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https://imaginary.org/film/mathlapse-monges-theorem | [
"# MathLapse - Monge's Theorem\n\nfilm\n\nMathLapse - Monge's Theorem\n\n## Credits\n\nPavel Boytchev\n_ghost\nMr_Yesterday\n\nThe Monge’s Circle Theorem as a silhouette of a higher dimension\n\nThe Monge’s Circle Theorem states that for any three circles, none of which is completely inside another, the intersection points of the common external tangents are colinear. Let’s try to make this theorem intuitivelly obvious.\n\nImagine this:\n\n• three spheres with equal sizes, but at different distances from us\n• the pairs of spheres are neatly placed in infinitely long tubes\n• the tubes converge into points somewhere on the horizon.\n\nOur intuition says that the horizon is only one. Whatever direction the tubes go, they will always vanish into that same horizon.\n\nNow, forget all about colours and shadings. Look only at the silhouettes. The three spheres become the three circles. The outlines of the three tubes are the external tangents. And the horizon … is the line where all tubes-tangents vanish into points.\n\nCould you find another theorem, which could be interpreted as a silhouette of a higher dimension?"
] | [
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https://legitwriting.com/name-lab-date/ | [
"### Welcome to Legit Writing",
null,
"",
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"Name Lab Date\nID Lab Group\nTA Name Day of the Week: M T W R F\nPartners’ Last Names\nEXPERIMENT 3\nProjectiles\nSAFETY: Be careful not to fire the projectile when anyone is in the line of fire and not\npaying attention. However, do have someone in your group prepared to stop the\nprojectile after it has hit the ground a few times to insure nobody else slips on a\nrolling, plastic ball. Always wear your safety glasses when firing the projectile.\nObjective: 1. To model motion in two dimensions using the concept of vector quantities.\n2. To determine the initial speed and angle of a projectile by measuring the\nhorizontal displacement and the time of flight.\nApparatus: Pasco apparatus, computer, and interface.\nTheory:\nThe velocity and acceleration of an object are both vector quantities. Once the velocity and the\n(constant) acceleration have been reduced to their respective components, an equation of motion\ncan be independently written for each of the x and y directions. This allows for the current\nposition of an object, with an initial position, initial velocity and acceleration, to be calculated at\nsome time, t, later. These general equations are\n2\n2\n1\n0 0\n2\n2\n1\n0 0\n( )\n( )\ny t y v t a t\nx t x v t a t\ny y\nx x\n? ? ?\n? ? ?\nBy definition, an object in projectile motion is subject to acceleration due to gravity alone. Thus\nthere is no acceleration in any other direction. The equations of motion of a projectile are\ntherefore given by (with ay = -g)\nx x v t ? 0\n? 0x\n\n2\n2\n1\ny y0\nv0\nt gt ? ?\ny\n?\nThese can be used to find the initial velocity components (assuming an angle of ?0 counterclockwise\nfrom the +x-axis)\nt\nx x\nv v x\n0\n0 0 0\ncos\n?\n? ? ? gt\nt\ny y\nv v y 2\n0 1\n0 0 0\nsin ?\n?\n? ? ?\nas well as the initial velocity and the launch angle, if t, x, x0, y, and y0 are measured.\n2\n0\n2\n0\n2\n0 x y\nv ? v ? v\nx\ny\nv\nv\nv\nv\n0\n0\n0 0\n0 0\n0\ncos\nsin\ntan ? ?\n?\n?\n?\nQuestion 1:\nAn object has a velocity of 5.0m/s and is travelling 30º above the\n+x-axis. Find the components of velocity.\nQuestion 2:\nAn object has an acceleration of 15m/s2\ndirected 25º clockwise from\nthe +y-axis. Find the components of acceleration.\nQuestion 3:\nWhat assumptions can be made (if any) about the value of the following quantities for an object\nin projectile motion:\na) vx\nb) vy\nc) ax\nd) ay\nRelevance to the Experiment: In this experiment, the time-of-flight, t, is measured by the\ncomputer, and x, x0, y0, and y are measured with a metre rule. Using the equations above, v0 and\n?0 are found.\nPRELAB CHECKPOINT\nMark: __________\nGet your TA’s initials before proceeding onto the next part. ____________\nProcedure:\nPart 1 – Data Collection (First Launch Angle):\n1. Make sure that nothing is plugged into the two USB ports on the front or right side of the\ncomputer.\n2. Turn on the computer and allow it to boot up. Log-on as Guest.\n3. Start the DataStudio program (the icon is on the desktop) and open the DataStudio workbook\nC:\\64-140\\Experiment 3 WB.ds.\n4. Look at the workbook pages by clicking the triangular arrows on each side of the page\nnumber at the bottom of each page. The workbook will show you the apparatus and the\nnames of the parts that you will use.\n5. Connect the two photogate heads to one photogate port, and connect the time of flight pad to\nthe second photogate port.\n6. Plug both photogate ports into the USB links, and plug the USB links into the USB ports.\n7. Clamp the base of the projectile launcher to the edge of the table, pointing it towards an open\nspace.\n8. Set the launcher to fire approximately horizontally, and measure the angle using the plumb\nbob and angle scale. Note this angle.\n? 0 = __________ ± __________\n9. Measure the height above the lab floor of the launch position of the ball. This will be\nindicated on one side of the launcher. Enter this value below as y0 .\n10. Measure the height of the centre of the ball above the ground when it is placed on the timeof-flight\npad. This is most easily done with the pad on the table. Enter this value as y.\n11. Estimate the errors in the height measurements.\ny0 = ?\ny = ?\n12. After checking that nobody is in the way, test fire a ball using the middle range position\non the launcher to see where the ball lands, and place the time-of-flight pad at this\nposition. Test fire again to ensure that the ball lands near the centre of the pad.\n13. Click the Start ( ) button, fire the ball, note exactly where the ball lands on the\npad, and click the Stop ( ) button.\n14. Measure the horizontal distance travelled by the ball and enter this value as x ? x0 in the\ntable below, together with t and v0 which are measured by the computer and shown in the\nworkbook table.\n15. Repeat these measurements three times.\n16. Calculate the average values for x ? x0, t and v0 for the chart and use these averages to fill\nin the section that follows.\n17. To calculate the error values, use the given excel spreadsheet.\nShot # x ? x0 v0 t\n1\n2\n3\n4\nAverage\nStandard\nDeviation\nStandard\nError\ny ? y0 = ? ?0\n? y?y = ____________\nt\nx x\nv x\n0\n0\n?\n?\n=\nx\nv ? 0\n=\ngt\nt\ny y\nv y 2\n0 1\n0\n?\n?\n?\n=\ny\nv ? 0\n=\nDerive the error formulas for v0x and v0y using (1.7) from the Lab Manual Intro document. Verify\nthe error in v0x and v0y using the excel sheet provided.\n2\n0\n2\n0\n2\n0 x y\nv ? v ? v\n= v0 = ____________ 0\n? v\n=\nv0 = __________ ± __________\nx\ny\nv\nv\n0\n0\n0\ntan? ?\n= ? 0 = ??0\n=\n? 0 = __________ ± __________\nQuestion 4: Compare this value of ? 0 with the value you measured setting up the launcher. Do\nthe two values agree within experimental error? Explain any disagreement below.\nCHECKPOINT 1\nMark: __________\nGet your TA’s initials before proceeding onto the next part. ____________\nPart 2 – Data Collection (Second Launch Angle):\nSet the angle of the launcher to approximately 30º. Measure the angle.\n? 0 = __________ ± __________\nRepeat all the measurements of Part 2. Measure the height of the first photosensor above the\nground.\ny0 = ?\nMake four shots using the short range firing position and measure the quantities as before,\nentering the data in the table below. To calculate the error values, use the given excel\nShot # x ? x0 v0 t\n1\n2\n3\n4\nAverage\nStandard\nDeviation\nStandard\nError\ny ? y0 =\n? ?0\n? y?y\n=\nt\nx x\nv x\n0\n0\n?\n?\n=\nx\nv ? 0\n=\nv0x = __________ ± __________\ngt\nt\ny y\nv y 2\n0 1\n0\n?\n?\n?\n=\ny\nv ? 0\n=\nv0y = __________ ± __________\n2\n0\n2\n0\n2\n0 x y\nv ? v ? v\n= v0 = _________\n0\n? v\n=\nv0 = __________ ± __________\nQuestion 5: Compare this value of v0 with the average value of the computer measurements in\nthe chart from Part 1. Do the two values agree within experimental error? Explain any\ndisagreement below.\nx\ny\nv\nv\n0\n0\n0\ntan? ?\n= ? 0 =\n??0\n=\n? 0 = __________ ± __________\nCHECKPOINT 2\nMark: __________\nGet your TA’s initials before proceeding onto the next part. ____________\nPart 3 – Data Collection (Third Launch Angle): Set the angle of the launcher to\napproximately 60º. Measure the angle.\n? 0 = __________ ± __________\nRepeat all the measurements of Part 2. Measure the height of the first photosensor above the\nground.\ny0 = ?\nMake four shots using the short range firing position and measure the quantities as before,\nentering the data in the table below. To calculate the error values, use the given excel\nShot # x ? x0 v0 t\n1\n2\n3\n4\nAverage\nStandard\nDeviation\nStandard\nError\ny ? y0 =\n? ?0\n? y?y\n=\nt\nx x\nv x\n0\n0\n?\n?\n=\nx\nv ? 0\n=\nv0x = __________ ± __________\ngt\nt\ny y\nv y 2\n0 1\n0\n?\n?\n?\n=\ny\nv ? 0\n=\nv0y = __________ ± __________\n2\n0\n2\n0\n2\n0 x y\nv ? v ? v\n= v0 = _________\n0\n? v\n=\nv0 = __________ ± __________\nx\ny\nv\nv\n0\n0\n0\ntan? ?\n= ? 0 =\n??0\n=\n? 0 = __________ ± __________\nCHECKPOINT 3\nMark: __________\nGet your TA’s initials before proceeding onto the next part. ____________\nQuestion 6: A ball is kicked from a rooftop that is 5m above a level soccer field. The ball\ntravels 20m horizontally before it hits the ground. This could be shown by the diagram below.\nIf this entire process takes 5 seconds to occur, indicate on the diagram the position of the ball 1s,\n2s, 3s and 4s after the ball was first kicked. Be sure to justify your answer below.\nCHECKPOINT 4\nMark: __________\nGet your TA’s initials before submitting the report and leaving the lab room. ___________\n\nAre you interested in this answer? Please click on the order button now to have your task completed by professional writers. Your submission will be unique and customized, so that it is totally plagiarism-free."
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https://quant.stackexchange.com/questions/57323/performance-measurement | [
"# Performance measurement\n\nWhen I regress the excess performance of a portfolio on the MKT Factor using daily data. I get a Beta of 0.95 and an alpha of 0.00011 that I annualize *252 = 2.77%\n\nI know that the annualized return of the MKT Factor is 8.5% for the period and the annualized performance of the excess return of the portfolio is 11%. When I add up 2.77% + 0.95*8.5% = 10.85% , I don't get the 11% annnualized performance of the portfolio. Why is that? Is my alpha correctly annualized?\n\nEdit : The return of the MKT is annualized using : (1+Return)^(252/Number of days)-1\n\nWhen Changing for 365 days instead of 252 days I go over the 11% return. Why is that? Alpha is annualized using 365 days and MKT return. Beta stays constant.\n\nI'm guessing you are regressing excess returns $$R_i$$ on asset $$i$$ (so returns $$r_i$$ minus the risk-free rate $$r_f$$). Then, for market excess returns $$R_M=r_M-r_f$$, we have: \\begin{align} R_i &= r_i - r_f = \\alpha_i + \\beta_i R_M + \\epsilon_i \\quad \\text{or} \\\\ r_i &= r_f + \\alpha_i + \\beta_i R_M + \\epsilon_i, \\\\ \\implies \\bar{R}_i &= \\hat\\alpha_i + \\hat\\beta_i \\bar{R}_M. \\end{align} So $$R_M$$ = 8.5%, $$\\hat\\beta$$ = 0.95, and $$\\hat\\alpha_i$$= 2.77%.\n\nThe one possible bit of wiggle room is in the values you have given. These are surely rounded off. If we consider the values that are possible for the numbers you gave, we can get an idea of how much round-off error might change the results.\n\n\\begin{align} \\text{Lower: } \\bar{R}_i &= 0.000105\\cdot252 + 0.945\\cdot 8.45\\% = 10.63\\% \\quad \\text{and} \\\\ \\text{Upper: } \\bar{R}_i &= 0.0001149\\cdot252 + 0.9549\\cdot 8.549\\% = 11.06\\%. \\end{align}\n\nSo, round-off error is a likely culprit.\n\n• By doing so, the return of the portfolio increases, so nothing changes – Circus_beta Aug 13 at 23:10\n• The 11% corresponds to Ri , it already has the rf substracted from it. If the risk free is removed the return of the portfolio increases – Circus_beta Aug 13 at 23:20\n• Ah! Yes, my comment was wrong; I will delete it so as not to mislead. Hmmm... I think this cold be due to round-off error. Will update my answer to show how. – kurtosis Aug 13 at 23:24\n• @Circus_beta Thanks for catching my mistake! Hopefully this explains things. I bet if you dig into a few more digits of your estimate, things will come out closer. – kurtosis Aug 13 at 23:33\n• I will look into the round off error. I doubt it's the error since the decimal value in excel goes up to 9. But thanks ! – Circus_beta Aug 13 at 23:37\n\nAlpha is a risk measurement & is not equal to excess return because of the beta."
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https://www.lessonplanet.com/search?ccss%5B%5D=197085 | [
"### We found145 reviewed resources\n\nVideos (Over 2 Million Educational Videos Available)",
null,
"5:34\nWomen's History Activator: Eleanor Roosevelt",
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"2:54\nPhonics Song 2 (new version)",
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"4:18\nThomas Jefferson - Author of The...\nOther Resource Types ( 145 )\n6 Items in Collection\nLesson Planet\n\n#### The Physics of Fluid Mechanics\n\nFor Students 9th - 12th Standards\nIn this collection, Pascal’s law, Archimedes’ Principle, and Bernoulli’s Principle aid young engineers in understanding the physics of fluid mechanics. Armed with the knowledge of how forces are applied to fluids, scholars design...\n10 Items in Collection\nLesson Planet\n\n#### Simple Machines\n\nFor Students 7th - 9th Standards\nA collection of five lessons and activities introduce young engineers to simple and compound machines. Groups use the engineering design process to design a Rube Goldberg machine to move a circus elephant into a rail car, a pulley system...\n22 Items in Collection\nLesson Planet\n\n#### Up, Up and Away! - Airplanes\n\nFor Students 5th - 7th Standards\nOh, if only Sir Isaac and Bernoulli could see us now. Even Orville and Wilbur would be amazed by what engineers have designed using their ideas. Next generation engineers put these principles to use as they use what they learn about...\nEngageNY\n\n#### EngageNY Algebra I Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs\n\nFor Teachers 9th - 12th Standards\nThe Algebra 1, Module 1 collection is designed to help students develop fluency in writing, interpreting, and solving linear equations The 30 lessons are divided into four topics: An Introduction to Functions , the Structure of...\n17 Items in Collection\nLesson Planet\n\n#### Solve Linear Equations and Inequalities in One Variable\n\nFor Teachers 9th - 12th Standards\nThe one thing you need to know in algebra is how to solve for an unknown variable. Start with the basics and practice solving linear equations and inequalities. Here are a variety of lessons and worksheets to hit your content.\n15 Items in Collection\nLesson Planet\n\n#### Solve Systems of Equations: Common Core High School Algebra\n\nFor Teachers 9th - 12th Standards\nWith many ways to solve a system, there is a method for everyone. This collection has the three methods of substitution, elimination, and graphing while finishing with solving linear/quadratic systems.\n35 Items in Collection\nLesson Planet\n\n#### Algebra: High School Common Core Math\n\nFor Teachers 9th - 12th Standards\nAll Algebra, all the time! Here is a compiled list of resources that address all of the standards regarding Algebra Common Core for high school. Each resource has the main standard it addresses under the notes section of the item.\nLesson Planet\n\n#### Out of Left Field\n\nFor Teachers 9th - 12th Standards\nA baseball trajectory and a parabola seem to make the best pair in real-world quadratic applications. Here is a current baseball resource with questions, discussions, and explorations regarding a quadratic function and home run...\nLesson Planet\n\nFor Teachers 8th - 10th Standards\nOne skill is sure to make young entrepreneurs successful: maximizing profit. Assess learning of relationships between quantities and reasoning using equations. A businessman sells two sizes of boomerangs for different prices, and...\n2 In 2 Collections\nLesson Planet\n\nFor Teachers 10th - 12th Standards\nThrough a variety of physical and theoretical situations, learners are led through the development of some of the deepest concepts in high school mathematics. Complex numbers, the fundamental theorem of algebra and rational exponents...\nLesson Planet\n\n#### Sonar & Echolocation\n\nFor Teachers 9th - 12th Standards\nA well-designed, comprehensive, and attractive slide show supports direct instruction on how sonar and echolocation work. Contained within the slides are links to interactive websites and instructions for using apps on a mobile device to...\n1 In 1 Collection\nLesson Planet\n\n#### Get Connected with Ohm's Law\n\nFor Teachers 5th - 12th Standards\nIdeal for your electricity unit, especially with middle schoolers, this lesson plan gets engineers using multimeters in electrical circuits to explore the relationships among voltage, current, and resistance. Older learners may even plot...\n1 In 1 Collection 3:45\nPD Learning Network\n\n#### Four Ways to Understand the Earth's Age\n\nFor Students 6th - 12th Standards\nCartoon children compare the earth's age to timescales that we understand:a calendar year, the thickness of a book, the human lifespan. This smart film clip is definitely worth adding to your geologic timescale lesson! If you subscribe...\nLesson Planet\n\n#### Finance: Depreciation (Double Declining)\n\nFor Teachers 6th - 12th Standards\nOf particular interest to a group of business and finance pupils, this lesson explores depreciation of automobile values by comparing the double declining balance to the straight line method. Mostly this is done through a slide...\nLesson Planet\n\n#### Sorting Equations of Circles 2\n\nFor Teachers 9th - 12th Standards\nHow much can you tell about a circle from its equation? This detailed lesson plan focuses on connecting equations and graphs of circles. Learners use equations to identify x- and y-intercepts, centers, and radii of circles. They also...\n3:35\nLesson Planet\n\n#### Literal Equation Application - Perimeter of a Rectangle (Example)\n\nFor Students 6th - 8th Standards\nNew ReviewDetermine the dimensions of a rectangle knowing the perimeter. Pupils find a method to determine the length of a rectangle given the perimeter. The video shows how to solve the perimeter equation for the length. Scholars see how to use...\nLesson Planet\n\n#### Money and Finance\n\nFor Teachers 9th - 12th Standards\nMake the connection between money and exponential equations. Pupils continue financial lessons as they learn about compound interest in savings accounts. They extend the investigation of savings by looking at annuities, and then...\nLesson Planet\n\n#### Prealgebra\n\nFor Students 5th - 9th Standards\nPre-algebra—all wrapped up in one place. The eBook contains everything needed to teach a typical Pre-Algebra course. Concepts in the course build upon previously learned concepts, allowing mathematicians to see the connections between...\nLesson Planet\n\n#### Proportional Representation\n\nFor Students 9th - 12th Standards\nSometimes the solution is all a matter of perspective. The short assessment task presents a problem to pupils that requires them to make sense of a diagram. Once learners see two similar triangles, the rest of the solution is solving a...\n2:54\nLesson Planet\n\n#### Write a Linear Relation as a Function (Example)\n\nFor Students 8th - 11th Standards\nSee how linear relations relate to functions. Viewers of a short video learn to write linear relations in functional form. Given a linear equation in standard form, they isolate one of the variables to rewrite it in terms of the other...\n6:09\nLesson Planet\n\n#### Density\n\nFor Students 7th - 12th Standards\nA short video introduces the triangle that illustrates how to find any value given the other two in the density formula. Class members use the triangle to work several problems to find the mass, volume, or density of an object.\n6:02\nLesson Planet\n\n#### Pressure\n\nFor Students 6th - 12th Standards\nDo not let the pressure get to you. After watching a video on calculating values using the pressure formula, pupils work problems on two worksheets. The problems range from finding the pressure, the force, or the area given the other two...\nLesson Planet\n\n#### Compound Measures\n\nFor Students 8th - 11th Standards\nCompounding is dividing units. Pupils practice using compound measures such as units for speed and density to solve problems that range from straightforward speed problems to those requiring conversions. The last few items challenge...\nLesson Planet\n\n#### Changing the Subject of a Formula\n\nFor Students 7th - 9th Standards\nWhat to do with more than one variable? Part of a larger series, the video shows how to solve a formula for another variable. Pupils solve equations with more than one variable for a given variable of interest. Problems include linear,..."
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"data:image/png;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8940078,"math_prob":0.81751704,"size":7010,"snap":"2020-45-2020-50","text_gpt3_token_len":1373,"char_repetition_ratio":0.12946047,"word_repetition_ratio":0.0018501388,"special_character_ratio":0.1853067,"punctuation_ratio":0.12859425,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9591732,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-06T00:23:17Z\",\"WARC-Record-ID\":\"<urn:uuid:0b147509-2904-411c-b0a3-f31258c9f83d>\",\"Content-Length\":\"130987\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f4d7310d-1cdf-4180-9c72-c7feb228fb06>\",\"WARC-Concurrent-To\":\"<urn:uuid:73494cdd-b29a-4733-ab4a-262146f76683>\",\"WARC-IP-Address\":\"18.223.202.35\",\"WARC-Target-URI\":\"https://www.lessonplanet.com/search?ccss%5B%5D=197085\",\"WARC-Payload-Digest\":\"sha1:W25IDQJWPFXI7VKM67NQDPLMXDKHFBHL\",\"WARC-Block-Digest\":\"sha1:PNHEW2LZTCDXA3WLFOZA22BIVJKIZT4N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141753148.92_warc_CC-MAIN-20201206002041-20201206032041-00535.warc.gz\"}"} |
https://www.wallstreetmojo.com/vba-transpose/ | [
"# VBA Transpose",
null,
"Article byJeevan A Y",
null,
"We have seen transpose function in excel worksheet when we paste any data table to the worksheet, what transpose does is that it changes the position of rows and columns i.e. rows become columns and columns becomes rows in a data table, now since it is a worksheet function in VBA we use it with the Application.worksheet method in VBA.\n\n## How to Transpose in VBA?\n\nSwitching rows and columns is one of the data manipulation techniques almost all the users do in excel. The process of converting horizontal data to vertical and vertical data to horizontal is called as “Transpose” in excel. I am sure you must be familiar with Transposing in a regular worksheet. In this article, we will show you how to use the transpose method in VBA coding.\n\nWe can transpose in VBA using two methods.\n\n1. Transpose Using TRANSPOSE Formula.\n2. Transpose Using Paste Special Method.\n\nWhen we are transposing, we are swapping rows to columns and columns to rows. For example, if the data is in a 4 X 3 array, then it becomes a 3 X 4 array.\n\nLet’s see some examples to transpose column to row in VBA.",
null,
"### #1 – VBA Transpose Using TRANSPOSE Formula\n\nLike how we use TRANSPOSE in excel similarly, we can use the TRANSPOSE formula in VBA too. We don’t have a TRANSPOSE formula in VBA, so we need to use it under the Worksheet Function class.\n\nFor example, look at the below data image.",
null,
"We will try to transpose this array of values. Follow the below steps to transpose the data.\n\nStep 1: Start the subprocedure.\n\nCode:\n\n```Sub Transpose_Example1()\n\nEnd Sub```",
null,
"Step 2: First, we need to decide where we are going to transpose the data. In this, I have chosen to transpose from cell D1 to H2. So, enter the VBA code as Range (“D1: H2”).Value =\n\nCode:\n\n```Sub Transpose_Example1()\n\nRange(\"D1:H2\").Value =\n\nEnd Sub```",
null,
"Step 3: Now, in the above-mentioned range, we need the value of range A1 to B5. To arrive at this open “Worksheet Function” class and select the “Transpose” formula.",
null,
"Step 4: In Arg 1, supply the data source range, i.e., Range (“A1: D5”).\n\nCode:\n\n```Sub Transpose_Example1()\n\nRange(\"D1:H2\").Value = WorksheetFunction.Transpose(Range(\"A1:D5\"))\n\nEnd Sub```",
null,
"Ok, we are done with TRANSPOSE formula coding. Now run the code to see the result in D1 to H2 range of cells.",
null,
"As we have seen in the above image, it has converted the range of cells from columns to rows.\n\n### #2 – VBA Transpose Using Paste Special Method\n\nWe can also transpose using Paste Special method. Consider the same data for this example as well.",
null,
"The first thing we need to do to transpose is to copy the data. So write the code as Range(“A1: B5”).Copy\n\nCode:\n\n```Sub Transpose_Example2()\n\nRange(\"A1:B5\").Copy\n\nEnd Sub```",
null,
"The next thing is we need to decide where we are going to paste the data. In this case, I have chosen D1 as the desired destination cell.\n\nCode:\n\n```Sub Transpose_Example2()\n\nRange(\"A1:B5\").Copy\n\nRange(\"D1\").\n\nEnd Sub```",
null,
"Once the desired destination cell is selected, we need to select “Paste Special Method.”",
null,
"With paste special, we can perform all the actions we have with regular paste special methods in a worksheet.\n\nIgnore all the parameters and select the last parameter, i.e., Transpose, and make this as TRUE.\n\nCode:\n\n```Sub Transpose_Example2()\n\nRange(\"A1:B5\").Copy\n\nRange(\"D1\").PasteSpecial Transpose:=True\n\nEnd Sub```",
null,
"This will also transpose the data like the previous method.",
null,
"Like this, we can use the TRANSPOSE formula or Paste Special method to transpose the data to switch rows to columns and columns to rows.\n\n### Things to Remember\n\n• If we are using the TRANSPOSE worksheet function, it is mandatory to calculate a number of rows and columns to transpose the data. If we have 5 rows and 3 columns, then while transposing, it becomes 3 rows and 5 columns.\n• If you want the same formatting while using paste special, then you have to use the Paste Type argument as “xlPasteFormats.”\n\nYou can download this VBA Transpose Excel Template from here – VBA Transpose Excel Template.\n\n### Recommended Articles\n\nThis has been a guide to VBA Transpose. Here we discussed how to transpose columns to row using VBA Code with examples and a downloadable excel template. Below are some useful articles related to VBA –\n\n• 3 Courses\n• 12 Hands-on Projects\n• 43+ Hours"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81823874,"math_prob":0.90710956,"size":4201,"snap":"2021-21-2021-25","text_gpt3_token_len":1014,"char_repetition_ratio":0.16273528,"word_repetition_ratio":0.030428769,"special_character_ratio":0.23470603,"punctuation_ratio":0.12763466,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99774504,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-18T03:50:05Z\",\"WARC-Record-ID\":\"<urn:uuid:c2615403-ccdc-438e-a008-d4efa9163d80>\",\"Content-Length\":\"162456\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:27b03624-1bf6-4d73-b941-4996cb57d4e0>\",\"WARC-Concurrent-To\":\"<urn:uuid:6cd78e88-5edf-4ea4-9253-3ddee262bb39>\",\"WARC-IP-Address\":\"34.211.154.233\",\"WARC-Target-URI\":\"https://www.wallstreetmojo.com/vba-transpose/\",\"WARC-Payload-Digest\":\"sha1:T22Z2MNO3SUJLXB4TCJFFH6EHB5CY4JZ\",\"WARC-Block-Digest\":\"sha1:U5V3XEGRAZCWF7KRDOWYCHBOBPWSUZ2J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989820.78_warc_CC-MAIN-20210518033148-20210518063148-00527.warc.gz\"}"} |
https://community.khronos.org/t/points-rotation/61255 | [
"",
null,
"# Points rotation?\n\nIs there anyway to constant rotate some points?!\n\nThe code:\nstatic void display(void)\n{\nglClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);\n\n``````glColor3d(0.0,0.0,1.0);\nglPushMatrix();\nglTranslatef(0.0,0.0,-5.0);\nglRotatef(angle,1.0,0.0,0.0);\n\nfor(int i=1;i<=36;i++)\n{\nglBegin(GL_POINTS);\nglVertex3f(sin(i),cos(i),0.0);\n}\n\nglPopMatrix();\nangle++;\nglutSwapBuffers();\n``````\n\n}\n\nHi saibot,\n\nIs there anyway to constant rotate some points?!\n\nOf course you can do it.\n\nIf the code you showed here is the real code it can’t give you any result because you have a lot of bugs there.\n\nI’ll try to fix it assuming you wanna rotate 36 points around the origin.\n\n``````\nstatic void display(void)\n{\nconst float PI = 3.1415; // It'll be needed to convert to radians\nglClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);\n\nglColor3d(0.0,0.0,1.0);\nglPushMatrix();\nglTranslatef(0.0,0.0,-5.0);\nglRotatef(angle,1.0,0.0,0.0);\n\nglBegin(GL_POINTS); // You must avoid if possible the inclusion of \"glBegin\" inside a loop\nfor (int i = 1; i <= 36; i++)\n{\n//glVertex3f(sin(i),cos(i),0.0); // Error!!! sin() and cos() expect the argument to be in radians not degrees\ndouble radian = i * PI / 180; // You must declare \"PI\" as a constant\nglVertex3f(sin(radian), cos(radian), 0); // In this case you can use glVertex2f instead. But it's ok like this.\n}\nglEnd(); // It was missing in your code.\n\nglPopMatrix();\nangle++;\nglutSwapBuffers();\n}\n\n``````\n\nThanks very much for the tips!"
] | [
null,
"https://community.khronos.org/uploads/default/original/2X/b/b557c879c88ec7b8104d755db805309010d5e1c0.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8286622,"math_prob":0.9841221,"size":1016,"snap":"2020-34-2020-40","text_gpt3_token_len":305,"char_repetition_ratio":0.100790516,"word_repetition_ratio":0.0,"special_character_ratio":0.32874015,"punctuation_ratio":0.2173913,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9898082,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-22T21:06:26Z\",\"WARC-Record-ID\":\"<urn:uuid:cd8bfb0c-181a-444c-818e-fd654bbe89da>\",\"Content-Length\":\"16604\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ac443711-1d69-4460-88fb-d4080b43027f>\",\"WARC-Concurrent-To\":\"<urn:uuid:f3d31d86-7b9b-46ec-b589-1312be480eed>\",\"WARC-IP-Address\":\"68.183.55.73\",\"WARC-Target-URI\":\"https://community.khronos.org/t/points-rotation/61255\",\"WARC-Payload-Digest\":\"sha1:IW5OKGRRP2UZHBZPUXKBOSPPCA7C4M6M\",\"WARC-Block-Digest\":\"sha1:IFTZKMYMXSCA3UK4F5MPTUC5HQ3JF57R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400206763.24_warc_CC-MAIN-20200922192512-20200922222512-00031.warc.gz\"}"} |
https://arxiv.org/abs/math-ph/0206001 | [
"math-ph\n\n# Title:Generalised quantum anharmonic oscillator using an operator ordering approach\n\nAbstract: We construct a generalised expression for the normal ordering of (a+a^{\\dagger})^{m} for integral values of m and use the result to study the quantum anharmonic oscillator problem in the Heisenberg approach. In particular, we derive generalised expressions for energy eigen values and frequency shifts for the Hamiltonian H=\\frac{x^{2}}{2}+\\frac{\\dot{x}^{2}}{2}+\\frac{\\lambda}{m}x^{m}. We also derive a closed form first order multi scale perturbation theoretic operator solution of this Hamiltonian with a view to generalise some recent results of Bender and Bettencourt .\n Comments: 9 pages, latex 2e, no figure Journal Reference: J. Phys. A 33 (2000) 5607-5613 Subjects: Mathematical Physics (math-ph); Quantum Algebra (math.QA) Cite as: arXiv:math-ph/0206001 (or for this version)\n\n## Submission history\n\nFrom: Anirban Pathak [view email]\n[v1] Sun, 2 Jun 2002 08:27:22 UTC (8 KB)"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.704425,"math_prob":0.7285384,"size":891,"snap":"2019-13-2019-22","text_gpt3_token_len":224,"char_repetition_ratio":0.091319054,"word_repetition_ratio":0.0,"special_character_ratio":0.2345679,"punctuation_ratio":0.084415585,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9733474,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-25T12:26:28Z\",\"WARC-Record-ID\":\"<urn:uuid:14f9735e-7671-4807-92a0-2f13b5543637>\",\"Content-Length\":\"16139\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e855a908-57b9-41b2-bc66-0cfdad72e598>\",\"WARC-Concurrent-To\":\"<urn:uuid:a2c8e9fe-69bb-48e5-97b0-24e48761f35d>\",\"WARC-IP-Address\":\"128.84.21.199\",\"WARC-Target-URI\":\"https://arxiv.org/abs/math-ph/0206001\",\"WARC-Payload-Digest\":\"sha1:ZBVYHY6FLIGNXD2GMITGIGMGRKXVB5G6\",\"WARC-Block-Digest\":\"sha1:Y2XYECJ4YRID32T25HKAHTVTJB3WIVDQ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912203947.59_warc_CC-MAIN-20190325112917-20190325134917-00437.warc.gz\"}"} |
https://www.aqua-calc.com/calculate/mass-molar-concentration/substance/borazole | [
"# Concentration of Borazole\n\n## borazole: convert between mass and molar concentration\n\n### Molar concentration per milliliter\n\n 0 mmol/ml 0.12 µmol/ml 124.22 nmol/ml 124 222.06 pmol/ml\n\n### Molar concentration per deciliter\n\n 0.01 mmol/dl 12.42 µmol/dl 12 422.21 nmol/dl 12 422 205.94 pmol/dl\n\n### Molar concentration per liter\n\n 0.12 mmol/l 124.22 µmol/l 124 222.06 nmol/l 124 222 059.35 pmol/l\n\n### Mass concentration per milliliter\n\n 1 × 10-5 g/ml 0.01 mg/ml 10 µg/ml 10 000 ng/ml 10 000 000 pg/ml\n\n### Mass concentration per deciliter\n\n 0 g/dl 1 mg/dl 1 000 µg/dl 1 000 000 ng/dl 1 000 000 000 pg/dl\n\n### Mass concentration per liter\n\n 0.01 g/l 10 mg/l 10 000 µg/l 10 000 000 ng/l 10 000 000 000 pg/l\n\n### Equivalent molar concentration per milliliter\n\n 0 meq/ml 0.12 µeq/ml 124.22 neq/ml 124 222.06 peq/ml\n\n### Equivalent molar concentration per deciliter\n\n 0.01 meq/dl 12.42 µeq/dl 12 422.21 neq/dl 12 422 205.94 peq/dl\n\n### Equivalent molar concentration per liter\n\n 0.12 meq/l 124.22 µeq/l 124 222.06 neq/l 124 222 059.35 peq/l\n• Borazole weighs 0.902 gram per cubic centimeter or 902 kilogram per cubic meter, i.e. density of borazole is equal to 902 kg/m³; at 25°C (77°F or 298.15K) at standard atmospheric pressure. In Imperial or US customary measurement system, the density is equal to 56.3 pound per cubic foot [lb/ft³], or 0.521 ounce per cubic inch [oz/inch³] .\n• Melting Point (MP), Borazole changes its state from solid to liquid at -58.2°C (-72.76°F or 214.95K)\n• Boiling Point (BP), Borazole changes its state from liquid to gas at 50.6°C (123.08°F or 323.75K)\n• Also known as: Borazane; Borazine; Inorganic benzene; s-Triazaborane; Triborine triamine; Triboron nitride.\n• Molecular formula: B3H6N3\n\nElements: Boron (B), Hydrogen (H), Nitrogen (N)\n\nMolecular weight: 80.501 g/mol\n\nMolar volume: 89.247 cm³/mol\n\nCAS Registry Number (CAS RN): 6569-51-3\n\n• Bookmarks: [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]\n• The units of amount of substance (e.g. mole) per milliliter, liter and deciliter are SI units of measurements of molar concentrations.\n• The units of molar concentration per deciliter:\n• millimole per deciliter [mm/dl], micromole per deciliter [µm/dl], nanomole per deciliter [nm/dl] and picomole per deciliter [pm/dl].\n• The units of molar concentration per milliliter:\n• millimole per milliliter [mm/ml], micromole per milliliter [µm/ml], nanomole per milliliter [nm/ml] and picomole per milliliter [pm/ml].\n• The units of molar concentration per liter:\n• millimole per liter [mm/l], micromole per liter [µm/l], nanomole per liter [nm/l] and picomole per liter [pm/l].\n• The units of mass per milliliter, liter and deciliter are non-SI units of measurements of mass concentrations still used in many countries.\n• The units of mass concentration per deciliter:\n• gram per deciliter [g/dl], milligram per deciliter [mg/dl], microgram per deciliter [µg/dl], nanogram per deciliter [ng/dl] and picogram per deciliter [pg/dl].\n• The units of mass concentration per milliliter:\n• gram per milliliter [g/ml], milligram per milliliter [mg/ml], microgram per milliliter [µg/ml], nanogram per milliliter [ng/ml] and picogram per milliliter [pg/ml].\n• The units of mass concentration per liter:\n• gram per liter [g/l], milligram per liter [mg/l], microgram per liter [µg/l], nanogram per liter [ng/l] and picogram per liter [pg/l].\n• The equivalent per milliliter, liter and deciliter are obsolete, non-SI units of measurements of molar concentrations still used in many countries. An equivalent is the number of moles of an ion in a solution, multiplied by the valence of that ion.\n• The units of equivalent concentration per deciliter:\n• milliequivalent per deciliter [meq/dl], microequivalent per deciliter [µeq/dl], nanoequivalent per deciliter [neq/dl] and picoequivalent per deciliter [peq/dl].\n• The units of equivalent concentration per milliliter:\n• milliequivalent per milliliter [meq/ml], microequivalent per milliliter [µeq/ml], nanoequivalent per milliliter [neq/ml] and picoequivalent per milliliter [peq/ml].\n• The units of equivalent concentration per liter:\n• milliequivalent per liter [meq/l], microequivalent per liter [µeq/l], nanoequivalent per liter [neq/l] and picoequivalent per liter [peq/l].\n\n#### Foods, Nutrients and Calories\n\nTRULY GOOD FOODS, BANANA SPLIT, UPC: 094184599375 contain(s) 467 calories per 100 grams or ≈3.527 ounces [ price ]\n\n#### Gravels, Substances and Oils\n\nCaribSea, Marine, Aragonite, Aragamax Sugar-Sized Sand weighs 1 537.8 kg/m³ (96.00172 lb/ft³) with specific gravity of 1.5378 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]\n\nPeat Reed-sedge, bulk weighs 588 kg/m³ (36.70764 lb/ft³) [ weight to volume | volume to weight | price | density ]\n\nVolume to weightweight to volume and cost conversions for Refrigerant R-422A, liquid (R422A) with temperature in the range of -40°C (-40°F) to 60°C (140°F)\n\n#### Weights and Measurements\n\nA dyne per are is a unit of pressure where a force of one dyne (dyn) is applied to an area of one are.\n\nEnergy is the ability to do work, and it comes in different forms: heat (thermal), light (radiant), motion (kinetic), electrical, chemical, nuclear and gravitational.\n\noz t/metric tbsp to lb/tsp conversion table, oz t/metric tbsp to lb/tsp unit converter or convert between all units of density measurement.\n\n#### Calculators\n\nCalculate volume of a dodecahedron and its surface area"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8053359,"math_prob":0.9765844,"size":1307,"snap":"2020-45-2020-50","text_gpt3_token_len":351,"char_repetition_ratio":0.10053722,"word_repetition_ratio":0.06635071,"special_character_ratio":0.27697015,"punctuation_ratio":0.12109375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9915333,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-27T12:13:42Z\",\"WARC-Record-ID\":\"<urn:uuid:13837867-8065-4a2c-9b07-39fca12c4729>\",\"Content-Length\":\"33775\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9ad69154-6942-42f9-8353-c46eb15af33e>\",\"WARC-Concurrent-To\":\"<urn:uuid:8dff50bf-f15f-4091-aede-1c72c64ff448>\",\"WARC-IP-Address\":\"69.46.0.75\",\"WARC-Target-URI\":\"https://www.aqua-calc.com/calculate/mass-molar-concentration/substance/borazole\",\"WARC-Payload-Digest\":\"sha1:PHFYRSC57YYPPOQNRUJKL625XF2SRFTR\",\"WARC-Block-Digest\":\"sha1:4AVOE6APFXN637OW6EYGVSK4V2JAKUKA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141191692.20_warc_CC-MAIN-20201127103102-20201127133102-00144.warc.gz\"}"} |
https://zbmath.org/?q=rf%3A5570809 | [
"## Odd, spoof perfect factorizations.(English)Zbl 1484.11232\n\nLet $$n$$ be an integer. An expression of the form $$n = \\prod_{i=1}^k x_i^{a_i},$$ where $$x_i$$ are integers and $$a_i$$ are positive integers, is called a factorization of $$n$$, and each $$x_i$$ a base of the factorization. A factorization is odd, when $$n$$ is odd; otherwise it is even. We define the following function to be evaluated on the collection of ordered pairs: $\\tilde{\\sigma}(\\{(x_i,a_i): 1 \\leq i \\leq k\\}) = \\prod_{i=1}^k \\left(\\sum_{j=1}^{a_i} x_i^j \\right).$\nA factorization as above is called spoof perfect, if $\\tilde{\\sigma}(\\{(x_i,a_i): 1 \\leq i \\leq k\\}) = 2n.$ A spoof perfect factorization $$\\prod_{ i=1}^k x_i^{a_i}$$ is said to be primitive, if for each proper subset $$S$$ of $$\\{1, 2, \\ldots, k\\}$$, the factorization $$\\prod_{i\\in S} x_i^{a_i}$$ is not spoof perfect. Furthermore, a spoof perfect factorization with a single base is called trivial.\nIn this paper, the spoof perfect factorizations are studied. More precisely, the trivial spoof perfect factorizations are characterized, and all nontrivial, odd, primitive spoof perfect factorizations with fewer than seven bases are computed. Moreover, it is proved that for each positive integer $$k$$, there are finitely many nontrivial, odd, primitive spoof perfect factorizations with $$k$$ bases. Finally, some interesting open questions are stated.\n\n### MSC:\n\n 11Y50 Computer solution of Diophantine equations 11D72 Diophantine equations in many variables 11A25 Arithmetic functions; related numbers; inversion formulas\nFull Text:\n\n### References:\n\n Banks, William D.; Güloğlu, Ahmet M.; Nevans, C. Wesley; Saidak, Filip, Descartes numbers, (Anatomy of Integers. Anatomy of Integers, CRM Proc. Lecture Notes, vol. 46 (2008), Amer. Math. Soc.: Amer. Math. Soc. Providence, RI), 167-173, MR 2437973 · Zbl 1186.11004 Dittmer, Samuel J., Spoof odd perfect numbers, Math. Comput., 83:289, 2575-2582 (2014), MR 3223347 · Zbl 1370.11005 Nagell, Trygve, Introduction to Number Theory (1964), Chelsea Publishing Co.: Chelsea Publishing Co. New York, MR 0174513 · Zbl 0042.26702 Nielsen, Pace P., Odd perfect numbers, Diophantine equations, and upper bounds, Math. Comput., 84:295, 2549-2567 (2015), MR 3356038 · Zbl 1325.11009 Voight, John, On the nonexistence of odd perfect numbers, (MASS Selecta (2003), Amer. Math. Soc.: Amer. Math. Soc. Providence, RI), 293-300, MR 2027187 · Zbl 1083.11008\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7461894,"math_prob":0.99353725,"size":3361,"snap":"2022-40-2023-06","text_gpt3_token_len":976,"char_repetition_ratio":0.13285671,"word_repetition_ratio":0.041666668,"special_character_ratio":0.31567985,"punctuation_ratio":0.24599709,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.998992,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-29T18:12:03Z\",\"WARC-Record-ID\":\"<urn:uuid:dfa4d72e-b272-4f71-b58d-1d54054e677c>\",\"Content-Length\":\"54904\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ad11d911-6a57-44f8-9744-9a37d3984133>\",\"WARC-Concurrent-To\":\"<urn:uuid:b58d71c8-4e75-4fd7-b616-ffc685622d53>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=rf%3A5570809\",\"WARC-Payload-Digest\":\"sha1:R6WTY3CTKK7XN2ZHWGHMYAZMGKATTBUY\",\"WARC-Block-Digest\":\"sha1:BKGJEZT7KH54SQDQ43FQQNFGNIJ3LB7O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335362.18_warc_CC-MAIN-20220929163117-20220929193117-00571.warc.gz\"}"} |
https://math.stackexchange.com/questions/877370/question-about-writing-proofs-for-limit | [
"# Question about writing proofs for limit\n\nI intuitively understand proof with limits, but I'm not sure on how to write a formal proof for this example.\n\nFor each $n \\in \\mathbb{N}$, let $a_n$, $b_n$ be real numbers. Also, let $a_{\\infty}$, $b_{\\infty}$ be real numbers. Suppose that $\\lim_{n\\to \\infty} a_n = a_{\\infty}$ and $\\lim_{n\\to \\infty} b_{n} = b_{\\infty}$.\n\nProve that $\\lim_{n\\to \\infty} (a_n b_n) = a_\\infty b_\\infty$.\n\n• Do you mean $\\displaystyle\\lim_{n \\to \\infty}(a_nb_n) = a_{\\infty}b_{\\infty}$ or $\\displaystyle\\lim_{n \\to \\infty}(a_n+b_n) = a_{\\infty}+b_{\\infty}$? – JimmyK4542 Jul 24 '14 at 22:51\n• I meant the former one. I changed it. – user164179 Jul 24 '14 at 22:54\n• Scroll down to \"Proof of the Product Rule for Limits\" en.wikibooks.org/wiki/Calculus/Proofs_of_Some_Basic_Limit_Rules – JimmyK4542 Jul 24 '14 at 22:56\n• $a_nb_n-a_{\\infty}b_{\\infty}=a_n(b_n-b_{\\infty})+(a_n-a_{\\infty})b_{\\infty}$ – Hamou Jul 24 '14 at 22:56\n\n## 1 Answer\n\nYou want to show that $\\lim_{n \\to \\infty} a_nb_n = ab$, where $\\lim_{n \\to \\infty} a_n = a$ and similarly $\\lim_{n \\to \\infty} b_n = b$. By the Triangle Inequality, we have that\n\n\\begin{align} |a_nb_n - ab| &= |(a_nb_n - a_nb) + (a_nb - ab)| \\\\ &\\leq |a_n(b_n - b)| + |b(a_n - a)| \\\\ &= |a_n||b_n - b| + |b||a_n - a|. \\end{align}\n\nSince $\\lim_{n \\to \\infty} a_n = a$, given $\\epsilon > 0$, there exists $M_1 > 0$ such that $|a_n| < M_1$ for all $n \\in \\mathbb{N}$. Set $M = \\sup\\{M_1,b\\}$. Then we have\n\n$$|a_nb_n - ab| \\leq M|b_n - b| + M|a_n - a|.$$\n\nNow from the convergence of the sequences $(a_n)$ and $(b_n)$, we have that if $\\epsilon > 0$ is given, there exist $K_1,K_2 \\in \\mathbb{N}$ such that if $n \\geq K_1$, $|a_n - a| < \\epsilon/2M$ and analogously, if $n \\geq K_2$ then $|b_n - b| < \\epsilon/2M$. Set $K = \\sup\\{K_1,K_2\\}$.\n\nWhat can you conclude on the quantity $|a_nb_n - ab|$ if $n \\geq K$?"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5071267,"math_prob":0.9999666,"size":904,"snap":"2019-35-2019-39","text_gpt3_token_len":388,"char_repetition_ratio":0.12888889,"word_repetition_ratio":0.01183432,"special_character_ratio":0.47013274,"punctuation_ratio":0.10555556,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.000006,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-24T20:24:28Z\",\"WARC-Record-ID\":\"<urn:uuid:9772e72a-7e7c-4129-8aa6-7a1ee86ad3e0>\",\"Content-Length\":\"134670\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3dc92d75-1b19-41ea-bed5-654ba5d17881>\",\"WARC-Concurrent-To\":\"<urn:uuid:51f50ffe-9c42-496c-9d1c-55a139f975a2>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/877370/question-about-writing-proofs-for-limit\",\"WARC-Payload-Digest\":\"sha1:XI6DXOEOAW4XPYMQNHSQJWZB4WFGDJMG\",\"WARC-Block-Digest\":\"sha1:26DPNXQIBJUQYWD7ESMDU3E3IAGJ6J73\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027321696.96_warc_CC-MAIN-20190824194521-20190824220521-00479.warc.gz\"}"} |
https://www.jmp.com/support/help/Factor_Analysis_Platform_Overview.shtml | [
"Consider a situation where you have ten observed variables, X1, X2, …, X10. Suppose that you want to model these ten variables in terms of two latent factors, F1 and F2. For convenience, it is assumed that the factors are uncorrelated and that each has mean zero and variance one. The model that you want to derive is of the form:\nIt follows that",
null,
". The portion of the variance of Xi that is attributable to the factors, the common variance or communality, is",
null,
". The remaining variance,",
null,
", is the specific variance, and is considered to be unique to Xi."
