jun 1 at 20:04 • Using user IDs is inherently unfair. – qwr Jun 2 at 14:47 # Write a program that mimics the output of the command env (with no arguments). The program should print (to standard output, or closest equivalent) a list of all current environment variables, with each variable on it's own line. The name and content of a variable should be separated by an equal sign. Here's a grammar: line ::= name "=" contents output ::= (line "\n")* No using the env command, or any 1 command that completes the challenge on its own. shortest code wins. • Can we use built-ins the get environment variables? Nov 15 '20 at 15:50 • @Adám If you mean "Can we use built-ins to get the environment variables", yes, I don't see a reason you couldn't, and otherwise it might be impossible to get them in some languages. Nov 15 '20 at 15:54 • And I assume (ba\|k\|c\|…)sh cannot participate? Nov 15 '20 at 15:56 • @Adám If they can find a way to get a list of environment variables without using env, I guess they could, unless I'm missing some other command that produces the same output. There's a lot of languages that can't complete the challenge (mostly esoteric ones), but they aren't technically banned. Nov 15 '20 at 23:28 • How about set on Windows/DOS? Nov 16 '20 at 8:07 • @Adám I have no experience with windows, does it have similar functionality to env? if so then no. Jun 3 at 16:46 • It is indeed. Note: As a beginner challenge writer (you've never posted a challenge before), you should avoid do X without Y challenges. Jun 3 at 17:28 • @Adám isn't "don't use a builtin that does exactly what you need" a standard loophole? Jun 3 at 19:32 • No, not at all. Jun 3 at 19:32 • @Adám Huh, I guess you're right. I just don't want to have the challenge solved before it's even posted, with the 3 character bash program env. Does trying to prevent 1 cheeky answer really put you in the same category as forbidding something random? Jun 3 at 20:50 • Does it really bother you that much that someone will post env or set as a solution? You're under no obligation to upvote or accept such an answer. Jun 3 at 20:54 • @Adám here's something you linked me: "A commonly used non-observable requirement is to avoid using a library function that solves the entire challenge. I believe these rules are OK. Like other non-observable requirements, one can find borderline cases of whether a built-in function solves all of the task or not, but the benefits of higher-quality answers outweigh the costs." this certainly seems to apply here. Jun 3 at 20:59 • @Adám I've never posted anything because whenever I post a question to the sandbox all I get is reasons why my question is bad. And putting any additional rules whatsoever gets the same "no X without Y" response. Then people don't elaborate on anything. Jun 3 at 21:10 • Don't be dismayed. We've all been there. Writing challenges is notoriously hard, especially for beginners. If you look at my account, I have lots of well-received challenges, but dig in, and you'll find that many of my early ones were flawed. I highly recommend answering a lot more existing challenges, before you attempt at writing your own. And when you do, decide on a simple unique task (this is hard, as a lot of obvious tasks have already been posted), with no special rules; just vanilla code golf. Jun 3 at 21:14 • Deadfish (at least, the original one) used i and d intead of + and -. Is there a reason you chose to use them? Jun 3 at 10:45 • @pxeger originally for the language I was going to have an input command so I swapped i and d for + and -, and I guess I never added them back. Jun 3 at 16:02 # branch golfing: sort a list by number of prime factors Specifically, first sorted by total number of prime factors, with any ties broken by number of unique prime factors (remove this?), then broken by size. The list will never contain 1. Test cases: [4,5,6,9] -> [5,4,9,6] [10,11,12] -> [11,10,12] [360 200 12] -> [12 200 360] [63 64 65] -> [65 63 64] • Please avoid challenges involving the prime numbers. Jun 3 at 20:42 • Please avoid challenges where the scoring criteria relies on non-observable program requirements. Jun 3 at 20:46 Jun 3 at 20:47 • Also, thank you for using the sandbox! :-) Jun 3 at 20:47 • @Adám Do you have any example of something where it isn't clear whether or not it is a conditional branch? Or any advice that isn't essentially just "this is bad"? Jun 3 at 20:56 • Maybe. Are there any branches in this APL solution? The code means: ⊢ the input ⊇⍨ permuted ∘⍋ according to a grade using ⍭ prime factors for each , followed by ⍥≢ when counted, ∘∪ the unique ¨ for each, ⍭ prime factors for each. Or in other words: Reorder according to a grading of prime factor count and then unique prime factor count. Jun 3 at 21:10 • From your description, ⍭ sounds like a loop construct, which I intended to include, but I could clarify. Jun 3 at 21:21 • If you count loop constructs, would you also count APL's + as a loop? It automatically loops over its arguments: Try it online! Besides, you state that "Builtin functions are presumed to be branchless." Jun 3 at 21:24 • No. It loops its own code, but can't be used to perform arbitrary code in a loop. Jun 3 at 21:28 • My usage of ⍭ was exactly parallel to any usage of +. They both auto-loop. However, ¨ is in fact an explicit loop. I could probably code around that too, though. An example would be extracting the odd numbers from a list. In many languages, you'd use a loop with a branch, but in APL you'd write List/⍨2\|List which means L where 2 divides the list ['s elements; that's implicit]. Looping? Branching? Anyway, you get the idea. It'll be next to impossible to determine for every language (there are lots!) what exactly constitutes an explicit loop, or for that sake a branch. Jun 3 at 21:32 • @Adám so they are analogous to functions that take 2 arrays as input and provide one as output? that doesn't seem like a branch to me. "I could probably code around that too, though." Yeah that's kinda the point of the challenge. Jun 3 at 21:41 • @Adám "but in APL you'd write List/⍨2\|List which means L where 2 divides the list ['s elements; that's implicit]." this explanation isn't very clear. this seems like this construct (or pair of constructs, which you are confusingly explaining as one) is some kind of predicate filter/map? Jun 3 at 22:01 Jun 3 at 22:01 # How much faster than qsort can you achieve? Sort 1 million 32 bit integers as quickly as possible. Your code must create 1 million random non-negative integers in the range 0 to 999999 and sort them. Only the sorting should be timed. # Score I will run your code and compare the timings to my default C code using qsort. That is: #include <stdio.h> #include <stdlib.h> #include <time.h> #include <stdint.h> int cmpfunc (const void * a, const void * b) { return ( (int)a - (int)b ); } int main(void) { int sz; srand(time(NULL)); printf("Enter the size of array::"); scanf("%d",&sz); uint32_t arr = malloc (sz sizeof(uint32_t)); int i; for(i=0;i<sz;i++) arr[i]=rand()%1000000; clock_t begin = clock(); qsort(arr, sz, sizeof(int), cmpfunc); clock_t end = clock(); double time_spent = (double)(end - begin) / CLOCKS_PER_SEC; printf("%f seconds\n", time_spent); printf("%d\n", arr[10]); return 0; } Your score will be the timing for my C code divided by the timing for your code. I will run both on my PC. This means I will need clear instructions for how to compile and run your code on linux. This means you will have to provide timing code in your solution so I am not timing the creation of the data. # Current timings On my PC: • The sample C code takes 0.11 seconds. • Python sorted takes 0.24 seconds. # Notes My code prints out one of the elements of the sorted array. This is to stop the opimizing compiler from removing the sorting code completely. You will have to do the same. Builins are allowed. As ever this is a challenge per language so if your Python answer is faster than any other Python answer you have won that mini-competition. • Obligatory question - builtins, yea or nay? Jun 7 at 14:31 • That's probably the wisest choice - probably want to edit the post to say that. Jun 7 at 14:35 • I strongly disagree with banning builtins. If any builtin is faster, either the builtin is interesting enough to be posted, or the challenge is boring. Banning builtins won't improve the latter, but it will hurt the former Jun 7 at 14:37 • @cairdcoinheringaahing I can go with that. Jun 7 at 14:37 • @StackMeter I deleted my comment as bultins are going to be allowed. Jun 7 at 14:39 • I would argue that "only time sorting should be timed" is a bit unclear. Consider my code is: int counts[MAX],i,j,k,sort(a,size){for(i=0;i<size;++i)++counts[a[i]];for(i=k=0;i<MAX;counts[i++]=0)for(j=counts[i];j--;)a[k++]=i;} (maybe some typos here, not tested) the initialize of counts array is required by sort. But C compiler will help me do so even begin main() entry and would probably not timed by my code. – tsh Jun 8 at 6:12 • @tsh what would you suggest? Jun 8 at 7:08 • Things to consider: 1) 32-bit signed or unsigned integers? Do I have a choice? 2) 32-bit (or any finite-sized) integers can be sorted using non-comparison sort algorithms, such as radix sort (which is asymptotically faster than any kind of comparison-based sort). Would you allow the answers to exploit this or not? Jun 14 at 1:24 # Does the regex match the string? Despite having over 80 challenges tagged , we don't have one to simply verify if a regex matches a string. Given a regular expression $$\R\$$ and a string $$\S\$$, output one of two distinct and consistent values to indicate whether the regex matches the string. You may assume: • The input and output will be in any convenient methods • The regular expression may be in any regex flavour that existed before this question was posted • $$\R\$$ will always be well formed, and will not error, regardless of the string matched against it • $$\R\$$ may be either a regular expression object (such as Javascript's RegExp, Python's re.compile etc.) or a plain string object. • The output will either be: • 2 distinct, consistent values, such as 1/0, true/false, "Hello"/5 • A truthy value and a falsey value, not necessarily consistent values This is , so the shortest code in bytes wins. # Meta • Changed the formatting of the "distinct, consistent values" section because I kept interpreting 1/0 as something to do with division by zero Jun 8 at 18:52 • I think the problem with a challenge like this is that there are only two types of solutions: generally low-effort calls to the built-in that does the task, and extremely tedious implementations of the built-in. For example, if I wanted to implement PCRE I would also need to handle the callout feature, which requires writing a whole API. You might be better off requiring a specific subset of behaviours to try to encourage answers that aren't just a built-in. Jun 9 at 17:28 # Smallest number of actions to change reputation by [x] points 8: Upvote + downvote 75: Bounty + upvote + accepted answer Draft • Potentially a dupe of this. Stoopid. Jun 9 at 4:55 • @lyxal This one's about positive or negative rep gains though, so it's more general Jun 9 at 4:57 # Finite Composition of Univariate Functions mathcode-golf ## Problem Given $$\mathbb{F}:=\{f_i \mid f_i : \mathbb{R} \to \mathbb{R}\}$$ the set of all the real maps, and the n-ary function composition $$\begin{gather} c_o(f_i) = f_1∘f_2∘f_3⋯∘f_n\\[8pt] ∀f_i ∈ \mathbb{F},~~~1≤i≤n∈\mathbb{N} \end{gather}$$ write a program which receives n functions of a real variable as the input and computes the in-order function composition, product of all the entered functions as a string. As standard, x is the name of the parameter variable in each function. Any other literal is taken as a constant. ## Examples ### Example 1 input number of functions: 3 f_1 = x + 1 f_2 = x - 10 f_3 = x^2 + 2 output c_o(f_1,f_2,f_3) = ((x^2 + 2) - 10) + 1 ### Example 2 input number of functions: 4 f_1 = 27x - sin(e) f_2 = -7(x^2 + 1)^(1/2) 4y^3 + 21 f_3 = x^(3/2) + 4 f_4 = ln(z + 1) (z, y, e are assumed as constants in the reals) output c_o(f_1,f_2,f_3,f_4) = 27(-7(((ln(z + 1))^(3/2) + 4)^2 + 1)^(1/2) 4y^3 + 21) - sin(e) • Any suggestion, improvement? please let me know. Thanks. Jun 8 at 20:08 • Thanks for using the sandbox! This seems like an interesting idea, but currently it is too underspecified to be clear. For example, the test cases imply that the input to each function is represented by x, but you haven't stated anywhere that we can assume that. Consider also a slight change to one of your cases, where one function has -y(x^2 + 1). Is y a function or a constant? There are too many specifics, so hopefully the general idea gets across that you must specify how you are representing the functions very precisely for this to work. Jun 9 at 17:41 • @FryAmTheEggman Thank you. I have edited it so that there's no ambiguity. Jun 9 at 18:19 • That is a good change, but doesn't address the ambiguity in separating functions like sin from the product of constants. It also isn't clear to me whether x^2 - 7 is a valid solution for the first example. These have some overlap in that they (probably?) won't affect a solution that just substitutes the functions appropriately, but will affect any symbolic parser based approach. Jun 9 at 19:53 • @FryAmTheEggman oh, yes, it's valid but it isn't required to go to such depths as to simplify symbolic expressions, that would be a problem on its own. I didn't specify so in the challenge and as for sin or those functions, since the problem to be addressed is the composition of functions any expression different from x should be "disregarded". I don't intend to make further computations with the output string. Jun 9 at 21:45 • What I'm trying to communicate is that your "point" for writing the challenge doesn't really matter - in some language it probably would be better to use a parser. If you just want to say "no other assumptions about the form of the function can be made" that is fine - but you have to actually say it, otherwise your challenge is unclear because I won't be able to tell if a solution is valid or not. Jun 9 at 23:13 # Introduction Unreadable is a programming language designed to be – as the name states – unreadable (in most fonts, anyway). Instructions are ' followed by a run of "s. So while in a code block, '""'""'""" looks just fine, it looks like '""'""'""" outside of one - it is extremely difficult to tell what is going on. The instructions are prefix, and are as so: Instruction Arity Behavior '" Unary Print; output the character with codepoint x, and return x. '"" Unary Increment; return x+1. '""" Nilary Unit; return 1. '"""" Binary Both; evaluate and disregard x, and return the result of evaluating y. '""""" Binary While; keep evaluating y until x is 0. Return the last result of evaluating y. '"""""" Binary Set; set the xth variable to y, and return y. '""""""" Unary Get; return the value of the xth value. Undefined is 0. '"""""""" Unary Decrement; return x-1. '""""""""" Ternary If-else; if x is 0, return z. Otherwise, return y. '"""""""""" Nilary Input; return the codepoint of the next character of input. EOF is -1. For example, a cat program would be: '"""""'""'""""""'"""'""""""""""'"'"""""""'""" while(++ setvar(1 = input())) {print(getvar(1))} Or, in its unreadable glory: '"""""'""'""""""'"""'""""""""""'"'"""""""'""" # Challenge Given a valid Unreadable program and its input separated by any non-' or " character, interpret the program as Unreadable code. # Test cases (uses ! as the separation character, I/O is represented in CP-437) Input -> Output '"'"""!this input does nothing lol -> ☺ '"""""'""'""""""'"""'""""""""""'"'"""""""'"""!This is a cat so I can output whatever I want. blah blah blah... -> This is a cat so I can output whatever I want. blah blah blah... '""""'""""'""""""'"""'""'""'""'""'""'""'""'""'""'"""'"""""'"""""""'"""'""""'""""""'""'"""'""'""'""'""'""'""'""'"""""""'""'"""'""""'""""""'""'""'"""'""'""'""'""'""'""'""'""'""'""'"""""""'""'""'"""'""""'""""""'""'""'""'"""'""'""'""'""'""'""'""'""'""'""'""'"""""""'""'""'""'"""'""""'""""""'""'""'""'""'"""'""'""'""'""'"""""""'""'""'""'""'"""'""""'""""""'""'""'""'""'""'"""'""'""'""'"""""""'""'""'""'""'""'"""'""""'""""""'""'""'""'""'""'""'"""'""'""'""'""'""'""'""'""'""'""'""'""'"""""""'""'""'""'""'""'""'"""'""""'""""""'""'""'""'""'""'""'""'"""'""'"""""""'""'""'""'""'""'""'""'"""'""""""'"""'""""""""'"""""""'"""'""""'"'""'""'"""""""'""'"""'""""'"'""'"""""""'""'""'"""'""""'"'""""""""'""""""""'"""""""'""'""'""'"""'""""'"'""""""""'""""""""'"""""""'""'""'""'"""'""""'"'""'"""""""'""'""'""'"""'""""'"'""'""'""'""'"""""""'""'""'""'""'"""'""""'"'""'""'"""""""'""'""'""'""'""'"""'""""'"'""""""""'"""""""'""'""'""'""'""'""'"""'""""'"'""'"""""""'""'""'""'"""'""""'"'""'""'""'""'"""""""'""'""'""'"""'""""'"'""""""""'""""""""'"""""""'""'""'""'"""'""""'"'"""""""'""'""'"""'""""'"'""'""'""'"""""""'""'""'""'""'""'"""'"'"""""""'""'""'""'""'""'""'""'""" -> Hello, world! '""""""'"""'""'""""""""""'""""""'""'"""'""""""""""'"'"""""'""""""'"""'""""""""'"""""""'"""'""""""'""'"""'""'"""""""'""'"""!A! -> b # Rules • Standard loopholes are forbidden. • A single trailing newline is allowed in the output, and no other trailing whitespace. • Please explain your answer. This is not necessary, but it makes it easier for others to understand. • Languages newer than the question are allowed. This means you could create your own language where it would be trivial to do this, but don't expect any upvotes. • This is , so shortest code in bytes wins! • A +100 rep bounty will be given to the first Unreadable answer. • This doesn't seem like a particularly interesting challenge to me, to be honest. Decoding the program into commands is fairly trivial (split by ' and find lengths of each), in both golfing and practical languages, and the actual interpreting part of the challenge is basically the same as any other language. Jun 10 at 1:20 • @RedwolfPrograms The interesting part is that commands are prefix, so it might take some thought to make an interpreter. Jun 10 at 1:55 # Can we measure time? You have 2 hourglasses with sand lying at bottom in both of hourglass, one that is $$\1\$$ sec hourglass(call it $$\A\$$) and other is $$\\sqrt2\$$ sec hourglass (call it $$\B\$$) (A hourglass is said to be n second hour glass if time taken by all of the sand of hourglass to fall from top to bottom is n second) Now we are given 2 integers $$\a\$$ and $$\b\$$ such that $$T=a+b\sqrt2 > 0$$ assume:$$\ 1000000\geq a \geq -1000000$$ $$\ 1000000\geq b \geq -1000000$$ we need to respond that can we measure this time $$\T\$$ with our hourglasses? ("YES" or "NO") Some clarifications and definitions: • We say we can measure time $$\t\$$ if at that instance in at least one of hourglass sand have just stopped falling. • We can make a hourglass upside down(any number of time) if and only if at least one of the hour glass has empty top. • Assume time taken to invert a hourglass is 0 sec.(by invert means upside down,i.e the sand that was in bottom is now at top and the sand at top is now at bottom). • If at least one of hourglass sand have just stopped falling we can report this instant as the amount of time measured. • If sand on both the hour glass is at bottom and we decided to do no any operation, then it is assumed that we are reporting the time measured at this instance and the process stops. Example 1: $$\a=0\$$ and $$\b=1\$$ Our answer is true, because initially both the hourglass is given with sand at bottom and at the start we can invert B, so when the sand from B has fallen all to bottom at that instance a total of $$\\sqrt2\$$ time has passed and we can report it,hence we have measured $$\\sqrt2\$$ time. Example 2: $$\ a=-1\$$ and $$\b=2\$$ $$\T = -1 +\sqrt2\$$ Our answer is true, we can obtain it as follows: At time 0, turn both A and B upside down, At 1 sec A will be emptied to bottom, turn A upside down again (B remain falling as it was), now when B fall completely,turn A upside down, now when A empties the total duration is $$\-1 +\sqrt2\$$. 1(All the A falls, at this instance B is left with $$\-1+\sqrt2\$$ ) $$+$$ $$\\sqrt2\$$-1 (at this duration after the previous one B have fallen completely and A is left with $$\1-(-1+\sqrt2)\$$ at top, but we have inverted it so that means A is actually left with $$\-1+\sqrt2\$$ at top) $$+$$ $$\-1+\sqrt2\$$ (at this duration after the previous one A have fallen completely) we can report this instant, the total time is: $$\(1)+(-1+\sqrt2)+(-1+\sqrt2)=-1 +2\sqrt2 \$$ And for following input it is not possible to measure and you have to output "NO" $$\a=11\$$ , $$\b=-5\$$ $$\a=35\$$ , $$\b=-21\$$ $$\a=-4\$$ , $$\b=4\$$ $$\a=-8815\$$ , $$\b=6261\$$ $$\a=-1\$$ , $$\b=1\$$ . . . ## code-golfmathnumber-theory The idea of this problem is not my original, it appeared in a coding competition some time back. • Thanks @pxeger for editing. May 31 at 9:10 • You should provide some impossible cases? – l4m2 Jun 3 at 14:09 • Thanks, I have added. Jun 3 at 14:53 # Convert permutations to integers and back again Convert a permutation of 0,...,n-1 to a number in the range 0,...,n!-1, and convert back again. Write a program containing two functions (or two programs). The first function should accept a permutation (i.e. an ordering) of the numbers 0,1,2,...,n-1 (where n is any positive integer) and output an integer in the range 0,1,...,n!-1 (where n! is n factorial). Distinct permutations of the same length n must produce distinct outputs, but there are no further restrictions. The second function should implement the inverse of the first function (except that it also needs to know n). It should accept an integer k in the range 0,1,...,n!-1 AND a positive integer n, and should output the permutation of 0,1,2,...,n-1 that yields k when sent to the first function. The time and space complexity of both functions MUST be at most polynomial in n, i.e. O(nc) for some fixed c. In particular, solutions that involving computing all n! permutations are NOT acceptable. (Unofficial bonus points for O(n)). Shortest solution (including both functions and any additional code) in each language wins. ## Examples Assuming the two functions are f and g: Input Output f([0]) 0 g(0,1) [0] f([0,1]), f([1,0]) 0 and 1 in either order g(0,2), g(1,2) [0,1] and [1,0] in either order f([3,4,1,0,2]) some integer in range 0,...,119 g(42,5) some permutation of 0,1,2,3,4 f(g(42,5)) 42 g(f([3,4,1,0,2]),5) [3,4,1,0,2] f([1,0,1]) not valid input (not a permutation) f([1,3,4]) not valid input (not a permutation of 0,1,2) g(42,4) not valid input (42 >= 4!) ## Notes Permutations can be represented in any reasonable form. You can assume that input(s) are valid (as described). You can assume that n! fits in the standard integer type in your language, but your algorithm should theoretically work for any n. Standard loopholes apply. # Boxes in boxes in boxes in boxes..... ASCII boxes are fun, but the box drawing questions seem to be limited insofar as that they seem to take a specific set of boxes. Today, I hope to fix that. The challenge here is to write a program that takes a list of sides of boxes and characters to draw them with, and outputs the boxes. To clarify some format issues: ## Input Input is flexible, and will look like 1) this 17 30 20 #$5 21 15$%^ 19 13 311 ... or 2) this [17, [30, 20, "#$5], [21, 15, "$%^], [19, 13, "311"], ...] You may support either or both of these types, but you must be consistent with this choice. Input is given in this order: The first number, a single number on one line, is the number of boxes to place. Then, for as many times as said first number, you will receive 2 numbers and three characters - the two numbers represent length and width, in any order. The three characters represent, in this order: 1. The character to use for the corners. 2. The character to use for the edges. 3. The character to fill the inside of the boxes. The boxes should not overlap, but you have the freedom to place the boxes wherever you wish. The allowed symbols are !\$%^&()_-+=#~/?\@s, where the s represents space and in order to aid those who want to split on whitespace. Input will be space separated within boxes and newline separated between boxes. ## Output Output should be a set of boxes, conforming to the regulations and conditions above - for example, %###% #**# ((* #*# ())( ## ())( %###% ((* for the input 2 5 5 %#* 4 4 () or [2, [5, 5, "%#"], [4, 4, "()"] ## Meta Any issues? Should the scoring be ? Is the challenge specified enough? • Your scoring shouldn't be pop-con. Also, you realise people are just going to draw all the boxes newline-separated? Be clearer that the input format is flexible. Jun 11 at 9:23 • @Ausername let me go fix that Jun 11 at 10:29 • @Ausername Should I add a bounding box to stop people just seperating the boxes by newlines Jun 11 at 10:33 • I'm not saying anything's wrong with that, just that's how people are going to do it because it's the simplest way. Jun 11 at 10:35 Given matrix $$\A\$$ and $$\A^n\$$, work out any possible positive $$\n\$$. Examples: • $$\\begin{pmatrix}1&2&3\\3&2&1\\1&0&1\end{pmatrix},\begin{pmatrix}36&32&44\\52&40&52\\12&8&12\end{pmatrix}\rightarrow 3\$$ • $$\\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}\rightarrow \text{any integer }\ge 3\$$ • $$\\begin{pmatrix}0&1\\1&0\end{pmatrix},\begin{pmatrix}0&1\\1&0\end{pmatrix}\rightarrow \text{any odd positive integer }\$$ # Sandbox Notes • Since this question likely fall into pure matmul as , will this be ? • If so, what size and $$\n\$$ is reasonable? • Should it multiply in a modulo-$$\p\$$ ring? • If someone provide enough extra info, they have right to post this question(actually that's why I posted in sandbox) • Can you give some examples? – math Jun 11 at 17:01 • Sometimes lots of different n can work. What should we output then? – xnor Jun 12 at 23:06 • @xnor "Any possible positive n". But I agree that it should be specified better. – user100690 Jun 14 at 11:22 • @ophact It's edited later than comment – l4m2 Jun 14 at 12:56 # I am not a dad! Given a non-empty string. Check if it starts with "I am"/"I'm" (case-insensitive) and if it is, output the dad joke. Otherwise, terminate P.S. There must be nothing in the STDERR. # Examples: "I am an idiot" -> "Hi an idiot, I am Dad" "I am rewuytheruty" -> "Hi rewuytheruty, I am Dad" "i Am case insensitive" -> "Hi case insensitive, I am Dad" "I'm A" -> "Hi A, I am Dad" "i'M B" -> "Hi B, I am Dad" "Iam together" -> terminated "Nothing here!" -> terminated # Rules • This is , so the answer with shortest bytes wins. • These loopholes are, obviously, forbidden. • Standard code-golf rules apply. • Please specify the language you are using and the amount of bytes. • It would be great if you would put a link to a sandbox where your code can be ran in action, such as TIO. • Explaining your code is very welcomed. • I don't think the exceptions add anything to this challenge to be honest. And if you do keep them, they need to be better specified. Jun 13 at 19:38 • Input validation is frowned upon (validation to check if any of the exceptions appear). As well as that, I observed that there are virtually no patterns or magic bitwise operations that can handle them, just hard-coding a few extra strings and some ternary operators. Doesn't really add anything to the other dad joke challenge to be honest. – user100690 Jun 14 at 12:19 • And if I would remove the exceptions, will the challenge become better? Will it be worthy publishing? Jun 14 at 12:52 Write a shortest bijective program from String to Object. You can specify two sets of chars $$\C_1\$$ and $$\C_2\$$, both of which consist of at least 4 different chars[1], and define: • String: string consisting of only chars in $$\C_1\$$ [C₁] • Name: non-empty string consisting of only chars in $$\C_2\$$ [C₂]+ • Object: a finite map from Name to Object, such that every way to go down finally go to empty Object(empty map) Reasonable I/O allowed. Outputting in Object in languages like JavaScript, Array/String with proper encoding, folder, shuld be fine. # Notes • [1] unary or binary is too easy. • I meant to map to folder but folder charset is bad, then decided to give you choice. # Sandbox Notes • What does proper encoding mean? Or do I just remove the allowance? # Largest number with no repeating substrings of length $$\l\$$ Inspired by this question, in turn inspired by this one. When I read the above questions, I thought: "Cool". And then I thought "What if we go further?" Therefore, in this challenge, you must find the largest number that has no repeating substrings of length n. This is : fewest bytes wins. # Meta Stuff This seems insanely hard - is it well specified enough? Any changes that seem reasonable to add? – user100690 Jun 22 at 9:34 • Is this the same as this? Jul 2 at 14:22 # Iterated ultimatum koth Each round, players are paired. One player is the Proposer and the other is the Receiver. The Proposer must choose a number x such that 0 < x < 100. The Receiver must then Accept or Reject this offer. If accepted, the Proposer gains x points and the Receiver gains 100 - x points. If rejected, neither player gains anything. This is repeated several times, then the Proposer and Receiver roles are reversed, then new pairings are made. Repeat. The idea is that the Proposer must try to maximise his own gain, while the Receiver can try to coerce the Proposer to give a larger share of the 100 points by rejecting unfair offers. https://en.wikipedia.org/wiki/Ultimatum_game ## Meta • Too similar to Iterated Prisoner's Dilemma? • Any other feedback? Solve Encode a Lenguage with score 0. Shortest code wins. Quine rule applies. I'll unlikely post it. • @pxeger Qunie rule. – l4m2 Jun 20 at 11:03 • Zsh, 120 bytes: Try it online!. Can probably be quite a lot shorter Jun 20 at 11:08 • Undeleted as potential reference – l4m2 Jul 1 at 17:53 # Smallest number of panels to represent $$\1\$$ through $$\n\$$ in base $$\b\$$ Inspired by this question. (I have taken a lot of inspiration from Puzzling recently). Because not all of you spend all your (non-code-golf) time looking at puzzles, let me explain the relevant details. The challenge involved figuring out how to express the numbers 1 through 8 in only 8 panels in bases 2, 3 and 4. Today, I want to do the converse. Your task is to, given a number $$\n\$$ and a base, $$\b\$$, find the fewest number of panels to represent 1 - n in base b. By that, I mean taking a single continuous slice of the set of panels, we can represent a number, and we can do that for 1 through n. To clarify, each panel can take a single base-$$\b\$$ number, and the goal is to find a way to represent each number from 1 through n as a base b number. A continuous slice is a set of panels such that all the panels are consecutive and in the same order as in the original number - for example, 221001120 - in which every two-digit and one-digit number is represented in reading order, left-to-right, as well as 100. Note that 101 and 111 are disallowed because they are not consecutive, with 0s in between, 110, 122 and 2110 are disallowed because they read from left to right, and 121 and 212 are disallowed because they reorder the panels. For a bigger example, let us take the base as 3 and the number as 18: a valid solution is 122121110221010011200, as every number between 1 and 18 can be represented in base 3 1: 122121110221010011200 2: 122121110221010011200 3: 122121110221010011200 4: 122121110221010011200 5: 122121110221010011200 6: 122121110221010011200 7: 122121110221010011200 8: 122121110221010011200 9: 122121110221010011200 10: 122121110221010011200 11: 122121110221010011200 12: 122121110221010011200 13: 122121110221010011200 14: 122121110221010011200 15: 122121110221010011200 16: 122121110221010011200 17: 122121110221010011200 18: 122121110221010011200 In order to verify your solution does work, provide an optimal answer by the number such that it can be checked. Input looks like this: n b, and output like this: number solution. Sample inputs and outputs to aid you (output is represented as o - yes I did work all of these out, and yes, I do believe it is the shortest to just find a place to append the number in): $$\b = 2\$$: n <= 8; o = n (11101000) n = 9, 10; o = 11 (10011101000) n = 11; o = 13 (1001011101000) n = 12, 13, 14; o = 14 (11001011101000) n = 15, 16; o = n (110010111010000) $$\b = 3\$$: n <= 9; o = n (221001120) n = 10; o = 11 (22101001120) n = 11; o = 13 (1022101001120) n = 12; o = 14 (11022101001120) n = 13, 14, 15; o = 15 (111022101001120) n = 16; o = 17 (12111022101001120) n = 17; o = 19 (12212111022101001120) n = 18; o = 20 (122121110221010011200) $$\b = 4\$$: n <= 10; o = n (1101220213) n = 11; o = 12 (231101220213) n = 12, 13; o = 13 (231101220213) n = 14, 15; o = n (33231101220213) n = 16, 17; o = 18 (10033231101220213) n = 18; o = 20 (1003323110122010213) n = 19, 20; o = 22 (100103323110122010213) (i'm not touching 5+ this took me an hour) # Are these strings within N bit flips of each other? code-golfstringbinarydecision-problem Given two strings a and b, and a positive integer n, determine if b can be reached by performing exactly n bit-flips on a. A bit-flip is defined as changing one of the bits in the character encoding of a string from a 0 to a 1 or from a 1 to a 0 The same bit may be flipped multiple times, so the 0110 can be reached in exactly 2 bit-flips of itself, as well as in 0 (e.g. 0110 -> 0100 -> 0110). ## Test-cases #### Truthy n a b ======================= 1 hello hullo 3 hello hullo 5 hello hullo 69 hello hullo 4 hello hello 1 x h 1 x p 1 x y 2 x q 2 x l 2 x x 4 x l 4 x x 4 x w 4 x e #### Falsey n a b ======================= 2 hello hullo 5 hello hello 1 x a 1 x n 1 x f 1 x x ## Rules • You may assume/require: • n > 0 • a and b are the same length • neither string is empty • both strings consist only of lowercase letters a to z (or uppercase A to Z instead, if you wish) • You should encode the strings in ASCII (or other ASCII-compatible encoding), but if your language cannot easily handle that, ask in the comments and I'll decide on special rules • Your code does not need to handle high n or long a or b, but it must work in theory • You may use any reasonable I/O method • Standard loopholes are forbidden • This is , so the shortest code in bytes wins # Meta • Better title suggestions? • I limited to ASCII to avoid people finding loopholes that trivialise the challenge. Do you have any other suggestions that would work better? • Test-case suggestions? • Is this a duplicate? • Is this clear enough? • Any other feedback? • This question is equals to: is n - (hamming distance of a and b) a non-negative even number? – tsh Jun 30 at 10:08 • @tsh yes, I've basically abandoned it Jun 30 at 12:18 # Maximum unique characters code-challenge Write a program which maximises the amount of unique characters in its output while minimising the size of its source. The program must halt. Functions for converting a character code to its character in any encoding are disallowed. Your score is output unique chars / size in bytes, and higher is better. ## Meta • Is this a duplicate? • Should character code functions be banned? • Do you mean "output" instead of "input" in the first line? Also, it sounds like "shortest code to print all chars having Unicode codepoint from 0 to 1114111", which is pretty boring. And you can't ban "character code functions" since it is simply not a thing in lots of languages. And you can't get most of the characters without it besides hardcoding the string itself anyway (or eval-ing), which makes it even more boring. Jun 22 at 7:29 • Maybe don't ban "character code functions" as they are pretty much the only way to produce an interesting solution to this challenge. – user100690 Jun 22 at 9:27 • @RecursiveCo. for(let i=0;i<[insert 9's here];i++){console.log(String.fromCharCode(i)} (in JS) I'm just really bad at coming up with ideas. I've given up on this idea. Jun 22 at 9:49 • What's to stop me making up a character encoding where every possible sequence of bytes in the range 0x48 to 0x57 represents one of infinite "characters" and then outputting all decimal numbers? Boom, infinite score. Jul 3 at 20:12 • What time in Friday? 12am? Jun 16 at 2:19 • @Razetime Any time in Friday – l4m2 Jun 16 at 8:57 # Extend the OEIS On the OEIS, there are a significant amount of sequences which do not have proper continuations to them, in other words, the length of the sequence was cut short by the methods at that point. An example of this is the sequence in this post before it was extended by this answer. A sequence that wouldn't qualify for this post would be the fibonacci sequence because while it may have a limit on the OEIS, the end is not due to the method used, but instead due to the large amount of current entries. #### Challenge: Find a sequence on the OEIS that follows the above guidelines and extend it by any amount. Your answer should include the code, an explanation of it, as well as the new entries. A1327807 previous largest n=8 number = 100000001 new highest n=12 number = 37812879128 // code // more code // explanation of code and how you figured it out Of course your answer doesn't need to follow that format exactly, it should just be close to it. #### Notes: You must write original code for this challenge. If there is already a program out there, and you're only contribution is running it on a better machine, that is a trivially extendable sequence. This also goes for answers, as the only thing contributed cannot be simply better hardware. This is for lack of a better tag, so good luck and have fun! • I think you should find one interesting sequence to extend (or some). Large part of this question is finding an extendable sequence that might be of interest (as popularity contest suggests). IMHO this part of research should be on asker's side. Jun 27 at 11:26 # Do I flip the side? ### Introduction When writing on a piece of paper folded $$\n\$$ times you have $$\2^{n+1}\$$ sides available to write on. When writing on those sides, you can imagine the following sequence of actions (for $$\n=2\$$): 1. write on the first side and fill the whole side, 2. flip the paper to the next empty side and write there until filled, 3. now we have to unfold the paper once (at least) to get to the next empty side and "flip" the fold, 4. flip the paper to the next empty side, 5. unfold the paper twice to get to the empty side and "flip" the fold, 6. etc. ### Challenge Given number of folds $$\n \geq 0 \$$ output the number of unfolds needed to get to the next clear side. Standard rules apply, so you can: • Take an index $$\k \leq 2^{n+1}-1\$$ and output the $$\k^{th}\$$ term, either 0 or 1 indexing. • Take a positive integer $$\k \leq 2^{n+1}-1\$$ and output the first $$\k\$$ terms. • Output whole sequence for given $$\n\$$. ### Test cases n \| sequence ---------------------------------- 0 \| 0 (no unfolds, just a flip) 1 \| 0 1 0 2 \| 0 1 0 2 0 1 0 3 \| 0 1 0 2 0 1 0 3 0 1 0 2 0 1 0 4 \| 0 1 0 2 0 1 0 3 0 1 0 2 0 1 0 4 0 1 0 2 0 1 0 3 0 1 0 2 0 1 0 ### Meta • Too easy? • Is this first $2^{n+1}-1$ items of oeis.org/A007814? – tsh Jun 28 at 8:41 • @tsh Yes, it is. Jun 28 at 15:02 • IMO this is a very near dupe of ABACABA sequence. Jul 7 at 6:06 # Is this a valid Temtem character name? Given a string of between 0 and 15 (inclusive) printable ASCII characters, determine whether the string represents an acceptable character name for the game Temtem. The following rules apply to character names: • Names must have at least 3 letters (a-z) • Names may have up to 2 total prefixes and/or suffixes of at least one letter • Prefixes and suffixes must be separated from the name by one of space, hyphen or apostrophe • The first letter of the name and each prefix may be capitalised but all other letters must be lower case. Truthy examples: Daddy's girl de Morgan Rip Van Winkle Spider-Man X-Men Falsy examples: McDonald's -XX- 'falsy' Note: Temtem may apply other checks to character names not included here. As this is , Output must be a single consistent value for all valid names. Output for invalid names must be consistent for a given invalid name, but may vary between different invalid names, so for instance a count of errors is an acceptable output format. This is , so the shortest program that breaks no standard loopholes wins! • Are Rip Van winkle and spider-Man valid? Jun 27 at 12:59 • @Adám They are for the purposes of this question, yes. – Neil Jun 27 at 13:43 # General Binary Counterman Jr Related General Binary Counterman Jr is the son of famed Mr. Binary Counterman. He’s continued in his father’s footsteps, counting strings of numbers in the family tradition. But binary has grown old; he’s not a computer. The General has waited for his father’s popularity to wane, and he has decided the time is nigh. The General is about to generalize! As a reminder, his father would count the evens and odds like this: 1 1 0 0 1 0 would be read as 1st odd, 2nd odd, 1st even, 2nd even, 3rd odd, 3rd even and it becomes 1 3 0 2 5 4. But not so the General. Evens and odds are just 0s and 1s mod 2, the General knows this. Instead of reducing mod 2, he likes to reduce strings down whatever mod he pleases. Once he does he counts the residues the same way as his father did, but in whatever mod he damn well wants. The General has read the algorithms that fans have sent his father. But he believes that his new method will yield sufficiently different strategies as to deserve its own page on the internet. ## Challenge Given two inputs representing a mod m and a list L, the nth occurrence of a given congruence class in the list should be output as the nth integer of that class. The I/O is very flexible and may take any reasonable format, so long as it doesn’t sidestep part of the problem. For instance, you may take the list backwards or as a string, but you can’t take it already reduced mod m. This is , so least bytes wins! Reference Implementation ## Example In my examples, I will represent the I/O as 3: 5 3 2 8 4 5 9 -> 1 3 2 4 5 6 to represent the reminder. This means m = 2 and L = 1 1 0 0 1 0. But again, you are not required to do it this way. In fact, you’re encouraged to find the I/O scheme that works best for your language and algorithm! Input: 3: 5 3 2 8 4 5 9 This gets counted mod 3 by its congruence class of [0], [1] or [2]: L m Count 5 %3 -> 1st [2] -> 2 3 %3 -> 1st [0] -> 0 2 %3 -> 2nd [2] -> 5 8 %3 -> 3rd [2] -> 8 4 %3 -> 1st [1] -> 1 5 %3 -> 4th [2] -> 11 9 %3 -> 2nd [0] -> 3 Output: 2 0 5 8 1 11 3 ## Test Cases The most fun part! How do you like the test cases? 2: 1 3 0 2 5 4 1 3 0 2 5 4 2: 1 1 0 0 1 0 1 3 0 2 5 4 5: 1 1 0 0 1 0 1 6 0 5 11 10 1: 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 100: 101 102 103 104 105 1 2 3 4 5 10: 3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 3 1 4 11 5 9 2 6 15 13 25 8 19 7 29 23 12 33 18 14 16 22 26 24 43 53 7: 10 20 30 40 50 60 70 80 90 100 3 6 2 5 1 4 0 10 13 9 8: 09 18 27 36 45 54 63 72 81 90 1 2 3 4 5 6 7 0 9 10 12: 11 22 33 44 55 66 77 88 99 110 121 132 11 10 9 8 7 6 5 4 3 2 1 0 11: 1 1 2 3 5 8 13 21 34 55 89 144 1 12 2 3 5 8 13 10 23 0 34 45 4: 2 4 8 16 32 64 128 256 512 1024 2048 2 0 4 8 12 16 20 24 28 32 36 ## Other Cheats In the interest of making this as flexible as possible for all the masochistic and impossible fun languages, here are other things you may skimp on: • You may assume list L will only contain single digit numbers. • You may assume mod m will be a single digit number. • You may do anything when m = 0, including throwing an error. • You may assume no leading 0s in the numbers. • It is less consistent with the pattern, but you may eliminate 0s from your output by starting the count for [0], and only the [0] class, at m instead. Then, in the example program, 1st [0] → 3 and 2nd [0] → 6. It would output 2 3 5 8 1 11 6 instead of 2 0 5 8 1 11 3 • Feel free to comment any other ideas and I will add them in if they don’t detract from the spirit of the challenge, and primarily serve to give the turing tarpits a hand. If the cheat would be used by mainstream and golfing langs, it’s probably not a good idea. # Will the hydra finally die? Part II ### Background This is a follow up question to the question: Will the Hydra finally die? As before a dangerous A Medusa have released a dangerous Hydra which is revived unless the exact number of heads it have is removed. The knights can remove a certain number of heads with each type of attack, and each attack causes a specific amount of heads to regrow. This time the knights are more impatient and having seen your previous abilities want you to** write a program or function that returns a list of hits which will leave the hydra 1 hit from death Note that this is fundamentally different from Become the Hydra Slayer. in 2 aspects. 1: We are not asking for the optimal solution 2: each attack causes a different number of heads to grow back. this radically changes the approach needed. For example: input: heads = 2, attacks = [1, 25, 62, 67], growths = [15, 15, 34, 25], output: [5, 1, 0, 0] Explanation: The Hydra has 10 heads to start with, we have 4 different attacks and for each attack, growth gives us the number of heads that grows back. hits gives us the number of times each attack is applied. So the number of heads the Hydra has after each attack is 2 -> 16 -> 30 -> 44 -> 58 -> 72 -> 62 Since 62 is a valid attack value (It lies in the attack list), we return True since the Hydra will die on the next attack (be left with 0 heads). Note that the order for when the attacks are done is irrelevant. 2 -> 16 -> 6 -> 20 -> 34 -> 48 -> 62 ### Input Input should contain heads (an integer), attacks (a list of how many heads can be removed), regrowths (how many heads grow back per attack) You may take input in any convenient method. This includes, but is not limited to • A list of tuples (1, 15, 5), (25, 15, 1), (62, 34, 0), (67, 25, 0) • Lists 2, [1, 25, 62, 67], [15, 15, 34, 25], [5, 1, 0, 0] • Reading values from STDIN 1 15 1 15 1 15 1 15 1 15 25 15 • A file of values ### Output • An array, or some way to easily indicate which hits the knights are to take. Example: 5 1 0 0 Note 1 Say that your input is attacks = [1, 25, 62, 67] and the hydra has 25 heads left, then you cannot output the answer as [1,0,0,0], [0,0,0,1] etc. Your output and input must be sorted similarly. Otherwise it will be very confusing for the Knights. Note 2: Your program can never leave the Hydra with a negative number of heads. Meaning if the Hydra has 15 heads, an attack removes 17 heads and regrows 38. You may not perform this attack. ### Scoring • This is a you will be scored as follows: the total sum of hits your code produces from the test data + length of answer in bytes • Note that how you input and parse the test data is not part of your code length. Your code should, as stated above simply take in some a headcount, attacks, regrowth's and return a list/representation of how many hits each attack has to be performed to leave the hydra exactly 1 hit from death. Lowest score wins! ### Test data attacks = [1, 25, 62, 67], growths = [15, 15, 34, 25], use every integer from 1 to 200 as the number of heads. An sample from the first 100 can be found below. Again your program does not have to return these values, it is merely an example of how the scoring would work. As seen below the total sum for each of these hits are 535 meaning my score would be 535+length of code in bytes 1 [0, 0, 0, 0] 2 [5, 1, 0, 0] 3 [3, 2, 0, 0] 4 [7, 4, 0, 0] 5 [5, 5, 0, 0] 6 [4, 0, 0, 0] 7 [2, 1, 0, 0] 8 [6, 3, 0, 0] 9 [4, 4, 0, 0] 10 [8, 6, 0, 0] 11 [1, 0, 0, 0] 12 [5, 2, 0, 0] 13 [3, 3, 0, 0] 14 [7, 5, 0, 0] 15 [5, 6, 0, 0] 16 [4, 1, 0, 0] 17 [2, 2, 0, 0] 18 [6, 4, 0, 0] 19 [4, 5, 0, 0] 20 [3, 0, 0, 0] 21 [1, 1, 0, 0] 22 [5, 3, 0, 0] 23 [3, 4, 0, 0] 24 [7, 6, 0, 0] 25 [0, 0, 0, 0] 26 [4, 2, 0, 0] 27 [2, 3, 0, 0] 28 [6, 5, 0, 0] 29 [4, 6, 0, 0] 30 [3, 1, 0, 0] 31 [1, 2, 0, 0] 32 [5, 4, 0, 0] 33 [3, 5, 0, 0] 34 [2, 0, 0, 0] 35 [0, 1, 0, 0] 36 [4, 3, 0, 0] 37 [2, 4, 0, 0] 38 [6, 6, 0, 0] 39 [2, 0, 0, 0] 40 [3, 2, 0, 0] 41 [1, 3, 0, 0] 42 [5, 5, 0, 0] 43 [3, 6, 0, 0] 44 [2, 1, 0, 0] 45 [0, 2, 0, 0] 46 [4, 4, 0, 0] 47 [2, 5, 0, 0] 48 [1, 0, 0, 0] 49 [2, 1, 0, 0] 50 [3, 3, 0, 0] 51 [1, 4, 0, 0] 52 [5, 6, 0, 0] 53 [1, 0, 0, 0] 54 [2, 2, 0, 0] 55 [0, 3, 0, 0] 56 [4, 5, 0, 0] 57 [2, 6, 0, 0] 58 [1, 1, 0, 0] 59 [2, 2, 0, 0] 60 [3, 4, 0, 0] 61 [1, 5, 0, 0] 62 [0, 0, 0, 0] 63 [1, 1, 0, 0] 64 [2, 3, 0, 0] 65 [0, 4, 0, 0] 66 [4, 6, 0, 0] 67 [0, 0, 0, 0] 68 [1, 2, 0, 0] 69 [2, 3, 0, 0] 70 [3, 5, 0, 0] 71 [1, 6, 0, 0] 72 [0, 1, 0, 0] 73 [1, 2, 0, 0] 74 [2, 4, 0, 0] 75 [0, 5, 0, 0] 76 [1, 0, 1, 0] 77 [0, 1, 0, 0] 78 [1, 3, 0, 0] 79 [2, 4, 0, 0] 80 [3, 6, 0, 0] 81 [1, 0, 1, 0] 82 [0, 2, 0, 0] 83 [1, 3, 0, 0] 84 [2, 5, 0, 0] 85 [0, 6, 0, 0] 86 [1, 1, 1, 0] 87 [0, 2, 0, 0] 88 [1, 4, 0, 0] 89 [2, 5, 0, 0] 90 [0, 0, 1, 0] 91 [1, 1, 1, 0] 92 [0, 3, 0, 0] 93 [1, 4, 0, 0] 94 [2, 6, 0, 0] 95 [0, 0, 1, 0] 96 [1, 2, 1, 0] 97 [0, 3, 0, 0] 98 [1, 5, 0, 0] 99 [2, 6, 0, 0] 100 [0, 1, 1, 0] 535 # Meta • What would a good scoring method for this type of question be? • My method outputs 2873 for the total count from 1 to 200, is this too high, too low? • Feel free to suggest other improvements to my testing data	2021-11-28 11:16:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 72, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3461514413356781, "perplexity": 1067.5460557658182}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358520.50/warc/CC-MAIN-20211128103924-20211128133924-00042.warc.gz"}
https://global-sci.org/intro/article_detail/jcm/9868.html	Volume 32, Issue 1 An Improved Non-Traditional Finite Element Formulation for Solving the Elliptic Interface Problems J. Comp. Math., 32 (2014), pp. 39-57. Published online: 2014-02 Cited by Export citation • Abstract We propose a non-traditional finite element method with non-body-fitting grids to solve the matrix coefficient elliptic equations with sharp-edged interfaces. All possible situations that the interface cuts the grid are considered. Both Dirichlet and Neumann boundary conditions are discussed. The coefficient matrix data can be given only on the grids, rather than an analytical function. Extensive numerical experiments show that this method is second order accurate in the $L^∞$ norm. • Keywords Elliptic equation, Sharp-edged interface, Jump condition, Matrix coefficient. 65N30. • BibTex • RIS • TXT @Article{JCM-32-39, author = {}, title = {An Improved Non-Traditional Finite Element Formulation for Solving the Elliptic Interface Problems}, journal = {Journal of Computational Mathematics}, year = {2014}, volume = {32}, number = {1}, pages = {39--57}, abstract = { We propose a non-traditional finite element method with non-body-fitting grids to solve the matrix coefficient elliptic equations with sharp-edged interfaces. All possible situations that the interface cuts the grid are considered. Both Dirichlet and Neumann boundary conditions are discussed. The coefficient matrix data can be given only on the grids, rather than an analytical function. Extensive numerical experiments show that this method is second order accurate in the $L^∞$ norm. }, issn = {1991-7139}, doi = {https://doi.org/10.4208/jcm.1309-m4207}, url = {http://global-sci.org/intro/article_detail/jcm/9868.html} } TY - JOUR T1 - An Improved Non-Traditional Finite Element Formulation for Solving the Elliptic Interface Problems JO - Journal of Computational Mathematics VL - 1 SP - 39 EP - 57 PY - 2014 DA - 2014/02 SN - 32 DO - http://doi.org/10.4208/jcm.1309-m4207 UR - https://global-sci.org/intro/article_detail/jcm/9868.html KW - Elliptic equation, Sharp-edged interface, Jump condition, Matrix coefficient. AB - We propose a non-traditional finite element method with non-body-fitting grids to solve the matrix coefficient elliptic equations with sharp-edged interfaces. All possible situations that the interface cuts the grid are considered. Both Dirichlet and Neumann boundary conditions are discussed. The coefficient matrix data can be given only on the grids, rather than an analytical function. Extensive numerical experiments show that this method is second order accurate in the $L^∞$ norm. Liqun Wang, Songming Hou & Liwei Shi. (1970). An Improved Non-Traditional Finite Element Formulation for Solving the Elliptic Interface Problems. Journal of Computational Mathematics. 32 (1). 39-57. doi:10.4208/jcm.1309-m4207 Copy to clipboard The citation has been copied to your clipboard	2023-01-31 03:52:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.34797394275665283, "perplexity": 1571.9554266000741}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499842.81/warc/CC-MAIN-20230131023947-20230131053947-00149.warc.gz"}
https://codegolf.meta.stackexchange.com/questions/10573/code-golf-jeopardy	# Code Golf Jeopardy I'm considering doing an original challenge that (at of right now) consists of golfed solutions in the languages C, Python, Jelly, Javascript, 05AB1E, Pyth, PHP, MATL. The challenge is to look at the golfed code and figure out what the intended challenge was. I think it could be a very interesting challenge, depending on the complexity of the intended challenge, but I fear that once someone figures out the solution (the intended challenge), the challenge is dead for everybody else, and that just makes it a challenge, which is not what I want. I would very much like any feedback on this type of challenge - would people be interested in doing this? (The reason this isn't posted in the sandbox is because it's a completely new type of challenge, and I'm not sure that comments are the best way to provide feedback to this type of challenge) • I like the idea, but I'm concerned that running the code will be more effective than trying to decipher what the code is actually doing. – xnor Nov 13 '16 at 9:26 • @xnor but that might not help you in figuring out the challenge. If you run the code and it outputs 21 how will you know what that means? – Daniel Nov 13 '16 at 10:20 • The code would take input, right? I'd try it for lots of inputs and try to puzzle out a relationship. It might still not be obvious, but nevertheless doable without analyzing the code. – xnor Nov 13 '16 at 10:23 • @xnor I think this strategy can be defeated by choosing a challenge that has many assumptions about the input, or arrays with a specific shape, etc. – feersum Nov 13 '16 at 21:20 • @Daniel are you still planning on doing this? If not, I'd love to post a challenge based on this in the sandbox. – Nathan Merrill Nov 17 '16 at 17:20 • @NathanMerrill I would love it if you did it. I think I need to work a lot more on my submission - can't wait to see yours :)! – Daniel Nov 17 '16 at 17:34 # Do it as a Cops n Robbers You'll need the following: 1. I wouldn't limit the language to those, but rather those with an Esolang page, or on wikipedia's list of languages. The language, however, must be specified by the cop. 2. I recommend explicitly saying that you must write your own submission 3. The challenge they choose must have a positive score and not be closed 4. The robber needs to find any challenge that the submission solves. Then, if you have your own code you want people to solve, you can simply add it! • So this would be a generalization of Programming Language Quiz. That was already a decent difficulty level for robbers, so free choice of challenge might seem to give too much advantage to cops. On the other hand, robbers may be able to generate cracks by finding an extremely simple challenge instead of the cop's intended one. – feersum Nov 13 '16 at 21:26 • Another possibility would be that the cops are required to disclose the challenge. – feersum Nov 13 '16 at 21:26 • The language would need to be disclosed – Nathan Merrill Nov 13 '16 at 22:28 • Ah, so it's the inverse of Programming Language Quiz: instead of known challenge and unknown language, this would be known language and unknown challenge. – DLosc Nov 15 '16 at 19:51 • Would this lead to very slow code that makes it impractical to run it against many inputs? (Not necessarily a bad thing - just wondering.) – trichoplax Nov 16 '16 at 22:45 • Would the challenges be limited to code-golf? Or could the answer be a "solution" to a KOTH or something? – mbomb007 Nov 17 '16 at 15:23 • I think that is up to the OP – Nathan Merrill Nov 17 '16 at 16:14 # Cops and robbers? people post a question, people post the answers. In this way, the challenge does not die immediately, and has a reasonable amount of time to it	2020-09-28 09:19:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24215491116046906, "perplexity": 786.6799704474155}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401598891.71/warc/CC-MAIN-20200928073028-20200928103028-00791.warc.gz"}