] | [
null,
"https://www.jmp.com/support/help/images/CR_04_Factor_Analysis_2.png",
null,
"https://www.jmp.com/support/help/images/CR_04_Factor_Analysis_3.png",
null,
"https://www.jmp.com/support/help/images/CR_04_Factor_Analysis_4.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9062528,"math_prob":0.98037034,"size":2105,"snap":"2020-10-2020-16","text_gpt3_token_len":392,"char_repetition_ratio":0.15706806,"word_repetition_ratio":0.012084592,"special_character_ratio":0.18194774,"punctuation_ratio":0.10054348,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9865151,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-28T09:36:07Z\",\"WARC-Record-ID\":\"<urn:uuid:df228074-be9d-42f3-bc89-e0cc1539e019>\",\"Content-Length\":\"12576\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bde8c0bd-d6f8-4677-b439-3c9c89ad1dae>\",\"WARC-Concurrent-To\":\"<urn:uuid:f19f3c30-30a9-4c7f-a22f-6973781e03a7>\",\"WARC-IP-Address\":\"23.13.155.135\",\"WARC-Target-URI\":\"https://www.jmp.com/support/help/Factor_Analysis_Platform_Overview.shtml\",\"WARC-Payload-Digest\":\"sha1:X32S7XWMLOBA2XLSXXEIMW3Y7KNLQ3ES\",\"WARC-Block-Digest\":\"sha1:63EU76JLYGSKLTIP3FPNVDQJEMMOK6QH\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370490497.6_warc_CC-MAIN-20200328074047-20200328104047-00458.warc.gz\"}"} |
https://kurzy.kpi.fei.tuke.sk/c-reference/en/cpp/numeric/math/cos.html | [
"std::cos\n\n< cpp | numeric | math\n\nC++\n Language Standard library headers Concepts Utilities library Strings library Containers library Algorithms library Iterators library Numerics library Input/output library Localizations library Regular expressions library (C++11) Atomic operations library (C++11) Thread support library (C++11)\n\nNumerics library\n Common mathematical functions Floating-point environment Complex numbers Numeric arrays Pseudo-random number generation Compile-time rational arithmetic (C++11) Generic numeric operations iota (C++11) accumulate inner_product adjacent_difference partial_sum\n\nCommon mathematical functions\nFunctions\nBasic operations\n abs(int)labsllabsimaxabs (C++11) abs(float)fabs divldivlldivimaxdiv (C++11) fmod\n remainder (C++11) remquo (C++11) fma (C++11) fmax (C++11) fmin (C++11) fdim (C++11) nannanfnanl (C++11)(C++11)(C++11)\nExponential functions\n exp exp2 (C++11) expm1 (C++11)\n log log10 log1p (C++11) log2 (C++11)\nPower functions\n sqrt cbrt (C++11)\n hypot (C++11) pow\nTrigonometric and hyperbolic functions\n sin cos tan asin acos atan atan2\n sinh cosh tanh asinh (C++11) acosh (C++11) atanh (C++11)\nError and gamma functions\n erf (C++11) erfc (C++11)\n lgamma (C++11) tgamma (C++11)\nNearest integer floating point operations\n ceil floor roundlroundllround (C++11)(C++11)(C++11)\n trunc (C++11) nearbyint (C++11) rintlrintllrint (C++11)(C++11)(C++11)\nFloating point manipulation functions\n ldexp scalbnscalbln (C++11)(C++11) ilogb (C++11) logb (C++11)\n frexp modf nextafternexttoward (C++11)(C++11) copysign (C++11)\nClassification/Comparison\n fpclassify (C++11) isfinite (C++11) isinf (C++11) isnan (C++11) isnormal (C++11) signbit (C++11)\n isgreater (C++11) isgreaterequal (C++11) isless (C++11) islessequal (C++11) islessgreater (C++11) isunordered (C++11)\nMacro constants\n HUGE_VALFHUGE_VALHUGE_VALL (C++11)(C++11) INFINITY (C++11) NAN (C++11)\n FP_NORMALFP_SUBNORMALFP_ZEROFP_INFINITEFP_NAN (C++11)(C++11)(C++11)(C++11)(C++11)\n\n Defined in header float cos( float arg ); (1) double cos( double arg ); (2) long double cos( long double arg ); (3) double cos( Integral arg ); (4) (since C++11)\n\nComputes the cosine of arg (measured in radians).\n\n4) A set of overloads or a function template accepting an argument of any integral type. Equivalent to 2) (the argument is cast to double).\n\nContents\n\nParameters\n\n arg - value representing angle in radians, of a floating-point or Integral type\n\nReturn value\n\nIf no errors occur, the cosine of arg (cos(arg)) in the range [-1 ; +1], is returned.\n\n The result may have little or no significance if the magnitude of arg is large (until C++11)\n\nIf a domain error occurs, an implementation-defined value is returned (NaN where supported)\n\nIf a range error occurs due to underflow, the correct result (after rounding) is returned.\n\nError handling\n\nErrors are reported as specified in math_errhandling\n\nIf the implementation supports IEEE floating-point arithmetic (IEC 60559),\n\n• if the argument is ±0, the result is 1.0\n• if the argument is ±∞, NaN is returned and FE_INVALID is raised\n• if the argument is NaN, NaN is returned"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.55785275,"math_prob":0.9862241,"size":2428,"snap":"2019-43-2019-47","text_gpt3_token_len":718,"char_repetition_ratio":0.11757426,"word_repetition_ratio":0.0051282053,"special_character_ratio":0.3360791,"punctuation_ratio":0.13763441,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99881387,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-23T04:44:51Z\",\"WARC-Record-ID\":\"<urn:uuid:b9ea6b8f-9fde-418c-9f8b-e49c0e36ceb5>\",\"Content-Length\":\"51637\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ad74634e-3491-4168-bf39-9c9431553cc0>\",\"WARC-Concurrent-To\":\"<urn:uuid:c0481a5a-4f7a-41da-895a-207b38f7b3b0>\",\"WARC-IP-Address\":\"147.232.41.75\",\"WARC-Target-URI\":\"https://kurzy.kpi.fei.tuke.sk/c-reference/en/cpp/numeric/math/cos.html\",\"WARC-Payload-Digest\":\"sha1:WXBFWXG5AB44FD5PO53HDXKRA4CKIT7K\",\"WARC-Block-Digest\":\"sha1:SD7HKMQ62Y7ULCW73CRJ7VZFULCXN2PW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987829458.93_warc_CC-MAIN-20191023043257-20191023070757-00204.warc.gz\"}"} |
https://www.flatbellybible.com/diet-tools/ | [
"The Diet, Nutrition and Fitness Tools section comprises useful calculators to help you reach your weight loss goals. Work out how many calories you need each day, or how much water you should be drinking for example.\n\nSee the current list of diet, nutrition and fitness calculators below, and try them out.\n\n## Calories Burned By Activity\n\nCalculate how many calories you burn when doing a variety of activities.\n\n## Resting Metabolic Rate (RMR) Calculator\n\nCalculate your resting metabolic rate (RMR).\n\n## Total Daily Energy Expenditure (TDEE) Calculator\n\nCalculate your total daily energy expenditure (TDEE).\n\n## Basal Metabolic Rate (BMR) Calculator\n\nBasal metabolic rate (BMR) is the amount of energy expended while at rest. Use this calculator to find out your BMR, determine your caloric needs, and lose or gain weight.\n\n## Activity Calorie Burn Calculator\n\nWant to find out how many calories you burn if you’re doing a particular sport or other everyday activity? The activity calorie burning calculator will tell you!\n\n## Body Fat Calculator (Navy Method)\n\nUse this calculator to figure out your body fat percentage using the Navy formula. Body fat percentage is simply the percentage of fat that your body contains.\n\n## Body Fat Calculator (Army Method)\n\nUse this calculator to estimate your body fat percent based on the formula used by the U.S. Army. It shows whether you pass the requirements of applicants and current service men, as well as how you fare relative to the Department of Defense goal.\n\n## Keto Calculator\n\nUse this keto macro calculator to easily calculate your mix of carbohydrates, fats and proteins per day depending on your goal: maintaining weight, gaining or losing weight. Calorie and macro balance is based on the ketogenic diet’s macronutrient recommendations.\n\n## Thermic Effect of Food Calculator\n\nUse this calculator to easily calculate the thermic effect of food (TEF), a.k.a. thermic effect of feeding, a.k.a. dietary induced thermogenesis (DIT) for diets with different amounts of calories and a different macronutrient mix.\n\n## Weight Loss Calculator\n\nUse this calorie calculator for weight loss to estimate how many calories you need to cut down on in order to achieve a given weight loss target, depending on whether or not you want to change your physical exercise level as well.\n\n## Body Mass Index (BMI) Calculator\n\nA good way to determine if your weight is healthy for your height is to calculate your body mass index (BMI).\n\n## Calories Burned by Walking Calculator\n\nWalking is one of the best and easiest ways to burn calories and lose weight. You can find out how many calories you actually burn with this calculator.\n\n## Daily Calorie Intake Calculator\n\nThis daily calorie calculator gives an approximation of the number of calories you need to eat per day to achieve your fitness goal, based on your height, weight and level of physical activity.\n\n## Daily Carbohydrate Intake Calculator\n\nUse this calculator to find out how much carbohydrate you should be taking on each day to achieve your fitness goal.\n\n## Daily Fat Intake Calculator\n\nCalculate your recommended daily fat intake based on your gender, height, weight and fitness goal.\n\n## Daily Macronutrient Intake Calculator\n\nUse this micronutrient calculator to get a guide on the amount of protein, carbs and fat you should be eating base don your height, weight, gender, activity level and fitness goal.\n\n## Daily Protein Intake Calculator\n\nCalculate how much protein you should be eating each day to achieve your fitness or weight loss goal.\n\n## Daily Water Intake Calculator\n\nHow much water should you drink every day? Find out with this daily water intake calculator.\n\n## Ideal Weight Calculator\n\nWhat is the ideal weight for your height? Use this calculator to find out."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8813826,"math_prob":0.9690631,"size":3900,"snap":"2023-14-2023-23","text_gpt3_token_len":795,"char_repetition_ratio":0.19199179,"word_repetition_ratio":0.036682617,"special_character_ratio":0.18641026,"punctuation_ratio":0.089871615,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9678708,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-25T18:22:20Z\",\"WARC-Record-ID\":\"<urn:uuid:92114267-f8be-4fb0-a8af-ebcb58aefde9>\",\"Content-Length\":\"123006\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8f3d02c7-ca99-4991-b8e9-0503449b6b09>\",\"WARC-Concurrent-To\":\"<urn:uuid:e698cf43-9465-497e-a1fb-60736f6e5626>\",\"WARC-IP-Address\":\"104.21.32.160\",\"WARC-Target-URI\":\"https://www.flatbellybible.com/diet-tools/\",\"WARC-Payload-Digest\":\"sha1:PL6OMEG4GO2CGMGFTS5SKSBLCKBDW4RA\",\"WARC-Block-Digest\":\"sha1:NCNKB3C3JSU7BMJXKB56LX4YXEOJGX5Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945368.6_warc_CC-MAIN-20230325161021-20230325191021-00591.warc.gz\"}"} |
https://mathemania.com/lesson/open-sets/ | [
"# Open sets\n\nIn the lesson Introduction to set theory you learned about set notation and various properties of sets. In addition, in this lesson we will talk about some specific sets that are subsets of a certain metric space.\n\n## Open ball\n\nDefinition: Let $x \\in \\mathbf{R^{n}}$ and $r>0$. A set\n\n$$K(x, r) = \\{ y \\in \\mathbf{R^{n}}: d(x,y)<r\\} = \\left \\{y \\in \\mathbf{R^{n}} : \\sqrt{\\sum_{i=1}^{n}(x_{i} – y_{i})^{2}}< r \\right \\}$$\n\nis called an open ball with center x and radius r.\n\nNote:Expression $d(x, y)$ represents the distance function. In other words, d is the standard Euclidean distance between two points.\n\nPictures below represent open balls in sets $\\mathbf{R}, \\mathbf{R^{2}}$ and $\\mathbf{R^{3}}$, respectively.",
null,
"",
null,
"",
null,
"## Open sets\n\nDefinition: We say that a set $A \\subseteq \\mathbf{R^{n}}$ is an open set if\n\n$$(\\forall x \\in A) \\ (\\exists r>0) \\ K(x, r) \\subseteq A.$$\n\nIn other words, every open set is the union of open balls.\n\nYou can think of it as a collection of elements that doesn’t include any limit points.\n\nExample 1: An open ball $K(x, r)$ is an open set.\n\nExample 2: An open interval $\\left<0, 1\\right>$ is an open set in $\\mathbf{R}$, but not in $\\mathbf{R^{2}}$. More precisely, it is not an open set in $\\mathbf{R^{2}}$ because we identify it with a set $\\{(x, 0): 0< x < 1 \\subset \\mathbf{R^{2}}\\}$.\n\nMoreover, any open interval is an open set.",
null,
"",
null,
"Example 3: $\\emptyset$ and $\\mathbf{R^{n}}$ are open sets.\n\n## Topology on a set\n\nTheorem: Let X be a metric space. Furthermore, let $\\mathcal{T}$ be a family of all open sets with properties:\n\n(1) $\\emptyset, X \\in \\mathcal{T}$\n\n(2) the union of any family from $\\mathcal{T}$ is from $\\mathcal{T}$, i.e.\n\n$$U_{\\alpha}\\in \\mathcal{T} \\rightarrow \\bigcup U_{\\alpha}\\in \\mathcal{T}$$\n\n(3) the intersection of any finite family of elements from $\\mathcal{T}$ is an element from $\\mathcal{T}$, i.e.\n\n$$U_{i} \\in \\mathcal{T}, \\ i = 1, \\cdots, m, \\rightarrow \\bigcap_{i=1}^{m} U_{i} \\in \\mathcal{T}.$$\n\nDefinition: We say that $\\mathcal{T}$ is a topological structure or simply a topology on set X. Furthermore, ordered pair $(X, \\mathcal{T})$ is called a topological space.\n\nElements of a set X are points and elements of a family $\\mathcal{T}$ are open sets of the topological space $(X, \\mathcal{T})$.\n\nExample 4: Let $X = \\{1, 2, 3\\}$ and $\\mathcal{T} = \\{\\emptyset, X, \\{1, 2\\}, \\{1, 3\\}\\}$. Is $\\mathcal{T}$ a topology on X?\n\nSolution: $\\mathcal{T}$ is not a topology on because $\\{1, 2\\} \\cap \\{1, 3\\} = \\{1\\} \\notin X$.\n\n## Interior of a set\n\nDefinition: Let $A \\subseteq \\mathbf{R}$. A point $x \\in A$ is the interior point of a set A if there exists an open set U such that $x \\in U \\subseteq A$.\n\nThe interior of a set A is a set of all interior points. In other words,\n\n$$IntA = \\{x \\in \\mathbf{R^{n}}: \\exists r >0, \\ K(x, r) \\subseteq A\\}.$$\n\nExample 5: Int A is an open set. Moreover, it is the largest open set in A.\n\nExample 6: Interior of a set $S = \\{(x, y) \\in \\mathbf{R}^{2} : 0< x \\leq 1\\}$ is $Int S = \\{(x, y) \\in \\mathbf{R}^{2} : 0 < x < 1\\}$.",
null,
"Namely, we can put all points $(x, y)$ (for which $0 < x < 1$ is valid) in an open set in S, so all that points are in Int S. On the other side, for a point $(1, y)$ every open set of radius r contains a point which is not an element of S. For instance, point $\\left (1 +\\frac{r}{2}, y \\right)$. Therefore, every open set which contains a point $(1, y)$, also contains a point which is not in S.",
null,
""
] | [
null,
"https://mathemania.com/wp-content/uploads/2019/06/open_1-e1559730502795-300x43.png",
null,
"https://mathemania.com/wp-content/uploads/2019/06/open_2-e1559730922271-300x261.png",
null,
"https://mathemania.com/wp-content/uploads/2019/06/open_3-e1559731505660-300x252.png",
null,
"https://mathemania.com/wp-content/uploads/2019/06/open_6-e1560788982739-300x86.png",
null,
"https://mathemania.com/wp-content/uploads/2019/06/open_7-e1560789392834-300x204.png",
null,
"https://mathemania.com/wp-content/uploads/2019/06/open_4-e1559808557455-300x231.png",
null,
"https://mathemania.com/wp-content/uploads/2019/06/open_5-e1559808747176-300x220.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.75942034,"math_prob":0.9999902,"size":3383,"snap":"2022-40-2023-06","text_gpt3_token_len":1142,"char_repetition_ratio":0.17312814,"word_repetition_ratio":0.016863406,"special_character_ratio":0.35560155,"punctuation_ratio":0.14285715,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000075,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-04T09:15:56Z\",\"WARC-Record-ID\":\"<urn:uuid:143fcdb0-1760-4a6f-966a-9e475edaac39>\",\"Content-Length\":\"100073\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:998b14c1-833d-4983-b4ea-891aea226231>\",\"WARC-Concurrent-To\":\"<urn:uuid:996723fa-55cc-4d23-8116-14f64c45a5e5>\",\"WARC-IP-Address\":\"137.184.39.193\",\"WARC-Target-URI\":\"https://mathemania.com/lesson/open-sets/\",\"WARC-Payload-Digest\":\"sha1:PR4JRJ3WIPOTIUKERUIJMCGQOM72WHYX\",\"WARC-Block-Digest\":\"sha1:ABUQRVXOMZ3XA3JHO2Q25W7F5WASIYAT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500095.4_warc_CC-MAIN-20230204075436-20230204105436-00555.warc.gz\"}"} |
https://appliednetsci.springeropen.com/articles/10.1007/s41109-020-00292-0/tables/1 | [
"# Table 1 Main indices of clique network used in this paper\n\nIndex\n\nDescription\n\nnq\n\nNumber of titles in the initial configuration.\n\nn\n\nNumber of network vertices in the final configuration.\n\nm\n\nNumber of edges in the final configuration.\n\nm0\n\nNumber of edges in the initial configuration.\n\nn0\n\nNumber of vertices in the initial configuration, n0n.\n\n#(vi)\n\nFrequency of vertex i in the initial configuration, i.e., the number of titles that contain vertex i (1≤#(vi)≤nq).\n\n#(i,j)\n\nFrequency of edge (i,j) in the initial configuration, i.e., the number of titles that contain the words i and j, 1≤#(i,j)≤nq, and i,j=1,2,...,n, with ij and (i,j)=(j,i).\n\nqi\n\nTitle size i. Number of vertices of title i in the initial configuration, (1≤inq).\n\nqmin.\n\nNumber of vertices of the smallest title in the initial configuration, (1≤qminn).\n\nqmax.\n\nNumber of vertices of the largest clique in the initial configuration, (1≤qmaxn).\n\nk\n\n$$\\langle k \\rangle =\\frac {\\sum _{1}^{n}k_{i}}{n}=\\frac {2m}{n}$$, where 〈k〉 is the average degree of an undirected network and ki is the degree of a vertex i, that is the number of edges incident on the vertex i.\n\n$$k_{i}^{hub}$$\n\n$$k_{i}^{hub}\\geq \\langle k \\rangle + 2\\sigma$$, are the degree values of the hubs, that is, vertices of very high degrees. σ is the standard deviation of the degree distribution.\n\n1. “Initial configuration” is related to the isolated cliques, and “final configuration” is related to the built “network of cliques”. The indices are valid for each time window considered",
null,
""
] | [
null,
"https://appliednetsci.springeropen.com/track/article/10.1007/s41109-020-00292-0",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.796375,"math_prob":0.9899306,"size":1343,"snap":"2023-14-2023-23","text_gpt3_token_len":401,"char_repetition_ratio":0.20612398,"word_repetition_ratio":0.09569378,"special_character_ratio":0.2732688,"punctuation_ratio":0.16442953,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984799,"pos_list":[0,1,2],"im_url_duplicate_count":[null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-10T11:44:21Z\",\"WARC-Record-ID\":\"<urn:uuid:7e03a6cd-918b-4131-adee-ce9b46dfe56d>\",\"Content-Length\":\"209082\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e574ecf7-92e2-416e-ae3e-e8cd23cfcb97>\",\"WARC-Concurrent-To\":\"<urn:uuid:3b54aac0-b3eb-42cd-9355-10a9766dfe9f>\",\"WARC-IP-Address\":\"146.75.36.95\",\"WARC-Target-URI\":\"https://appliednetsci.springeropen.com/articles/10.1007/s41109-020-00292-0/tables/1\",\"WARC-Payload-Digest\":\"sha1:NH7XQSJUE67WNKEHBOQFVNRPODVBEAJR\",\"WARC-Block-Digest\":\"sha1:MG47T6MKBHKMEPUBXGH3DLOVIJTGBHY6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224657169.98_warc_CC-MAIN-20230610095459-20230610125459-00209.warc.gz\"}"} |
https://po.kattis.com/problems/fargrobot | [
"# Färgrobot\n\nA robot can move along a row of colored squares. Every square can be either red, green or blue. A command to the robot is a color. The robot responds by moving to the right until it stands on the color specified by the command. Write a program that, for a given sequence of squares and a number $N$, writes out the sequence of $N$ commands that makes the robot go as far to the right as possible.",
null,
"Figure 1: An illustration of the third sample case.\n\n## Input\n\nThe first line containes an integer $N$, the number of commands to give to the robot. The second line contains a string of letters, R , G, or B, which specifies the color of each square in the row, from left to right. The string contains less than 200 characters, and is always sufficiently long that no sequence of $N$ commands may cause the robot to reach outside the row of squares.\n\n## Output\n\nThe program should output a string of $N$ characters, each being R, G, or B, which constitutes the sequence of commands you should give to the robot to make it move as far as possible along the row of squares. If there are several sequences that accomplish this, you may output anyone of them.\n\n## Limits\n\n1. In one test case worth 20 points, $N=4$ and the sequence is ’RGBGGBGRBRBRGRGRB’ (i.e. the case shown on this year’s PO poster)\n\n2. In other test cases worth 30 points, $1 \\le N \\le 4$.\n\n3. In test cases worth 20 points, $5 \\le N \\le 10$.\n\n4. In test cases worth 30 points, $20 \\le N \\le 25$.\n\nSample Input 1 Sample Output 1\n1\nRRBBBGG\n\nG\n\nSample Input 2 Sample Output 2\n2\nGBBRGRGBRG\n\nRB\n\nSample Input 3 Sample Output 3\n3\nRGBGRRBGBBRG\n\nBBR\n\nSample Input 4 Sample Output 4\n4\nRBRBRBBRBRBRGRGGRGGGRBBRBRRBBRBRBRGBGBGBBGBGBGBBBGBGR\n\nGBGR"
] | [
null,
"https://po.kattis.com/problems/fargrobot/file/statement/en/img-0001.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8660422,"math_prob":0.994676,"size":1650,"snap":"2020-45-2020-50","text_gpt3_token_len":466,"char_repetition_ratio":0.13304982,"word_repetition_ratio":0.012987013,"special_character_ratio":0.25939393,"punctuation_ratio":0.10119048,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9872034,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T14:10:54Z\",\"WARC-Record-ID\":\"<urn:uuid:48280fb1-1a7d-4782-a679-b919910292a9>\",\"Content-Length\":\"18884\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:15a6614b-55dc-4d90-bd74-30384a452c3e>\",\"WARC-Concurrent-To\":\"<urn:uuid:9388eafe-97e6-42aa-940f-5534e8292ed4>\",\"WARC-IP-Address\":\"104.26.13.234\",\"WARC-Target-URI\":\"https://po.kattis.com/problems/fargrobot\",\"WARC-Payload-Digest\":\"sha1:QZXVGH6RP6CEL7SVM5YU7S4FFA3RJQGL\",\"WARC-Block-Digest\":\"sha1:KH4JBKLWNGFFFNDF6O2V6D77FZ3IH7VM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107876500.43_warc_CC-MAIN-20201021122208-20201021152208-00332.warc.gz\"}"} |
https://calcforme.com/percentage-calculator/113-is-67-percent-of-what | [
"# 113 is 67 Percent of what?\n\n## 113 is 67 Percent of 168.66\n\n%\n\n113 is 67% of 168.66\n\nCalculation steps:\n\n113 ÷ ( 67 ÷ 100 ) = 168.66\n\n### Calculate 113 is 67 Percent of what?\n\n• F\n\nFormula\n\n113 ÷ ( 67 ÷ 100 )\n\n• 1\n\nPercent to decimal\n\n67 ÷ 100 = 0.67\n\n• 2\n\n113 ÷ 0.67 = 168.66 So 113 is 67% of 168.66\n\nExample"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7264648,"math_prob":0.9959513,"size":181,"snap":"2022-27-2022-33","text_gpt3_token_len":87,"char_repetition_ratio":0.20903955,"word_repetition_ratio":0.2,"special_character_ratio":0.6906077,"punctuation_ratio":0.1590909,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99883074,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-25T02:25:23Z\",\"WARC-Record-ID\":\"<urn:uuid:175796df-c078-4c34-a6ad-f44cbb663c49>\",\"Content-Length\":\"14565\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a33d3701-fa52-4079-b128-0269e7eae576>\",\"WARC-Concurrent-To\":\"<urn:uuid:231ae2c0-7e73-486b-9532-163e57458181>\",\"WARC-IP-Address\":\"76.76.21.22\",\"WARC-Target-URI\":\"https://calcforme.com/percentage-calculator/113-is-67-percent-of-what\",\"WARC-Payload-Digest\":\"sha1:LSQ4DEEIZFJFFPVOZNA53OPK2QQM6AO3\",\"WARC-Block-Digest\":\"sha1:P3I4XSYVSCUY5QYWYWLSNIU3JVDZ3NQ6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103033925.2_warc_CC-MAIN-20220625004242-20220625034242-00789.warc.gz\"}"} |
http://unitconverter.io/square-yard/square-miles/8475 | [
"",
null,
"# 8,475 square yards to square miles\n\nto\n\n8,475 Square yards = 0.0027 Square miles\n\nThis conversion of 8,475 square yards to square miles has been calculated by multiplying 8,475 square yards by 0.000000322830578510643253918366 and the result is 0.0027 square miles."
] | [
null,
"http://unitconverter.io/img/area.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93761104,"math_prob":0.9985643,"size":306,"snap":"2019-13-2019-22","text_gpt3_token_len":82,"char_repetition_ratio":0.2615894,"word_repetition_ratio":0.0,"special_character_ratio":0.375817,"punctuation_ratio":0.13333334,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96861136,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-19T16:48:41Z\",\"WARC-Record-ID\":\"<urn:uuid:16bdac30-7832-4c51-a4f8-5898ad883d76>\",\"Content-Length\":\"16353\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b5b3ce9e-a59b-4e14-a326-f83baac12441>\",\"WARC-Concurrent-To\":\"<urn:uuid:5f833346-8f7c-42ae-b792-ce7351fc4009>\",\"WARC-IP-Address\":\"23.239.25.168\",\"WARC-Target-URI\":\"http://unitconverter.io/square-yard/square-miles/8475\",\"WARC-Payload-Digest\":\"sha1:GT5Z7HXYHCSZW2HBP4ESORI5NC2IEEYY\",\"WARC-Block-Digest\":\"sha1:UJCMAXJ6TVZEJOLGU274LBLKLYHI6WDP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202003.56_warc_CC-MAIN-20190319163636-20190319185636-00464.warc.gz\"}"} |
https://replit.com/talk/learn/PHP-Tutorial-No-5-Functions/15595 | [
"Learn to Code via Tutorials on Repl.it!\n\n← Back to all posts\n\n# PHP Tutorial No. 5: Functions\n\nFunctions are named, encapsulated blocks of code that ideally do one thing. They can take input (arguments), but do not have to, and return something, but do not have to. Given the same input they should always return the same value.\n\nPHP has many, many built in functions. For example count, which takes one argument, an array, and returns a whole number, the number of elements in the array. Another is rand. It takes in two integer arguments - the start and end of the range of numbers to randomly return a number from - and returns a whole number.\n\n``````\\$arr = [1,2,3,4,5,6];\n\\$num_numbers = count(\\$arr); // \\$num_numbers is 6\n\n\\$number = rand(1, 10); // Resultant is an integer in the range 1 to 10 inclusive ``````\n\nCustom functions can also be created. These can contain other functions. This includes built in ones. Below is an example.\n\n``````\\$a_animals = [\n'Aardvark',\n'Abyssinian',\n'Affenpinscher',\n'Akbash',\n'Akita',\n'Albatross',\n'Alligator',\n'Alpaca',\n'Angelfish',\n'Ant',\n'Anteater',\n'Antelope'\n];\n\n// This function takes in an array and returns a randomly chosen item from it\nfunction getRandomItem(\\$array) {\n\\$array_length = count(\\$array);\n// Because rand returns a number inclusive of the value of the end item\n// 1 must be subtracted, otherwise the index returned may be greater\n// than the biggest index value\n\\$random_index = rand(0, \\$array_length - 1);\nreturn \\$array[\\$random_index];\n}\n\n// Use the function to help display 10 randomly chosen names\n\\$num_names = 10;\n\\$counter = 0;\nwhile (\\$counter < \\$num_names) {\n\\$random_name = getRandomItem(\\$a_animals);\necho \"<p>\" . \\$random_name . \"</p>\";\n\\$counter += 1;\n}``````\n\nHere is a calculator program that uses simple functions.\n\n``````function add(\\$n1, n2) {\nreturn \\$n1 + \\$n2;\n}\n\nfunction subtract(\\$n1, n2) {\nreturn \\$n1 - \\$n2;\n}\n\nfunction multiply(\\$n1, n2) {\nreturn \\$n1 * \\$n2;\n}\n\nfunction divide(\\$n1, n2) {\nreturn \\$n1 / \\$n2;\n}\n\\$operators = [\"+\", \"-\", \"*\", \"/\"];\n\n/*\nThis is a contrived example. In some way two numbers and\nan operator is got from the user e.g. via text boxes. input prompt.\nExample values:\n\\$n1 = 10;\n\\$n2 = 20;\n\\$op = \"*\";\n*/\n// Use basic checking: if \\$op is not in the operators array stop\n// any more of the program code from executing\nif (!in_array(\\$op, \\$operators)) {\nexit(\"Operator is not valid\");\n}\n\n\\$result = NULL;\nswitch (\\$op) {\ncase \"+\": \\$result = add(\\$n1, \\$n2);\nbreak;\ncase \"-\": \\$result = subtract(\\$n1, \\$n2);\nbreak;\ncase \"*\": \\$result = multiply(\\$n1, \\$n2);\nbreak;\ncase \"/\": \\$result = divide(\\$n1, \\$n2);\nbreak;\ndefault: echo \"Not a valid option\";\n}\n\necho \"<p>The result is \\$result</p>\";``````\n##### Comments\nhotnewtop\nPavanBaddi (1)\n\nIn latest PHP version 7.X, You can use Scalar type declarations, return type and some other useful features.\nMentioned in Official PHP website\n\nmalvoliothegood (852)\n\n@PavanBaddi Thanks for the link. How about you updating the example code so it uses Scalar type declarations?"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6480201,"math_prob":0.97976553,"size":2883,"snap":"2021-04-2021-17","text_gpt3_token_len":793,"char_repetition_ratio":0.12400139,"word_repetition_ratio":0.008403362,"special_character_ratio":0.31182796,"punctuation_ratio":0.20181818,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.993045,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-23T00:24:20Z\",\"WARC-Record-ID\":\"<urn:uuid:1e2cfca5-8aa3-4f7e-a758-0031f547ae12>\",\"Content-Length\":\"348262\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:244a4592-5a47-40af-8c42-97f271e52524>\",\"WARC-Concurrent-To\":\"<urn:uuid:af249e0c-226e-46ed-b54f-013179f13ce2>\",\"WARC-IP-Address\":\"104.18.12.38\",\"WARC-Target-URI\":\"https://replit.com/talk/learn/PHP-Tutorial-No-5-Functions/15595\",\"WARC-Payload-Digest\":\"sha1:BRD4LX4JIYEWUCA42GMOBIUK66FTACSR\",\"WARC-Block-Digest\":\"sha1:IZXL56V5JNI7XNK3Q7D7C46EERJGDKKD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039563095.86_warc_CC-MAIN-20210422221531-20210423011531-00435.warc.gz\"}"} |
https://ouachitahigh.opsb.net/415473_3 | [
"DISTRICT CAMPUS\n\n## GT Geometry\n\nTest #14 (through Lesson 70 & Investigation 7)\n\n1. Perimeter of a Regular Polygon (Lesson 66)\n2. Angles Interior to a Circle (Lesson 64)\n3. Parallel Lines (Lesson 37)\n4. Trig Ratios (Lesson 68)\n5. Proportional Segments (Lesson 60)\n6. Explain why a figure is/is not a rectangle (Lesson 65)\n7. Surface Area of Cylinder (Lesson 62)\n8. Perimeter of Polygon (Lesson 61)\n9. OMIT\n10. Volume of Prism (Lesson 59)\n11. Area of Square (Lesson 53)\n12. Lateral Area of Pyramid (Lesson 70)\n13. Identify Transformation (Lesson 67)\n14. Area of Triangle on Grid (Lesson 57)\n15. Geometric Mean (Lesson 50)\n16. Distance from Point to Line (Lesson 42)\n17. Midsegment of Trapezoid (Lesson 69)\n18. Quadrilateral based on given Information (Lesson 58)\n19. Isosceles Triangle (Lesson 51)\n20. Vector Magnitude (Lesson 63)\n\nTest #13 (through Lesson 65)\n\n1. Area of Sector (Lesson 35)\n2. Perimeter of Equiangular Triangle (Lesson 51)\n3. Similar Polygons & Perimeter (Lesson 44)\n4. Midsegment of Triangle (Lesson 55)\n5. Tangent to a Circle (Lesson 64)\n6. Prove a Given Line is a Tangent (Lesson 58)\n7. Properties of Parallelograms (Lesson 61)\n8. Volume of Prism (Lesson 59)\n9. Area of Parallelogram (Lesson 22)\n10. Properties of Rhombus (Lesson 65)\n11. Area of Triangle (Lesson 56)\n12. Line Perpendicular to Given Line (Lesson 37)\n13. Lateral Area of Cylinder (Lesson 62)\n14. Similar Polygons (Lesson 41)\n15. V-E+F=2 (Lesson 49)\n16. Are Lines Parallel in Figure? (Lesson 60)\n17. Perimeter of Figure (Lesson 57)\n18. Do the given sides make a right triangle? (Lesson 33)\n19. Draw Valid Conclusion & Identify the Law Used (Lesson 21)\n20. Name Vector & Terminal Point (Lesson 63)\n\nTest #12 (through Lesson 60 & Investigation 6)\n\n1. Distance from Point to Line (Lesson 42)\n2. 300-600-900 Triangle (Lesson 56)\n3. Midsegment of Triangle Coordinates (Lesson 55)\n4. Isosceles Triangle & Angles (Lesson 51)\n5. Area of Shaded Region (Lesson 40)\n6. Coordinate Proof (Lesson 45)\n7. Proportionality Theorem (Lesson 60)\n8. Two-Point Perspective (Lesson 54)\n9. Perimeter from Coordinates (Lesson 57)\n10. Chord in a Circle (Lesson 43)\n11. Properties/Definition of Rectangles (Lesson 52)\n12. Determine if Given Side Lengths Create Triangle (Lesson 39)\n13. Lateral Area for Prism (Lesson 59)\n14. Geometric Mean (Lesson 50)\n15. Similar Polygons (Lesson 44)\n16. Prove Similar Triangles (Lesson 46)\n17. Solve a Proportion (Lesson 41)\n18. Angles in a Circle (Lesson 58)\n19. Spinner Angles (Investigation 6)\n20. Perimeter of Triangle (Lesson 53)\n\nTest #11 (through Lesson 55)\n\n1. Periemter of Octagon (Lesson 44)\n2. Chord in Circle (Lesson 43)\n3. Equiangular Triangle & Perimeter (Lesson 51)\n4. Order Angles/Sides of Triangle (Lesson 39)\n5. Draw a Solid in One-Point Perspective (Lesson 54)\n6. Show that figures are Similar (Lesson 46)\n7. Arc Length (Lesson 35)\n8. V – E + F = 2 (Lesson 49)\n9. Geometric Mean w/ Triangle (Lesson 50)\n10. Indirect Proof (Lesson 48)\n11. Rectangle Properties (Lesson 52)\n12. Distance from Point to Line (Lesson 42)\n13. Angle Bisector (Lesson 38)\n14. Midsegment of Triangle (Lesson 55)\n15. Coordinate Proof (Lesson 45)\n16. Conjunction of Statements & Truth Table (Lesson 20)\n17. Inscribed Angle of Circle (Lesson 47)\n18. Write Ratios Comparing Side Lengths (Lesson 41)\n19. Area of Trapezoid (Lesson 22)\n20. 450-450-900 Triangle (Lesson 53)\n\nTest #10 (through Lesson 50 & Investigation 5)\n\n1. Distance from a point to a Line (Lesson 42)\n2. Pythagorean Theorem (Lesson 33)\n3. Measure of Arc of a Circle (Lesson 26)\n4. Solve Proportion (Lesson 41)\n5. Parallel or Perpendicular Lines (Lesson 37)\n6. Indirect Proof (Lesson 48)\n7. Area of Sector (Lesson 35)\n8. Included Side and Angle (Lesson 28)\n9. Pythagorean Triple (Lesson 29)\n10. Show Triangles Similar and Find Side Length (Lesson 46)\n11. Classify Solid (Lesson 49)\n12. Find length of chord (Lesson 43)\n13. Range of Side Lengths for Triangle (Lesson 39)\n14. Draw a net (Investigation 5)\n15. Slope between Two Points (Lesson 16)\n16. Prove Triangles are Similar (Lesson 44)\n17. Inscribed Angle (Lesson 47)\n18. Graph a Given Triangle (Lesson 45)\n19. Geometric Mean (Lesson 50)\n20. Find Perimeter (Lesson 40)\n\nTest #9 (through Lesson 45)\n\n1. Circumference (Lesson 23)\n2. Line Parallel to Given Line through Given Point (Lession 37)\n3. Converse: If p, then q becomes If q, then p (Lesson 20)\n4. Perimeter of Composite Figure (Lesson 40)\n5. Two-Column Proof (Lesson 30)\n6. Distance from Point to a Line (Lesson 42)\n7. Similar Figures (Lesson 44)\n8. Write a Proportion to Show Triangles are Similar (Lesson 41)\n9. Centroid of a Triangle (Lesson 32)\n10. Classify a Triangle by Side Lengths & Angles (Lesson 33)\n11. Algebraic Proof (Lesson 24)\n12. Pythagorean Triple (Lesson 29)\n13. Prove Triangles Congruent (Lesson 36)\n14. Circumcenter (Lesson 38)\n15. Tangent Line and Point of Tangency (Lesson 43)\n16. Sum of Interior Angles (Investigation 3)\n17. Angles in a Parallelogram (Lesson 34)\n18. Order Sides Lengths of Triangle (Lesson 39)\n19. Included Side (Lesson 28)\n20. Graph a Triangle with Given Information (Lesson 45)\n\nBenchmark Test #2 (through Lesson 40 & Investigation 4)\n\n1. Interior/Exterior Angle Measures (Investigation 3)\n2. Included Angle/Side (Lesson 28)\n3. Midpoint (Lesson 11)\n4. Identify a Property (Lesson 24)\n5. Area of Triangle (Lesson 13)\n6. Angle Bisector (Lesson 3)\n7. Congruent Triangles (Lesson 36)\n8. Congruent Triangles (Lesson 30)\n9. Distance Formula (Lesson 9)\n10. Equation of a Line Perpendicular to Given Line (Lesson 37)\n11. Segment Addition Postulate (Lesson 2)\n12. Range of Values (Investigation 4)\n13. Definition of Terms (Lesson 1)\n14. Perimeter of Rectangle (Lesson 8)\n15. Area of Quadrilaterals (Lesson 22)\n16. Properties of Parallelogram (Lesson 34)\n17. Complementary/Supplementary Angles (Lesson 6)\n18. Six Congruency Statements (Lesson 25)\n19. Sketch a Triangle with Given Information (Lesson 18)\n20. Circle Properties (Lessons 23, 26, 33)\n\nTest #7 (through Lesson 35)\n\n1. Pythagorean Theorem (Lesson 34)\n2. Measure of an Arc (Lesson 26)\n3. Corresponding Sides of Triangles (Lesson 30)\n4. Area of Parallelogram (Lesson 22)\n5. Midpoint of Two Points (Lesson 11)\n6. Two-Column Proof (Proving Theorem 6-3) (Lesson 27)\n7. Arc Length (Lesson 35)\n8. Remote Interior Angles Theorem (Lesson 18)\n9. Contrapositive (Lesson 17)\n10. Flowchart to Two-Column Proof (Lesson 31)\n11. Area of a Rectangle—Algebraic Proof (Lesson 24)\n12. Are triangle congruent? (Lesson 28)\n13. Hypothesis, Conclusion, Counterexample (Lesson 14)\n14. Centroid (Lesson 32)\n15. Concave or Convex Polygon (Lesson 15)\n16. Pythagorean Triple (Lesson 29)\n17. Area of Triangle (Lesson 13)\n18. Area of Circle (Lesson 23)\n19. Law of Detachment (Lesson 21)\n20. Pythagorean Theorem (Lesson 33)\n\nTest #6 (through Lesson 30 & Investigation 3)\n\n1. Law of Detachment to write a Statement (Lesson 21)\n2. Area of Trapezoid (Lesson 22)\n3. Circles—central angle, minor arc, major arc, and semicircle (Lesson 26)\n4. Conjecture (Lesson 10)\n5. Counterexample (Lesson 14)\n6. Exterior Angles Formula (Investigation 3)\n7. Congruence Statement (Lesson 25)\n8. Circumference of Circle (Lesson 23)\n9. TWO COLUMN PROOF—REASONS ONLY (Lesson 27)\n10. Properties (Lesson 2)\n11. Conjecture (Lesson 7)\n12. Included Side & Angle (Lesson 28)\n13. Hypothesis & Conclusion & Negation of Each (Lesson 17)\n14. Name a Ray (Lesson 3)\n15. Area of Triangle (Lesson 30)\n16. Triangle Angle Sum Theorem (Lesson 18)\n17. Area of a Rectangle—Algebraic Proof (Lesson 24)\n18. Pythagorean Theorem (Lesson 29)\n19. Sketch a quadrilateral (Lesson 19)\n20. Using Area Formula of a Rectangle (Lesson 8)\n\nTest #5 (through Lesson 25)\n\n1. Deductive Reasoning (Lesson 21)\n2. Circle Properties—Name, Diameter, Radius, & Center (Lesson 23)\n3. Three Collinear & Three Noncollinear Points (Lesson 1)\n4. Conjecture (Lesson 7)\n5. Counterexample (Lesson 14)\n6. Prove Lines Parallel (Lesson 12)\n7. Angle Addition Postulate (Lesson 3)\n8. Area of Parallelogram (Lesson 22)\n9. Interior & Exterior Angles (Lesson 15)\n10. Fahrenheit to Celsius Formula (Lesson 8)\n11. Midpoint (Lesson 2)\n12. Corresponding Parts to Triangles (Lesson 25)\n13. Hypothesis, Conclusion, & Converse (Lesson 17)\n14. Triangle Angle Sum Theorem (Lesson 18)\n15. Distance on a Number Line (Lesson 9)\n16. Perimeter of a Triangle (Lesson 13)\n17. Algebraic Proof (Lesson 24) 2(x+2) = x+6\n18. Slope of a Line (Lesson 16)\n19. Converse & Truth Value (Lesson 20)\n20. Perimeter & Area of Rectangle (Lesson 19)\n\nBenchmark Test #1 (through Lesson 20 & Investigation 2)\n\n1. Parallel & Perpendicular Lines (Lesson 5)\n2. Distance Formula (Lesson 9)\n3. Midpoint Formula (Lesson 11)\n4. Angle Addition Postulate (Lesson 3)\n5. Counterexample (Lesson 14)\n6. Intersection of a Line & Plane (Lesson 1)\n7. Vertical Angles (Lesson 6)\n8. Converse (Lesson 17)\n9. Perimeter of Square (Lesson 8)\n10. Triangle Angle Sum Theorem (Lesson 18)\n11. Equiangular, Equilateral, Concave, Convex, Regular, Irregular (Lesson 15)\n12. Equation through a Given Point with a Given Slope (Lesson 16)\n13. Determine Truth Value of Mathematical Statements (Lesson 10)\n14. Pythagorean Theorem (Investigation 2)\n15. Angle Pairs (Investigation 1)\n16. Segment Addition Postulate (Lesson 2)\n17. Perimeter of a Rectangle (Lesson 19)\n18. Inductive Reasoning Number Patterns (Lesson 7)\n19. Draw a Figure (Lesson 4)\n20. Triangle—classify, perimeter, area (Investigation 2 & Lesson 13)\n\nTest #3 (through Lesson 15)\n\n1. Angle Addition Postulate (Lesson 3)\n2. Complementary & Supplementary Angles (Lesson 6)\n3. Coplanar & Noncoplanar Lines (Lesson 1)\n4. Counterexample (Lesson 14)\n5. Area of a Triangle (Lesson 13)\n6. Proving Lines Parallel (Lesson 12)\n7. Inductive Reasoning (Lesson 7)\n8. Area of a Rectangle (Lesson 8)\n9. Proving Lines Parallel (Lesson 12)\n10. Distance Formula (Lesson 9)\n11. Conditional Statement (Lesson 10)\n12. Using Angle Pairs (Investigation 1)\n13. Midpoint (Lesson 11)\n14. Find the Diagonal of a Rectangle (Lesson 8)\n15. Classify an Angle & Measure it with a Protractor (Lesson 3)\n16. Midpoint (Lesson 11)\n17. Convex or Concave Polygon (Lesson 15)\n18. Perimeter & Area of a Triangle (Lesson 13)\n19. Hypothesis & Conclusion of Conditional Statement (Lesson 10)\n20. Counterexample (Lesson 14)\n\nTest #2 (through Lesson 10 & Investigation 1)\n\n1. Relationship between Parallel & Perpendicular Lines (Lesson 5)\n2. Supplementary Angles (Lesson 6)\n3. When are lines coplanar? (Lesson 1)\n4. Hypothesis & Conclusion (Lesson 10)\n5. Identify Pairs of Angles (Investigation 1)\n6. Segment Addition Postulate (Lesson 2)\n7. Inductive Reasoning Number Patterns (Lesson 7)\n8. Angle Addition Postulate (Lesson 3)\n9. Where do two planes intersect? (Lesson 4)\n10. Angles formed by Perpendicular Lines (Lesson 5)\n11. Area of a Triangle (Lesson 8)\n12. Complementary & Supplementary Angles (Lesson 6)\n13. Inductive Reasoning Number Patterns (Lesson 7)\n14. How many points define a line, plane, and space? (Lesson 1)\n15. Segment Addition Postulate (Lesson 2)\n16. Perimeter of a Triangle (Lesson 8)\n17. Distance Formula (Lesson 9)\n18. Conditional Statement True/False (Lesson 10)\n19. Angle between hands on a clock (Lesson 3)\n20. Identify Collinear Points (Lesson 4)\n\nTest #1 (through Lesson 5)\n\n1. Name a line (Lesson 1)\n2. Name a plane (Lesson 1)\n3. Identify Coplanar and Noncoplanar Lines (Lesson 1)\n4. Intersection of two lines (Lesson 1)\n5. Identify Property (Lesson 2)\n6. Finding Midpoint of a Distance (Lesson 2)\n7. Segment Addition Postulate (Lesson 2)\n8. Finding Distance on a Number Line (Lesson 2)\n9. Use a protractor to find an amount from circle graph (Lesson 3)\n10. Classify an angle and give angle measure (Lesson 3)\n11. Angle Addition Postulate (Lesson 3)\n12. Name Rays (Lesson 3)\n13. Intersection of Planes (Lesson 4)\n14. Explain why tripod is best! (Lesson 4)\n15. Name planes, lines, and points (Lesson 4)\n16. Postulate question (Lesson 4)\n17. Classify Pairs of Lines (Lesson 5)\n18. Identify two parallel lines (Lesson 5)\n19. Draw parallel lines through a point (Lesson 5)\n20. Transitive Property of Parallel Lines (Lesson 5)\n\nTest 14 (through Lesson 70 & Investigation 7)\n\n1. Perimeter of a regular polygon (lesson 66)\n2. Angles inside of Circles (lesson 64)\n3. Line parallel to given line going through specific point (lesson 37)\n4. Sine Cosine & Tangent ratios of a triangle (lesson 68)\n5. Parallel lines cut by transversal (lesson 60)\n6. Is the given parallelogram a rectangle? Explain!!! (Lesson 65)\n7. Surface Area of Cylinder (lesson 62)\n8. Perimeter of a polygon (lesson 61)\n9. Trig Ratios (Investigation 7)\n10. Volume of a Prism (lesson 59)\n11. Area of a polygon (lesson 53)\n12. Lateral Area of a pyramid (lesson 70)\n13. Transformation—Identify the type (lesson 67)\n14. Area of a triangle—coordinate proof type (lesson 57)\n15. Geometric mean—not with a triangle (lesson 50)\n16. Distance from a point to a line (lesson 42)\n17. Midsegment length (lesson 69)\n18. Classify a quadrilateral (lesson 58)\n19. Isoscles trangle (lesson 51)\n20. Vector magnitude (lesson 63)"
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https://www.hindawi.com/journals/jfs/2014/398673/ | [
"/ / Article\nSpecial Issue\n\n## Ulam’s Type Stability and Fixed Points Methods\n\nView this Special Issue\n\nResearch Article | Open Access\n\nVolume 2014 |Article ID 398673 | https://doi.org/10.1155/2014/398673\n\nChun Wang, Tian-Zhou Xu, \"Hyers-Ulam Stability of Differentiation Operator on Hilbert Spaces of Entire Functions\", Journal of Function Spaces, vol. 2014, Article ID 398673, 6 pages, 2014. https://doi.org/10.1155/2014/398673\n\n# Hyers-Ulam Stability of Differentiation Operator on Hilbert Spaces of Entire Functions\n\nAccepted16 Jun 2014\nPublished06 Jul 2014\n\n#### Abstract\n\nWe investigate the Hyers-Ulam stability of differentiation operator on Hilbert spaces of entire functions. We give a necessary and sufficient condition in order that the operator has the Hyers-Ulam stability and also show that the best constant of Hyers-Ulam stability exists.\n\n#### 1. Introduction\n\nIn 1940, the first stability problem concerning group homomorphisms was raised by Ulam . Let be a group and let be a metric group with a metric . Given any , does there exist a such that if a function satisfies the inequality for all , then there exists a homomorphism with for all ?\n\nIn the following years, Hyers affirmatively answered the question of Ulam for the case where and are Banach spaces (see ). Furthermore, the result of Hyers has been generalized by Rassias (see ).\n\nSince then, the stability of many algebraic, differential, integral, operatorial, functional equations have been extensively investigated (see and the references therein).\n\nIn this paper, we discuss the Hyers-Ulam stability of differentiation operator on Hilbert spaces of entire functions and give a necessary and sufficient condition in order that the operator has the Hyers-Ulam stability, and we show that the best constant of Hyers-Ulam stability exists.\n\n#### 2. Hilbert Spaces of Entire Functions\n\nIn this section, we describe the Hilbert spaces of entire functions in which the rest of our work is set and record their most basic properties. About the function spaces, we recommend the research papers [18, 19]. For the sake of coherency we recall a few basic definitions, notions, and theorems from , and we also give some typical examples; in particular Fock space in these examples is a very important tool for quantum stochastic calculus in the case of quantum probability (see ).\n\nLet us call an entire function a comparison function if for each , and the sequence of ratios decreases to zero as increases to . For each comparison function , we define to be the Hilbert space of power series for which\n\nIt is easy to check that each element of is an entire function and that every sequence convergent in the norm of the space is uniformly convergent on compact subsets of the plane. In this case, the inner product of is given by and the functions form an orthonormal basis for . We can see that the polynomials are dense in .\n\nExample 1. We consider the comparison function ; that is, ; by a simple calculation, we can see that , and then is the famous Fock space.\n\nExample 2. If we put comparison function , that is, , we can see that on .\n\nThroughout this paper, let be the differentiation operator defined by\n\nAn important result about is the following theorem.\n\nTheorem 3 (see ). The operator is bounded on if and only if the sequence is bounded, where .\n\nBy Theorem 3, we can obtain that the operator is unbounded on Fock space , and it is bounded on .\n\nThroughout this paper, we suppose that the sequence is bounded.\n\n#### 3. Hyers-Ulam Stability of Differentiation Operator\n\nLet be normed spaces and consider a mapping . The following definition can be found in .\n\nDefinition 4. We say that has the Hyers-Ulam stability property (briefly, is HU-stable) if there exists a constant such that, for any , , and with , there exists an with and . The number is called a Hyers-Ulam stability constant (briefly HUS-constant) and the infimum of all HUS constants of is denoted by ; generally, is not a HUS constant of (see [9, 10]).\n\nTheorem 5. Let be the differentiation operator on the Hilbert spaces of entire functions . Then the following statements are equivalent:(a)has Hyers-Ulam stability on ;(b)the sequence is bounded, where .\n\nProof. (b)⇒(a). Suppose that the sequence is bounded, and let . Since the polynomials are dense in , we just need to show that has Hyers-Ulam stability on the polynomials dense subspace. For each , we take any two polynomials and that satisfy , and can be represented by the orthonormal basis. Then if , , we have For any nonnegative integers and such that , we can get Hence,\nLet be the function defined by\nIt is easy to check that , also; from (8), we obtain Next, assume that . It follows that Hence, Let Then ; by (12), we have Finally, if , we get Hence, Thus, from (16), it follows that Therefore, (a) holds.\n(a)⇒(b). Suppose that is stable with Hyers-Ulam stability constant . For any nonnegative integer , let . Then , so there exists such that and . Hence, and consequently for any nonnegative integer . This completes the proof.\n\nExample 6. We consider the comparison function ; that is, ; by a simple calculation, we have and (), and hence operator has Hyers-Ulam stability on the .\n\nExample 7. If we put comparison function , that is, , we have and (), is unbounded, and hence operator is not Hyers-Ulam stable on the .\n\nExample 8. We consider the comparison function ; we get , , (), and by a simple calculation, we have , , () and , , (), where is bounded; hence, operator has Hyers-Ulam stability on the .\n\nRemark 9. From Theorem 5 and Examples 68, we can see that the Hyers-Ulam stability of differentiation operator on Hilbert spaces of entire functions depends on the comparison functions . It shows that the comparison functions affects the behaviors of the operators and the functions in the corresponding Hilbert space .\n\nNext, we will show that the best constant of Hyers-Ulam stability exists.\n\nTheorem 10. If differentiation operator has the Hyers-Ulam stability on Hilbert spaces of entire functions , then and is a HUS constant of .\n\nProof. Suppose that has Hyers-Ulam stability on . By the proof of Theorem 5, we know that is a constant of the Hyers-Ulam stability of differentiation operator . Next, we show that it is the infimum of all the Hyers-Ulam stability constants. Let be an arbitrary Hyers-Ulam stability constant for ; put , and for any nonnegative integer , we can obtain , so there exists , such that and , and hence and so .\n\n#### Conflict of Interests\n\nThe authors declare that there is no conflict of interests regarding the publication of this paper.\n\n#### Acknowledgments\n\nThe authors are very grateful to the editor and reviewers for their valuable comments and suggestions on the paper. This work is supported by the National Natural Science Foundation of China (Grant no. 11171022).\n\n1. S. M. Ulam, A Collection of Mathematical Problems, Interscience, New York, NY, USA, 1960. View at: MathSciNet\n2. D. H. Hyers, “On the stability of the linear functional equation,” Proceedings of the National Academy of Sciences of the United States of America, vol. 27, pp. 222–224, 1941. View at: Publisher Site | Google Scholar | MathSciNet\n3. Th. M. Rassias, “On the stability of the linear mapping in Banach spaces,” Proceedings of the American Mathematical Society, vol. 72, no. 2, pp. 297–300, 1978.\n4. J. Brzdęk and S.-M. Jung, “A note on stability of an operator linear equation of the second order,” Abstract and Applied Analysis, vol. 2011, Article ID 602713, 15 pages, 2011. View at: Publisher Site | Google Scholar | MathSciNet\n5. J. Brzdek, D. Popa, and B. Xu, “On approximate solutions of the linear functional equation of higher order,” Journal of Mathematical Analysis and Applications, vol. 373, no. 2, pp. 680–689, 2011. View at: Publisher Site | Google Scholar | MathSciNet\n6. K. Ciepliński, “Generalized stability of multi-additive mappings,” Applied Mathematics Letters, vol. 23, no. 10, pp. 1291–1294, 2010. View at: Publisher Site | Google Scholar | MathSciNet\n7. K. Ciepliński, “Applications of fixed point theorems to the Hyers-Ulam stability of functional equations—a survey,” Annals of Functional Analysis, vol. 3, no. 1, pp. 151–164, 2012. View at: Publisher Site | Google Scholar | MathSciNet\n8. P. Găvruţa, “A generalization of the Hyers-Ulam-Rassias stability of approximately additive mappings,” Journal of Mathematical Analysis and Applications, vol. 184, no. 3, pp. 431–436, 1994. View at: Publisher Site | Google Scholar | MathSciNet\n9. O. Hatori, K. Kobayashi, T. Miura, H. Takagi, and S. E. Takahasi, “On the best constant of Hyers-Ulam stability,” Journal of Nonlinear and Convex Analysis, vol. 5, no. 3, pp. 387–393, 2004. View at: Google Scholar | MathSciNet\n10. G. Hirasawa and T. Miura, “Hyers-Ulam stability of a closed operator in a Hilbert space,” Bulletin of the Korean Mathematical Society, vol. 43, no. 1, pp. 107–117, 2006.\n11. T. Miura, G. Hirasawa, and S. E. Takahasi, “Ger-type and Hyers-Ulam stabilities for the first-order linear differential operators of entire functions,” International Journal of Mathematics and Mathematical Sciences, no. 22, pp. 1151–1158, 2004. View at: Publisher Site | Google Scholar | MathSciNet\n12. D. Popa and I. Raşa, “The Fréchet functional equation with application to the stability of certain operators,” Journal of Approximation Theory, vol. 164, no. 1, pp. 138–144, 2012. View at: Publisher Site | Google Scholar | MathSciNet\n13. D. Popa and I. Raşa, “On the stability of some classical operators from approximation theory,” Expositiones Mathematicae, vol. 31, no. 3, pp. 205–214, 2013. View at: Publisher Site | Google Scholar | MathSciNet\n14. H. Takagi, T. Miura, and S. E. Takahasi, “Essential norms and stability constants of weighted composition operators on C(X),” Bulletin of the Korean Mathematical Society, vol. 40, no. 4, pp. 583–591, 2003. View at: Publisher Site | Google Scholar | MathSciNet\n15. T. Z. Xu, “Approximate multi-Jensen, multi-Euler-Lagrange additive and quadratic mappings in $n$-Banach spaces,” Abstract and Applied Analysis, vol. 2013, Article ID 648709, 12 pages, 2013. View at: Publisher Site | Google Scholar | MathSciNet\n16. T. Z. Xu, “On the stability of multi-Jensen mappings in $\\beta$-normed spaces,” Applied Mathematics Letters, vol. 25, no. 11, pp. 1866–1870, 2012. View at: Publisher Site | Google Scholar | MathSciNet\n17. T. Z. Xu, Z. Yang, and J. M. Rassias, “Direct and fixed point approaches to the stability of an {AQ}-functional equation in non-Archimedean normed spaces,” Journal of Computational Analysis and Applications, vol. 17, no. 4, pp. 697–706, 2014.\n18. K. C. Chan and J. H. Shapiro, “The cyclic behavior of translation operators on Hilbert spaces of entire functions,” Indiana University Mathematics Journal, vol. 40, no. 4, pp. 1421–1449, 1991.\n19. G. A. Chacón, G. R. Chacón, and J. Giménez, “Composition operators on spaces of entire functions,” Proceedings of the American Mathematical Society, vol. 135, no. 7, pp. 2205–2218, 2007. View at: Publisher Site | Google Scholar | MathSciNet\n20. V. Crismale, “Quantum stochastic calculus on interacting Fock spaces: semimartingale estimates and stochastic integral,” Communications on Stochastic Analysis, vol. 1, no. 2, pp. 321–341, 2007. View at: Google Scholar | MathSciNet\n21. B. K. Das, J. M. Lindsay, and O. Tripak, “Sesquilinear quantum stochastic analysis in Banach space,” Journal of Mathematical Analysis and Applications, vol. 409, no. 2, pp. 1032–1051, 2014. View at: Publisher Site | Google Scholar | MathSciNet\n22. U. C. Ji and N. Obata, “Quantum stochastic integral representations of Fock space operators,” Stochastics, vol. 81, no. 3-4, pp. 367–384, 2009. View at: Publisher Site | Google Scholar | MathSciNet"
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https://boundaryvalueproblems.springeropen.com/articles/10.1186/s13661-015-0510-6 | [