https://readtiger.com/wkp/en/Lightning_Network	Lightning Network Routing through an idealized mesh network of payment channels. The Lightning Network is a "second layer" payment protocol that operates on top of a blockchain (most commonly Bitcoin). It enables fast transactions between participating nodes and has been touted as a solution to the bitcoin scalability problem. It features a peer-to-peer system for making micropayments of digital cryptocurrency through a network of bidirectional payment channels without delegating custody of funds. Lightning Network implementation simplifies atomic swaps. Normal use of the Lightning Network consists of opening a payment channel by committing a funding transaction to the relevant blockchain, followed by making any number of Lightning transactions that update the tentative distribution of the channel's funds without broadcasting to the blockchain, optionally followed by closing the payment channel by broadcasting the final version of the transaction to distribute the channel's funds. To perform as intended, Lightning Network requires a transaction malleability fix, such as Segregated Witness (SegWit) in Bitcoin.[1] History Joseph Poon and Thaddeus Dryja published the Lightning white paper in 2015. More detailed specifications are being developed by multiple parties, including a non-commercial implementation by MIT Digital Currency Initiative,[2] and Bitcoin Core/bitcoind implementations by Blockstream, Lightning Labs, and ACINQ. In December 2017, a series of interoperable test transactions were performed on Bitcoin Core implementations. In January 2018 Blockstream launched a payment processing system for web retailers. Blockstream noted that Lightning was live on mainnet with 60 nodes operating as of January 18, 2018, and should be considered "in testing." On March 15, 2018, Lightning Labs CEO Elizabeth Stark announced the initial release of lnd 0.4-beta for developers with the intent on making it available for testing purposes on the main bitcoin network with Litecoin support. In March 2018 the Stellar CTO Jed McCaleb announced that the Stellar Network would be implementing a protocol inspired by the Lightning Network. The Lightning Network concept was endorsed by mobile payment entrepreneur Jack Dorsey in March 2018.[3] From April 2018 to August 2018, the Bitcoin Lightning Network had a monthly growth rate of about 15%. The number of nodes increased from 1,500 to 3,000, and the number of channels increased from 4,000 to 11,000.[4] In April 2018, the eclair wallet was removed from the Google Play store for a few days because the app developer lost the private signing key making it unable to be updated for critical bugs. 2018 DDoS attacks On March 20, 2018, Lightning Network nodes faced a Distributed Denial of Service (DDoS) attack that sent around 200 nodes offline, down from around 1,050 to 870. The exploit used as many node connections as possible to prevent any new connections.[5] A month earlier in February, Bitcoin Core developer Peter Todd said the Lightning protocol could very well "prove to be vulnerable to DoS [denial of service] attacks in its current incarnation." According to the cryptographer, this poses danger to both the peer-to-peer as well as the blockchain level of the project. Design A Lightning Network overview The payment channels allow participants to transfer money to each other without having to make all their transactions public on the blockchain. This is done by penalizing uncooperative participants. When opening a channel, participants must commit an amount (in a funding transaction, which is on the blockchain). Time-based script extensions like CheckSequenceVerify and CheckLockTimeVerify make the penalties possible. If we presume a large network of channels on the Bitcoin blockchain, and all Bitcoin users are participating on this graph by having at least one channel open on the Bitcoin blockchain, it is possible to create a near-infinite amount of transactions inside this network. The only transactions that are broadcast on the Bitcoin blockchain prematurely are with uncooperative channel counterparties. Lightning Paper[6] The CheckSequenceVerify (CSV) Bitcoin Improvement Proposal details how Hash Time-Locked Contracts are implemented with CSV and used in Lightning.[7] Commitment transactions If Alice and Bob have a payment channel, both of them also have a "latest" commitment transaction. A commitment transaction divides the funds from the funding transaction according to the correct allocation between Alice and Bob. For example, if Alice owns 1.0 mBTC (equal to 0.001 bitcoins or 100000 satoshis) and Bob owns 1.0 mBTC in the channel, the commitment transactions divide the total channel funds in that way. Commitment transactions allow multiple users to participate in a single transaction (and thus act as a single entity), using a multi-key system. Determination of congestion falls primarily on the miners, so this network rests on the assumption that honest miners will not organize a 51% attack. Since commitment transactions spend the funding transaction, they must be signed by both partners. Commitment transactions are actually a pair of asymmetrical transactions. Alice's commitment transaction contains two outputs: one which pays Bob outright, and another which is a timelocked, revocable output that eventually pays Alice. The revocable output may be revoked by Bob if Bob knows the revocation key. Bob's commitment transaction is the reverse: it pays outright to Alice, but pays Bob's share under a timelocked, revocable output; if Alice knows the revocation key to Bob's commitment transaction, she can revoke it. Initially, Alice holds the ${\displaystyle A_{1}}$ commitment transaction, and Bob holds the ${\displaystyle B_{1}}$ commitment transaction. The revocation key for ${\displaystyle A_{1}}$, ${\displaystyle R_{A_{1}}}$, is known by Alice, but not by Bob; the revocation key for ${\displaystyle B_{1}}$, ${\displaystyle R_{B_{1}}}$, is likewise known only by Bob. Suppose Alice decides to pay Bob 0.25 mBTC (before this, each owns 1 mBTC): 1. Alice creates a new Bob's transaction, ${\displaystyle B_{2}}$, which allocates 0.75 mBTC to Alice and 1.25 mBTC to Bob. 2. Alice signs ${\displaystyle B_{2}}$ and sends to Bob. 3. Bob receives ${\displaystyle B_{2}}$, signs it, and keeps it. 4. Bob creates a new Alice's transaction, ${\displaystyle A_{2}}$, which allocates 0.75 mBTC to Alice and 1.25 mBTC to Bob. 5. Bob signs ${\displaystyle A_{2}}$ and sends to Alice. 6. Alice receives ${\displaystyle A_{2}}$, signs it, and keeps it. 7. Alice provides ${\displaystyle R_{A_{1}}}$, invalidating ${\displaystyle A_{1}}$; she can then delete ${\displaystyle A_{1}}$. 8. Bob provides ${\displaystyle R_{B_{1}}}$, invalidating ${\displaystyle B_{1}}$; he can then delete ${\displaystyle B_{1}}$. Limitations The Lightning Network is made up of bidirectional payment channels between two nodes which combined create smart contracts. If at anytime either party drops the channel, the channel will close and be settled on the blockchain. Due to the nature of the Lightning Network's dispute mechanism which requires all users to watch the blockchain constantly for fraud, the concept of a "watchtower" has been developed, where trust can be outsourced to watchtower nodes to monitor for fraud. Routing The original whitepaper in reference to routing suggests that "eventually, with optimizations, the network will look a lot like the correspondent banking network, or Tier-1 ISPs". Implementations The specification for the commercial varieties of the lightning network is available on GitHub, and its largest contributor is Rusty Russell of Blockstream.[1] The non-commercial lightning implementation called 'LIT' is available through the MIT-DCI Github. The MIT-DCI is the largest contributor to this version of the lightning network. The public alpha release of lnd was made on 10 January 2017. The Release Candidate (RC1) of the Lightning protocol specification was released on 6 December 2017. References 1. ^ a b "lightning-rfc: Lightning Network Specifications". 25 September 2017 – via GitHub. 2. ^ "The Digital Currency Initiative at the MIT Media Lab". MIT Digital Currency Initiative. Retrieved 2018-07-16. 3. ^ 4. ^ "Statistics". 1ml.com. 5. ^ "Lightning Network DDoS Sends 20% of Nodes Down". News Article. TrustNodes. 2018-03-21. 6. ^ "The Bitcoin Lightning Network: Scalable Off-Chain Instant Payments" (PDF). lightning.network. January 14, 2016. 7. ^ "bips: Bitcoin Improvement Proposals". 25 September 2017 – via GitHub.	2018-11-16 03:05:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 18, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.26587024331092834, "perplexity": 5953.3495662792175}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039742970.12/warc/CC-MAIN-20181116025123-20181116051123-00361.warc.gz"}
https://brilliant.org/problems/for-my-friend-nihar-mahajan-happy-birthday-man/	For My Friend Nihar Mahajan (Happy Birthday Man) Algebra Level 5 $\large{{ x }_{ n+1 }=\left\lfloor \sqrt [ 3 ]{ { x }_{ n }^{ 3 }-\left\lfloor \sqrt [ 3 ]{ { x }_{ n } } \right\rfloor } \right\rfloor }$ Let there be a sequence defined by above rule for $$n \ge 1$$. For $$n \ge k$$ the value of $$a_{n}$$ will be $$0$$, for some positive integer $$k$$. Find the value of $$k$$, if $$a_{1}=839475687$$. ×	2017-05-22 23:32:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8132408857345581, "perplexity": 536.9970611747656}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463607242.32/warc/CC-MAIN-20170522230356-20170523010356-00567.warc.gz"}
https://kb.osu.edu/dspace/handle/1811/10620	# INTENSITIES AND HALF-WIDTHS MEASURED AT DIFFERENT TEMPERATURES FOR THE $CO_{2}$ BAND $201_{III} \leftarrow 000$ Please use this identifier to cite or link to this item: http://hdl.handle.net/1811/10620 Files Size Format View 1978-TE-03.jpg 97.39Kb JPEG image Title: INTENSITIES AND HALF-WIDTHS MEASURED AT DIFFERENT TEMPERATURES FOR THE $CO_{2}$ BAND $201_{III} \leftarrow 000$ Creators: Valero, Francisco P. J.; Suarez, Carlos B.; Boese, Robert W. Issue Date: 1978 Publisher: Ohio State University Abstract: We report measurements performed at 197, 233, and $294^{\circ}\,K$ of line intensities and broadening coefficients for the $CO_{2}$ transition $201_{III} \leftarrow 000$ around $4854 cm^{-1}$. A total of 109 spectra were obtained for this research by using a Fourier Transform interferometer and a cooled absorption cell. Each of these spectra resulted from co-adding an average of 44 interferograms. The purely vibrational transition moment $\|R\|$, the Herman-Wallis vibration-rotation Interaction factor F(m) and the total band intensity Band are deduced from equivalent width measurements. Some of the values obtained are: $\begin{array}{l} \|R^{201}_{000}III \| = (0.2064 \pm 0.0017) \times 10^{-2} Debye\\ F(m) = 1+ (0.22 \pm 0.07) \times 10^{-2}m + (0.56 \pm 0.16) \times 10^{-4}m^{2}\\ S_{Band} = 21 329 \pm 69 cm^{-1}km^{-1}atm^{-1}{_{STP}} \end{array}$ Description: Author Institution: NASA-Ames Research Center, Moffett Field URI: http://hdl.handle.net/1811/10620 Other Identifiers: 1978-TE-03	2018-06-24 07:01:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7440809607505798, "perplexity": 6329.663283585321}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267866888.20/warc/CC-MAIN-20180624063553-20180624083553-00612.warc.gz"}
https://jpmccarthymaths.com/category/math7019/	You are currently browsing the category archive for the ‘MATH7019’ category. I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at [email protected] and I will add you to the mailing list. Manuals The manuals are available in the Copy Centre. Please purchase ASAP. More information has been sent via email. Telegram App I am exploring the possibility of setting up a classroom group chat that would have the functionality of WhatsApp without having your phone number made public. If this is something you might be interested in, please download the Telegram app and set yourself up with an @ handle. FAO: Erasmus Students — Calculators The Student Resources tab above contains some information about calculators. Here is a list of some allowed and not allowed calculators. If you have to purchase a calculator, my recommendation is that you purchase something like a Casio fx-83GT PLUS. This might be available in the CIT shop. Tutorials Tutorials, which are absolutely vital, start next week. There may be a split but this might not occur until Week 3. Week 1 In week one we had one and a half classes. One half class was given over to a general overview of MATH7019 and we spent about an hour introducing the topic of Curve Fitting including Lagrange Interpolation. Week 2 We will start talking about Least Squares curve fitting. I would urge anyone having any problems with material that isn’t being addressed in the tutorials to use the Academic Learning Centre. If you are a little worried about your maths this semester, perhaps after the Quick Test or in general, you need to be aware of this resource. The timetable should be up some time next week. You will get best results if you come to the helpers there with specific questions. Next week, some students will receive slips detailing areas of maths that they should brush up on. Assessment 1 Assessment 1 will have a hand-in date the Friday of Week 5, 11 October. The Assignment is in the manual but I must also send on your personal data sets next week. Study Please feel free to ask me questions about the exercises via email or even better on this webpage — especially those of us who struggled in the test. Student Resources Talk delivered to the Conversations on Teaching and Learning Winter Programme 2018/19, organised by the Teaching & Learning Unit in CIT (click link for slides): Contexts and Concepts: A Case Study of Mathematics Assessment for Civil & Environmental Engineering • The first piece of advice is to read questions carefully. Don’t glance at a question and go off writing: take a moment to understand what you have been asked to do. • Don’t use tippex; instead draw a simple line(s) through work that you think is incorrect. • For equations, check your solution by substituting your solution into the original equation. If your answer is wrong and you know it is wrong: write that on your script. If you do have time at the end of the exam, go through each of your answers and ask yourself: 2. does my answer make sense? If no, say so, and then try and fix your solution. 3. check your answer (e.g. for a fitted curve or beam function, input values and see do they make sense; substitute your solution into equations; check your answer against a rough estimate; or what a picture is telling you; etc). If your answer is wrong, say so, and then try and fix your solution. Student Feedback You are invited to give your feedback on my teaching and this module here. Assessment 2 I am working my way through these. I promise you your final CA results before Monday morning. Week 12 We finished off Chapter 4 by looking at Error Analysis, including rounding error. We had a tutorial for a lecture on Wednesday, and Thursday might allow some tutorial time. Week 13 We will go through last year’s exam on the board and then I will answer your questions if there are any. If there are none I will help one-to-one. Usual class times and locations. We will also have tutorials on Friday 14 December at the usual times and venue. If there aren’t too many students present (max 18), you can attend both tutorials. Past exam papers (Winter and Autumn) may be found here. Study Please feel free to ask me questions about the exercises via email or even better on this webpage — especially those of us who struggled in the test. Student Resources Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, past exam papers, etc. Student Feedback You are invited to give your feedback on my teaching and this module here. Assessment 2 I have started correcting these: I won’t make any promises but will do my utmost to have them to you as soon as possible, certainly sooner than the end of next week. Week 11 We looked at more general Taylor Series: not just near $a=0$ — and also for functions of several variables. Here are some reasons why an engineer might be interested in Taylor Series of functions of several variables. Week 12 We will finish off Chapter 4 by looking at Error Analysis, including rounding error. We should be able to hold an additional tutorial or two. Week 13 We will go through last year’s exam on the board and then I will answer your questions if there are any. If there are none I will help one-to-one. Usual class times and locations. We will also have tutorials on Friday 14 December at the usual times and venue. Study Please feel free to ask me questions about the exercises via email or even better on this webpage — especially those of us who struggled in the test. Assessment 2 Assessment 2 is on p.136. • The hand in deadline is 16:00 this Monday 26 November 2018. • Hand it at tutorial today — or 13:00 on Monday. Otherwise drop it into my office, A283. I should be here Monday 11:00-13:00, 14:00-15:00. Outside these times there should be someone in A283 who can let you in and leave your assignment on my desk. • Hand in whatever you have done by the deadline: work handed in late will be assigned a mark of zero. • Email me your Excel work. • Print off a hard copy of your excel and submit this with any other written work. • Further instructions in the manual. Week 10 We began Chapter 4 with a Revision of Differentiation, and on Wednesday you had a tutorial to get up to grips with calculating derivatives. We went on to look at at Maclaurin Series. Week 11 We will look at more general Taylor Series: not just near $a=0$ — and also for functions of several variables. Assessment 2 Assessment 2 is on p.136. It has a hand-in time of 16:00 Monday 26 November. As suggested in class, I would advise you to — if possible — complete this assignment early if you can, freeing up time in your tutorial to get work done on Chapter 3: Probability & Statistics. Week 9 We completed Chapter 3 by looking at Sampling and Hypothesis Testing. We had a tutorial for a lecture on Wednesday. Week 10 We will begin Chapter 4 with a Revision of Differentiation and go on to look at then at Maclaurin Series and Taylor Series. We might hold an extra tutorial for a lecture. Assessment 1 – Results I will have the assignments with me tomorrow if you want to see your work. Assessment 2 Assessment 2 is on p.136. It has a hand-in time of 16:00 Monday 26 November. As suggested in class, I would advise you to — if possible — complete this assignment early if you can, freeing up time in your tutorial to get work done on Chapter 3: Probability & Statistics. Week 8 We looked at the Poisson distribution, the Normal distribution, and started discussing Sampling. Week 9 We will complete Chapter 3 by looking at Sampling and Hypothesis Testing. We may have an extra tutorial during one of the lectures. Assessment 1 – Results I will have the assignments with me tomorrow and next Friday if you want to see your work. Assessment 2 Assessment 2 is on p.136. It has a hand-in time of 16:00 Monday 26 November. Week 7 We finished looking at Chapter 2 by looking at the Three Term Taylor Method for approximating solutions of ordinary differential equations. We started Chapter 3 (Probability and Statistics) by looking at some general concepts in probability and then we looked at random variables with a binomial distribution. Week 8 We will look at the Poisson distribution and perhaps the Normal distribution. Assessment 1 – Results I am starting corrections today and will get the results to you as soon as I can. I cannot give an accurate day at this stage: it could be Monday but just as easily could be a few days after this – I can’t make any promises. Assessment 2 Assessment 2 is on p.136. It has a hand-in date of Monday 26 November and we have already covered everything that will be asked and so you have over five weeks to complete the assignment. MicDrop Project On Monday you will be sent a 15 minute survey that you will take on a mobile internet device — such as your mobile phone — during Monday’s lecture. This survey is part of a larger project the Mathematics Department is undertaking — Mathematics in Context: Developing Relevancy-Orientated Problems — in an effort to improve our teaching. If you do not have an internet ready device you may leave class early. Maths Classes will be going full steam ahead on Monday 22 October as well as Wednesday, Thursday, Friday 1, 2, 3 November. I will call the next two weeks by Week 7. Week 6 In Week 6 we finished looking at cantilvers and then summarised what we learnt about beams. We had one lecture as a tutorial but then looked at numerical approximations to solutions of differential equations that we cannot solve exactly. After the storm last year I recorded some examples. If you missed some classes this week you could do worse than watch this cantilever example and this summary of beams to catch up Week 7 In Week 7 we will look at the Three Term Taylor Method and begin Chapter 3 on Probability and Statistics. Assessment 1 If you haven’t started this you seriously need to get cracking. • The hand in deadline is 17:30 on Tuesday 16 October 2018. • Hand it in class at 13:00 on Monday or else drop it into my office, A283. I should be here Monday 14:00-16:00, Tuesday 11:00-15:00 and 17:00-17:30. • Hand in whatever you have done by the deadline: work handed in late will be assigned a mark of zero. • Email me your Excel work. • Print off a hard copy of your excel and submit this with any other written work. • Further instructions in the manual. The Lee Fields Medal — CIT Maths Challenge Maths Competition on next Wednesday with cash prizes. Poster below. Week 5 In Week 5 we finished looking at simply supported beams. We then looked at fixed end beams and cantilevered beams. Week 6 In Week 6 we will finish looking at cantilvers and then summarise what we learnt about beams but then look at numerical approximations to solutions of differential equations that we cannot solve exactly. J.P. McCarthy on MATH6040: Spring 2019, Week… Student on MATH6040: Spring 2019, Week… J.P. McCarthy on MATH6040: Spring 2019, Week… Student on MATH6040: Spring 2019, Week… J.P. McCarthy on MATH6040: Spring 2019, Week…	2019-09-20 23:30:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1856454312801361, "perplexity": 1410.3289226455045}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574084.88/warc/CC-MAIN-20190920221241-20190921003241-00499.warc.gz"}
http://mathoverflow.net/questions/10335/about-hilbert-and-siegel-modular-varieties-forms?sort=oldest	about Hilbert and Siegel modular varieties (forms) It seems to me that Hilbert modular varieties (forms) are generalization from Q to totally real fields. While Siegel modular varieties (forms) are generalization from 1 dimensional to higher dimensional abelian varieties. But they should both be some kind of Shimura variety (automorphic forms), right? According to Milne's note of Shimura varieties, Siegel modular varieties are Shimura varieties coming from the Shimura datum (G, X) where G is the symplectic similitude group of a symplectic space (V, \phi). So what is the corresponding Shimura datum for the Hilbert modular variety? Or am I asking a wrong question? Also, in the definition of Shimura varieties G(Q)\G(A_f)X/K, why we only consider Q and its adele group? why not general number fields and their adeles? (again, maybe a wrong question) Thank you. - 2 Answers Both of your questions have the same answer: if $K$ is an arbitrary number field, it is not necessary to consider separately reductive groups over $K$, because if $G_{/K}$ is a reductive group, then $Res_{K/\mathbb{Q}} G$ is a reductive group over Q. Here $Res_{K/\mathbb{Q}}$ denotes Weil restriction. In particular, if K is a totally real field, then the reductive group whose Shimura variety is the corresponding Hilbert modular variety is $Res_{K/\mathbb{Q}} \operatorname{GL}_2$. - The Shimura datum for a Hilbert modular variety is the pair (G,X) where G is the group over Q obtained from GL(2) over a totally real number field F by (Weil) restriction of scalars and X is a product of copies of C minus the real axis indexed by the real embeddings of F. In the definition of Shimura varieties we work only over Q by convention: we could work over number fields, but this gives nothing new (one can always take Weil restriction). However, in some cases, e.g., Hilbert modular varieties, it seems (to me to be) more natural to work over a number field other than Q. -	2015-03-06 12:59:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8432239890098572, "perplexity": 151.97009588746482}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936468546.71/warc/CC-MAIN-20150226074108-00035-ip-10-28-5-156.ec2.internal.warc.gz"}
https://tex.stackexchange.com/questions/202425/how-to-get-the-text-equally-separated-from-the-headrule-and-footrule-with-geomet	# How to get the text equally separated from the headrule and footrule with geometry? In the following MWE I'd like to define a dimen, usep, and do the blue margins (uhsep and ufsep) be equal to it, and ubsep (in green) to the half of it, when footnotes are to be printed. Otherwise, uhsep and ubsep should be equal to usep. Note that the distances should be measured from the headrule, footnoterule and footrule. Nor the headrule neither the footrule should be moved. After the comments, I've achieved the basic layout (uhsep=ubsep=usep) with headsep, bottom and footskip: %pages without footnotes \def\usep{1cm} \newdimen\hsep \setlength{\hsep}{\dimexpr\usep+.3\baselineskip\relax} \newdimen\fskip \setlength{\fskip}{\dimexpr\usep+.9\baselineskip\relax} \newgeometry[left=3.5cm, right=1.5cm, top=2.5cm, bottom=\bmar, headsep=\hsep, footskip=\fskip, showframe]{geometry} \documentclass[a4paper,titlepage,11pt,twoside,openright]{report} \usepackage[left=3.5cm, right=1.5cm, top=2.5cm, bottom=2.5cm, showframe]{geometry} \usepackage{lipsum} \usepackage[usenames,dvipsnames,svgnames,table]{xcolor} \usepackage[pagestyles,explicit]{titlesec} \makeatletter \renewcommand\setfootrule[1]{% \ifdim#1=\z@ \let\makefootrule\@empty \else \def\makefootrule{\color{red}\rule[.9\baselineskip]{\linewidth}{#1}} \fi} \makeatother {\color{red}\sffamily Section title}{}{} %odd \setfoot[\color{red}\sffamily\thepage][][\color{red}\sffamily Degree] %even {\color{red}\sffamily Theis title}{}{\color{red}\sffamily\thepage} %odd } \begin{document} %Change seps, skips... \pagestyle{main} \lipsum[1-2] \footnote{This is a footnote} \lipsum[3-4] \footnote{This is another footnote} \lipsum[5-7] \end{document} Solution using titlesec and geometry which provides a command \updategeometry{<tbmargin>}{<sep>}, where <tbmargin> is the top and bottom margin, measured to the main head and foot rules, and is the separation from the rules to the text. \def\chapv{1cm} \def\usep{0cm} \def\tbmargin{\dimexpr\usep+\chapv\relax} \def\hsep{\dimexpr.3\baselineskip+\usep\relax} \def\fskip{\dimexpr.9\baselineskip+\usep\relax} \newcommand\updategeometry[2]{ \def\chapv{#1} \def\usep{#2} \newgeometry{left=3.5cm, right=1.5cm, top=\tbmargin, bottom=\tbmargin, headsep=\hsep, footskip=\fskip} } • This will result in uneven pages because you'll have more space at the bottom of some pages than others. Why do you wish to do this? – cfr Sep 21 '14 at 22:45 • I'd like to consider the footnotes as 'part of the footer' rather than 'part of the text' (ubsep may be rather small when footnotes are shown). On the other hand, when footnotes are not displayed, I want the body of the text to be smaller in pages when the Table of Contents and Lists of Figures/Tables are shown. Having uhsep and ubsep equal centers the content in the page. – U.Martinez-Corral Sep 21 '14 at 22:48 • But when you have a longer footnote, it will use part of the space otherwise used by the body. So it cannot look like 'part of the footer', can it? There's no problem using a different layout for the toc, tof, tot etc. if you really wish to. The footnote issue is a lot trickier. I don't say it cannot be done. But I do think it will not be at all straightforward. – cfr Sep 21 '14 at 22:52 • Do you mean getting to know in which pages the footnotes are going to be shown is difficult? I can understand it, since the layout affects placing them and vice versa. If so, could you help me just to have ubsep equal to uhsep? Should it be something like \addtolength{\textheight}{-(\usep-\footskip)}? – U.Martinez-Corral Sep 21 '14 at 23:01 • Hang on. I'll have a look. That should be doable. – cfr Sep 21 '14 at 23:03 This is an answer to the modified question in the comments - not to your original question. I think the original question would be far from straightforward, if it can be done at all. I think this is right but I'm not sure. You need to set \ulength to half of the value you want for the dimension you mentioned. geometry is then used to set the headsep and footskip and fancyhdr is used to manage the headers and footers: \documentclass[a4paper,titlepage,11pt,twoside,openright]{report} \usepackage{geometry} \usepackage{lipsum} \usepackage[usenames,dvipsnames,svgnames,table]{xcolor} \usepackage[pagestyles,explicit]{titlesec} \newlength{\ulength} \setlength{\ulength}{5mm} \usepackage{fancyhdr} \fancypagestyle{main}{% \fancyhf{}% \fancyhf[loh]{\color{red}\sffamily Section title}% \fancyhf[reh]{\color{red}\sffamily Chapter title}% \fancyhf[lof]{\color{red}\sffamily Thesis title}% \fancyhf[rof,lef]{\color{red}\sffamily\thepage}% \fancyhf[ref]{\color{red}\sffamily Degree}% \renewcommand{\footrulewidth}{.4pt}% \renewcommand{\footrule}{% \begin{document} %Change seps, skips... \pagestyle{main} \lipsum[1-2] \footnote{This is a footnote} \lipsum[3-4] \footnote{This is another footnote} \lipsum[5-7] \end{document} Note that I used fancyhdr because it is easier to layout the page consistently. Moreover, fancyhdr tells you if things are going to go wrong because there is not enough space and so on. I used geometry to set footskip etc. because, if you are using the package at all, you need to use it to make changes to the page layout. Otherwise, it doesn't know you've changed things and cannot take those changes into account when calculating lengths. So even when you are specifying 'native' dimensions such as footskip, make sure to do it using geometry. # Update To get the chapter and section titles automatically included in the header, I would do something like the following. This is partly because I'm just not very familiar with titlesec and have never used it for headers and footers. I mainly used fancyhdr because it tells you if the size of the header or footer is too small and tells you how large it needs to be. This was useful in laying out the page appropriately. However, I cannot see any reason why you could not substitute the facilities of titleps and drop fancyhdr if you prefer. \documentclass[a4paper,titlepage,11pt,twoside,openright]{report} \usepackage{geometry} \usepackage{lipsum} \usepackage[usenames,dvipsnames,svgnames,table]{xcolor} \usepackage{titlesec} \newlength{\ulength} \setlength{\ulength}{5mm} \usepackage{fancyhdr} \fancypagestyle{main}{% \fancyhf{}% \fancyhf[loh]{\color{red}\sffamily \rightmark}% \fancyhf[reh]{\color{red}\sffamily \leftmark}% \fancyhf[lof]{\color{red}\sffamily Thesis title}% \fancyhf[rof,lef]{\color{red}\sffamily\thepage}% \fancyhf[ref]{\color{red}\sffamily Degree}% \renewcommand{\footrulewidth}{.4pt}% \renewcommand{\footrule}{% \begin{document} %Change seps, skips... \pagestyle{main} \renewcommand{\chaptermark}[1]{% \markboth{#1}{}} \renewcommand{\sectionmark}[1]{% \markright{#1}} \chapter{Chapter title} \lipsum[1-2] \footnote{This is a footnote} \section{Section title} \lipsum[3-4] \footnote{This is another footnote} \lipsum[5-12] \end{document} • Thanks for this solution. I'm using titlesec to set the titles of chapters and sections. Although I've used fancyhdr in other documents, I'd prefer to avoid loading more packages (if possible). However, I'll keep this as a reference, in case I can't get it done just with titlesec and geometry. – U.Martinez-Corral Sep 22 '14 at 3:22 • If I use fancyhdr to define pagestyles instead of titlesec, \sectiontitle and \chaptertile won't work, even if I load \usepackage[pagestyles,explicit]{titlesec}. I'd like to show those instead of 'Chapter title' and 'Section title' in your example. Is it possible or should change \leftmark and \rightmark? – U.Martinez-Corral Sep 22 '14 at 16:13 • @U.Martinez-Corral See my edit. I imagine you could substitute titleps if you wish. I'm just not very familiar with that package and don't have time to look at it right now so I've given you a quick solution with fancyhdr. – cfr Sep 22 '14 at 16:42 • That's enough. It was just curiosity about how those are defined, since they seem clearer to me than leftmark and rightmark. Thanks again for your replies. – U.Martinez-Corral Sep 22 '14 at 16:49	2021-08-03 05:34:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7537558674812317, "perplexity": 1644.545590113781}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154420.77/warc/CC-MAIN-20210803030201-20210803060201-00617.warc.gz"}
http://docs.itascacg.com/pfc700/pfc/pfcmodule/doc/manual/ball_manual/ball_commands/cmd_ball.accumulate-stress.html	# ball accumulate-stress command Syntax ball accumulate-stress b Specify that stresses are accumulated and stored during cycling. This slows performance by ~15% but allows the current stress state to be queried and used in contact models during force-displacement computations. By default, the ball stresses are not accumulated but are computed when queried.	2021-10-18 16:49:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6599735617637634, "perplexity": 8695.694507729548}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585204.68/warc/CC-MAIN-20211018155442-20211018185442-00020.warc.gz"}
https://hackage-origin.haskell.org/package/summoner-1.2.0/candidate/docs/Summoner-Text.html	summoner-1.2.0: Tool for scaffolding completely configured production Haskell projects. Summoner.Text Synopsis # Documentation Creates module name from the name of the package Ex: my-lovely-projectMyLovelyProject intercalateMap :: Text -> (a -> Text) -> [a] -> Text Source # Converts every element of list into Text and then joins every element into single Text like intercalate. tconcatMap :: (a -> Text) -> [a] -> Text Source # Convert every element of a list into text, and squash the results	2022-10-07 12:44:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21839125454425812, "perplexity": 8154.598924585832}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030338073.68/warc/CC-MAIN-20221007112411-20221007142411-00277.warc.gz"}
http://markun.cs.shinshu-u.ac.jp/mirror/mizar/JFM/Vol8/yellow_3.html	Journal of Formalized Mathematics Volume 8, 1996 University of Bialystok Copyright (c) 1996 Association of Mizar Users ## Cartesian Products of Relations and Relational Structures Artur Kornilowicz Institute of Mathematics, Warsaw University, Bialystok ### Summary. In this paper the definitions of cartesian products of relations and relational structures are introduced. Facts about these notions are proved. This work is the continuation of formalization of [8]. This work was partially supported by Office of Naval Research Grant N00014-95-1-1336. #### MML Identifier: YELLOW_3 The terminology and notation used in this paper have been introduced in the following articles [9] [7] [12] [13] [15] [14] [5] [10] [6] [1] [11] [2] [3] [4] #### Contents (PDF format) 1. Preliminaries 2. Properties of Cartesian Products of Relational Structures #### Bibliography [1] Grzegorz Bancerek. Curried and uncurried functions. Journal of Formalized Mathematics, 2, 1990. [2] Grzegorz Bancerek. Complete lattices. Journal of Formalized Mathematics, 4, 1992. [3] Grzegorz Bancerek. Bounds in posets and relational substructures. Journal of Formalized Mathematics, 8, 1996. [4] Grzegorz Bancerek. Directed sets, nets, ideals, filters, and maps. Journal of Formalized Mathematics, 8, 1996. [5] Czeslaw Bylinski. Functions and their basic properties. Journal of Formalized Mathematics, 1, 1989. [6] Czeslaw Bylinski. Functions from a set to a set. Journal of Formalized Mathematics, 1, 1989. [7] Czeslaw Bylinski. Some basic properties of sets. Journal of Formalized Mathematics, 1, 1989. [8] G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M. Mislove, and D.S. Scott. \em A Compendium of Continuous Lattices. Springer-Verlag, Berlin, Heidelberg, New York, 1980. [9] Andrzej Trybulec. Tarski Grothendieck set theory. Journal of Formalized Mathematics, Axiomatics, 1989. [10] Andrzej Trybulec. Tuples, projections and Cartesian products. Journal of Formalized Mathematics, 1, 1989. [11] Wojciech A. Trybulec. Partially ordered sets. Journal of Formalized Mathematics, 1, 1989. [12] Zinaida Trybulec. Properties of subsets. Journal of Formalized Mathematics, 1, 1989. [13] Edmund Woronowicz. Relations and their basic properties. Journal of Formalized Mathematics, 1, 1989. [14] Edmund Woronowicz. Relations defined on sets. Journal of Formalized Mathematics, 1, 1989. [15] Edmund Woronowicz and Anna Zalewska. Properties of binary relations. Journal of Formalized Mathematics, 1, 1989.	2017-11-18 06:31:38	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9540648460388184, "perplexity": 9646.744907786782}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934804666.54/warc/CC-MAIN-20171118055757-20171118075757-00017.warc.gz"}
http://mathematica.stackexchange.com/questions/51478/i-cannot-get-tables-names-and-columns-names-from-database	# I cannot get tables names and columns names from database I am working with database using OpenSQLConnection-JDBC. I am able to fetch the data perfectly from the tables but I cannot get nor the names of the tables (using SQLTableNames) neither the column names (using SQLColumnNames). I tried also to work with the help examples; but I got same problem. what could be the reason? conn = OpenSQLConnection[ JDBC["MySQL(Connector/J)", "IP/databasename"], eventsdata = SQLSelect[ conn, "tablename", SQLColumn["columnname"] == somevalue] where: IP= Ip address of the database, databasename=database name that I am connecting to, tablename=table name which I have it before, columnname=columns names of the table that I have it before, somevalue=value so that I can fetch some of the data. - What command are you using to fetch the data? Can you show us the code that works? – mfvonh Jun 25 '14 at 4:22 @mfvonh I have add part of the code that works. – Algohi Jun 25 '14 at 4:50 What output do you get when you try these: SQLTables[conn], SQLTableNames[conn], SQLTableInformation[conn]? – mfvonh Jun 25 '14 at 4:52 I got all empty list {}. – Algohi Jun 25 '14 at 4:53 Try SQLExecute[conn, "use tablename"] and then SQLTables[conn]. (with a real table name of course) – mfvonh Jun 25 '14 at 4:58 Module[{dbConn, sqlAct, records},	2016-05-26 10:47:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4483705461025238, "perplexity": 3360.9875087489936}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049275835.98/warc/CC-MAIN-20160524002115-00105-ip-10-185-217-139.ec2.internal.warc.gz"}
https://github.com/pilif/Minecraft-Overviewer	publicpilif/Minecraft-Overviewer forked from overviewer/Minecraft-Overviewer Subversion checkout URL You can clone with HTTPS or Subversion. Render large resolution images of a Minecraft map with a Google Maps powered interface This branch is 0 commits ahead and 0 commits behind master Fetching latest commit… Cannot retrieve the latest commit at this time textures .gitignore COPYING.txt README.rst blockcounter.py chunk.py gmap.py nbt.py quadtree.py setup.py template.html textures.py util.py world.py Minecraft Overviewer By Andrew Brown and contributors http://github.com/brownan/Minecraft-Overviewer Generates large resolution images of a Minecraft map. In short, this program reads in Minecraft world files and renders very large resolution images. It performs a similar function to the existing Minecraft Cartographer program but with a slightly different goal in mind: to generate large resolution images such that one can zoom in and see details. See some examples here! http://github.com/brownan/Minecraft-Overviewer/wiki/Map-examples (To contact me, send me a message on Github) Features • Renders large resolution images of your world, such that you can zoom in and see details • Customizable textures! Pulls textures straight from your installed texture pack! • Outputs a Google Map powered interface that is memory efficient, both in generating and viewing. • Renders efficiently in parallel, using as many simultaneous processes as you want! • Utilizes 2 levels of caching to speed up subsequent renderings of your world. • Throw the output directory up on a web server to share your Minecraft world with everyone! Requirements This program requires: I develop and test this on Linux, but need help testing it on Windows and Mac. If something doesn't work, let me know. Using the Overviewer Disclaimers Before you dive into using this, just be aware that, for large maps, there is a lot of data to parse through and process. If your world is very large, expect the initial render to take at least an hour, possibly more. (Since Minecraft maps are practically infinite, the maximum time this could take is also infinite!) If you press ctrl-C, it will stop. The next run will pick up where it left off. Once your initial render is done, subsequent renderings will be MUCH faster due to all the caching that happens behind the scenes. Just use the same output directory and it will only update the tiles it needs to. There are probably some other minor glitches along the way, hopefully they will be fixed soon. See the Bugs section below. Textures The Overviewer uses actual textures to render your world. However, I don't include textures in the package. You will need to do one of two things before you can use the Overviewer: • Make sure the Minecraft client is installed. The Overviewer will find the installed minecraft.jar and extract the textures from it. • Install a texture file yourself. This file is called "terrain.png" and is normally found in your minecraft.jar file (not "Minecraft.jar", the launcher, but rather the file that's downloaded by the launcher and installed into a hidden directory). You can also get this file from any of the third party texture packs out there. Running To generate a set of Google Map tiles, use the gmap.py script like this: python gmap.py [OPTIONS] <World Number / Path to World> <Output Directory> The output directory will be created if it doesn't exist. This will generate a set of image tiles for your world in the directory you choose. When it's done, you will find an index.html file in the same directory that you can use to view it. The Overviewer will put a cached image for every chunk directly in your world directory by default. If you do not like this behavior, you can specify another location with the --cachedir option. See below for details. Options -h, --help Shows the list of options and exits --cachedir=CACHEDIR By default, the Overviewer will save in your world directory one image file for every chunk in your world. If you do backups of your world, you may not want these images in your world directory. Use this option to specify an alternate location to put the rendered chunk images. You must specify this same directory each rendering so that it doesn't have to render every chunk from scratch every time. Example: python gmap.py --cachedir= --imgformat=FORMAT Set the output image format used for the tiles. The default is 'png', but 'jpg' is also supported. Note that regardless of what you choose, Overviewer will still use PNG for cached images to avoid recompression artifacts. -p PROCS, --processes=PROCS Adding the "-p" option will utilize more cores during processing. This can speed up rendering quite a bit. The default is set to the same number of cores in your computer, but you can adjust it. Example to run 5 worker processes in parallel: python gmap.py -p 5 -z ZOOM, --zoom=ZOOM The Overviewer by default will detect how many zoom levels are required to show your entire map. This is equivilant to the dimensions of the highest zoom level, in tiles. A zoom level of z means the highest zoom level of your map will be 2^z by 2^z tiles. The -z option will set the zoom level manually. This could be useful if you have some outlier chunks causing your map to be too large. This will render your map with 7 zoom levels: python gmap.py -z 7 Remember that each additional zoom level adds 4 times as many tiles as the last. This can add up fast, zoom level 10 has over a million tiles. Tiles with no content will not be rendered, but they still take a small amount of time to process. -d, --delete This option changes the mode of execution. No tiles are rendered, and instead, cache files are deleted. Explanation: The Overviewer keeps two levels of cache: it saves each chunk rendered as a png, and it keeps a hash file along side each tile in your output directory. Using these cache files allows the Overviewer to skip rendering of any tile image that has not changed. By default, the chunk images are saved in your world directory. This example will remove them: python gmap.py -d You can also delete the tile cache as well. This will force a full re-render, useful if you've changed texture packs and want your world to look uniform. Here's an example: python gmap.py -d <# / path> Be warned, this will cause the next rendering of your map to take significantly longer, since it is having to re-generate the files you just deleted. --chunklist=CHUNKLIST Use this option to specify manually a list of chunks to consider for updating. Without this option, every chunk is checked for update and if necessary, re-rendered. If this option points to a file containing, 1 per line, the path to a chunk data file, then only those in the list will be considered for update. It's up to you to build such a list. On Linux or Mac, try using the "find" command. You could, for example, output all chunk files that are older than a certain date. Or perhaps you can incrementally update your map by passing in a subset of chunks each time. It's up to you! Viewing the Results Within the output directory you will find two things: an index.html file, and a directory hierarchy full of images. To view your world, simply open index.html in a web browser. Internet access is required to load the Google Maps API files, but you otherwise don't need anything else. You can throw these files up to a web server to let others view your map. You do not need a Google Maps API key (as was the case with older versions of the API), so just copying the directory to your web server should suffice. You are, however, bound by the Google Maps API terms of service. Crushing the Output Tiles Image files taking too much disk space? Try using pngcrush. On Linux and probably Mac, if you have pngcrush installed, this command will go and crush all your images in the given destination. This took the total disk usage of the render for my world from 85M to 67M. find /path/to/destination -name ".png" -exec pngcrush {} {}.crush \; -exec mv {}.crush {} \; Or if you prefer a more parallel solution, try something like this: find /path/to/destination -print0 \| xargs -0 -n 1 -P <nr_procs> sh -c 'pngcrush $0 temp.$$&& mv temp.$$$0' If you're on Windows, I've gotten word that this command line snippet works provided pngout is installed and on your path. Note that the % symbols will need to be doubled up if this is in a batch file. FOR /R c:\path\to\tiles\folder %v IN (.png) DO pngout %v /y Bugs This program has bugs. They are mostly minor things, I wouldn't have released a completely useless program. However, there are a number of things that I want to fix or improve. For a current list of issues, visit http://github.com/brownan/Minecraft-Overviewer/issues Feel free to comment on issues, report new issues, and vote on issues that are important to you, so I can prioritize accordingly. An incomplete list of things I want to do soon is: • Improve efficiency • Rendering non-cube blocks, such as torches, flowers, mine tracks, fences, doors, and the like. Right now they are either not rendered at all, or rendered as if they were a cube, so it looks funny.	2014-07-13 02:47:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20199596881866455, "perplexity": 3258.5780489755984}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1404776435842.8/warc/CC-MAIN-20140707234035-00086-ip-10-180-212-248.ec2.internal.warc.gz"}
http://nrich.maths.org/6451&part=	### Mathematical Issues for Chemists A brief outline of the mathematical issues faced by chemistry students. ### Reaction Rates Explore the possibilities for reaction rates versus concentrations with this non-linear differential equation ### Catalyse That! Can you work out how to produce the right amount of chemical in a temperature-dependent reaction? # Bond Angles ##### Stage: 5 Challenge Level: This problem makes use of vectors, coordinates and scalar products. It is a very useful fact that if two vectors ${\bf u}$ and ${\bf v}$ have coordinates $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ then the cosine of the angle $\theta$ they make is $$\cos(\theta) = \frac{{\bf u}\cdot {\bf v}}{\|{\bf u}\|\|{\bf v}\|}\quad\,,\mbox{ where } {\bf u}\cdot {\bf v}=x_1x_2+y_1y_2+z_1z_2$$ This is useful, because if we can label the coordinates of the atoms in a molecule or crystal then we can easily work out the angles between the bonds. In a perfect tetrahedral molecule there is a central atom attached to four other atoms which lie on the vertices of a perfect tetrahedron. If the central atom is at the coordinate origin and the molecule rests on the plane $z=-h$ then what would be the coordinates of the other atoms, assuming a bond length of $1$ unit? Find the scalar products between the vectors joining the origin to each atom and hence the value of $h$. It is usually stated that the angles each of these bonds make is $109.5^\circ$; however, this is only an approximation. What is the exact value of the bond angle in a perfect tetrahedron? If the perfect tetrahedron is deformed slightly, how many of the bond angles could be exactly $109.5^\circ$? How would it be deformed to achieve this? Would the other bond angles increase or decrease under such a deformation? A perfect trigonal pyramidal molecule is the same as a perfect tetrahedral molecule with a single outer atom removed. Ammonia NH$_3$ is approximately a trigonal pyramid with bond angle $107.5$. Does this correspond to a lengthening or shortening of the bonds relative to a perfect structure? By what percentage? If the H atoms were fixed and a vertical force were applied downwards to the N atom in the diagram, would the bond angles increase or decrease? What would be the maximum possible angle mathematically?	2016-07-01 15:18:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6209506988525391, "perplexity": 384.9394246995191}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783403502.46/warc/CC-MAIN-20160624155003-00165-ip-10-164-35-72.ec2.internal.warc.gz"}
https://www.jiskha.com/display.cgi?id=1353090004	Calculus posted by . A 12' ladder is leaning against a wall making an angle theta with the ground. The ladder begins to slide down the wall. what is the rate of change of the height of the ladder, h, with respect to the change in the angle of the ladder with the floor when the ladder reaches a height of 9 feet? • Calculus - if the distance from wall is x, x^2 + h^2 = 144 x = √(144-h^2) when h=9, x = √(144-81) = √63 sinθ = h/12 the wording of the question is odd. Usually they ask for the rate of change of x when h=9. In that case, 2x dx/dt + 2h dh/dt = 0 but we don't have dh/dt. So, it appears they want dh/dθ h/12 = sinθ h = 12sinθ dh/dθ = 12cosθ = 12(x/12) = x = √63 however, rather than asking for "the rate of change of h with respect to θ" you say they want "rate of change of h with respect to the change of θ". Don't know how to interpret that Similar Questions 1. Calculus related rates: a ladder, 12 feet long, is leaning against a wall. if the foot of the ladder slides away from the wall along level ground,what is the rate of change of the top of the ladder with respect to the distance of the foot of … 2. Math A ladder 20 ft long rests against a vertical wall. Let \theta be the angle between the top of the ladder and the wall and let x be the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from … 3. Math A ladder 20 ft long rests against a vertical wall. Let \theta be the angle between the top of the ladder and the wall and let x be the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from … A ladder 20 ft long rests against a vertical wall. Let \theta be the angle between the top of the ladder and the wall and let x be the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from … 5. math A ladder 14 ft long rests against a vertical wall. Let \theta be the angle between the top of the ladder and the wall and let x be the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from … 6. calculus A ladder 14 ft long rests against a vertical wall. Let \theta be the angle between the top of the ladder and the wall and let x be the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from … 7. CALCULUS A 25 ft ladder is leaning against a vertical wall. At what rate (with respect to time) is the angle theta between the ground and the ladder changing, if the top of the ladder is sliding down the wall at the rate of r inches per second, … 8. maths a 4m ladder is leaning against a wall when its base starts to slide away. By the time the base is 2m from the wall, the base is moving at the rate of 1m/sec. find a) the speed at which the top of the ladder is sliding down the wall … 9. Calculus 1a. 15 ft ladder is placed against a vertical wall. the bottom of the ladder is sliding away from the wall at a rate of 2 ft/sec. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 9 ft from the … 10. calculus A 17- foot ladder is leaning against a wall, and begins to slide. The top of the ladder is falling at a rate of 2 feet per second at the instant that the bottom of the ladder is 8 feet from the wall. What is the rate of change of the … More Similar Questions	2018-01-22 10:39:51	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8104047179222107, "perplexity": 190.93236602420606}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891277.94/warc/CC-MAIN-20180122093724-20180122113724-00292.warc.gz"}
http://atozmath.com/Default.aspx?q1=Is%20Scalar%20matrix%20%5B%5B2%2C0%2C0%5D%2C%5B0%2C2%2C0%5D%2C%5B0%2C0%2C2%5D%5D%60519&do=1	Home > Matrix Algebra calculators > is Scalar Matrix calculator Solve any problem (step by step solutions) Input table (Matrix, Statistics) Mode : SolutionHelp Solution Find Is Scalar matrix [[2,0,0],[0,2,0],[0,0,2]] Solution: Your problem -> Is Scalar matrix [[2,0,0],[0,2,0],[0,0,2]] A square matrix, in which all diagonal elements are equal to same scalar and all other elements are zero, is called a scalar matrix. Or A diagonal matrix, in which all diagonal elements are equal to same scalar, is called a scalar matrix. A = 2 0 0 0 2 0 0 0 2 Here, all diagonal elements are equal to same scalar(2) and all other elements are zero, so it is a scalar matrix. Solution provided by AtoZmath.com Any wrong solution, solution improvement, feedback then Submit Here Want to know about AtoZmath.com and me	2019-04-20 20:37:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6447698473930359, "perplexity": 1300.738843554935}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578530040.33/warc/CC-MAIN-20190420200802-20190420222802-00319.warc.gz"}
https://denisegaskins.com/tag/my-math-books/page/2/	## Math Journals for Elementary and Middle School This fall, my homeschool co-op math class will play with math journaling. But my earlier dot-grid notebooks were designed for adults. Too thick, too many pages. And the half-cm dot grid made lines too narrow for young writers. So I created a new series of paperback dot-grid journals for my elementary and middle school students. I hope you enjoy them, too! ## Math Journaling Prompts So, what can your kids do with a math journal? Here are a few ideas: I’m sure we’ll use several of these activities in my homeschool co-op math class this fall. ### Noticing and Wondering Learning math requires more than mastering number facts and memorizing rules. At its heart, math is a way of thinking. So more than anything else, we need to teach our kids to think mathematically — to make sense of math problems and persevere in figuring them out. Whenever your children need to learn a new idea in math, or whenever they get stuck on a tough homework problem, that’s a good time to step back and make sense of the math. Kids can write their noticings and wonderings in the math journal. Or you can act as the scribe, writing down (without comment) everything child says. For more tips on teaching students to brainstorm about math, check out these online resources from The Math Forum: Problem-solving is a habit of mind that you and your children can learn and grow in. Help your kids practice slowing down and taking the time to fully understand a problem situation. ### Puzzles Are Math Experiments Almost anything your child notices or wonders can lead to a math experiment. For example, one day my daughter played an online math game… A math journal can be like a science lab book. Not the pre-digested, fill-in-the-blank lab books that some curricula provide. But the real lab books that scientists write to keep track of their data, and what they’ve tried so far, and what went wrong, and what finally worked. Here are a few open-ended math experiments you might try: ##### Explore Shapes • Pick out a 3×3 set of dots. How many different shapes can you make by connecting those dots? Which shapes have symmetry? Which ones do you like the best? • What if you make shapes on isometric grid paper? How many different ways can you connect those dots? • Limit your investigation to a specific type of shape. How many different triangles can you make on a 3×3 set of dots? How many different quadrilaterals? What if you used a bigger set of dots? ##### Explore Angles • On your grid paper, let one dot “hold hands” with two others. How many different angles can you make? Can you figure out their degree without measuring? • Are there any angles you can’t make on your dot grid? If your paper extended forever, would there be any angles you couldn’t make? • Does it make a difference whether you try the angle experiments on square or isometric grid paper? ##### Explore Squares • How many different squares can you draw on your grid paper? (Don’t forget the squares that sit on a slant!) How can you be sure that they are perfectly square? • Number the rows and columns of dots. Can you find a pattern in the corner positions for your squares? If someone drew a secret square, what’s the minimum information you would need to duplicate it? • Does it make a difference whether you try the square experiments on square or isometric grid paper? ### Or Try Some Math Doodles Create math art. Check out my math doodling collection on Pinterest and my Dot Grid Doodling blog post. Can you draw an impossible shape? ## How Would YOU Use a Math Journal? I’d love to hear your favorite math explorations or journaling tips! P.S.: Do you have a blog? If you’d like to feature a math journal review and giveaway, I’ll provide the prize. Send a message through my contact form or leave a comment below, and we’ll work out the details. I’ve been working on my next Playful Math Singles book, based on the popular Things to Do with a Hundred Chart post. My hundred chart list began many years ago as seven ideas for playing with numbers. Over the years, it grew to its current 30+ activities. Now, in preparing the new book, my list has become a monster. I’ve collected almost 70 ways to play with numbers, shapes, and logic from preschool to middle school. Just yesterday I added activities for fraction and decimal multiplication, and also tips for naming complex fractions. Wow! Gonna have to edit that cover file… In the “Advanced Patterns” chapter, I have a section on math debates. The point of a math debate isn’t that one answer is “right” while the other is “wrong.” You can choose either side of the question — the important thing is how well you support your argument. Here’s activity #69 in the current book draft. ### Have a Math Debate: Adding Fractions When you add fractions, you face a problem that most people never consider. Namely, you have to decide exactly what you are talking about. For instance, what is one-tenth plus one-tenth? Well, you might say that: $\frac{1}{10}$ of one hundred chart + $\frac{1}{10}$ of the same chart = $\frac{2}{10}$ of that hundred chart But, you might also say that: $\frac{1}{10}$ of one chart + $\frac{1}{10}$ of another chart = $\frac{2}{20}$ of the pair of charts That is, you started off counting on two independent charts. But when you put them together, you ended up with a double chart. Two hundred squares in all. Which made each row in the final set worth $\frac{1}{20}$ of the whole pair of charts. So what happens if you see this question on a math test: $\frac{1}{10}$ + $\frac{1}{10}$ = ? If you write the answer “$\frac{2}{20}$”, you know the teacher will mark it wrong. Is that fair? Why, or why not? CREDITS: Feature photo (above) by Thor/geishaboy500 via Flickr (CC BY 2.0). “One is one … or is it?” video by Christopher Danielson via TED-Ed. This math debate was suggested by Marilyn Burns’s blog post Can 1/3 + 1/3 = 2/6? It seemed so! ## Let’s Play Math in Korean Ooo, look at my shiny new book! Let’s Play Math is now out in Korean. How cool is that? You can find the book at these two major bookstores: And probably in other places where Korean education or parenting books are sold. I’m sorry to say I can’t read Korean — but I did play math there a couple years back. My daughter teaches English through EPIK, and I had a wonderful visit with her in Jeju. If you’re interested, you can see a few of my photos here, and my fraction-math sidewalk puzzle here. And if you know a Korean-speaking family who wants to play math with their kids, I’d be honored if you share my book. ## New Book: Word Problems from Literature The posts on my Let’s Play Math blog are, for the most part, first-draft material. Of course, I’ve proofread each post — many times! because I’m a perfectionist that way, and yet I still miss typos — but these articles haven’t gotten the sort of feedback that polishes a book manuscript. Well, now I’m taking some of the best of my old blog posts, expanding them with a few new games or activities, and giving them that book-quality polish. Let me introduce my newest series, the Playful Math Singles. ### Under Construction … The Playful Math Singles from Tabletop Academy Press will be short, topical books featuring clear explanations and ready-to-play activities. I’m hoping to finish up two or three of these this year. Watch for them at your favorite online bookstore. The first one is done … ### Word Problems from Literature: An Introduction to Bar Model Diagrams You can help prevent math anxiety by giving your children the mental tools they need to conquer the toughest story problems. Young children expect to look at a word problem and instantly see the answer. But as they get older, their textbook math problems also grow in difficulty, so this solution-by-intuitive-leap becomes impossible. Too often the frustrated child concludes, “I’m just not good at math.” But with guided practice, any student can learn to master word problems. Word Problems from Literature features math puzzles for elementary and middle school students from classic books such as Mr. Popper’s Penguins and The Hobbit. For each puzzle, I demonstrate step by step how to use the problem-solving tool of bar model diagrams, a type of pictorial algebra. For children who are used to playing with Legos or other blocks — or with computer games like Minecraft — this approach reveals the underlying structure of a math word problem. Students can make sense of how each quantity in the story relates to the others and see a path to the solution. And when you finish the puzzles in this book, I’ll show you how to create your own word problems from literature, based in your children’s favorite story worlds. If you’re using these word problems with your children, consider buying them the paperback companion Word Problems from Literature Student Workbook. ##### … and People Like It! A screen shot from this past weekend: “I found this method really clarified for me what was going on visually and conceptually. Particularly when it came to more complex questions, for which I would normally write out an equation, I felt that thinking about what was going on with the bars actually made more sense … This is a wonderful book for those who want to support their children in finding better ways to work on word problems.” —Miranda Jubb, Amazon customer reviewer ## How to Break In Your New Math Journal I love my new paperback math journal series. The books are sturdy, inexpensive, and fit nicely in my purse. But as with any paperback book, these have one problem. How do I use them without cracking the spine? When we exercise, we need to warm up our bodies with a bit of stretching to prevent injury. In the same way, we need to warm up a new book to protect it. The process is called “breaking it in.” It only takes a few minutes to break in a paperback book: ### Step by Step Never force the book but help it limber up gradually, and it will serve you well. Because my journals are working books, I take the breaking-in process a bit further than shown in the video: (1) Set the book on its back and follow the process above. Press down each cover, but not completely flat — let it bend at the fold line, about 1 cm from the actual spine. Then press a couple pages at a time, alternating front and back, down flat on each cover. (2) Flip through the pages of the book forward and backward to limber them up. (3) Repeat the steps of the video. This time, gently lean the main part of the book away from the part you are pressing down. Aim for a 130–140 degree angle. (4) Flip through the pages again. Even roll the book back and forth a bit — curving the cover and pages as if you’re trying to fold the book in half — to encourage flexibility. (5) Repeat the breaking-in process one more time. This time, fold each section back as close to 180 degrees as it will go. And you’re done! The pages will still curve in at the fold line, where they connect to the spine of the book. You want that because it makes the book strong. But now they’ll also open up to provide a nice, wide area for writing or math doodling. ## Dot Grid Doodling ### What can you DO with a page full of dots? Yesterday, I mentioned my new series of paperback dot grid notebooks, and I promised to share a few ideas for mathematical doodling. Doodling gives our minds a chance to relax, wander, and come back to our work refreshed. And though it goes against intuition, doodling can help us remember more of what we learn. Math doodles let us experiment with geometric shapes and symmetries. We can feel our way into math ideas gradually, through informal play. Through doodles, our students will explore a wide range of mathematical structures and relationships. Our own school experiences can make it hard for us to teach. What we never learned in school was the concept of playing around with math, allowing ideas to “percolate,” so to speak, before mastery occurs, and that process may take time. —Julie Brennan I like to doodle on dotty grid paper, like the pages in my math journals, but there’s No Purchase Necessary! You can design your own printable dot page at Incompetech’s PDF generator, or download my free coloring book (which includes several pages of printable dot and graph paper). ### Patterns in Shape and Angle To make a faceted mathematical gemstone, start with any shape you like. Then build other shapes around it. What do you notice? Does your pattern grow outward from its center? Or flow around the corner of your page? How is each layer similar, and how is it different? Arbitrary constraints can lead to mathematically interesting doodles. For instance, create a design out of 45-45-90 triangles by coloring exactly half of every grid square. How many variations can you find? ### Symmetry Challenge Play a symmetry puzzle game. Draw a line of symmetry and fill in part of the design. Then trade with a partner to finish each other’s doodles. Make more complex symmetry puzzles with additional reflection lines. • Who can talk about mathematical doodling without mentioning Vi Hart? If you’ve never seen her “Doodling in Math Class” video series, you’re in for a treat! • See if you can draw a rotational-symmetry design, like Don’s “Order 4” graphs. • Or experiment with the more flexible rules in John’s “Knot Fun” lesson. • And my latest obsession: the “ultimate” tutorial series on Celtic Knotwork, which explores the link between knots and their underlying graphs. Also available through: Before you start doodling: How to Break In Your New Math Journal. Feature photo (top): Sommermorgen (Alte Holzbrücke in Pretzfeld) by Curt Herrmann, via Wikimedia Commons. [Public domain]	2019-06-25 11:30:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 10, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.27828842401504517, "perplexity": 2038.1338222721001}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999838.23/warc/CC-MAIN-20190625112522-20190625134522-00033.warc.gz"}
https://www.martinkysel.com/hackerrank-hackerrank-in-a-string-solution/	# HackerRank ‘HackerRank in a String!’ Solution ##### Short Problem Definition: We say that a string contains the word hackerrank if a subsequence of its characters spell the word hackerrank. For example, if string s = haacckkerrannkk it does contain hackerrank, but s = haacckkerannk does not. In the second case, the second r is missing. If we reorder the first string as , it no longer contains the subsequence due to ordering. HackerRank in a String! ##### Complexity: time complexity is O(N) space complexity is O(1) ##### Execution: Keep two pointers. One to the expected string (needle) and one to the input string. If you find the needle in the haystack before you run out of characters, you are good. ##### Solution: def hackerrankInString(s): needle = 'hackerrank' idx_in_needle = 0 for c in s: if c == needle[idx_in_needle]: idx_in_needle += 1 if idx_in_needle == len(needle): break if idx_in_needle == len(needle): return "YES" else: return "NO" If you enjoyed this post, then make sure you subscribe to my Newsletter and/or Feed.	2018-11-14 20:19:17	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.32971274852752686, "perplexity": 4260.925128280263}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039742263.28/warc/CC-MAIN-20181114191308-20181114212734-00028.warc.gz"}
https://en.wikipedia.org/wiki/Talk:Riemann%E2%80%93Lebesgue_lemma	# Talk:Riemann–Lebesgue lemma WikiProject Mathematics (Rated B-class, Low-importance) This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks. Mathematics rating: B Class Low Importance Field: Analysis ## which "special cases" in proof? The first sentence of the proof is confusing, and it is not clear which "special cases" it refers to, and which if any of them is the "first one". Perhaps it could be replaced with something like the following: "The proof can be organized into steps, proving increasingly general special cases; the 4th step extends the result to the original formulation." but personally I think that it might be better to remove that sentence altogether --AmitAronovitch (talk) 18:35, 21 May 2010 (UTC) The statement of the Theorem seems to be flawed. f is assumed to be a measurable function from R to C. But the proof deals with an interval [a,b]. Should we say f:[a,b] to C? or should the proof omit [a,b]? —Preceding unsigned comment added by 69.142.45.52 (talk) 01:22, 29 April 2011 (UTC) Also the proof is only for f on R, while the statement of the theorem has f on R^d — Preceding unsigned comment added by Bzhao2017 (talkcontribs) 04:10, 2 December 2019 (UTC) ## Proof seems to be a bit unclear You have written: We'll focus on the one-dimensional case, the proof in higher dimensions is similar. Suppose first that f is a compactly supported smooth function. Then integration by parts in each variable yields ${\displaystyle \left\|\int f(x)e^{-izx}dx\right\|=\left\|\int {\frac {1}{iz}}f'(x)e^{-izx}dx\right\|\leq {\frac {1}{\|z\|}}\int \|f'(x)\|dx\rightarrow 0{\mbox{ as }}z\rightarrow \pm \infty .}$ How do you know that if f(x) is L1 function then its derivative is L1 function too? Consider: ${\displaystyle f(x)={\frac {\sin(e^{x})}{1+x^{2}}}}$. or ${\displaystyle f(x)={\begin{cases}x^{2}e^{-x^{2}}\sin(1/x^{2})&\|x\|\in (0,1]\\0&{\text{otherwise}}\end{cases}}}$ Concluding: If you just write that you take into account only the situation where f(x) is differentiable and additionally f'(x) is L1 function too,then proof is more clear and universal. — Preceding unsigned comment added by 89.79.154.60 (talk) 19:50, 27 February 2014 (UTC) Consider : ${\displaystyle f(x)={\frac {1_{\mathbb {Q} }2-1}{1+x^{2}}}}$ It's L1${\displaystyle \scriptstyle \mathrm {(} \mathbb {R} ),}$ integrable function, but ${\displaystyle \lim _{\|\omega \|\to +\infty }\int \limits _{\mathbb {R} }f(x)e^{-i\omega x}dx\neq 0}$ — Preceding unsigned comment added by 89.76.155.25 (talk) 18:05, 24 December 2014 (UTC) What is ${\displaystyle {1_{\mathbb {Q} }2}}$? If it's a complex number, I think we're okay. 178.39.163.55 (talk) 11:50, 5 July 2015 (UTC) ## Article in "References" section (Self-promotion?) The Researchgate article included in the references appears to have been added by User:Anilped. User:Anilped claims to be "Prof Anil Pedgaonkar", which also happens to be the alias of that article's author. I've made a brief skim of the article, and it seems inconsequential. But, then again, I've never had cause to use the lemma (yet). Maybe the article is of value, in a way I can't see. Can someone with a background in this area check that his generalizations are actually of sufficient novelty and power that we ought be directing readers there? (If not, note that we should probably remove the last comment in "Other Versions" too.) 2601:240:C400:D60:902F:C0CF:8076:BA68 (talk) 03:21, 17 July 2017 (UTC) ## Abstract measure spaces The proof in the research gate article is wrong, the reference was removed. — Preceding unsigned comment added by 81.243.243.97 (talk) 21:10, 21 November 2018 (UTC)	2020-02-22 17:07:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7569137215614319, "perplexity": 1141.3846494040426}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145708.59/warc/CC-MAIN-20200222150029-20200222180029-00465.warc.gz"}
https://math.stackexchange.com/questions/1027871/logarithmic-equation-solve-for-x	# Logarithmic Equation: Solve for $x$ $$\log_{3x}81 = 2$$ How would I go about solving this? This is what I tried: $$\log_{3x}81 = 2$$ $$\frac{\log81}{\log 3 + \log x }= 2$$ Where do I go from here? If I isolate $\log x$ on one side, how do I get rid of the log? • $81=3^4$, so $\log81=4\log3$. – Akiva Weinberger Nov 18 '14 at 17:57 • Or, do $(3x)^\textrm{both sides}$ from the start. – Akiva Weinberger Nov 18 '14 at 17:59 • Do you mind showing me your steps for this? Are you starting at where I left off? – McB Nov 18 '14 at 18:00 • The second thing is starting from the start. Remember that $(3x)^{\log_{3x}(\textrm{stuff})}=\textrm{stuff}$. – Akiva Weinberger Nov 18 '14 at 18:01 • Aside from the simple way of doing this that others have already posted, you can also solve the equation $\dfrac{\log81}{\log 3 + \log x }= 2$ for $x$. I explained how in an answer posted below. ${}\qquad{}$ – Michael Hardy Nov 18 '14 at 18:16 $\log_{3x}(81)=2$ is equivalent to $$(3x)^2=9x^2=81$$ by the definition of the logarithm. $$9x^2=81 \Leftrightarrow x^2=9$$ This gives solutions $x=3$ and $x=-3$, but only $x=3$ is a solution, since the base of a logarithm must be greater than zero. • Very helpful, thanks. For some reason I failed to see that I could isolate x very easily once I "booted the log". – McB Nov 18 '14 at 18:06 For any two real numbers $b$ and $x$ where $b$ is positive and $b ≠ 1$, $$y=b^z\Leftrightarrow z=\log_b(y)$$ so for $\log_{3x}81=2$ we have $$(3x)^2=81=(3\cdot 3)^2\Rightarrow x=3$$ Aside from the simpler way of solving this that someone else has already posted, here's something else you can do: $$\frac{\log81}{\log 3 + \log x }= 2$$ $$\frac{\log 3 + \log x}{\log 81} = \frac 1 2$$ $$\log 3 + \log x = \frac 1 2 \log 81$$ $$\log x = \frac 1 2 \log 81 - \log 3 = \frac 1 2 \log(3^4) - \log 3$$ $$=\frac 1 2\cdot 4 \log 3 - \log 3 = \left(\frac 1 2\cdot 4 - 1\right)\log 3 = \log 3.$$ Now you have $$\log x = \log 3,$$ and since logarithmic functions are one-to-one, you can conclude that $x=3$. • Thank-you, that is very helpful. – McB Nov 18 '14 at 18:18	2019-07-23 05:42:55	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8932347297668457, "perplexity": 288.07259457329087}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195528869.90/warc/CC-MAIN-20190723043719-20190723065719-00365.warc.gz"}
https://socratic.org/questions/how-do-you-find-the-slope-that-is-perpendicular-to-the-line-7x-6y-13	# How do you find the slope that is perpendicular to the line 7x-6y=13? May 14, 2017 $- \frac{6}{7}$ #### Explanation: First of all solve for $y$ $7 x - 6 y = 13 \implies - 6 y = 13 - 7 x$ $y = \frac{13 - 7 x}{-} 6 = - \frac{13}{6} + \frac{7 x}{6}$ $\textcolor{red}{y = \frac{\textcolor{b l u e}{7}}{\textcolor{g r e e n}{6}} x - \frac{13}{6}}$ If the line is perpendicular to this line then its slope will be the negative reciprocal of the slope of is line So the slope is $\textcolor{red}{-} \frac{\textcolor{g r e e n}{6}}{\textcolor{b l u e}{7}}$	2021-11-30 15:29:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8337810039520264, "perplexity": 282.9684875131847}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964359037.96/warc/CC-MAIN-20211130141247-20211130171247-00253.warc.gz"}
https://getpractice.com/subjects/physics/sound-waves?page=4	### Sound Waves goals Velocity (v) of sound in air, by vibrating resonating column is found by ______ (${\ell}_1$, ${\ell}_2$ and n are first second resonating lengths and frequency of tuning fork used respectively). Ultrasound waves have : A cars horn produced a sound wave of a constant frequency. Car is moving to the right as shown above. The five points A, B, C, D and E are marked, at which point would a man listen a lower pitch than pitch produced when car is at rest? The velocity of sound in a gas in which two waves of wavelengths $50$ cm and $50.5$ cm produces $6$ beats per second is? A supersonic jet produces waves in air. The wave front is	2020-10-29 22:10:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23675134778022766, "perplexity": 905.7994220594293}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107905965.68/warc/CC-MAIN-20201029214439-20201030004439-00670.warc.gz"}
https://ncatlab.org/nlab/show/spectrum+of+a+commutative+ring	# nLab spectrum of a commutative ring This entry is about the formal dual topological space of a commutative ring. For the very different notion of a similar name in higher algebra see at ring spectrum. For more see at spectrum - disambiguation. # Contents ## Idea Given a commutative ring $R$, its spectrum is the topological space whose points are the prime ideals of $R$ and whose topology is the Zariski topology on these prime ideals. (The prime spectrum) ## References Revised on May 6, 2017 07:31:01 by Anonymous (87.165.118.230)	2018-03-17 22:05:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5735688209533691, "perplexity": 615.5355565432449}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257645362.1/warc/CC-MAIN-20180317214032-20180317234032-00523.warc.gz"}
https://timodenk.com/blog/tag/graph/	## LaTeX Plot Snippets The LaTeX package tikz contains a set of commands that can render vector based graphs and plots. This post contains several examples which are intended to be used as a cut-and-paste boilerplate. Each sample comes with a screenshot and a snippet that contains the relevant parts of the LaTeX source code. The entire source code can be examined by clicking onto the individual images. Continue reading LaTeX Plot Snippets ## Graph Theory Overview In the field of computer science, a graph is an abstract data type that is widely used to represent connections and relationships. This post gives an overview about a selection of definitions, terms, and algorithms, which are related to graphs. The content was put together during preparation for a theoretical computer science test at Cooperative State University Baden-Württemberg and is mostly taken from either Wikipedia or lecture notes. Continue reading Graph Theory Overview	2019-02-21 10:34:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6506783962249756, "perplexity": 737.2904700221748}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247503844.68/warc/CC-MAIN-20190221091728-20190221113728-00620.warc.gz"}
https://mathstodon.xyz/@btcprox	Article on Nature outlining a framework for mathematicians to use to discover insights. with case studies in knot theory + representation theory: nature.com/articles/s41586-021 Yet another discussion on the origin of (discovered or invented?), this time involving William Shatner + Stanford U's Keith Devlin: Celebrating 10 years of by examining the number 10 (and talking about friendly numbers) 🎂🎉 The recording of "The Great Big and Gameshow", hosted by Dr Tom Crawford on 13 October at IF Oxford, is out on YouTube: "Irish Dara Ó Briain has received an award for his contribution to raising public awareness of . This is an award that is presented each year during Maths Week – the annual all-island festival of maths and numeracy, which is currently running with events across the country and online." siliconrepublic.com/innovation Just thought of signal boosting the Azimuth Project: azimuthproject.org/azimuth/sho Seeing the latest strip, wondering what unsolved problems you would considered "cursed" 🥴 xkcd.com/2529/ It's pretty amusing that a big part of Matt Parker's legacy is having "Parker" being synonymous with being "almost correct" or a "near miss" (with big help from Brady Haran), to the point of being referenced in an actual maths preprint Another video from Matt Parker, this time exploring the squircle (particularly the shape defined by \|𝑥\|⁴+\|𝑦\|⁴=1) and the complications from trying to work out its area, featuring some lemniscate biscuits, and a lot of interactions between different versions of himself The classic discussion on whether exists in our reality for us to discover, or exists only as man-made inventions to describe said reality (among others), is once again explored in this panel discussion (with a 10+ minute intro from moderator Brian Greene) No one: $$\mathbb{R}\setminus\{1\}$$ Absolutely no one: $$[0,\infty)\setminus\{1\}$$ Newest Numberphile video features Tom Crawford breaking down an approach to an ex-Oxford Admissions Question: given the task of completely filling up a square with N non-overlapping rectangles of any size (not necessarily all uniform), as long as each rectangle has one side twice the length of the other, for which values of N is this task possible? Matt Parker's latest video is a nice basic (and self-admittedly simplified) exposure to some of the concepts of : Saw this joke online: log😅 = 💧log😄 From /r/math: "what is the most enraging/funniest 'how is this useful' question someone has asked you?" reddit.com/r/math/comments/oqr Imaginary numbers seem to often get harshly judged by their label and ridiculed for being "imaginary maths" Would the stigma be lessened if they were called something else? Can they even be called anything else since the "real numbers" had already established prominence? "...we obtain the first consistent mathematical description of multiple wave dynamics and its inter-wave strolling regime. Our results are tested and calibrated against the pandemic data. Because of the simplicity of our approach that is organized around symmetry principles, our discovery amounts to a paradigm shift in the way epidemiological data are mathematically modelled." frontiersin.org/articles/10.33 Not really sticking to any particular kind of note-taking system/ideology (e.g. Zettelkasten), instead focusing on just generating the content and saving the refactoring for later Dabbling with yet another : Obsidian () also does the whole note linking + graphs + tags, but at least there's a free desktop client extendable by plugins, and I could do my own free syncing + versioning: obsidian.md/ Currently incorporated some notes from previous online courses + using it for ongoing notes for a training program, but will probably try redoing notes for past uni modules + outlining books/articles Math prof Jordan Ellenberg, author of How Not To Be Wrong, just did a Reddit AMA:	2021-12-07 08:42:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2626871168613434, "perplexity": 3550.062764382276}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363337.27/warc/CC-MAIN-20211207075308-20211207105308-00467.warc.gz"}
http://simbad.u-strasbg.fr/simbad/sim-ref?querymethod=bib&simbo=on&submit=submit+bibcode&bibcode=2012A%26A...543A..83S	# 2012A&A...543A..83S other querymodes : Identifierquery Coordinatequery Criteriaquery Referencequery Basicquery Scriptsubmission TAP Outputoptions Help Query : 2012A&A...543A..83S 2012A&A...543A..83S - Astronomy and Astrophysics, volume 543A, 83-83 (2012/7-1) Modelling the black hole silhouette in Sagittarius A* with ion tori. STRAUB O., VINCENT F.H., ABRAMOWICZ M.A., GOURGOULHON E. and PAUMARD T. Abstract (from CDS): We calculate the observed at infinity'' image and spectrum of the accretion structure in Sgr A, by modelling it as an optically thin, constant angular momentum ion torus in hydrodynamic equilibrium. The physics we consider includes a two-temperature plasma, a toroidal magnetic field, as well as radiative cooling by bremsstrahlung, synchrotron, and inverse Compton processes. Our relativistic model has the virtue of being fully analytic and very simple, depending only on eight tunable parameters: the black hole spin and the inclination of the spin axis to our line of sight, the torus angular momentum, the polytropic index, the magnetic to total pressure ratio, the central values of density and electron temperature, and the ratio of electron to ion temperatures. The observed image and spectrum are calculated numerically using the ray-tracing code GYOTO. Our results demonstrate that the ion torus model is able to account for the main features of the accretion structure surrounding Sgr A. Journal keyword(s): black hole physics - accretion, accretion disks - Galaxy: center Full paper Number of rows : 4 N Identifier Otype ICRS (J2000) RA ICRS (J2000) DEC Mag U Mag B Mag V Mag R Mag I Sp type #ref 1850 - 2022 #notes 1 NAME Galactic Center reg 17 45 39.60213 -29 00 22.0000 ~ 13084 0 2 NAME Sgr A* X 17 45 40.03599 -29 00 28.1699 ~ 3929 3 3 [EG97] S2 * 17 45 40.0442 -29 00 27.975 B0/2V 331 1 4 HD 226868 HXB 19 58 21.6757355952 +35 12 05.784512688 9.38 9.72 8.91 8.42 O9.7Iabpvar 4122 0 To bookmark this query, right click on this link: simbad:objects in 2012A&A...543A..83S and select 'bookmark this link' or equivalent in the popup menu 2022.08.10-06:23:58 © Université de Strasbourg/CNRS	2022-08-10 04:23:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.441051185131073, "perplexity": 7357.50403462196}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571147.84/warc/CC-MAIN-20220810040253-20220810070253-00372.warc.gz"}
http://askubuntu.com/questions/186649/t-string-with-sed	# \t string with sed I'm using sed to replace the string "\UPo" by the string "\textit{Old (User's preferences)}". The command I'm using is: find /home/rom/Desktop/nodeRemoval/report -name \*.tex -exec sed -i "s/\\UPo/\\textit{Old (User's preferences)}/g" {} \; I want to escape the backslashes as they are part of the string. But instead of a backslash it gives me a tab! The output is: \[TAB]extit{Old (User's preferences)} What's wrong? Thanks in advance, Rom - add comment ## 1 Answer Since your expression is in double quotes, the string gets evaluated by bash (which replaces \\ with \). In fact, the regex passed on to sed looks like that: s/\UPo/\textit{Old (User's preferences)}/g Either escape each backslash twice: sed -i "s/\\\\UPo/\\\\textit{Old (User's preferences)}/g" or, better, use single quotes. Anything in single quotes will not get evaluated by bash, but beware: a single quote may not appear between single quotes, so you need to finish the single quote, escape single quote, start single quote again: sed -i "s/\\UPo/\\textit{Old (User'\''s preferences)}/g" This is all a bit silly, which is why we have Perl. - It works great! Thank you. Rom. – rom Sep 12 '12 at 9:14 add comment	2013-12-06 00:30:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9100892543792725, "perplexity": 12673.67211713499}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163048688/warc/CC-MAIN-20131204131728-00094-ip-10-33-133-15.ec2.internal.warc.gz"}