"We’d like to understand how you use our websites in order to improve them. Register your interest.\n\n# Infinitely many solutions for p-biharmonic equation with general potential and concave-convex nonlinearity in $$\\mathbb{R}^{N}$$\n\n## Abstract\n\nIn this paper, we study the existence of multiple solutions to a class of p-biharmonic elliptic equations, $$\\Delta^{2}_{p}u-\\Delta_{p}u+V(x)|u|^{p-2}u =\\lambda h_{1}(x)|u|^{m-2}u+h_{2}(x)|u|^{q-2}u$$, $$x\\in \\mathbb{R}^{N}$$, where $$1< m< p< q< p_{*}=\\frac {pN}{N-2p}$$, $$\\Delta^{2}_{p} u=\\Delta(|\\Delta u|^{p-2}\\Delta u)$$ is a p-biharmonic operator and $$\\Delta_{p}u=\\operatorname{div}(|\\nabla u|^{p-2}\\nabla u)$$. The potential function $$V(x)\\in C{({\\mathbb{R}}^{N})}$$ satisfies $$\\inf_{x\\in{\\mathbb{R}} ^{N}}V(x)>0$$. By variational methods, we obtain the existence of infinitely many solutions for a p-biharmonic elliptic equation in $${\\mathbb{R}}^{N}$$.\n\n## Introduction\n\nIn this paper, we are interested in the existence of solutions to the following p-biharmonic elliptic equation:\n\n$$\\Delta^{2}_{p}u-\\Delta_{p}u+V(x)|u|^{p-2}u =f(x,u),\\quad x\\in\\mathbb{R}^{N},$$\n(1.1)\n\nwhere $$2<2p<N$$, $$f(x,u)=\\lambda h_{1}(x)|u|^{m-2}u+h_{2}(x)|u|^{q-2}u$$, $$1< m< p< q< p_{*}=\\frac{pN}{N-2p}$$, $$\\Delta^{2}_{p} u=\\Delta(|\\Delta u|^{p-2}\\Delta u)$$ is a p-biharmonic operator and $$\\Delta _{p}u=\\operatorname{div}(|\\nabla u|^{p-2}\\nabla u)$$. The potential function $$V(x)\\in C{({\\mathbb{R}}^{N})}$$ satisfies $$\\inf_{x\\in{\\mathbb{R}}^{N}}V(x)>0$$.\n\nRecently, the nonlinear biharmonic equation in an unbounded domain has been extensively investigated, we refer the reader to and the references therein. For the whole space $${\\mathbb{R}}^{N}$$ case, the main difficulty of this problem is the lack of compactness for the Sobolev embedding theorem. In order to overcome this difficulty, the authors always assumed the potential $$V(x)$$ has some special characteristic. For example, in , Yin and Wu studied the following fourth-order elliptic equation:\n\n$$\\left \\{ \\textstyle\\begin{array}{l} \\Delta^{2} u-\\Delta u+V(x)u =f(x,u), \\quad x\\in\\mathbb{R}^{N}, \\\\ u(x)\\in H^{2}({{\\mathbb{R}}^{N}}), \\end{array}\\displaystyle \\right .$$\n(1.2)\n\nwhere the potential $$V(x)$$ satisfied\n\n(V0)::\n\n$$V\\in C({\\mathbb{R}}^{N})$$ satisfies $$\\inf_{x\\in{\\mathbb {R}}^{N}}V(x)>0$$, and for each $$M>0$$, $$\\operatorname{meas}\\{x\\in \\mathbb{R}^{N}\\leq M\\}<+\\infty$$.\n\nThis assumption guarantees that the embedding $$H^{2}\\hookrightarrow L^{s}({\\mathbb{R}}^{N})$$ is compact for each $$s\\in[2,\\frac{2N}{N-4})$$ and obeys the coercivity condition: $$V(x)\\to\\infty$$ as $$|x|\\to\\infty$$. Hence, under various sets of assumptions on the nonlinear term $$f(x, t)$$ (subcriticality, superquadraticity, etc.), the authors proved the existence of infinitely many solutions to problem (1.2) by using the variational techniques in a standard way. In , Liu et al. considered the following fourth-order elliptic equation:\n\n$$\\left \\{ \\textstyle\\begin{array}{l} \\Delta^{2} u-\\Delta u+\\lambda V(x)u =f(x,u),\\quad x\\in\\mathbb{R}^{N}, \\\\ u(x)\\in H^{2}({{\\mathbb{R}}^{N}}), \\end{array}\\displaystyle \\right .$$\n(1.3)\n\nwhere the potential $$V(x)$$ satisfied a weaker condition than (V0), that is,\n\n(V1)::\n\n$$V\\in C({\\mathbb{R}}^{N})$$ satisfies $$\\inf_{x\\in{\\mathbb {R}}^{N}}V(x)>0$$, there exists some $$M>0$$, $$\\operatorname{meas}\\{x\\in \\mathbb{R}^{N}\\leq M\\}<+\\infty$$.\n\nUnder the assumption (V1), the compactness of the embedding is lost and this renders variational techniques more delicate. With the aid of the parameter $${\\lambda}>0$$, they proved that the variational functional satisfies $$(\\mathit{PS})$$ condition, and then they showed the existence and multiplicity results of problem (1.3). A natural question is whether the existence results still holds if we assume a more general potential $$V(x)$$ than (V0), (V1), namely,\n(V)::\n\n$$V(x)\\in C({\\mathbb{R}}^{N})$$ satisfies $$\\inf_{x\\in {\\mathbb{R}}^{N}}V(x)>0$$.\n\nIn the present paper, we will answer this interesting question. We consider the existence of solutions to the p-biharmonic problem (1.1) with a more general potential $$V(x)$$. To prove that the $$(\\mathit{PS})$$ sequence weakly converges to a critical point of the corresponding functional, we adapt ideas developed by and then by variational methods, we establish the existence of infinitely many high-energy solutions to problem (1.1) with a concave-convex nonlinearity, i.e., $$f(x,u)=\\lambda h_{1}(x)|u|^{m-2}u+h_{2}(x)|u|^{q-2}u$$, $$1< m< p< q< p_{*}=\\frac{pN}{N-2p}$$. To the best of our knowledge, little has been done for p-biharmonic problems with this type of nonlinearity. Here, we give our assumptions on the weight functions $$h_{1}(x)$$ and $$h_{2}(x)$$:\n\n(H1):\n\n$$h_{1}\\in L^{\\sigma}({\\mathbb{R}}^{N})$$ with $$\\sigma=\\frac {p}{p-m}$$;\n\n(H2):\n\n$$h_{2}(x)\\geq0$$ (0), $$h_{2}(x)\\in L^{\\infty}({\\mathbb{R}}^{N})$$.\n\nThe main result in this paper is as follows.\n\n### Theorem 1.1\n\nLet $$2<2p<N$$, $$1< m< p< q< p_{*}=\\frac{pN}{N-2p}$$. Assume (V), (H1), and (H2) hold. Then there exists $$\\lambda_{0}>0$$ such that for all $$\\lambda\\in [0,\\lambda_{0}]$$, problem (1.1) admits infinitely many high-energy solutions in $${\\mathbb{R}}^{N}$$.\n\nThis paper is organized as follows. In Section 2, we build the variational framework for problem (1.1) and establish a series of lemmas, which will be used in the proof of Theorem 1.1. In Section 3, we prove Theorem 1.1 by the mountain pass theorem .\n\n## Preliminaries\n\nIn order to apply the variational setting, we assume the solutions of (1.1) belong to the following subspace of $$\\mathcal {D}^{2,p}({\\mathbb{R}}^{N})$$:\n\n$$E=\\biggl\\{ u\\in\\mathcal{D}^{2,p}\\bigl({ \\mathbb{R}}^{N}\\bigr)\\Big| \\int_{{\\mathbb {R}}^{N}}|\\Delta u|^{p}+|\\nabla u|^{p}+V(x)|u|^{p} \\,dx< \\infty\\biggr\\}$$\n(2.1)\n\nendowed with the norm\n\n$$\\|u\\|_{E}=\\biggl( \\int_{{\\mathbb{R}}^{N}}\\bigl(|\\Delta u|^{p}+|\\nabla u|^{p}+V(x)|u|^{p}\\bigr)\\, dx\\biggr)^{1/p},$$\n(2.2)\n\nwhere $$\\mathcal{D}^{2,p}({\\mathbb{R}}^{N})=\\{u\\in L^{p_{*}}({\\mathbb {R}}^{N}) |\\Delta u\\in L^{p}({\\mathbb{R}}^{N})\\}$$, $$\\|\\cdot\\|_{s}$$ means the norm in $$L^{s}({\\mathbb{R}}^{N})$$.\n\nWe denote by $$S_{*}$$ the Sobolev constant, that is,\n\n$$S_{*}=\\inf_{u\\in\\mathcal{D}^{2,p}\\setminus\\{0\\}}\\frac{\\int _{{\\mathbb{R}}^{N}}|\\Delta u|^{p}\\,dx}{(\\int_{{\\mathbb {R}}^{N}}|u|^{p_{*}}\\, dx)^{p/p_{*}}}$$\n(2.3)\n\nand\n\n$$S_{*}\\biggl( \\int_{{\\mathbb{R}}^{N}}|u|^{p_{*}}\\, dx\\biggr)^{p/p_{*}}\\leq \\int_{{\\mathbb {R}}^{N}}|\\Delta u|^{p}\\,dx,\\quad \\forall u\\in \\mathcal{D}^{2,p}\\bigl({\\mathbb{R}}^{N}\\bigr),$$\n(2.4)\n\nwhere $$S_{*}$$ is obtained by a positive and radially symmetric function; see for instance .\n\n### Definition 2.1\n\nA function $$u\\in E$$ is said to be a weak solution of (1.1) if, for any $$\\varphi\\in E$$, we have\n\n\\begin{aligned}& \\int_{{\\mathbb{R}}^{N}}\\bigl(|\\Delta u|^{p-2}\\Delta u\\Delta\\varphi+| \\nabla u|^{p-2}\\nabla u\\varphi+V|u|^{p-2}u\\varphi\\bigr)\\, dx \\\\& \\quad = \\int_{{\\mathbb{R}} ^{N}}\\bigl(\\lambda h_{1}(x)|u|^{m-2}u+h_{2}(x)|u|^{q-2}u \\bigr)\\varphi\\, dx. \\end{aligned}\n(2.5)\n\nLet $$J(u):E\\to{\\mathbb{R}}$$ be the energy functional associated with problem (1.1) defined by\n\n$$J(u)=\\frac{1}{p}\\|u\\|^{p}_{E}- \\frac{\\lambda}{m} \\int_{{\\mathbb{R}} ^{N}}h_{1}|u|^{m}\\,dx-\\frac{1}{q} \\int_{{\\mathbb{R}}^{N}}h_{2}|u|^{q}\\,dx.$$\n(2.6)\n\nFrom the embedding inequality (2.4) and the assumptions in Theorem 1.1, we see the functional $$J\\in C^{1}(E,{\\mathbb{R}})$$ and its Gateaux derivative is given by\n\n\\begin{aligned} J'(u)\\varphi =& \\int_{{\\mathbb{R}}^{N}}\\bigl(|\\Delta u|^{p-2}\\Delta u\\Delta \\varphi +|\\nabla u|^{p-2}\\nabla u\\varphi+V(x)|u|^{p-2}u\\varphi\\bigr)\\, dx \\\\ &{}- \\int_{{\\mathbb{R}}^{N}}\\bigl(\\lambda h_{1}(x)|u|^{m-2}u+h_{2}(x)|u|^{q-2}u \\bigr)\\varphi\\, dx. \\end{aligned}\n(2.7)\n\nTo prove the existence of infinitely many solutions to problem (1.1), we need to prove that the functional J defined by (2.6) satisfies the $$(\\mathit{PS})$$ condition. Recall that a sequence $$\\{u_{n}\\}$$ in E is called a $$(\\mathit{PS})_{c}$$ sequence of J if\n\n$$J(u_{n})\\to c, \\qquad J'(u_{n}) \\to0 \\quad \\mbox{in }E^{*}\\mbox{ as }n\\to\\infty.$$\n(2.8)\n\nThe functional J satisfies the $$(\\mathit{PS})$$ condition if any $$(\\mathit{PS})_{c}$$ sequence possesses a convergent subsequence in E.\n\n### Lemma 2.1\n\nAssume (V), (H1), and (H2) hold. If $$\\{u_{n}\\}\\subset E$$ is a $$(\\mathit{PS})_{c}$$ sequence of J, then $$\\{u_{n}\\}$$ is bounded in E.\n\n### Proof\n\nIt follows from Hölder’s inequality that\n\n\\begin{aligned} \\int_{{\\mathbb{R}}^{N}}|h_{1}||u_{n}|^{m}\\,dx \\leq&V_{0}^{-\\frac{m}{p}}\\biggl( \\int _{{\\mathbb{R}} ^{N}}|h_{1}|^{\\sigma}\\, dx \\biggr)^{\\frac{1}{\\sigma}}\\biggl( \\int_{{\\mathbb{R}} ^{N}}V|u_{n}|^{p}\\,dx \\biggr)^{\\frac{m}{p}} \\\\ \\leq&a_{1}\\|u_{n}\\|^{m}_{E}, \\end{aligned}\n(2.9)\n\nwhere $$a_{1}=V_{0}^{-{m}/{p}}\\|h_{1}\\|_{\\sigma}$$. Choose $$t\\in(0,1)$$ such that $$q=pt+(1-t)p_{*}$$, then\n\n\\begin{aligned} \\int_{{\\mathbb{R}}^{N}}h_{2}|u_{n}|^{q}\\,dx \\leq&\\biggl( \\int_{{\\mathbb {R}}^{N}}V|u_{n}|^{p}\\,dx \\biggr)^{t}\\biggl( \\int_{{\\mathbb{R}} ^{N}}|u_{n}|^{p_{*}}h_{2}^{-{\\frac{1}{1-t}}}V^{-\\frac{t}{1-t}} \\, dx\\biggr)^{1-t} \\\\ \\leq& a_{2}\\biggl( \\int_{{\\mathbb{R}}^{N}}V|u_{n}|^{p}\\,dx \\biggr)^{t}\\|\\Delta u_{n}\\| _{p}^{{(1-t)}p_{*}}\\leq a_{2}\\|u_{n}\\|^{q}_{E}, \\end{aligned}\n(2.10)\n\nwhere $$a_{2}=S_{*}^{-{p_{*}(q-p)}/{p(p_{*}-p)}}V_{0}^{-t}\\|h_{2}\\|_{\\infty}$$. Thus,\n\n\\begin{aligned} c+1+\\|u_{n}\\|_{E} \\geq& J(u_{n})-q^{-1}J'(u_{n})u_{n} \\\\ \\geq&\\biggl(\\frac{1}{p}-\\frac{1}{q}\\biggr)\\|u\\|_{E}^{p}- \\lambda\\biggl(\\frac{1}{m}-\\frac {1}{q}\\biggr) \\int_{{\\mathbb{R}}^{N}}|h_{1}||u|^{m}\\,dx \\\\ \\geq&\\biggl(\\frac{1}{p}-\\frac{1}{q}\\biggr)\\|u_{n} \\|^{p}_{E}-\\lambda\\biggl(\\frac {1}{m}-\\frac{1}{q} \\biggr)a_{1}\\|u_{n}\\|^{m}_{E}. \\end{aligned}\n(2.11)\n\nSince $$1< m< p< q$$, we conclude that $$\\|u\\|_{E}$$ is bounded and the proof is complete. □\n\nIn the following, we shall show that $$\\{u_{n}\\}$$ has a convergent subsequence in E. Since the sequence $$\\{u_{n}\\}$$ given by (2.8) is a bounded sequence in E, there exist a subsequence of $$\\{u_{n}\\}$$ (still denoted by $$\\{u_{n}\\}$$) and $$v\\in E$$ such that $$\\|u_{n}\\|_{E}\\leq M$$, $$\\|v\\|_{E}\\leq M$$, and\n\n\\begin{aligned}& u_{n}\\rightharpoonup v \\quad \\text{weakly in }E, \\\\& u_{n}\\to v \\quad \\text{in } L^{s}_{\\mathrm{loc}}\\bigl({ \\mathbb{R}}^{N}\\bigr), 1< s< p_{*}, \\\\& u_{n}(x)\\to v(x)\\quad \\text{a.e. in } {\\mathbb{R}}^{N}. \\end{aligned}\n(2.12)\n\n### Lemma 2.2\n\nAssume (V), (H1), and (H 2) hold. If the sequence $$\\{u_{n}\\}$$ is bounded in E satisfying (2.12), then\n\n1. (i)\n\n$$\\lim_{n\\to\\infty}\\int_{{\\mathbb{R}} ^{N}}h_{1}(x)|u_{n}|^{m}\\,dx=\\int_{{\\mathbb{R}}^{N}}h_{1}(x)|v|^{m}\\,dx$$, $$\\lim_{n\\to \\infty}\\int_{{\\mathbb{R}}^{N}}h_{1}(x)|u_{n}-v|^{m}\\,dx=0$$;\n\n2. (ii)\n\n$$\\lim_{n\\to\\infty}\\int_{{\\mathbb{R}} ^{N}}h_{2}(x)|u_{n}|^{q}\\,dx=\\int_{{\\mathbb{R}}^{N}}h_{2}(x)|v|^{q}\\,dx$$, $$\\lim_{n\\to \\infty}\\int_{{\\mathbb{R}}^{N}}h_{2}(x)|u_{n}-v|^{q}\\,dx=0$$.\n\n### Proof\n\n(i) In fact, from $$h\\in L^{\\sigma}({\\mathbb{R}}^{N})$$ and (2.12), we obtain, for any $$r>0$$,\n\n$$\\int_{B_{r}}h_{1}(x)|u_{n}|^{m}\\,dx \\to \\int_{B_{r}}h_{1}(x)|v|^{m}\\,dx \\quad \\mbox{as }n\\to \\infty,$$\n(2.13)\n\nwhere and in the sequel $$B_{r}=\\{x\\in{\\mathbb{R}}^{N}:|x|< r\\}$$, $$B^{c}_{r}={\\mathbb{R}}^{N}\\setminus \\overline{B}_{r}$$. On the other hand, we see from the Hölder inequality that\n\n\\begin{aligned} \\begin{aligned}[b] \\int_{B^{c}_{r}}|h_{1}||u_{n}|^{m}\\,dx& \\leq V_{0}^{-\\frac{m}{p}}\\biggl( \\int _{B^{c}_{r}}|h_{1}|^{\\sigma}\\, dx \\biggr)^{\\frac{1}{\\sigma}}\\biggl( \\int _{B^{c}_{r}}V|u_{n}|^{p}\\,dx \\biggr)^{\\frac{m}{p}} \\\\ &\\leq V_{0}^{-\\frac{m}{p}}\\|h_{1}\\|_{L^{\\sigma}(B^{c}_{r})} \\|u_{n}\\|^{m}_{E}\\leq V_{0}^{-\\frac{m}{p}}M^{m} \\|h_{1}\\|_{L^{\\sigma}(B^{c}_{r})}\\to0 \\end{aligned} \\end{aligned}\n(2.14)\n\nas $$r\\to\\infty$$. By Fatou’s lemma, we see that, as $$n\\to\\infty$$,\n\n$$\\int_{B^{c}_{r}}|h_{1}||v|^{m}\\,dx\\leq\\liminf _{n\\to\\infty} \\int _{B^{c}_{r}}|h_{1}||u_{n}|^{m}\\,dx \\leq V_{0}^{-\\frac{m}{p}}M^{m}\\|h_{1} \\|_{L^{\\sigma}(B^{c}_{r})}\\to0.$$\n(2.15)\n\nThen, the application of (2.13)-(2.15) gives the first limit of (i). Furthermore, by the Brezis-Lieb lemma in , we have the second limit of (i).\n\n(ii) To prove the conclusion (ii), we follow the argument used in . Here, we give a detailed proof for the reader’s convenience.\n\nSince $$p< p< p_{*}$$, it easy to see that, for any small $$\\varepsilon>0$$, there exist $$S_{0}>s_{0}>0$$ such that $$|s|^{q}<\\varepsilon|s|^{p}$$ if $$|s|\\leq s_{0}$$ and $$|s|^{q}\\leq\\varepsilon|s|^{p_{*}}$$, if $$|s|\\geq S_{0}$$. This shows that\n\n$$|s|^{q}\\leq\\varepsilon\\bigl(|s|^{p}+|s|^{p_{*}} \\bigr)+\\chi _{[s_{0},S_{0}]}\\bigl(\\vert s\\vert \\bigr)|s|^{q}, \\quad \\forall s\\in{\\mathbb{R}}.$$\n(2.16)\n\nDenote $$A_{n}=\\{x\\in{\\mathbb{R}}^{N};s_{0}\\leq|u_{n}(x)|\\leq S_{0}\\}$$. It follows from (2.16), (2.4), and (2.12) that\n\n\\begin{aligned} \\int_{B^{c}_{r}}|h_{2}||u_{n}|^{q}\\,dx \\leq&\\|h_{2}\\|_{\\infty}\\int_{B^{c}_{r}} \\bigl(\\varepsilon\\bigl(|u_{n}|^{p}+|u_{n}|^{p_{*}} \\bigr)+\\chi_{[s_{0},S_{0}]}\\bigl(\\vert u_{n}\\vert \\bigr)|u_{n}|^{q} \\bigr)\\, dx \\\\ \\leq&\\varepsilon\\|h_{2}\\|_{\\infty}\\biggl( \\int_{{\\mathbb{R}} ^{N}}V_{0}^{-1}V(x)|u_{n}|^{p} \\,dx+S^{-\\frac{p_{*}}{p}}\\|\\Delta u\\|^{p_{*}}_{p}\\biggr) \\\\ &{} +S_{0}^{q}\\|h_{2}\\|_{\\infty}\\operatorname{meas}\\bigl(A_{n}\\cap B^{c}_{r}\\bigr) \\\\ \\leq& M_{1}\\varepsilon+S_{0}^{q} \\|h_{2}\\|_{\\infty}\\operatorname{meas}\\bigl(A_{n}\\cap B^{c}_{r}\\bigr) \\end{aligned}\n(2.17)\n\nwith some constant $$M_{1}>0$$, and\n\n$$|s_{0}|^{p_{*}}|A_{n}|\\leq \\int_{{\\mathbb{R}}^{N}}|u_{n}|^{p_{*}}\\, dx\\leq M_{1},\\quad \\forall n\\in {\\mathbb{N}},$$\n(2.18)\n\nwhere $$|A_{n}|=\\operatorname{meas}(A_{n})$$. Equation (2.18) implies that $$\\sup_{n\\in{\\mathbb{N}}}|A_{n}|\\leq M_{1}|s_{0}|^{-p_{*}}<\\infty$$, so it is easy to see that\n\n$$\\lim_{r\\to\\infty}\\operatorname{meas} \\bigl(A_{n}\\cap B_{r}^{c}\\bigr)=0,\\quad \\mbox{for all } n\\in{\\mathbb{N}}.$$\n(2.19)\n\nIn the following, we show that $$\\lim_{r\\to\\infty }\\operatorname{meas}(A_{n}\\cap B_{r}^{c})=0$$ uniformly in $$n\\in{\\mathbb{N}}$$.\n\nIn fact, it follows from (2.12) that $$v\\in L^{p}({\\mathbb{R}}^{N})$$ and $$u_{n}(x)\\to v(x)$$ a.e. $${\\mathbb{R}}^{N}$$. Therefore, for any small $$\\varepsilon >0$$, there exists $$r_{0}>1$$ such that $$r\\geq r_{0}$$,\n\n$$\\int_{B_{r}^{c}}|v|^{p}\\,dx\\leq\\varepsilon.$$\n\nFor this ε, we choose $$t_{1}=r_{0}$$, $$t_{j}\\uparrow\\infty$$ such that $$D_{j}=B^{c}_{t_{j}}\\setminus\\overline{B}^{c}_{t_{j+1}}$$, $$B^{c}_{r_{0}}=\\bigcup^{\\infty}_{j=1}D_{j}$$ and\n\n$$\\int_{D_{j}}|v|^{p}\\, dx\\leq\\frac{\\varepsilon}{2^{j}},\\quad \\forall j\\in{\\mathbb{N}}.$$\n\nObviously, for every fixed $$j\\in N$$, $$D_{j}$$ is a bounded domain and $$D_{j}\\cap D_{i}=\\emptyset$$ ($$j\\neq i$$). Furthermore, $$s_{0}\\leq|u_{n}|\\leq S_{0}$$ in $$D_{j}\\cap A_{n}$$. By Fatou’s lemma, we have, for every $$j\\in {\\mathbb{N}}$$,\n\n$$\\limsup_{n\\to\\infty} \\int_{D_{j}\\cap A_{n}}|u_{n}|^{p}\\,dx\\leq \\int _{D_{j}}\\limsup_{n\\to\\infty}|u_{n}|^{p} \\,dx\\leq \\int _{D_{j}}|v|^{p}\\,dx\\leq\\frac{\\varepsilon}{2^{j}}.$$\n\nThen, for $$s_{1}=2^{1-q}s_{0}^{q}$$, we obtain\n\n\\begin{aligned} s_{1}\\limsup_{n\\to\\infty}\\bigl\\vert A_{n}\\cap B_{r_{0}}^{c}\\bigr\\vert \\leq&\\limsup _{n\\to\\infty} \\int_{B_{r_{0}}^{c}\\cap A_{n}}|u_{n}|^{p}\\,dx \\\\ =&\\limsup _{n\\to\\infty}\\sum^{\\infty}_{j=1} \\int_{D_{j}\\cap A_{n}}|u_{n}|^{p}\\,dx \\\\ \\leq&\\sum^{\\infty}_{j=1}\\limsup _{n\\to\\infty} \\int _{D_{j}\\cap A_{n}}|u_{n}|^{p}\\,dx \\\\ \\leq&\\sum ^{\\infty}_{j=1} \\int _{D_{j}}|v|^{p}\\,dx\\leq\\sum ^{\\infty}_{j=1}\\frac{\\varepsilon }{2^{j}}=\\varepsilon. \\end{aligned}\n(2.20)\n\nNotice that, for any $$r\\geq r_{0}$$ and $$n\\in{\\mathbb{N}}$$, we have $$(A_{n}\\cap B^{c}_{r})\\subset(A_{n}\\cap B_{r_{0}}^{c})$$. Therefore, the application of (2.19) and (2.20) yields $$\\lim_{r\\to\\infty}|A_{n}\\cap B_{r}^{c}|=0$$ uniformly in $$n\\in{\\mathbb{N}}$$. Thus, for any $$\\varepsilon>0$$, there exists $$r_{0}\\geq1$$ such that $$\\operatorname{meas}(A_{n}\\cap B^{c}_{r})<\\frac{\\varepsilon}{S_{0}^{q}\\|h_{2}\\| _{\\infty}}$$, for $$r\\geq r_{0}$$. Then it follows from (2.17) that\n\n$$\\int_{B^{c}_{r}}h_{2}|u_{n}|^{q}\\,dx \\leq\\max\\{M_{1},1\\}\\varepsilon,\\quad \\forall n\\in {\\mathbb{N}}, r\\geq r_{0}$$\n(2.21)\n\nand\n\n$$\\int_{B^{c}_{r}}h_{2}|v|^{q}\\,dx\\leq\\liminf _{n\\to\\infty} \\int _{B^{c}_{r}}h_{2}|u_{n}|^{q}\\,dx \\leq\\max\\{M_{1},1\\}\\varepsilon, \\quad r\\geq r_{0}.$$\n(2.22)\n\nMoreover, we derive from (2.12) that\n\n$$\\int_{B_{r}}h_{2}(x)|u_{n}|^{q}\\,dx \\to \\int_{B_{r}}h_{2}(x)|v|^{q}\\,dx.$$\n(2.23)\n\nTherefore, using (2.21) and (2.22), and the application of Brezis-Lieb lemma in we conclude the second limit of (ii). Then the proof is complete. □\n\n### Lemma 2.3\n\nLet $$\\{u_{n}\\}$$ be a $$(\\mathit{PS})_{c}$$ sequence satisfying (2.12), then $$u_{n}\\to v$$ in E, that is, the functional J satisfies the $$(\\mathit{PS})$$ condition.\n\n### Proof\n\nDenote\n\n\\begin{aligned} P_{n} =&J'(u_{n}) (u_{n}-v) \\\\ =& \\int_{{\\mathbb{R}}^{N}} \\bigl(|\\Delta u_{n}|^{p-2}\\Delta u_{n} \\Delta (u_{n}-v)+|\\nabla u_{n}|^{p-2} \\nabla u_{n}\\nabla(u_{n}-u) \\\\ &{}+V(x)|u_{n}|^{p-2}u_{n}(u_{n}-v) \\bigr)\\, dx \\\\ &{} - \\int_{{\\mathbb{R}}^{N}}\\bigl(\\lambda h_{1}(x)|u_{n}|^{m-2}u_{n}+h_{2}(x)|u_{n}|^{q-2}u_{n} \\bigr) (u_{n}-v) \\, dx. \\end{aligned}\n\nThen the fact $$J'(u_{n})\\to0$$ in $$E^{*}$$ shows that $$P_{n}\\to0$$ as $$n\\to \\infty$$. Moreover, the fact $$u_{n}\\rightharpoonup v$$ in E implies $$Q_{n}\\to0$$, where\n\n$$Q_{n}= \\int_{{\\mathbb{R}}^{N}}\\bigl(|\\Delta v|^{p-2}\\Delta v \\Delta (u_{n}-v)+|\\nabla u|^{p-2}\\nabla u\\nabla(u_{n}-u)+V(x)|v|^{p-2}v(u_{n}-v) \\bigr)\\, dx.$$\n\nIt follows from the Hölder inequality and the limit (i) in Lemma 2.2 that\n\n\\begin{aligned} \\int_{{\\mathbb{R}}^{N}}\\bigl\\vert h_{1}(x)\\bigr\\vert \\vert u_{n}\\vert ^{m-1}\\vert u_{n}-v\\vert \\, dx \\leq&\\biggl( \\int _{{\\mathbb{R}} ^{N}}\\bigl\\vert h_{1}(x)\\bigr\\vert \\vert u_{n}-v\\vert ^{m}\\,dx\\biggr)^{\\frac{1}{m}}\\biggl( \\int_{{\\mathbb{R}} ^{N}}\\bigl\\vert h_{1}(x)\\bigr\\vert \\vert u_{n}\\vert ^{m}\\,dx\\biggr)^{\\frac{m-1}{m}} \\\\ \\to&0. \\end{aligned}\n(2.24)\n\nSimilarly, we can derive from the limit (ii) in Lemma 2.2 that\n\n\\begin{aligned} \\int_{{\\mathbb{R}}^{N}}h_{2}(x)|u_{n}|^{q-1}|u_{n}-v| \\, dx \\leq&\\biggl( \\int_{{\\mathbb{R}} ^{N}}h_{2}(x)|u_{n}-v|^{q} \\,dx\\biggr)^{\\frac{1}{q}}\\biggl( \\int_{{\\mathbb {R}}^{N}}h_{2}(x)|u_{n}|^{q}\\,dx \\biggr)^{\\frac{q-1}{q}} \\\\ \\to&0. \\end{aligned}\n(2.25)\n\nThen (2.24) and (2.25) show that as $$n\\to\\infty$$\n\n\\begin{aligned} o_{n}(1) =&P_{n}-Q_{n} \\\\ =& \\int_{{\\mathbb{R}}^{N}}\\bigl(\\bigl(|\\Delta u_{n}|^{p-2} \\Delta u_{n}-|\\Delta v|^{p-2}\\Delta v\\bigr) \\Delta(u_{n}-v) \\\\ &{}+\\bigl(|\\nabla u_{n}|^{p-2}\\nabla u_{n}-|\\nabla v|^{p-2}\\nabla v\\bigr) \\nabla (u_{n}-v) \\\\ &{}+V(x) \\bigl(|u_{n}|^{p-2}u_{n}-|v|^{p-2}v\\bigr) (u_{n}-v)\\bigr)\\, dx. \\end{aligned}\n(2.26)\n\nThen we have $$\\|u_{n}-v\\|_{E}\\to0$$ as $$n\\to\\infty$$. Thus $$J(u)$$ satisfies the $$(\\mathit{PS})$$ condition on E and the proof is completed. □\n\n## Proof of Theorem 1.1\n\nIn this section, we will give the proof of Theorem 1.1. We assume that all conditions in the theorem hold. The proof mainly relies on the mountain pass theorem.\n\n### Lemma 3.1\n\n()\n\nLet E be an infinite dimensional real Banach space, $$J\\in C^{1}(E,{\\mathbb{R}})$$ be even and satisfies the $$(\\mathit{PS})$$ condition and $$J(0)=0$$. Assume $$E=Y\\oplus Z$$, Y is finite dimensional, and J satisfies:\n\n(J1):\n\nThere exist constants $$\\rho,\\alpha>0$$ such that $$J(u)\\leq \\alpha$$ on $$\\partial B_{\\rho}\\cap Z$$.\n\n(J2):\n\nFor each finite dimensional subspace $$E_{0}\\subset E$$, there is an $$R_{0}=R_{0}(E_{0})$$ such that $$J(u)\\leq0$$ on $$E_{0}\\setminus B_{R_{0}}$$, where $$B_{r}=\\{u\\in E: \\|u\\|_{E}< r\\}$$.\n\nThen J possesses an unbounded sequence of critical values.\n\n### Proof of Theorem 1.1\n\nClearly, the functional J defined by (2.6) is even in E. By Lemma 2.2 in Section 2, the functional satisfies the $$(\\mathit{PS})$$ condition. Next, we prove that J satisfies (J1) and (J2). From (2.9) and (2.10), it follows that\n\n$$J(u)\\geq\\frac{1}{p}\\|u\\|^{p}_{E}-\\lambda \\frac{a_{1}}{m}\\|u\\|^{m}_{E}-\\frac {a_{2}}{q}\\|u \\|^{q}_{E}, \\quad u\\in E.$$\n\nDenote\n\n$$\\phi(z)=z^{p}\\biggl(\\frac{1}{p}-\\lambda\\frac{a_{1}}{m}z^{m-p}- \\frac {a_{2}}{q}z^{q-p}\\biggr),\\quad z>0.$$\n\nThen, there exist $$\\lambda_{0},z_{1},\\alpha>0$$ such that $$\\phi(z_{1})\\geq \\alpha$$ for any $$\\lambda\\in[0,\\lambda_{0}]$$. Let $$\\rho=z_{1}$$, we have $$J(u)\\geq\\alpha$$ with $$\\|u\\|_{E}=\\rho$$ and $$\\lambda\\in[0,\\lambda _{0}]$$. So the condition (J1) is satisfied.\n\nWe now verify (J2). For any finite dimensional subspace $$E_{0}\\subset E$$, we assert that there is a constant $$R_{0}>\\rho$$ such that $$J<0$$ on $$E_{0}\\setminus B_{R_{0}}$$. Otherwise, there exists a sequence $$\\{u_{n}\\} \\subset E_{0}$$ such that $$\\|u\\|_{n}\\to\\infty$$ and $$J(u_{n})\\geq0$$. Hence\n\n$$\\frac{1}{p}\\|u_{n}\\|^{p}_{E} \\geq\\frac{\\lambda}{m} \\int _{R^{N}}h_{1}|u_{n}|^{m}\\,dx+ \\frac{1}{q} \\int_{{\\mathbb{R}}^{N}}h_{2}\\|u_{n}\\|^{q}\\,dx.$$\n(3.1)\n\nSet $$\\omega_{n}=\\frac{u_{n}}{\\|u_{n}\\|_{E}}$$. Then up to a sequence, we can assume $$\\omega_{n}\\rightharpoonup\\omega$$ in E, $$\\omega_{n}\\rightarrow \\omega$$ a.e. in $${\\mathbb{R}}^{N}$$. Denote $$\\Omega=\\{x\\in{\\mathbb {R}}^{N}:\\omega(x)\\neq 0\\}$$. Assume $$|\\Omega|>0$$. Clearly, $$u_{n}(x)\\to\\infty$$ in Ω. It follows from (2.8) and (2.9) that\n\n$$\\|u\\|^{-p}_{E} \\int_{\\Omega}|h_{1}||u_{n}|^{m}\\,dx \\leq a_{1}\\|u_{n}\\|^{m-p}_{E}\\to 0 \\quad \\mbox{as }n\\to\\infty.$$\n\nOn the other hand, we derive\n\n$$\\|u\\|^{-p}_{E} \\int_{\\Omega}|h_{2}||u_{n}|^{q}\\,dx \\leq a_{2}\\|u_{n}\\|^{q-p}_{E}\\to \\infty \\quad \\mbox{as }n\\to\\infty.$$\n\nTherefore, multiplying (3.1) by $$\\|u\\|_{E}^{-p}$$ and passing to the limit as $$n\\to\\infty$$ show that $$\\frac{1}{p}\\geq\\infty$$. This is impossible. So $$|\\Omega|=0$$ and $$\\omega(x)=0$$ a.e. on $${\\mathbb {R}}^{N}$$. By the equivalence of all norms in $$E_{0}$$, there exists a constant $$\\beta >0$$ such that\n\n$$\\int_{{\\mathbb{R}}^{N}}|h_{2}||u|^{q}\\,dx\\geq \\beta^{q}\\|u\\|^{q}_{E}, \\quad \\forall u\\in E_{0} \\quad \\mbox{and}\\quad \\int_{{\\mathbb{R}}^{N}}|h_{2}||u_{n}|^{q}\\,dx \\geq\\beta^{q}\\|u_{n}\\| ^{q}_{E}, \\quad \\forall n\\in{\\mathbb{N}}.$$\n\nHence\n\n$$0< \\beta^{q}\\leq\\lim\\sup_{n\\to\\infty} \\int_{{\\mathbb{R}} ^{N}}|h_{2}||\\omega_{n}|^{q} \\,dx\\leq \\int_{{\\mathbb{R}}^{N}}\\limsup_{n\\to \\infty}\\bigl(|h_{2}|| \\omega_{n}|^{q}\\bigr)\\, dx= \\int_{{\\mathbb{R}}^{N}}\\bigl(|h_{2}||\\omega|^{q}\\bigr)\\, dx=0.$$\n\nThis is a contradiction. So there exists a constant $$R_{0}$$ such that $$J<0$$ on $$E_{0}\\setminus B_{R_{0}}$$. Therefore, the existence of infinitely many solutions $$\\{u_{n}\\}$$ for problem (1.1) follows from Lemma 3.1 and we finish the proof of Theorem 1.1. □\n\n## References\n\n1. 1.\n\nCarrião, PC, Demarque, R, Miyagaki, OH: Nonlinear biharmonic problems with singular potentials. Commun. Pure Appl. Anal. 13, 2141-2154 (2014)\n\n2. 2.\n\nLiu, J, Chen, SX, Wu, X: Existence and multiplicity of solutions for a class of fourth-order elliptic equations in $${\\mathbb{R}}^{N}$$. J. Math. Anal. Appl. 395, 608-615 (2012)\n\n3. 3.\n\nChabrowski, J, Marcos do Ó, J: On some fourth-order semilinear elliptic problems in $${\\mathbb{R}}^{N}$$. Nonlinear Anal. 49, 861-884 (2002)\n\n4. 4.\n\nYin, Y, Wu, X: High energy solutions and nontrivial solutions for fourth-order elliptic equations. J. Math. Anal. Appl. 375, 699-705 (2011)\n\n5. 5.\n\nYe, YW, Tang, CL: Infinitely many solutions for fourth-order elliptic equations. J. Math. Anal. Appl. 394, 841-854 (2012)\n\n6. 6.\n\nYe, YW, Tang, CL: Existence and multiplicity of solutions for fourth-order elliptic equations in $${\\mathbb{R}}^{N}$$. J. Math. Anal. Appl. 406, 335-351 (2013)\n\n7. 7.\n\nZhang, W, Tang, X, Zhang, J: Infinitely many solutions for fourth-order elliptic equations with general potentials. J. Math. Anal. Appl. 407, 359-368 (2013)\n\n8. 8.\n\nZhang, G, Costa, DG: Existence result for a class of biharmonic equations with critical growth and singular potential in $${\\mathbb{R}}^{n}$$. Appl. Math. Lett. 29, 7-12 (2014)\n\n9. 9.\n\nBhakta, M, Musina, R: Entire solutions for a class of variational problems involving the biharmonic operator and Rellich potentials. Nonlinear Anal. 75, 3836-3848 (2012)\n\n10. 10.\n\nChen, CS: Multiple solutions for a class of quasilinear Schrödinger equations in $$\\mathbb{R}^{N}$$. J. Math. Phys. 56, 071507 (2015)\n\n11. 11.\n\nChen, CS: Infinitely many solutions to a class of quasilinear Schrödinger system in $$\\mathbb{R}^{N}$$. Appl. Math. Lett. 52, 176-182 (2016)\n\n12. 12.\n\nChen, CS, Chen, Q: Infinitely many solutions for p-Kirchhoff equation with concave-convex nonlinearities in $$\\mathbb{R}^{N}$$. Math. Methods Appl. Sci. (2015). doi:10.1002/mma.3583\n\n13. 13.\n\nRabinowitz, PH: Minimax Methods in Critical Point Theory with Application to Differential Equations. CBMS Reg. Conf. Ser. Math., vol. 65. Am. Math. Soc., Providence (1986)\n\n14. 14.\n\nSwanson, CA: The best Sobolev constant. Appl. Anal. 47, 227-239 (1992)\n\n15. 15.\n\nBrezis, H, Lieb, EH: A relation between pointwise convergence of functions and convergence of functionals. Proc. Am. Math. Soc. 88, 486-490 (1983)\n\n## Acknowledgements\n\nThe authors wish to express their gratitude to the referees for valuable comments and suggestions. This work was supported by the Project of Innovation in Scientific Research for Graduate Students of Jiangsu Province (No. CXZZ13-0263) and the Fundamental Research Funds for the Central Universities of China (2015B31014).\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Lihua Liu.\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n### Authors’ contributions\n\nEach of the authors contributed to each part of this study equally. All authors read and approved the final vision of the manuscript.\n\n## Rights and permissions",
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"https://boundaryvalueproblems.springeropen.com/track/article/10.1186/s13661-015-0510-6",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.757887,"math_prob":1.0000042,"size":16238,"snap":"2020-24-2020-29","text_gpt3_token_len":5849,"char_repetition_ratio":0.12923494,"word_repetition_ratio":0.0599233,"special_character_ratio":0.3857002,"punctuation_ratio":0.14753157,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000091,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-25T18:52:47Z\",\"WARC-Record-ID\":\"<urn:uuid:92d813c6-c094-4394-b46a-21c8ddd45970>\",\"Content-Length\":\"192674\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1c4a984a-2e7a-47bc-b030-ea65bb2d29d1>\",\"WARC-Concurrent-To\":\"<urn:uuid:76aaeea0-cffa-45da-9a37-e9699e758d2b>\",\"WARC-IP-Address\":\"199.232.64.95\",\"WARC-Target-URI\":\"https://boundaryvalueproblems.springeropen.com/articles/10.1186/s13661-015-0510-6\",\"WARC-Payload-Digest\":\"sha1:VKGXNC4J5WMYMOBIJILIC2W2VEX5EIJ6\",\"WARC-Block-Digest\":\"sha1:WUDJVGG6IBZQEGSKMNL4N6QH5KABMYUO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347389309.17_warc_CC-MAIN-20200525161346-20200525191346-00088.warc.gz\"}"} |
https://www.eegate.com/2014/10/electrical-resonance.html | [
"Electrical resonance - Electrical Engineering Gate\n\n# Electrical Engineering Gate\n\nOnline Electrical Engineering Study Site\n\n## Sunday, October 19, 2014\n\nElectrical resonance occurs in an electric circuit at a particular resonance frequency when the imaginary parts of impedances or admittances of circuit elements cancel each other. In some circuits this happens when the impedance between the input and output of the circuit is almost zero and the transfer function is close to one\n\nresonance: In an electrical circuit, the condition that exists when the inductive reactance and the capacitive reactance are of equal magnitude, causing electrical energy to oscillate between the magnetic field of the inductor and the electric field of the capacitor\n\nLC circuits\nResonance of a circuit involving capacitors and inductors occurs because the collapsing magnetic field of the inductor generates an electric current in its windings that charges the capacitor, and then the discharging capacitor provides an electric current that builds the magnetic field in the inductor. This process is repeated continually\n\nRLC circuits\nAn RLC circuit (or LCR circuit) is an electrical circuit consisting of a resistor, an inductor, and a capacitor, connected in series or in parallel. The RLC part of the name is due to those letters being the usual electrical symbols for resistanceinductance and capacitance respectively. The circuit forms a harmonic oscillator for current and resonatessimilarly to an LC circuit. The main difference stemming from the presence of the resistor is that any oscillation induced in the circuit decays over time if it is not kept going by a source. This effect of the resistor is called damping. The presence of the resistance also reduces the peak resonant frequency. Some resistance is unavoidable in real circuits, even if a resistor is not specifically included as a component. A pure LC circuit is an ideal that only exists in theory"
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https://testbook.com/question-answer/which-one-of-the-following-relations-with-usual-no--5e7c87bcf60d5d178e393513 | [
"Which one of the following relations with usual notations will hold good in a dynamic vibration absorber system under tuned conditions?\n\nThis question was previously asked in\nESE Mechanical 2019: Official Paper\nView all UPSC IES Papers >\n1. k1k2 = m1m2\n2. k1m2 = m1k2\n3. k1m1 = k2m2\n4. k1 + k2 = m1 + m2\n\nOption 2 : k1m2 = m1k2\nFree\nCT 1: Building Materials\n1209\n10 Questions 20 Marks 12 Mins\n\nDetailed Solution\n\nConcept:\n\nFor dynamics vibration absorber system under tuned condition,\n\nFrequency is equal i.e.\n\nf1 = f2\n\n$$\\sqrt {\\frac{{{k_1}}}{{{m_1}}} = } \\sqrt {\\frac{{{k_2}}}{{{m_2}}}}$$\n\n∴ k1m2 = m1k2"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.88826144,"math_prob":0.99597174,"size":4932,"snap":"2022-05-2022-21","text_gpt3_token_len":1277,"char_repetition_ratio":0.12094156,"word_repetition_ratio":0.2742857,"special_character_ratio":0.2570965,"punctuation_ratio":0.0990807,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9767586,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-28T16:01:09Z\",\"WARC-Record-ID\":\"<urn:uuid:e0ac9736-32a5-4a15-876c-63d846eb063a>\",\"Content-Length\":\"123963\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:23fcef8c-4893-4b9b-b15f-7de708aaa48e>\",\"WARC-Concurrent-To\":\"<urn:uuid:cda6e4ac-c35b-4201-ae58-69a187a8e279>\",\"WARC-IP-Address\":\"104.22.45.238\",\"WARC-Target-URI\":\"https://testbook.com/question-answer/which-one-of-the-following-relations-with-usual-no--5e7c87bcf60d5d178e393513\",\"WARC-Payload-Digest\":\"sha1:JFBYU2KIJY5DFCT3KKRYVSPFE36DVS6V\",\"WARC-Block-Digest\":\"sha1:V75YRC3QSJWU447JUVSBH2HAVAJL425G\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320306301.52_warc_CC-MAIN-20220128152530-20220128182530-00413.warc.gz\"}"} |
https://jiaocheng.hxsd.com/course/content/11381/ | [
"U I 设计\n\nIT 学院\n\n# 大气渲染技术解析！如何在游戏中模拟出千变万化的天空？\n\n1.1 介绍",
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"Figure 1: Top, a real photograph of the Earth from outer space. Bottom, a simulation using Nishita's model. These images have been copied from the paper \"Display of the Earth taking into account Atmospheric Scattering\" written by Nishita in 1993 (c) Siggraph.\n\n1.2 大气模型",
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"",
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"Figure 2: aerosols present in the atmosphere are responsible for haze.",
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"Figure 3: a simple graphical representation of the atmospheric model we will be using in this lesson. It is defined by the radius of the planet (Re) and the radius of the atmosphere (Ra). The atmosphere is made of aerosols (mainly present at low altitude) and air molecules. Density of particles decreases exponentially with altitude. This diagram is not to scale.",
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"Figure 4: the sun is so far away from the Earth, that each light ray reaching the Earth's atmosphere can be considered as being parallel to each other.\n\n1.3 瑞利散射",
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"Nishita的论文和我们在本篇中提到的其他论文的主要问题之一是他们要么没有给出N，n等的值来生成论文中显示的图像。在使用测量数据时，论文的作者通常不会引用它们的来源。在海平面上寻找分子密度的科学数据并不容易，如果你有任何关于此主题的有用链接或信息，你可以向我们提供，我们希望收到你的来信。我们在本篇中使用的海平面散射系数可以在两篇论文中找到：\"Efficient Rendering of Atmospheric Phenomena\", by Riley et al. and \"Precomputed Atmospheric Scattering\" by Bruneton et al.",
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"",
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"1.4 米氏散射",
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"1.5 光学深度的概念",
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"Figure 6: the camera can either be inside the atmosphere or outside. When it is inside, we are only interested in the intersection point between the camera ray and the atmosphere. When the camera is outside it can intersect the atmosphere in two points.",
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"Figure 7: single scattering is responsible for the sky color. A viewer is rarely directly looking at the sun (which is dangerous). However when we are looking away from the sun the atmosphere has a color which is the result of blue light from the sunlight being scattered in the direction of the observer's eyes.",
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"Figure 8: to compute the sky color we first trace a ray from Pc to Pa and then integrate the amount of light coming from the sun (with direction L) that is reflected back along that ray (V).",
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"Figure 9: as light (Lb) travels from Pb to Pa, it gets attenuated because of out-scattering and absorption. The result is La. Transmittance is the amount of light received at Pa from Pb, after it was attenuated while traveling through the atmosphere (that is T=La/Lb).",
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"1.6 将阳光加起来",
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"1.7 计算天空颜色",
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"",
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"vec2 uv =(2.*fragCoord/iResolution.xy1.)*vec2(iResolution.x/iResolution.y,1.);\n\nfloat viewScale = 1.;\n\nvec2 p = uv*viewScale;\n\n//单位正方形内的点映射到半球面上\n\nfloat z2 = p.x*p.x+p.y*p.y;\n\nfloat phi = atan(p.y,p.x);\n\nfloat theta = acos(1.-z2);\n\nvec3 dir = vec3(sin(theta)*cos(phi),\n\ncos(theta),\n\nsin(theta)*sin(phi));\n\ncolor = getIncidentLight(r);\n\nfragColor = vec4(color,1.);\n\n//sun position\n\nfloat rotSpeedFactor = 3.;\n\nmat3 sunRot = rotate_around_x(-abs(sin(u_time/rotSpeedFactor))*90.);\n\nsunDir *= sunRot;\n\n//绕x轴的旋转矩阵\n\nmat3 rotate_around_x(const in float degrees)\n\n{\n\nfloat _sin = sin(angle);\n\nfloat _cos = cos(angle);\n\nreturn mat3(1.,0.,0.,\n\n0.,_cos,-_sin,\n\n0.,_sin,_cos);\n\n}\n\n#define PI 3.14159265359\n\n#define USE_HENYEY 1\n\n#define u_res iResolution\n\n#define u_time iTime\n\nvec3 sunDir = vec3(0.,1.,0.);\n\nfloat sunPower = 20.;\n\nconst int numSamples = 16;\n\nconst int numSamplesLight = 8;//光线与大气层顶端交点到采样点的采样数量\n\nconst float hR = 7994.;//rayleigh\n\nconst float hM = 1200.;//mie\n\nconst vec3 betaR = vec3(5.5e-6,13.0e-6,22.4e-6);//rayleigh\n\nconst vec3 betaM = vec3(21e-6);//mie\n\nstruct ray\n\n{\n\nvec3 origin;\n\nvec3 direction;\n\n};\n\nstruct sphere\n\n{\n\nvec3 center;\n\n};\n\n//表示大气层的球\n\n//计算射线与球是否相交以及交点\n\nbool raySphereIntersect(const in ray r,const in sphere s,inout float t0,inout float t1)\n\n{\n\nvec3 oc = r.origin-s.center;\n\nfloat a = dot(r.direction,r.direction);\n\nfloat b = dot(oc,r.direction);\n\nfloat discriminant = b*b-(a*c);\n\nif(discriminant>0.0)\n\n{\n\nfloat tmp = sqrt(discriminant);\n\nt0 = -b-tmp;\n\nt1 = -b+tmp;\n\nreturn true;\n\n}\n\nreturn false;\n\n}\n\n//----------------phase function----------------\n\nfloat rayleighPhaseFunc(float mu)\n\n{ return 3.*(1.+mu*mu)/(16.*PI);\n\n}\n\nconst float g = 0.76;\n\nfloat henyeyGreensteinPhaseFunc(float mu)\n\n{\n\nreturn (1.-g*g)/((4.*PI)*pow(1.+g*g-2.*g*mu,1.5));\n\n}\n\nconst float k = 1.55*g-0.55*(g*g*g);\n\nfloat schlickPhaseFunc(float mu)\n\n{\n\nreturn (1.-k*k)/(4.*PI*(1.+k*mu)*(1.+k*mu));\n\n}\n\n//----------------phase function----------------",
null,
"//计算光学深度\n\nbool getSunLight(const in ray r,inout float opticalDepthR,inout float opticalDepthM)\n\n{\n\nfloat t0,t1;\n\nraySphereIntersect(r,atmosphere,t0,t1);\n\nfloat marchPos = 0.;\n\nfloat marchStep = t1/float(numSamplesLIght);\n\nfor(int i = 0;i\n\n{\n\n//相邻两个采样点的中点\n\nvec3 s =r.origin+r.direction*(marchPos+0.5*marchStep);\n\nif(height<0.)\n\nreturn false;\n\nopticalDepthR += exp(-height/hR)*marchStep;\n\nopticalDepthM += exp(-height/hM)*marchStep;\n\nmarchPos += marchStep;\n\n}\n\nreturn true;\n\n}\n\nvec3 getIncidentLight(const in ray r)\n\n{\n\nfloat t0,t1;\n\nif(!raySphereIntersect(r,atmosphere,t0,t1))\n\n{\n\nreturn vec3(0.);\n\n}\n\nfloat marchStep = t1/float(numSamples);\n\nfloat mu = dot(r.direction,sunDir);\n\nfloat phaseR = rayleighPhaseFunc(mu);\n\nfloat phaseM =\n\n#if USE_HENYEY\n\nhenyeyGreensteinPhaseFunc(mu);\n\n#else\n\nschlickPhaseFunc(mu);\n\n#endif\n\nfloat opticalDepthR = 0.;\n\nfloat opticalDepthM = 0.;\n\nvec3 sumR = vec3(0.);\n\nvec3 sumM = vec3(0.);\n\nfloat marchPos = 0.;\n\nfor(int i=0;i\n\n{\n\nvec3 s = r.origin+r.direction*(marchPos+0.5*marchStep);\n\n//计算光学深度累加和\n\nfloat hr = exp(-height/hR)*marchStep;\n\nfloat hm = exp(-height/hM)*marchStep;\n\nopticalDepthR += hr;\n\nopticalDepthM += hm;\n\nray lightRay = ray(s,sunDir);\n\nfloat opticalDepthLightR = 0.;\n\nfloat opticalDepthLightM = 0.;\n\nbool bOverGround = getSunLight(lightRay,opticalDepthLightR,opticalDepthLightM);\n\nif(bOverGround)\n\n{\n\nvec3 t = betaR*(opticalDepthR+opticalDepthLightR)+\n\nbetaM*1.1*(opticalDepthM+opticalDepthLightM);\n\nvec3 attenuation = exp(-t);\n\nsumR += hr*attenuation;\n\nsumM += hm*attenuation;\n\n}\n\nmarchPos += marchStep;\n\n}\n\nreturn sunPower*(sumR*phaseR*betaR+sumM*phaseM*betaM);\n\n}",
null,
"",
null,
"•",
null,
"2101期学员李思庭作品\n\n•",
null,
"2104期学员林雪茹作品\n\n•",
null,
"2107期学员赵凌作品\n\n•",
null,
"2107期学员赵燃作品\n\n•",
null,
"2106期学员徐正浩作品\n\n•",
null,
"2106期学员弓莉作品\n\n•",
null,
"2105期学员白羽新作品\n\n•",
null,
"2107期学员王佳蕊作品\n\n• c语言更难一些，其实Python的底层就是c语言实现的，如果你想深入的话，建议先学c语言，在学习Python。如果只是想掌握一......\n\n• 在我国目前很多数字媒体专业在国内外都有很多学校开设，尤其是近几年来很多院校都开设了数字媒体专业的课程，不过目前很多同学除了可以......\n\n• 在当下信息爆炸时代，很多碎片化信息无法被大家精准捕捉到，很多投资商想要吸引到大家都会通过广告的方式，但是广告的种类有很多，每个......\n\n• 每个人都有一个成为UP主的心，但是对于成为一个UP主需要掌握的专业知识和技能却了解不够的深入，想要制作一个精美的视频，剪辑是首......\n\n• 原画是一个很受年轻人喜欢的行业，也是当前发展比较好的方向，不管是游戏原画还是影视原画，都是极为火热且人才缺口大的职业。而想要成......\n\n• 随着动画行业的发展，cg三维动画因为不受时间、空间、地点、条件、对象的限制，能够运用各种表现形式把复杂、抽象的节目内容、科学原......",
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https://www.numberempire.com/101761?number=101761 | [
"Home | Menu | Get Involved | Contact webmaster",
null,
"",
null,
"",
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"",
null,
"",
null,
"# Number 101761\n\none hundred one thousand seven hundred sixty one\n\n### Properties of the number 101761\n\n Factorization 11 * 11 * 29 * 29 Divisors 1, 11, 29, 121, 319, 841, 3509, 9251, 101761 Count of divisors 9 Sum of divisors 115843 Previous integer 101760 Next integer 101762 Is prime? NO Previous prime 101749 Next prime 101771 101761st prime 1324429 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? square(319) Binary 11000110110000001 Octal 306601 Duodecimal 4aa81 Hexadecimal 18d81 Square 10355301121 Square root 319 Natural logarithm 11.53038220557 Decimal logarithm 5.0075813661144 Sine -0.99484708814177 Cosine 0.10138674082858 Tangent -9.8123983472729\nNumber 101761 is pronounced one hundred one thousand seven hundred sixty one. Number 101761 is a composite number. Factors of 101761 are 11 * 11 * 29 * 29. Number 101761 has 9 divisors: 1, 11, 29, 121, 319, 841, 3509, 9251, 101761. Sum of the divisors is 115843. Number 101761 is not a Fibonacci number. It is not a Bell number. Number 101761 is not a Catalan number. Number 101761 is not a regular number (Hamming number). It is a not factorial of any number. Number 101761 is a deficient number and therefore is not a perfect number. Number 101761 is a square number with n=319. Binary numeral for number 101761 is 11000110110000001. Octal numeral is 306601. Duodecimal value is 4aa81. Hexadecimal representation is 18d81. Square of the number 101761 is 10355301121. Square root of the number 101761 is 319. Natural logarithm of 101761 is 11.53038220557 Decimal logarithm of the number 101761 is 5.0075813661144 Sine of 101761 is -0.99484708814177. Cosine of the number 101761 is 0.10138674082858. Tangent of the number 101761 is -9.8123983472729"
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https://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-r-section-r-5-factoring-polynomials-r-5-assess-your-understanding-page-57/49 | [
"## College Algebra (10th Edition)\n\n$(x-8)(x-2)$\nRECALL: A trinomial of the form $x^2+bx+c$ can be factored if there are integers $d$ and $e$ such that $c=de$ and $b=d+e$. The trinomial's factored form will be: $x^2+bx+c=(x+d)(x+e)$ The given trinomial has $b=-10$ and $c=16$. Note that $16=(-8)(-2)$ and $-10=(-8)+(-2)$. This means that $d=-8$ and $e=-2$ Thus, the factored form of the trinomial is: $=[x+(-8)] [x+(-2)] \\\\=(x-8)(x-2)$"
] | [
null
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https://testbook.com/question-answer/the-height-of-a-right-circular-cone-is-5-cm-and-it--5e75d63cf60d5d7f3bfc4b10 | [
"# The height of a right circular cone is 5 cm and its base radius is 12 cm. What is the curved surface area of the cone?\n\nThis question was previously asked in\nSSC MTS Previous Paper 3 (Held On: 2 August 2019 Shift 3)\nView all SSC MTS Papers >\n1. 156π cm2\n2. 132π cm2\n3. 143π cm2\n4. 168π cm2\n\nOption 1 : 156π cm2\n\n## Detailed Solution\n\nh = 5 cm, r = 12 cm\n\nLet the lateral height be l,\n\n⇒ l2 = r2 + h2\n\n⇒ l2 = 144 + 25 = 169\n\n⇒ l = 13 cm\n\n⇒ The curved surface area of the cone = π × r × l\n\n⇒ The curved surface area of the cone = π × 12 × 13 = 156 π cm2\n\n∴ The curved surface area of the cone is 156 π cm2.\n\nFree\nSSC MTS Full Mock Test\n258670\n100 Questions 100 Marks 90 Mins"
] | [
null
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https://www.chegg.com/homework-help/use-figurewrite-expression-area-region-chapter-14.7-problem-3e-solution-9780395919323-exc | [
"Solutions\nAlgebra and Geometry, Grade 8\n\n# Algebra and Geometry, Grade 8 (0th Edition) Edit edition Problem 3E from Chapter 14.7: Use the figure.Write an expression for the area of each region.\n\nWe have solutions for your book!\nChapter: Problem:\n\nUse the figure.",
null,
"Write an expression for the area of each region.\n\nStep-by-step solution:\nChapter: Problem:\n• Step 1 of 5\n\nDistributive property: it states that when we multiply a number by a group of numbers which are added together is the same as doing each multiplication separately.\n\nConsider the following figure.",
null,
"Since the region 1 constitute a rectangle.\n\nSo, area of region 1",
null,
"Where l and",
null,
"represents the breadth and the vertical height of the triangle respectively.\n\nHere,",
null,
"And",
null,
"Hence, area of region 1",
null,
"Use distributive property",
null,
"Thus, area of region 1",
null,
"",
null,
"Therefore, expression for the area of region 1 is",
null,
"• Chapter , Problem is solved.\nCorresponding Textbook",
null,
"Algebra and Geometry, Grade 8 | 0th Edition\n9780395919323ISBN-13: 0395919320ISBN: LarsonAuthors:\nThis is an alternate ISBN. View the primary ISBN for: Passport to Algebra and Geometry, Grade 8 0th Edition Textbook Solutions"
] | [
null,
"https://mgh-images.s3.amazonaws.com/9780395879887/14580-14.7-3IE1.png",
null,
"https://chegg-html-solutions.s3.amazonaws.com/9780395879887/14580-14.7-3E-i1.png",
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"https://chegg-html-solutions.s3.amazonaws.com/9780395879887/14580-14.7-3E-i2.png",
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"https://chegg-html-solutions.s3.amazonaws.com/9780395879887/14580-14.7-3E-i3.png",
null,
"https://chegg-html-solutions.s3.amazonaws.com/9780395879887/14580-14.7-3E-i4.png",
null,
"https://chegg-html-solutions.s3.amazonaws.com/9780395879887/14580-14.7-3E-i5.png",
null,
"https://chegg-html-solutions.s3.amazonaws.com/9780395879887/14580-14.7-3E-i6.png",
null,
"https://chegg-html-solutions.s3.amazonaws.com/9780395879887/14580-14.7-3E-i7.png",
null,
"https://chegg-html-solutions.s3.amazonaws.com/9780395879887/14580-14.7-3E-i8.png",
null,
"https://chegg-html-solutions.s3.amazonaws.com/9780395879887/14580-14.7-3E-i9.png",
null,
"https://chegg-html-solutions.s3.amazonaws.com/9780395879887/14580-14.7-3E-i10.png",
null,
"https://cs.cheggcdn.com/",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.85266167,"math_prob":0.73856515,"size":639,"snap":"2019-13-2019-22","text_gpt3_token_len":143,"char_repetition_ratio":0.16692914,"word_repetition_ratio":0.08571429,"special_character_ratio":0.20813772,"punctuation_ratio":0.12598425,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.998341,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-22T18:09:55Z\",\"WARC-Record-ID\":\"<urn:uuid:030314ca-5ca4-417f-b650-29b9ac935aa1>\",\"Content-Length\":\"134517\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7ed654ce-af27-45ba-9f42-2f2c8f40427d>\",\"WARC-Concurrent-To\":\"<urn:uuid:1cbb8e1c-236d-4a1e-b800-9f98e97045cf>\",\"WARC-IP-Address\":\"35.162.211.150\",\"WARC-Target-URI\":\"https://www.chegg.com/homework-help/use-figurewrite-expression-area-region-chapter-14.7-problem-3e-solution-9780395919323-exc\",\"WARC-Payload-Digest\":\"sha1:MZLCQG2E5ZGOILXONDDXCF6ISC6RCKAS\",\"WARC-Block-Digest\":\"sha1:K4CIJOB27SNC74QN56432RP7GGW4B3EK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202688.89_warc_CC-MAIN-20190322180106-20190322202106-00425.warc.gz\"}"} |
https://discuss.codechef.com/t/random-network-topology-generator/7952 | [
"",
null,
"# Random Network Topology Generator\n\nNetwork Topology:\nInput: Number of nodes\nMaximum Edges\nMinimum Edges\nIn this first we pick a random node form the given input which is the root node then we should pick another node from the given input and connect to the root node whose edge is 1 and pick another node from the input and connect to the root. This process is continued until the edges are equal to the maximum and minimum edges. In this way we pick all the nodes from the network and generate the topology.\nOutput: network with connected edges"
] | [
null,
"https://s3.amazonaws.com/discourseproduction/original/3X/7/f/7ffd6e5e45912aba9f6a1a33447d6baae049de81.svg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8803663,"math_prob":0.9261877,"size":528,"snap":"2022-40-2023-06","text_gpt3_token_len":112,"char_repetition_ratio":0.15076336,"word_repetition_ratio":0.04347826,"special_character_ratio":0.19507575,"punctuation_ratio":0.067961164,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9724532,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T01:52:42Z\",\"WARC-Record-ID\":\"<urn:uuid:52cf389e-911d-4556-93af-9c735159c5f7>\",\"Content-Length\":\"11275\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7ac70099-80f3-4e4d-84b3-b20fbf48666e>\",\"WARC-Concurrent-To\":\"<urn:uuid:a54433c8-7adb-49bf-a510-b7cf16fed025>\",\"WARC-IP-Address\":\"52.54.174.7\",\"WARC-Target-URI\":\"https://discuss.codechef.com/t/random-network-topology-generator/7952\",\"WARC-Payload-Digest\":\"sha1:ZUOHZYPRM6O545JBXX3VNJHABPL7T2GW\",\"WARC-Block-Digest\":\"sha1:7UMVMTJAD6SRTYIR4XULRBNZFHUP3GIG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337531.3_warc_CC-MAIN-20221005011205-20221005041205-00086.warc.gz\"}"} |
https://www.fujiitoshiki.com/improvesociety/?p=2439 | [
"# How to calculate appropriate sample size in Cox proportional hazard analysis with cross tabulation?\n\nPocket\n\nIn this article, I’d like to describe how to calculate appropriate sample size in Cox proportional analysis with cross tabulation, a error and b error. a error is called as statistical significance or type 1 error and b error is called as type 2 error, respectively. 1 – b is called as statistical power. a is usually configured at 0.05 (two-tailed) and b is configured at 0.2 (one-sided), respectively. As a result, Za/2 is 1.96 and Zb is 0.84, respectively.\n\nI’d like to assume that S1 is survival rate of risk group or intervention group and S0 is survival rate of control group, without risk or intervention. q is ratio of logarithm of them.",
null,
"$\\displaystyle LN(S_1) = Exp(B)LN(S_0)\\vspace{0.1in}\\\\\\theta = Exp(B) = \\frac{LN(S_1)}{LN(S_0)}$\n\nI’d like to use cross tabulation here. You can replace endpoint with death or failure.\n\n ENDPOINT CENSOR Marginal total POSITIVE a b a + b NEGATIVE c d c + d Marginal total a + c b + d N",
null,
"$\\displaystyle S_1 = \\frac{b}{a+b}\\vspace{0.1in}\\\\S_0 = \\frac{d}{c+d}$\n\nYou can calculate estimated number of death (e) in both group as following formula by Freedman’s approximate calculation.",
null,
"$\\displaystyle e = \\left(\\frac{\\theta+1}{\\theta-1}\\right)^2(Z_{\\alpha/2}+Z_\\beta)^2$\n\nYou can calculate entry size (n) in each group, as following formula.",
null,
"$\\displaystyle e = n(1-S_0)+n(1-S_1)\\vspace{0.1in}\\\\n = \\frac{e}{2 - S_0 - S_1}$\n\nYou have to correct entry size with drop-out rate (w) as following formula. Throughout trial, two times of n is needed.",
null,
"$\\displaystyle n = \\frac{e}{(2 - S_0 - S_1)(1-w)}$\n\nPocket",
null,
""
] | [
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://secure.gravatar.com/avatar/91ab61b5e6905e4505ecc130ce15522d",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.94123137,"math_prob":0.9989716,"size":1156,"snap":"2020-34-2020-40","text_gpt3_token_len":289,"char_repetition_ratio":0.09375,"word_repetition_ratio":0.0,"special_character_ratio":0.24653979,"punctuation_ratio":0.1092437,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998894,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-04T04:56:27Z\",\"WARC-Record-ID\":\"<urn:uuid:7ea3be23-0432-4700-9d13-62f31db766b5>\",\"Content-Length\":\"56193\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:06cabc37-54d8-40a9-b390-aaa22e3d8365>\",\"WARC-Concurrent-To\":\"<urn:uuid:190a2ca4-1c97-4280-a0f8-b3fe86901c9b>\",\"WARC-IP-Address\":\"183.90.240.38\",\"WARC-Target-URI\":\"https://www.fujiitoshiki.com/improvesociety/?p=2439\",\"WARC-Payload-Digest\":\"sha1:TIN2LZSPNMAGKR5W7GEFUSWFRV4S7KHD\",\"WARC-Block-Digest\":\"sha1:SSWXPM6MEDIDGJJVDYWUAB2JWHLNSIK2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735860.28_warc_CC-MAIN-20200804043709-20200804073709-00038.warc.gz\"}"} |
https://math.stackexchange.com/questions/2763434/obtaining-a-positive-definite-covariance-matrix-of-order-statistics | [
"Obtaining a positive definite covariance matrix of order statistics\n\nSuppose $X_1,\\dots,X_n$ are independent samples from some distribution with known absolutely continuous CDF $F:\\mathbb{R}\\rightarrow[0,1]$. Let $X_{(1)},\\dots,X_{(n)}$ denote the order statistics, i.e. the ordered sample. Defining the column vector $X_{(\\cdot)}=[X_{(1)},\\dots,X_{(n)}]'$, we want to numerically calculate $\\mathbb{E}[(X_{(\\cdot)}-\\mathbb{E}X_{(\\cdot)})(X_{(\\cdot)}-\\mathbb{E}X_{(\\cdot)})']$.\n\nThe most obvious approach uses the fact that for $j,k\\in\\{1,\\dots,n\\}$, $j<k$:\n\n$$\\mathbb{E}X_{(k)}=\\int{x\\frac{n! [F(x)]^{k-1}[1-F(x)]^{n-k}}{(k-1)!(n-k)!} dF(x)},$$ $$\\mathbb{E}X_{(k)}^2=\\int{x^2\\frac{n! [F(x)]^{k-1}[1-F(x)]^{n-k}}{(k-1)!(n-k)!} dF(x)},$$ $$\\mathbb{E}X_{(j)}X_{(k)}=\\int{\\int{xy\\frac{n! [F(x)]^{j-1}[F(y)-F(x)]^{k-1-j}[1-F(y)]^{n-k}}{(j-1)!(k-j-1)!(n-k)!} 1[x\\le y] dF(x) } dF(y)},$$\n\nwhere $1[\\cdot]$ is the indicator function. See e.g. Wikipedia here for an informal proof.\n\nUsing these formulae with standard numerical integration methods works well for small $n$. However, for large $n$ the resulting covariance matrix often ends up non-positive definite unless implausibly many integration nodes are used. This is despite the individual elements of the covariance matrix usually being (loosely) close to a covariance matrix generated via a naïve Monte Carlo approach that guarantees positive definiteness (i.e. draw such a sample of length $n$, then sort it, repeat this lots of times, take the covariance).\n\nIs it possible to express these integrals in such a way that the resulting covariance matrix is guaranteed to be positive definite?\n\nE.g. is it the case that:\n\n$$\\mathbb{E}[(X_{(\\cdot)}-\\mathbb{E}X_{(\\cdot)})(X_{(\\cdot)}-\\mathbb{E}X_{(\\cdot)})']=\\int{\\int{g(u,v) dF(u)}dF(v)},$$\n\nfor some function $g:\\mathbb{R}^2\\rightarrow \\mathbb{R}^{n\\times n}$ where $g(u,v)$ is positive semi-definite for all $u,v\\in\\mathbb{R}$.\n\n• Nice question. Did you try to go further when $F$ is uniform on $[0,1]$? – Did May 8 '18 at 15:42\n• Is there a reason why you don't want to use the Monte Carlo approach? – Mike Hawk May 8 '18 at 15:48\n• @Did I don't think the uniform case is actually any easier. The $F$ is not really the source of difficulty. I will have a play though. – cfp May 8 '18 at 16:02\n• @MikeHawk The Monte Carlo approach requires you to sample from a space of dimension $n$. While with $T$ samples the error is on the order of $\\frac{1}{\\sqrt{T}}$, in practice with a large $n$ (e.g. 2000), the constant is huge, and obtaining e.g. 1e-8 precision is essentially impossible. – cfp May 8 '18 at 16:05\n• Well, at least when $F$ is uniform, one knows every $E(X_{(k)})$, and, even though these do not appear in your computations, they seem to be needed to compute the covariance matrix. – Did May 8 '18 at 16:22"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6832954,"math_prob":0.99983555,"size":1852,"snap":"2019-43-2019-47","text_gpt3_token_len":637,"char_repetition_ratio":0.14069264,"word_repetition_ratio":0.0,"special_character_ratio":0.34989202,"punctuation_ratio":0.13625304,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999895,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-18T04:22:58Z\",\"WARC-Record-ID\":\"<urn:uuid:09e72383-cd69-4882-bd58-768a476a7481>\",\"Content-Length\":\"137156\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:11661dac-5421-4a50-a48a-7591c1e2bca0>\",\"WARC-Concurrent-To\":\"<urn:uuid:3bbedcbd-0d1e-40d8-9bbe-7f75270215a0>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/2763434/obtaining-a-positive-definite-covariance-matrix-of-order-statistics\",\"WARC-Payload-Digest\":\"sha1:IJO352HREUZN6CIRJFVYMPODF5G23JUC\",\"WARC-Block-Digest\":\"sha1:U7LJKZXSY45FZDWLAQJUYTU67IPK4WR7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986677884.28_warc_CC-MAIN-20191018032611-20191018060111-00082.warc.gz\"}"} |
https://www.nagwa.com/en/videos/178143989742/ | [
"# Video: Arranging Fractions with the Same Denominator\n\nComplete 4/9 ＿ 7/9 using <, = or >.\n\n02:35\n\n### Video Transcript\n\nComplete four-ninths what seven-ninths using the symbol for is less than, is equal to, or is greater than.\n\nIn this question, we’re being asked to compare two fractions, four-ninths and seven-ninths. And we can see in between them in the question there’s a gap. We need to complete this gap using a symbol that compares both fractions together. Is four-ninths less than seven-ninths, equal to seven-ninths, or greater than seven-ninths? To help us choose the right symbol, let’s take a moment to look more closely at these fractions. What do we notice?\n\nWell, the first thing we can see is that they both show a number of ninths. The denominator in both fractions is the same. We know that a denominator shows the number of equal parts as a whole that has been split into. So if they’re both split into the same amount, we can say that the size of these parts is the same. And because, as we’ve said, the size of the parts in these fractions is the same, all we have to do is to compare the numerators, to compare the number of these parts.\n\nComparing fractions when they both have the same denominator is quite straightforward. We know that four is less than seven. So four-ninths is less as a fraction than seven-ninths. Let’s show this using our diagram. Here’s what four-ninths looks like, four out of nine. And here is what seven-ninths looks like, seven out of a possible nine. And this diagram shows us visually that four-ninths is less than seven-ninths. So the symbol we need to use in between both of these fractions is the one that represents is less than.\n\nBecause our two fractions have the same denominator, we compared them just by looking at the numerator. Having the same denominator meant that the size of each part was the same. So we simply needed to look at the number of parts that each fraction was referring to. Four parts is less than seven parts. So we can say that four-ninths is less than seven-ninths.\n\nThe correct symbol to compare these fractions is the one that represents is less than."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9589394,"math_prob":0.97528756,"size":2030,"snap":"2019-51-2020-05","text_gpt3_token_len":462,"char_repetition_ratio":0.17423494,"word_repetition_ratio":0.04481793,"special_character_ratio":0.2093596,"punctuation_ratio":0.09859155,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9971767,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-08T21:50:55Z\",\"WARC-Record-ID\":\"<urn:uuid:e1afe6a8-e8d5-4e77-819a-074161d4f049>\",\"Content-Length\":\"25607\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4f60dee6-ea3c-440d-8d88-6a507ca94f61>\",\"WARC-Concurrent-To\":\"<urn:uuid:50a18799-7bf4-4b3b-b934-27d60b75dea3>\",\"WARC-IP-Address\":\"52.87.1.166\",\"WARC-Target-URI\":\"https://www.nagwa.com/en/videos/178143989742/\",\"WARC-Payload-Digest\":\"sha1:BAOOLWMOR3HRMFKVHY64RSBKEDQWPGOJ\",\"WARC-Block-Digest\":\"sha1:JGXXHQ6BCOHBZNEJZPB26MQIKARG5IFT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540514893.41_warc_CC-MAIN-20191208202454-20191208230454-00064.warc.gz\"}"} |
https://urho3d.github.io/documentation/1.7.1/class_urho3_d_1_1_matrix4.html | [
"Urho3D::Matrix4 Class Reference\n\n4x4 matrix for arbitrary linear transforms including projection. More...\n\n`#include <Urho3D/Math/Matrix4.h>`\n\nCollaboration diagram for Urho3D::Matrix4:\n[legend]\n\n## Public Member Functions\n\nMatrix4 ()\nConstruct an identity matrix.\n\nMatrix4 (const Matrix4 &matrix)\nCopy-construct from another matrix.\n\nMatrix4 (const Matrix3 &matrix)\nCopy-construct from a 3x3 matrix and set the extra elements to identity.\n\nMatrix4 (float v00, float v01, float v02, float v03, float v10, float v11, float v12, float v13, float v20, float v21, float v22, float v23, float v30, float v31, float v32, float v33)\nConstruct from values.\n\nMatrix4 (const float *data)\nConstruct from a float array.\n\nMatrix4operator= (const Matrix4 &rhs)\nAssign from another matrix.\n\nMatrix4operator= (const Matrix3 &rhs)\nAssign from a 3x3 matrix. Set the extra elements to identity.\n\nbool operator== (const Matrix4 &rhs) const\nTest for equality with another matrix without epsilon.\n\nbool operator!= (const Matrix4 &rhs) const\nTest for inequality with another matrix without epsilon.\n\nVector3 operator* (const Vector3 &rhs) const\nMultiply a Vector3 which is assumed to represent position.\n\nVector4 operator* (const Vector4 &rhs) const\nMultiply a Vector4.\n\nMatrix4 operator+ (const Matrix4 &rhs) const\nAdd a matrix.\n\nMatrix4 operator- (const Matrix4 &rhs) const\nSubtract a matrix.\n\nMatrix4 operator* (float rhs) const\nMultiply with a scalar.\n\nMatrix4 operator* (const Matrix4 &rhs) const\nMultiply a matrix.\n\nMatrix4 operator* (const Matrix3x4 &rhs) const\nMultiply with a 3x4 matrix.\n\nvoid SetTranslation (const Vector3 &translation)\nSet translation elements.\n\nvoid SetRotation (const Matrix3 &rotation)\nSet rotation elements from a 3x3 matrix.\n\nvoid SetScale (const Vector3 &scale)\nSet scaling elements.\n\nvoid SetScale (float scale)\nSet uniform scaling elements.\n\nMatrix3 ToMatrix3 () const\nReturn the combined rotation and scaling matrix.\n\nMatrix3 RotationMatrix () const\nReturn the rotation matrix with scaling removed.\n\nVector3 Translation () const\nReturn the translation part.\n\nQuaternion Rotation () const\nReturn the rotation part.\n\nVector3 Scale () const\nReturn the scaling part.\n\nVector3 SignedScale (const Matrix3 &rotation) const\nReturn the scaling part with the sign. Reference rotation matrix is required to avoid ambiguity.\n\nMatrix4 Transpose () const\nReturn transposed.\n\nbool Equals (const Matrix4 &rhs) const\nTest for equality with another matrix with epsilon.\n\nvoid Decompose (Vector3 &translation, Quaternion &rotation, Vector3 &scale) const\nReturn decomposition to translation, rotation and scale.\n\nMatrix4 Inverse () const\nReturn inverse.\n\nconst float * Data () const\nReturn float data.\n\nfloat Element (unsigned i, unsigned j) const\nReturn matrix element.\n\nVector4 Row (unsigned i) const\nReturn matrix row.\n\nVector4 Column (unsigned j) const\nReturn matrix column.\n\nString ToString () const\nReturn as string.\n\n## Static Public Member Functions\n\nstatic void BulkTranspose (float *dest, const float *src, unsigned count)\nBulk transpose matrices.\n\nfloat m00_\n\nfloat m01_\n\nfloat m02_\n\nfloat m03_\n\nfloat m10_\n\nfloat m11_\n\nfloat m12_\n\nfloat m13_\n\nfloat m20_\n\nfloat m21_\n\nfloat m22_\n\nfloat m23_\n\nfloat m30_\n\nfloat m31_\n\nfloat m32_\n\nfloat m33_\n\n## Static Public Attributes\n\nstatic const Matrix4 ZERO\nZero matrix.\n\nstatic const Matrix4 IDENTITY\nIdentity matrix.\n\n## Detailed Description\n\n4x4 matrix for arbitrary linear transforms including projection.\n\nThe documentation for this class was generated from the following files:\n• Source/Urho3D/Math/Matrix4.h\n• Source/Urho3D/Math/Matrix4.cpp"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.54234755,"math_prob":0.9621799,"size":3864,"snap":"2019-35-2019-39","text_gpt3_token_len":1004,"char_repetition_ratio":0.21606217,"word_repetition_ratio":0.071813285,"special_character_ratio":0.27432713,"punctuation_ratio":0.13043478,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.993389,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-23T23:03:09Z\",\"WARC-Record-ID\":\"<urn:uuid:f0149111-99ea-41bf-ab6c-7dc8db7ad158>\",\"Content-Length\":\"49462\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4a3126c0-9bd9-458d-aaf7-153dc1baba48>\",\"WARC-Concurrent-To\":\"<urn:uuid:f69b6546-5e97-49ca-9b13-e8d96362c18e>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://urho3d.github.io/documentation/1.7.1/class_urho3_d_1_1_matrix4.html\",\"WARC-Payload-Digest\":\"sha1:HQXAOSJURWOF7Y4M57J4HN7IMCZXRN24\",\"WARC-Block-Digest\":\"sha1:3I7AIPNRBNB46GCPMOE32WWE64XB2VHR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027319082.81_warc_CC-MAIN-20190823214536-20190824000536-00290.warc.gz\"}"} |
https://justaaa.com/economics/89857-consider-the-equation-of-exchange-quantity-theory | [
"Question\n\n# Consider the equation of exchange (Quantity Theory of Money). Imagine that it is true that velocity...\n\nConsider the equation of exchange (Quantity Theory of Money). Imagine that it is true that velocity is fixed. Show that money demand does not depend on interest rates. If this is true, draw a graph of Money Demand.\n\nVelocity refers to measure of how often money “turns over” in a period; and is computed as nominal GDP divided by the nominal money supply. When quantity theory of money assumes that velocity is fixed, which indicates that real money demand is proportional to real income and is unaffected by the real interest rate, thus quantity equation turns to theory of the effects of money, called the quantity theory of money. Since velocity is constant, a change in the quantity of money (M) must cause a same change in nominal GDP (PY). Thus quantity of money determines the money value of the economy’s output.\n\nSince demand for money does not depend on the interest rate, thus the LM curve is vertical (because the demand now will equal to money supply only at the specific level of income, Y for which that is true for all r).",
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"https://imgs.justaaa.com/questions/da928ef0-76a3-11ed-9059-097d5af0751d.png",
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https://www.groundai.com/project/on-the-complexity-and-approximability-of-optimal-sensor-selection-for-kalman-filtering/ | [
"On the Complexity and Approximability of Optimal Sensor Selection for Kalman Filtering\n\n# On the Complexity and Approximability of Optimal Sensor Selection for Kalman Filtering\n\nLintao Ye, Sandip Roy and Shreyas Sundaram This research was supported by NSF grant CMMI-1635014. Lintao Ye and Shreyas Sundaram are with the School of Electrical and Computer Engineering at Purdue University. Email: {ye159,sundara2}@purdue.edu. Sandip Roy is with the School of Electrical and Computer Engineering and Computer Science at Washington State University. Email: [email protected].\n###### Abstract\n\nGiven a linear dynamical system, we consider the problem of selecting (at design-time) an optimal set of sensors (subject to certain budget constraints) to minimize the trace of the steady state error covariance matrix of the Kalman filter. Previous work has shown that this problem is NP-hard for certain classes of systems and sensor costs; in this paper, we show that the problem remains NP-hard even for the special case where the system is stable and all sensor costs are identical. Furthermore, we show the stronger result that there is no constant-factor (polynomial-time) approximation algorithm for this problem. This contrasts with other classes of sensor selection problems studied in the literature, which typically pursue constant-factor approximations by leveraging greedy algorithms and submodularity of the cost function. Here, we provide a specific example showing that greedy algorithms can perform arbitrarily poorly for the problem of design-time sensor selection for Kalman filtering.\n\n## I Introduction\n\nSelecting an appropriate set of actuators or sensors in order to achieve certain performance requirements is an important problem in control system design (e.g., , , ). For instance, in the case of linear Gauss-Markov systems, researchers have studied techniques to select sensors dynamically (at run-time) or statically (at design-time) in order to minimize certain metrics of the error covariance of the corresponding Kalman Filter. These are known as sensor scheduling problems (e.g., , , ) and design-time sensor selection problems (e.g., , , , ), respectively. These problems are NP-hard in general (e.g., ), and various approximation algorithms have been proposed to solve them. For example, the concept of submodularity has been widely used to analyze the performance of greedy algorithms for sensor scheduling and selection (e.g., , , , ).\n\nIn this paper, we consider the design-time sensor selection problem for optimal filtering of discrete-time linear dynamical systems. We study the problem of choosing a subset of sensors (under given budget constraints) to optimize the steady state error covariance of the corresponding Kalman filter. We refer to this problem as the Kalman filtering sensor selection (KFSS) problem. We summarize some related work as follows.\n\nIn , the authors considered the design-time sensor selection problem of a sensor network for discrete-time linear dynamical systems, also known as dynamic data-reconciliation problems. They assumed that each sensor measures one component of the system state and the measured and unmeasured states are related via network defined mass-balance functions. The objective is to minimize the cost of implementing the network configuration subject to certain performance criteria. They transformed the problem into convex optimization problems, but did not give complexity analysis of the problem. In contrast, we consider the problem of minimizing the estimation error under a cardinality constraint on the chosen sensors without network configuration and analyze the complexity of the problem.\n\nIn , the authors studied the design-time sensor selection problem for discrete-time linear time-varying systems over a finite time horizon, under the assumption that each sensor measures one component of the system state vector. The objective is to minimize the number of chosen sensors while guaranteeing a certain level of performance (or alternatively, to minimize the estimation error with a cardinality constraint on the chosen sensors). In contrast, we consider general measurement matrices and aim to minimize the steady state estimation error.\n\nThe papers and considered the same design-time sensor selection as the one we consider here. In , the authors expressed the problem as a semidefinite program (SDP). However, they did not provide theoretical guarantees on the performance of the proposed algorithm. The paper showed that the problem is NP-hard and gave examples showing that the cost function is not submodular in general. The authors also provided upper bounds on the performance of algorithms for the problem; these upper bounds were functions of the system matrices. Although showed via simulations that greedy algorithms performed well for several randomly generated systems, the question of whether such algorithms (or other polynomial-time algorithms) could provide constant-factor approximation ratios for the problem was left open.\n\nOur contributions to this problem are as follows. First, we show that the KFSS problem is NP-hard even for the special case when the system is stable and all sensors have the same cost. This complements and strengthens the complexity result in , which only showed NP-hardness for two subclasses of problem instances: (1) when the system is unstable and the sensor costs are identical, and (2) when the system is stable but the sensor costs are arbitrary. The NP-hardness of those cases followed in a relatively straightforward manner via reductions from the minimal controllability and knapsack problems, respectively. In contrast, the stronger NP-hardness proof that we provide here requires a more careful analysis, and makes connections to finding sparse solutions to linear systems of equations, yielding new insights into the problem.\n\nAfter establishing NP-hardness of the problem as above, our second (and most significant) contribution is to show that there is no constant factor approximation algorithm for this problem (unless ). In other words, there is no polynomial-time algorithm that can find a sensor selection that is always guaranteed to yield a mean square estimation error (MSEE) that is within any constant finite factor of the MSEE for the optimal selection. This stands in stark contrast to other sensor selection problems studied in the literature, which leveraged submodularity of their associated cost functions to provide greedy algorithms with constant-factor approximation ratios .\n\nOur inapproximability result above immediately implies that greedy algorithms cannot provide constant-factor guarantees for our problem. Our third contribution in this paper is to explicitly show how greedy algorithms can provide arbitrarily poor performance even for very small instances of the KFSS problem (i.e., in systems with only three states and three sensors to choose from).\n\nThe rest of this paper is organized as follows. In Section II, we formulate the KFSS problem. In Section III, we analyze the complexity of the KFSS problem. In Section IV, we study a greedy algorithm for the KFSS problem and analyze its performance. In Section V, we conclude the paper.\n\n### I-a Notation and terminology\n\nThe set of natural numbers, integers, real numbers, rational numbers, and complex numbers are denoted as , , , and , respectively. For any , denote as the least integer greater than or equal to . For a square matrix , let , , , and be its transpose, determinant, set of eigenvalues and set of singular values, respectively. The set of eigenvalues of are ordered with nondecreasing magnitude, i.e., ; the same order applies to the set of singular values . Denote as the element in the th row and th column of . A positive definite (resp. positive semi-definite) matrix is denoted as (resp. ), and if . The set of by positive definite (resp. positive semi-definite) matrices is denoted as (resp. ). The identity matrix with dimension is denoted as . For a vector , denote as the th element of and let be its support, where . Denote the Euclidean norm of by . Define to be a row vector where the th element is and all the other elements are zero; the dimension of the vector can be inferred from the context. For a random variable , let be its expectation. For a set , let be its cardinality.\n\n## Ii Problem Formulation\n\nConsider the discrete-time linear system\n\n x[k+1]=Ax[k]+w[k], (1)\n\nwhere is the system state, is a zero-mean white Gaussian noise process with for all , and is the system dynamics matrix. We assume throughout this paper that the pair is stabilizable.\n\nConsider a set consisting of sensors. Each sensor provides a measurement of the system in the form\n\n yi[k]=Cix[k]+vi[k], (2)\n\nwhere is the state measurement matrix for sensor , and is a zero-mean white Gaussian noise process. We further define , and . Thus, the output provided by all sensors together is given by\n\n y[k]=Cx[k]+v[k], (3)\n\nwhere and . We denote and consider , .\n\nConsider that there are no sensors initially deployed on the system. Instead, the system designer must select a subset of sensors from the set to install. Each sensor has a cost ; define the cost vector . The designer has a budget , representing the total cost that can be spent on sensors from .\n\nAfter a set of sensors is selected and installed, the Kalman filter is then applied to provide an optimal estimate of the states using the measurements from the installed sensors (in the sense of minimizing the MSEE). We define a vector as the indicator vector of the selected sensors, where if and only if sensor is installed. Denote as the measurement matrix of the installed sensors indicated by , i.e., , where . Similarly, denote as the measurement noise covariance matrix of the installed sensors, i.e., , where . Let and denote the a priori error covariance matrix and the a posteriori error covariance matrix of the Kalman filter at time step , respectively, when the sensors indicated by are installed. We will use the following result .\n\n###### Lemma 1\n\nSuppose the pair is stabilizable. For a given indicator vector , both and will converge to finite limits and , respectively, as if and only if the pair is detectable.\n\nThe limit satisfies the discrete algebraic Riccati equation (DARE) :\n\n Σ(μ)=AΣ(μ)AT+W−AΣ(μ)C(μ)T(C(μ)Σ(μ)C(μ)T+V(μ))−1C(μ)Σ(μ)AT. (4)\n\nApplying the matrix inversion lemma , we can rewrite Eq. (4) as\n\n Σ(μ)=W+A(Σ−1(μ)+R(μ))−1AT, (5)\n\nwhere is the sensor information matrix corresponding to sensor selection indicated by . Note that the inverses in Eq. and Eq. are interpreted as pseudo-inverses if the arguments are not invertible. For the case when , the matrix inverse lemma does not hold under pseudo-inverse (unless ), we compute via Eq. .\n\nThe limits and are coupled as :\n\n Σ(μ)=AΣ∗(μ)AT+W. (6)\n\nFor the case when the pair is not detectable, we define the limit . The Kalman filter sensor selection (KFSS) problem is defined as follows.\n\n###### Problem 1\n\n(KFSS Problem) Given a system dynamics matrix , a measurement matrix containing all of the individual sensor measurement matrices, a system noise covariance matrix , a sensor noise covariance matrix , a cost vector and a budget , the Kalman filtering sensor selection problem is to find the sensor selection , i.e., the indicator vector of the selected sensors, that solves\n\n minμ trace(Σ(μ))s.t. bTμ≤Bμ∈{0,1}q\n\nwhere is given by Eq. if the pair is detectable, and , otherwise.\n\n## Iii Complexity Analysis\n\nAs described in the Introduction, the KFSS problem was shown to be NP-hard in for two classes of systems and sensor costs. First, when the matrix is unstable, the set of chosen sensors must cause the resulting system to be detectable in order to obtain a finite steady state error covariance matrix. Thus, for systems with unstable and identical sensor costs, provided a reduction from the “minimal controllability” (or minimal detectability) problem considered in to the KFSS problem. Second, when the matrix is stable (so that all sensor selections cause the system to be detectable), showed that when the sensor costs can be arbitrary, the knapsack problem can be encoded as a special case of the KFSS problem, thereby again showing NP-hardness of the latter problem.\n\nIn this section, we provide a stronger result and show that the KFSS problem is NP-hard even for the special case where the matrix is stable and all sensors have the same cost. Hereafter, it will suffice for us to consider the case when , , i.e., each sensor corresponds to one row of matrix , and the sensor selection cost vector is , i.e., each sensor has cost equal to .\n\nWe will use the following results in our analysis (the proofs are provided in the appendix).\n\n###### Lemma 2\n\nConsider a discrete-time linear system as defined in and . Suppose the system dynamics matrix is of the form with , , the system noise covariance matrix is diagonal, and the sensor noise covariance matrix is . Then, the following holds for all sensor selections .\n\n1. , satisfies\n\n Wii≤(Σ(μ))ii≤Wii1−λ2i. (7)\n2. If such that , then .\n\n3. If such that , then .\n\n4. If such that and the th column of is zero, then .\n\n5. If such that , then .\n\n###### Lemma 3\n\nConsider a discrete-time linear system as defined in Eq. and Eq. . Suppose the system dynamics matrix is of the form , where , the measurement matrix , where , the system noise covariance matrix , and the sensor noise covariance matrix . Then, the MSEE of state , i.e., , satisfies\n\n Σ11=1+α2λ21−α2+√(α2−α2λ21−1)2+4α22, (8)\n\nwhere . Moreover, if we view as a function of , denoted as , then is a strictly increasing function of with .\n\n### Iii-a NP-hardness of the KFSS problem\n\nTo prove the KFSS problem (Problem 1) is NP-hard, we relate it to the problem described below.\n\n###### Definition 1\n\nGiven a finite set with and a collection of -element subsets of , an exact cover for is a subcollection such that every element of occurs in exactly one member of .\n\n###### Remark 1\n\nSince each member in is a subset of with exactly elements, if there exists an exact cover for , then it must consist of exactly members of .\n\nWe will use the following result .\n\n###### Lemma 4\n\nGiven a finite set with and a collection of -element subsets of , the problem to determine whether contains an exact cover for is NP-complete.\n\nWe are now in place to prove the following result.\n\n###### Theorem 1\n\nThe KFSS problem is NP-hard when the system dynamics matrix is stable and each sensor has identical cost.\n\n{proof}\n\nWe give a reduction from to KFSS. Consider an instance of to be a finite set with , and a collection of -element subsets of , where . For each element , define the column vector to encode which elements of are contained in . In other words, for and , if element of set is in , and otherwise. Define the matrix . Furthermore, define . Thus has a solution such that has nonzero entries if and only if the answer to the instance of is “yes” .\n\nGiven the above instance of , we then construct an instance of KFSS as follows. We define the system dynamics matrix as , where .111We take for the proof. The set is defined to contain sensors with the collective measurement matrix\n\n C=[1dT0GT], (9)\n\nwhere and are defined, based on the given instance of , as above. The system covariance matrix is set to be , and the measurement noise covariance matrix is set to be . Finally, the cost vector is set as , and the sensor selection budget is set as . Note that the sensor selection vector for this instance is denoted by . For the above construction, since the only nonzero eigenvalue of is , we know from Lemma 2(c) that for all sensor selections .\n\nWe claim that the solution to the constructed instance of the KFSS problem satisfies if and only if the answer to the given instance of the problem is “yes”.\n\nSuppose that the answer to the instance of the problem is “yes”. Then has a solution such that has nonzero entries. Denote the solution as and denote . Define as the sensor selection vector that indicates selecting the first and the th to the th sensors, i.e., sensors that correspond to rows , from (9). Since , we have for as defined in Eq. . Noting that , it then follows that . Hence, we know from Lemma 2(a) and Lemma 2(e) that , which is also the minimum value of among all possible sensor selections . Since always holds as argued above, we have and is the optimal sensor selection, i.e., .\n\nConversely, suppose that the answer to the problem is “no”. Then, for any union of () subsets in , denoted as , there exist elements in that are not covered by , i.e., for any and , has zero rows, for some . We then show that for all sensor selections that satisfy the budget constraint. First, for any possible sensor selection that does not select the first sensor, we have the first column of is zero (from the form of as defined in Eq. ) and we know from Lemma 2(d) that , which implies that . Thus, consider sensor selections that select the first sensor, denote , where and define . We then have\n\n C(μ)=[1dT0G(μ)T], (10)\n\nwhere has zero columns, for some . As argued in Lemma 5 in the appendix, there exists an orthogonal matrix of the form (where is also an orthogonal matrix) such that\n\n ~C(μ)≜C(μ)T=[1γβ00~G(μ)T].\n\nIn the above expression, is of full column rank, where . Furthermore, and of its elements are ’s, and . We perform a similarity transformation on the system with the matrix (which does not change the trace of the error covariance matrix), and perform additional elementary row operations to transform into the matrix\n\n ~C′(μ)=[1γ000~G(μ)T]. (11)\n\nSince and are both diagonal, and , we can obtain from Eq. that the steady state error covariance corresponding to the sensing matrix is of the form\n\n ~Σ′(μ)=[~Σ′1(μ)00~Σ′2(μ)],\n\nwhere , denoted as for simplicity, satisfies\n\n Σ=A1ΣAT1+W1−A1ΣCT1(C1ΣCT1)−1C1ΣAT1,\n\nwhere , and . We then know from Lemma 3 that since . Hence, we have .\n\nThis completes the proof of the claim above. Suppose that there is an algorithm that outputs the optimal solution to the instance of the KFSS problem defined above. We can call algorithm to solve the problem. Specifically, if the algorithm outputs a solution such that , then the answer to the instance of is “yes”; otherwise, the answer is “no”.\n\nHence, we have a reduction from to the KFSS problem. Since is NP-complete and KFSS , we conclude that the KFSS problem is NP-hard.\n\n### Iii-B Inapproximability of the KFSS Problem\n\nIn this section, we analyze the achievable performance of algorithms for the KFSS problem. Specifically, consider any given instance of the KFSS problem. For any given algorithm , we define the following ratio:\n\n rA(Σ)≜trace(ΣA)trace(Σopt), (12)\n\nwhere is the optimal solution to the KFSS problem and is the solution to the KFSS problem given by algorithm .\n\nIn , the authors showed that there is an upper bound for for any sensor selection algorithm , in terms of the system matrices. However, the question of whether it is possible to find an algorithm that is guaranteed to provide an approximation ratio that is independent of the system parameters has remained open up to this point. In particular, it is typically desirable to find constant-factor approximation algorithms, where the ratio is upper-bounded by some (system-independent) constant. Here, we provide a strong negative result and show that for the KFSS problem, there is no constant-factor approximation algorithm in general, i.e., for all polynomial-time algorithms and , there are instances of the KFSS problem where .\n\n###### Theorem 2\n\nIf , then there is no polynomial-time constant-factor approximation algorithm for the KFSS problem.\n\n{proof}\n\nSuppose that there exists such a (polynomial-time) approximation algorithm , i.e., such that for all instances of the KFSS problem, where is as defined in Eq. . We will show that can be used to solve the problem as described in Lemma 4. Given an arbitrary instance of the problem (with a base set containing elements and a collection of -element subsets of ), we construct a corresponding instance of the KFSS problem in a similar way to that described in the proof of Theorem 1. Specifically, the system dynamics matrix is set as , where and satisfies . The set contains sensors with collective measurement matrix\n\n C=[1εdT0GT], (13)\n\nwhere , depend on the given instance of and are as defined in the proof of Theorem 1. The constant is chosen as . The system noise covariance matrix is set to , and the measurement noise covariance matrix is set to be . The sensor cost vector is set as , and the sensor selection budget is set as . Note that the sensor selection vector is given by .\n\nWe claim that there exists a sensor selection vector such that if and only if the answer to the problem is “yes”.\n\nSuppose that the answer to the problem is “yes”. We know from Theorem 1 that there exists a sensor selection such that .\n\nConversely, suppose that the answer to the problem is “no”. Then, for any union of () subsets in , denoted as , there exist elements in that are not covered by . We follow the discussion in Theorem 1. First, for any sensor selection that does not select the first sensor, we have . Hence, by our choice of , we have , which implies since for all possible sensor selections. Thus, consider sensor selections that include the first sensor. As argued in the proof of Theorem 1 leading up to Eq. , we can perform an orthogonal similarity transformation on the system, along with elementary row operations on the measurement matrix to obtain a measurement matrix of the form\n\n ~C′(μ)=[1εγ000~G(μ)T], (14)\n\nwhere elements of are ’s and . Then, we have . Moreover, we obtain from Lemma 3\n\n (Σ(μ))11=1+α2λ21−α2+√(α2−α2λ21−1)2+4α22.\n\nIf we view as a function of , denoted as , we know from Lemma 3 that is an increasing function of . Hence, we have , i.e.,\n\n Σ11(α2)≥1+ε2λ21−ε2+√(ε2−ε2λ21−1)2+4ε22.\n\nSince , we have , which implies . Hence, we have .\n\nThis completes the proof of the claim above. Hence, if algorithm for the KFSS problem has for all instances, it is clear that can be used to solve the problem by applying it to the above instance. Specifically, if the answer to the instance is “yes”, then the optimal sensor selection would yield a trace of , and thus the algorithm would yield a trace no larger than . On the other hand, if the answer to the instance is “no”, all sensor selections would yield a trace larger than , and thus so would the sensor selection provided by . In either case, the solution provided by could be used to find the answer to the given instance. Since is NP-complete, there is no polynomial-time algorithm for it if , and we get a contradiction. This completes the proof of the theorem.\n\n## Iv Greedy Algorithm\n\nOur result in Theorem 2 indicates that no polynomial-time algorithm can be guaranteed to yield a solution that is within any constant factor of the optimal solution. In particular, this result applies to the greedy algorithms that are often studied for sensor selection in the literature , where sensors are iteratively selected in order to produce the greatest decrease in the error covariance at each iteration. In particular, it was shown via simulations in that such algorithms work well in practice (e.g., for randomly generated systems). In this section, we provide an explicit example showing that greedy algorithms for KFSS can perform arbitrarily poorly, even for small systems (containing only three states). We will focus on the simple greedy algorithm for the KFSS problem defined as Algorithm 1, for instances where all sensor costs are equal to , and the sensor budget for some (i.e., up to sensors can be chosen). For any such instance of the KFSS problem, define , where is the solution of the DARE corresponding to the sensors selected by Algorithm 1.\n\n###### Example 1\n\nConsider an instance of the KFSS problem with matrices , , and , defined as\n\n A=⎡⎢⎣λ100000000⎤⎥⎦,C=⎡⎢⎣1hh10h011⎤⎥⎦,\n\nwhere , and . In addition, we have the set of candidate sensors , the selection budget and the cost vector .\n\n###### Theorem 3\n\nFor the instance of the KFSS problem defined in Example 1, the ratio satisfies\n\n limh→∞rgre(Σ)=23+13(1−λ21). (15)\n{proof}\n\nWe first prove that the greedy algorithm defined as Algorithm 1 selects sensor and sensor in its first and second iterations. Since the only nonzero eigenvalue of is , we know from Lemma 2(c) that and , , which implies that and . Hence, we focus on determining .\n\nIn the first iteration of the greedy algorithm, suppose the first sensor, i.e., sensor corresponding to , is selected. Then, using the result in Lemma 3, we obtain , denoted as , to be\n\n σ1=2√(1−λ21−12h2)2+2h2+1−λ21−12h2.\n\nSimilarly, if the second sensor, i.e., the sensor corresponding to , is selected in the first iteration of the greedy algorithm, we have , denoted as , to be\n\n σ2=2√(1−λ21−1h2)2+4h2+1−λ21−1h2.\n\nIf the third sensor, i.e., the sensor corresponding to , is selected in the first iteration of the greedy algorithm, the first column of is zero, which implies based on Lemma 2(d). If we view as a function of , denoted as , we have . Since we know from Lemma 3 that is an increasing function of and upper bounded by , we obtain , which implies that the greedy algorithm selects the second sensor in its first iteration.\n\nIn the second iteration of the greedy algorithm, if the first sensor is selected, we have , on which we perform elementary row operations and obtain . By direct computation from Eq. , we obtain . If the third sensor is selected, we have . By direct computation from Eq. , we obtain , denoted as , to be\n\n σ23=2√(1−λ21−2h2)2+8h2+1−λ21−2h2.\n\nSimilar to the argument above, we have and , where , which implies that the greedy algorithm selects the third sensor in its second iteration. Hence, we have .\n\nIf , then and thus we know from Lemma 2(a) and Lemma 2(e) that , which is also the minimum value of among all possible sensor selections . Combining the results above and taking the limit as , we obtain the result in Eq. .\n\nExamining Eq. (15), we see that for the given instance of KFSS, we have as and . Thus, can be made arbitrarily large by choosing the parameters in the instance appropriately. It is also useful to note that the above behavior holds for any algorithm that outputs a sensor selection that contains sensor for the above example.\n\n## V Conclusions\n\nIn this paper, we studied the KFSS problem for linear dynamical systems. We showed that this problem is NP-hard and has no constant-factor approximation algorithms, even under the assumption that the system is stable and each sensor has identical cost. We provided an explicit example showing how a greedy algorithm can perform arbitrarily poorly on this problem, even when the system only has three states. Our results shed new insights into the problem of sensor selection for Kalman filtering and show, in particular, that this problem is more difficult than other variants of the sensor selection problem that have submodular cost functions. Future work on characterizing achievable (non-constant) approximation ratios, or identifying classes of systems that admit near-optimal approximation algorithms, would be of interest.\n\n## Appendix\n\n### Proof of Lemma 2:\n\nSince and are diagonal, the system represents a set of scalar subsystems of the form\n\n xi[k+1]=λixi[k]+wi[k],∀i∈{1,…,n},\n\nwhere is the th state of and is a zero-mean Gaussian noise process with variance . As is stable, the pair is detectable and the pair is stabilizable for all sensor selections . Thus, the limit exists (based on Lemma 1), and is denoted as .\n\nProof of (a). Since and are diagonal, we know from Eq. that\n\n (Σ(μ))ii=λ2i(Σ∗(μ))ii+Wii,\n\nwhich implies , . Moreover, it is easy to see that , . Since , we obtain from Eq.\n\n Σ(0)=AΣ(0)AT+W. (16)\n\nwhich implies that since is diagonal. Hence, , .\n\nProof of (b). Letting in inequality , we obtain .\n\nProof of (c). Letting in inequality , we obtain .\n\nProof of (d). Assume without loss of generality that the first column of is zero, since we can simply renumber the states to make this the case without affecting the trace of the error covariance matrix. Hence, we have of the form\n\n C(μ)=[0C1(μ)].\n\nMoreover, since and are diagonal and , we can obtain from Eq. that is of the form\n\n Σ(μ)=[Σ1(μ)00Σ2(μ)],\n\nwhere and satisfies\n\n (Σ(μ))11=λ2i(Σ(μ))11+W11,\n\nwhich implies .\n\nProof of (e). Similar to the proof of (d), we can assume without loss of generality that . If we further perform elementary row operations on ,222Note that since we assume , it is easy to see that the Kalman filter gives the same results if we perform any elementary row operation on . we get a matrix of the form\n\n ~C(μ)=[100~C1(μ)].\n\nMoreover, since and are diagonal and , we can obtain from Eq. that is of the form\n\n Σ(μ)=[Σ1(μ)00Σ2(μ)],\n\nwhere and satisfies\n\n (Σ(μ))11=λ21(Σ(μ))11+W11<"
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http://msgf.aavt.pw/coterminal-and-reference-angles.html | [
"# Coterminal And Reference Angles\n\nLESSON 4 COTERMINAL ANGLES Topics in this lesson: 1. Find a positive angle that is coterminal to -30°. On the other end of the spectrum, to find the reference angle for 960 degrees: Determine the quadrant in which the terminal side lies. The reference angle is the acute angle formed by the terminal side of a given angle and the x-axis. You can put this solution on YOUR website! Find a coterminal angle for Q = 15pi/4 terminal side of angle 15pi/4 is 3π/4 so the co-terminal angle is -5π/4 note that both angle together cover 2π radians. The quiz contains questions on finding coterminal angles, reference angles and the values of the six trig functions using right triangles and the unit circle. A reference angle is the smallest angle between the terminal side of an angle and the x-axis. Reference angle. Sometimes, using a negative angle rather than a positive angle is more convenient, or the answer to an application may involve more than one. asked by Kaylen on February 12, 2016; eakin elementary. Relevant page. Solve the problems & select the right answers. Best Answer: Coterminal angles are angles which are the \"same\", that is, they share the same terminal side. 10270 Apr 159:07 AM Sep 104:19 PM Nov 1111:21 AM Reference Angle Cheat sheet: Positive, acute, in all quadrants (how far is it. Byju's Coterminal Angle Calculator is a tool which makes calculations very simple and interesting. 3 3 3 1 O A 3 2π cot = =− =− b. When an angle is not between 0 and 360° ( ), find a coterminal angle that is within that range. In general, when someone is looking for a reference angle, they have a given angle theta to start with. Be sure to show the terminal ray and label the reference angle in your diagram. Coterminal angles are angles formed by different rotations but with the same initial and terminal sides. (a) θ=−2587. Positive Angle. Worksheets are Coterminal angles and reference angles, Coterminal angles, 1, Infinite algebra 2, Draw an angle with the given measure in standard, Angles and angle measure date period, Lesson 4 coterminal angles, 0 1. Improve your math knowledge with free questions in \"Coterminal and reference angles\" and thousands of other math skills. Start studying Coterminal and Reference Angles. Reference angle: _____. Draw the following angles in standard position, then find the reference angle. Just add OR subtract! Section 6. Coterminal Angle Formula. Reference Angles and Coterminal Angles Mazes. -3750 77t 12. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. The angle 11 4 πis. The word \"coterminal\" is slightly confusing, but all it's meant to denote is angles that terminate at the same point. What is the terminal side?. Find a coterminal angle between 0° and 360°. THE DEFINITION AND EXAMPLES OF REFERENCE ANGLES Definition The reference angle of the angle T, denoted by T'. Best Answer: 445 what? Degrees? The \"next\" to me implies that it is greater than 445 degrees but, because of \"smallest,\" you want the one that isn't greater than one revolution greater than 445 degrees. -1-Find the measure of each angle. You can sketch the angles and often tell just form looking at them if they are coterminal. Reference angles must be. Using the unit circle below, explain how you can find the sine of any given angle, both by the ratio (opposite over hypotenuse) and by the ordered pairs. D: 450 and -270 with 450, we subtract 360 to get 90, which will satisfy the coterminal angles. All of these angles are coterminal angles. a) 4 11 b) 395° c) 3 2 d) -45° 15. 777 Reference Angles an acute angle (< 90), that is formed by the terminal side & xaxis. The reference angle $$\\text{ must be } 90^{\\circ}$$ In radian measure, the reference angle $$\\text{ must be } \\frac{\\pi}{2}$$ Basically, any angle on the x-y plane has a reference angle, which is always between 0 and 90 degrees. 1: Coterminal & Reference Angles PreCalculus August 20, 2015 Usefulness of coterminal angles: Find an angle between 0 and 360 that is coterminal with each angle. When an angle is not between 0 and 360° ( ), find a coterminal angle that is within that range. An angle's reference angle is the measure of the smallest, positive, acute angle t formed by the terminal side of the angle t. This article explains what a reference angle is, providing a reference angle definition. +/- angles, coterminal angles and values of trig functions What are some tips or tricks you would give to someone who is just starting to learn about positive and negative angles, coterminal and reference angles, and the values of trig functions?. (a) -135o (b) 1200o Reference Angles REFERENCE ANGLE: a positive acute angle made by the terminal side of an angle and the x-axis. Answers can vary. For any angle α, the positive coterminal angle can be found by: α + 360°∙n , if α is given in degrees, where n=0,1,2,3,. It must be less than 90 degree, and always positive. 510 degrees 3. I N THE RADIAN SYSTEM of angular measurement, the measure of one revolution is 2 π. Examples: • 60° and 30° are complementary angles. (For negative angles add 360 instead). 10270 Apr 159:07 AM Sep 104:19 PM Nov 1111:21 AM Reference Angle Cheat sheet: Positive, acute, in all quadrants (how far is it. 2 radians (see Figure 6a) and the reference angle for −137 is 180 ( ) + −137 =43 (see Figure 6b). To find the coterminal angle of -690° we could add 360 because it is negative (subtract if positive) so: -690+360 =-330 and -330+360=30° So the smallest positve coterminal angle of -690° is 30°. asked by tj on October 6, 2010; Pre-Calc. Relevant page. e To find a NEGATIVE coterminal (if in deg) (if in rads) (I call these \"swirlies\") e Coterminal angles can contain Example 2a: Find a positive and negative coterminal angle for the given angle 0. (In the next Topic, Arc Length, we will see the actual definition of radian measure. #trigonometry #anglesintrigonometry Category. Quadrant Angle Measure (radians) Angle Measure (degrees) Coordinate on Terminal Side of Angle 1 2 3 4. 2 Angles and Radian Measure 471 Finding Coterminal Angles In Example 1(b), the angles 500° and 140° are coterminal because their terminal sides coincide. For example: Find the standard coterminal angle for a. moomoomath. Reference Angle Worksheet Give the reference angle for each angle below. The reference angle is the positive acute angle that can represent an angle of any measure. kcl e Unit Circle: Right Triangle Trigonometry D4 Notes Reference Angles & Exact Values Reference Angles: Special Right Triangles: 300 Name: Coterminal Angles: ABIes CO-MC. W C eAtlhlS lrxipgZhztRsN orAeYsreMrsvwegdz. The first example we did was: Find θ, given that tan θ = 0. h T JMKaHdMeD jwTistyh0 cI7n7fliKnQidtGeu cAzlggPerbar qab 62p. Fab Five for Trigonometry Level Three: Reference Angles, Coterminal Angles, Linear Speed and Angular Speed - Kindle edition by Kathryn Gniadek. It is reflects Algebra 2 (algebra ii) level exercises. The reference angle is the smallest angle between the terminal side of the angle and the x-axis. asked by Kaylen on February 12, 2016; eakin elementary. 13) −330 ° 30 ° 14) −435 ° 285 ° 15) 640 ° 280 ° 16) −442 ° 278 ° Find a coterminal angle between 0 and 2222ππππ for each given angle. Mathematics. Reference Angles and Triangles - Independent Practice Worksheet Complete all the problems. Trig Worksheet-Day 2 (Coterminal Angles & Angle Conversions) Determine two coterminal angles (one positive and one negative) for each angle. Coterminal angles: are angles in standard position (angles with the initial side on the positive x-axis) that have a common terminal side. 13-2 Angles of Rotation 937 Coterminal angles are angles in standard position with £Óä Ó{äÂ Þ Ý the same terminal side. Best Answer: Coterminal angle in radians = (angle in radians) +- (k)(2π) Since k can be any integer, therefore there can be an infinite number of coterminal angles for any angle. NOTE: Only your test content will print. There are an infinite number of coterminal angles that can be found. Converting between degrees and radians. Following this procedure, all coterminal angles can be found. Find a coterminal angle between 0° and 360°. Upon this discovery, I asked them how they could tell if two angles were coterminal based on this. Solving Triangles a. To find the coterminal angle of an angle, simply add or subtract radians, or 360. The reference angle is always positive. Solve advanced problems in Physics, Mathematics and Engineering. Start below. EXERCISES: 1. We can list all angles coterminal to the angle x using the shorthand notation x + 2n, n an integer. Therefore, the reference angle is always coterminal with the original angle θ. If an input is given then it can easily show the result for the given number. 1 COTERMINAL AND REFERENCE ANGLES. By simply. 21) −330 ° 30 ° 22) −435 ° 285 ° 23) 640 ° 280 ° 24) −442 ° 278 ° Find a coterminal angle between 0 and 2222 πππ for each given angle. Coterminal angles. Given the angle Quadrant Coterminal Angle(-) find the following: Il. asked by tj on October 6, 2010; Pre-Calc. Give your answers in radians. For example, 100° and 460° are coterminal for this reason, as is −260°. If you examine the equation above, you will see that we are just adding or subtracting 2π radians (=360º) to the angle to find it's coterminal. 13) −330 ° 30 ° 14) −435 ° 285 ° 15) 640 ° 280 ° 16) −442 ° 278 ° Find a coterminal angle between 0 and 2222ππππ for each given angle. Two Angles are Complementary when they add up to 90 degrees (a Right Angle). (look below). So I now give the student the angle 3pi/5 to find a reference angle. Give 2 coterminal angles, one positive and one negative for each of the following. 342 62/87,21 All angles measuring are coterminal with a 342 DQJOH Sample answer: Let n = 1 and í1. 8along with the Quotient and Reciprocal Identities in Theorem10. Reference and Coterminal Angles Find a positive and a negative coterminal angle for each given angle. This trigonometry video tutorial explains how to find a positive and a negative coterminal angle given another angle in degrees or in radians using the unit circle. Are reference angles and radians the same? No. They can be positive or negative integers. Start below. Fab Five for Trigonometry Level Three: Reference Angles, Coterminal Angles, Linear Speed and Angular Speed - Kindle edition by Kathryn Gniadek. 4 in your book. THE REFERENCE ANGLE THEOREM Reference Angle Theorem: If θis an angle, in standard position,that lies in a quadrant and αis. Worksheet – Coterminal, Reference, and Special Angles Find two angles (one in a clockwise direction and the other in a counterclockwise direction, from the given angle) that are coterminal to the following:. (a) -135o (b) 1200o Reference Angles REFERENCE ANGLE: a positive acute angle made by the terminal side of an angle and the x-axis. 212° - ° 6. There are an infinite number of coterminal angles that can be found. Coterminal Angles are two angles in standard position with the same terminal side. THE DEFINITION AND EXAMPLES OF COTERMINAL ANGLES Definition Two angles are said to be coterminal if their terminal sides are the same. 10270 Apr 159:07 AM Sep 104:19 PM Nov 1111:21 AM Reference Angle Cheat sheet: Positive, acute, in all quadrants (how far is it. Coterminal and reference angles calculator keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website. When a circular cake is cut up into four equal pieces, each piece has a right angle at the center. 3 General Angles. If the measure of the original angle is given in degrees, its reference angle must also be in degrees. Learn the basics of trigonometry: What are sine, cosine, and tangent? How can we use them to solve for unknown sides and angles in right triangles?. 3 Coterminal Angles and Evaluating Trig Function for Angles Lesson 8. 12 12 117t 15 os IS Sketch a picture and determine the Reference Angle for each find coterminal angles if needed). _ Coterminal Angles 2. 25) 11 π 3 5π 3 26) − 35 π 18 π 18 27) 15 π 4 7π 4 28) − 19 π 12 5π 12 Find a positive and a negative coterminal angle for each given. Find the reference angle for the following. Let n represent any integer. Find one positive angle and one negative angle that are coterminal with each angle. Coterminal Angles are two angles in standard position with the same terminal side. If the angle is in the first quadrant, then the reference angle is the original angle. In degrees: You can add or subtract $360˚$ however many times you want, and the new angle will always be coterminal to the original angle. (More about the reference angle in Objective 5. Any angle θ is coterminal with θ + 360° -- because we are just going around the circle. Coterminal angle calculator that returns exact values and steps given either a degree or radian value, Trigonometry Calculator Identity Reference. Angles in quadrant I. 2 Angles and Radian Measure 471 Finding Coterminal Angles In Example 1(b), the angles 500° and 140° are coterminal because their terminal sides coincide. Determine the cosine of 23pi/4, 13pi/3, and 37pi/6 using reference angles. There are an infinite number of coterminal angles that can be found. Standard Coterminal Angles If you are asked to find a standard coterminal angle, that means the answer needs to be between 00 and 3600. Best Answer: Coterminal angle in radians = (angle in radians) +- (k)(2π) Since k can be any integer, therefore there can be an infinite number of coterminal angles for any angle. Reference Angle Ordered Pair Degrees O Coterminal Angle(-) find the following: I. 3 04 Reference Angle: Coterminal Angle: Determine the reference angle and at least one coterminal angle for each angle drawn below. Trig Quiz Chapter 5 Sections 1-3. Best Answer: 445 what? Degrees? The \"next\" to me implies that it is greater than 445 degrees but, because of \"smallest,\" you want the one that isn't greater than one revolution greater than 445 degrees. Leave your answer in the form that was given (radians or degrees). It is reflects Algebra 2 (algebra ii) level exercises. Notes Coterminal Angles April 27, 2015 Coterminal Angles Lesson objectives Teachers' notes Identify the quadrant of a terminal side. o o o A positive angle is A negative angle is. Coterminal angles are angles which share the same initial side and terminal sides. It is reflects Algebra 2 (algebra ii) level exercises. The reference angle is †§ = 180° º 150° = 30°. -1500 a:5!:l_ ~ Find two angles (one inaclockwise direction and the other inacounterclockwise. This article explains what a reference angle is, providing a reference angle definition. These two angles are coterminal; see Fig. Answers need to be in the same measure as the given angle. Exercises on Angles of Rotation - Coterminal Angles. Algebra -> Trigonometry-basics-> SOLUTION: what is the coterminal angles of the following: 1. Trigonometric ratios of angles in radians. Coterminal angles are two angles that are drawn in the standard position (so their initial sides are on the positive x-axis) and have the same terminal side like 110° and -250° Another way to describe coterminal angles is that they are two angles in the standard position and one angle is a multiple of 360 degrees larger or smaller than the other. Refer Reference Angles; 7-5; Reference Angles; Reference Angles; me and sarah are currently in the apple store, and Reference Angles; Unit circles and reference angles. Coterminal Angles Find two coterminal angles for each given angle. Coterminal angles: are angles in standard position (angles with the initial side on the positive x-axis) that have a common terminal side. This is the acute angle formed by the terminal side of the given angle and the X-axis. Reference Angles (Guided) PC-Trigonometry- Rotations and the Unit Circle- Finding Reference Angles. Since there are an infinite number of coterminal angles, this calculator finds the one whose size is between 0 and 360 degrees or between 0 and 2π depending on the unit of the given angle. For each angle in standard position, find two positive angles and two negative angles that are coterminal with the given angle. The tangent function is negative in Quadrant II, so you can write: tan (º210°) = ºtan 30° = º b. A calculator to find the exact value of a coterminal angle to a given trigonometric angle. For example 30 ° , − 330 ° and 390 ° are all coterminal. Reference Angle. Coterminal definition, having a common boundary; bordering; contiguous. How to Compute the Measure of the Reference Angle of an Angle in Standard Position Terminal Side In Quadrant I Quadrant II Quadrant III Quadrant IV Reference Angle (θin degrees) θ Reference Angle (θin radians) θ 180q T S T T 180q 360q T T S 2S T. I begin working with special angles because students will need to use these angles the most through the rest of the year. - 310° Find the complement and supplement for the given angles. Angles, Angular Conversions, and Trigonometric Functions. How do you write an expression to represent every angle coterminal to s? Given: s in a unit circle with point (-1/2 , -sqrt(3)/2) I got that it is quadrant 3, and the trig functions are those of angle 60°. QUESTION: Find the reference angle for. Reference angle is the smallest positive acute angle formed by the x-axis and the terminal side. I don't want students to think that the numerator is always just pi on a reference angle. Finding the reference angle. About This Quiz & Worksheet. Coterminal Angles are angles that start and stop in the same place. Answers can vary. THE REFERENCE ANGLE THEOREM Reference Angle Theorem: If θis an angle, in standard position,that lies in a quadrant and αis. SOLUTION a. Coterminal angles are angles formed by different rotations but with the same initial and terminal sides. REFERENCE ANGLES ARE ALWAYS POSITIVE! To find a reference angle for angles outside the interval 0o < θ < 360o or 0π < θ < 2π you must first find a corresponding coterminal angle in this interval. Reinforce the concept of reference and coterminal angles with the multiple response worksheets featured here. Then find and draw one positive and one negative angle coterminal with the given angle. a) b) c) d)38 27 72 58 a) b) c) d) a) b) c) d) a) b) c) d) What!are!the!possible!positive!and!negative!coterminal!angles!of!320 ? 680 ,!±40 1040 ,!±400 675 ,!±45. All angles throughout this unit will be drawn in standard position. Best Answer: 445 what? Degrees? The \"next\" to me implies that it is greater than 445 degrees but, because of \"smallest,\" you want the one that isn't greater than one revolution greater than 445 degrees. 1) −5° 2) 5° 3) 45 ° 4) −45 ° 5) 5 π 2 6) −π 7) 17 π 9 8) 13 π 18 State if the two given angles are coterminal or not. Find all coterminal angles for each angle such that -360° < θ < 360°. 5 1radians is 7. cs/coterminal-angles. There are an infinite number of ways to draw an angle on the coordinate axes. Determine if you are moving clockwise or. Note how the reference angle always remain less than or equal to 90°, even for large angles. 6 Coterminal Angles Notes. Tt from positive coterminal angle 44 of 45 16. Reference Angles and Coterminal Angles MazesStudents will practice identifying reference angles and coterminal angles given angles in degrees and radians. The key to solving a reference angle is to understand which quadrant the angle lies in. There are two methods that can. One way to find the measure of an angle that is coterminal with an. Find the reference angle for the following. coterminal angles is as simple as adding or subtracting 360 ° or 2π to each angle, depending on whether the given angle is in degrees or radians. -1- ©z 62f0q1i2J XKuugt6aa ASTo1fntVwuaKrDeB WLfLxCm. In this calculus worksheet, learners identify terminal and coterminal angles using the unit circle. 1) 326 ° 686 ° and −34 ° 2) 530 ° 170 ° and −190 ° 3) −215 ° 145 ° and −575 ° 4) −84 ° 276 ° and −444 ° 5) 215 ° 575 ° and −145 ° 6) 255 ° 615 ° and −105 ° 7) −660 ° 60 ° and −300 ° 8) −255 °. Identify all angles that are coterminal with each angle. Be sure to show the terminal ray and label the reference angle in your diagram. Sketch both angles on the unit circle provided. If you examine the equation above, you will see that we are just adding or subtracting 2π radians (=360º) to the angle to find it's coterminal. 45o All angles have a measure of 45 + 360k, where k is an integer, are coterminal with 45. We are experts in trigonometry. Math Expression Renderer, Plots, Unit Converter, Equation Solver, Complex Numbers, Calculation History. between 0 and 360°. Quadrant I Quadrant II Quadrant III Quadrant IV reference reference 180 reference 180 reference 360 State the quadrant of each given. 2_practice_solutions. 62/87,21 All angles measuring are coterminal with a UDGLDQDQJOH Sample answer: Let n = 1 and í1. and the closest xaxis is its reference angle. Find the reference angle. The previous examples suggest a pattern in finding reference angles based on their quadrant. Angles in Standard Position: C-øu. An angle's reference angle is the size of the smallest acute angle,$\\,{t}^{\\prime },$formed by the terminal side of the angle$\\,t\\,$and the horizontal axis. Just add OR subtract! Section 6. Best Answer: 2pi/3 a full cycle is 2pi (360 degrees), so to find a coterminal angle, add 2pi or subtract 2pi to the angles you are given, so Add 2pi to find a positive angle coterminal to it 2pi/3 + 2pi = 8pi/3 Then you could add another 2pi to find another coterminal angle, so 8pi/3 + 2pi = 14pi/3. Sometimes, using a negative angle rather than a positive angle is more convenient, or the answer to an application may involve more than one. You can see that a negative angle can be coterminal with a positive one. 7) -470° x. Coterminal Angles are angles who share the same initial side and terminal sides. 21) −330 ° 30 ° 22) −435 ° 285 ° 23) 640 ° 280 ° 24) −442 ° 278 ° Find a coterminal angle between 0 and 2222 πππ for each given angle. USING A REFERENCE ANGLE TO FIND THE VALUE OF THE SIX TRIGONOMETRIC FUNCTIONS 1. Trigonometry Finding Angles, Coterminal Angles, Reference Angles (TFACARA) Find the measure of each angle. Finding Coterminal Angles algebraically is a little more challenging: First, we have to decipher what the question is asking. Using the picture shown below of a non-special angle on the Unit Circle. There are an infinite number of ways to draw an angle on the coordinate axes. Coterminal Angles Two angles that are in standard position and share a common terminal side are said to be coter-minal angles. Mathematics. asked by tj on October 6, 2010; Pre-Calc. Find a coterminal angle between 0° and 360°. Draw each angle, θ, in standard position. 6 5π − is coterminal with 6 7π which terminates in Quadrant III. Fun maths practice! Improve your skills with free problems in 'Coterminal and reference angles' and thousands of other practice lessons. Students use their solutions to navigate through the maze. For any angle α, the positive coterminal angle can be found by: α + 360°∙n , if α is given in degrees, where n=0,1,2,3,. And, most important, each right angle is half. This will help your child to distinguish between the various types of angles. Best Answer: 445 what? Degrees? The \"next\" to me implies that it is greater than 445 degrees but, because of \"smallest,\" you want the one that isn't greater than one revolution greater than 445 degrees. A useful tool for finding values of trigonometric functions of certain angles (you’ll learn how to do that later) is the reference angle. Consider the following two angles: 210° and -150°. 4 Coterminal and Reference Angles Objectives: 1. Take your time, as you have all the time in the world. Otherwise, for each angle do the following: If the angle is positive, keep subtracting 360 from it until the result is between 0 and +360. To find the coterminal angles, simply add or subtract 360 degrees as many times as needed from the reference angle. Download it once and read it on your Kindle device, PC, phones or tablets. Trigonometric Angles formulas list online. This smart calculator is provided by wolfram alpha. Answers can vary. For each angle in standard position, find two positive angles and two negative angles that are coterminal with the given angle. Reference and Coterminal Angles Find a positive and a negative coterminal angle for each given angle. Finding coterminal angles is as simple as adding or subtracting 360° or 2π to each angle, depending on whether the given angle is in degrees or radians. Remove Ads. find the least positive degree of an angle that is coterminal with an angle of the following measurment. What is the terminal side?. Scholars explore the relationships between the angle and reference angles along with coterminal angles. A negative angle is 45° 360°( 2) or 675°. The coterminal angles can be positive or negative. Learn More. Trigonometry: Reference Angles Science and Mathematics Education Research Group Supported by UBC Teaching and Learning Enhancement Fund 2012-2013 Department of Curriculum and Pedagogy a place of mind F A C U L T Y O F E D U C A T I O N. Every angle, θ (the Greek letter theta), has a reference angle, , associated with it. (look below). Coterminal angles are found by adding/subtracting 360 degrees (for degree angle measure) or 2pi (for radian angle measure) to/from the given angle. To determine coterminal angles for a given angle. Coterminal definition, having a common boundary; bordering; contiguous. An angle's reference angle is the measure of the smallest, positive, acute angle t formed by the terminal side of the angle t. Ex 2: Find one positive and one negative angle that are coterminal to 112º. If θ is an acute angle is standard position and B (-3, 4) is a point on the terminal side of the angle, what is the value of. Following this procedure, all coterminal angles can be found. FINDING COTERMINAL ANGLES 3. Coterminal angles: are angles in standard position (angles with the initial side on the positive x-axis) that have a common terminal side. For example: Find the standard coterminal angle for a. Intro to Video: Reference Triangles and Sine, Cosine, and Tangent; Overview of Reference Triangles, Reference Angles, and SOH-CAH-TOA; Theorem for Reference Triangles and Trigonometric Functions with Examples. Use the below online coterminal angle calculator to find out the positive and negative coterminal angles for the given angle by entering angle value in the input field. Using our calculators, we found that θ = tan-1 0. 68% average accuracy. So suppose the given angle has this property, i. Draw the angle in standard position b. For example 30°, -330° and 390° are all coterminal. -3750 77t 12. Coterminal Angles are angles who share the same initial side and terminal sides. Coterminal Angles. All of the angles in the following figure are coterminal with an angle of degree measure of 45°. It also shows you how to. Draw each angle, θ, in standard position. Reference Angles and Coterminal Angles Draw an angle with the given measure in standard position. To preview this test, click on the File menu and select Print Preview. -xative coterminal angle. I understand perfectly how to find the reference angle for a degree, such as 150 degress = reference angle of 30 degrees, because the 2nd quadrant goes from 90 to 180 degrees, so you simply subtract 150 from 180 to come up with 30. Function Values. Four Examples – Find a coterminal angle within a specified domain; Reference Triangles. Identify all angles coterminal with the given angle. Reference angles are typically used in trigonometric theorem problems. The first example we did was: Find θ, given that tan θ = 0. Scroll down the page for more examples and solutions. Draw the angle in standard position Reference angle. Name four angles between 00 and 3600 with a reference angle of 200 If each angle is in standard position, determine a coterminal angle that is between 00 and 3600. But if an angle is less than 90 degrees, so, for example, both of these angles that we started our discussion with are less than 90 degrees, we call them \"acute angles. the clockwise direction, this angle has a measure of –270°. What I've done so far. So that is an acute angle, and that is an acute angle right over here. My question. Find positive and negative coterminal angles. Find the reference angle, α. Jan 299:53 AM. If angle is measured in degrees, add or subtract 360 If angle is measured in radians, add or subtract 2 Examples: Find a positive and a negative angle coterminal with the given angle. a) 460° b) 350° 0. A reference angle is the smallest angle between the terminal side of an angle and the x-axis. Trig Worksheet-Day 2 (Coterminal Angles & Angle Conversions) Determine two coterminal angles (one positive and one negative) for each angle. 1400/ Convert to radians. For each of the following angles,. In figure let the initial side is OX. The formulas above would only yield one of each co terminal angles. There are four unique mazes included. 1) -100° y 2) 1020° y x x 60° 80° 3) -130° y 4) 160° y 70° x x 50° 5) -290° y 6) -780° y 20° x x 60° Draw an angle with the given measure in standard position. Let n represent any integer. Learn the basics of trigonometry: What are sine, cosine, and tangent? How can we use them to solve for unknown sides and angles in right triangles?. Coterminal Angles. REFERENCE ANGLES ARE ALWAYS POSITIVE! To find a reference angle for angles outside the interval 0o < θ < 360o or 0π < θ < 2π you must first find a corresponding coterminal angle in this interval. Unit Circle Trigonometry Labeling Special Angles on the Unit Circle Labeling Special Angles on the Unit Circle We are going to deal primarily with special angles around the unit circle, namely the multiples of 30o, 45o, 60o, and 90o. Download it once and read it on your Kindle device, PC, phones or tablets."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8212873,"math_prob":0.9540417,"size":30167,"snap":"2020-10-2020-16","text_gpt3_token_len":7755,"char_repetition_ratio":0.25471604,"word_repetition_ratio":0.28131613,"special_character_ratio":0.255511,"punctuation_ratio":0.1186787,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9985889,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-07T17:09:50Z\",\"WARC-Record-ID\":\"<urn:uuid:21010239-00b0-4ff6-8627-7d05204f7033>\",\"Content-Length\":\"33079\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c50d50a7-c3b4-4392-bab4-444ec5b073e4>\",\"WARC-Concurrent-To\":\"<urn:uuid:fdfd88cd-8be8-4b05-9a32-e8168eefc86e>\",\"WARC-IP-Address\":\"104.18.49.180\",\"WARC-Target-URI\":\"http://msgf.aavt.pw/coterminal-and-reference-angles.html\",\"WARC-Payload-Digest\":\"sha1:QSJPTDYPRZAR34PLIIRHWELNK6HCZ7UC\",\"WARC-Block-Digest\":\"sha1:5VXTYJABWVY3K6DN5HFLEJSCDDRWIEUD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371803248.90_warc_CC-MAIN-20200407152449-20200407182949-00012.warc.gz\"}"} |
https://studyqas.com/the-formal-definition-of-a-polygon-states-that-a-polygon/ | [
"# The formal definition of a polygon states that a polygon is a closed plane figure formed by or more\n\nThe formal definition of a polygon states that a polygon is a closed plane figure formed by or more segments called sides. each side intersects exactly two other noncollinear sides, one at each endpoint.\n\na. four\nb. one\nc. zero\nd. two\ne. three\n\n## This Post Has 4 Comments\n\n1.",
null,
"lakiyalundy57 says:\n\nNo\n\nStep-by-step explanation:\n\nCollinear is if points are on the same line.\n\n2 points are not collinear, as they form a line.\n\nIf 3 points were on the same line, then it would be considered collinear.\n\n2.",
null,
"memeE15 says:\n\n1. Two points coplanar with the point D are the points A and B which are on the same plane, K, as the point D\n\n2. Another name for the line P is a straight straight angle AEB\n\n3. The intersection of line r and plane K is the point X\n\n4. A point non-coplanar to the plane K is the point W\n\nStep-by-step explanation:\n\n3.",
null,
"junkmailemail42 says:\n\nTwo non co-linear segments with a slope of 3 in a coordinate plane would mean that the two segments are parallel to each other and not attached to each other. Having the same slope is the definition of being parallel in the same plane.\n\n4.",
null,
"shyra94 says:\n\nThe answer would be three. The reasons are clear. You can't form a closed figure when you just have two sides. All they would do is just intersect each other. Also, one side must intersect with two other noncollinear sides. This means that a polygon has to have at least three sides."