https://cs.stackexchange.com/tags/arrays/hot	# Tag Info ## Hot answers tagged arrays 82 You have at least two options, depending on what problem you want to solve. If you want innocent readers of your code to not get the answers inadvertently, or you at least want to make it a bit difficult so that users are not tempted, you can encrypt the solutions and store the key as part of your code, perhaps a result of some computation (to make it even ... 46 Fixed-size queues are often implemented using what some people call circular buffers. If you remove the protection against it being full, you get the desired behaviour. Of course, no actual pushing will happen in the array -- that would be too expensive -- but it will look like it from the outside. 41 Here is the answer which elaborates upon the algorithm from the paper linked by Joe: http://arxiv.org/abs/0805.1598 First let us consider a $\Theta(n \log n)$ algorithm which uses divide and conquer. 1) Divide and Conquer We are given $$a_1, a_2, \dots , b_1, b_2, \dots b_n$$ Now to use divide and conquer, for some $m = \theta(n)$, we try to get the ... 31 No, it depends on your application. The measures of sortedness are often refered to as measures of disorder, which are functions from $N^{<N}$ to $\mathbb{R}$, where $N^{<N}$ is the collection of all finite sequences of distinct nonnegative integers. The survey by Estivill-Castro and Wood [1] lists and discusses 11 different measures of disorder in the ... 31 I see four main ways to solve this problem, with different running times: $O(n^2)$ solution: this would be the solution that you propose. Note that, since the arrays are unsorted, deletion takes linear time. You carry out $n$ deletions; therefore, this algorithm takes quadratic time. $O(n \: log \: n)$ solution: sort the arrays beforehand; then, perform a ... 28 You have two three options: Keep the answers separate from the rest of the source code If you want your code to be open source, however don't want the answers to be open source, then you open source the code for the application without the questions & answers, with the questions & answers being a separate closed source "plugin" or data file. Your ... 26 The solution in fade2black's answer is the standard one, but it uses $O(n)$ space. You can improve this to $O(1)$ space as follows: Let the array be $A[1],\ldots,A[n]$. For $d=1,\ldots,5$, compute $\sigma_d = \sum_{i=1}^n A[i]^d$. Compute $\tau_d = \sigma_d - \sum_{i=1}^{n-5} i^d$ (you can use the well-known formulas to compute the latter sum in $O(1)$). ... 25 The array indexing operation a[i] gains its meaning from the following features of C The syntax a[i] is equivalent to (a + i). Thus it is valid to say 5[a] to get at the 5th element of a. Pointer-arithmetic says that given a pointer p and an integer i, p + i the pointer p advanced by i sizeof(*p) bytes The name of an array a very quickly devolves to a ... 23 We count the number of array element reads and writes. To do bubble sort, you need $1 + 4n$ accesses (the initial write to the end, then, in the worst case, two reads and two writes to do $n$ swaps). To do the binary search, we need $2\log n + 2n + 1$ ($2\log n$ for binary search, then, in the worst case, $2n$ to shift the array elements to the right, then 1 ... 22 You could create an additional array $B$ of size $n$. Initially set all elements of the array to $0$. Then loop through the input array $A$ and increase $B[A[i]]$ by 1 for each $i$. After that you simply check the array $B$: loop over $A$ and if $B[A[i]] > 1$ then $A[i]$ is repeated. You solve it in $O(n)$ time at the cost of memory which is $O(n)$ and ... 18 I'm pretty sure I've found an algorithm that does not rely on number theory or cycle theory. Note that there are a few details to work out (possibly tomorrow), but I'm quite confident they will work out. I handwave as I'm supposed to be sleeping, not because I'm trying to hide problems :) Let A be the first array, B the second, \|A\| = \|B\| = N and assume N=2^... 16 The $\Theta(n)$ difference-of-sums solution proposed by Tobi and Mario can in fact be generalized to any other data type for which we can define a (constant-time) binary operation $\oplus$ that is: total, such that for any values $a$ and $b$, $a \oplus b$ is defined and of the same type (or at least of some appropriate supertype of it, for which the ... 15 This is a very general trick, which can be used for other purposes than hashing. Below I give an implementation (in pseudo-code). Let three uninitialized vectors $A$, $P$ and $V$ of size $n$ each. We will use these to do the operations requested by our data structure. We also maintain a variable $pos$. The operations are implemented as following: init: ... 15 If you don't know anything about the contents of the matrix (such as some kind of monotonicity property), linear time is the best you can do for a one-off search with a deterministic algorithm by a simple adversary argument: if you don't look at everything, then you can't distinguish between the cases where the maximum component is/isn't one of the ones you ... 15 Element = Sum(Array2) - Sum(Array1) I sincerely doubt this is the most optimum algorithm. But it's another way to solve the problem, and is the simplest way to solve it. Hope it helps. If the number of added elements is more than one, this won't work. My answer has the same run time complexity for best, worst, and average case, EDIT After some thinking, ... 14 Arrays are simply laid out as contiguous chunks of memory. An array access such as a[i] is converted to an access to memory location addressOf(a)+i. This the code a[-1] is perfectly understandable, it simply refers to the address one before the start of the array. This may seem crazy, but there are many reasons why this is allowed: it is expensive to check ... 14 I'd post this as a comment on Tobi's answer, but I don't have the reputation yet. As an alternative to calculating the sum of each list (especially if they are large lists or contain very large numbers that might overflow your data type when summed) you can use xor instead. Just calculate the xor-sum (i.e. x[0]^x[1]^x[2]...x[n]) of each list and then xor ... 12 By a simple "adversary argument", you have to check each element (in some way): Suppose you have missed some element $x$ and get an answer "The sum is even": the adversary can modify $x$ (if it's odd, make it even; if it's even, make it odd), which will change the correct result but not your computation. The adversary argument tells that in theory you have ... 11 A note on methodology I thought a bit about this problem, and came to a solution. When I read Saeed Amiri's answer, I realized that what I came up with was a specialized version of the standard longest subsequence finding algorithm for a sequence of length 3. I'm posting the way I came up with the solution, because I think it is an interesting example of ... 11 Some work has been done that matches your description. For instance: Compiler-directed array interleaving for reducing energy in multi-bank memories. by Delaluz, V. Design Automation Conference, 2002. Proceedings of ASP-DAC 2002. 7th Asia and South Pacific and the 15th International Conference on VLSI Design. Proceedings. describes a such an optimization. 11 An oracle is basically a magic black box that does something (e.g. query the entries of an array), usually in constant time. How the oracle does this is abstracted away from (and sometimes an oracle might even do something impossible like solve the halting problem). It's just a way of saying "presume we had these computational abilities, then ...". 11 Skiena didn't say "applicative order", just "applicative". This is sometimes used as meaning something like "purely functional", or a language that evaluates via the application of functions as opposed to the execution of state manipulating commands. Mathematica (at least nowadays) isn't purely functional, but it does seem that it strongly encourages a ... 11 Let's use $+$ to denote the start of a segment and $-$ to denote the end. For each segment, create two pairs, one for each endpoint: Segment1: (-2, +), (3, -) Segment2: (1, +), (5, -) Segment3: (-3, +), (1, -) Sort the $2N$ pairs by their first coordinate (in case of equality, put + before -). You can do this in time $O(N \log N)$ with any reasonable ... 10 Strategy The following linear-time algorithm adopts the strategy of hovering around $0$, by choosing either positive or negative numbers based on the sign of the partial sum. It preprocesses the list of numbers; it computes the permutation of the input on-the-fly, while performing the addition. Algorithm Partition $a_1, \ldots, a_n$ into a two lists, ... 10 This sounds more like the makings of a computational model rather than a programming language, as such, perhaps in the same way that the quantum computation can form the basis of a programming language such as the quantum lambda calculus. Ask yourself: What kinds of computation are you trying to perform? How can these computations be composed? How can the ... 10 If the elements need not be distinct, then you cannot have an $O(\log n)$ time algorithm. Consider the sorted array $[0,0, \dots, 1]$ which has been cyclic shifted $k$ (unknown) times and you need to find where the $1$ appears. This needs $\Omega(n)$ time, as you need to examine at least $n-1$ elements. However, if you assume the elements are distinct, ... 10 Mannila [1] axiomatizes presortedness (with a focus on comparison-based algorithms) as follows (paraphrasing). Let $\Sigma$ a totally ordered set. Then a mapping $m$ from $\Sigma^{\star}$ (the sequences of distinct elements from $\Sigma$) to the naturals is a measure of presortedness if it satisfies below conditions. If $X \in \Sigma^{\star}$ is ... 9 Here is an algorithm which satisfies your conditions: for i in 0,...,n-1: A'[i] = 0 Clearly your conditions are lacking. Correction of a sorting program, like any other program, is a combination of two condition: termination and partial correction. Termination just states that if the preconditions are satisfied then the program always terminates (in ... 9 Let's assume that adding two strings of lengths $a,b$ takes time $a+b$. Consider the following strategy to convert a list of $n$ characters into a list: Read the list in chunks of $k$, convert them to strings, and sum the chunks. Creating each chunk takes time $\Theta(1+2+\cdots+k) = \Theta(k^2)$, and we do this $n/k$ times, for a total of $\Theta(nk)$. ... Only top voted, non community-wiki answers of a minimum length are eligible	2019-05-19 15:37:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6829007267951965, "perplexity": 524.9547921929053}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232254889.43/warc/CC-MAIN-20190519141556-20190519163556-00047.warc.gz"}
https://www.physicsforums.com/threads/is-ln-x-differentiable-at-negative-x-axis.851955/	# Is ln(x) differentiable at negative x-axis Tags: 1. Jan 12, 2016 ### Miraj Kayastha Since lnx is defined for positive x only shouldnt the derivative of lnx be 1/x, where x is positive. My books does not specify that x must be positive, so is lnx differentiable for all x? 2. Jan 12, 2016 ### Staff: Mentor No. If a function isn't defined on some interval, its derivative isn't defined there, either. The real function ln(x) is defined only for x > 0, as you are aware, so the domain for the derivative is x > 0, as well. As it turns out, the function y = 1/x is defined for any $x \ne 0$, but the left-hand branch does not represent the derivative of the natural log function. 3. Jan 13, 2016 ### lightarrow I assume you are referring to real numbers (you use "x" for the ind. variable and you say that lnx is defined for positive x only) because in the complex field it's all another story... -- lightarrow 4. Jan 13, 2016 ### mathman ln(x) can be defined for x < 0, using $x=-xe^{\pi i}$. Therefore $ln(x)=ln(-x)+\pi i$. 5. Jan 14, 2016 ### Svein In fact, $\frac{d}{dx}ln(\lvert x \rvert)=\frac{1}{x}$.	2017-08-16 14:36:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9306164383888245, "perplexity": 1133.1423297346628}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886101966.48/warc/CC-MAIN-20170816125013-20170816145013-00470.warc.gz"}
https://www.jobilize.com/physics/course/5-2-drag-forces-further-applications-of-newton-s-laws-friction-by-open	# 5.2 Drag forces Page 1 / 6 • Express mathematically the drag force. • Discuss the applications of drag force. • Define terminal velocity. • Determine the terminal velocity given mass. Another interesting force in everyday life is the force of drag on an object when it is moving in a fluid (either a gas or a liquid). You feel the drag force when you move your hand through water. You might also feel it if you move your hand during a strong wind. The faster you move your hand, the harder it is to move. You feel a smaller drag force when you tilt your hand so only the side goes through the air—you have decreased the area of your hand that faces the direction of motion. Like friction, the drag force always opposes the motion of an object. Unlike simple friction, the drag force is proportional to some function of the velocity of the object in that fluid. This functionality is complicated and depends upon the shape of the object, its size, its velocity, and the fluid it is in. For most large objects such as bicyclists, cars, and baseballs not moving too slowly, the magnitude of the drag force ${F}_{\text{D}}$ is found to be proportional to the square of the speed of the object. We can write this relationship mathematically as ${F}_{\text{D}}\propto \phantom{\rule{0.15em}{0ex}}{v}^{2}$ . When taking into account other factors, this relationship becomes ${F}_{\text{D}}=\frac{1}{2}C\rho {\text{Av}}^{2}\text{,}$ where $C$ is the drag coefficient, $A$ is the area of the object facing the fluid, and $\rho$ is the density of the fluid. (Recall that density is mass per unit volume.) This equation can also be written in a more generalized fashion as ${F}_{\text{D}}={\text{bv}}^{2}$ , where $b$ is a constant equivalent to $0\text{.5}\mathrm{C\rho A}$ . We have set the exponent for these equations as 2 because, when an object is moving at high velocity through air, the magnitude of the drag force is proportional to the square of the speed. As we shall see in a few pages on fluid dynamics, for small particles moving at low speeds in a fluid, the exponent is equal to 1. ## Drag force Drag force ${F}_{\text{D}}$ is found to be proportional to the square of the speed of the object. Mathematically ${F}_{\text{D}}\propto \phantom{\rule{0.25em}{0ex}}{v}^{2}$ ${F}_{\text{D}}=\frac{1}{2}\mathrm{C\rho }{\text{Av}}^{2}\text{,}$ where $C$ is the drag coefficient, $A$ is the area of the object facing the fluid, and $\rho$ is the density of the fluid. Athletes as well as car designers seek to reduce the drag force to lower their race times. (See [link] ). “Aerodynamic” shaping of an automobile can reduce the drag force and so increase a car’s gas mileage. The value of the drag coefficient, $C$ , is determined empirically, usually with the use of a wind tunnel. (See [link] ). The drag coefficient can depend upon velocity, but we will assume that it is a constant here. [link] lists some typical drag coefficients for a variety of objects. Notice that the drag coefficient is a dimensionless quantity. At highway speeds, over 50% of the power of a car is used to overcome air drag. The most fuel-efficient cruising speed is about 70–80 km/h (about 45–50 mi/h). For this reason, during the 1970s oil crisis in the United States, maximum speeds on highways were set at about 90 km/h (55 mi/h). what is Andromeda what is velocity displacement per unit time Murlidhar the ratec of displacement over time Jamie the rate of displacement over time Jamie the rate of displacement over time Jamie did you need it right now Pathani up to tomorrow Santosh i need a description and derivation of kinetic theory of gas Santosh pls the sum of change in kinetic and potential energy is always what ? Faith i need a description and derivation of kinetic theory of gas Santosh did you need it right now Pathani A few grains of table salt were put in a cup of cold water kept at constant temperature and left undisturbed. eventually all the water tasted salty. this is due to? Aunt Faith,please i am thinking the dissolution here from the word "solution" exposed the grains of salt to be dissolved in the water.Thankyou Junior Junior Aunt Faith,please i am thinking the dissolution here from the word "solution" exposed the grains of salt to be dissolved in the water.Thankyou Junior it is either diffusion or osmosis. just confused Faith due to solvation.... Pathani what is solvation pls Faith water molecule surround the salt molecules . solute solute attraction break in the same manner solvent solvent interaction also break. as a result solute and solvent attraction took place. Pathani okay thanks Faith my pleasure Pathani what is solvation pls Faith water act as a solvent and salt act as solute Pathani okay thanks Faith its ok Pathani due to solvation.... Pathani water molecule surround the salt molecules . solute solute attraction break in the same manner solvent solvent interaction also break. as a result solute and solvent attraction took place. Pathani what is magnetism physical phenomena arising from force caused by magnets is the phenomenon of attracting magnetic substance like iron, cobalt etc. Faith what is heat Heat is a form of energy where molecules move saran topic-- question Salman I know this is unrelated to physics, but how do I get the MCQs and essay to work. they arent clickable. 20cm3 of 1mol/dm3 solution of a monobasic acid HA and 20cm3 of 1mol/dm3 solution of NaOH are mixed in a calorimeter and a temperature rise of 274K is observed. If the heat capacity of the calorimeter is 160J/K, calculate the enthalpy of neutralization of the acid.(SHCw=4.2J/g/K) Formula. (mscs+C)T why is a body moving at a constant speed able to accelerate 20cm3 of 1mol/dm3 solution of a monobasic acid HA and 20cm3 of 1mol/dm3 solution of NaOH are mixed in a calorimeter and a temperature rise of 274K is observed. If the heat capacity of the calorimeter is 160J/K, calculate the enthalpy of neutralization of the acid.(SHCw=4.2J/g/K) Formula. (mscs+C)T Lilian because it changes only direction and the speed is kept constant Justice Why is the sky blue...? It's filtered light from the 2 forms of radiation emitted from the sun. It's mainly filtered UV rays. There's a theory titled Scatter Theory that covers this topic Mike A heating coil of resistance 30π is connected to a 240v supply for 5min to boil a quantity of water in a vessel of heat capacity 200jk. If the initial temperature of water is 20°c and it specific heat capacity is 4200jkgk calculate the mass of water in a vessel A thin equi convex lens is placed on a horizontal plane mirror and a pin held 20 cm vertically above the lens concise in position with its own image the space between the undersurface of d lens and the mirror is filled with water (refractive index =1•33)and then to concise with d image d pin has to Be raised until its distance from d lens is 27cm find d radius of curvature Azummiri what happens when a nuclear bomb and atom bomb bomb explode add the same time near each other A monkey throws a coconut straight upwards from a coconut tree with a velocity of 10 ms-1. The coconut tree is 30 m high. Calculate the maximum height of the coconut from the top of the coconut tree? Can someone answer my question v2 =u2 - 2gh 02 =10x10 - 2x9.8xh h = 100 ÷ 19.6 answer = 30 - h. 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https://encyclopediaofmath.org/index.php?title=Acceptance-rejection_method&oldid=50204	# Acceptance-rejection method (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) Introduced by J. von Neumann [a8], this is the most adaptable method for sampling from complicated distributions (cf. also Sample; Distribution). It works as follows. Let $f ( x )$ be a given probability density (cf. also Density of a probability distribution), and let $h ( x )$ be a function such that $f ( x ) \leq h ( x )$ within the range of $f ( x )$. If the integral of $h ( x )$ over this range is a finite number $\alpha$, then $g ( x ) = h ( x ) / \alpha$ is a probability density function, and the following procedure is valid: 1) Take a random sample $X$ from the distribution with probability density $g ( x ) = h ( x ) / \alpha$. 2) Generate a uniform random deviate $U$ between zero and one. If $U \leq f ( X ) / h ( X )$, accept $X$ as a sample from the distribution $f ( x )$. Otherwise reject $X$ and go back to Step 1. The ease of the method depends on the following properties of the hat function $h ( x )$: A) One has to select a hat function $h ( x )$ from which it is easy to sample. Examples are given below. B) The parameters of the hat function have to be determined in such a way that the area $\alpha$ below $h ( x )$ becomes minimal. Optimal hat functions can be calculated by analytical methods. Assume that the hat function touches $f ( x )$ at two points $L$ (left) and $R$ (right), where $L < R$. Furthermore, suppose that the hat function depends on two parameters, called $m$ and $s$. Thus $$\tag{a1} f ( L ) = \alpha g ( L ; m , s ) , f ( R ) = \alpha g ( R ; m , s ),$$ and $f ( x ) \leq \alpha g ( x ; m , s )$ for all other $x$. Since $L$ and $R$ are local maxima of $f ( x ) / \operatorname { g } ( x ; m , s )$, one has the necessary conditions $$\tag{a2} \frac { f ^ { \prime } ( L ) } { f ( L ) } = \frac { g ^ { \prime } ( L ; m , s ) } { g ( L ; m , s ) }$$ and $$\tag{a3} \frac { f ^ { \prime } ( R ) } { f ( R ) } = \frac { g ^ { \prime } ( R ; m , s ) } { g ( R ; m , s ) }.$$ If $L$ and $R$ are uniquely determined, they should satisfy the sufficient conditions \begin{equation} \frac { f ^ { \prime } ( L ) } { f ( L ) } < \frac { g ^ { \prime } ( L ; m , s ) } { g ( L ; m , s ) } , \frac { f ^ { \prime } ( R ) } { f ( R ) } < \frac { g ^ { \prime } ( R ; m , s ) } { g ( R ; m , s ) }. \end{equation} Otherwise, the first derivative of $\operatorname { ln } ( f ( x ) / g ( x ; m , s ) )$ has to be discussed in detail. Equations (a1), (a2) and (a3) are four equations for the determination of $L$, $R$, $m$, and $s$. Assuming that $L$, $R$ and $m$ can be expressed as functions of $s$, one has to minimize $$\tag{a4} \alpha ( s ) = \frac { f ( L ( s ) ) } { g ( L ( s ) ; m ( s ) , s ) } = \frac { f ( R ( s ) ) } { g ( R ( s ) ; m ( s ) , s ) }.$$ This leads to the necessary conditions \begin{equation} - \frac { d } { d s } \operatorname { ln } \alpha ( s ) = - \frac { d } { d L } \operatorname { ln } \frac { f ( L ) } { g ( L ; m , s ) } \frac { d L } { d s } + \end{equation} \begin{equation} + \frac { d } { d m } \operatorname { ln } g ( L ; m , s ) \frac { d m } { d s } + \frac { d } { d s } \operatorname { ln } g ( L ; m , s ) = 0 , - \frac { d } { d s } \operatorname { ln } \alpha ( s ) = - \frac { d } { d R } \operatorname { ln } \frac { f ( R ) } { g ( R ; m , s ) } \frac { d R } { d s }+ \end{equation} \begin{equation} + \frac { d } { d m } \operatorname { ln } g ( R ; m , s ) \frac { d m } { d s } + \frac { d } { d s } \operatorname { ln } g ( R ; m , s ) = 0. \end{equation} The first expression in each line is zero, by (a2) and (a3). Solving both equations for $dm / \ ds$, and comparing, yields the fundamental relation $$\tag{a5} \frac { d \operatorname { ln } g ( L ; m , s ) } { d m } \frac { d \operatorname { ln } g ( R ; m , s ) } { d s }=$$ \begin{equation} = \frac { d \operatorname { ln } g ( R ; m , s ) } { d m } \frac { d \operatorname { ln } g ( L ; m , s ) } { d s }. \end{equation} (a1), (a2), (a3) and (a5) contain five conditions for finding candidates $L$, $R$, $m$, $s$ and $\alpha$. ## Examples. ### Triangular hat functions. \begin{equation} g ( x ; m , s ) = \left\{ \begin{array} { l l } { \frac { 1 } { s } - \frac { m - x } { s ^ { 2 } } } & { \text { if } m - s \leq x \leq m, } \\ { \frac { 1 } { s } - \frac { x - m } { s ^ { 2 } } } & { \text { if } m \leq x \leq m + s. } \end{array} \right. \end{equation} Samples may be obtained as $X \leftarrow m + s ( U _ { 1 } + U _ { 2 } - 1 )$. The fundamental identity (a5) leads to $s = R - L$. With its help all constants $L$, $R$, $m$, $s$, and $\alpha$ can be determined. ### Double exponential hat functions. The double exponential (or Laplace) distribution is given by \begin{equation} g ( x ; m , s ) = \left\{ \begin{array} { l l } { \frac { 1 } { 2 s } \operatorname { exp } ( \frac { x - m } { s } ) } & { \text { for } x \leq m, } \\ { \frac { 1 } { 2 s } \operatorname { exp } ( \frac { m - x } { s } ) } & { \text { for } x \geq m. } \end{array} \right. \end{equation} Samples are obtained as $X \leftarrow m + T s E$, where $E$ is a standard exponential deviate and $T$ a random sign $\pm$. The fundamental identity (a5) leads to $2 s = R - L$. ### Student-$t$ hat functions. Its probability density function is given as \begin{equation} t _ { n } ( x ) = \frac { c _ { n } } { s } \left( 1 + \frac { ( x - m ) ^ { 2 } } { s ^ { 2 } n } \right) ^ { - ( n + 1 ) / 2 } \end{equation} for $- \infty < x < \infty$, $n \geq 1$, where \begin{equation} c _ { n } = \frac { 1 } { \sqrt { n } B \left( \frac { n } { 2 } , \frac { 1 } { 2 } \right) } = \frac { \Gamma \left( \frac { n + 1 } { 2 } \right) } { \sqrt { n \pi } \Gamma \left( \frac { n } { 2 } \right) }. \end{equation} The $t$-family contains the Cauchy distribution for $n = 1$ and the normal distribution for $n \rightarrow \infty$ as extreme cases. For $n = 2$ and $n = 3$ special sampling methods are available: If $U$ denotes a $( 0,1 )$-uniform random variable, then $X \leftarrow ( U - 1 / 2 ) / ( \sqrt { ( U - U ^ { 2 } ) } / 2 )$ samples from $t_2$ and values from the $t_3$-distribution can be generated efficiently by the ratio-of-uniforms method of A.J. Kinderman and J.F. Monahan [a5], which will be discussed subsequently. The fundamental identity (a5) yields $s ^ { 2 } = ( R - m ) ( m - L )$. As before, $L$, $R$, $s$, $m$, and $\alpha$ can be determined explicitly. The ratio-of-uniforms method was introduced by Kinderman and Monahan for sampling from a density $f ( x )$. First the table mountain-function is constructed: Let $a$, $b$ and $c$ be real numbers and let $k$ be equal to $1$, but $k = 2$ might be another possible choice for some densities. Then \begin{equation} y = \left\{ \begin{array} { l l } { \left( \frac { c } { a - x } \right) ^ { k + 1 } } & { \text { for } x \in ( - \infty , a - c ], } \\ { 1 } & { \text { for } x \in [ a - c , a - c + b ], } \\ { \left( \frac { b - c } { x - a } \right) ^ { k + 1 } } & { \text { for } x \in [ a - c + b , \infty ]. } \end{array} \right. \end{equation} Samples from the area below the table mountain-function are obtained by the transformation \begin{equation} X = \alpha + \frac { b V - c } { U ^ { 1 / k } } , Y = U ^ { 1 / k }. \end{equation} The table mountain-function is taken as a hat function for $f ( x ) / f$, where $f = \operatorname { max } f ( x )$. In this case the fundamental identity leads to $f ( L ) = f ( R )$ for calculating optimal constants. ### Squeeze functions. Step 2 of the acceptance-rejection method can be improved if some lower bound \begin{equation} b ( x ) \leq q ( x ) = \frac { f ( x ) } { h ( x ) } , \text { for all } - \infty < x < \infty, \end{equation} is known and easy to calculate. Step 2 then changes to: 2') Generate a uniform random deviate $U$ between zero and one. If $U \leq b ( X )$, accept $X$ as a sample from the target distribution $f ( x )$. If $U \geq f ( X ) / h ( X )$, reject $X$ and go back to $1$. Otherwise accept $X$. Squeeze functions have been constructed in many procedures. See [a4] for some theory. #### References [a1] J.H. Ahrens, U. Dieter, "Computer methods for sampling from gamma, beta, Poisson and binomial distributions" Computing , 12 (1974) pp. 223–246 [a2] L. Devroye, "Non-uniform random variate generation" , Springer (1986) [a3] U. Dieter, "Optimal acceptance-rejection methods for sampling from various distributions" P.R. Nelson (ed.) , The Frontiers of Statistical Computation, Simulation, & Modeling , Amer. Ser. Math. Management Sci. (1987) [a4] U. Dieter, "Mathematical aspects of various methods for sampling from classical distributions" , Proc. 1989 Winter Simulation Conference (1989) pp. 477–483 [a5] A.J. Kinderman, J.F. Monahan, "Computer generation of random variables using the ratio of uniform deviates" ACM Trans. Math. Software , 3 (1977) pp. 257–260 [a6] D.E. Knuth, "The art of computer programming" , 2: Seminumerical algorithms , Addison-Wesley (1998) (Edition: Third) [a7] J. Leydold, "Automatic sampling with the ratio-of-uniform method" ACM Trans. Math. Software , 26 (2000) pp. 78–88 [a8] J. von Neumann, "Various techniques used in connection with random digits. Monte Carlo methods" Nat. Bureau Standards , 12 (1951) pp. 36–38 [a9] E. Stadlober, "Sampling from Poisson, binomial and hypergeometric distributions: Ratio of uniforms as a simple and fast alternative" Math.–Statist. Sekt. 303 Forschungsgesellschaft Joanneum, Graz, Austria (1989) [a10] E. Stadlober, "The ratio of uniforms approach for generating discrete random varates" J. Comput. Appl. Math. , 31 : 1 (1990) pp. 181–189 How to Cite This Entry: Acceptance-rejection method. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Acceptance-rejection_method&oldid=50204 This article was adapted from an original article by U. Dieter (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article	2020-12-04 08:57:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 5, "x-ck12": 0, "texerror": 0, "math_score": 0.9928451180458069, "perplexity": 374.6054827202076}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141735395.99/warc/CC-MAIN-20201204071014-20201204101014-00597.warc.gz"}
http://dieben.blogspot.com/2017/07/a-letter-to-winston-ewert.html	## Monday, July 17, 2017 ### A letter to Winston Ewert Winston Ewert, Wiliam Dembski, and Robert Marks have written a new book "Introduction to Evolutionary Informatics" Fair to say, I do not like it very much - so I wrote a letter to Winston Ewert, the most accessible of the "humble authors"... Dear Winston, congratulations for publishing your first book! It took me some time to get to read it (though I'm always interested in the output of the Evo Lab). Over the last couple of weeks I've discussed your oeuvre on various blogs. I assume that some of you are aware of the arguments at UncommonDescent and TheSkepticalZone, but as those are not peer reviewed papers, the debates may have been ignored. Fair to say, I'm not a great fan of your new book. I'd like to highlight my problems by looking into two paragraphs which irked me during the first reading: In your section about "Loaded Die and Proportional Betting", you write on page 77: The performance of proportional betting is akin to that of a search algorithm. For proportional betting, you want to extract the maximum amount of money from the game in a single bet. In search, you wish to extract the maximum amount of information in a single query. The mathematics is identical" This is at odds with the previous paragraphs: proportional betting doesn't optimize a single bet, but a sequence of bets - as you have clearly stated before. I'm well aware of Cover's and Thomas's "Elements of Information Theory", but I fail to say how their chapter on "Gambling and Data Compression" is applicable to your idea of a search. I tried to come up with an example, but if I have to search two equally sized subsets $\Omega_1$ and $\Omega_2$, and the target is to be found in $\Omega_1$ with a probability bigger than to be found in $\Omega_2$, proportional betting isn't the optimal way to go! Does proportional betting really extract the maximum of information in a single guess? Then there is this following paragraph on page 173: One’s first inclination is to use an S4S search space populated by different search algorithms such as particle swarm, conjugate gradient descent or Levenberg-Marquardt search. Every search algorithm, in turn, has parameters. Search would not only need to be performed among the algorithms, but within the algorithms over a range of different parameters and initializations. Performing an S4S using this approach looks to be intractable. We note, however, the choice of an algorithm along with its parameters and initialization imposes a probability distribution over the search space. Searching among these probability distributions is tractable and is the model we will use. Our S4S search space is therefore populated by a large number of probability distributions imposed on the search space. Identifying/representing/translating/imposing a search and a probability distribution is central to your theory. It's quite disappointing that you are glossing over it in your new book! While you give generally a quite extensive bibliography, it is surprising that you do not quote any mechanism which translates the algorithm in a probability distribution. Therefore I do not know whether you are thinking about the mechanism as described in "Conservation of Information in Search: Measuring the Cost of Success": this one results in every exhaustive search finding its target. Or are you talking about the "representation" in "A General Theory of Information Cost Incurred by Successful Search": here, all exhaustive searches will do on average at best as a single guess (and yes, I think that this in counter-intuitive). As you are talking about $\Omega$ and not any augmented space, I suppose you have the latter in mind... But if two of your own "representations" result in such a difference between probabilities ($1$ versus $1/\|\Omega\|$), how can you be comfortable with making such a wide-reaching claim like "each search algorithm imposes a probability distribution over the search space" without further corroboration? Could you - for example - translate the damping parameters of the Levenberg-Marquardt search into such a probability distribution? I suppose that any attempt to do so would show a fundamental flaw in your model: the separation between the optimum of the function and the target.... I'd appreciate if you could address my concerns - at UD, TSZ, or my blog. Thanks, Yours Di$\dots$ Eb$\dots$ P.S.: I have to add that I find the bibliographies quite annoying: why can't you add the number of the page if you are citing a book? Sometimes the terms which are accompanied by a footnote cannot be found at all in the given source! It is hard to imagine what the "humble authors" were thinking when they send their interested readers on such a futile search!	2019-02-22 03:49:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6603767275810242, "perplexity": 586.3117878325066}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247513222.88/warc/CC-MAIN-20190222033812-20190222055812-00111.warc.gz"}