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"https://secure.gravatar.com/avatar/",
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"https://secure.gravatar.com/avatar/",
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"https://secure.gravatar.com/avatar/",
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https://www.doglink.pt/russell-banks-hyb/1bqsn1i.php?tag=d18aef-huber-loss-python | [
"Here's how I implemented Huber Loss for Keras (note that I'm using Keras from Tensorflow 1.5). reduction: Type of reduction to apply to loss. weights: Optional Tensor whose rank is either 0, or the same rank as labels, and must be broadcastable to labels (i.e., all dimensions must be either 1, or the same as the corresponding losses dimension). Python chainer.functions.huber_loss() Examples The following are 13 code examples for showing how to use chainer.functions.huber_loss(). Now that we can start coding, letâs import the Python dependencies that we need first: ''' Keras model demonstrating Huber loss ''' from keras.datasets import boston_housing from keras.models import Sequential from keras.layers import Dense from keras.losses import huber_loss import numpy as np import matplotlib.pyplot as plt. import numpy as np import tensorflow as tf ''' ' Huber loss. Subscribe to the Fritz AI Newsletter to learn more about this transition and how it can help scale your business. Note: When beta is set to 0, this is equivalent to L1Loss.Passing a negative value in for beta will result in an exception. delta: float, the point where the huber loss function changes from a quadratic to linear. 5. The Huber Loss¶ A third loss function called the Huber loss combines both the MSE and MAE to create a loss function that is differentiable and robust to outliers. Quantile Loss. These examples are extracted from open source projects. The add_loss() API. scope: The scope for the operations performed in computing the loss. I came here with the exact same question. The Huber loss can be used to balance between the MAE (Mean Absolute Error), and the MSE (Mean Squared Error). x x x and y y y arbitrary shapes with a total of n n n elements each the sum operation still operates over all the elements, and divides by n n n.. beta is an optional parameter that defaults to 1. From the probabilistic point of view the least-squares solution is known to be the maximum likelihood estimate, provided that all $\\epsilon_i$ are independent and normally distributed random variables. When writing the call method of a custom layer or a subclassed model, you may want to compute scalar quantities that you want to minimize during training (e.g. Loss functions applied to the output of a model aren't the only way to create losses. More than 50 million people use GitHub to discover, fork, and contribute to over 100 million projects. Python code for Huber and Log-cosh loss functions: Machine learning is rapidly moving closer to where data is collected â edge devices. Such formulation is intuitive and convinient from mathematical point of view. regularization losses). predictions: The predicted outputs. Returns: Weighted loss float Tensor. The accepted answer uses logcosh which may have similar properties, but it isn't exactly Huber Loss. Args; labels: The ground truth output tensor, same dimensions as 'predictions'. loss_collection: collection to which the loss will be added. You can use the add_loss() layer method to keep track of such loss terms. GitHub is where people build software.\n2020 huber loss python"
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https://www.climate-policy-watcher.org/energy-sources/energy-units.html | [
"## Energy Units\n\nThe basic energy unit is the erg, defined as the work done by a force of 1 dyne moving a distance of 1 cm. The dyne is the force which, acting on a mass of 1 gram produce an acceleration of 1 cm per sec. A joule (J) is 10 ergs, and is also defined as the kinetic energy of a mass of 1 kg moving at 1 metre per second. Since this is very small for practical purposes, large multiples of the joule are frequently used, particularly the megajoule (MJ) (10(6)J), the gigajoule (GJ) (10(9)J), the pentajoule (10(15)J) and the exajoule (EJ) (10 (18) J). In the oil industry, the unit is the tonne of oil equivalent (TOE). A tonne is 1000 kg. 1 EJ = 22.7 TOE, or 1 TOE = 44 GJ. Also, 7.3 barrels = 12 tonnes of oil. One barrel of oil per day is 50 TOE per year. Rates of heat production are measured in watts. A watt is the rate of working of one joule per second, so the watt has 'per second' built into it. A kilowatt (kW) is 1000 watts, a megawatt (MW) is 10 (6) watts, a gigawatt (GW) is 10 (9) watts and a terawatt (TW) is 10 (12) watts. One kilowatt hour (kWh) is 3.6 MJ. 1 EJ per year is 32.2 GJ per second or 32.2 gigawatts. An energy unit used in nuclear physics is the electron volt (eV). One eV is 1.6 x 10 (-19) J. A million electron volts (1 MeV) is 1.6 x 10 (13) J. Each fission of a uranium nucleus releases 200 MeV = 3.2 x 10 (-13) J. One gram of uranium undergoing fission releases 82,000 MJ.\n\nIn discussion of energy the terms 'energy' and 'power' are often used indiscriminately. It is important to note that energy is power multiplied by time. Thus for example if an electric light bulb has a power of 100 watts and it is switched on for one hour it delivers 100Wh of energy."
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https://www.colorhexa.com/ffc3f0 | [
"# #ffc3f0 Color Information\n\nIn a RGB color space, hex #ffc3f0 is composed of 100% red, 76.5% green and 94.1% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 23.5% magenta, 5.9% yellow and 0% black. It has a hue angle of 315 degrees, a saturation of 100% and a lightness of 88.2%. #ffc3f0 color hex could be obtained by blending #ffffff with #ff87e1. Closest websafe color is: #ffccff.\n\n• R 100\n• G 76\n• B 94\nRGB color chart\n• C 0\n• M 24\n• Y 6\n• K 0\nCMYK color chart\n\n#ffc3f0 color description : Very pale pink.\n\n# #ffc3f0 Color Conversion\n\nThe hexadecimal color #ffc3f0 has RGB values of R:255, G:195, B:240 and CMYK values of C:0, M:0.24, Y:0.06, K:0. Its decimal value is 16761840.\n\nHex triplet RGB Decimal ffc3f0 `#ffc3f0` 255, 195, 240 `rgb(255,195,240)` 100, 76.5, 94.1 `rgb(100%,76.5%,94.1%)` 0, 24, 6, 0 315°, 100, 88.2 `hsl(315,100%,88.2%)` 315°, 23.5, 100 ffccff `#ffccff`\nCIE-LAB 85.293, 28.452, -13.923 76.481, 66.584, 91.256 0.326, 0.284, 66.584 85.293, 31.676, 333.925 85.293, 32.088, -26.743 81.599, 24.508, -9.188 11111111, 11000011, 11110000\n\n# Color Schemes with #ffc3f0\n\n• #ffc3f0\n``#ffc3f0` `rgb(255,195,240)``\n• #c3ffd2\n``#c3ffd2` `rgb(195,255,210)``\nComplementary Color\n• #f0c3ff\n``#f0c3ff` `rgb(240,195,255)``\n• #ffc3f0\n``#ffc3f0` `rgb(255,195,240)``\n• #ffc3d2\n``#ffc3d2` `rgb(255,195,210)``\nAnalogous Color\n• #c3fff0\n``#c3fff0` `rgb(195,255,240)``\n• #ffc3f0\n``#ffc3f0` `rgb(255,195,240)``\n• #d2ffc3\n``#d2ffc3` `rgb(210,255,195)``\nSplit Complementary Color\n• #c3f0ff\n``#c3f0ff` `rgb(195,240,255)``\n• #ffc3f0\n``#ffc3f0` `rgb(255,195,240)``\n• #f0ffc3\n``#f0ffc3` `rgb(240,255,195)``\n• #d2c3ff\n``#d2c3ff` `rgb(210,195,255)``\n• #ffc3f0\n``#ffc3f0` `rgb(255,195,240)``\n• #f0ffc3\n``#f0ffc3` `rgb(240,255,195)``\n• #c3ffd2\n``#c3ffd2` `rgb(195,255,210)``\n• #ff77dd\n``#ff77dd` `rgb(255,119,221)``\n• #ff90e3\n``#ff90e3` `rgb(255,144,227)``\n• #ffaaea\n``#ffaaea` `rgb(255,170,234)``\n• #ffc3f0\n``#ffc3f0` `rgb(255,195,240)``\n• #ffddf6\n``#ffddf6` `rgb(255,221,246)``\n• #fff6fd\n``#fff6fd` `rgb(255,246,253)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\nMonochromatic Color\n\n# Alternatives to #ffc3f0\n\nBelow, you can see some colors close to #ffc3f0. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #ffc3ff\n``#ffc3ff` `rgb(255,195,255)``\n• #ffc3fa\n``#ffc3fa` `rgb(255,195,250)``\n• #ffc3f5\n``#ffc3f5` `rgb(255,195,245)``\n• #ffc3f0\n``#ffc3f0` `rgb(255,195,240)``\n• #ffc3eb\n``#ffc3eb` `rgb(255,195,235)``\n• #ffc3e6\n``#ffc3e6` `rgb(255,195,230)``\n• #ffc3e1\n``#ffc3e1` `rgb(255,195,225)``\nSimilar Colors\n\n# #ffc3f0 Preview\n\nThis text has a font color of #ffc3f0.\n\n``<span style=\"color:#ffc3f0;\">Text here</span>``\n#ffc3f0 background color\n\nThis paragraph has a background color of #ffc3f0.\n\n``<p style=\"background-color:#ffc3f0;\">Content here</p>``\n#ffc3f0 border color\n\nThis element has a border color of #ffc3f0.\n\n``<div style=\"border:1px solid #ffc3f0;\">Content here</div>``\nCSS codes\n``.text {color:#ffc3f0;}``\n``.background {background-color:#ffc3f0;}``\n``.border {border:1px solid #ffc3f0;}``\n\n# Shades and Tints of #ffc3f0\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #12000e is the darkest color, while #fffeff is the lightest one.\n\n• #12000e\n``#12000e` `rgb(18,0,14)``\n• #26001d\n``#26001d` `rgb(38,0,29)``\n• #3a002b\n``#3a002b` `rgb(58,0,43)``\n• #4d003a\n``#4d003a` `rgb(77,0,58)``\n• #610049\n``#610049` `rgb(97,0,73)``\n• #750057\n``#750057` `rgb(117,0,87)``\n• #880066\n``#880066` `rgb(136,0,102)``\n• #9c0075\n``#9c0075` `rgb(156,0,117)``\n• #af0084\n``#af0084` `rgb(175,0,132)``\n• #c30092\n``#c30092` `rgb(195,0,146)``\n• #d700a1\n``#d700a1` `rgb(215,0,161)``\n• #ea00b0\n``#ea00b0` `rgb(234,0,176)``\n• #fe00be\n``#fe00be` `rgb(254,0,190)``\n• #ff12c4\n``#ff12c4` `rgb(255,18,196)``\n• #ff26c9\n``#ff26c9` `rgb(255,38,201)``\n• #ff3ace\n``#ff3ace` `rgb(255,58,206)``\n• #ff4dd3\n``#ff4dd3` `rgb(255,77,211)``\n• #ff61d7\n``#ff61d7` `rgb(255,97,215)``\n• #ff75dc\n``#ff75dc` `rgb(255,117,220)``\n• #ff88e1\n``#ff88e1` `rgb(255,136,225)``\n• #ff9ce6\n``#ff9ce6` `rgb(255,156,230)``\n• #ffafeb\n``#ffafeb` `rgb(255,175,235)``\n• #ffc3f0\n``#ffc3f0` `rgb(255,195,240)``\n• #ffd7f5\n``#ffd7f5` `rgb(255,215,245)``\n• #ffeafa\n``#ffeafa` `rgb(255,234,250)``\n• #fffeff\n``#fffeff` `rgb(255,254,255)``\nTint Color Variation\n\n# Tones of #ffc3f0\n\nA tone is produced by adding gray to any pure hue. In this case, #e3dfe2 is the less saturated color, while #ffc3f0 is the most saturated one.\n\n• #e3dfe2\n``#e3dfe2` `rgb(227,223,226)``\n• #e6dce3\n``#e6dce3` `rgb(230,220,227)``\n• #e8dae4\n``#e8dae4` `rgb(232,218,228)``\n``#ead8e6` `rgb(234,216,230)``\n• #edd5e7\n``#edd5e7` `rgb(237,213,231)``\n• #efd3e8\n``#efd3e8` `rgb(239,211,232)``\n• #f1d1e9\n``#f1d1e9` `rgb(241,209,233)``\n• #f3cfea\n``#f3cfea` `rgb(243,207,234)``\n• #f6cceb\n``#f6cceb` `rgb(246,204,235)``\n• #f8caed\n``#f8caed` `rgb(248,202,237)``\n• #fac8ee\n``#fac8ee` `rgb(250,200,238)``\n• #fdc5ef\n``#fdc5ef` `rgb(253,197,239)``\n• #ffc3f0\n``#ffc3f0` `rgb(255,195,240)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #ffc3f0 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.51618755,"math_prob":0.91512954,"size":3701,"snap":"2021-21-2021-25","text_gpt3_token_len":1641,"char_repetition_ratio":0.14417095,"word_repetition_ratio":0.011090573,"special_character_ratio":0.51364493,"punctuation_ratio":0.23006834,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9572251,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-15T21:56:14Z\",\"WARC-Record-ID\":\"<urn:uuid:277cd332-3515-41b9-9780-14a46f2984e4>\",\"Content-Length\":\"36335\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2f784850-2ef2-4247-9064-fe949cab2cc3>\",\"WARC-Concurrent-To\":\"<urn:uuid:6575b5d4-3243-4e7a-970a-80d3c47a8765>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/ffc3f0\",\"WARC-Payload-Digest\":\"sha1:QE5R4MEATGBNNW7O2MUKOIHPVLD6VQAJ\",\"WARC-Block-Digest\":\"sha1:2QGEGQNZ5TS4DUXARBGMCOVL63OMT6J6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991378.52_warc_CC-MAIN-20210515192444-20210515222444-00148.warc.gz\"}"} |
https://astarmathsandphysics.com/ib-maths-notes/sequences-and-series/4959-sum-of-a-telescoping-surd-series.html | [
"## Sum of a Telescoping Surd Series\n\nConsider the sequence\n$\\frac{1}{\\sqrt{2}+\\sqrt{3}}, \\: \\frac{1}{\\sqrt{3}+\\sqrt{4}}, \\frac{1}{\\sqrt{4}+\\sqrt{5}},..., \\: \\frac{1}{\\sqrt{n}+\\sqrt{n+1}}$\n.\nWe can find the sum of this series if\n$n$\nis finite by rationalising the denominator of each time.\n$\\frac{1}{\\sqrt{k}+\\sqrt{k+1}}= \\frac{1}{\\sqrt{k}+\\sqrt{k+1}} \\times \\frac{\\sqrt{k}-\\sqrt{k+1}}{\\sqrt{k}-\\sqrt{k+1}}=\\sqrt{k+1}-\\sqrt{k}$\n.\nThe sequence becomes\n$\\sqrt{3}-\\sqrt{2}, \\: \\sqrt{4}-\\sqrt{3}, \\: \\sqrt{5}-\\sqrt{4},..., \\: \\sqrt{n+1}-\\sqrt{n}$\n.\nNow we can add the series.\n$S_n = ( \\sqrt{3}-\\sqrt{2})+ (\\sqrt{4}-\\sqrt{3})+( \\sqrt{5}-\\sqrt{4})+...+( \\sqrt{n+1}-\\sqrt{n})$\n.\nThe last term in each bracket cancels with the first term in the next bracket except for the second term in the first bracket second term in the first bracket. The sum of the series is\n$S_n = -\\sqrt{2})+ \\sqrt{n+1}=\\sqrt{n+1}-\\sqrt{2}$\n.",
null,
""
] | [
null,
"https://astarmathsandphysics.com/component/jcomments/captcha/49691.html",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5882888,"math_prob":0.9999807,"size":888,"snap":"2022-40-2023-06","text_gpt3_token_len":350,"char_repetition_ratio":0.24660634,"word_repetition_ratio":0.021505376,"special_character_ratio":0.4403153,"punctuation_ratio":0.14851485,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000091,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-03T19:41:06Z\",\"WARC-Record-ID\":\"<urn:uuid:823e9208-b644-46db-b547-0d543e90dd65>\",\"Content-Length\":\"31434\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:918d458e-9982-4840-a727-8578e348144c>\",\"WARC-Concurrent-To\":\"<urn:uuid:40cc3c5c-0a78-49c8-8c0d-c7def4e14210>\",\"WARC-IP-Address\":\"172.67.134.107\",\"WARC-Target-URI\":\"https://astarmathsandphysics.com/ib-maths-notes/sequences-and-series/4959-sum-of-a-telescoping-surd-series.html\",\"WARC-Payload-Digest\":\"sha1:FHE7SJDBLIJKAGRGJW5BW3LMN4IND6ZY\",\"WARC-Block-Digest\":\"sha1:PUR2EXXU5HNP4ZBQ25BG4O7JD4YXBHPN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500074.73_warc_CC-MAIN-20230203185547-20230203215547-00210.warc.gz\"}"} |
https://research-portal.uea.ac.uk/en/publications/remarks-on-singular-cayley-graphs-and-vanishing-elements-of-simpl | [
"# Remarks on singular Cayley graphs and vanishing elements of simple groups\n\nResearch output: Contribution to journalArticle\n\n3 Citations (Scopus)\n\n## Abstract\n\nLet Γ be a finite graph and let A(Γ) be its adjacency matrix. Then Γ is singular if A(Γ) is singular. The singularity of graphs is of certain interest in graph theory and algebraic combinatorics. Here we investigate this problem for Cayley graphs Cay(G,H) when G is a finite group and when the connecting set H is a union of conjugacy classes of G. In this situation, the singularity problem reduces to finding an irreducible character χ of G for which ∑h∈Hχ(h)=0. At this stage, we focus on the case when H is a single conjugacy class hG of G; in this case, the above equality is equivalent to χ(h)=0 . Much is known in this situation, with essential information coming from the block theory of representations of finite groups. An element h∈G is called vanishing if χ(h)=0 for some irreducible character χ of G. We study vanishing elements mainly in finite simple groups and in alternating groups in particular. We suggest some approaches for constructing singular Cayley graphs.\nOriginal language English 379-401 23 Journal of Algebraic Combinatorics 50 4 30 Nov 2018 https://doi.org/10.1007/s10801-018-0860-0 Published - Dec 2019\n\n## Keywords\n\n• Singular Cayley graphs\n• Vertex transitive graphs\n• Vanishing elements\n• Block theory of symmetric and alternating groups"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.85578716,"math_prob":0.5979042,"size":1534,"snap":"2021-31-2021-39","text_gpt3_token_len":370,"char_repetition_ratio":0.09803922,"word_repetition_ratio":0.0,"special_character_ratio":0.22294654,"punctuation_ratio":0.073529415,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9800472,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-22T15:29:52Z\",\"WARC-Record-ID\":\"<urn:uuid:3cdf9b67-9d82-478c-acfc-3525217b0b4c>\",\"Content-Length\":\"48429\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:634a91ff-abb5-4649-ae7c-f835cae8035d>\",\"WARC-Concurrent-To\":\"<urn:uuid:d5d12861-df8f-4228-90f9-b51c52cc3315>\",\"WARC-IP-Address\":\"34.248.98.230\",\"WARC-Target-URI\":\"https://research-portal.uea.ac.uk/en/publications/remarks-on-singular-cayley-graphs-and-vanishing-elements-of-simpl\",\"WARC-Payload-Digest\":\"sha1:LPJ3DRX3MFCT3UCQ63H7QSEJQMC4AIFD\",\"WARC-Block-Digest\":\"sha1:6Q44YWPA6BXH5UUJDKFSLZHG5HPKMYJ3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057366.40_warc_CC-MAIN-20210922132653-20210922162653-00249.warc.gz\"}"} |
https://books.google.com.jm/books?id=IQQFAAAAQAAJ&lr= | [
"# An introduction to geometry, consisting of Euclid's Elements, book i, accompanied by numerous explanations, questions, and exercises, by J. Walmsley. [With] Answers, Volume 1\n\n1884\n0 Reviews\nReviews aren't verified, but Google checks for and removes fake content when it's identified\n\n### What people are saying -Write a review\n\nWe haven't found any reviews in the usual places.\n\n### Popular passages\n\nPage 132 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.\nPage 86 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.\nPage 139 - If two triangles have two angles of the one equal to two angles of the other, each to each, and also one side of the one equal to the corresponding side of the other, the triangles are congruent.\nPage 133 - The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.\nPage 134 - Prove that parallelograms on the same base and between the same parallels are equal in area.\nPage 134 - To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line ; it is required to draw a straight line through the point A, parallel to the straight hue BC.\nPage 50 - if two straight lines\" &c. QED COR. 1. From this it is manifest, that if two straight lines cut one another, the angles which they make at the point where they cut, are together equal to four right angles.\nPage 20 - PROB. from a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line : it is required to draw from the point A a straight line equal to BC.\nPage 96 - Parallelograms upon the same base and between the same parallels, are equal to one another.\nPage 49 - If at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles ; then these two straight lines shall be in one and the same straight line."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.855443,"math_prob":0.99281496,"size":3201,"snap":"2023-14-2023-23","text_gpt3_token_len":689,"char_repetition_ratio":0.17766656,"word_repetition_ratio":0.12915128,"special_character_ratio":0.20774758,"punctuation_ratio":0.09931507,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9972791,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-03T05:10:37Z\",\"WARC-Record-ID\":\"<urn:uuid:87e13e81-768c-4a1b-b263-b95c99395957>\",\"Content-Length\":\"54941\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:46216de7-ad26-4432-890e-0b716e98b306>\",\"WARC-Concurrent-To\":\"<urn:uuid:a9ef6db9-3098-40fe-9b62-5926015bc6f2>\",\"WARC-IP-Address\":\"172.253.63.101\",\"WARC-Target-URI\":\"https://books.google.com.jm/books?id=IQQFAAAAQAAJ&lr=\",\"WARC-Payload-Digest\":\"sha1:YOCNOYMOZHDLF76FQEPM6TLPBUVQV6HL\",\"WARC-Block-Digest\":\"sha1:A5HCN3D2BADUKFGLUKUN4366IHA5FADT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649105.40_warc_CC-MAIN-20230603032950-20230603062950-00182.warc.gz\"}"} |
https://www.transtutors.com/questions/the-energy-band-diagram-for-a-forward-biased-si-pn-junction-diode-maintained-under-s-777867.htm | [
"# The energy band diagram for a forward-biased Si pn-junction diode maintained under steady-state...\n\nThe energy band diagram for a forward-biased Si pn-junction diode maintained under steady-state conditions at room temperature (T = 300 K) is pictured in Fig. P5.6. Note that EFn - Ej = Ej - EFp = EG/4.",
null,
"(a) For the particular situation pictured in Fig. P5.6, and assuming single-level R-G center statistics, Tn = Tp = T, and ET \" = Ej , show that the steady-state net recombination rate inside the electrostatic depletion region (-xp ≤ X ≤ xn) can be simplified to",
null,
"(b) Plot R/(n/r) versus x for -xp ≤ X ≤ Xn\" Assume the Ei variation between x = -xp and x = Xn is approximately linear in constructing the plot.\n\n(c) What is the purpose or point of this problem?",
null,
"## Plagiarism Checker\n\nSubmit your documents and get free Plagiarism report\n\nFree Plagiarism Checker"
] | [
null,
"https://files.transtutors.com/book/qimage/image04012015052.png",
null,
"https://files.transtutors.com/book/qimage/image04012015053.png",
null,
"https://files.transtutors.com/resources/images/answer-blur.webp",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86523485,"math_prob":0.9758922,"size":954,"snap":"2021-04-2021-17","text_gpt3_token_len":243,"char_repetition_ratio":0.07684211,"word_repetition_ratio":0.0,"special_character_ratio":0.25262055,"punctuation_ratio":0.09139785,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.988422,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-17T10:04:09Z\",\"WARC-Record-ID\":\"<urn:uuid:f4150c83-4864-4ba2-8661-de0cd8e158be>\",\"Content-Length\":\"77364\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1aab651a-82d6-4d10-a7a2-be7b0e9c31ab>\",\"WARC-Concurrent-To\":\"<urn:uuid:7f36b9da-00f5-4896-b7b7-177a97e067fc>\",\"WARC-IP-Address\":\"34.224.249.39\",\"WARC-Target-URI\":\"https://www.transtutors.com/questions/the-energy-band-diagram-for-a-forward-biased-si-pn-junction-diode-maintained-under-s-777867.htm\",\"WARC-Payload-Digest\":\"sha1:X46OF7ZTFZICJKOUP7RKNDCM4RKCMGCI\",\"WARC-Block-Digest\":\"sha1:HOUPL2FNJOQWUK27TNITK3L7UAMC3ZIM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703511903.11_warc_CC-MAIN-20210117081748-20210117111748-00600.warc.gz\"}"} |
https://statisticsglobe.com/extract-numerator-denominator-fraction-r | [
"# Extract Numerator & Denominator of Fraction in R (Example)\n\nIn this R programming tutorial, you’ll learn how to extract numerators and denominators of fractions.\n\nThe tutorial is structured as follows:\n\nLet’s take a look at some R codes in action.\n\n## Example Data\n\nBefore all else, we have to define some data that we can use in the examples below.\n\n`fractions <- c(\"5/3\", \"6/4\", \"7/2\") # define fraction vector`\n\nAs seen, we defined a vector of fractions called `fractions`.\n\nPlease be aware that the fractions are in the string format here. If you are interested in numeric fractions, see my tutorial Convert Decimals to Reduced Fractions in R.\n\nIn the following example, we will see how to extract the numerators and denominators of the given fractions.\n\n## Example: Get Numerator & Denominator of Fractions\n\nIn this example, we will use the strplit() function, which splits the elements of a string vector into substrings according to the matching substring. In our case, this substring is the slash symbol `\"/\"`.\n\n```split_fract <- strsplit(fractions, \"/\") # split fraction into two parts\nsplit_fract\n# []\n# \"5\" \"3\"\n\n# []\n# \"6\" \"4\"\n\n# []\n# \"7\" \"2\"```\n\nThe previous output of the RStudio console shows that `strsplit()` split the fractions into parts of numerators and denominators. All split fractions are stored in a separate vector within the output list `split_fract`.\n\nIn order to extract the numerators, we should retrieve the first elements of these vectors. Let’s see how it is done!\n\n```num<- as.integer(sapply(split_fractions, \"[\", 1)) # extract numerators\nnum # print numerators\n# 5 6 7```\n\nIn the script above, `\"[\"` is the indexing function, which access to the indexes of a list or a vector. `1` is the parameter indicating which index will be accessed.\n\nThe sapply() function is employed to apply this indexing function to each vector of the `split_fract` list. Next, we will extract the denominator in a similar manner.\n\n```denom <- as.integer(sapply(split_fractions, \"[\", 2)) # extract denominators\ndenom # print denominators\n# 3 4 2```\n\nAs promised, all denominators are extracted calling the second index of the vectors of `split_fract`.\n\n## Video, Further Resources & Summary\n\nDo you want to know more about the extraction of numerator and denominators? Then you may want to have a look at the following video on my YouTube channel. I’m explaining the R code of this tutorial in the video.\n\nSummary: In this article, I have demonstrated how to get numerators and denominators in R programming. Let me know in the comments, in case you have further questions.\n\nThis page was created in collaboration with Cansu Kebabci. Have a look at Cansu’s author page to get more information about her professional background, a list of all his tutorials, as well as an overview on her other tasks on Statistics Globe.\n\nSubscribe to the Statistics Globe Newsletter"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8438735,"math_prob":0.7178964,"size":3053,"snap":"2023-40-2023-50","text_gpt3_token_len":718,"char_repetition_ratio":0.1554608,"word_repetition_ratio":0.007858546,"special_character_ratio":0.2358336,"punctuation_ratio":0.10676157,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9954523,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-21T23:28:44Z\",\"WARC-Record-ID\":\"<urn:uuid:b269ecbd-e033-4592-9888-8c56b1f16a58>\",\"Content-Length\":\"206603\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:931aa5d0-40f3-4835-91c1-f1d5146c3185>\",\"WARC-Concurrent-To\":\"<urn:uuid:355b7ebf-feb6-4973-b78f-60835bb73b9c>\",\"WARC-IP-Address\":\"217.160.0.159\",\"WARC-Target-URI\":\"https://statisticsglobe.com/extract-numerator-denominator-fraction-r\",\"WARC-Payload-Digest\":\"sha1:VEGIF6SN7NV2OLBPEFJ7YRUGM6Y4B2F6\",\"WARC-Block-Digest\":\"sha1:T7UBLIL23TYTYPENE5QXLXNXYXS6WQAV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506045.12_warc_CC-MAIN-20230921210007-20230922000007-00504.warc.gz\"}"} |
https://fmhelp.filemaker.com/help/15/fmp/en/FMP_Help/financial-functions.html | [
"Reference > Functions reference > Financial functions\n\nFinancial functions\nFinancial functions calculate financial information, such as net present value and payments. For example, you can calculate the monthly payments required to buy a car at a certain loan rate using the PMT function.\nClick a function name for details.\n\n This function Returns FV The future value (FV) of an initial investment, based on a constant interest rate and payment amount for the number of periods in months. NPV The net present value (NPV) of a series of unequal payments made at regular intervals, assuming a fixed interest rate per interval. PMT The payment (PMT) required by the term, interest rate, and principal. PV The present value (PV) of a series of equal payments made at regular intervals, assuming a fixed interest rate per interval.\nRelated topics"
] | [
null
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https://issues.apache.org/jira/browse/SIS-455 | [
"",
null,
"# Compute length of cubic Bézier curve\n\nXMLWordPrintableJSON\n\n#### Details\n\n•",
null,
"Improvement\n• Status: Closed\n•",
null,
"Major\n• Resolution: Won't Fix\n• None\n• None\n\n#### Description\n\nWe need a way to estimate the length of a cubic Bézier curve from its starting point at t=0 to an arbitrary t value where t[0…1]. Conversely, we need to estimate the t parameter for a given length since the starting point. There is no exact solution for this problem, but we may estimate the length using Legendre-Gauss approach documented in A Primer on Bézier Curves page. The accuracy is determined by the number Gaussian Quadrature Weights and Abscissae used. For example with 3 terms:\n\n```w₁ = 0.8888888888888888; a₁ = 0;\nw₂ = 0.5555555555555556; a₂ = -0.7745966692414834;\nw₃ = 0.5555555555555556; a₃ = +0.7745966692414834;\nlength(t) ≈ t/2 * (w₁*f(a₁*t/2 - t/2) + w₂*f(a₂*t/2 - t/2) + w₃*f(a₃*t/2 - t/2))\n```\n\nwith f(t) defined as hypot(x′(t), y′(t)) and with x′(t) and y′(t) the first derivatives of Bézier equations for x(t) and y(t).\n\nOnce we have the length for a given t value, we can try to find the converse by using an iterative approach as described in the Moving Along a Curve with Specified Speed paper from geometric tools.\n\nOnce we are able to estimate the t parameters for a given length, we should delete the Bezier.isValid(x, y) method (and consequently remove its use and the TransformException in the curve method). Instead, given the geodesic distance from Bézier start point to ¼ of the distance from start point to end point, estimate the t parameter at that position. It should be a value close but not identical to t≈¼. We can then compute the (x, y) coordinates of the point on that curve at that t parameter value and compare with the expected coordinates. It should (hopefully) be a point closer to expected than the point computed at exactly t=¼, thus removing the need for the Bezier.isValid(x,y) hack.\n\nAlternatively, all the above is a complicated way to say that we want the shortest distance between a point on the geodesic path and a point on the curve which is known to be at position close to (but not exactly at) t≈¼ and t≈¾.\n\n#### People",
null,
"Martin Desruisseaux",
null,
"Martin Desruisseaux"
] | [
null,
"https://issues.apache.org/jira/secure/projectavatar",
null,
"https://issues.apache.org/jira/secure/viewavatar",
null,
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null,
"https://issues.apache.org/jira/secure/useravatar",
null,
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https://x-largeclothing.com/how-to-435 | [
"",
null,
"",
null,
"# Apps to help with math problems\n\nLooking for an app to help with math problems? Check out our list of the best apps to help with everything from basic arithmetic to calculus.\n\n• Decide math question\n• Fast solutions\n• Decide mathematic equation\nx\n\n## 8 Best Math Solver Apps for iOS and Android\n\n• 1\n\nProvide multiple ways\n\nThere are many ways to skin a cat, and each person has their own method that works best for them.\n\n• 2\n\nPassing Quality\n\nmath is the study of numbers, shapes, and patterns. It is used in everyday life, from counting to measuring to more complex calculations.\n\n• 3\n\nMath can be confusing, but there are ways to make it easier. One way is to clear up the equations.\n\n• 4\n\nDo math equation\n\nNo matter what question you have, you can always find an answer with a quick online search.\n\n## This app doesn't just do your homework for you, it shows you",
null,
"## 9 Best Apps to Do Math Homework for You\n\nDecide math equation\n\nClear up math problem\n\nExplain math problems\n\nHomework Support Solutions\n\n## Photomath: Home",
null,
"Solve mathematic\n\nThe answer to the equation is 4.",
null,
"Do math problems\n\nTo pass quality, the sentence must be free of errors and meet the required standards.",
null,
"Clarify mathematic problems\n\nDoing math problems can be a great way to improve your math skills.",
null,
"## 5 Best Math Problem Solver Apps to Help Students with Their\n\nLooking for some help with math? Check out our list of the best apps to help with math problems.\nGet Started",
null,
"",
null,
"",
null,
"`"
] | [
null,
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null,
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null,
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null,
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http://www.silencelee.com/518.html | [
"HTML 表单\n\n1. 表单是一个包含表单元素的区域。\n3. 表单使用表单标签 <form> 来设置.\n<form>\n.\ninput 元素\n.\n</form>\n\nHTML 表单 - 输入元素\n\n单行文本域（Text Fields）\n\n<form>\nFirst name: <input type=\"text\" name=\"firstname\"><br>\nLast name: <input type=\"text\" name=\"lastname\">\n</form>\n\n多行文本域(Textarea)\n\n<textarea rows=\"10\" cols=\"30\">\n\n</textarea>\n\n<form>\n</form>\n\n<form>\n</form>\n\n复选框（Checkboxes）\n\n<input type=\"checkbox\"> 定义了复选框. 用户需要从若干给定的选择中选取一个或若干选项。\n\n<form>\n<input type=\"checkbox\" name=\"vehicle\" value=\"Bike\">I have a bike<br>\n<input type=\"checkbox\" name=\"vehicle\" value=\"Car\">I have a car\n</form>\n\n预选下拉框\n\n<form action=\"\">\n<select name=\"cars\">\n<option value=\"volvo\">Volvo</option>\n<option value=\"saab\">Saab</option>\n<option value=\"fiat\" selected>Fiat</option>\n<option value=\"audi\">Audi</option>\n</select>\n</form>\n\n提交按钮(Submit Button)\n\n<input type=\"submit\"> 定义了提交按钮. 当用户单击确认按钮时，表单的内容会被传送到另一个文件。表单的动作属性定义了目的文件的文件名。由动作属性定义的这个文件通常会对接收到的输入数据进行相关的处理。\n\n<!-- action定义这个form表单要提交的处理的程序，method定义了处理程序接受这个表单的传输方法 -->\n<form name=\"input\" action=\"html_form_action.php\" method=\"get\">\n<!-- value值是提交按钮上显示的文字 -->\n<input type=\"submit\" value=\"Submit\">\n</form>\n\n其他类型以及新标签\n\n<button>\n\n<button> 标签定义一个按钮。 在 <button> 元素内部，您可以放置内容，比如文本或图像。这是该元素与使用 <input> 元素创建的按钮之间的不同之处。\n\nHTML5 <datalist> 新标签\n\n1. <datalist> 标签规定了 <input> 元素可能的选项列表。\n2. <datalist> 标签被用来在为 <input> 元素提供\"自动完成\"的特性。用户能看到一个下拉列表，里边的选项是预先定义好的，将作为用户的输入数据。\n\n<form action=\"demo-form.php\" method=\"get\">\n<input list=\"browsers\" name=\"browser\">\n<datalist id=\"browsers\">\n<option value=\"Internet Explorer\">\n<option value=\"Firefox\">\n<option value=\"Chrome\">\n<option value=\"Opera\">\n<option value=\"Safari\">\n</datalist>\n<input type=\"submit\">\n</form>\n\nHTML5 <output> 标签\n\n<output> 标签作为计算结果输出显示(比如执行脚本的输出)\n\n<form oninput=\"x.value=parseInt(a.value)+parseInt(b.value)\">0\n<input type=\"range\" id=\"a\" value=\"50\">100\n+<input type=\"number\" id=\"b\" value=\"50\">\n=<output name=\"x\" for=\"a b\"></output>\n</form>\n\n<output> 标签属性以及属性值\n\nfor element_id 描述计算中使用的元素与计算结果之间的关系。\nform form_id 定义输入字段所属的一个或多个表单。\nnameN name 定义对象的唯一名称（表单提交时使用）。\n\n表单应用实例\n\n带边框的表单\n<form action=\"\">\n<fieldset>\n<legend>Personal information:</legend>\nName: <input type=\"text\" size=\"30\"><br>\nE-mail: <input type=\"text\" size=\"30\"><br>\nDate of birth: <input type=\"text\" size=\"10\">\n</fieldset>\n</form>\n\n发送邮件表单\n<h3>发送邮件到 [email protected]:</h3>\n<form action=\"MAILTO:[email protected]\" method=\"post\" enctype=\"text/plain\">\nName:<br>\nE-mail:<br>\nComment:<br>\n<input type=\"text\" name=\"comment\" value=\"your comment\" size=\"50\"><br><br>\n<input type=\"submit\" value=\"发送\">\n<input type=\"reset\" value=\"重置\">\n</form>\n\nform标签相关\n\n<form> 定义供用户输入的表单\n<input> 定义输入域\n<textarea> 定义文本域 (一个多行的输入控件)\n<label> 定义了 <input> 元素的标签，一般为输入标题\n<fieldset> 定义了一组相关的表单元素，并使用外框包含起来\n<legend> 定义了 <fieldset> 元素的标题\n<select> 定义了下拉选项列表\n<optgroup> 定义选项组\n<option> 定义下拉列表中的选项\n<button> 定义一个点击按钮\n<datalist> H5 指定一个预先定义的输入控件选项列表\n<keygen> H5 废弃 定义了表单的密钥对生成器字段\n<output> H5 定义一个计算结果"
] | [
null
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https://www.marsja.se/pandas-excel-tutorial-how-to-read-and-write-excel-files/?share=twitter | [
"Last updated on November 13, 2019\n\nIn this Pandas Excel tutorial, we will learn how to work with Excel files in Python. It will provide an overview of how to use Pandas to load xlsx files and write spreadsheets to Excel.\n\nIn the first section, we will go through, with examples, how to use Pandas read_excel to; read an Excel file, read specific columns from a spreadsheet, read multiple spreadsheets and combine them to one dataframe. Furthermore, we are going to learn how to read many Excel files, and how to convert data according to specific datatypes (e.g., using Pandas dtypes).\n\nWhen we have done this, we will continue by learning how to use Pandas to write Excel files; how to name the sheets and how to write to multiple sheets.\n\n## How to Install Pandas\n\nBefore we continue with this Pandas read and write Excel files tutorial there is something we need to do; installing Pandas (and Python, of course, if it’s not installed). We can install Pandas using Pip, given that we have Pip installed, that is. See here how to install pip.\n\n``````# Linux Users\npip install pandas\n\n# Windows Users\npython pip install pandas``````\n\nNote, if pip is telling us that there’s a newer version of pip, we may want to upgrade it. In a recent post, we cover how this is done:\n\n### Installing Anaconda Scientific Python Distribution\n\nAnother great option is to consider is to install the Anaconda Python distribution. This is really an easy and fast way to get started with computer science. No need to worry about installing the packages you need to do computer science separately.\n\nBoth of the above methods are explained in this tutorial.\n\n## How to Read Excel Files to Pandas Dataframes:\n\nCan Pandas read xlsx files? The short answer is, of course, “yes”. In this section, we are going to learn how to read Excel files and spreadsheets to Pandas dataframe objects. All examples in this Pandas Excel tutorial use local files. Note, that read_excel also can also load Excel files from a URL to a dataframe. As always when working with Pandas, we have to start by importing the module:\n\n``import pandas as pd``\n\nNow it’s time to learn how to use Pandas read_excel to read in data from an Excel file. The easiest way to use this method is to pass the file name as a string. If we don’t pass any other parameters, such as sheet name, it will read the first sheet in the index. In the first example we are not going to use any parameters:\n\n``````df = pd.read_excel('MLBPlayerSalaries.xlsx')\n\nHere, Pandas read_excel method read the data from the Excel file into a Pandas dataframe object. We then stored this dataframe into a variable called df.\n\nWhen using read_excel Pandas will, by default, assign a numeric index or row label to the dataframe, and as usual, when int comes to Python, the index will start with zero. We may have a reason to leave the default index as it is.\n\nFor instance, if your data doesn’t have a column with unique values that can serve as a better index. In case there is a column that would serve as a better index, we can override the default behavior.\n\nThis is done by setting the index_col parameter to a column. It takes a numeric value for setting a single column as index or a list of numeric values for creating a multi-index. In the example below, we use the column ‘Player’ as indices. Note, these are not unique and it may, thus, not make sense to use these values as indices.\n\n``df = pd.read_excel('MLBPlayerSalaries.xlsx', sheet_names='MLBPlayerSalaries', index_col='Player')``\n\n### Importing an Excel File to Pandas in Two Easy Steps:\n\nTime needed: 1 minute.\n\nHere’s a quick answer to the How do you import an Excel file into Python using Pandas? Importing an Excel file into a Pandas dataframe basically only requires two steps, given that we know the path, or URL, to the Excel file:\n\n1. Import Pandas\n\nIn the script type import pandas as pd\n\nNext step is to type df = pd.read_excel(FILE_PATH_OR_URL)\nRemember to change FILE_PATH_OR_URL to the path or the URL of the Excel file.\n\nNow that we know how easy it is to load an Excel file into a Pandas dataframe we are going to continue with learning more about the read_excel method.\n\nWhen using Pandas read_excel we will automatically get all columns from an Excel file. If we, for some reason, don’t want to parse all columns in the Excel file, we can use the parameter usecols. Let’s say we want to create a dataframe with the columns PlayerSalary, and Position, only. We can do this by adding 1, 3, and 4 in a list:\n\n``````cols = [1, 2, 3]\n\nAccording to the read_excel documentation, we should be able to put in a string. For instance, cols=’Player:Position’ should give us the same results as above.\n\n### Missing Data\n\nIf our data has missing values in some cells and these missing values are coded in some way, like “Missing” we can use the na_values parameter.\n\n#### Pandas Read Excel Example with Missing Data\n\nIn the example below, we are using the parameter na_values and we are putting in a string (i.e., “Missing’):\n\n``````df = pd.read_excel('MLBPlayerSalaries_MD.xlsx', na_values=\"Missing\", sheet_names='MLBPlayerSalaries', usecols=cols)\n\n### How to Skip Rows when Reading an Excel File\n\nNow we will learn how to skip rows when loading an Excel file using Pandas. For this read excel example, we will use data that can be downloaded here.\n\nIn the following Pandas read_excel example we load the sheet ‘session1’, which contains rows that we need to skip (these rows contain some information about the dataset).\n\nWe will use the parameter sheet_name=’Session1′ to read the sheet named ‘Session1’ (the example data contains more sheets; e.g., ‘Session2’ will load that sheet). Note, the first sheet will be read if we don’t use the sheet_name parameter. In this example, the important part is the parameter skiprow=2. We use this to skip the first two rows:\n\n``````df = pd.read_excel('example_sheets1.xlsx', sheet_name='Session1', skiprows=2)\n\nAnother way to get Pandas read_excel to read from the Nth row is by using the header parameter. In the example Excel file, we use here, the third row contains the headers and we will use the parameter header=2 to tell Pandas read_excel that our headers are on the third row.\n\n``df = pd.read_excel('example_sheets1.xlsx', sheet_name='Session1', header=2)``\n\nNow, if we want Pandas read_excel to read from the second row, we change the number in the skiprows and header arguments to 2, and so on.\n\n## Reading Multiple Excel Sheets to Pandas Dataframes\n\nIn this section of the Pandas read excel tutorial, we are going to learn how to read multiple sheets. Our Excel file, example_sheets1.xlsx’, has two sheets: ‘Session1’, and ‘Session2.’ Each sheet has data from an imagined experimental session. In the next example we are going to read both sheets, ‘Session1’ and ‘Session2’. Here’s how to use Pandas read_excel to read multiple sheets:\n\n``df = pd.read_excel('example_sheets1.xlsx', sheet_name=['Session1', 'Session2'], skiprows=2)``\n\nBy using the parameter sheet_name, and a list of names, we will get an ordered dictionary containing two dataframes:\n\n``df``\n\nWhen working with Pandas read_excel we may want to join the data from all sheets (in this case sessions). Merging Pandas dataframes are quite easy; we just use the concat function and loop over the keys (i.e., sheets):\n\n``df2 = pd.concat(df[frame] for frame in data.keys())``\n\nNow in the example Excel file, there is a column identifying the dataset (e.g., session number). However, maybe we don’t have that kind of information in our Excel file. To merge the two dataframes and adding a column depicting which session we can use a for loop:\n\n``````dfs = []\nfor framename in data.keys():\ntemp_df = data[framename]\ntemp_df['Session'] = framename\ndfs.append(temp_df)\n\ndf = pd.concat(dfs)``````\n\nIn the code above, we start by creating a list and continue by looping through the keys in the list of dataframes. Finally, we create a temporary dataframe and take the sheet name and add it in the column ‘Session’.\n\nNow, it is, of course, possible that when we want to read multiple sheets we also want to read all the sheets in the Excel file. That is, if we want to use read_excel to load all sheets from an Excel file to a dataframe it is possible. When reading multiple sheets and we want all sheets we can set the parameter sheet_name to None.\n\n``all_sheets_df = pd.read_excel('example_sheets1.xlsx', sheet_name=None)``\n\nIn this section, of the Pandas read excel tutorial, we will learn how to load many files into a Pandas dataframe because, in some cases, we may have a lot of Excel files containing data from, let’s say, different experiments. In Python, we can use the modules os and fnmatch to read all files in a directory. Finally, we use list comprehension to use read_excel on all files we found:\n\n``````import os, fnmatch\nxlsx_files = fnmatch.filter(os.listdir('.'), '*concat*.xlsx')\n\ndfs = [pd.read_excel(xlsx_file) for xlsx_file in xlsx_files]``````\n\nIf it makes sense we can, again, use the function concat to merge the dataframes:\n\n``df = pd.concat(dfs, sort=False)``\n\nThere are other methods for reading many Excel files and merging them. We can, for instance, use the module glob together with Pandas concat to read multiple xlss files:\n\n``````import glob\nlist_of_xlsx = glob.glob('./*concat*.xlsx')\ndf = pd.concat(list_of_xlsx)``````\n\n## Setting the Data type for data or columns\n\nWe can also, if we like, set the data type for the columns. Let’s use Pandas to read the example_sheets1.xlsx again. In the Pandas read_excel example below we use the dtype parameter to set the data type of some of the columns.\n\n``````df = pd.read_excel('example_sheets1.xlsx',sheet_name='Session1',\n'Mean':int, 'Session':str})``````\n\nWe can use the method info to see the what data types the different columns have:\n\n``df.info()``\n\n## Writing Pandas Dataframes to Excel\n\nExcel files can, of course, be created in Python using Pandas to_excel method. In this section of the post, we will learn how to create an excel file using Pandas. First, before writing an Excel file, we will create a dataframe containing some variables. Before that, we need to import Pandas:\n\n``import pandas as pd``\n\nThe next step is to create the dataframe. We will create the dataframe using a dictionary. The keys will be the column names and the values will be lists containing our data:\n\n``````df = pd.DataFrame({'Names':['Andreas', 'George', 'Steve',\n'Sarah', 'Joanna', 'Hanna'],\n'Age':[21, 22, 20, 19, 18, 23]})``````\n\nIn this Pandas write to Excel example, we will write the dataframe to an Excel file using the to_excel method. Noteworthy, when using Pandas to_excel in the code chunk below, we don’t use any parameters.\n\n``df.to_excel('NamesAndAges.xlsx')``\n\nIn the Excel file created when using Pandas to_excel is shown below. Evidently, if we don’t use the parameter sheet_name we get the default sheet name, ‘Sheet1’. Now, we can also see that we get a new column in our Excel file containing numbers. These are the index from the dataframe.\n\nIf we want our sheet to be named something else and we don’t want the index column we can add the following argument and parameters when we use Pandas to write to Excel:\n\n``df.to_excel('NamesAndAges.xlsx', sheet_name='Names and Ages', index=False)``\n\n### Writing Multiple Pandas Dataframes to an Excel File:\n\nIn this section, we are going to use Pandas ExcelWriter and Pandas to_excel to write multiple Pandas dataframes to one Excel file. That is if we happen to have many dataframes that we want to store in one Excel file but on different sheets, we can do this easily. However, we need to use Pandas ExcelWriter now:\n\n``````df1 = pd.DataFrame({'Names': ['Andreas', 'George', 'Steve',\n'Sarah', 'Joanna', 'Hanna'],\n'Age':[21, 22, 20, 19, 18, 23]})\n\ndf2 = pd.DataFrame({'Names': ['Pete', 'Jordan', 'Gustaf',\n'Sophie', 'Sally', 'Simone'],\n'Age':[22, 21, 19, 19, 29, 21]})\n\ndf3 = pd.DataFrame({'Names': ['Ulrich', 'Donald', 'Jon',\n'Jessica', 'Elisabeth', 'Diana'],\n'Age':[21, 21, 20, 19, 19, 22]})\n\ndfs = {'Group1':df1, 'Group2':df2, 'Group3':df3}\nwriter = pd.ExcelWriter('NamesAndAges.xlsx', engine='xlsxwriter')\n\nfor sheet_name in dfs.keys():\ndfs[sheet_name].to_excel(writer, sheet_name=sheet_name, index=False)\n\nwriter.save()``````\n\nIn the code above, we create 3 dataframes and then we continue to put them in a dictionary. Note, the keys are the sheet names and the cell names are the dataframes. After this is done we create a writer object using the xlsxwriter engine. We then continue by looping through the keys (i.e., sheet names) and add each sheet. Finally, the file is saved. Note, the final step is important as leaving this out will not give you the intended results.\n\nOf course, there are other ways to store data. One of them is using JSON files. See the latest tutorial on how to read and write JSON files using Pandas to learn about one way to load and save data in the JSON format.\n\nMore resources on how to load data in different formats:\n\n## Summary: How to Work With Excel Files using Pandas\n\nThat was it! In this post, we have learned a lot! We have, among other things, learned how to:\n\n• Load Excel files to dataframes:\n• Read Excel sheets and skip rows\n• Merging many sheets to a dataframe\n• Write a dataframe to an Excel file\n• Taking many dataframes and writing them to one Excel file with many sheets\n\nLeave a comment below if you have any requests or suggestions on what should be covered next! Check the post A Basic Pandas Dataframe Tutorial for Beginners to learn more about working with Pandas dataframe. That is after you have loaded them from a file (e.g., Excel spreadsheets)\n\n1. Christian\n\nGood article! Thank you very much!\n\n• Thank Christian! Glad you liked it! Also thanks for letting me know about the link. It should work now! Again, thanks. /Erik\n\n2. Hervé\n\nThanks Christian,\nJust want to point a typo (I think – anyway strangely I did not manage to read another sheet than the first one for now) at sheet_nameS here\n\n• Hervé\n\nI just found that “sheetname” works for me whereas other sites write sheet_name…\n\n• Hey Hervé,\n\nThanks for your comment. I just checked my code and using sheet_name works for me. Did you get an error using sheet_name?\n\nBest,\n\nErik\n\n3. Hervé\n\nI did not get any error but the data from the 1st sheet are always displayed.\nHowever if I use “sheetname=’…'”, then I can access other sheets.\nAre there different versions of pandas?\n\n• Ok. I use Pandas 2.5.1, right now. If you want to find out which version you have installed; open up a terminal (e.g., Anaconda Prompt on Windows) and type `conda list pandas`. You can also use `print(pd.__version__)` after you have imported Pandas (`import pandas as pd`).\n\nErik\n\n4. Herve\n\nMy pandas version is 0.20.1.\nShould I update it?\n\n• I would suggest you upgrade Pandas,\n\nBest,\n\nErik\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
] | [
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https://blog.andyselle.com/2014/11/01/circuit-simulation-in-javascript-part-i/ | [
"# Circuit simulation in Javascript Part I\n\nAs I’ve been trying to learn more about electronics, circuit simulation is an interesting route to go. No messy wires, can easily see idealized properties on a virtual oscilloscope, don’t have to wait for parts. Obviously there are limitations to models and nothing beats getting the real thing to work.\n\nThere are many SPICE derivatives that one can select. Perhaps the easiest to use is LTspice. The main innovation in the original spice paper was a practical bipolar transistor model that was suitable for simulation. The basic technique of SPICE varies based on what type of simulation. Currently I am interested in the transient analysis which gives a time varying definition of the voltage and current for any point in the circuit. There are other modes and even the simplest operating point analysis (tutorial on the eevblog) is powerful.\n\n## The project\n\nSPICE is nice and all, but since I want to learn how things work, I thought I’d write a circuit simulator from scratch in JavaScript. It shouldn’t be that hard, and it should really hone my intuition about how circuits work in a way that is much funner than solving basic circuits. Ideally I’ll do this in phases. For the first version described here I’ll support\n\n• Ideal voltage sources\n• Manual hard-coded circuits in the javascript file\n• Resistors and capacitors (linear passives)\n• Oscilloscope like view\n\nIn a later version these features would be nice to have\n\n• Inductors and diodes (non-linear passives)\n• Ideal opamps\n• Transistors and FETs\n• Transformers\n\n## Theory on solving circuits\n\nThe basic method we are going to use is Kirchoff’s Current Law that states that the net incoming and outgoing current is zero. So, basically for any component that has k nodes, it will contribute to k KCL equations in the circuit. Basically we need to solve all of the KCL simultaneously, so this should basically imply we are solving a big matrix. So let’s do a simple circuit example:",
null,
"Notice how I have specified a GND to the circuit. This is essential, if you don’t do this you can’t solve for voltages because they won’t be relative to any zero. Now that we have our circuit, we need to write the KCL laws for all the nodes which will form some of the constraints of our circuit solution.\n\nLet’s denote voltage at A, B, GND to be v(A), v(B), v(GND), respectively. The KCL at B is clearly",
null,
"$\\frac{v(A)-v(B)}{100} - \\frac{v(B)-v(GND)}{100} = 0$. Note I have used Ohm’s law for R1 and R2 separately and each component incident on B contributed to the KCL for B.\n\nNow the slightly tricky part, writing the KCL laws for GND and A, because we don’t directly know the current going through V1. There is no Ohm’s law to help us here. Probably I should have started this endeavear with current sources, but let’s just go with it. The major observation is that we need to introduce an unknown variable for the current through V1, we’ll call it i(V1). That gives us",
null,
"$i(V1) - \\frac{v(A)-v(B)}{100} = 0$. Notice how this equation’s second term is the negative of the first term of the KCL at B. That is because these circuit elements contribute symmetrically to the linear system and so we will get a symmetric matrix if we do things right. This is a great property to have because there are special matrix solvers that are much more efficient but only apply to positive definite and symmetric matrices (i.e. conjugate gradient).\n\nNext we should write KCL for the GND node. However, that really wouldn’t be particularly interesting, because we already know what v(GND) will be — zero. So we will use",
null,
"$v(\\textrm{GND})=0$.\n\nSo now we have three equations but four unknowns v(A),v(B),v(G),i(V1). We need one more equation to have a fully determined linear system,. One fact we haven’t included is that the voltage between A and G should be 5. So that gives us",
null,
"$v(A)-v(G)-5=0$. If we write these all as a system we get",
null,
"$\\begin{matrix} \\frac{v(A)-v(B)}{100} - \\frac{v(B)-v(G)}{100} & = & 0 \\\\ v(G) & = & 0 \\\\ i(V1) - \\frac{v(A)-v(B)}{100} & = & 0 \\\\ v(A)-v(G) - 5 & = & 0 \\end{matrix}$\n\nWe can rewrite this in matrix form (Ax=b) and we get",
null,
"The rows can be reordered to reflect the symmetry better.",
null,
"If we plug this into our favorite linear solver (in this case I’m using SAGE)\n\nfrom numpy import arange, eye, linalg\na=1/100;\nA=matrix([[-a,a,0,1],[a,-2*a,a,0],[0,0,1,0],[1,0,-1,0]]);\nb=matrix([,,,]);\nlinalg.solve(A,b)\n\n\narray([[ 5. ],\n[ 2.5 ],\n[ 0. ],\n[ 0.025]])\n\nIn other words v(A)=5 volts, v(B)=2.5 volts (go voltage divider), v(G)=0 volts, i(V1)=0.025 A. Everything is as we expected. In the next part we see how to code this into something that doesn’t require all this manual effort.\n\n## One Reply to “Circuit simulation in Javascript Part I”\n\n1.",
null,
"Irvan says:\n\nHi Andrew,\n\nI am just learning javascript and plans to make the visualization of simple electric circuit. I think it really helped me to start.. Thank you very much!"
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"https://s0.wp.com/latex.php",
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https://github.com/nd7141/graph_datasets | [
"# nd7141/graph_datasets\n\nData for \"Understanding Isomorphism Bias in Graph Data Sets\" paper.\nPython\nLatest commit 98f80fb Nov 5, 2019\nType Name Latest commit message Commit time\nFailed to load latest commit information.",
null,
"compact/MUTAG Oct 18, 2019",
null,
"datasets Oct 17, 2019",
null,
"edgelist/MUTAG Oct 18, 2019",
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"graphml/MUTAG Oct 18, 2019",
null,
"README.md Nov 5, 2019",
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"preprocessing.py Oct 18, 2019\n\n## Graph Classification Data Sets\n\nThis repo contains manually curated list of graph datasets for evaluation graph classification methods. These data sets are results of removing isomorphic copies of graphs from the original data sets. There are at the moment 54 data sets. The code to generate data sets is available here (https://github.com/nd7141/iso_bias).\n\n### Citation\n\nIf you found our work useful, please consider citing our work.\n\n``````@misc{ivanov2019understanding,\ntitle={Understanding Isomorphism Bias in Graph Data Sets},\nauthor={Sergei Ivanov and Sergei Sviridov and Evgeny Burnaev},\nyear={2019},\neprint={1910.12091},\narchivePrefix={arXiv},\nprimaryClass={cs.LG}\n}\n``````\n\n### Getting graphs for a data set\n\nAll datasets are zipped. There is a class `GraphDataset` that extracts, transforms, and save graphs to necessary formats.\n\n``````dataset = GraphDataset()\n\n# extract dataset\ndataset_path = 'datasets/'\nd = 'MUTAG'\noutput = 'compact/'\ndataset.extract_folder(dataset_path + d + '.zip', output)\n``````\n\nAfter the dataset was extracted locally, we can read graphs as a list of `GraphStruct` object, where each graph is a collection of nodes, edges, labels/attributes in simple python data structures (list of dict). `GraphStruct` also allows creating networkx graphs.\n\n``````# read graphs\ngraphs = dataset.read_graphs(output + d + '/')\n``````\n\nWe can additionally save graphs in `graphml` format, which will preserve node/edge labels/attributes.\n\n``````# save graphml\noutput = 'graphml/'\ndataset.save_graphs_graphml(graphs, output + d + '/')\n``````\n\nWe can also save graphs in `edgelist` format, which purely keeps topology of a graph.\n\n``````# save edgelist\noutput = 'edgelist/'\ndataset.save_graphs_edgelist(graphs, output + d + '/')\n``````\n\n### Data Sets in PyTorch-Geometric\n\nYou can find the same data sets in the PyTorch-Geometric library. To get clean version of the data sets, use parameter `cleaned=True` in `TUDataset` class. For example, to train a model on MUTAG data set:\n\n``````root = './'\ndataset = TUDataset(root, 'MUTAG', cleaned=True)\nprint(dataset)\n>>> MUTAG(135)\n``````\n\n### Format of Data Sets\n\nCompact format of data sets is described in https://ls11-www.cs.tu-dortmund.de/staff/morris/graphkerneldatasets and is efficient for storing large number of graphs. Each data set contains necessarily three files with `_A.txt`, `_graph_indicator.txt`, and `_graph_labels.txt`.\n\n`_A.txt` is edge list of all graphs in a data set. All nodes consecutive, and no node_ids are the same for two graphs.\n\n`_graph_indicator.txt` contains mapping between node_id and graph_id, so that lines correspond to nodes and content of lines correspond to graph. For example, if line 35 has 2, it means that node_id = 35 belongs to graph 2.\n\nFinally, `_graph_labels.txt` contains mapping between graph_id and its target label. graph_id corresponds to a line in a final, and target label corresponds to the content of a line. For example if line 45 contains 2, it means that graph 45 has label 2.\n\nAdditionally, folder may include `node/edge_labels/attributes.txt` files that provide additional information about the graphs.\n\nGraphml format contains each graph in its separate file in graphml format, that includes all meta-information about the graphs.\n\nEdgelist format contains each graph in its seperate file that provides edge list, without any label/attribute information.\n\nIn graphml and edgelist formats, the target labels are not generated again and one can use `_graph_labels.txt` to see the mapping.\n\n### Data Set Stats",
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http://www.jxlz.gov.cn/gplz/201808/t20180828_89944.htm | [
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\"+(e.bdUrl||n.getPageUrl()))})},c=function(){window.print()},h=function(){window._bd_share_main.F.use(\"trans/trans_bdxc\",function(e){e&&e.run()})},p=function(e){window._bd_share_main.F.use(\"trans/trans_bdysc\",function(t){t&&t.run(e)})},d=function(e){window._bd_share_main.F.use(\"trans/trans_weixin\",function(t){t&&t.run(e)})},v=function(e){o(e)};t.run=v,i()});\n\n九江市常态化、制度化开展警示教育活动，结合党校教学、参观警示教育馆、开展“三会一书两公开”等多种形式确保警示教育全覆盖。同时，按照市纪委监委统一部署，各地各单位结合班子民主生活会、理论中心组学习会、党员领导干部警示教育会，在规定时间、一定范围内组织观看《为了政治生态的山清水秀》。\n\n817日，九江市委举行中心组学习会，深入学习省委十四届六次全会精神，与会人员观看了警示教育片《为了政治生态的山清水秀》。市委书记林彬杨指出，“再次观看全省党员领导干部警示教育片，说明了建设山清水秀政治生态的极端重要性和必要性。”\n\n该市将肃清苏荣案余毒纳入市委党校主体班进行专题教学。20187月市委党校举办了“市委党校主体班肃清苏荣余毒补训班”，开展专题讲课，播放了《蝇贪之害》《造绿之殃》等录像，进行专题研讨。在九江市党员干部廉政教育馆播放苏荣忏悔录，在专门展厅集中展示典型案例，通过当事人忏悔影像等资料开展警示教育。今年以来，教育馆已累计接待参观246场，11839人次。\n\n九江市纪委监委注重运用查处的典型案例开展警示教育，做好执纪审查“后半篇文章”。根据全市首例留置案，以及德安县丰林镇原党委书记李汉民等违纪违法案例，拍摄制作了《村官的堕落》《两面人》等警示教育片，供各单位开展警示教育活动观看。同时，将制作的警示教育片定期在廉政教育馆进行展播，充分发挥反面典型警示震慑作用，让警示教育触及灵魂。（九江市纪委监委宣传部 许久）\n\n• 版权说明 | 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https://math.libretexts.org/Courses/Monroe_Community_College/MTH_225_Differential_Equations/z10%3A_Linear_Systems_of_Differential_Equations/10.5%3A_Constant_Coefficient_Homogeneous_Systems_II | [
"# 10.5: Constant Coefficient Homogeneous Systems II\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$$$\\newcommand{\\AA}{\\unicode[.8,0]{x212B}}$$\n\n$$\\newcommand{\\place}{\\bigskip\\hrule\\bigskip\\noindent} \\newcommand{\\threecol}{\\left[\\begin{array}{r}#1\\\\#2\\\\#3\\end{array}\\right]} \\newcommand{\\threecolj}{\\left[\\begin{array}{r}#1\\$1\\jot]#2\\\\[1\\jot]#3\\end{array}\\right]} \\newcommand{\\lims}{\\,\\bigg|_{#1}^{#2}} \\newcommand{\\twocol}{\\left[\\begin{array}{l}#1\\\\#2\\end{array}\\right]} \\newcommand{\\ctwocol}{\\left[\\begin{array}{c}#1\\\\#2\\end{array}\\right]} \\newcommand{\\cthreecol}{\\left[\\begin{array}{c}#1\\\\#2\\\\#3\\end{array}\\right]} \\newcommand{\\eqline}{\\centerline{\\hfill\\displaystyle#1\\hfill}} \\newcommand{\\twochar}{\\left|\\begin{array}{cc} #1-\\lambda\\\\#3-\\lambda\\end{array}\\right|} \\newcommand{\\twobytwo}{\\left[\\begin{array}{rr} #1\\\\#3\\end{array}\\right]} \\newcommand{\\threechar}{\\left[\\begin{array}{ccc} #1-\\lambda\\\\#4-\\lambda\\\\#7 -\\lambda\\end{array}\\right]} \\newcommand{\\threebythree}{\\left[\\begin{array}{rrr} #1\\\\#4\\\\#7 \\end{array}\\right]} \\newcommand{\\solutionpart}{\\vskip10pt\\noindent\\underbar{\\color{blue}\\sc Solution({\\bf #1})\\ }} \\newcommand{\\Cex}{\\fbox{\\textcolor{red}{C}}\\, } \\newcommand{\\CGex}{\\fbox{\\textcolor{red}{C/G}}\\, } \\newcommand{\\Lex}{\\fbox{\\textcolor{red}{L}}\\, } \\newcommand{\\matfunc}{\\left[\\begin{array}{cccc}#1_{11}(t)_{12}(t)&\\cdots _{1#3}(t)\\\\#1_{21}(t)_{22}(t)&\\cdots_{2#3}(t)\\\\\\vdots& \\vdots&\\ddots&\\vdots\\\\#1_{#21}(t)_{#22}(t)&\\cdots_{#2#3}(t) \\end{array}\\right]} \\newcommand{\\col}{\\left[\\begin{array}{c}#1_1\\\\#1_2\\\\\\vdots\\\\#1_#2\\end{array}\\right]} \\newcommand{\\colfunc}{\\left[\\begin{array}{c}#1_1(t)\\\\#1_2(t)\\\\\\vdots\\\\#1_#2(t)\\end{array}\\right]} \\newcommand{\\cthreebythree}{\\left[\\begin{array}{ccc} #1\\\\#4\\\\#7 \\end{array}\\right]} 1 \\ newcommand {\\ dy} {\\ ,\\ mathrm {d}y} \\ newcommand {\\ dx} {\\ ,\\ mathrm {d}x} \\ newcommand {\\ dyx} {\\ ,\\ frac {\\ mathrm {d}y}{\\ mathrm {d}x}} \\ newcommand {\\ ds} {\\ ,\\ mathrm {d}s} \\ newcommand {\\ dt }{\\ ,\\ mathrm {d}t} \\ newcommand {\\dst} {\\ ,\\ frac {\\ mathrm {d}s}{\\ mathrm {d}t}}$$ We saw in Section 10.4 that if an $$n\\times n$$ constant matrix $$A$$ has $$n$$ real eigenvalues $$\\lambda_1$$, $$\\lambda_2$$, …, $$\\lambda_n$$ (which need not be distinct) with associated linearly independent eigenvectors $${\\bf x}_1$$, $${\\bf x}_2$$, …, $${\\bf x}_n$$, then the general solution of $${\\bf y}'=A{\\bf y}$$ is \\[{\\bf y}=c_1{\\bf x}_1e^{\\lambda_1t}+c_2{\\bf x}_2e^{\\lambda_2 t} +\\cdots+c_n{\\bf x}_ne^{\\lambda_n t}. \\nonumber$\n\nIn this section we consider the case where $$A$$ has $$n$$ real eigenvalues, but does not have $$n$$ linearly independent eigenvectors. It is shown in linear algebra that this occurs if and only if $$A$$ has at least one eigenvalue of multiplicity $$r>1$$ such that the associated eigenspace has dimension less than $$r$$. In this case $$A$$ is said to be defective. Since it is beyond the scope of this book to give a complete analysis of systems with defective coefficient matrices, we will restrict our attention to some commonly occurring special cases.\n\n##### Example 10.5.1\n\nShow that the system\n\n$\\label{eq:10.5.1} {\\bf y}'=\\left[\\begin{array}{cc}{11}&{-25}\\\\{4}&{-9}\\end{array} \\right] {\\bf y}$\n\ndoes not have a fundamental set of solutions of the form $$\\{{\\bf x}_1e^{\\lambda_1t},{\\bf x}_2e^{\\lambda_2t}\\}$$, where $$\\lambda_1$$ and $$\\lambda_2$$ are eigenvalues of the coefficient matrix $$A$$ of Equation \\ref{eq:10.5.1} and $${\\bf x}_1$$, and $${\\bf x}_2$$ are associated linearly independent eigenvectors.\n\n###### Solution\n\nThe characteristic polynomial of $$A$$ is\n\n\\begin{align*} \\twochar{11}{-25}4{-9} &=(\\lambda-11)(\\lambda+9)+100\\\\ &=\\lambda^2-2\\lambda+1 \\\\ &=(\\lambda-1)^2.\\end{align*}\\nonumber\n\nHence, $$\\lambda=1$$ is the only eigenvalue of $$A$$. The augmented matrix of the system $$(A-I){\\bf x}={\\bf 0}$$ is\n\n$\\left[\\begin{array}{rrcr}10&-25&\\vdots&0\\\\4& -10&\\vdots&0\\end{array}\\right],\\nonumber$\n\nwhich is row equivalent to\n\n$\\left[\\begin{array}{cccc}{1}&{-\\frac{5}{2}}&{\\vdots }&{0}\\\\{0}&{0}&{\\vdots }&{0} \\end{array} \\right] \\nonumber$\n\nHence, $$x_1=5x_2/2$$ where $$x_2$$ is arbitrary. Therefore all eigenvectors of $$A$$ are scalar multiples of $$\\bf {x}_1=\\left[\\begin{array}{c}{5}\\\\{2}\\end{array} \\right]$$, so $$A$$ does not have a set of two linearly independent eigenvectors.\n\nFrom Example 10.5.1 , we know that all scalar multiples of $$\\bf {y}_1= \\twocol52 e^t$$ are solutions of Equation \\ref{eq:10.5.1}; however, to find the general solution we must find a second solution $$\\bf {y}_2$$ such that $$\\{ \\bf {y}_1, \\bf {y}_2\\}$$ is linearly independent. Based on your recollection of the procedure for solving a constant coefficient scalar equation\n\n$ay''+by'+cy=0\\nonumber$\n\nin the case where the characteristic polynomial has a repeated root, you might expect to obtain a second solution of Equation \\ref{eq:10.5.1} by multiplying the first solution by $$t$$. However, this yields $${\\bf y}_2=\\twocol52 te^t$$, which does not work, since\n\n${\\bf y}_2'=\\twocol52(te^t+e^t),\\quad \\text{while} \\quad \\left[\\begin{array}{cc}{11}&{-25}\\\\{4}&{-9} \\end{array} \\right] {\\bf y}_2=\\twocol52te^t.\\nonumber$\n\nThe next theorem shows what to do in this situation.\n\n##### Theorem 10.5.1\n\nSuppose the $$n\\times n$$ matrix $$A$$ has an eigenvalue $$\\lambda_1$$ of multiplicity $$\\ge2$$ and the associated eigenspace has dimension $$1;$$ that is$$,$$ all $$\\lambda_1$$-eigenvectors of $$A$$ are scalar multiples of an eigenvector $${\\bf x}.$$ Then there are infinitely many vectors $${\\bf u}$$ such that\n\n$\\label{eq:10.5.2} (A-\\lambda_1I){\\bf u}={\\bf x}.$\n\nMoreover$$,$$ if $${\\bf u}$$ is any such vector then\n\n$\\label{eq:10.5.3} {\\bf y}_1={\\bf x}e^{\\lambda_1t}\\quad\\mbox{and }\\quad {\\bf y}_2={\\bf u}e^{\\lambda_1t}+{\\bf x}te^{\\lambda_1t}$\n\nare linearly independent solutions of $${\\bf y}'=A{\\bf y}.$$\n\nA complete proof of this theorem is beyond the scope of this book. The difficulty is in proving that there’s a vector $${\\bf u}$$ satisfying Equation \\ref{eq:10.5.2}, since $$\\det(A-\\lambda_1I)=0$$. We’ll take this without proof and verify the other assertions of the theorem. We already know that $${\\bf y}_1$$ in Equation \\ref{eq:10.5.3} is a solution of $${\\bf y}'=A{\\bf y}$$. To see that $${\\bf y}_2$$ is also a solution, we compute\n\n\\begin{align*} {\\bf y}_2'-A{\\bf y}_2&=\\lambda_1{\\bf u}e^{\\lambda_1t}+{\\bf x} e^{\\lambda_1t} +\\lambda_1{\\bf x} te^{\\lambda_1t}-A{\\bf u}e^{\\lambda_1t}-A{\\bf x} te^{\\lambda_1t}\\\\ &=(\\lambda_1{\\bf u}+{\\bf x} -A{\\bf u})e^{\\lambda_1t}+(\\lambda_1{\\bf x} -A{\\bf x} )te^{\\lambda_1t}.\\end{align*}\\nonumber\n\nSince $$A{\\bf x}=\\lambda_1{\\bf x}$$, this can be written as\n\n${\\bf y}_2'-A{\\bf y}_2=- \\left((A-\\lambda_1I){\\bf u}-{\\bf x}\\right)e^{\\lambda_1t},\\nonumber$\n\nand now Equation \\ref{eq:10.5.2} implies that $${\\bf y}_2'=A{\\bf y}_2$$. To see that $${\\bf y}_1$$ and $${\\bf y}_2$$ are linearly independent, suppose $$c_1$$ and $$c_2$$ are constants such that\n\n$\\label{eq:10.5.4} c_1{\\bf y}_1+c_2{\\bf y}_2=c_1{\\bf x}e^{\\lambda_1t}+c_2({\\bf u}e^{\\lambda_1t} +{\\bf x}te^{\\lambda_1t})={\\bf 0}.$\n\nWe must show that $$c_1=c_2=0$$. Multiplying Equation \\ref{eq:10.5.4} by $$e^{-\\lambda_1t}$$ shows that\n\n$\\label{eq:10.5.5} c_1{\\bf x}+c_2({\\bf u} +{\\bf x}t)={\\bf 0}.$\n\nBy differentiating this with respect to $$t$$, we see that $$c_2{\\bf x}={\\bf 0}$$, which implies $$c_2=0$$, because $${\\bf x}\\ne{\\bf 0}$$. Substituting $$c_2=0$$ into Equation \\ref{eq:10.5.5} yields $$c_1{\\bf x}={\\bf 0}$$, which implies that $$c_1=0$$, again because $${\\bf x}\\ne{\\bf 0}$$\n\n##### Example 10.5.2\n\nUse Theorem 10.5.1 to find the general solution of the system\n\n$\\label{eq:10.5.6} {\\bf y}'=\\left[\\begin{array}{cc}{11}&{-25}\\\\{4}&{-9}\\end{array} \\right]{\\bf y}$\n\nconsidered in Example 10.5.1 .\n\n###### Solution\n\nIn Example 10.5.1 we saw that $$\\lambda_1=1$$ is an eigenvalue of multiplicity $$2$$ of the coefficient matrix $$A$$ in Equation \\ref{eq:10.5.6}, and that all of the eigenvectors of $$A$$ are multiples of\n\n${\\bf x}=\\twocol52.\\nonumber$\n\nTherefore\n\n${\\bf y}_1=\\twocol52e^t\\nonumber$\n\nis a solution of Equation \\ref{eq:10.5.6}. From Theorem 10.5.1 , a second solution is given by $${\\bf y}_2={\\bf u}e^t+{\\bf x}te^t$$, where $$(A-I){\\bf u}={\\bf x}$$. The augmented matrix of this system is\n\n$\\left[\\begin{array}{rrcr}10&-25&\\vdots&5\\\\4&-10&\\vdots&2\\end{array}\\right],\\nonumber$\n\nwhich is row equivalent to\n\n$\\left[\\begin{array}{rrcr}1&-\\frac{5}{2}&\\vdots&\\frac{1}{2}\\\\0&0&\\vdots&0\\end{array}\\right],\\nonumber$\n\nTherefore the components of $${\\bf u}$$ must satisfy\n\n$u_1-{5\\over2}u_2={1\\over2},\\nonumber$\n\nwhere $$u_2$$ is arbitrary. We choose $$u_2=0$$, so that $$u_1=1/2$$ and\n\n${\\bf u}=\\twocol{1\\over2}0.\\nonumber$\n\nThus,\n\n${\\bf y}_2=\\twocol10{e^t\\over2}+\\twocol52te^t.\\nonumber$\n\nSince $${\\bf y}_1$$ and $${\\bf y}_2$$ are linearly independent by Theorem 10.5.1 , they form a fundamental set of solutions of Equation \\ref{eq:10.5.6}. Therefore the general solution of Equation \\ref{eq:10.5.6} is\n\n${\\bf y}=c_1\\twocol52e^t+c_2\\left(\\twocol10{e^t\\over2}+\\twocol52te^t\\right).\\nonumber$\n\nNote that choosing the arbitrary constant $$u_2$$ to be nonzero is equivalent to adding a scalar multiple of $${\\bf y}_1$$ to the second solution $${\\bf y}_2$$ (Exercise 10.5.33).\n\n##### Example 10.5.3\n\nFind the general solution of\n\n$\\label{eq:10.5.7} {\\bf y}'=\\left[\\begin{array}{ccc}{3}&{4}&{-10}\\\\{2}&{1}&{-2}\\\\{2}&{2}&{-5}\\end{array} \\right] {\\bf y}.$\n\n###### Solution\n\nThe characteristic polynomial of the coefficient matrix $$A$$ in Equation \\ref{eq:10.5.7} is\n\n$\\left|\\begin{array}{ccc} 3-\\lambda & 4 & -10\\\\ 2 & 1-\\lambda & -2\\\\ 2 & 2 &-5-\\lambda\\end{array}\\right| =- (\\lambda-1)(\\lambda+1)^2.\\nonumber$\n\nHence, the eigenvalues are $$\\lambda_1=1$$ with multiplicity $$1$$ and $$\\lambda_2=-1$$ with multiplicity $$2$$. Eigenvectors associated with $$\\lambda_1=1$$ must satisfy $$(A-I){\\bf x}={\\bf 0}$$. The augmented matrix of this system is\n\n$\\left[\\begin{array}{rrrcr} 2 & 4 & -10 &\\vdots & 0\\\\ 2& 0 & -2 &\\vdots & 0\\\\ 2 & 2 & -6 & \\vdots & 0\\end{array}\\right],\\nonumber$\n\nwhich is row equivalent to\n\n$\\left[\\begin{array}{rrrcr} 1 & 0 & -1 &\\vdots& 0\\\\ 0 & 1 & -2 &\\vdots& 0\\\\ 0 & 0 & 0 &\\vdots&0\\end{array}\\right].\\nonumber$\n\nHence, $$x_1 =x_3$$ and $$x_2 =2 x_3$$, where $$x_3$$ is arbitrary. Choosing $$x_3=1$$ yields the eigenvector\n\n${\\bf x}_1=\\threecol121.\\nonumber$\n\nTherefore\n\n${\\bf y}_1 =\\threecol121e^t\\nonumber$\n\nis a solution of Equation \\ref{eq:10.5.7}. Eigenvectors associated with $$\\lambda_2 =-1$$ satisfy $$(A+I){\\bf x}={\\bf 0}$$. The augmented matrix of this system is\n\n$\\left[\\begin{array}{rrrcr} 4 & 4 & -10 &\\vdots & 0\\\\ 2 & 2 & -2 & \\vdots & 0\\\\2 & 2 & -4 &\\vdots & 0\\end{array}\\right],\\nonumber$\n\nwhich is row equivalent to\n\n$\\left[\\begin{array}{rrrcr} 1 & 1 & 0 &\\vdots& 0\\\\ 0 & 0 & 1 &\\vdots& 0 \\\\ 0 & 0 & 0 &\\vdots&0\\end{array}\\right].\\nonumber$\n\nHence, $$x_3=0$$ and $$x_1 =-x_2$$, where $$x_2$$ is arbitrary. Choosing $$x_2=1$$ yields the eigenvector\n\n${\\bf x}_2=\\threecol{-1}10,\\nonumber$\n\nso\n\n${\\bf y}_2 =\\threecol{-1}10e^{-t}\\nonumber$\n\nis a solution of Equation \\ref{eq:10.5.7}. Since all the eigenvectors of $$A$$ associated with $$\\lambda_2=-1$$ are multiples of $${\\bf x}_2$$, we must now use Theorem 10.5.1 to find a third solution of Equation \\ref{eq:10.5.7} in the form\n\n$\\label{eq:10.5.8} {\\bf y}_3={\\bf u}e^{-t}+\\threecol{-1}10te^{-t},$\n\nwhere $${\\bf u}$$ is a solution of $$(A+I){\\bf u=x}_2$$. The augmented matrix of this system is\n\n$\\left[\\begin{array}{rrrcr} 4 & 4 & -10 &\\vdots & -1\\\\ 2 & 2 & -2 & \\vdots & 1\\\\ 2 & 2 & -4 &\\vdots & 0\\end{array}\\right],\\nonumber$\n\nwhich is row equivalent to\n\n$\\left[\\begin{array}{rrrcr} 1 & 1 & 0 &\\vdots& 1\\\\ 0 & 0 & 1 &\\vdots& {1\\over2} \\\\ 0 & 0 & 0 &\\vdots&0\\end{array}\\right].\\nonumber$\n\nHence, $$u_3=1/2$$ and $$u_1 =1-u_2$$, where $$u_2$$ is arbitrary. Choosing $$u_2=0$$ yields\n\n${\\bf u} =\\threecol10{1\\over2},\\nonumber$\n\nand substituting this into Equation \\ref{eq:10.5.8} yields the solution\n\n${\\bf y}_3=\\threecol201{e^{-t}\\over2}+\\threecol{-1}10te^{-t}\\nonumber$\n\nof Equation \\ref{eq:10.5.7}. Since the Wronskian of $$\\{{\\bf y}_1,{\\bf y}_2,{\\bf y}_3\\}$$ at $$t=0$$ is\n\n$\\left|\\begin{array}{rrr} 1&-1&1\\\\2&1&0\\\\1&0&1\\over2\\end{array}\\right|={1\\over2},\\nonumber$\n\n$$\\{{\\bf y}_1,{\\bf y}_2,{\\bf y}_3\\}$$ is a fundamental set of solutions of Equation \\ref{eq:10.5.7}. Therefore the general solution of Equation \\ref{eq:10.5.7} is\n\n${\\bf y}=c_1\\threecol121e^t+c_2\\threecol{-1}10e^{-t}+c_3\\left (\\threecol201{e^{-t}\\over2}+\\threecol{-1}10te^{-t}\\right).\\nonumber$\n\n##### Theorem 10.5.2\n\nSuppose the $$n\\times n$$ matrix $$A$$ has an eigenvalue $$\\lambda_1$$ of multiplicity $$\\ge 3$$ and the associated eigenspace is one–dimensional$$;$$ that is$$,$$ all eigenvectors associated with $$\\lambda_1$$ are scalar multiples of the eigenvector $${\\bf x}.$$ Then there are infinitely many vectors $${\\bf u}$$ such that\n\n$\\label{eq:10.5.9} (A-\\lambda_1I){\\bf u}={\\bf x},$\n\nand, if $${\\bf u}$$ is any such vector$$,$$ there are infinitely many vectors $${\\bf v}$$ such that\n\n$\\label{eq:10.5.10} (A-\\lambda_1I){\\bf v}={\\bf u}.$\n\nIf $${\\bf u}$$ satisfies Equation \\ref{eq:10.5.9} and $${\\bf v}$$ satisfies Equation \\ref{eq:10.5.10}, then\n\n\\begin{aligned} {\\bf y}_1 &={\\bf x} e^{\\lambda_1t},\\\\ {\\bf y}_2&={\\bf u}e^{\\lambda_1t}+{\\bf x} te^{\\lambda_1t},\\mbox{ and }\\\\ {\\bf y}_3&={\\bf v}e^{\\lambda_1t}+{\\bf u}te^{\\lambda_1t}+{\\bf x} {t^2e^{\\lambda_1t}\\over2}\\end{aligned}\\nonumber\n\nare linearly independent solutions of $${\\bf y}'=A{\\bf y}$$.\n\nAgain, it is beyond the scope of this book to prove that there are vectors $${\\bf u}$$ and $${\\bf v}$$ that satisfy Equation \\ref{eq:10.5.9} and Equation \\ref{eq:10.5.10}. Theorem 10.5.1 implies that $${\\bf y}_1$$ and $${\\bf y}_2$$ are solutions of $${\\bf y}'=A{\\bf y}$$. We leave the rest of the proof to you (Exercise 10.5.34).\n\n##### Example 10.5.4\n\nUse Theorem 10.5.2 to find the general solution of\n\n$\\label{eq:10.5.11} {\\bf y}'=\\left[\\begin{array}{ccc}1&1&1 \\\\ 1&3&-1 \\\\ 0&2&2 \\end{array} \\right] {\\bf y}.$\n\n###### Solution\n\nThe characteristic polynomial of the coefficient matrix $$A$$ in Equation \\ref{eq:10.5.11} is\n\n$\\left|\\begin{array}{ccc} 1-\\lambda & 1 & \\phantom{-}1\\\\ 1 & 3-\\lambda & -1\\\\ 0 & 2 & 2-\\lambda\\end{array}\\right| =-(\\lambda-2)^3.\\nonumber$\n\nHence, $$\\lambda_1=2$$ is an eigenvalue of multiplicity $$3$$. The associated eigenvectors satisfy $$(A-2I){\\bf x=0}$$. The augmented matrix of this system is\n\n$\\left[\\begin{array}{rrrcr} -1 & 1 & 1 &\\vdots & 0\\\\ 1& 1 & -1 &\\vdots & 0\\\\ 0 & 2 & 0 & \\vdots & 0\\end{array}\\right],\\nonumber$\n\nwhich is row equivalent to\n\n$\\left[\\begin{array}{rrrcr} 1 & 0 &- 1 &\\vdots& 0\\\\ 0 & 1 & 0 &\\vdots& 0 \\\\ 0 & 0 & 0 &\\vdots&0\\end{array}\\right].\\nonumber$\n\nHence, $$x_1 =x_3$$ and $$x_2 = 0$$, so the eigenvectors are all scalar multiples of\n\n${\\bf x}_1=\\threecol101.\\nonumber$\n\nTherefore\n\n${\\bf y}_1=\\threecol101e^{2t}\\nonumber$\n\nis a solution of Equation \\ref{eq:10.5.11}. We now find a second solution of Equation \\ref{eq:10.5.11} in the form\n\n${\\bf y}_2={\\bf u}e^{2t}+\\threecol101te^{2t},\\nonumber$\n\nwhere $${\\bf u}$$ satisfies $$(A-2I){\\bf u=x}_1$$. The augmented matrix of this system is\n\n$\\left[\\begin{array}{rrrcr} -1 & 1 & 1 &\\vdots & 1\\\\ 1& 1 & -1 &\\vdots & 0\\\\ 0 & 2 & 0 & \\vdots & 1\\end{array}\\right],\\nonumber$\n\nwhich is row equivalent to\n\n$\\left[\\begin{array}{rrrcr} 1 & 0 &- 1 &\\vdots& -{1\\over2}\\\\ 0 & 1 & 0 &\\vdots& {1\\over2}\\\\ 0 & 0 & 0 &\\vdots&0\\end{array}\\right].\\nonumber$\n\nLetting $$u_3=0$$ yields $$u_1=-1/2$$ and $$u_2=1/2$$; hence,\n\n${\\bf u}={1\\over2}\\threecol{-1}10\\nonumber$\n\nand\n\n${\\bf y}_2=\\threecol{-1}10{e^{2t}\\over2}+\\threecol101te^{2t}\\nonumber$\n\nis a solution of Equation \\ref{eq:10.5.11}. We now find a third solution of Equation \\ref{eq:10.5.11} in the form\n\n${\\bf y}_3={\\bf v}e^{2t}+\\threecol{-1}10{te^{2t}\\over2}+\\threecol101{t^2e^{2t}\\over2}\\nonumber$\n\nwhere $${\\bf v}$$ satisfies $$(A-2I){\\bf v}={\\bf u}$$. The augmented matrix of this system is\n\n$\\left[\\begin{array}{rrrcr} -1 & 1 & 1 &\\vdots &-{1\\over2}\\\\ 1& 1 & -1 &\\vdots & {1\\over2}\\\\ 0 & 2 & 0 & \\vdots & 0\\end{array}\\right],\\nonumber$\n\nwhich is row equivalent to\n\n$\\left[\\begin{array}{rrrcr} 1 & 0 &- 1 &\\vdots& {1\\over2}\\\\ 0 & 1 & 0 &\\vdots& 0\\\\ 0 & 0 & 0 &\\vdots&0\\end{array}\\right].\\nonumber$\n\nLetting $$v_3=0$$ yields $$v_1=1/2$$ and $$v_2=0$$; hence,\n\n${\\bf v}={1\\over2}\\threecol100.\\nonumber$\n\nTherefore\n\n${\\bf y}_3=\\threecol100{e^{2t}\\over2}+ \\threecol{-1}10{te^{2t}\\over2}+\\threecol101{t^2e^{2t}\\over2}\\nonumber$\n\nis a solution of Equation \\ref{eq:10.5.11}. Since $${\\bf y}_1$$, $${\\bf y}_2$$, and $${\\bf y}_3$$ are linearly independent by Theorem 10.5.2 , they form a fundamental set of solutions of Equation \\ref{eq:10.5.11}. Therefore the general solution of Equation \\ref{eq:10.5.11} is\n\n\\begin{aligned} {\\bf y} = c_{1}\\left[\\begin{array}{c}{1}\\\\{0}\\\\{1}\\end{array} \\right]e^{2t}+c_{2}\\left(\\left[ \\begin{array}{c}{-1}\\\\{1}\\\\{0}\\end{array} \\right]\\frac{e^{2t}}{2}+\\left[\\begin{array}{c}{1}\\\\{0}\\\\{1}\\end{array} \\right] te^{2t} \\right) + c_{3}\\left(\\left[\\begin{array}{c}{1}\\\\{0}\\\\{0}\\end{array} \\right]\\frac{e^{2t}}{2}+\\left[\\begin{array}{c}{-1}\\\\{1}\\\\{0}\\end{array} \\right]\\frac{te^{2t}}{2}+\\left[\\begin{array}{c}{1}\\\\{0}\\\\{1}\\end{array} \\right]\\frac{t^{2}e^{2t}}{2} \\right) \\end{aligned}\\nonumber\n\n##### Theorem 10.5.3\n\nSuppose the $$n\\times n$$ matrix $$A$$ has an eigenvalue $$\\lambda_1$$ of multiplicity $$\\ge 3$$ and the associated eigenspace is two–dimensional; that is, all eigenvectors of $$A$$ associated with $$\\lambda_1$$ are linear combinations of two linearly independent eigenvectors $${\\bf x}_1$$ and $${\\bf x}_2$$$$.$$ Then there are constants $$\\alpha$$ and $$\\beta$$ $$($$not both zero$$)$$ such that if\n\n$\\label{eq:10.5.12} {\\bf x}_3=\\alpha{\\bf x}_1+\\beta{\\bf x}_2,$\n\nthen there are infinitely many vectors $${\\bf u}$$ such that\n\n$\\label{eq:10.5.13} (A-\\lambda_1I){\\bf u}={\\bf x}_3.$\n\nIf $${\\bf u}$$ satisfies Equation \\ref{eq:10.5.13}, then\n\n$\\label{eq:10.5.14} \\begin{array}{ll} {\\mathbf{y}_{1}}&{=\\mathbf{x}_{1}e^{\\lambda _{1}t}}\\\\ {\\mathbf{y}_{2}}&{=\\mathbf{x}_{2}e^{\\lambda _{1}t},\\ \\text{and}}\\\\ {\\mathbf{y}_{3}}&{=\\mathbf{u}e^{\\lambda _{1}t}+\\mathbf{x}_{3}te^{\\lambda _{1}t},} \\end{array}$\n\nare linearly independent solutions of $${\\bf y}'=A{\\bf y}.$$\n\nWe omit the proof of this theorem.\n\n##### Example 10.5.5\n\nUse Theorem 10.5.3 to find the general solution of\n\n$\\label{eq:10.5.15} {\\bf y}'=\\left[\\begin{array}{ccc}{0}&{0}&{1}\\\\{-1}&{1}&{1}\\\\{-1}&{0}&{2}\\end{array} \\right]{\\bf y}.$\n\n###### Solution\n\nThe characteristic polynomial of the coefficient matrix $$A$$ in Equation \\ref{eq:10.5.15} is\n\n$\\left|\\begin{array}{ccc} -\\lambda & 0 & 1\\\\ -1 & 1-\\lambda & 1\\\\ -1 & 0 & 2-\\lambda\\end{array}\\right| =-(\\lambda-1)^3.\\nonumber$\n\nHence, $$\\lambda_1=1$$ is an eigenvalue of multiplicity $$3$$. The associated eigenvectors satisfy $$(A-I){\\bf x=0}$$. The augmented matrix of this system is\n\n$\\left[\\begin{array}{rrrcr} -1 & 0 & 1 &\\vdots & 0\\\\ -1& 0 & 1 &\\vdots & 0\\\\ -1 & 0 & 1 & \\vdots & 0\\end{array}\\right],\\nonumber$\n\nwhich is row equivalent to\n\n$\\left[\\begin{array}{rrrcr} 1 & 0 &- 1 &\\vdots& 0\\\\ 0 & 0 & 0 &\\vdots& 0 \\\\ 0 & 0 & 0 &\\vdots&0\\end{array}\\right].\\nonumber$\n\nHence, $$x_1 =x_3$$ and $$x_2$$ is arbitrary, so the eigenvectors are of the form\n\n${\\bf x}_1=\\threecol{x_3}{x_2}{x_3}=x_3\\threecol101+x_2\\threecol010.\\nonumber$\n\nTherefore the vectors\n\n$\\label{eq:10.5.16} {\\bf x}_1 =\\threecol101\\quad\\mbox{and }\\quad {\\bf x}_2=\\threecol010$\n\nform a basis for the eigenspace, and\n\n${\\bf y}_1 =\\threecol101e^t \\quad \\text{and} \\quad {\\bf y}_2=\\threecol010e^t\\nonumber$\n\nare linearly independent solutions of Equation \\ref{eq:10.5.15}. To find a third linearly independent solution of Equation \\ref{eq:10.5.15}, we must find constants $$\\alpha$$ and $$\\beta$$ (not both zero) such that the system\n\n$\\label{eq:10.5.17} (A-I){\\bf u}=\\alpha{\\bf x}_1+\\beta{\\bf x}_2$\n\nhas a solution $${\\bf u}$$. The augmented matrix of this system is\n\n$\\left[\\begin{array}{rrrcr} -1 & 0 & 1 &\\vdots &\\alpha\\\\ -1& 0 & 1 &\\vdots &\\beta\\\\ -1 & 0 & 1 & \\vdots &\\alpha\\end{array}\\right],\\nonumber$\n\nwhich is row equivalent to\n\n$\\label{eq:10.5.18} \\left[\\begin{array}{rrrcr} 1 & 0 &- 1 &\\vdots& -\\alpha\\\\ 0 & 0 & 0 &\\vdots&\\beta-\\alpha\\\\ 0 & 0 & 0 &\\vdots&0\\end{array} \\right].$\n\nTherefore Equation \\ref{eq:10.5.17} has a solution if and only if $$\\beta=\\alpha$$, where $$\\alpha$$ is arbitrary. If $$\\alpha=\\beta=1$$ then Equation \\ref{eq:10.5.12} and Equation \\ref{eq:10.5.16} yield\n\n${\\bf x}_3={\\bf x}_1+{\\bf x}_2= \\threecol101+\\threecol010=\\threecol111,\\nonumber$\n\nand the augmented matrix Equation \\ref{eq:10.5.18} becomes\n\n$\\left[\\begin{array}{rrrcr} 1 & 0 &- 1 &\\vdots& -1\\\\ 0 & 0 & 0 &\\vdots& 0\\\\ 0 & 0 & 0 &\\vdots&0\\end{array} \\right].\\nonumber$\n\nThis implies that $$u_1=-1+u_3$$, while $$u_2$$ and $$u_3$$ are arbitrary. Choosing $$u_2=u_3=0$$ yields\n\n${\\bf u}=\\threecol{-1}00.\\nonumber$\n\nTherefore Equation \\ref{eq:10.5.14} implies that\n\n${\\bf y}_3={\\bf u}e^t+{\\bf x}_3te^t=\\threecol{-1}00e^t+\\threecol111te^t\\nonumber$\n\nis a solution of Equation \\ref{eq:10.5.15}. Since $${\\bf y}_1$$, $${\\bf y}_2$$, and $${\\bf y}_3$$ are linearly independent by Theorem 10.5.3 , they form a fundamental set of solutions for Equation \\ref{eq:10.5.15}. Therefore the general solution of Equation \\ref{eq:10.5.15} is\n\n${\\bf y}=c_1\\threecol101e^t+c_2\\threecol010e^t +c_3\\left(\\threecol{-1}00e^t+\\threecol111te^t\\right).\\bbox\\nonumber$\n\n## Geometric Properties of Solutions when $$n=2$$\n\nWe’ll now consider the geometric properties of solutions of a $$2\\times2$$ constant coefficient system\n\n$\\label{eq:10.5.19} \\twocol{y_1'}{y_2'}=\\left[\\begin{array}{cc}a_{11}&a_{12}\\\\a_{21}&a_{22} \\end{array}\\right]\\twocol{y_1}{y_2}$\n\nunder the assumptions of this section; that is, when the matrix\n\n$A=\\left[\\begin{array}{cc}a_{11}&a_{12}\\\\a_{21}&a_{22} \\end{array}\\right]\\nonumber$\n\nhas a repeated eigenvalue $$\\lambda_1$$ and the associated eigenspace is one-dimensional. In this case we know from Theorem 10.5.1 that the general solution of Equation \\ref{eq:10.5.19} is\n\n$\\label{eq:10.5.20} {\\bf y}=c_1{\\bf x}e^{\\lambda_1t}+c_2({\\bf u}e^{\\lambda_1t}+{\\bf x}te^{\\lambda_1t}),$\n\nwhere $${\\bf x}$$ is an eigenvector of $$A$$ and $${\\bf u}$$ is any one of the infinitely many solutions of\n\n$\\label{eq:10.5.21} (A-\\lambda_1I){\\bf u}={\\bf x}.$\n\nWe assume that $$\\lambda_1\\ne0$$.\n\nLet $$L$$ denote the line through the origin parallel to $${\\bf x}$$. By a half-line of $$L$$ we mean either of the rays obtained by removing the origin from $$L$$. Equation \\ref{eq:10.5.20} is a parametric equation of the half-line of $$L$$ in the direction of $${\\bf x}$$ if $$c_1>0$$, or of the half-line of $$L$$ in the direction of $$-{\\bf x}$$ if $$c_1<0$$. The origin is the trajectory of the trivial solution $${\\bf y}\\equiv{\\bf 0}$$.\n\nHenceforth, we assume that $$c_2\\ne0$$. In this case, the trajectory of Equation \\ref{eq:10.5.20} can’t intersect $$L$$, since every point of $$L$$ is on a trajectory obtained by setting $$c_2=0$$. Therefore the trajectory of Equation \\ref{eq:10.5.20} must lie entirely in one of the open half-planes bounded by $$L$$, but does not contain any point on $$L$$. Since the initial point $$(y_1(0),y_2(0))$$ defined by $${\\bf y}(0)=c_1{\\bf x}_1+c_2{\\bf u}$$ is on the trajectory, we can determine which half-plane contains the trajectory from the sign of $$c_2$$, as shown in Figure . For convenience we’ll call the half-plane where $$c_2>0$$ the positive half-plane. Similarly, the-half plane where $$c_2<0$$ is the negative half-plane. You should convince yourself (Exercise 10.5.35) that even though there are infinitely many vectors $${\\bf u}$$ that satisfy Equation \\ref{eq:10.5.21}, they all define the same positive and negative half-planes. In the figures simply regard $${\\bf u}$$ as an arrow pointing to the positive half-plane, since wen’t attempted to give $${\\bf u}$$ its proper length or direction in comparison with $${\\bf x}$$. For our purposes here, only the relative orientation of $${\\bf x}$$ and $${\\bf u}$$ is important; that is, whether the positive half-plane is to the right of an observer facing the direction of $${\\bf x}$$ (as in Figures 10.5.2 and 10.5.5 ), or to the left of the observer (as in Figures 10.5.3 and 10.5.4 ).\n\nMultiplying Equation \\ref{eq:10.5.20} by $$e^{-\\lambda_1t}$$ yields\n\n$e^{-\\lambda_1t}{\\bf y}(t)=c_1{\\bf x}+c_2{\\bf u}+c_2t {\\bf x}.\\nonumber$\n\nSince the last term on the right is dominant when $$|t|$$ is large, this provides the following information on the direction of $${\\bf y}(t)$$:\n\n1. Along trajectories in the positive half-plane ($$c_2>0$$), the direction of $${\\bf y}(t)$$ approaches the direction of $${\\bf x}$$ as $$t\\to\\infty$$ and the direction of $$-{\\bf x}$$ as $$t\\to-\\infty$$.\n2. Along trajectories in the negative half-plane ($$c_2<0$$), the direction of $${\\bf y}(t)$$ approaches the direction of $$-{\\bf x}$$ as $$t\\to\\infty$$ and the direction of $${\\bf x}$$ as $$t\\to-\\infty$$.\n\nSince $\\lim_{t\\to\\infty}\\|{\\bf y}(t)\\|=\\infty\\quad \\text{and} \\quad \\lim_{t\\to-\\infty}{\\bf y}(t)={\\bf 0}\\quad \\text{if} \\quad \\lambda_1>0,\\nonumber$\n\nor\n\n$\\lim_{t-\\to\\infty}\\|{\\bf y}(t)\\|=\\infty \\quad \\text{and} \\quad \\lim_{t\\to\\infty}{\\bf y}(t)={\\bf 0} \\quad \\text{if} \\quad \\lambda_1<0,\\nonumber$ there are four possible patterns for the trajectories of Equation \\ref{eq:10.5.19}, depending upon the signs of $$c_2$$ and $$\\lambda_1$$. Figures 10.5.2 - 10.5.5 illustrate these patterns, and reveal the following principle:\n\nIf $$\\lambda_1$$ and $$c_2$$ have the same sign then the direction of the traectory approaches the direction of $$-{\\bf x}$$ as $$\\|{\\bf y} \\|\\to0$$ and the direction of $${\\bf x}$$ as $$\\|{\\bf y}\\|\\to\\infty$$. If $$\\lambda_1$$ and $$c_2$$ have opposite signs then the direction of the trajectory approaches the direction of $${\\bf x}$$ as $$\\|{\\bf y} \\|\\to0$$ and the direction of $$-{\\bf x}$$ as $$\\|{\\bf y}\\|\\to\\infty$$."