https://datascience.stackexchange.com/questions/9509/predicting-when-a-partition-in-oracle-database-will-be-archived/9530	# Predicting when a partition in Oracle database will be archived We have an Oracle database in which the main table "A1" is partitioned by the hour of insertion of the row. A row once inserted, may be updated depending upon data in 6 other tables (B,C,D,E,F,G) in the same database. A partition in table A1 typically gets updated for a few hours after it starts receiving data. After the updates to the partition are finished, the partition gets archived, i.e, it is moved from production to an archive database. The time of creation of a partition and the time of its archival are logged. Now, my question is given the past history of creation and archival times of partitions, how can I predict when a partition created now in A1 will be archived? What kind of model (regression, etc.) is the best to answer this question? How can I set up a model that tracks the dependence of the archival time of a partition on the attributes in other tables (B,C,D,E,F,G)? I'd suggest first visualize your data. The time to archive a partition could be defined as a difference between the partition start (first possible insert) and archiving time (last possible update). Plot the total distribution of this data for some time interval - check if it follows some distribution. Is the average, median or sd meaningful? Make a boxplot for each hour to see inter quartile interval and the outliers. Plot it as a time series - does it have a trend or daily/weekly pattern? In your simple set up (with one predictor and one response) this will give you impression about the accuracy that you can expect from a model. The choose of the model depends on the result of the visualization. 1. all partition archived after $N$ hours; 2. independent model for each hour predicting say $.9$ quantile of some history interval;	2021-05-07 13:06:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.48473644256591797, "perplexity": 1173.1354120770416}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988793.99/warc/CC-MAIN-20210507120655-20210507150655-00605.warc.gz"}
https://codereview.stackexchange.com/questions/222554/collect-maximum-fruits-in-two-baskets-of-at-most-2-types-of-fruits	# Collect maximum fruits in two baskets of at most 2 types of fruits is taken from LeetCode In a row of trees, the i-th tree produces fruit with type tree[i]. You start at any tree of your choice, then repeatedly perform the following steps: 2. Move to the next tree to the right of the current tree. If there is no tree to the right, stop. Note that you do not have any choice after the initial choice of starting tree: you must perform step 1, then step 2, then back to step 1, then step 2, and so on until you stop. You have two baskets, and each basket can carry any quantity of fruit, but you want each basket to only carry one type of fruit each. What is the total amount of fruit you can collect with this procedure? Example 1: Input: [1,2,1] Output: 3 // Explanation: We can collect [1,2,1]. Example 2: Input: [0,1,2,2] Output: 3 // Explanation: We can collect [1,2,2]. // If we started at the first tree, we would only collect [0, 1]. Example 3: Input: [1,2,3,2,2] Output: 4 // Explanation: We can collect [2,3,2,2]. // If we started at the first tree, we would only collect [1, 2]. Example 4: Input: [3,3,3,1,2,1,1,2,3,3,4] Output: 5 // Explanation: We can collect [1,2,1,1,2]. // If we started at the first tree or the eighth tree, we would only collect 4 fruits. Note: 1. 1 <= tree.length <= 40000 2. 0 <= tree[i] < tree.length My solution: has time and space complexity of $$\O(n)\$$. At first I thought it was easy. But then, I got confused with the one of the test cases (i.e. I = [3,3,3,1,2,1,1,2,3,3,4];) and everything inside the else-block is a bit hacky afterwards. Maybe there is a more elegant solution to that. /** * @param {number[]} tree * @return {number} / var totalFruit = function(tree) { const set = new Set(tree); if (set.size <= 2) { return tree.length; } const fruits = new Set(); let i = 0; let j = 0; let max = 0; let count = 0; while (j < tree.length) { if (fruits.size <= 2 && !fruits.has(tree[j])) { } if (fruits.size <= 2) { count++; max = Math.max(max, count); j++; } else { fruits.delete(tree[i]); const lastIndex = tree.slice(i, j - 1).lastIndexOf(tree[i]); i += lastIndex + 1; count-= lastIndex + 1; } } return max; }; let I = [1,2,1]; I = [0,1,2,2]; I = [3,3,3,1,2,1,1,2,3,3,4]; console.log(totalFruit(I)); Update: I made a mistake. This should be the accurate solution: /* * @param {number[]} tree * @return {number} */ var totalFruit = function(tree) { let max = 0, count = 0; for (let i = 0, first = 0, second = -1; i < tree.length; i++) { count++; if (tree[i] === tree[first]) { first = i; } else if (second === -1 \|\| tree[i] === tree[second]) { second = i; } else { max = Math.max(count - 1, max); count = Math.abs(first - second) + 1; first = i - 1; second = i; } } return Math.max(count, max); }; Time complexity $$\O(n)\$$ and space complexity $$\O(1)\$$ • $O(n)$ space seems like an overkill. I see no reason to keep the entire set of trees. You only really care of the two fruits you currently carry. Three indices (two for the last time each fruit was picked, and one for a current position) shall do it in constant space. – vnp Jun 19 at 1:48 Your second solution is good and fast, but you asked about more elegant solution, so I am suggesting mine. Firstly, I solved this task by Python, then convert all logic into Javascript. It is a little slower, than your (20 ms) and use more memory, but I think it more straightforward and understandable. The code only answers do not liked on this site, so I add some comparisons: • Algorithm. Both algorithms are similar but: • Mine: keeps track of every number changing positions. The start1 position changes every time the number was changed, so I always know the position, where the previous number was started. The start2 position changes only when third number occurs, so I just subtract the start2 from the current index and get the needed two number sequence length. • Your: keeps track of last occurrences the first and second numbers, so you miss their start positions, and thus, you should use the count variable for storing the length of the current two number sequence. When the third number appears, you need to calculate the value of the last uninterruptible one number sequence by Math.abs(first - second). Also, you don't know which number was last - first or second, so the Math.abs function is needed. • Mine: uses the iterator - for...of statement and Array.entries(). It relieves us from tree[i] and tree[second] like constructions. • Your: Uses counter and array indexes to access the needed item. The Javascript code: var totalFruit = function(tree) { let n1 = -1; let n2 = -1; let start1 = 0; let start2 = 0; let maxim = 1; # Add extra element in the end of array (which is not occured in array) # to get rid of the second 'Math.max(maxim, k - start2)' call tree.push(-2); for (let [k, num] of tree.entries()) { if (num !== n1) { if (num !== n2) { maxim = Math.max(maxim, k - start2); start2 = start1; } n2 = n1; n1 = num; start1 = k; } } return maxim; } The original Python code: class Solution: def totalFruit(self, tree): n1 = -1 n2 = -1 start1 = 0 start2 = 0 maxim = 1 tree.append(-2) for k, num in enumerate(tree): if num != n1: if num != n2: maxim = max(maxim, k - start2) start2 = start1 n2 = n1 n1 = num start1 = k return maxim	2019-12-08 19:34:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3109146058559418, "perplexity": 4471.993319895522}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540514475.44/warc/CC-MAIN-20191208174645-20191208202645-00204.warc.gz"}
http://mathoverflow.net/revisions/106148/list	2 deleted 1 characters in body; edited title # Chern classes of ideal sheaf of an analytic subetsubset Hello, Here's my question : Let $X$ be a Kähler manifold of dimension $n$, and $Z \subset X$ an analytic subset of codimension $k$. I have read in a paper the following result, a proof of which I cannot find: $$c_k(\mathcal{I}_Z) = (-1)^k(k-1)![Z]$$ The form of the expression suggests using GRR, but I cannot figure out how. Does anyone know a proof for this (or at least an online reference). ? Thanks. 1 # Chern classes of ideal sheaf of an analytic subet Hello, Here's my question : Let $X$ be a Kähler manifold of dimension $n$, and $Z \subset X$ an analytic subset of codimension $k$. I have read in a paper the following result, a proof of which I cannot find: $$c_k(\mathcal{I}_Z) = (-1)^k(k-1)![Z]$$ The form of the expression suggests using GRR, but I cannot figure out how. Does anyone know a proof for this (or at least an online reference). Thanks.	2013-06-18 05:13:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9142331480979919, "perplexity": 97.1412568082478}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368706934574/warc/CC-MAIN-20130516122214-00067-ip-10-60-113-184.ec2.internal.warc.gz"}
https://www.physicsforums.com/threads/why-must-this-expression-for-the-curl-be-wrong.913566/	Why must this expression for the curl be wrong? 1. May 4, 2017 scotty_le_b 1. The problem statement, all variables and given/known data Without explicit calculation, argue why the following expression cannot be correct: $$\nabla \times (\mathbf{c} \times \mathbf{r}) = c_{2}\mathbf{e_{1}}+c_{1}\mathbf{e_{2}}+c_{3}\mathbf{e_{3}}$$ where $\mathbf{c}$ is a constant vector and $\mathbf{r}$ is the position vector. 2. Relevant equations 3. The attempt at a solution So I can do the explicit calculation to see that in fact the curl should be parallel to the vector $\mathbf{c}$ but then I struggle to provide an argument for why this should be so without the calculation. I think that the incorrect solution has flipped the vector $\mathbf{c}$ in the x-y plane but left the z component unchanged. The position vector treats all directions equally so it seems strange that the z-component of $\mathbf{c}$ should be unchanged by this operation. However, I am unable to explain why this solution can't be true. 2. May 4, 2017 BvU Hi, Can you show us in detail ? What does 'parallel' mean to you ? Are you allowed to use / familiar with the triple product expansion ? 3. May 4, 2017 scotty_le_b Hi, So I used the formula $\nabla \times(\mathbf{c}\times\mathbf{r}) = (\nabla \cdot \mathbf{r})\mathbf{c}+(\mathbf{r}\cdot\nabla)\mathbf{c}-(\nabla\cdot\mathbf{c})\mathbf{r}-(\mathbf{c}\cdot\nabla)\mathbf{r}$. Then the terms where $\nabla$ acts on $\mathbf{c}$ will be zero since $\mathbf{c}$ is constant. Also $\nabla \cdot \mathbf{r}=3$ and $(\mathbf{c}\cdot\nabla)\mathbf{r}=\mathbf{c}$ so the whole expression reduces to $3\mathbf{c}-\mathbf{c}=2\mathbf{c}$ which is why I though that the answer should then be parallel to $\mathbf{c}$. However, I think the point of the question was to justify this intuitively without explicitly doing the calculation above. And that is where I'm unsure. Thanks 4. May 4, 2017 BvU I see. 'Parallel' in the sense of 'linearly dependent'. I was under the impresssion you worked out the components of $\vec c \times\vec r$ and then applied the $\vec \nabla \times$ to the result. That, to me, is an explicit calculation. I tried it and I think it yields $2\,\vec c$ as you found. So you are fine. However, with the triple product expansion expression in the link I gave, I managed to confuse myself: the Lagrange formula reads $${\bf a}\times\left ( {\bf b} \times {\bf c} \right ) = {\bf b} \left ( {\bf a} \cdot {\bf c} \right ) - {\bf c} \left ( {\bf a} \cdot {\bf b} \right )$$so that $$\nabla \times(\mathbf{c}\times\mathbf{r}) = \mathbf{c} (\nabla \cdot \mathbf{r}) - \mathbf{r} (\nabla\cdot \mathbf{c}\ ) \ ,$$ only two terms, and yielding $3\bf c$.... Perhaps some math expert can put me right ?	2018-01-21 17:18:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8403396010398865, "perplexity": 246.86986341540106}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084890795.64/warc/CC-MAIN-20180121155718-20180121175718-00694.warc.gz"}
https://curriculum.illustrativemathematics.org/k5/teachers/grade-5/unit-3/lesson-5/lesson.html	# Lesson 5 Multiply a Unit Fraction by a Non-unit Fraction ## Warm-up: Estimation Exploration: Shaded Rectangle (10 minutes) ### Narrative The purpose of this Estimation Exploration is for students to estimate the area of a shaded region. In a previous Estimation Exploration, students looked at a shaded region where the length was represented by a unit fraction and the width could be estimated with a unit fraction. For the diagram presented here, the length is a unit fraction but the width can not be since it is greater than $$\frac{1}{2}$$. This prepares students for the work of this lesson which is to consider products of a unit fraction and a non-unit fraction. ### Launch • Groups of 2 • Display the image. • “What is an estimate that’s too high? Too low? About right?” • 1 minute: quiet think time ### Activity • 1 minute: partner discussion • Record responses. ### Student Facing What is the area of the shaded region? Record an estimate that is: too low about right too high $$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$ $$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$ $$\phantom{\hspace{2.5cm} \\ \hspace{2.5cm}}$$ ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis • “Is the area of the shaded region more or less than $$\frac{1}{4}$$ square unit? How do you know?” (It’s more because $$\frac{1}{2}$$ of $$\frac{1}{2}$$ is $$\frac{1}{4}$$, and more than that is shaded.) ## Activity 1: Write Equations (15 minutes) ### Narrative The purpose of this activity is for students to write expressions for and find the area of shaded regions whose side lengths are a unit fraction and a non-unit fraction. Students look at a series of diagrams with an increasingly large shaded region so they can look for and make use of structure (MP7). It is important to relate the work here to what students have already learned about the product of a unit fraction and a unit fraction and this is the goal of the synthesis. For example, students have seen that $$\frac{1}{5} \times \frac{1}{3} = \frac{1}{15}$$. Since $$\frac{6}{5}$$ is 6 $$\frac{1}{5}$$s, the product $$\frac{6}{5} \times \frac{1}{3}$$ will be 6 $$\frac{1}{15}$$s or $$\frac{6}{15}$$. MLR8 Discussion Supports. Prior to solving the problems, invite students to make sense of the situations and take turns sharing their understanding with their partner. Listen for and clarify any questions about the context. ### Launch • Groups of 2 • Display image A from the student workbook. • “What is a multiplication expression that represents the shaded region?”($$\frac{2}{5} \times \frac{1}{3}$$, $$2 \times \frac{1}{15}$$) • “How does the diagram represent your expression?”(There is $$\frac{2}{5}$$ of the first row shaded and that row is $$\frac{1}{3}$$ of the square. There are 2 shaded pieces and each is $$\frac{1}{15}$$ of the whole square.) ### Activity • 3–5 minutes: independent work time • 3–5 minutes: partner discussion • Monitor for students who: • notice that the product of the numerators represents how many pieces of the square are shaded. • notice that the product of the denominators represents the number of pieces in the whole square. ### Student Facing 1. Write a multiplication expression that represents the shaded region in each diagram. 2. What patterns do you notice in the multiplication expressions? 3. Han wrote this equation to represent the area of the shaded region. Explain how the diagram represents the equation. $$\phantom{2.5cm} \\ \frac {6}{5} \times \frac {1}{3} = \frac {6}{15}$$ ### Student Response For access, consult one of our IM Certified Partners. If students write mathematically correct multiplication expressions that do not represent the product of a unit fraction and a non-unit fraction, write this type of multiplication expression that represents the diagram and ask “How does this expression represent the area of the shaded region in the diagram?” ### Activity Synthesis • Ask previously selected students to share the patterns they noticed in the table. • Display the expression from the last problem: $$\frac {6}{5} \times \frac {1}{3} = \frac {6}{15}$$. • “How does the diagram show $$\frac {1}{3}$$?” (The first row of pieces in a square is $$\frac{1}{3}$$ of the square.) • “How does the diagram show $$\frac {6}{5}$$?” (There are 6 pieces shaded and each one is $$\frac{1}{5}$$ of the row.) • “How does the diagram show $$\frac {6}{5} \times \frac {1}{3}$$?” (There is $$\frac{6}{5}$$ of a row shaded and that row is $$\frac{1}{3}$$ of a square unit.) ## Activity 2: Estimate With Expressions (20 minutes) ### Narrative The purpose of this activity is for students to write multiplication expressions to estimate the area of a shaded region. This builds on the warm-up and the activity launches by asking students to write an expression for the shaded region in the image they considered in the warm-up. This work in this activity combines the skill of estimation with an understanding that the area of a rectangle relates to its length and width. So far, students have only calculated these areas as products of fractions when at least one side length is a unit fraction. Students may write products of two non-unit fractions. While they have not yet learned how to calculate these products and relate them to areas, these answers are valid when the estimates are reasonable and students will learn in the next lesson how to find the value of these expressions. Action and Expression: Internalize Executive Functions. Synthesis: Invite students to plan a strategy, including the tools they will use, for estimating the area of the shaded rectangle. If time allows, invite students to share their plan with a partner before they begin. Supports accessibility for: Conceptual Processing, Language ### Launch • Groups of 2 • Display the image from the warm-up: “What multiplication expression might represent the area of the shaded region?” ($$\frac {2}{3} \times \frac {1}{2}$$, $$\frac {3}{5} \times \frac{1}{2}$$) • “Why do those expressions make sense?” (We can see that the shaded region is a fraction of $$\frac {1}{2}$$. It is more than $$\frac {1}{2} \times \frac {1}{2}$$.) • “We are going to look at more diagrams and write multiplication expressions that might represent the area of the shaded regions in each one.” ### Activity • 3–5 minutes: independent work time • 3–5 minutes: partner discussion • Monitor for students who: • use a unit fraction to help them determine reasonable expressions. • reason about the size of the shaded region before writing an expression. • draw lines to partition the squares. ### Student Facing Write a multiplication expression that might represent the area of the shaded region. Be prepared to explain your reasoning. 1. 2. 3. 4. ### Student Response For access, consult one of our IM Certified Partners. ### Activity Synthesis • Ask previously selected students to share their solutions. • Display the first diagram. • “How does thinking about $$\frac {1}{2}$$ help us estimate the area of the shaded region?” (We know the area is less than $$\frac {1}{2}$$ because only part of $$\frac{1}{2}$$ is shaded.) • “About what fraction of $$\frac {1}{2}$$ is shaded?” (More than $$\frac {1}{2}$$ of $$\frac {1}{2}$$. Maybe $$\frac {4}{5}$$ of $$\frac {1}{2}$$ or $$\frac {3}{4}$$ of $$\frac {1}{2}$$.) ## Lesson Synthesis ### Lesson Synthesis “Today we multiplied a unit fraction by a non-unit fraction.” Display the diagram from Activity 1 showing a shaded region with side lengths $$\frac{6}{5}$$ and $$\frac{1}{3}$$ and the following explanation: “I think the area of the shaded region is $$\frac {6}{30}$$ because $$\frac {6}{10}$$ of $$\frac {1}{3}$$ of the whole thing is shaded.”	2021-10-20 01:48:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5997932553291321, "perplexity": 1308.4995013703438}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585290.83/warc/CC-MAIN-20211019233130-20211020023130-00696.warc.gz"}
https://physics.stackexchange.com/questions/307607/field-operators-for-states-with-spin	# Field operators for states with spin I'm just starting to learn about field operators. The construction I saw was the following: We want an operator $\Psi^\dagger(x)$ which, acting on the vacuum state $\vert 0 \rangle$ gives the state with well defined position $\vert x \rangle$ (in one dimension). Well, we have $$\vert x\rangle=\underset{i}{\sum}\vert\phi_{i}\rangle\langle\phi_{i}\vert x\rangle=\underset{i}{\sum}\phi_{i}^{}(x)a_i^{\dagger}\vert0\rangle$$ (with $\{\vert\phi_i\rangle\}$ an eigenbasis of some observable $A$). So $$\Psi^{\dagger}(x)=\underset{i}{\sum}\phi_{i}^{}(x)a_i^{\dagger}$$ Now I want to do the same for the case when one has to consider the spin. Thus, I want an operator $\Psi_\sigma^\dagger(x)$ which, acting on the vacuum state $\vert 0 \rangle$ gives the state $\vert x \,\sigma \rangle$. In this case: $$\vert x\,\sigma\rangle=\underset{i}{\sum}\vert\Phi_{i}\rangle\langle\Phi_{i}\vert x\,\sigma\rangle=\underset{i}{\sum}\Phi_{i\sigma}^{}(x)\,a_{i}^{\dagger}\vert0\rangle$$ (with $\{\vert\Phi_i\rangle\}$ an eigenbasis of some observable $A$ acting on the state space $\mathcal{E}=\mathcal{E_x}\otimes\mathcal{E_\sigma}$). So $$\Psi_{\sigma}^{\dagger}(x)=\underset{i}{\sum}\Phi_{i\sigma}^{} (x)\,a_{i}^{\dagger}$$ I now try something a little bit different: instead of using the closure relation of the basis $\{\vert\Phi_i\rangle\}$, I use the one of $\{\vert\phi_i\rangle\}$: $$\vert x\,\sigma\rangle=(\underset{i}{\sum}\vert\phi_{i}\rangle\langle\phi_{i}\vert x\rangle)\otimes\vert\sigma\rangle=\underset{i}{\sum}\langle\phi_{i}\vert x\rangle(\vert\phi_{i}\rangle\otimes\vert\sigma\rangle)=\underset{i}{\sum}\phi_{i}^{}(x)a_{i\sigma}^{\dagger}\vert0\rangle$$ (Note: I am not completely sure of my second equal sign). So $$\Psi_{\sigma}^{\dagger}(x)=\underset{i}{\sum}\phi_{i}^{}(x)a_{i\sigma}^{\dagger}$$ I would like you to please confirm this (the last one in particular, which I haven't seen anywhere), since I'm not completely sure of it. • In your first formula, where does the $a^\dagger\lvert 0\rangle$ come from? Should it be $a_i^\dagger\lvert 0\rangle$, and if yes, how do you know such an operator exists? Same question for the $a_{i\sigma}^\dagger$. – ACuriousMind Jan 26 '17 at 0:22 • Yes, you're right, I'll correct it. Well honestly I don't know the answer to that. In my classes, we motivated their existence considering an analogy with a system of N oscillators, but it was hardly rigorous and I did not understand it completely. – Soap Jan 26 '17 at 0:30 • @ACuriousMind (I forgot to identify you in the previous comment). – Soap Jan 26 '17 at 1:07	2019-10-20 22:03:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9639531373977661, "perplexity": 124.13681476877048}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986726836.64/warc/CC-MAIN-20191020210506-20191020234006-00373.warc.gz"}
https://imathworks.com/tex/tex-latex-itemize-environment-within-a-tabular-environment/	# [Tex/LaTex] Itemize environment within a tabular environment itemizetables I've got the following code: \begin{tabular}{r\|l} {Column 1 content}&{\begin{itemize} \item Column 2 content \end{itemize}} \end{tabular} But for some reason, the editor I'm using to code – Latexian – keeps giving me errors at lines 3 and 4 of this code, stating LaTeX error: Something's wrong – perhaps a missing \item Line 59: \item C In the PDF preview, the list displays without bullet points, but the table appears properly formatted. What does this error mean, and how can it be fixed? There are at least two ways of solving your problem. 1. Using a p type column. 2. Using a minipage environment. Which you want to use will depend on the context. But perhaps the later one might be better when you consider vertical alignments. Here are the solutions and corresponding outputs. Using p column. \begin{tabular}{r\|p{0.4\textwidth}} Column 1 content & \begin{itemize} \item Column 2 content 1 \item Column 2 content 2 \end{itemize} \end{tabular} Using a minipage. \begin{tabular}{r\|l} Column 1 content & \begin{minipage}[t]{0.4\textwidth} \begin{itemize} \item Column 2 content 1 \item Column 2 content 2 \end{itemize} \end{minipage} \end{tabular}	2022-09-26 10:32:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8495303988456726, "perplexity": 4332.153080077953}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334855.91/warc/CC-MAIN-20220926082131-20220926112131-00720.warc.gz"}
https://blog.givewell.org/category/malaria-charity/	Intermittent preventive treatment in infants (IPTi) for malaria provides preventive antimalarial medicine to children under 12 months old. It is among the most promising programs we’ve identified in our active pipeline of new interventions. It’s also underutilized, and the population it targets is especially vulnerable to malaria. That implies potential to open up large amounts of room for more funding if IPTi begins to be used more widely—our crude estimate is between $50 million and$200 million globally once it’s scaled—which is something we’re increasingly thinking about as we aim to direct $1 billion in cost-effective funding by 2025. In September 2021, we recommended a small grant to Malaria Consortium and PATH to assess the feasibility and cost-effectiveness of implementing IPTi at national scale in two countries. We’re hopeful that this scoping exercise will answer some of our many open questions about IPTi, and that this intervention continues to look promising as we learn more. Read More # Initial thoughts on malaria vaccine approval The World Health Organization (WHO) recently recommended the widespread use of the malaria vaccine RTS,S/AS01 for children. It provides an additional, effective tool to fight malaria. This is great news! We’ve been following this vaccine’s development for years and, in the last few months, have been speaking with organizations involved in its development and potential wider rollout. Our work on RTS,S (and other malaria vaccines) is ongoing, and we might significantly update our views in the near future. But because we’ve been following its progress, we’re sharing some initial thoughts. Read More # GiveWell donors supported more than direct delivery: AMF and new net research Supporters of the Against Malaria Foundation in recent years may have had even more impact than they expected. The Against Malaria Foundation (AMF) is a GiveWell top charity because we believe its program to distribute insecticide-treated nets prevents people from dying of malaria. AMF-supported net distributions are incredibly cost-effective; we estimate that a$2,000-3,000 donation averts one death.[1] AMF’s work is important in and of itself to fund. Not all AMF donations, however, just support typical net distributions. In recent years, AMF supported research on a new type of insecticide-treated net, the piperonyl butoxide (PBO) net. This research itself (i.e., researcher time, equipment, and administrative costs) was funded by a small number of AMF donors who explicitly agreed to support it. The research was conducted on nets that were contributed by a broad group of AMF donors. Preliminary data suggest that PBO nets are more effective at preventing malaria than standard insecticide-treated nets in areas where mosquitoes have developed insecticide resistance. We think it is likely that AMF sped up the completion of a large-scale, high-quality study of these new nets. We’re thrilled to recommend charities that contribute research in the fields in which they operate. AMF doesn’t just prevent deaths from malaria by distributing nets—it has improved our and others’ understanding of which nets can work best in the future. This post is to share this story with our donors, whose contributions make this work possible. ## Summary In this post, we’ll discuss: • Insecticide resistance and the potential of PBO nets. (More) • AMF’s role in PBO net research. (More) # Our recent visit to Burkina Faso GiveWell staff recently visited Burkina Faso to meet with staff of one of our top charities, Malaria Consortium’s seasonal malaria chemoprevention (SMC) program, and observe its work. Through its SMC program, Malaria Consortium distributes preventative anti-malarial medication at a time of year when it is needed most. As I write below, GiveWell donors have directed more than \$37 million to Malaria Consortium over the last 18 months at our recommendation. We expect that this will provide preventative treatments to 4.8 million children and avert over 16,000 deaths. We’re so appreciative of the support of our community in enabling this tremendous impact. We originally sent a version of the following message to supporters of Malaria Consortium’s SMC program in late August. We received positive feedback on this message and decided to share it more broadly on our blog. We plan to publish more information about the 2019 Burkina Faso site visit in the future. Hello from Burkina Faso! I’m here on a site visit of Malaria Consortium, one of our recommended charities, to see its malaria prevention program in action. This visit helped me relate more deeply to the program by getting to know some of the people who run it and some of the people who benefit from it. I wanted to share my experience with you. # Why we don’t use subnational malaria mortality estimates in our cost-effectiveness models ### Summary We recently completed a small project to determine whether using subnational baseline malaria mortality estimates would make a difference to our estimates of the cost-effectiveness of two of our top charities, the Against Malaria Foundation and Malaria Consortium. We ultimately decided not to include these adjustments because they added complexity to our models and would require frequent updating, while only making a small difference (a 3-4% improvement) to our bottom line. Though this post is on a fairly narrow topic, we believe this example illustrates the principles we use to make decisions about what to include in our cost-effectiveness model.	2022-10-05 09:27:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.18433792889118195, "perplexity": 4165.744006294373}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337595.1/warc/CC-MAIN-20221005073953-20221005103953-00726.warc.gz"}