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https://pubs.vmware.com/hyperic-58/topic/com.vmware.hyperic.monitoring.management.doc/GUID-FA25CDB8-67FA-4E34-9682-A9F1F7DF8307.html | [
"This topic provides information about understanding full-page charts.\n\nEach bar on a chart represents a timeslice that is 1/60 of the selected metric display range. A chart can plot a maximum of 60 points. The value of the Y-axis at the top of the bar is the observed metric value for that timeslice. What the value represents varies by the nature of the metric:\n\nStatic\n\nFor a static metric, one whose value does not vary over time (for instance a timestamp), the plotted value corresponds to a single measurement.\n\nDynamic\n\nFor a dynamic metric, one whose value may go up or down over time:\n\n ■ o The plotted point is the average of the values collected over the timeslice that ended at the time shown for the bar in the X-axis. The length of time over which the value was averaged depends on the current metric display range. For a display range of eight hours, each charted point represents the average of the preceding eight-minute period (8 hours / 60 time slices along the X-axis = 8 minutes). If the metric is collected every 60 seconds, and the chart's display range is 60 minutes, each of the 60 plotted points on the graph represents the single, observed value for the metric at a single point in time. ■ The I-bar superimposed on the vertical bar shows the range of values collected during the timeslice.\n\n• Trends Up and Trends Down\n\nFor a cumulative metric, one whose values either always increases or decreases, (such as bytes served, uptime, minimum response time), the plotted point is the maximum (or minimum) value for the time slice. The value shown for a metric that for trends up or down metrics is not the average over the timeslice.\n\nCharts are divided into a maximum of 60 slices, where each slice represents the same amount of time, over any display range the user chooses. For example:\n\n ■ If the charted metric's collection interval is 5 minutes and the selected metric display range is one hour, during which the metric value was collected 12 times, the chart will contain 12 time slices, and each slice will represent a single observed metric value. ■ If the metric display range is 5 hours, during which the metric value was collected 60 times, the chart will display 60 time slices, each will represent a single observed metric value. ■ If the selected display range is 10 hours long, during which the metric value was collected 120 times, the chart will display 60 timeslices - the maximum - and each bar will summarize the 2 data points collected during the timeslice. For dynamic metrics, data points are averaged. For trends up or down metrics the highest/lowest value is plotted.\n\nBy default, three colored horizontal lines appear on a metric chart that show the peak (pink), average (green) and low (blue) values collected for the metric for duration plotted.\n\nIn vCenter Hyperic, this additional information may appear on a chart:\n\n ■ Baseline — The baseline value for the metric is shown as a tan horizontal line. ■ Low Range — A pink horizontal bar appears across the chart, between the low range value and the maximum value shown on the chart's Y-axis. ■ High Range — A green horizontal bar appears across the chart, between the high range value and the minumum value shown on the chart's Y-axis."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86053616,"math_prob":0.9919509,"size":2325,"snap":"2019-35-2019-39","text_gpt3_token_len":500,"char_repetition_ratio":0.16846187,"word_repetition_ratio":0.10591133,"special_character_ratio":0.22021505,"punctuation_ratio":0.0952381,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97556114,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-16T12:42:49Z\",\"WARC-Record-ID\":\"<urn:uuid:c631db50-e805-4263-ae56-a8975372bed9>\",\"Content-Length\":\"22641\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d91e547f-f2c5-4bee-80b7-104684670ea8>\",\"WARC-Concurrent-To\":\"<urn:uuid:e765b942-5d0d-4b1c-a3ee-fea220d8641e>\",\"WARC-IP-Address\":\"23.50.112.225\",\"WARC-Target-URI\":\"https://pubs.vmware.com/hyperic-58/topic/com.vmware.hyperic.monitoring.management.doc/GUID-FA25CDB8-67FA-4E34-9682-A9F1F7DF8307.html\",\"WARC-Payload-Digest\":\"sha1:HYNMUMPXJZ65555FC2EGZPPSJS26AJBS\",\"WARC-Block-Digest\":\"sha1:CIJHYNMKQJZIORJDZYP7IEKP7E554ADN\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514572556.54_warc_CC-MAIN-20190916120037-20190916142037-00421.warc.gz\"}"} |
https://artofproblemsolving.com/wiki/index.php?title=2016_JBMO_Problems&oldid=94155 | [
"# 2016 JBMO Problems\n\n(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)\n\n## Problem 1\n\nA trapezoid",
null,
"$ABCD$ (",
null,
"$AB || CF$,",
null,
"$AB > CD$) is circumscribed.The incircle of the triangle",
null,
"$ABC$ touches the lines",
null,
"$AB$ and",
null,
"$AC$ at the points",
null,
"$M$ and",
null,
"$N$,respectively.Prove that the incenter of the trapezoid",
null,
"$ABCD$ lies on the line",
null,
"$MN$.\n\n## Problem 2\n\nLet",
null,
"$a,b,c$ be positive real numbers.Prove that",
null,
"$\\frac{8}{(a+b)^2 + 4abc} + \\frac{8}{(b+c)^2 + 4abc} + \\frac{8}{(a+c)^2 + 4abc} + a^2 + b^2 + c ^2 \\ge \\frac{8}{a+3} + \\frac{8}{b+3} + \\frac{8}{c+3}$.\n\n## Problem 3\n\nFind all triplets of integers",
null,
"$(a,b,c)$ such that the number",
null,
"$$N = \\frac{(a-b)(b-c)(c-a)}{2} + 2$$\n\n\nis a power of",
null,
"$2016$.\n\n(A power of",
null,
"$2016$ is an integer of form",
null,
"$2016^n$,where",
null,
"$n$ is a non-negative integer.)\n\n## Problem 4\n\nA",
null,
"$5 \\times 5$ table is called regular f each of its cells contains one of four pairwise distinct real numbers,such that each of them occurs exactly one in every",
null,
"$2 \\times 2$ subtable.The sum of all numbers of a regular table is called the total sum of the table.With any four numbers,one constructs all possible regular tables,computes their total sums and counts the distinct outcomes.Determine the maximum possible count.\n\n## See also\n\n 2016 JBMO (Problems • Resources) Preceded by2015 JBMO Problems Followed by2017 JBMO Problems 1 • 2 • 3 • 4 All JBMO Problems and Solutions\n\nInvalid username\nLogin to AoPS"
] | [
null,
"https://latex.artofproblemsolving.com/f/9/e/f9efaf9474c5658c4089523e2aff4e11488f8603.png ",
null,
"https://latex.artofproblemsolving.com/b/9/2/b925177b8cc5c668006d4d330be8d48aebc2ca1d.png ",
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"https://latex.artofproblemsolving.com/7/8/0/78045f207b8b1b88e062db9ba7d3563c6232981b.png ",
null,
"https://latex.artofproblemsolving.com/e/2/a/e2a559986ed5a0ffc5654bd367c29dfc92913c36.png ",
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"https://latex.artofproblemsolving.com/5/7/e/57ee5125358c0606c9b588580ddfa66f83e607b7.png ",
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"https://latex.artofproblemsolving.com/a/1/7/a179ff2638e4799cadd820db205c2beff6299ce9.png ",
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"https://latex.artofproblemsolving.com/5/d/1/5d1e4485dc90c450e8c76826516c1b2ccb8fce16.png ",
null,
"https://latex.artofproblemsolving.com/f/c/9/fc97ef67268cd4e91bacdf12b8901d7036c9a056.png ",
null,
"https://latex.artofproblemsolving.com/f/9/e/f9efaf9474c5658c4089523e2aff4e11488f8603.png ",
null,
"https://latex.artofproblemsolving.com/b/b/c/bbcfabd65745f2b759dea1840f9a0753b2f992e8.png ",
null,
"https://latex.artofproblemsolving.com/a/5/b/a5b29b358e825defa3e11b7d903d43ec31e5909a.png ",
null,
"https://latex.artofproblemsolving.com/7/3/8/738b50c1464997293a7bc215a2b91ccd33e40b5f.png ",
null,
"https://latex.artofproblemsolving.com/f/3/d/f3d9e22097e20a1249b5332202b24941a4bf9e1f.png ",
null,
"https://latex.artofproblemsolving.com/4/9/6/496f7460105427895e3c88cb3d99ffadfb21bc41.png ",
null,
"https://latex.artofproblemsolving.com/4/b/7/4b781fdf1cd172b567be38e2ffa6f20f0b8390c0.png ",
null,
"https://latex.artofproblemsolving.com/4/b/7/4b781fdf1cd172b567be38e2ffa6f20f0b8390c0.png ",
null,
"https://latex.artofproblemsolving.com/7/d/f/7dfa2dc389103eac034705404f922458e2fa4edc.png ",
null,
"https://latex.artofproblemsolving.com/1/7/4/174fadd07fd54c9afe288e96558c92e0c1da733a.png ",
null,
"https://latex.artofproblemsolving.com/1/9/7/197151a067f49bd4daa1952d1a4fac322409f0a6.png ",
null,
"https://latex.artofproblemsolving.com/e/4/c/e4c044326e2e84b768dafef07060bd42b0a5cb21.png ",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82205653,"math_prob":0.9997563,"size":966,"snap":"2021-21-2021-25","text_gpt3_token_len":239,"char_repetition_ratio":0.13825364,"word_repetition_ratio":0.0,"special_character_ratio":0.23084886,"punctuation_ratio":0.092391305,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999591,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,null,null,5,null,9,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,5,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-15T18:52:41Z\",\"WARC-Record-ID\":\"<urn:uuid:ebd3d64d-571f-477a-85fe-e54660201cdb>\",\"Content-Length\":\"40208\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6ec23d47-4220-443f-87d2-f631173a0636>\",\"WARC-Concurrent-To\":\"<urn:uuid:f75e25d7-9090-4397-952e-b3118f725382>\",\"WARC-IP-Address\":\"104.26.10.229\",\"WARC-Target-URI\":\"https://artofproblemsolving.com/wiki/index.php?title=2016_JBMO_Problems&oldid=94155\",\"WARC-Payload-Digest\":\"sha1:I73DYMD5QUXNXE7UJQJ3FP2I4UFFOHCX\",\"WARC-Block-Digest\":\"sha1:JYVNBNR5BYLY4OXST4LI6GPDKNOSGISR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487621519.32_warc_CC-MAIN-20210615180356-20210615210356-00516.warc.gz\"}"} |
https://itprospt.com/num/6890542/the-drug-oxycontin-is-used-to-treat-patients-with-severe | [
"5\n\n# The drug OxyContin is used to treat patients with severe pain. However, it is extremely addictive and can be lethal. In a clinical trial, 216 subjects were given Ox...\n\n## Question\n\n###### The drug OxyContin is used to treat patients with severe pain. However, it is extremely addictive and can be lethal. In a clinical trial, 216 subjects were given OxyContin as part of their treatment Of those 216 treated, 56 developed nausea. Another 123 patients were given a placebo and of those, subjects developed nausea_5. (Submit a screenshot of your written or typed answer:) Use a 0.05 significance level to test the claim that the population proportion of those taking the placebo and gettin\n\nThe drug OxyContin is used to treat patients with severe pain. However, it is extremely addictive and can be lethal. In a clinical trial, 216 subjects were given OxyContin as part of their treatment Of those 216 treated, 56 developed nausea. Another 123 patients were given a placebo and of those, subjects developed nausea_ 5. (Submit a screenshot of your written or typed answer:) Use a 0.05 significance level to test the claim that the population proportion of those taking the placebo and getting nauseous is smaller than the population proportion of those who are nauseous and take OxyContin: Use the provided information to find point estimate for the population proportion of subjects with nausea and were given the placebo, P1, and the point estimate for the population of subjects with nausea and were given OxyContin pz- b. State the null and alternative hypotheses. Indicate the significance level: Use the Brock Excel calculator to test the claim. Submit a screenshot of that sheet; including all input values and output results. d_ What is the P-Value from the test? Explain whether or not the null hypothesis is rejected. Write a concluding statement that addresses the claim_ Explain what it would mean in this context to make a Type error g. Explain what it would mean in this context to make a Type I| error h Based on the conclusion of the hypothesis test; which error(s) could have been made? Explain_",
null,
"",
null,
"#### Similar Solved Questions\n\n##### (4pts) Consider the two planes 4x 3y + 22 = 12 and € + Sy 2 = 25Find vector parallel to the line of intersection of these two planes.Find an equation of the plane through the point (~2,0,1) which is perpendicular to the line of intersection of the planes in Part (a)\n(4pts) Consider the two planes 4x 3y + 22 = 12 and € + Sy 2 = 25 Find vector parallel to the line of intersection of these two planes. Find an equation of the plane through the point (~2,0,1) which is perpendicular to the line of intersection of the planes in Part (a)...\n##### Stroboscope is set to flash every 6.06x10-5 s_ What is the frequency of the flashes in kHz?\nstroboscope is set to flash every 6.06x10-5 s_ What is the frequency of the flashes in kHz?...\n##### 3. Given that 9 = ~2.00 KC and d = 15.5 cm, find the magnitude of the nct clcctrostatic force exerted 0n thc point charge 42 in the figure. 1.4968-209+30\n3. Given that 9 = ~2.00 KC and d = 15.5 cm, find the magnitude of the nct clcctrostatic force exerted 0n thc point charge 42 in the figure. 1.4968 -209 +30...\n##### Tr %&aln Pracilce sml 4Prevlous hextYou have anstycred Oul 0f 4 parts. catrectly;wire of lengch 17 inchcs picccs which are then bent into the hpa = ofa circle of radius and = squarc ofsidc %, Thcn thc tocal Atc4 cncloscd by thc circle and squarc thc following function of and Arca Zur+ 45 If wc solvc for [ChISOt rcexprcss chis Arca _ following frc ton of r alone: A(-) 'The Hamnn Ac Occun when\nTr %&aln Pracilce sml 4 Prevlous hext You have anstycred Oul 0f 4 parts. catrectly; wire of lengch 17 inchcs picccs which are then bent into the hpa = ofa circle of radius and = squarc ofsidc %, Thcn thc tocal Atc4 cncloscd by thc circle and squarc thc following function of and Arca Zur+ 45 If w...\n##### KO CEN:Reaction Type:| Acidic or Basic Conditions?\nKO CEN: Reaction Type:| Acidic or Basic Conditions?...\n##### Find the extreme values of f (x) on the interval [f(r) =r _ 212 +5 On I = [-2,2]. flr) =\"V4-2 on [ = [-2,2.\nFind the extreme values of f (x) on the interval [ f(r) =r _ 212 +5 On I = [-2,2]. flr) =\"V4-2 on [ = [-2,2....\n##### (CH3)zck 26) (3pts) Cumipound? How man Rerlay crtonuuctuln 24 Aonn ntne odone=27) Cpts)E CHCHMCOR\" CH;CHFCOz- strongcr base? Explain ouchoicProductsfollunr acid-tise reaction;HCO2HI _ ~NHz29) (3pts) When _ small amowng of CH3(CH2HCOzH (pKa-4.8) is added {0 4 separatory funne] that contains cihet ayer AS and water with # pH = [2.0,it is found mainly Jin!ether; CH3(CHz)4CO2\" B) water; CH3(CH2HCOz\" ether; CH;(CHzHCOzH DJ) water: CHJ(CH2HCOzH E) none of the abovefollowing compounds h\n(CH3)zck 26) (3pts) Cumipound? How man Rerlay crton uuctuln 24 Aonn ntne odone= 27) Cpts)E CHCHMCOR\" CH;CHFCOz- strongcr base? Explain ouchoic Products follunr acid-tise reaction; HCO2HI _ ~NHz 29) (3pts) When _ small amowng of CH3(CH2HCOzH (pKa-4.8) is added {0 4 separatory funne] that contai...\n##### Calculate the concentrations of all species in a $0.100 \\mathrm{M}$ $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ solution.\nCalculate the concentrations of all species in a $0.100 \\mathrm{M}$ $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ solution....\n##### Reduce each rational expression to lowest terms.$$rac{x^{2}+x-20}{4-x}$$\nReduce each rational expression to lowest terms. $$\\frac{x^{2}+x-20}{4-x}$$...\n##### 1Mlch pulCntdaltnnLaLs |pdlTumoutnrawodr100t0 0Cornpite Uhc Ir-brxly durrana fr tho {0 slluatkea xwrn Anntan Oun 7J0 pulJwt '74) AMTTMLEF mi UANox Docun g1Dacumenimmchdld\n1 Mlch pul Cntdaltnn LaLs |pdl Tumoutn rawodr 100t 0 0 Cornpite Uhc Ir-brxly durrana fr tho {0 slluatkea xwrn Anntan Oun 7J0 pul Jwt '74) AMTTML EF mi UA Nox Docun g1 Dacumeni mmchdld...\n##### How many positive integers between 1000 and 9999 inclusivea) are divisible by 9?b) are even?c) have distinct digits?d) are not divisible by 3?e) are divisible by 5 or 7?f ) are not divisible by either 5 or 7?g) are divisible by 5 but not by 7?h) are divisible by 5 and 7?\nHow many positive integers between 1000 and 9999 inclusive a) are divisible by 9? b) are even? c) have distinct digits? d) are not divisible by 3? e) are divisible by 5 or 7? f ) are not divisible by either 5 or 7? g) are divisible by 5 but not by 7? h) are divisible by 5 and 7?...\n##### Exercise 2.3.9: IfSCR is a set, then x € Ris a cluster point if for every & > 0, the set (x-&,x+e)nsl {x} is not empty That is, if there are points of S arbitrarily close to x For example, S := {In:n € N} has a unique (only one) cluster point 0, but 0 # $Prove the following version of the Bolzano-Weierstrass theorem: Theorem: Let S € R be a bounded infinite set; then there exists at least one cluster point of$. Hint: If S is infinite, then S contains & countably infinite s\nExercise 2.3.9: IfSCR is a set, then x € Ris a cluster point if for every & > 0, the set (x-&,x+e)nsl {x} is not empty That is, if there are points of S arbitrarily close to x For example, S := {In:n € N} has a unique (only one) cluster point 0, but 0 # \\$ Prove the following v...\n##### Applicalhttps / /leain snhu eduld2ifle/content/224369/viewContent/5381336MiewaOption 2:Write mathematical scenario to describe each of the scatter plots.Answer prompts made by fellow students with the following information: Discuss an alternative scenario to represent the data in the scatter plots. In this scenario_ assume there is correlation but where it would be inappropriate to conclude causation.\nApplical https / /leain snhu eduld2ifle/content/224369/viewContent/5381336Miewa Option 2: Write mathematical scenario to describe each of the scatter plots. Answer prompts made by fellow students with the following information: Discuss an alternative scenario to represent the data in the scatter plo...\n##### A8ed IXaNazed snolaxd0z jo 8 a8ed'sjeuuppap asn J0u 0Q 'Xoq e Ul JUEJSUO) 4jE? Jalu? 01 3uns 3xe1 :01 Ienba Ajanujadsau '5ja3a1ul aJe (C/SE)D pue (c/oz-) ua41 'Ajanujadsaj8 >x> [ #! € I>x>0 J xfJo suoisuaixa Dipojuad u3^a pue ppo a41 aq (ro pue (r)J 137(1uiod T) 8 uousanoOz J0 8 a3eda3ed IXONa8ed snoiald\na8ed IXaN azed snolaxd 0z jo 8 a8ed 'sjeuuppap asn J0u 0Q 'Xoq e Ul JUEJSUO) 4jE? Jalu? 01 3uns 3xe1 :01 Ienba Ajanujadsau '5ja3a1ul aJe (C/SE)D pue (c/oz-) ua41 'Ajanujadsaj 8 >x> [ #! € I>x>0 J xf Jo suoisuaixa Dipojuad u3^a pue ppo a41 aq (ro pue (r)J 137 ...\n##### An air column open at both ends. The frequency of certaln harmonic (n) ix 410 Hz, and the frequency of the nert higher harmonic is 492 HzDetermine the Ewo harmonics_b) Determine the length ofthe column ifthe speed of sound in the nir column T0 m/a\nAn air column open at both ends. The frequency of certaln harmonic (n) ix 410 Hz, and the frequency of the nert higher harmonic is 492 Hz Determine the Ewo harmonics_ b) Determine the length ofthe column ifthe speed of sound in the nir column T0 m/a..."
] | [
null,
"https://cdn.numerade.com/ask_images/0a9e82d7fb234313a2746556f2d6e5fd.jpg ",
null,
"https://cdn.numerade.com/previews/133b02e2-209c-4bce-993f-5192cf3e0973_large.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9699239,"math_prob":0.914318,"size":4264,"snap":"2022-27-2022-33","text_gpt3_token_len":906,"char_repetition_ratio":0.18309858,"word_repetition_ratio":0.020539153,"special_character_ratio":0.20239212,"punctuation_ratio":0.079905994,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97658646,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-14T06:23:14Z\",\"WARC-Record-ID\":\"<urn:uuid:756a976e-fd50-46f5-9626-cbe63b1f5fc7>\",\"Content-Length\":\"84906\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3f56402c-e1ce-47e7-8b80-14747b001134>\",\"WARC-Concurrent-To\":\"<urn:uuid:dd80e984-a149-4e8a-9e0c-311e0a5cf0c6>\",\"WARC-IP-Address\":\"172.67.73.211\",\"WARC-Target-URI\":\"https://itprospt.com/num/6890542/the-drug-oxycontin-is-used-to-treat-patients-with-severe\",\"WARC-Payload-Digest\":\"sha1:TBDUMAJLVXCGZNEUYK22CPRBE2RGS4AQ\",\"WARC-Block-Digest\":\"sha1:2DF2IEEKNXVZXDKSAWSKCTJ5PKQ43NLQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571996.63_warc_CC-MAIN-20220814052950-20220814082950-00658.warc.gz\"}"} |
https://www.hyatt.com/zh-HK/hotel/brazil/grand-hyatt-sao-paulo/saogh/rooms | [
"• 「嘉賓軒」貴賓廳使用禮遇\n• 私人禮賓服務\n• 家庭影院系統\n• 凱悅豪華睡床\n• 43 吋平面電視\n• 電吹風\n• 浴袍\n• 電熱咖啡壺\n• 迷你冰箱\n• 熨斗和燙衣板\n• 客房內保險箱\n\n• 「嘉賓軒」貴賓廳使用禮遇\n• 凱悅豪華睡床\n• 43 吋平面電視\n• 電吹風\n• 浴袍\n• 電熱咖啡壺\n• 迷你冰箱\n• 熨斗和燙衣板\n• 客房內保險箱\n\n• 「嘉賓軒」貴賓廳使用禮遇\n• 私人禮賓服務\n• 家庭影院系統\n• 凱悅豪華睡床\n• 43 吋平面電視\n• 電吹風\n• 浴袍\n• 電熱咖啡壺\n• 迷你冰箱\n• 熨斗和燙衣板\n• 客房內保險箱\n\n• 43 吋平面電視\n• 獨立暖氣及空調控制\n• 遮光窗簾\n• 熨斗和燙衣板\n• 客房內保險箱\n• 可應要求提供嬰兒床\n• 電吹風\n• 電熱咖啡壺\n• 迷你冰箱\n\n• 43 吋平面電視\n• 獨立暖氣及空調控制\n• 遮光窗簾\n• 熨斗和燙衣板\n• 客房內保險箱\n• 可應要求提供嬰兒床\n• 電吹風\n• 電熱咖啡壺\n• 迷你冰箱\n\n• 43 吋平面電視\n• 獨立暖氣及空調控制\n• 遮光窗簾\n• 熨斗和燙衣板\n• 客房內保險箱\n• 可應要求提供嬰兒床\n• 電吹風\n• 電熱咖啡壺\n• 迷你冰箱\n\n• 43 吋平面電視\n• 獨立暖氣及空調控制\n• 遮光窗簾\n• 熨斗和燙衣板\n• 客房內保險箱\n• 可應要求提供嬰兒床\n• 電吹風\n• 電熱咖啡壺\n• 迷你冰箱\n\n• 43 吋平面電視\n• 獨立暖氣及空調控制\n• 遮光窗簾\n• 熨斗和燙衣板\n• 客房內保險箱\n• 可應要求提供嬰兒床\n• 電吹風\n• 電熱咖啡壺\n• 迷你冰箱\n\n• 43 吋平面電視\n• 獨立暖氣及空調控制\n• 遮光窗簾\n• 熨斗和燙衣板\n• 客房內保險箱\n• 可應要求提供嬰兒床\n• 電吹風\n• 電熱咖啡壺\n• 迷你冰箱\n\n• 43 吋平面電視\n• 獨立暖氣及空調控制\n• 遮光窗簾\n• 熨斗和燙衣板\n• 客房內保險箱\n• 可應要求提供嬰兒床\n• 電吹風\n• 電熱咖啡壺\n• 迷你冰箱\n\n• 43 吋平面電視\n• 獨立暖氣及空調控制\n• 遮光窗簾\n• 熨斗和燙衣板\n• 客房內保險箱\n• 可應要求提供嬰兒床\n• 電吹風\n• 電熱咖啡壺\n• 迷你冰箱\n\n• 43 吋平面電視\n• 獨立暖氣及空調控制\n• 遮光窗簾"
] | [
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.9775372,"math_prob":0.45005494,"size":1312,"snap":"2020-45-2020-50","text_gpt3_token_len":1778,"char_repetition_ratio":0.14296636,"word_repetition_ratio":0.0,"special_character_ratio":0.22256097,"punctuation_ratio":0.020725388,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96123415,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-27T16:27:58Z\",\"WARC-Record-ID\":\"<urn:uuid:f24eed08-4cbf-49b7-aafa-6a2f762e685a>\",\"Content-Length\":\"216929\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6bd47f07-80ab-4d07-9e21-f6d1af9ce569>\",\"WARC-Concurrent-To\":\"<urn:uuid:f7608463-0110-4a47-bb11-3c78b4ead715>\",\"WARC-IP-Address\":\"104.108.115.52\",\"WARC-Target-URI\":\"https://www.hyatt.com/zh-HK/hotel/brazil/grand-hyatt-sao-paulo/saogh/rooms\",\"WARC-Payload-Digest\":\"sha1:6VY3SDPITD5YUEWNX6XCGIEJWIGC2FGT\",\"WARC-Block-Digest\":\"sha1:IJAJOJXITPFMPNBXI3FPPBTZ2MCPDDEO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107894203.73_warc_CC-MAIN-20201027140911-20201027170911-00273.warc.gz\"}"} |
https://getrevising.co.uk/revision-tests/all-about-maths?game_type=flashcards | [
"# All About Maths!\n\n• Created by: Abc12398\n• Created on: 21-04-19 16:52\na mathematical phrase that can contain ordinary numbers, variables\nexpression\n1 of 6\na single number or variable, or numbers and variables multiplied together.\nterm\n2 of 6\na mathematical statement that two things are equal. It consists of two expressions, one on each side of an 'equals' sign.\nequation\n3 of 6\ntwo sides of the equation are equal\nidentity\n4 of 6\nit has more than one letter and one equals sign\nformula\n5 of 6\na symbol usually a letter that could be worth anything\nvariable\n6 of 6\n\n## Other cards in this set\n\n### Card 2\n\n#### Front\n\na single number or variable, or numbers and variables multiplied together.\n\nterm\n\n### Card 3\n\n#### Front\n\na mathematical statement that two things are equal. It consists of two expressions, one on each side of an 'equals' sign.\n\n#### Back",
null,
"### Card 4\n\n#### Front\n\ntwo sides of the equation are equal\n\n#### Back",
null,
"### Card 5\n\n#### Front\n\nit has more than one letter and one equals sign\n\n#### Back",
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""
] | [
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"https://getrevising.co.uk/revision-tests/all-about-maths/2/b/preview.png",
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"https://getrevising.co.uk/revision-tests/all-about-maths/3/b/preview.png",
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"https://getrevising.co.uk/revision-tests/all-about-maths/4/b/preview.png",
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https://mr-mathematics.com/product/inverse-functions/ | [
"# Inverse Functions\n\nIn this mathematics lesson students learn how to find the inverse of a function using function machines. As learning progresses students use transposition of formulae and recognise inverse functions as a reflection of its original in y = x.\n##### Differentiated Learning Objectives\n• All students should be able to use function machines to find an inverse function.\n• Most students should be able to find an inverse function using algebraic methods.\n• Some students should be able to draw the graph of an inverse function using a reflection on y = x.\n\n### Mr Mathematics Blog\n\n#### Trigonometric Identities Sin, Cos and Tan\n\nHow to introduce the sin, cos and tan trigonometric identities.\n\n#### Calculating a Reverse Percentage\n\nHow to teach calculating the original amount after a percentage change.\n\n#### Comparing Datasets using the Mean and Range\n\nThe importance of the range when comparing comparing datasets."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8119901,"math_prob":0.9119746,"size":303,"snap":"2020-34-2020-40","text_gpt3_token_len":60,"char_repetition_ratio":0.16053511,"word_repetition_ratio":0.06122449,"special_character_ratio":0.19471948,"punctuation_ratio":0.057692308,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9974805,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-29T06:29:57Z\",\"WARC-Record-ID\":\"<urn:uuid:f76a9e2a-9eff-48fd-a1d5-477de8f451cd>\",\"Content-Length\":\"75416\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ad10f093-09dd-4707-9644-75c3b6668e41>\",\"WARC-Concurrent-To\":\"<urn:uuid:332e4349-ee8f-42bd-8875-82deed0fd6b4>\",\"WARC-IP-Address\":\"68.66.200.219\",\"WARC-Target-URI\":\"https://mr-mathematics.com/product/inverse-functions/\",\"WARC-Payload-Digest\":\"sha1:UQDOBENDEROKDC4CTIREHEQCMAR5G2WG\",\"WARC-Block-Digest\":\"sha1:CCRJJTNKAE4R47IB4Z3S5MW3BQ3HN6HO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401632671.79_warc_CC-MAIN-20200929060555-20200929090555-00175.warc.gz\"}"} |
https://projecteuclid.org/journals/annals-of-statistics/volume-48/issue-1/Adaptive-risk-bounds-in-univariate-total-variation-denoising-and-trend/10.1214/18-AOS1799.short | [
"Translator Disclaimer\nFebruary 2020 Adaptive risk bounds in univariate total variation denoising and trend filtering\nAdityanand Guntuboyina, Donovan Lieu, Sabyasachi Chatterjee, Bodhisattva Sen\nAnn. Statist. 48(1): 205-229 (February 2020). DOI: 10.1214/18-AOS1799\n\n## Abstract\n\nWe study trend filtering, a relatively recent method for univariate nonparametric regression. For a given integer $r\\geq1$, the $r$th order trend filtering estimator is defined as the minimizer of the sum of squared errors when we constrain (or penalize) the sum of the absolute $r$th order discrete derivatives of the fitted function at the design points. For $r=1$, the estimator reduces to total variation regularization which has received much attention in the statistics and image processing literature. In this paper, we study the performance of the trend filtering estimator for every $r\\geq1$, both in the constrained and penalized forms. Our main results show that in the strong sparsity setting when the underlying function is a (discrete) spline with few “knots,” the risk (under the global squared error loss) of the trend filtering estimator (with an appropriate choice of the tuning parameter) achieves the parametric $n^{-1}$-rate, up to a logarithmic (multiplicative) factor. Our results therefore provide support for the use of trend filtering, for every $r\\geq1$, in the strong sparsity setting.\n\n## Citation\n\nAdityanand Guntuboyina. Donovan Lieu. Sabyasachi Chatterjee. Bodhisattva Sen. \"Adaptive risk bounds in univariate total variation denoising and trend filtering.\" Ann. Statist. 48 (1) 205 - 229, February 2020. https://doi.org/10.1214/18-AOS1799\n\n## Information\n\nReceived: 1 February 2017; Revised: 1 June 2018; Published: February 2020\nFirst available in Project Euclid: 17 February 2020\n\nzbMATH: 07196536\nMathSciNet: MR4065159\nDigital Object Identifier: 10.1214/18-AOS1799\n\nSubjects:\nPrimary: 62G08 , 62J05 , 62J07\n\nKeywords: Adaptive splines , discrete splines , fat shattering , higher order total variation regularization , metric entropy bounds , nonparametric function estimation , risk bounds , subdifferential , tangent cone\n\nRights: Copyright © 2020 Institute of Mathematical Statistics\n\nJOURNAL ARTICLE\n25 PAGES",
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"",
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] | [
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"https://projecteuclid.org/Content/themes/SPIEImages/Share_black_icon.png",
null,
"https://projecteuclid.org/images/journals/cover_aos.jpg",
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http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff;f=assignment9.mdwn;h=c8d2889f4f16f69752416f1938310d2bde927037;hp=b991c525d39e0e405ac411007e91eb82bfb4af61;hb=4afb5894348b73e779d6a8f88b835b96f18c282e;hpb=d08c54cb6bf8c3d2513708ba18460b5077b7041e | [
"index b991c52..c8d2889 100644 (file)\n@@ -4,7 +4,6 @@ Using continuations to solve the same-fringe problem\nThe problem\n-----------\n\n-We've seen two solutions to the same fringe problem so far.\nThe problem, recall, is to take two trees and decide whether they have\nthe same leaves in the same order.\n\n@@ -26,47 +25,48 @@ let tc = Node (Leaf 1, Node (Leaf 3, Leaf 2));;\nSo `ta` and `tb` are different trees that have the same fringe, but\n`ta` and `tc` are not.\n\n+We've seen two solutions to the same fringe problem so far.\nThe simplest solution is to map each tree to a list of its leaves,\nthen compare the lists. But because we will have computed the entire\nfringe before starting the comparison, if the fringes differ in an\nearly position, we've wasted our time examining the rest of the trees.\n\nThe second solution was to use tree zippers and mutable state to\n-simulate coroutines (see [[coroutines and aborts]]). In that\n-solution, we pulled the zipper on the first tree until we found the\n-next leaf, then stored the zipper structure in the mutable variable\n-while we turned our attention to the other tree. This solution is\n-efficient: the zipper doesn't visit any leaves beyond the first\n-mismatch.\n+simulate coroutines (see [[coroutines and aborts]], and\n+[[assignment8]]). In that solution, we pulled the zipper on the first\n+tree until we found the next leaf, then stored the zipper structure in\n+a mutable variable while we turned our attention to the other tree.\n+This solution is efficient: the zipper doesn't visit any leaves beyond\n+the first mismatch.\n\nSince zippers are just continuations reified, we expect that the\nsolution in terms of zippers can be reworked using continuations, and\nthis is indeed the case. Your assignment is to show how.\n\n-TO-DO LIST for solving the problem\n-----------------------------------\n+The first step is to review your answer to [[assignment8]], and make\n+sure you understand what is going on.\n+\n\n-1. Review the simple but inefficient solution (easy).\n-2. Understand the zipper/mutable state solution in [[coroutines and aborts]] (harder).\n+Two strategies for solving the problem\n+--------------------------------------\n\n-3. Two obvious approaches:\n\n-a. Review the list-zipper/list-continuation example given in\n+1. Review the list-zipper/list-continuation example given in\nclass in [[from list zippers to continuations]]; then\nfigure out how to re-functionalize the zippers used in the zipper\nsolution.\n\n-b. Review how the continuation-flavored tree\\_monadizer managed to\n-map a tree to a list of its leaves, in [[manipulating trees with\n-monads]]. Spend some time trying to understand exactly what it does:\n-compute the tree-to-list transformation for a tree with two leaves,\n-performing all beta reduction by hand using the definitions for\n-bind\\_continuation, unit\\_continuation and so on. If you take this\n-route, study the description of streams (a particular kind of data\n-structure) below. The goal will be to arrange for the\n-continuation-flavored tree_monadizer to transform a tree into a stream\n-instead of into a list. Once you've done that, completing the\n-same-fringe problem will be easy.\n+2. Review how the continuation-flavored tree\\_monadizer managed to\n+ map a tree to a list of its leaves, in [[manipulating trees with monads]].\n+ Spend some time trying to understand exactly what it\n+ does: compute the tree-to-list transformation for a tree with two\n+ leaves, performing all beta reduction by hand using the\n+ definitions for bind\\_continuation, unit\\_continuation and so on.\n+ If you take this route, study the description of streams (a\n+ particular kind of data structure) below. The goal will be to\n+ arrange for the continuation-flavored tree_monadizer to transform\n+ a tree into a stream instead of into a list. Once you've done\n+ that, completing the same-fringe problem will be easy.\n\n-------------------------------------\n\n@@ -104,9 +104,9 @@ class notes for [[week6]] on thunks, as well as [[assignment5]]).\n\nThere is a special stream called `End` that represents a stream that\ncontains no (more) elements, analogous to the empty list `[]`.\n-Streams that are not empty contain a first object paired with a\n+Streams that are not empty contain a first object, paired with a\nthunked stream representing the rest of the series. In order to get\n-access to the next element in the stream, we must forced the thunk by\n+access to the next element in the stream, we must *force* the thunk by\napplying it to the unit. Watch the behavior of this stream in detail.\nThis stream delivers the natural numbers, in order: 1, 2, 3, ...\n\n@@ -118,14 +118,14 @@ val make_int_stream : int -> int stream = [fun]\nval int_stream : int stream = Next (1, [fun]) (* First element: 1 *)\n\n# let tail = match int_stream with Next (i, rest) -> rest;;\n-val tail : unit -> int stream = <fun> (* Tail: a thunk *)\n+val tail : unit -> int stream = [fun] (* Tail: a thunk *)\n\n(* Force the thunk to compute the second element *)\n# tail ();;\n-- : int stream = Next (2, [fun])\n+- : int stream = Next (2, [fun]) (* Second element: 2 *)\n\n# match tail () with Next (_, rest) -> rest ();;\n-- : int stream = Next (3, <fun>)\n+- : int stream = Next (3, [fun]) (* Third element: 3 *)\n</pre>\n\nYou can think of `int_stream` as a functional object that provides"
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https://www.arxiv-vanity.com/papers/math/0407238/ | [
"# On convexified packing and entropy duality\n\nS. Artstein, V. Milman, S. Szarek, N. Tomczak-Jaegermann\n\n1. Introduction. If and are subsets of a vector space (or just a group, or even a homogeneous space), the covering number of by , denoted , is the minimal number of translates of needed to cover . Similarly, the packing number is the maximal number of disjoint translates of by elements of . The two concepts are closely related; we have . If is a ball in a normed space (or in an appropriate invariant metric) and a subset of that space (the setting and the point of view we will usually employ), these notions reduce to considerations involving the smallest -nets or the largest -separated subsets of .\n\nBesides the immediate geometric framework, packing and covering numbers appear naturally in numerous subfields of mathematics, ranging from classical and functional analysis through probability theory and operator theory to information theory and computer science (where a code is typically a packing, while covering numbers quantify the complexity of a set). As with other notions touching on convexity, an important role is played by considerations involving duality. The central problem in this area is the 1972 “duality conjecture for covering numbers” due to Pietsch which has been originally formulated in the operator-theoretic context, but which in the present notation can be stated as\n\n###### Conjecture 1\n\nDo there exist numerical constants such that for any dimension and for any two symmetric convex bodies in one has\n\n b−1logN(B∘,aK∘)≤logN(K,B)≤blogN(B∘,a−1K∘)? (1)\n\nAbove and in what follows is the polar body of ; “symmetric” is a shorthand for “symmetric with respect to the origin” and, for definiteness, all logarithms are to the base 2. In our preferred setting of a normed space , the proper generality is achieved by considering for , where is the unit ball and – a generic (convex, symmetric) subset of . The polars should then be thought of as subsets of , with the unit ball of that space. With minimal care, infinite-dimensional spaces and sets may be likewise considered. To avoid stating boundedness/compactness hypotheses, which are peripheral to the phenomena in question, it is convenient to allow , etc. to take the value .\n\nThe quantity has a clear information-theoretic interpretation: it is the complexity of , measured in bits, at the level of resolution with respect to the metric for which is the unit ball. Accordingly, (1) asks whether the complexity of is controlled by that of the ball in the dual space with respect to (the gauge of , i.e., the norm whose unit ball is ), and vice versa, at every level of resolution. [The original formulation of the conjecture involved relating – in a quantitative way – compactness of an operator to that of its adjoint.] In a very recent paper Conjecture 1 has been verified in the special yet most important case where is an ellipsoid (or, equivalently, when is a subset of a Hilbert space); the reader is referred to that article for a more exhaustive discussion of historical and mathematical background and for further references.\n\nIn the present note we introduce a new notion, which we call “convex separation” (or “convexified packing”) and prove a duality theorem related to that concept. This will lead to a generalization of the results from to the setting requiring only mild geometric assumptions about the underlying norm. [Both the definition and the generalization are motivated by an earlier paper .] For example, we now know that Conjecture 1 holds – in the sense indicated in the paragraph following (1) – in all - and -spaces (classical or non-commutative) for , with constants depending only on and uniformly bounded if stays away from 1 and , and similarly in all uniformly convex and all uniformly smooth spaces.\n\nAt the same time, and perhaps more importantly, the new approach “demystifies” duality results for usual covering/packing numbers in that it splits the proof into two parts. One step is a duality theorem for convex separation, which predictably is a consequence of the Hahn-Banach theorem. The other step involves geometric considerations relating convex separation to the usual separation; while often delicate and involved, they are always set in a given normed space and reflect properties of that space without appealing to duality, and thus are conceptually simpler.\n\nIn the next two sections we shall give the definition of convex separation and prove the corresponding duality theorem. Then, in section 4, we shall state several estimates for the convexified packing/convex separation numbers, in particular those that compare them to the usual packing/covering numbers. In section 5 we state the generalization of the duality result from alluded to above, and give some hints at its proof. We include details only for the proof of Theorem 2 (duality for convex separation) which is the part we consider conceptually new. Although we see our Theorem 5 as an essential progress towards Conjecture 1, in this announcement we omit the proofs; while non-trivial and technically involved, they require tools which were developed and used in our previous papers.\n\n2. Defining convex separation. The following notion plays a central role in this note. For a set and a symmetric convex body we define\n\n ^M(K,B) := sup{N:∃ x1,…,xN∈K such that (2) (xj+intB)∩conv{xi,i\n\nwhere “” stands for the interior of a set.111When defining packing in convex geometry, it is customary to require that only the interiors of the translates of be disjoint; we follow that convention here even though it is slightly unsound in the categorical sense. We shall refer to any sequence satisfying the condition (2) as -convexly separated. Leaving out the convex hull operation “” leads to the usual -separated set, which is the same as -packing. Thus we have . We emphasize that, as opposed to the usual notions of packing and covering, the order of the points is important here.\n\nThe definition (2) is very natural from the point of view of complexity theory and optimization. A standard device in constructing geometric algorithms is a “separation oracle” (cf. ): if is a convex set then, for a given , the oracle either attests that or returns a functional efficiently separating from . It is arguable that quantities of the type correctly describe complexities of the set with respect to many such algorithms.\n\nSince, as pointed out above and in the preceding section, packings, coverings and separated sets are very closely connected, and the corresponding “numbers” are related via two sided estimates involving (at most) small numerical constants, in what follows we shall use all these terms interchangeably.\n\n3. Duality for . While it is still an open problem whether Conjecture 1 holds in full generality, the corresponding duality statement for convex separation is fairly straightforward.\n\n###### Theorem 2\n\nFor any pair of convex symmetric bodies one has\n\n ^M(K,B)≤^M(B∘,K∘/2)2.\n\nProof of Theorem 2 Let ; i.e., is the radius of with respect to the gauge of . We will show that\n(i)\n(ii) + 1\n\nOnce the above are proved, Theorem 2 readily follows. To show (i), denote and let be a -convexly separated sequence in . Then, by (the elementary version of) the Hahn-Banach theorem, there exist separating functionals such that\n\n 1≤i\n\na condition which is in fact equivalent to being -convexly separated. Now and imply that , and hence dividing into subintervals of length we may deduce that one of these subintervals contains of the numbers . To simplify the notation, assume that this occurs for , that is\n\n 1≤i,j≤M⇒−1/2≤⟨yi,xi⟩−⟨yj,xj⟩≤1/2. (4)\n\nCombining (3) and (4) we obtain for\n\n ⟨yi−yj,xi⟩=⟨yi,xi⟩−⟨yj,xj⟩+⟨yj,xj⟩−⟨yj,xi⟩≥−1/2+1=1/2,\n\nwhich is again a condition of type (3) and thus shows that the sequence (in this order!) is -convexly separated. Hence , which is exactly the conclusion of (i).\n\nTo show (ii) we note that is also the radius of with respect to the gauge of . Since we are in a finite-dimensional space, that radius is attained and so there is a segment with . This implies that . However, in dimension one separated and convexly separated sets coincide; this allows to conclude that , as required.\n\n4. Separation vs. convex separation. It appears at the first sight that the notion of convex separation is much more restrictive than that of usual separation and, consequently, that – except for very special cases such as that of one-dimensional sets mentioned above – should be significantly smaller than or . However, we do not have examples when that happens. On the other hand, for several interesting classes of sets we do have equivalence for not-so-trivial reasons. Here we state two such results.\n\n###### Theorem 3\n\nThere exist numerical constants such that, for any and any pair of ellipsoids one has\n\n logM(E,B)≤Clog^M(E,cB).\n\nWhile Theorem 3 deals with purely Euclidean setting, the next result holds under rather mild geometric assumptions about the underlying norm. It requires -convexity, a property which goes back to and is well known to specialists; we refer to for background and properties. While many interesting descriptions of that class are possible, here we just mention that -convexity is equivalent to the absence of large subspaces well-isomorphic to finite-dimensional -spaces and that it can be quantified, i.e., there is a parameter called the -convexity constant and denoted , which can be defined both for finite and infinite dimensional spaces, and which has good permanence properties with respect to standard functors of functional analysis. For example, as hinted in the introduction, all -spaces for are -convex (with constants depending only on ), and similarly all uniformly convex and all uniformly smooth spaces.\n\n###### Theorem 4\n\nLet be a normed space which is -convex with and let be its unit ball. Then for any bounded set one has\n\n logM(T,B)≤βlog^M(T,B/2),\n\nwhere depends only on and the diameter of . Similarly, if and if is a symmetric convex subset of with , then\n\n logM(B,U)≤β′log^M(B,U/2)\n\nwith depending only on and .\n\nThe proof of Theorem 3 is non-trivial but elementary. The proof of Theorem 4 is based on the so called Maurey’s lemma (see ) and the ideas from . The details of both arguments will be presented elsewhere.\n\n5. Duality of covering and packing numbers in -convex spaces. If is the unit ball in a -convex space , then, combining Theorems 2 and 4, we obtain for any bounded symmetric convex set\n\n logM(T,B)≤βlog^M(T,B/2)≤2βlog^M(B∘,T∘/4)≤2βlogM(B∘,T∘/8),\n\nwhere depends only on the diameter of (and on ) and, similarly, . To show the latter, we apply the second part of Theorem 4 to , the dual of , and to (see the comments following (1)), and use the known fact that .\n\nOn the other hand, an iteration scheme developed in can be employed to show that if, for some normed space , duality in the sense of the preceding paragraph holds – with some constants – for, say, all , then it also holds for all sets with constants depending only on . We thus have\n\n###### Theorem 5\n\nLet be a normed space which is -convex with and let be its unit ball. Then for any symmetric convex set and any one has\n\n b−1logM(B∘,aϵT∘)≤logM(T,ϵB)≤blogM(B∘,a−1ϵT∘),\n\nwhere depend only on .\n\nTheorem 5 is related to in a similar way as the results of were related to : in both cases a statement concerning duality of covering numbers is generalized from the Hilbertian setting to that of a -convex space. A more detailed exposition of its proof will be presented elsewhere.\n\n6. Final remarks. While the approach via convexly separated sets does not yet settle Conjecture 1 in full generality, it includes, in particular, all cases for which the Conjecture has been previously verified. One such special case, which does not follow directly from the results included in the preceding sections, was settled in and subsequently generalized in : If, for some , , then with depending only on . In the same direction, we have\n\n###### Theorem 6\n\nLet and let be symmetric convex bodies with . Then\n\n log^M(K,B/α)≥β−1logM(K,B), (5)\n\nwhere depends only on and is a universal constant.\n\nThe result from , mentioned above is then an easy corollary: combine (5) with Theorem 2 to obtain , and the later expression is “trivially” . In turn, Theorem 6 can be derived by applying Theorem 3 to the so-called -ellipsoids of and . This is admittedly not the simplest argument, but it subscribes to our philosophy of minimizing the role of duality considerations and sheds additional light on the relationship between and .\n\nWe conclude the section and the note by pointing out that a slightly different – but equally natural from the algorithmic point of view – definition of convex separation is possible. Namely, we may consider sets of points in verifying (2) with a modified condition . For the so modified convex separation number – let us call it – the statement (i) from the proof of Theorem 2 remains true (by an almost identical argument), and so for “bounded” sets we do have duality. However, the statement (ii) is false already for the simple case of a segment.\n\nAcknowledgement. This research has been supported in part by grants from the U.S.-Israel Binational Science Foundation for the first three named authors and from the National Science Foundation (U.S.A.) for the third-named author. The fourth-named author holds the Canada Research Chair in Geometric Analysis. The first and second-named authors thank respectively University Paris VI and IHES for their hospitality during the time when this work was in progress.\n\n## References\n\n• S. Artstein, V. D. Milman and S. J. Szarek, Duality of Metric Entropy. Annals of Math., to appear.\n• J. Bourgain, A. Pajor, S. J. Szarek and N. Tomczak-Jaegermann, On the duality problem for entropy numbers of operators. Geometric aspects of functional analysis (1987–88), Lecture Notes in Math., 1376, Springer, Berlin-New York (1989) 50–63.\n• M. Grötschel, L. Lovász and A. Schrijver, Geometric algorithms and combinatorial optimization. Algorithms and Combinatorics, 2. Springer-Verlag, Berlin, 1993.\n• H. König and V. D. Milman, On the covering numbers of convex bodies, Geometric aspects of functional analysis (1985–86), Lecture Notes in Math., vol. 1267, Springer, Berlin-New York (1987) 82–95.\n• B. Maurey and G. Pisier, Séries de variables aléatoires vectorielles indépendantes et propriétés géométriques des espaces de Banach. (French) Studia Math. 58 (1976), no. 1, 45–90.\n• A. Pietsch, Theorie der Operatorenideale (Zusammenfassung), Friedrich-Schiller-Universität Jena, 1972.\n• G. Pisier, Remarques sur un résultat non publié de B. Maurey. (French) Séminaire d’Analyse Fonctionnelle, 1980–1981, Exp. No. V, 13 pp. École Polytechnique, Palaiseau, 1981.\n• G. Pisier, The Volume of Convex Bodies and Banach Space Geometry, Cambridge Tracts in Mathematics, 94, Cambridge University Press, Cambridge (1989).\n• G. Pisier, A new approach to several results of V. Milman, J. Reine Angew. Math. 393 (1989), 115-131.\n• N. Tomczak-Jaegermann, Dualité des nombres d’entropie pour des opérateurs à valeurs dans un espace de Hilbert, C. R. Acad. Sci. Paris Sér. I Math. 305 (1987), no. 7, 299–301.\n• \n\nS. Artstein, V. D. Milman\n\nSchool of Mathematical Sciences, Tel Aviv University, Tel Aviv 69978, Israel\n\nE-mail: ,\n\nS. J. Szarek\n\nDepartment of Mathematics, Case Western Reserve University,Cleveland, OH 44106-7058, U.S.A.\n\nand\n\nEquipe d’Analyse Fonctionnelle, B.C. 186, Université Paris VI, 4 Place Jussieu, 75252 Paris, France\n\nE-mail:\n\nN. Tomczak-Jaegermann\n\nDepartment of Mathematical and Statistical Sciences, University of Alberta,"
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https://inches-to-cm.appspot.com/682-inches-to-cm.html | [
"Inches To Centimeters\n\n# 682 in to cm682 Inches to Centimeters\n\nin\n=\ncm\n\n## How to convert 682 inches to centimeters?\n\n 682 in * 2.54 cm = 1732.28 cm 1 in\nA common question is How many inch in 682 centimeter? And the answer is 268.503937008 in in 682 cm. Likewise the question how many centimeter in 682 inch has the answer of 1732.28 cm in 682 in.\n\n## How much are 682 inches in centimeters?\n\n682 inches equal 1732.28 centimeters (682in = 1732.28cm). Converting 682 in to cm is easy. Simply use our calculator above, or apply the formula to change the length 682 in to cm.\n\n## Convert 682 in to common lengths\n\nUnitUnit of length\nNanometer17322800000.0 nm\nMicrometer17322800.0 µm\nMillimeter17322.8 mm\nCentimeter1732.28 cm\nInch682.0 in\nFoot56.8333333333 ft\nYard18.9444444444 yd\nMeter17.3228 m\nKilometer0.0173228 km\nMile0.0107638889 mi\nNautical mile0.0093535637 nmi\n\n## What is 682 inches in cm?\n\nTo convert 682 in to cm multiply the length in inches by 2.54. The 682 in in cm formula is [cm] = 682 * 2.54. Thus, for 682 inches in centimeter we get 1732.28 cm.\n\n## 682 Inch Conversion Table",
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"## Alternative spelling\n\n682 Inches to Centimeter, 682 Inches in Centimeter, 682 Inch to Centimeter, 682 Inch in Centimeter, 682 in to cm, 682 in in cm, 682 in to Centimeters, 682 in in Centimeters, 682 Inch to Centimeters, 682 Inch in Centimeters, 682 Inches to cm, 682 Inches in cm, 682 in to Centimeter, 682 in in Centimeter"
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https://www.1911forum.com/threads/determining-recoil-spring-specs-mechanical-engineers-welcome.1036578/ | [
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"1 - 20 of 25 Posts\n\n#### Glynn863\n\n·\n##### Registered\nJoined\n·\n83 Posts\nDiscussion Starter · ·\nOkay, in recent purchase of various 1911 parts, I have gotten two unknown 1911 recoil springs. I'm trying to determine their Force / rating. I have a NOS USGI spring for comparison. Below are the details, as measured:\n\nUSGI\nFree length: 6.55\"; Coil outer diameter: 0.438\"; Wire diameter: 0.044\"; # of coils: 30\nUsually listed as a \"16-lb\" spring\n\nUnknown #1\nFree length: 5.88\"; Coil outer diameter: 0.435\"; Wire diameter: 0.046\"; # of coils: 34\n\nUnknown #2\nFree length: 6.88\", Coil outer diameter: 0.420\"; Wire diameter: 0.038\"; # of coils: 44\n\n·\n##### Registered\nJoined\n·\n701 Posts\nAre the unknown springs new? The one being about an inch shorter than the USGI could indicate used, having taken a set, or, maybe it's a Commander spring.\n\nAssuming similar temper, you can figure the smaller diameter wire is a lighter spring, and the larger, heavier. Is the coil spacing even, or are some spaced smaller on one end? This could indicate a variable rate spring.\n\nSince we don't know the material or temper, the only real way to tell is by testing. Such as, how much weight is needed to compress the spring(s) one inch? If not a variable spring, this would give you a value. Ther4eby, a 16-lb spring (non-variable) should show one inch compression with 16 pounds of weight.\n\n#### megafiddle\n\n·\n##### Registered\nJoined\n·\n610 Posts\nThe spring power for the unknown springs can be calculated by comparing to a spring with known power and dimensions.\n\nUsing a 16 lb Wolff govt model spring for reference, this spring has the following relevant dimensions:\nfree length 6.5\"\nactive coils 31\nwire diameter .045\"\nmean coil diameter .386\"\n\nUsing a compressed length of 1.625\", the rate is simply given by compressed tension / (free length - compressed length)\nwhich equals 16.0 / (6.5 - 1.625)\nwhich equals 3.28 lb / inch\n\nThe spring rate K, of the unknown springs is calculated by using the proportional relationship of three parameters:\n\nK is proportional to the wire diameter (d) raised to the 4th power\nK is inversely proportional to the mean coil diameter (D) raised to the 3rd power\nK is inversely proportional to the number of free coils (N)\n\nFor your first spring, the wire diameter changes d in proportion to (.044 / .045) ^ 4 which equals .914\nThe mean coil diameter changes D in inverse proportion to (.394 / .386) ^ 3 which equals 1.06\nThe number of free coils changes N in inverse proportion to (30 / 31) which equals .968\n\nSo the unknown spring rate equals 3.28 x .914 / (1.06 x .968) which equals 2.92 lb / inch\n\nThe unknown spring power is then 2.92 x (6.55 - 1.625) which equals 14.4 lb\nInteresting. Sounds like an ordnance power spring.\n\nSimple, no?\n\nI am assuming that one end of your springs are closed, and that you are counting active coils?\nConfirm, and I will do the others.\n\n-\n\n•",
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"japsco113\n\n#### megafiddle\n\n·\n##### Registered\nJoined\n·\n610 Posts\nAre the unknown springs new? The one being about an inch shorter than the USGI could indicate used, having taken a set, or, maybe it's a Commander spring.\n\nAssuming similar temper, you can figure the smaller diameter wire is a lighter spring, and the larger, heavier. Is the coil spacing even, or are some spaced smaller on one end? This could indicate a variable rate spring.\n\nSince we don't know the material or temper, the only real way to tell is by testing. Such as, how much weight is needed to compress the spring(s) one inch? If not a variable spring, this would give you a value. Ther4eby, a 16-lb spring (non-variable) should show one inch compression with 16 pounds of weight.\nRecoil spring power is not measured at 1 inch of compression. That value would be spring rate.\n\nThe 16 lbs of tension occurs at about 4.9\" of compression.\n\nThe shear modulus used for spring rate is based on the modulus of elasticity, which doesn't vary much for common spring wire at about 29,000,000 lbs / square inch.\nIt's safe to say that spring wire is spring wire, at least for our purposes.\n\n-\n\n·\n##### Registered\nJoined\n·\n701 Posts\nTrue. I was unclear and implied 1 inch would be 16 pounds and that is incorrect. I had intended to say that the force applied by 1 inch compression would show differences in the springs, if any. The military print lists 2.88 lb/in for the spring rate.\n\nThanks for the clear explanation and setting things straight.\n\n#### megafiddle\n\n·\n##### Registered\nJoined\n·\n610 Posts\nMeasuring the tension at 1 inch of deflection is certainly one way do it. You can then find the spring power from the free length and the compressed length of 1.625\".\n\nNote that the actual spring power also depends on the free length. A spring could have a lower spring rate, and a longer free length, and still have the same compressed tension at 1.625\".\n\nAnd of course you could simply measure the tension at the compressed length of 1.625\" directly, if you have a good pull scale.\n\n-\n\n#### rickgman\n\n·\n##### Registered\nJoined\n·\n850 Posts\nMy rule of thumb is that when it comes to springs is if I can’t easily identify the specific spring and it’s condition (new or used) I simply throw it away. Springs are cheap but the damage that can be done when an inappropriate spring is used can be great. Recoil springs are a consumable so throwing them in the trash isn’t exactly a big deal.\n\n#### Glynn863\n\n·\n##### Registered\nJoined\n·\n83 Posts\nDiscussion Starter · ·\nOP here - Thanks for the quick replies. I don't intend on using these because I have a few extra NOS USGI springs. I was just curious; someone might want to use them.\n\nI'll look at them closer later today. The USGI spring's ends are open coil; the other two may be closed, but I'll need to verify.\n\n#### Jim Watson\n\n·\n##### Registered\nJoined\n·\n22,403 Posts\nEmpirical test. Put the unknowns in a gun, check that they are not stacking \"solid\" in full recoil, and shoot.\n\n#### Black sunshine\n\n·\n##### Registered\nJoined\n·\n164 Posts\nHere's a calculator that will get you in the ballpark. As stated above, the specific material isn't all that important...if it's some flavor steel, its elasticity will be about the same. A change from steel to aluminum or titanium, for example, would result in a significant change in elasticity....not at all likely you would have any material other than some variety of steel or sst.\n\n#### 1saxman\n\n·\nJoined\n·\n16,301 Posts\nEmpirical test. Put the unknowns in a gun, check that they are not stacking \"solid\" in full recoil, and shoot.\nYes, a very important step, and one that is frequently omitted. If it stacks, its the bushing that is now stopping the recoil of the slide - that usually doesn't last too long.\n\n#### japsco113\n\n·\n##### Registered\nJoined\n·\n405 Posts\nThe spring power for the unknown springs can be calculated by comparing to a spring with known power and dimensions.\n\nUsing a 16 lb Wolff govt model spring for reference, this spring has the following relevant dimensions:\nfree length 6.5\"\nactive coils 31\nwire diameter .045\"\nmean coil diameter .386\"\n\nUsing a compressed length of 1.625\", the rate is simply given by compressed tension / (free length - compressed length)\nwhich equals 16.0 / (6.5 - 1.625)\nwhich equals 3.28 lb / inch\n\nThe spring rate K, of the unknown springs is calculated by using the proportional relationship of three parameters:\n\nK is proportional to the wire diameter (d) raised to the 4th power\nK is inversely proportional to the mean coil diameter (D) raised to the 3rd power\nK is inversely proportional to the number of free coils (N)\n\nFor your first spring, the wire diameter changes d in proportion to (.044 / .045) ^ 4 which equals .914\nThe mean coil diameter changes D in inverse proportion to (.394 / .386) ^ 3 which equals 1.06\nThe number of free coils changes N in inverse proportion to (30 / 31) which equals .968\n\nSo the unknown spring rate equals 3.28 x .914 / (1.06 x .968) which equals 2.92 lb / inch\n\nThe unknown spring power is then 2.92 x (6.55 - 1.625) which equals 14.4 lb\nInteresting. Sounds like an ordnance power spring.\n\nSimple, no?\n\nI am assuming that one end of your springs are closed, and that you are counting active coils?\nConfirm, and I will do the others.\n\n-\nWhen I have have and arithmetic problem I going to need your help for sure. All I can say is WOW.\n\n#### Jim Watson\n\n·\n##### Registered\nJoined\n·\n22,403 Posts\nYes, a very important step, and one that is frequently omitted. If it stacks, its the bushing that is now stopping the recoil of the slide - that usually doesn't last too long.\nI didn't do arithmetic to see if more turns of finer wire led to a longer stacked length, but it certainly must be checked. Or the springs simply discarded as somebody said he would do.\n\n#### Evil_Ed\n\n·\nJoined\n·\n131 Posts\nNot sure if it's helpful but per Brent (formerly at Colt), Colt does not rate by weight, but by wire diameter and coil count (direct quote)..so comparing measurements to a known factory and known good Colt supplied recoil spring could at least get you a ballpark?\n\n•",
null,
"Black sunshine\n\n#### sevenL4\n\n·\nJoined\n·\n3,063 Posts\nI made a recoil spring gauge out of a piece of 1/2 inch PVC pipe and a few pieces of hardware from my rust collection. The most expensive part was the fish scale that cost about \\$15.00. There's plenty of examples online.\n\n#### log man\n\n·\nJoined\n·\n14,943 Posts\nFor some reason spring discussions can become testy, lol. The way gun springs are currently rated as a #16, for example is the force it produces at the full recoil position of the slide. So a 5\"GM #16 will register #16 at 1.6125\", a Commander spring will register at 1.125\", A Kimber Pro .940\" an OACP at .700\" a Detonics at .500\" .\n\nGlock flat wire springs are rated on the full recoil space of either the 19, or 17 as most Gocks no matter the model fall under one or the other of available recoil compression spaces, which are 1.105\" for the 17, and .920\" for the 19. Note, no one to date is making flat wire recoil springs specifically for the 1911, however they can be used to great advantage as they have more coils.\n\nSomething to consider also is the number of coils relates to the compressible length, not to the compression strength, as each coil of equal helix, or the distance of each coils spacing being equal, produce the same resistance at full compression. When a full size 1911 #16 recoil spring is compressed to 1.625\" it is very near stacking and producing as much force as it is capable, #16.\n\nSo, realize also that if you cut, say a #18 so it registers #16 at the full compressed recoil space of 1.625\" it will differ from the uncut #16 when in battery, and be less. So always use a springs whose force at full recoil allows full function, and has the most number of coils so when going to battery it also has sufficient force to smoothly chamber and close.\n\nTo clarify and rate recoil springs I highly recommend building a spring tester, you will be able to compare one to another in both the full recoil space as well as the in battery space. The in battery space for the GM it is 3.6875\", Commander 3.130\", OACP 2.5625\", Kimber Pro 2.815\", and the Detonics 2.1875\". This is without a buffer which will increase the in battery force, but not the full recoil force.\n\nLOG\n\nA full size 5\" GM #16 recoil spring compressed to the full recoil space of a 1911.\n\n•",
null,
"japsco113 and Heavy Industries\n\n#### Glynn863\n\n·\n##### Registered\nJoined\n·\n83 Posts\nDiscussion Starter · ·\nOP here - the USGI spring has open ends. Unknown #1 has one open end and one closed end (not ground). Unknown #2 has closed ends (not ground).\n\nHere's a calculator that will get you in the ballpark. As stated above, the specific material isn't all that important...if it's some flavor steel, its elasticity will be about the same. A change from steel to aluminum or titanium, for example, would result in a significant change in elasticity....not at all likely you would have any material other than some variety of steel or sst.\n\nI went to this website and played with some numbers. Since the USGI spring is rated for 16 lbs, I plugged in its data. The website said it has 10.74 lb max load. That was the only result in lbs. The USGI spring had a rate (k) of 3.036.\n\nIf I do a little math interpolation (since I know the USGI spec), I figure Unknown #1 is an 18 lb spring, and Unknown #2 is an 11 lb spring.\n\n•",
null,
"Black sunshine\n\n#### megafiddle\n\n·\n##### Registered\nJoined\n·\n610 Posts\nOP here - the USGI spring has open ends. Unknown #1 has one open end and one closed end (not ground). Unknown #2 has closed ends (not ground).\n\nI went to this website and played with some numbers. Since the USGI spring is rated for 16 lbs, I plugged in its data. The website said it has 10.74 lb max load. That was the only result in lbs. The USGI spring had a rate (k) of 3.036.\n\nIf I do a little math interpolation (since I know the USGI spec), I figure Unknown #1 is an 18 lb spring, and Unknown #2 is an 11 lb spring.\nThe ordnance recoil spring is spec'd at 13.55 lbs at 1.81\" compressed length. Wolff, and many here, use a compressed length of 1.625\" for govt model recoil springs. At a compressed length of 1.625\", the ordnance spring is about 14.2 lbs.\n\nUsing the same method of comparison (as I used earlier) to calculate your USGI spring from ordnance spec, I get about 14.3 lbs. I can say your USGI spring is not 16 lb, but closer to the original ordnance weight of 14.2 lb.\n\nI also tried the spring calculator. It reported a spring rate of 2.935 lb / inch. This is very close to the 2.92 lb / inch that I calculated earlier.\n\nYou cannot use the maximum load reported there. It's not measured (calculated) at the compressed length you are interested in, which is 1.625\". The compressed length they are using is far short of compressing down to 1.625\". Their compressed length can also vary with different springs, so you can't simply scale their maximum loads to a known spring. (I believe scaling is what you meant by \"interpolation\").\n\nYou can use the spring rate though, which does appear to be pretty accurate. Use the calculator to the find the spring rate for each spring. Then simply multiply (free length - 1.625) by the spring rate. You should get a pretty accurate spring power value that way.\n\nYou may already know, but be careful counting active coils. It's easy to be off by one.\n\nAlso, fully compress the spring several times before measuring free length. A new spring will shorten a bit and \"set\" after being compressed the first few times.\n\n-\n\n•",
null,
"japsco113\n\n#### megafiddle\n\n·\n##### Registered\nJoined\n·\n610 Posts\nI used the spring calculator at thespringstore.com to verify the Wolff 16 lb spring specs:\n\n16 lb Wolff govt model spring:\n\nfree length 6.5\"\nactive coils 31\nwire diameter .045\"\ncoil diameter .431\"\nmean coil diameter .386\"\n\nspring rate from rated power and free length 3.28 lb / in\nspring rate from the spring calculator 3.304 lb / in\n\nspring power from the Wolff spec 16.0 lb\nspring power from spring calculator rate 16.1 lb\n\nThat's pretty good agreement.\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -\n\nThe calculator at thespringstore.com has been unaccesible at times. I only have comparisons for your USGI spring so far:\n\nUSGI:\n\nfree length 6.55\"\nactive coils 30\nwire diameter .044\"\ncoil diameter .438\"\nmean coil diameter .394\" (coil OD - wire diameter)\n\nd = (.044 / .045) ^ 4 = .914\nD = (.394 / .386) ^ 3 = 1.06\nN = (30 / 31) = .968\n\nas calculated in proportion to the Wolff 16 lb:\n\nspring rate = 3.28 x .914 / (1.06 x .968) = 2.92 lb / in\nspring power = 2.92 x (6.55 - 1.625) = 14.4 lb\n\nas calculated from the spring calculator rate:\n\nspring rate = 2.935 lb / in\nspring power = 2.935 x (6.55 - 1.625) = 14.5 lb\n\nAlso pretty good agreement.\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -\n\nFor the other two springs, these are only in proportion to the Wolff 16 lb.\nIf I can access the spring calculator again, I expect it will also be in agreement:\n\nUnknown #1:\n\nfree length 5.88\"\nactive coils 34\nwire diameter .046\"\ncoil diameter .435\"\nmean coil diameter .389\"\n\nd = (.046 / .045) ^ 4 = 1.09\nD = (.389 / .386) ^ 3 = 1.02\nN = (34 / 31) = 1.10\n\nspring rate = 3.28 x 1.09 / (1.02 x 1.10) = 3.19 lb / in\nspring power = 3.19 x (5.88 - 1.625) = 13.6 lb\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -\n\nUnknown #2:\n\nfree length 6.88\"\nactive coils 44\nwire diameter .038\"\ncoil diameter .420\"\nmean coil diameter .382\"\n\nd = (.038 / .045) ^ 4 = .508\nD = (.382 / .386) ^ 3 = .969\nN = (44 / 31) = 1.42\n\nspring rate = 3.28 x .508 / (.969 x 1.42) = 1.21 lb / in\nspring power = 1.21 x (6.88 - 1.625) = 6.36 lb\n\n-\n\n#### Glynn863\n\n·\n##### Registered\nJoined\n·\n83 Posts\nDiscussion Starter · ·\nOP here - Thanks for that.\n\nSo a USGI spring, by US Ordnance values / specs, is actually closer to a 14-lb spring, rather than 16-lb?\n\nI have never changed a spring in any of my 1911's. My 1983-vintage Series 80 Government Model is still stock. I bought a NOS Colt Government spring from Midway for a Remington R1 I'm pseudo-custom building. Unknown #1 came from a Remington Rand M1911A1 as I received it, along with a 2-piece long guide rod. I bought a trio of NOS USGI springs (contract date 1984) off eBay, and have used one in the RR, so I have two spares.\n\nI already knew that the Unknown #2 was a lighter spring just by a \"hands-on\" examination; I didn't know of a numerical value to associate it with.\n\n1 - 20 of 25 Posts"
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https://in.mathworks.com/matlabcentral/cody/problems/361-energy-of-a-photon/solutions/1296376 | [
"Cody\n\n# Problem 361. Energy of a photon\n\nSolution 1296376\n\nSubmitted on 18 Oct 2017 by Angelo Gelmini\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nF = 1; E_correct = 3/10^15; assert(photon_energy(F)>E_correct)\n\n2 Pass\nF = 100; E_correct = 500/10^15; assert(photon_energy(F)<E_correct)\n\n3 Pass\nF = 500; E_correct = 2100/10^15; assert(photon_energy(F)<E_correct)\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
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https://socratic.org/questions/how-many-2p-orbitals-are-there-in-an-atom | [
"# How many 2p orbitals are there in an atom?\n\nThere are three $2 p$ orbitals in an atom.\nThere are three $2 p$ orbitals in an atom, and they are:\n${p}_{x}$, ${p}_{y}$ and ${p}_{z}$\nThese orbitals are oriented toward the three axis: $x , y \\text{ and } z$."