https://www.tutorialexample.com/understand-tf-reduce_mean-with-examples-for-beginners-tensorflow-tutorial/	# Understand tf.reduce_mean with Examples for Beginners – TensorFlow Tutorial By \| June 27, 2019 tf.reduce_mean() can allow us to compute the mean value of a tensor in tensorflow. This function is widely used in tensorflow applications. However, to use this function correctly, we must concern how this function compute the mean of a tensor and how about the result. ## Key 1. tf.reduce_mean computes the average of a tensor along axis. (1) If we do not set axis value, tf.reduce_mean() will compute the mean of all elements in tensor. For example: We create a tensor with shape [5, 10, 9, 8], then compute the mean without axis. import tensorflow as tf import numpy as np x = tf.Variable(tf.random_uniform([5, 10, 9, 8], -1, 1), name='x') avg_x_4 = tf.reduce_mean(x) init = tf.global_variables_initializer() init_local = tf.local_variables_initializer() with tf.Session() as sess: sess.run([init, init_local]) x0= (sess.run([avg_x_4])) print x0 Will get the value: -0.008919376 (2) if we set the value of axis, tf.reduce_mean() will compute along the axis. For example: We set axis = 0, we will compute the mean like this: import tensorflow as tf import numpy as np x = tf.Variable(tf.random_uniform([5, 10, 9, 8], -1, 1), name='x') avg_x_4 = tf.reduce_mean(x, axis = 0) init = tf.global_variables_initializer() init_local = tf.local_variables_initializer() with tf.Session() as sess: sess.run([init, init_local]) x0 = (sess.run([avg_x_4])) print x.shape The value of shape is: (10, 9, 8) Here we will find an interesting thing, the shape of tensor x is reduced. Key 2. If we do not set keepdims=true, the shape of x tensor will reduce one dimension, which is the axis you set. For examle: We create a [5, 10, 9, 8] tensor and compute its mean along aix = 0, 1, 2, 3, then we print each shape. import tensorflow as tf import numpy as np x = tf.Variable(tf.random_uniform([5, 10, 9, 8], -1, 1), name='x') avg_x_0 = tf.reduce_mean(x, 0) avg_x_1 = tf.reduce_mean(x, 1) avg_x_2 = tf.reduce_mean(x, 2) avg_x_3 = tf.reduce_mean(x, 3) init = tf.global_variables_initializer() init_local = tf.local_variables_initializer() with tf.Session() as sess: sess.run([init, init_local]) x, x0, x1, x2, x3= (sess.run([x, avg_x_0, avg_x_1, avg_x_2, avg_x_3])) print 'x tensor shape = ' print x.shape print 'x tensor mean along axis= 0 ' print x0.shape print 'x tensor mean along axis= 1 ' print x1.shape print 'x tensor mean along axis= 2 ' print x2.shape print 'x tensor mean along axis= 3 ' print x3.shape The output is: x tensor shape = (5, 10, 9, 8) x tensor mean along axis= 0 (10, 9, 8) x tensor mean along axis= 1 (5, 9, 8) x tensor mean along axis= 2 (5, 10, 8) x tensor mean along axis= 3 (5, 10, 9) We will find: If we compute the mean of x tensor along axis = 0, the first dimension of x (5) will be removed. Similarly, if we compute along axis = 1, 2, 3. the dimension of x (10, 9, 8) will be removed. So axis = 0, [5, 10, 9, 8] will be [10, 9, 8] axis = 1, [5, 10, 9, 8] will be [5, 9, 8] axis = 2, [5, 10, 9, 8] will be [5, 10, 8] axis = 3, [5, 10, 9, 8] will be [5, 10, 9]	2022-05-17 05:00:45	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8028379082679749, "perplexity": 6899.688142124206}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662515501.4/warc/CC-MAIN-20220517031843-20220517061843-00778.warc.gz"}
http://mathhelpforum.com/trigonometry/7558-please-help-me-graph-equation.html	Hi everyone, How do I write this equation in the standard form and then graph it. I don't how I could get a standard circle or a hyperbola equation from the following. I am to graph the equation after writing it but I'm sure I can if I can write it in the correct standrd form. Here it is. 9x^2 - y^2 = -16 Thanks 2. Originally Posted by coopsterdude Hi everyone, How do I write this equation in the standard form and then graph it. I don't how I could get a standard circle or a hyperbola equation from the following. I am to graph the equation after writing it but I'm sure I can if I can write it in the correct standrd form. Here it is. 9x^2 - y^2 = -16 Thanks You can solve equation for $y$: $\begin{array}{l} 9x^2 - y^2 = - 16 \\ y^2 = 9x^2 + 16 \\ y = \sqrt {9x^2 + 16} \\ \end{array} $ Then you can graph function: $f(x) = \sqrt {9x^2 + 16}$ Here is graph: 3. Originally Posted by OReilly You can solve equation for $y$: $\begin{array}{l} 9x^2 - y^2 = - 16 \\ y^2 = 9x^2 + 16 \\ y = \sqrt {9x^2 + 16} \\ \end{array} $ Then you can graph function: $f(x) = \sqrt {9x^2 + 16}$ Here is graph: Note though that when you take the square root you have to provide BOTH possibilities: + and -: $y = \pm \sqrt{9x^2+16}$ -Dan 4. Originally Posted by topsquark Note though that when you take the square root you have to provide BOTH possibilities: + and -: $y = \pm \sqrt{9x^2+16}$ -Dan Yes, minus for me...	2016-07-29 06:02:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8933153748512268, "perplexity": 424.3155739944344}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257829972.19/warc/CC-MAIN-20160723071029-00025-ip-10-185-27-174.ec2.internal.warc.gz"}
https://en.m.wikipedia.org/wiki/Equicontinuous	Equicontinuity (Redirected from Equicontinuous) In mathematical analysis, a family of functions is equicontinuous if all the functions are continuous and they have equal variation over a given neighbourhood, in a precise sense described herein. In particular, the concept applies to countable families, and thus sequences of functions. Equicontinuity appears in the formulation of Ascoli's theorem, which states that a subset of C(X), the space of continuous functions on a compact Hausdorff space X, is compact if and only if it is closed, pointwise bounded and equicontinuous. As a corollary, a sequence in C(X) is uniformly convergent if and only if it is equicontinuous and converges pointwise to a function (not necessarily continuous a-priori). In particular, the limit of an equicontinuous pointwise convergent sequence of continuous functions fn on either metric space or locally compact space[1] is continuous. If, in addition, fn are holomorphic, then the limit is also holomorphic. The uniform boundedness principle states that a pointwise bounded family of continuous linear operators between Banach spaces is equicontinuous. Equicontinuity between metric spaces Let X and Y be two metric spaces, and F a family of functions from X to Y. We shall denote by d the respective metrics of these spaces. The family F is equicontinuous at a point x0 ∈ X if for every ε > 0, there exists a δ > 0 such that d(ƒ(x0), ƒ(x)) < ε for all ƒ ∈ F and all x such that d(x0x) < δ. The family is pointwise equicontinuous if it is equicontinuous at each point of X.[2] The family F is uniformly equicontinuous if for every ε > 0, there exists a δ > 0 such that d(ƒ(x1), ƒ(x2)) < ε for all ƒ ∈ F and all x1, x2 ∈ X such that d(x1x2) < δ.[3] For comparison, the statement 'all functions ƒ in F are continuous' means that for every ε > 0, every ƒ ∈ F, and every x0 ∈ X, there exists a δ > 0 such that d(ƒ(x0), ƒ(x)) < ε for all x ∈ X such that d(x0x) < δ. • For continuity, δ may depend on ε, f, and x0. • For uniform continuity, δ may depend on ε and ƒ. • For pointwise equicontinuity, δ may depend on ε and x0. • For uniform equicontinuity, δ may depend only on ε. More generally, when X is a topological space, a set F of functions from X to Y is said to be equicontinuous at x if for every ε > 0, x has a neighborhood Ux such that ${\displaystyle d_{Y}(f(y),f(x))<\epsilon }$ for all yUx and ƒ ∈ F. This definition usually appears in the context of topological vector spaces. When X is compact, a set is uniformly equicontinuous if and only if it is equicontinuous at every point, for essentially the same reason as that uniform continuity and continuity coincide on compact spaces. Used on its own, the term "equiconituity" may refer to either the pointwise or uniform notion, depending on the context. On a compact space, these notions coincide. Some basic properties follow immediately from the definition. Every finite set of continuous functions is equicontinuous. The closure of an equicontinuous set is again equicontinuous. Every member of a uniformly equicontinuous set of functions is uniformly continuous, and every finite set of uniformly continuous functions is uniformly equicontinuous. Examples • A set of functions with a common Lipschitz constant is (uniformly) equicontinuous. In particular, this is the case if the set consists of functions with derivatives bounded by the same constant. • Uniform boundedness principle gives a sufficient condition for a set of continuous linear operators to be equicontinuous. • A family of iterates of an analytic function is equicontinuous on the Fatou set.[4][5] Counterexamples • The sequence of functions fn(x) = arctan(nx), is not equicontinuous because the definition is violated at x0=0 Equicontinuity of maps valued in topological groups Suppose that T is a topological space and Y is an additive topological group (i.e. a group endowed with a topology making its operations continuous). Topological vector spaces are prominent examples of topological groups and every topological group has an associated canonical uniformity. Definition:[6] A family H of maps from T into Y is said to be equicontinuous at tT if for every neighborhood V of 0 in Y, there exists some neighborhood U of t in T such that h(U) ⊆ h(t) + V for every hH. We say that H is equicontinuous if it is equicontinuous at every point of T. Note that if H is equicontinuous at a point then every map in H is continuous at the point. Clearly, every finite set of continuous maps from T into Y is equicontinuous. Equicontinuous linear operators Note that every topological vector space (TVS) is a topological group so the definition of an equicontinuous family of maps given for topological groups transfers to TVSs without change. Characterization of equicontinuous linear operators Notation: If H is a family of maps and U is a set then let H(U) := h(U). Let X and Y be topological vector spaces (TVSs) and H be a family of linear operators from X into Y. Then the following are equivalent: 1. H is equicontinuous; 2. H is equicontinuous at every point of X; 3. H is equicontinuous at some point of X; 4. H is equicontinuous at 0; • i.e. for every neighborhood V of 0 in Y, there exists a neighborhood U of 0 in X such that H(U) ⊆ V (or equivalently, h(U) ⊆ V for every hH). 5. for every neighborhood V of 0 in Y, h−1(V) is a neighborhood of 0 in X; 6. the closure of H in Lσ(X; Y) is equicontinuous; • Lσ(X; Y) denotes L(X; Y) endowed with the topology of point-wise convergence; 7. the balanced hull of H is equicontinuous; while if Y is locally convex then we may add to this list: 1. the convex hull of H is equicontinuous;[7] 2. the convex balanced hull of H is equicontinuous;[8][7] while if X and Y are locally convex then we may add to this list: 1. for every continuous seminorm q on Y, there exists a continuous seminorm p on X such that qhp for all hH;[7] • Here, qhp means that q(h(x)) ≤ p(x) for all xX. while if X is barreled and Y is locally convex then we may add to this list: 1. H is bounded in Lσ(X; Y);[9] 2. H is bounded in L𝛽(X; Y);[9] • L𝛽(X; Y) denotes L(X; Y) endowed with the topology of bounded convergence (i.e. uniform convergence on bounded subsets of X; while if X and Y are Banach spaces then we may add to this list: 1. ${\displaystyle \sup \left\{\left\\|T\right\\|:T\in H\right\}<\infty }$ (that is, H is uniformly bounded in the operator norm). Characterization of equicontinuous linear functionals Let X be a topological vector space (TVS) with continuous dual space X'. For any subset H of X', the following are equivalent:[7] 1. H is equicontinuous; 2. H is equicontinuous at the origin; 3. H is equicontinuous at some point of X; 4. H is contained in the polar of some neighborhood of 0 in X;[8] 5. the (pre)polar of H is a neighborhood of 0 in X; 6. the weak* closure of H in X' is equicontinuous; 7. the balanced hull of H is equicontinuous; 8. the convex hull of H is equicontinuous; 9. the convex balanced hull of H is equicontinuous;[8] while if X is normed then we may add to this list: 1. H is a strongly bounded subset of X';[8] while if X is barreled then we may add to this list: 1. H is relatively compact in the weak* topology on X';[9] 2. H is weak* bounded (i.e. H is σ(X', X)-bounded in X');[9] 3. H is bounded in the topology of bounded convergence (i.e. H is 𝛽(X', X)-bounded in X').[9] Properties of equicontinuous linear maps The uniform boundedness principle (also known as the Banach–Steinhaus theorem) states that a set H of linear maps between Banach spaces is equicontinuous if it is pointwise bounded; i.e., sup { \|\|h(x)\|\| : h ∈ H} < ∞ for each xX. The result can be generalized to a case when Y is locally convex and X is a barreled space.[10] Properties of equicontinuous linear functionals Alaoglu's theorem implies that the weak-* closure of an equicontinuous subset of ${\displaystyle X^{\prime }}$ is weak-* compact; thus that every equicontinuous subset is weak-* relatively compact.[11][7] If X is any locally convex TVS, then the family of all barrels in X and the family of all subsets of X' that are convex, balanced, closed, and bounded in ${\displaystyle X_{\sigma }^{\prime }}$ , correspond to each other by polarity (with respect to X, X#).[12] It follows that a locally convex TVS X is barreled if and only if each bounded subset of ${\displaystyle X_{\sigma }^{\prime }}$ is equicontinuous.[12] Theorem — Suppose that X is a separable TVS. Then every closed equicontinuous subset of ${\displaystyle X_{\sigma }^{\prime }}$ is a compact metrizable space (under the subspace topology). If in addition X is metrizable then ${\displaystyle X_{\sigma }^{\prime }}$ is separable.[12] Equicontinuity and uniform convergence Let X be a compact Hausdorff space, and equip C(X) with the uniform norm, thus making C(X) a Banach space, hence a metric space. Then Arzelà–Ascoli theorem states that a subset of C(X) is compact if and only if it is closed, uniformly bounded and equicontinuous. This is analogous to the Heine–Borel theorem, which states that subsets of Rn are compact if and only if they are closed and bounded. As a corollary, every uniformly bounded equicontinuous sequence in C(X) contains a subsequence that converges uniformly to a continuous function on X. In view of Arzelà–Ascoli theorem, a sequence in C(X) converges uniformly if and only if it is equicontinuous and converges pointwise. The hypothesis of the statement can be weakened a bit: a sequence in C(X) converges uniformly if it is equicontinuous and converges pointwise on a dense subset to some function on X (not assumed continuous). Proof — Suppose fj is an equicontinuous sequence of continuous functions on a dense subset D of X. Let ε > 0 be given. By equicontinuity, for each zD, there exists a neighborhood Uz of z such that ${\displaystyle \|f_{j}(x)-f_{j}(z)\|<\epsilon /3}$ for all j and xUz. By denseness and compactness, we can find a finite subset D′D such that X is the union of Uz over zD′. Since fj converges pointwise on D′, there exists N > 0 such that ${\displaystyle \|f_{j}(z)-f_{k}(z)\|<\epsilon /3}$ whenever zD′ and j, k > N. It follows that ${\displaystyle \sup _{X}\|f_{j}-f_{k}\|<\epsilon }$ for all j, k > N. In fact, if xX, then xUz for some zD′ and so we get: ${\displaystyle \|f_{j}(x)-f_{k}(x)\|\leq \|f_{j}(x)-f_{j}(z)\|+\|f_{j}(z)-f_{k}(z)\|+\|f_{k}(z)-f_{k}(x)\|<\epsilon }$ . Hence, fj is Cauchy in C(X) and thus converges by completeness. This weaker version is typically used to prove Arzelà–Ascoli theorem for separable compact spaces. Another consequence is that the limit of an equicontinuous pointwise convergent sequence of continuous functions on a metric space, or on a locally compact space, is continuous. (See below for an example.) In the above, the hypothesis of compactness of X cannot be relaxed. To see that, consider a compactly supported continuous function g on R with g(0) = 1, and consider the equicontinuous sequence of functions {ƒn} on R defined by ƒn(x) = g(xn). Then, ƒn converges pointwise to 0 but does not converge uniformly to 0. This criterion for uniform convergence is often useful in real and complex analysis. Suppose we are given a sequence of continuous functions that converges pointwise on some open subset G of Rn. As noted above, it actually converges uniformly on a compact subset of G if it is equicontinuous on the compact set. In practice, showing the equicontinuity is often not so difficult. For example, if the sequence consists of differentiable functions or functions with some regularity (e.g., the functions are solutions of a differential equation), then the mean value theorem or some other kinds of estimates can be used to show the sequence is equicontinuous. It then follows that the limit of the sequence is continuous on every compact subset of G; thus, continuous on G. A similar argument can be made when the functions are holomorphic. One can use, for instance, Cauchy's estimate to show the equicontinuity (on a compact subset) and conclude that the limit is holomorphic. Note that the equicontinuity is essential here. For example, ƒn(x) = arctan nx converges to a multiple of the discontinuous sign function. Generalizations Equicontinuity in topological spaces The most general scenario in which equicontinuity can be defined is for topological spaces whereas uniform equicontinuity requires the filter of neighbourhoods of one point to be somehow comparable with the filter of neighbourhood of another point. The latter is most generally done via a uniform structure, giving a uniform space. Appropriate definitions in these cases are as follows: A set A of functions continuous between two topological spaces X and Y is topologically equicontinuous at the points xX and yY if for any open set O about y, there are neighborhoods U of x and V of y such that for every fA, if the intersection of f[U] and V is nonempty, f[U] ⊆ O. Then A is said to be topologically equicontinuous at xX if it is topologically equicontinuous at x and y for each yY. Finally, A is equicontinuous if it is equicontinuous at x for all points xX. A set A of continuous functions between two uniform spaces X and Y is uniformly equicontinuous if for every element W of the uniformity on Y, the set { (u,v) ∈ X × X: for all fA. (f(u),f(v)) ∈ W } is a member of the uniformity on X Introduction to uniform spaces We now briefly describe the basic idea underlying uniformities. The uniformity 𝒱 is a non-empty collection of subsets of Y × Y where, among many other properties, every V ∈ 𝒱, V contains the diagonal of Y (i.e. {(y, y) ∈ Y}). Every element of 𝒱 is called an entourage. Uniformities generalize the idea (taken from metric spaces) of points that are "r-close" (for r > 0), meaning that their distance is < r. To clarify this, suppose that (Y, d) is a metric space (so the diagonal of Y is the set {(y, z) ∈ Y × Y : d(y, z) = 0}) For any r > 0, let Ur = {(y, z) ∈ Y × Y : d(y, z) < r} denote the set of all pairs of points that are r-close. Note that if we were to "forget" that d existed then, for any r > 0, we would still be able to determine whether or not two points of Y are r-close by using only the sets Ur. In this way, the sets Ur encapsulate all the information necessary to define things such as uniform continuity and uniform convergence without needing any metric. Axiomatizing the most basic properties of these sets leads to the definition of a uniformity. Indeed, the sets Ur generate the uniformity that is canonically associated with the metric space (Y, d). The benefit of this generalization is that we may now extend some important definitions that make sense for metric spaces (e.g. completeness) to a broader category of topological spaces. In particular, to topological groups and topological vector spaces. A weaker concept is that of even continuity A set A of continuous functions between two topological spaces X and Y is said to be evenly continuous at xX and yY if given any open set O containing y there are neighborhoods U of x and V of y such that f[U] ⊆ O whenever f(x) ∈ V. It is evenly continuous at x if it is evenly continuous at x and y for every yY, and evenly continuous if it is evenly continuous at x for every xX. Stochastic equicontinuity Stochastic equicontinuity is a version of equicontinuity used in the context of sequences of functions of random variables, and their convergence.[13] Notes 1. ^ More generally, on any compactly generated space; e.g., a first-countable space. 2. ^ Reed & Simon (1980), p. 29; Rudin (1987), p. 245 3. ^ Reed & Simon (1980), p. 29 4. ^ Alan F. Beardon, S. Axler, F.W. Gehring, K.A. Ribet : Iteration of Rational Functions: Complex Analytic Dynamical Systems. Springer, 2000; ISBN 0-387-95151-2, ISBN 978-0-387-95151-5; page 49 5. ^ Joseph H. Silverman : The arithmetic of dynamical systems. Springer, 2007. ISBN 0-387-69903-1, ISBN 978-0-387-69903-5; page 22 6. ^ Narici & Beckenstein 2011, pp. 133–136. 7. Narici & Beckenstein 2011, pp. 225–273. 8. ^ a b c d Trèves 2006, pp. 335–345. 9. Trèves 2006, pp. 346–350. 10. ^ Schaefer 1966, Theorem 4.2. 11. ^ Schaefer 1966, Corollary 4.3. 12. ^ a b c Schaefer & Wolff 1999, pp. 123–128. 13. ^ de Jong, Robert M. (1993). "Stochastic Equicontinuity for Mixing Processes". Asymptotic Theory of Expanding Parameter Space Methods and Data Dependence in Econometrics. Amsterdam. pp. 53–72. ISBN 90-5170-227-2.	2021-09-28 05:37:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 11, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9375976920127869, "perplexity": 478.12391952817785}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780060201.9/warc/CC-MAIN-20210928032425-20210928062425-00374.warc.gz"}
http://math.stackexchange.com/questions/695073/infinite-series-for-partial-sums-of-square-roots	# Infinite series for partial sums of square roots. Can you prove these infinite series for partial sums of square roots? $$\sqrt{1}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$$ $$\sqrt{1}+\sqrt{2}=\sum\limits_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}-\frac{2}{\sqrt{n+2}}\right)$$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}-\frac{3}{\sqrt{n+3}}\right)$$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}=\sum _{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\frac{1}{\sqrt{n+3}}-\frac{4}{\sqrt{n+4}}\right)$$ $$\cdots$$ And is there some easy cancellation that I have missed on the right hand side? Mathematica: Clear[s, i, n, j] s = 1/2; i = 1; j = 0; Sum[1/(n + 0)^s - 1/(n + 1)^s, {n, 1, Infinity}] N[%, 20] Sum[1/(n + 1)^s - 2/(n + 2)^s + 1/(n + 0)^s, {n, 1, Infinity}] N[%, 20] Sum[1/(n + 1)^s + 1/(n + 2)^s - 3/(n + 3)^s + 1/(n + 0)^s, {n, 1, Infinity}] N[%, 20] Sum[1/(n + 1)^s + 1/(n + 2)^s + 1/(n + 3)^s - 4/(n + 4)^s + 1/(n + 0)^s, {n, 1, Infinity}] N[%, 20] N[Accumulate[Sqrt[Range[4]]], 20] - Try writing out the terms. – John Habert Mar 1 at 14:37 The problem is that the sequence of numbers on the right hand side within the sum, is so long that I can't spot the possible cancellations. – Mats Granvik Mar 1 at 14:42 For the first one at least, it should be very obvious after about the first 3 terms or so. The others will have similar style cancellation though a bit different. – John Habert Mar 1 at 14:43 Let's see how we find the first sum which's known as telescoping sum and the other sums are almost the same: the idea is to change the index and then cancel most of the terms $$\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\lim_{N\to\infty}\sum\limits_{n=1}^{N} \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\lim_{N\to\infty}\sum\limits_{n=1}^{N} \frac{1}{\sqrt{n}}-\sum\limits_{n=2}^{N+1}\frac{1}{\sqrt{n}}\\=\lim_{N\to\infty}\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{N+1}}=\frac{1}{\sqrt{1}}$$ - Excellent, Sami! – amWhy Mar 2 at 14:33 \begin{align} \sum_{n=1}^\infty\left(\left(\sum_{k=0}^{m-1}\frac1{\sqrt{n+k}}\right)-\frac{m}{\sqrt{n+m}}\right) &=\lim_{N\to\infty}\sum_{n=1}^N\sum_{k=0}^{m-1}\left(\frac1{\sqrt{n+k}}-\frac1{\sqrt{n+m}}\right)\\ &=\lim_{N\to\infty}\sum_{k=0}^{m-1}\sum_{n=1}^N\left(\frac1{\sqrt{n+k}}-\frac1{\sqrt{n+m}}\right)\\ &=\lim_{N\to\infty}\sum_{k=0}^{m-1}\left(\sum_{n=k+1}^{k+N}\frac1{\sqrt{n}}-\sum_{n=m+1}^{m+N}\frac1{\sqrt{n}}\right)\\ &=\lim_{N\to\infty}\sum_{k=0}^{m-1}\left(\sum_{n=k+1}^m\frac1{\sqrt{n}}-\sum_{n=k+N+1}^{m+N}\frac1{\sqrt{n}}\right)\\ &=\sum_{k=0}^{m-1}\sum_{n=k+1}^m\frac1{\sqrt{n}}\\ &=\sum_{n=1}^m\sum_{k=0}^{n-1}\frac1{\sqrt{n}}\\ &=\sum_{n=1}^m\sqrt{n} \end{align} - Once more, a beautiful and elegant answer ! – Claude Leibovici Mar 1 at 16:03 You have a very elegant demonstration given by Sami Ben Romdhane. If, instead of summing to $\infty$, you sum to $m$, you will get the following formulas for the different sums $$\sqrt{1}-\frac{1}{\sqrt{m+1}}$$ $$\sqrt{1}+\sqrt{2}+H_m^{\left(\frac{1}{2}\right)}+H_{m+1}^{\left(\frac{1}{2}\right)}-2 H_{m+2}^{\left(\frac{1}{2}\right)}$$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}+H_m^{\left(\frac{1}{2}\right)}+H_{m+1}^{\left(\frac{1}{2}\right)}+H_{m+2}^{\left(\frac{1}{2}\right)}-3 H_{m+3}^{\left(\frac{1}{2}\right)}$$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+H_m^{\left(\frac{1}{2}\right)}+H_{m+1}^{\left(\frac{1}{2}\right)}+H_{m+2}^{\left(\frac{1}{2}\right)}+H_{m+3}^{\left(\frac{1}{2}\right)}-4 H_{m+4}^{\left(\frac{1}{2}\right)}$$ Now, push $m$ to $\infty$ and use the properties of the harmonic numbers. - For the first one, \begin{align} \sum_{n=1}^\infty \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right) &= \sum_{n=1}^\infty \frac{1}{\sqrt{n}}-\sum_{n=1}^\infty\frac{1}{\sqrt{n+1}} \\ &= \sum_{n=1}^\infty \frac{1}{\sqrt{n}}-\sum_{n=2}^\infty\frac{1}{\sqrt{n}} \\ &= \frac{1}{\sqrt{1}}\\ &= \sqrt{1}. \end{align} For the second, \begin{align} \sum_{n=1}^\infty \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}-\frac{2}{\sqrt{n+2}}\right) &=\sum_{n=1}^\infty \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right) +\sum_{n=1}^\infty\left(\frac{2}{\sqrt{n+1}}-\frac{2}{\sqrt{n+2}}\right)\\ &= \sqrt{1} + 2\sum_{n=2}^\infty\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\\ &= \sqrt{1} + \frac{2}{\sqrt{2}} \\ &= \sqrt{1} + \sqrt{2}. \end{align}	2014-07-24 08:54:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9999836683273315, "perplexity": 576.9985230774108}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997888216.78/warc/CC-MAIN-20140722025808-00020-ip-10-33-131-23.ec2.internal.warc.gz"}
https://userpage.fu-berlin.de/soga/300/307000_time_series/307081_Working_data_set.html	For the purpose of comparison with STL decomposition we showcase the elimination of a linear trend and seasonal effects on carbon dioxide (CO2) data taken at the Mauna Loa Observatory in Hawaii First, we load the Keeling Curve. The data is provided by Dr. Pieter Tans, NOAA/ESRL and Dr. Ralph Keeling, Scripps Institution of Oceanography. Revisit the section on Data sets used to remind yourself how to download and extract the data set of interest. library(xts) load(url("https://userpage.fu-berlin.de/soga/300/30100_data_sets/KeelingCurve.Rdata")) For the purpose of demonstration we focus on the period from 1990 to 2015. We subset the time series by applying the window() function. co2.1990.2015 <- window(co2, start = "1990-01-01", end = "2015-12-31") plot(co2.1990.2015, type = 'o', cex = 0.5, ylab = expression("CO"[2]" ppm"), main = expression("CO"[2]" Concentration at Mauna Loa Observatory, Hawaii (1990-2015)"), cex.main = 0.85) Finally, we store the time series data set in a .RData file for further processing. save(file = "KeelingCurve_1990-2015.RData", co2.1990.2015)	2019-06-27 09:02:06	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5698865056037903, "perplexity": 4151.615613125824}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628001014.85/warc/CC-MAIN-20190627075525-20190627101525-00052.warc.gz"}

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