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https://numberworld.info/12115 | [
"Number 12115\n\nProperties of number 12115\n\nCross Sum:\nFactorization:\nDivisors:\n1, 5, 2423, 12115\nCount of divisors:\nSum of divisors:\nPrime number?\nNo\nFibonacci number?\nNo\nBell Number?\nNo\nCatalan Number?\nNo\nBase 2 (Binary):\nBase 3 (Ternary):\nBase 4 (Quaternary):\nBase 5 (Quintal):\nBase 8 (Octal):\n2f53\nBase 32:\nbqj\nsin(12115)\n0.85144148424633\ncos(12115)\n0.52444961522\ntan(12115)\n1.623495297807\nln(12115)\n9.4021996332506\nlg(12115)\n4.0833234184735\nsqrt(12115)\n110.06816070054\nSquare(12115)\n\nNumber Look Up\n\nLook Up\n\n12115 which is pronounced (twelve thousand one hundred fifteen) is a very amazing number. The cross sum of 12115 is 10. If you factorisate the figure 12115 you will get these result 5 * 2423. The figure 12115 has 4 divisors ( 1, 5, 2423, 12115 ) whith a sum of 14544. The figure 12115 is not a prime number. The figure 12115 is not a fibonacci number. The number 12115 is not a Bell Number. The number 12115 is not a Catalan Number. The convertion of 12115 to base 2 (Binary) is 10111101010011. The convertion of 12115 to base 3 (Ternary) is 121121201. The convertion of 12115 to base 4 (Quaternary) is 2331103. The convertion of 12115 to base 5 (Quintal) is 341430. The convertion of 12115 to base 8 (Octal) is 27523. The convertion of 12115 to base 16 (Hexadecimal) is 2f53. The convertion of 12115 to base 32 is bqj. The sine of 12115 is 0.85144148424633. The cosine of the number 12115 is 0.52444961522. The tangent of 12115 is 1.623495297807. The root of 12115 is 110.06816070054.\nIf you square 12115 you will get the following result 146773225. The natural logarithm of 12115 is 9.4021996332506 and the decimal logarithm is 4.0833234184735. that 12115 is very impressive number!"
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https://topics.libra.titech.ac.jp/xc/search/DA01348607?os%5Bua%5D=DA01348607 | [
"",
null,
"1.\n\n[by] C. H. Edwards, Jr\n2.\n\nC.H. Edwards, Jr\n 出版情報: New York ; Heidelberg ; Berlin ; Tokyo : Springer-Verlag, c1979 xii, 351 p. ; 25 cm 所蔵情報: loading…\n\n Area Number and Limit Concepts in Antiquity / 1： Archimedes / 2： Twilight, Darkness and Dawn / 3： Early Indivisibles and Infinitesimal Techniques / 4： Early Tangent Constructions / 5： Napier's Wonderful Logarithms / 6： The Arithmetic of the Infinite / 7： The Calculus According to Newton / 8： The Calculus According to Leibniz / 9： The Age of Euler / 10： The Calculus According to Cauchy, Riemann and Weierstrass / 11： Postscript: The Twentieth Century / 12： Index\n Area Number and Limit Concepts in Antiquity / 1： Archimedes / 2： Twilight, Darkness and Dawn / 3：\n3.\n\nedited by J.C. Cantrell and C.H. Edwards, Jr.\n 出版情報: Chicago : Markham, c1970 xiv, 514 p ; 24 cm シリーズ名: Markham mathematics series 所蔵情報: loading…\n4.\n\nC. Henry Edwards, David E. Penney\n 出版情報: Upper Saddle River, N.J. : Prentice Hall International, c2001 xiii, 684, 48, 6 p. ; 24 cm. 所蔵情報: loading…\n\n First-Order Differential Equations / 1： Differential Equations and Mathematical Model Integrals as General and Particular Solutions Direction Fields and Solution Curves Separable Equations and Applications Linear First-Order Equations Substitution Methods and Exact Equations Mathematical Models and Numerical Methods / 2： Population Models Equilibrium Solutions and Stability Acceleration-Velocity Models Numerical Approximation: Euler's Method A Closer Look at the Euler Method The Runge-Kutta Method Linear Systems and Matrices / 3： Introduction to Linear Systems Matrices and Gaussian Elimination Reduced Row-Echelon Matrices Matrix Operations Inverses of Matrices Determinants Linear Equations and Curve Fitting Vector Spaces / 4： The Vector Space R The Vector Space Rn and Subspaces Linear Combinations and Independence of Vectors Bases and Dimension for Vector Spaces General Vector Spaces Linear Equations of Higher Order / 5： Introduction: Second-Order Linear Equations General Solutions of Linear Equations Homogeneous Equations with Constant Coefficients Mechanical Vibrations Undetermined Coefficients and Variation of Parameters Forced Oscillations and Resonance Eigenvalues and Eigenvectors / 6： Introduction to Eigenvalues Diagonalization of Matrices Applications Involving Powers of Matrices Linear Systems of Differential Equations / 7： First-Order Systems and Applications Matrices and Linear Systems The Eigenvalue Method for Linear Systems Second-Order Systems and Mechanical Applications Multiple Eigenvalue Solutions Numerical Methods for Systems Matrix Exponential Methods / 8： Matrix Exponentials and Linear Systems Nonhomogeneous Linear Systems Spectral Decomposition Methods Nonlinear Systems and Phenomena / 9： Stability and the Phase Plane Linear and Almost Linear Systems Ecological Models: Predators and Competitors Nonlinear Mechanical Systems Laplace Transform Methods / 10： Laplace Transforms and Inverse Transforms Transformation of Initial Value Problems Translation and Partial Fractions Derivatives, Integrals, and Products of Transforms Periodic and Piecewise Continuous Forcing Functions Power Series Methods / 11： Introduction and Review of Power Series Power Series Solutions Frobenius Series Solutions Bessel's Equation References for Further Study The Existence and Uniqueness of Solutions / Appendix A： Theory of Determinants / Appendix B： Answers to Selected Problems Index\n First-Order Differential Equations / 1： Differential Equations and Mathematical Model Integrals as General and Particular Solutions",
null,
"文献複写・貸借依頼"
] | [
null,
"https://topics.libra.titech.ac.jp/sites/all/modules/xc/xc_search/images/alert-close.png",
null,
"https://topics.libra.titech.ac.jp/sites/all/modules/xc/xc_search/images/univ.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6432328,"math_prob":0.8747971,"size":2624,"snap":"2020-24-2020-29","text_gpt3_token_len":569,"char_repetition_ratio":0.1908397,"word_repetition_ratio":0.07339449,"special_character_ratio":0.17606707,"punctuation_ratio":0.01510574,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9760182,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-29T22:01:58Z\",\"WARC-Record-ID\":\"<urn:uuid:5046a2fe-3d5f-4a83-a75c-aeff9d83077a>\",\"Content-Length\":\"68857\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f32fe51a-97f1-4f8a-819c-6a41f4c3b9ab>\",\"WARC-Concurrent-To\":\"<urn:uuid:74a81e1d-c07c-462b-a7fd-653a3c06f8d8>\",\"WARC-IP-Address\":\"133.130.93.108\",\"WARC-Target-URI\":\"https://topics.libra.titech.ac.jp/xc/search/DA01348607?os%5Bua%5D=DA01348607\",\"WARC-Payload-Digest\":\"sha1:IU5IIJNB7CWTQAARUMCXGOP46T3G5XEG\",\"WARC-Block-Digest\":\"sha1:FITBDNUMLDEZ5WNL3LWTYUJ5DJYAO54N\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347406785.66_warc_CC-MAIN-20200529214634-20200530004634-00579.warc.gz\"}"} |
https://metanumbers.com/102869 | [
"# 102869 (number)\n\n102,869 (one hundred two thousand eight hundred sixty-nine) is an odd six-digits composite number following 102868 and preceding 102870. In scientific notation, it is written as 1.02869 × 105. The sum of its digits is 26. It has a total of 3 prime factors and 8 positive divisors. There are 92,160 positive integers (up to 102869) that are relatively prime to 102869.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 6\n• Sum of Digits 26\n• Digital Root 8\n\n## Name\n\nShort name 102 thousand 869 one hundred two thousand eight hundred sixty-nine\n\n## Notation\n\nScientific notation 1.02869 × 105 102.869 × 103\n\n## Prime Factorization of 102869\n\nPrime Factorization 13 × 41 × 193\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 102869 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 102,869 is 13 × 41 × 193. Since it has a total of 3 prime factors, 102,869 is a composite number.\n\n## Divisors of 102869\n\n8 divisors\n\n Even divisors 0 8 8 0\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 114072 Sum of all the positive divisors of n s(n) 11203 Sum of the proper positive divisors of n A(n) 14259 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 320.732 Returns the nth root of the product of n divisors H(n) 7.21432 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 102,869 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 102,869) is 114,072, the average is 14,259.\n\n## Other Arithmetic Functions (n = 102869)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 92160 Total number of positive integers not greater than n that are coprime to n λ(n) 960 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 9812 Total number of primes less than or equal to n r2(n) 32 The number of ways n can be represented as the sum of 2 squares\n\nThere are 92,160 positive integers (less than 102,869) that are coprime with 102,869. And there are approximately 9,812 prime numbers less than or equal to 102,869.\n\n## Divisibility of 102869\n\n m n mod m 2 3 4 5 6 7 8 9 1 2 1 4 5 4 5 8\n\n102,869 is not divisible by any number less than or equal to 9.\n\n## Classification of 102869\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n• Square Free\n\n### Other numbers\n\n• LucasCarmichael\n• Sphenic\n\n## Base conversion (102869)\n\nBase System Value\n2 Binary 11001000111010101\n3 Ternary 12020002222\n4 Quaternary 121013111\n5 Quinary 11242434\n6 Senary 2112125\n8 Octal 310725\n10 Decimal 102869\n12 Duodecimal 4b645\n20 Vigesimal ch39\n36 Base36 27dh\n\n## Basic calculations (n = 102869)\n\n### Multiplication\n\nn×y\n n×2 205738 308607 411476 514345\n\n### Division\n\nn÷y\n n÷2 51434.5 34289.7 25717.2 20573.8\n\n### Exponentiation\n\nny\n n2 10582031161 1088562963500909 111979383492375007921 11519207200477124689825349\n\n### Nth Root\n\ny√n\n 2√n 320.732 46.8556 17.909 10.0567\n\n## 102869 as geometric shapes\n\n### Circle\n\n Diameter 205738 646345 3.32444e+10\n\n### Sphere\n\n Volume 4.55976e+15 1.32978e+11 646345\n\n### Square\n\nLength = n\n Perimeter 411476 1.0582e+10 145479\n\n### Cube\n\nLength = n\n Surface area 6.34922e+10 1.08856e+15 178174\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 308607 4.58215e+09 89087.2\n\n### Triangular Pyramid\n\nLength = n\n Surface area 1.83286e+10 1.28288e+14 83992.2\n\n## Cryptographic Hash Functions\n\nmd5 2fe2186da00dfaa4bb4829cd7655ea0d 4628e4076d154a2fb9d1b45e08f16dd9eaccbba6 983cc86d1c720874c2640dc328f363cd9c7ed2092d840638b24c1c237e3d81b3 b872e6174ce160a8d8db4f196edd00c52b96d0c64e6014121f1691e90b5f864bab3daa38390f5133ba9b9d26318ef71b6d78036e033797fbeb94cebe17b0b8bb 088667ee2fe7b3ebac154341cd24d6fd9fc4588a"
] | [
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https://lucatrevisan.wordpress.com/tag/public-key-encryption/ | [
"# CS276 Lecture 19: Trapdoor Permutations\n\nScribed by Cynthia Sturton\n\nSummary\n\nToday we continue to discuss number-theoretic constructions of CPA-secure encryption schemes.\n\nFirst, we return to the Decision Diffie Hellman assumption (the one under which we proved the security of the El Gamal encryption scheme) and we show that it fails for",
null,
"${{\\mathbb Z}^*_p}$, although it is conjectured to hold in the subgroup of quadratic residues of",
null,
"${{\\mathbb Z}^*_p}$.\n\nThen we introduce the notion of trapdoor permutation and show how to construct CPA-secure public-key encryption from any family of trapdoor permutations. Since RSA is conjectured to provide a family of trapdoor permutations, this gives a way to achieve CPA-secure encryption from RSA.\n\n# CS276 Lecture 23 (draft)\n\nSummary\n\nToday we show how to construct an efficient CCA-secure public-key encryption scheme in the random oracle model using RSA.\n\nAs we discussed in the previous lecture, a cryptographic scheme defined in the random oracle model is allowed to use a random function",
null,
"${H: \\{ 0,1 \\}^n \\rightarrow \\{ 0,1 \\}^m}$ which is known to all the parties. In an implementation, usually a cryptographic hash function replaces the random oracle. In general, the fact that a scheme is proved secure in the random oracle model does not imply that it is secure when the random oracle is replaced by a hash function; the proof of security in the random oracle model gives, however, at least some heuristic confidence in the soundness of the design.\n\n# CS276 Lecture 17: El Gamal\n\nScribed by Matt Finifter\n\nSummary\n\nToday we begin to talk about public-key cryptography, starting from public-key encryption.\n\nWe define the public-key analog of the weakest form of security we studied in the private-key setting: message-indistinguishability for one encryption. Because of the public-key setting, in which everybody, including the adversary, has the ability to encrypt messages, this is already equivalent to CPA security.\n\nWe then describe the El Gamal cryptosystem, which is message-indistinguishable (and hence CPA-secure) under the plausible Decision Diffie-Hellman assumption.\n\n# CS276 Lecture 17 (draft)\n\nSummary\n\nToday we begin to talk about public-key cryptography, starting from public-key encryption.\n\nWe define the public-key analog of the weakest form of security we studied in the private-key setting: message-indistinguishability for one encryption. Because of the public-key setting, in which everybody, including the adversary, has the ability to encrypt messages, this is already equivalent to CPA security.\n\nWe then describe the El Gamal cryptosystem, which is message-indistinguishable (and hence CPA-secure) under the plausible Decision Diffie-Hellman assumption."
] | [
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
null,
"https://s0.wp.com/latex.php",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8904765,"math_prob":0.82346445,"size":2589,"snap":"2021-43-2021-49","text_gpt3_token_len":510,"char_repetition_ratio":0.13462283,"word_repetition_ratio":0.41450778,"special_character_ratio":0.17883353,"punctuation_ratio":0.087053575,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97508603,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-20T14:35:04Z\",\"WARC-Record-ID\":\"<urn:uuid:350d67b9-a2dd-4ead-9d3b-ab66ad1e74b1>\",\"Content-Length\":\"131450\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0d9cf977-d1f7-437c-af95-d9a0096ad662>\",\"WARC-Concurrent-To\":\"<urn:uuid:a8ec9933-a85f-48bb-af19-9b12b377245e>\",\"WARC-IP-Address\":\"192.0.78.13\",\"WARC-Target-URI\":\"https://lucatrevisan.wordpress.com/tag/public-key-encryption/\",\"WARC-Payload-Digest\":\"sha1:KHQKQCE3EEZFOHLEPPNFJ7FJ3D4GHGY3\",\"WARC-Block-Digest\":\"sha1:SPMBKHFCIPXUWPSDGUBOQFZYI75IADS5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585321.65_warc_CC-MAIN-20211020121220-20211020151220-00131.warc.gz\"}"} |
https://www.vladonai.com/online-manual?app-name=all-my-notes-organizer&platform=windows&edition=deluxe&version=3.31&topic=how-to-use-in-text-calculator-math | [
"# Support ‹ AllMyNotes Organizer manual ‹ Topic: How to use the Calculator/Math\n\n## Calculator\n\nAllMyNotes Organizer has a Calculator built-in to the text editor. It can automatically evaluate math expressions typed anywhere in the text editor. Just type 1+1= and 2 will appear automatically. It's that easy.\n\n## Supported math Operations and Functions:\n\n- - Substraction operation.\n\n* or x - Multiplication operation.\n\n/ - Division operation.\n\n^ - Power operation.\n\n() and {} - Parentheses can be used to Group terms as in a standard expression.\n\nABS - Absolute Value function.\n\nSQRT - Square Root function.\n\nLOG - Log base 10 function.\n\nLN - Natural Log function.\n\nEXP - Exponential function.\n\nSIN - Sine function\n\nCOS - Cosine function\n\nTAN - Tangent function.\n\nCOT - Cotangent function.\n\nARCSIN - Arcsine function.\n\nARCCOS - Arccosine function.\n\nARCTAN - Arctangent function.\n\nSINH - Hyperbolic Sine function.\n\nCOSH - Hyperbolic Cosine function.\n\nTANH - Hyperbolic Tangent function.\n\nARSINH - Inverse Hyperbolic Sine function.\n\nARCOSH - Inverse Hyperbolic Cosine function.\n\nARTANH - Inverse Hyperbolic Tangent function.\n\nINT - Integer Rounding function.\n\nRAD - the Arc of a given angle function.\n\nDEG - the angle of an arithmetic expression in degrees function.\n\nSIGN - Sign function.\n\nPI - π constant (equal to 3.14159265358979323846264338327950288419716939937510).\n\nPHI - golden ratio constant (φ is equal to 1.618033988749894848204586834).\n\n## Examples:\n\n1 + 2 = 3\n\n(2 + 2) x 2 = 8\n\n2 + 2 x 2 = 6\n\ncos(PI) / sin(PI / 2) = -1\n\nint(1.98) = 1"
] | [
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https://excelkid.com/not-function/ | [
"# NOT function\n\nThe Excel NOT function returns a reversed boolean value of a given logical or boolean value. NOT converts TRUE to FALSE and vice verse.\n\n## Syntax and arguments, return value\n\nThe syntax is simple:\n\n``=NOT(logical)``\n\nArgument: The function uses one argument: “Logical”, the value that the function will convert to TRUE or FALSE. Make sure you use a logical or numeric value as an argument; elsewhere, the formula will return a #VALUE! Error.\n\nReturn value: The return value is a boolean expression, TRUE or FALSE.\n\n## NOT function Examples\n\nAs stated above, the LEN function returns the source cell’s opposite, containing a logical or Boolean value. You can use the NOT function to get the reverse of a logical expression.\n\n• NOT returns TRUE if the logical value is FALSE\n• NOT returns FALSE if the logical value is TRUE\n\n### Example #1: Using NOT and logical functions in a formula\n\nIn Excel, we frequently use logical functions (AND, OR) with the IF and NOT to find records where the given condition is not TRUE. In the example, we have a table that wants to display the records where the t-shirt size is not “L” and not “M”.\n\nFind cells that do not meet specific criteria:\n\nThe formula in D3 is the following:\n\n``=NOT(OR(C3=\"M\",C3=\"L\"))``\n\nWe use text values in the test, so do not forget to use double quotes.\n\nExplanation: The formula uses the following conditions: “the size is NOT “L” AND “M”. The formula will test all records in the range and returns TRUE if the size in column B is equal to “L” or “M”. So it is an OR relationship between the two conditions. I the t-shirt size is not “L” or “M” the formula will return FALSE.\n\nIf you do not want to return booleans as output, make the formula easy to read using the IF function. Use a descriptive output for TRUE and FALSE results.\n\nFormula:\n\n``=IF(NOT(OR(C3=\"M\",C3=\"L\")),\"match\",\"no match\")``\n\nExplanation: The IF function helps us create a descriptive name as an output. If the logical test is TRUE (we find the t-shirts where the size is not “M” or “L”), write “match”. Else, use the “no-match” string.\n\n### Example #2: Count non-blank cells in a range\n\nIf you want to count blank cells in a range, you can use the COUNTA function. In case of a cell status check, use the ISBLANK function. What if we want to highlight cells that are NOT blank?\n\nCombine NOT and ISBLANK!\n\nFor the sake of simplicity, we use the same table:\n\n``=NOT(ISBLANK(C3))``\n\nIn this case, you can add the IF function if you want to display “blank” or “non-blank” outputs instead of TRUE or FALSE.",
null,
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"https://secure.gravatar.com/avatar/de3e5e3f08ca8d6395363d2e8fac3af0",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7429172,"math_prob":0.94307286,"size":2501,"snap":"2022-27-2022-33","text_gpt3_token_len":599,"char_repetition_ratio":0.15018022,"word_repetition_ratio":0.013636364,"special_character_ratio":0.23790483,"punctuation_ratio":0.11596958,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99387014,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-19T07:54:09Z\",\"WARC-Record-ID\":\"<urn:uuid:57d7affd-9d13-4425-8e7e-ca62fc708f3b>\",\"Content-Length\":\"111692\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1e2f70c-5f7f-47b8-af47-aa131346c1c9>\",\"WARC-Concurrent-To\":\"<urn:uuid:4c374968-2b4c-4198-b7ba-efaa615e0df9>\",\"WARC-IP-Address\":\"104.26.4.66\",\"WARC-Target-URI\":\"https://excelkid.com/not-function/\",\"WARC-Payload-Digest\":\"sha1:TI5FXQR7AR35YFP4HBM35AS5PTDR74E6\",\"WARC-Block-Digest\":\"sha1:7UNL23MXF4BX57ZQWSDQP45CAAMI4KEX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573630.12_warc_CC-MAIN-20220819070211-20220819100211-00576.warc.gz\"}"} |
https://www.programmingtunes.com/dynamic-memory-allocation-in-c/ | [
"# Dynamic Memory allocation in C++\n\n0\n209",
null,
"It’s the concept of dynamic memory allocation at the runtime, it is simply the use of “new” and “delete” operators. Normally we cannot specify at the runtime that what will be the size of our array or what will be our memory allocation after the definition of size of array at the runtime? so it’s the solution to that, we can allocate memory for an array by defining its size while the program is running,simply its all about dynamic memory allocation at runtime. We use the “new” and “delete ” operators as:\n\ne.g. int number;\n\ncin>>number;\n\nint *ptr = new int [number];\n\nthis new operator provides the dynamic memory allocation.\n\nNow if we must have to delete this memory:\n\nDelete[] ptr;\n\nHere the “delete” operator finishes the memory allocation.\n\nNow in this case “ptr” is an array of the size equal to “number”. It will be clear from the following example:\n\n```//dynamic memery allocation in C++\n#include\nusing namespace std;\nint main()\n{\nfloat *p;\nint i;\np = new float ; // get a 10-element array\nif(!p) {\ncout << \"Allocation Failuren\";\nreturn 1;\n}\n// assign the values 100 through 109\nfor(i=0; i<10; i++) p[i] = 100.00 + i;\n\n// display the contents of the array\nfor(i=0; i<10; i++) cout << p[i] << \" \";\n\ndelete [] p; // delete the entire array\n\nreturn 0;\n}\n```\n\nIn the above example first we defined a pointer and then we set a memory by using a “new” operator, then we used the pointer as in the method of Arrays, and at the end we deleted the dynamic memory allocation."
] | [
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https://math.libretexts.org/Courses/Monroe_Community_College/MTH_210_Calculus_I_(Seeburger)/04%3A_Applications_of_Derivatives/4.06%3A_Applied_Optimization_Problems | [
"# 4.6: Applied Optimization Problems\n\n•",
null,
"• OpenStax\n• OpenStax\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$$$\\newcommand{\\AA}{\\unicode[.8,0]{x212B}}$$\n\n##### Learning Objectives\n• Set up and solve optimization problems in several applied fields.\n\nOne common application of calculus is calculating the minimum or maximum value of a function. For example, companies often want to minimize production costs or maximize revenue. In manufacturing, it is often desirable to minimize the amount of material used to package a product with a certain volume. In this section, we show how to set up these types of minimization and maximization problems and solve them by using the tools developed in this chapter.\n\n## Solving Optimization Problems over a Closed, Bounded Interval\n\nThe basic idea of the optimization problems that follow is the same. We have a particular quantity that we are interested in maximizing or minimizing. However, we also have some auxiliary condition that needs to be satisfied. For example, in Example $$\\PageIndex{1}$$, we are interested in maximizing the area of a rectangular garden. Certainly, if we keep making the side lengths of the garden larger, the area will continue to become larger. However, what if we have some restriction on how much fencing we can use for the perimeter? In this case, we cannot make the garden as large as we like. Let’s look at how we can maximize the area of a rectangle subject to some constraint on the perimeter.\n\n##### Example $$\\PageIndex{1}$$: Maximizing the Area of a Garden\n\nA rectangular garden is to be constructed using a rock wall as one side of the garden and wire fencing for the other three sides (Figure $$\\PageIndex{1}$$). Given $$100\\,\\text{ft}$$ of wire fencing, determine the dimensions that would create a garden of maximum area. What is the maximum area?",
null,
"Figure $$\\PageIndex{1}$$: We want to determine the measurements $$x$$ and $$y$$ that will create a garden with a maximum area using $$100\\,\\text{ft}$$ of fencing.\n\nSolution\n\nLet $$x$$ denote the length of the side of the garden perpendicular to the rock wall and $$y$$ denote the length of the side parallel to the rock wall. Then the area of the garden is\n\n$$A=x⋅y.$$\n\nWe want to find the maximum possible area subject to the constraint that the total fencing is $$100\\,\\text{ft}$$. From Figure $$\\PageIndex{1}$$, the total amount of fencing used will be $$2x+y.$$ Therefore, the constraint equation is\n\n$$2x+y=100.$$\n\nSolving this equation for $$y$$, we have $$y=100−2x.$$ Thus, we can write the area as\n\n$$A(x)=x⋅(100−2x)=100x−2x^2.$$\n\nBefore trying to maximize the area function $$A(x)=100x−2x^2,$$ we need to determine the domain under consideration. To construct a rectangular garden, we certainly need the lengths of both sides to be positive. Therefore, we need $$x>0$$ and $$y>0$$. Since $$y=100−2x$$, if $$y>0$$, then $$x<50$$. Therefore, we are trying to determine the maximum value of $$A(x)$$ for $$x$$ over the open interval $$(0,50)$$. We do not know that a function necessarily has a maximum value over an open interval. However, we do know that a continuous function has an absolute maximum (and absolute minimum) over a closed interval. Therefore, let’s consider the function $$A(x)=100x−2x^2$$ over the closed interval $$[0,50]$$. If the maximum value occurs at an interior point, then we have found the value $$x$$ in the open interval $$(0,50)$$ that maximizes the area of the garden.\n\nTherefore, we consider the following problem:\n\nMaximize $$A(x)=100x−2x^2$$ over the interval $$[0,50].$$\n\nAs mentioned earlier, since $$A$$ is a continuous function on a closed, bounded interval, by the extreme value theorem, it has a maximum and a minimum. These extreme values occur either at endpoints or critical points. At the endpoints, $$A(x)=0$$. Since the area is positive for all $$x$$ in the open interval $$(0,50)$$, the maximum must occur at a critical point. Differentiating the function $$A(x)$$, we obtain\n\n$$A′(x)=100−4x.$$\n\nTherefore, the only critical point is $$x=25$$ (Figure $$\\PageIndex{2}$$). We conclude that the maximum area must occur when $$x=25$$.",
null,
"Figure $$\\PageIndex{2}$$: To maximize the area of the garden, we need to find the maximum value of the function $$A(x)=100x−2x^2$$.\n\nThen we have $$y=100−2x=100−2(25)=50.$$ To maximize the area of the garden, let $$x=25\\,\\text{ft}$$ and $$y=50\\,\\text{ft}$$. The area of this garden is $$1250\\, \\text{ft}^2$$.\n\n##### Exercise $$\\PageIndex{1}$$\n\nDetermine the maximum area if we want to make the same rectangular garden as in Figure $$\\PageIndex{2}$$, but we have $$200\\,\\text{ft}$$ of fencing.\n\nHint\n\nWe need to maximize the function $$A(x)=200x−2x^2$$ over the interval $$[0,100].$$\n\nThe maximum area is $$5000\\, \\text{ft}^2$$.\n\nNow let’s look at a general strategy for solving optimization problems similar to Example $$\\PageIndex{1}$$.\n\n##### Problem-Solving Strategy: Solving Optimization Problems\n1. Introduce all variables. If applicable, draw a figure and label all variables.\n2. Determine which quantity is to be maximized or minimized, and for what range of values of the other variables (if this can be determined at this time).\n3. Write a formula for the quantity to be maximized or minimized in terms of the variables. This formula may involve more than one variable.\n4. Write any equations relating the independent variables in the formula from step $$3$$. Use these equations to write the quantity to be maximized or minimized as a function of one variable.\n5. Identify the domain of consideration for the function in step $$4$$ based on the physical problem to be solved.\n6. Locate the maximum or minimum value of the function from step $$4.$$ This step typically involves looking for critical points and evaluating a function at endpoints.\n\nNow let’s apply this strategy to maximize the volume of an open-top box given a constraint on the amount of material to be used.\n\n##### Example $$\\PageIndex{2}$$: Maximizing the Volume of a Box\n\nAn open-top box is to be made from a $$24\\,\\text{in.}$$ by $$36\\,\\text{in.}$$ piece of cardboard by removing a square from each corner of the box and folding up the flaps on each side. What size square should be cut out of each corner to get a box with the maximum volume?\n\nSolution\n\nStep 1: Let $$x$$ be the side length of the square to be removed from each corner (Figure $$\\PageIndex{3}$$). Then, the remaining four flaps can be folded up to form an open-top box. Let $$V$$ be the volume of the resulting box.",
null,
"Figure $$\\PageIndex{3}$$: A square with side length $$x$$ inches is removed from each corner of the piece of cardboard. The remaining flaps are folded to form an open-top box.\n\nStep 2: We are trying to maximize the volume of a box. Therefore, the problem is to maximize $$V$$.\n\nStep 3: As mentioned in step 2, are trying to maximize the volume of a box. The volume of a box is\n\n$V=L⋅W⋅H \\nonumber, \\nonumber$\n\nwhere $$L,\\,W,$$and $$H$$ are the length, width, and height, respectively.\n\nStep 4: From Figure $$\\PageIndex{3}$$, we see that the height of the box is $$x$$ inches, the length is $$36−2x$$ inches, and the width is $$24−2x$$ inches. Therefore, the volume of the box is\n\n\\begin{align*} V(x) &=(36−2x)(24−2x)x \\\\[4pt] &=4x^3−120x^2+864x \\end{align*}. \\nonumber\n\nStep 5: To determine the domain of consideration, let’s examine Figure $$\\PageIndex{3}$$. Certainly, we need $$x>0.$$ Furthermore, the side length of the square cannot be greater than or equal to half the length of the shorter side, $$24\\,\\text{in.}$$; otherwise, one of the flaps would be completely cut off. Therefore, we are trying to determine whether there is a maximum volume of the box for $$x$$ over the open interval $$(0,12).$$ Since $$V$$ is a continuous function over the closed interval $$[0,12]$$, we know $$V$$ will have an absolute maximum over the closed interval. Therefore, we consider $$V$$ over the closed interval $$[0,12]$$ and check whether the absolute maximum occurs at an interior point.\n\nStep 6: Since $$V(x)$$ is a continuous function over the closed, bounded interval $$[0,12]$$, $$V$$ must have an absolute maximum (and an absolute minimum). Since $$V(x)=0$$ at the endpoints and $$V(x)>0$$ for $$0<x<12,$$ the maximum must occur at a critical point. The derivative is\n\n$$V′(x)=12x^2−240x+864.$$\n\nTo find the critical points, we need to solve the equation\n\n$$12x^2−240x+864=0.$$\n\nDividing both sides of this equation by $$12$$, the problem simplifies to solving the equation\n\n$$x^2−20x+72=0.$$\n\nUsing the quadratic formula, we find that the critical points are\n\n\\begin{align*} x &=\\dfrac{20±\\sqrt{(−20)^2−4(1)(72)}}{2} \\\\[4pt] &=\\dfrac{20±\\sqrt{112}}{2} \\\\[4pt] &=\\dfrac{20±4\\sqrt{7}}{2} \\\\[4pt] &=10±2\\sqrt{7} \\end{align*}. \\nonumber\n\nSince $$10+2\\sqrt{7}$$ is not in the domain of consideration, the only critical point we need to consider is $$10−2\\sqrt{7}$$. Therefore, the volume is maximized if we let $$x=10−2\\sqrt{7}\\,\\text{in.}$$ The maximum volume is\n\n$V(10−2\\sqrt{7})=640+448\\sqrt{7}≈1825\\,\\text{in}^3. \\nonumber$\n\nas shown in the following graph.",
null,
"Figure $$\\PageIndex{4}$$: Maximizing the volume of the box leads to finding the maximum value of a cubic polynomial.\n##### Exercise $$\\PageIndex{2}$$\n\nSuppose the dimensions of the cardboard in Example $$\\PageIndex{2}$$ are $$20\\,\\text{in.}$$ by $$30\\,\\text{in.}$$ Let $$x$$ be the side length of each square and write the volume of the open-top box as a function of $$x$$. Determine the domain of consideration for $$x$$.\n\nHint\n\nThe volume of the box is $$L⋅W⋅H.$$\n\n$$V(x)=x(20−2x)(30−2x).$$ The domain is $$[0,10]$$.\n\n##### Example $$\\PageIndex{3}$$: Minimizing Travel Time\n\nAn island is $$2$$ mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that is $$6$$ mi west of that point. The visitor is planning to go from the cabin to the island. Suppose the visitor runs at a rate of $$8$$ mph and swims at a rate of $$3$$ mph. How far should the visitor run before swimming to minimize the time it takes to reach the island?\n\nSolution\n\nStep 1: Let $$x$$ be the distance running and let $$y$$ be the distance swimming (Figure $$\\PageIndex{5}$$). Let $$T$$ be the time it takes to get from the cabin to the island.",
null,
"Figure $$\\PageIndex{5}$$: How can we choose $$x$$ and $$y$$ to minimize the travel time from the cabin to the island?\n\nStep 2: The problem is to minimize $$T$$.\n\nStep 3: To find the time spent traveling from the cabin to the island, add the time spent running and the time spent swimming. Since Distance = Rate × Time $$(D=R×T),$$ the time spent running is\n\n$$T_{running}=\\dfrac{D_{running}}{R_{running}}=\\dfrac{x}{8}$$,\n\nand the time spent swimming is\n\n$$T_{swimming}=\\dfrac{D_{swimming}}{R_{swimming}}=\\dfrac{y}{3}$$.\n\nTherefore, the total time spent traveling is\n\n$$T=\\dfrac{x}{8}+\\dfrac{y}{3}$$.\n\nStep 4: From Figure $$\\PageIndex{5}$$, the line segment of $$y$$ miles forms the hypotenuse of a right triangle with legs of length $$2$$ mi and $$6−x$$ mi. Therefore, by the Pythagorean theorem, $$2^2+(6−x)^2=y^2$$, and we obtain $$y=\\sqrt{(6−x)^2+4}$$. Thus, the total time spent traveling is given by the function\n\n$$T(x)=\\dfrac{x}{8}+\\dfrac{\\sqrt{(6−x)^2+4}}{3}$$.\n\nStep 5: From Figure $$\\PageIndex{5}$$, we see that $$0≤x≤6$$. Therefore, $$[0,6]$$ is the domain of consideration.\n\nStep 6: Since $$T(x)$$ is a continuous function over a closed, bounded interval, it has a maximum and a minimum. Let’s begin by looking for any critical points of $$T$$ over the interval $$[0,6].$$ The derivative is\n\n\\begin{align*} T′(x) &=\\dfrac{1}{8}−\\dfrac{1}{2}\\dfrac{[(6−x)^2+4]^{−1/2}}{3}⋅2(6−x) \\\\[4pt] &=\\dfrac{1}{8}−\\dfrac{(6−x)}{3\\sqrt{(6−x)^2+4}} \\end{align*}\n\nIf $$T′(x)=0,$$, then\n\n$\\dfrac{1}{8}=\\dfrac{6−x}{3\\sqrt{(6−x)^2+4}} \\label{ex3eq1}$\n\nTherefore,\n\n$3\\sqrt{(6−x)^2+4}=8(6−x). \\label{ex3eq2}$\n\nSquaring both sides of this equation, we see that if $$x$$ satisfies this equation, then $$x$$ must satisfy\n\n$9[(6−x)^2+4]=64(6−x)^2,\\nonumber$\n\nwhich implies\n\n$55(6−x)^2=36. \\nonumber$\n\nWe conclude that if $$x$$ is a critical point, then $$x$$ satisfies\n\n$(x−6)^2=\\dfrac{36}{55}. \\nonumber$\n\n[Note that since we are squaring, $$(x-6)^2 = (6-x)^2.$$]\n\nTherefore, the possibilities for critical points are\n\n$x=6±\\dfrac{6}{\\sqrt{55}}.\\nonumber$\n\nSince $$x=6+6/\\sqrt{55}$$ is not in the domain, it is not a possibility for a critical point. On the other hand, $$x=6−6/\\sqrt{55}$$ is in the domain. Since we squared both sides of Equation \\ref{ex3eq2} to arrive at the possible critical points, it remains to verify that $$x=6−6/\\sqrt{55}$$ satisfies Equation \\ref{ex3eq1}. Since $$x=6−6/\\sqrt{55}$$ does satisfy that equation, we conclude that $$x=6−6/\\sqrt{55}$$ is a critical point, and it is the only one. To justify that the time is minimized for this value of $$x$$, we just need to check the values of $$T(x)$$ at the endpoints $$x=0$$ and $$x=6$$, and compare them with the value of $$T(x)$$ at the critical point $$x=6−6/\\sqrt{55}$$. We find that $$T(0)≈2.108\\,\\text{h}$$ and $$T(6)≈1.417\\,\\text{h}$$, whereas\n\n$T(6−6/\\sqrt{55})≈1.368\\,\\text{h}. \\nonumber$\n\nTherefore, we conclude that $$T$$ has a local minimum at $$x≈5.19$$ mi.\n\n##### Exercise $$\\PageIndex{3}$$\n\nSuppose the island is $$1$$ mi from shore, and the distance from the cabin to the point on the shore closest to the island is $$15$$ mi. Suppose a visitor swims at the rate of $$2.5$$ mph and runs at a rate of $$6$$ mph. Let $$x$$ denote the distance the visitor will run before swimming, and find a function for the time it takes the visitor to get from the cabin to the island.\n\nHint\n\nThe time $$T=T_{running}+T_{swimming}.$$\n\n$$T(x)=\\dfrac{x}{6}+\\dfrac{\\sqrt{(15−x)^2+1}}{2.5}$$\n\nIn business, companies are interested in maximizing revenue. In the following example, we consider a scenario in which a company has collected data on how many cars it is able to lease, depending on the price it charges its customers to rent a car. Let’s use these data to determine the price the company should charge to maximize the amount of money it brings in.\n\n##### Example $$\\PageIndex{4}$$: Maximizing Revenue\n\nOwners of a car rental company have determined that if they charge customers $$p$$ dollars per day to rent a car, where $$50≤p≤200$$, the number of cars $$n$$ they rent per day can be modeled by the linear function $$n(p)=1000−5p$$. If they charge $$50$$ per day or less, they will rent all their cars. If they charge $$200$$ per day or more, they will not rent any cars. Assuming the owners plan to charge customers between $$50$$ per day and $$200$$ per day to rent a car, how much should they charge to maximize their revenue?\n\nSolution\n\nStep 1: Let $$p$$ be the price charged per car per day and let $$n$$ be the number of cars rented per day. Let $$R$$ be the revenue per day.\n\nStep 2: The problem is to maximize $$R.$$\n\nStep 3: The revenue (per day) is equal to the number of cars rented per day times the price charged per car per day—that is, $$R=n×p.$$\n\nStep 4: Since the number of cars rented per day is modeled by the linear function $$n(p)=1000−5p,$$ the revenue $$R$$ can be represented by the function\n\n\\begin{align*} R(p) &=n×p \\\\[4pt] &=(1000−5p)p \\\\[4pt] &=−5p^2+1000p.\\end{align*}\n\nStep 5: Since the owners plan to charge between $$50$$ per car per day and $$200$$ per car per day, the problem is to find the maximum revenue $$R(p)$$ for $$p$$ in the closed interval $$[50,200]$$.\n\nStep 6: Since $$R$$ is a continuous function over the closed, bounded interval $$[50,200]$$, it has an absolute maximum (and an absolute minimum) in that interval. To find the maximum value, look for critical points. The derivative is $$R′(p)=−10p+1000.$$ Therefore, the critical point is $$p=100$$. When $$p=100, R(100)=50,000.$$ When $$p=50, R(p)=37,500$$. When $$p=200, R(p)=0$$.\n\nTherefore, the absolute maximum occurs at $$p=100$$. The car rental company should charge $$100$$ per day per car to maximize revenue as shown in the following figure.",
null,
"Figure $$\\PageIndex{6}$$: To maximize revenue, a car rental company has to balance the price of a rental against the number of cars people will rent at that price.\n##### Exercise $$\\PageIndex{4}$$\n\nA car rental company charges its customers $$p$$ dollars per day, where $$60≤p≤150$$. It has found that the number of cars rented per day can be modeled by the linear function $$n(p)=750−5p.$$ How much should the company charge each customer to maximize revenue?\n\nHint\n\n$$R(p)=n×p,$$ where $$n$$ is the number of cars rented and $$p$$ is the price charged per car.\n\nThe company should charge $$75$$ per car per day.\n\n##### Example $$\\PageIndex{5}$$: Maximizing the Area of an Inscribed Rectangle\n\nA rectangle is to be inscribed in the ellipse\n\n$\\dfrac{x^2}{4}+y^2=1. \\nonumber$\n\nWhat should the dimensions of the rectangle be to maximize its area? What is the maximum area?\n\nSolution\n\nStep 1: For a rectangle to be inscribed in the ellipse, the sides of the rectangle must be parallel to the axes. Let $$L$$ be the length of the rectangle and $$W$$ be its width. Let $$A$$ be the area of the rectangle.",
null,
"Figure $$\\PageIndex{7}$$: We want to maximize the area of a rectangle inscribed in an ellipse.\n\nStep 2: The problem is to maximize $$A$$.\n\nStep 3: The area of the rectangle is $$A=LW.$$\n\nStep 4: Let $$(x,y)$$ be the corner of the rectangle that lies in the first quadrant, as shown in Figure $$\\PageIndex{7}$$. We can write length $$L=2x$$ and width $$W=2y$$. Since $$\\dfrac{x^2}{4}+y^2=1$$ and $$y>0$$, we have $$y=\\sqrt{1-\\dfrac{x^2}{4}}$$. Therefore, the area is\n\n$$A=LW=(2x)(2y)=4x\\sqrt{1-\\dfrac{x^2}{4}}=2x\\sqrt{4−x^2}$$\n\nStep 5: From Figure $$\\PageIndex{7}$$, we see that to inscribe a rectangle in the ellipse, the $$x$$-coordinate of the corner in the first quadrant must satisfy $$0<x<2$$. Therefore, the problem reduces to looking for the maximum value of $$A(x)$$ over the open interval $$(0,2)$$. Since $$A(x)$$ will have an absolute maximum (and absolute minimum) over the closed interval $$[0,2]$$, we consider $$A(x)=2x\\sqrt{4−x^2}$$ over the interval $$[0,2]$$. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval.\n\nStep 6: As mentioned earlier, $$A(x)$$ is a continuous function over the closed, bounded interval $$[0,2]$$. Therefore, it has an absolute maximum (and absolute minimum). At the endpoints $$x=0$$ and $$x=2$$, $$A(x)=0.$$ For $$0<x<2$$, $$A(x)>0$$.\n\nTherefore, the maximum must occur at a critical point. Taking the derivative of $$A(x)$$, we obtain\n\n\\begin{align*} A'(x) &=2\\sqrt{4−x^2}+2x⋅\\dfrac{1}{2\\sqrt{4−x^2}}(−2x) \\\\[4pt] &=2\\sqrt{4−x^2}−\\dfrac{2x^2}{\\sqrt{4−x^2}} \\\\[4pt] &=\\dfrac{8−4x^2}{\\sqrt{4−x^2}} . \\end{align*}\n\nTo find critical points, we need to find where $$A'(x)=0.$$ We can see that if $$x$$ is a solution of\n\n$\\dfrac{8−4x^2}{\\sqrt{4−x^2}}=0, \\label{ex5eq1}$\n\nthen $$x$$ must satisfy\n\n$8−4x^2=0. \\nonumber$\n\nTherefore, $$x^2=2.$$ Thus, $$x=±\\sqrt{2}$$ are the possible solutions of Equation \\ref{ex5eq1}. Since we are considering $$x$$ over the interval $$[0,2]$$, $$x=\\sqrt{2}$$ is a possibility for a critical point, but $$x=−\\sqrt{2}$$ is not. Therefore, we check whether $$\\sqrt{2}$$ is a solution of Equation \\ref{ex5eq1}. Since $$x=\\sqrt{2}$$ is a solution of Equation \\ref{ex5eq1}, we conclude that $$\\sqrt{2}$$ is the only critical point of $$A(x)$$ in the interval $$[0,2]$$.\n\nTherefore, $$A(x)$$ must have an absolute maximum at the critical point $$x=\\sqrt{2}$$. To determine the dimensions of the rectangle, we need to find the length $$L$$ and the width $$W$$. If $$x=\\sqrt{2}$$ then\n\n$y=\\sqrt{1−\\dfrac{(\\sqrt{2})^2}{4}}=\\sqrt{1−\\dfrac{1}{2}}=\\dfrac{1}{\\sqrt{2}}.\\nonumber$\n\nTherefore, the dimensions of the rectangle are $$L=2x=2\\sqrt{2}$$ and $$W=2y=\\dfrac{2}{\\sqrt{2}}=\\sqrt{2}$$. The area of this rectangle is $$A=LW=(2\\sqrt{2})(\\sqrt{2})=4.$$\n\n##### Exercise $$\\PageIndex{5}$$\n\nModify the area function $$A$$ if the rectangle is to be inscribed in the unit circle $$x^2+y^2=1$$. What is the domain of consideration?\n\nHint\n\nIf $$(x,y)$$ is the vertex of the square that lies in the first quadrant, then the area of the square is $$A=(2x)(2y)=4xy.$$\n\n$$A(x)=4x\\sqrt{1−x^2}.$$ The domain of consideration is $$[0,1]$$.\n\n## Solving Optimization Problems when the Interval Is Not Closed or Is Unbounded\n\nIn the previous examples, we considered functions on closed, bounded domains. Consequently, by the extreme value theorem, we were guaranteed that the functions had absolute extrema. Let’s now consider functions for which the domain is neither closed nor bounded.\n\nMany functions still have at least one absolute extrema, even if the domain is not closed or the domain is unbounded. For example, the function $$f(x)=x^2+4$$ over $$(−∞,∞)$$ has an absolute minimum of $$4$$ at $$x=0$$. Therefore, we can still consider functions over unbounded domains or open intervals and determine whether they have any absolute extrema. In the next example, we try to minimize a function over an unbounded domain. We will see that, although the domain of consideration is $$(0,∞),$$ the function has an absolute minimum.\n\nIn the following example, we look at constructing a box of least surface area with a prescribed volume. It is not difficult to show that for a closed-top box, by symmetry, among all boxes with a specified volume, a cube will have the smallest surface area. Consequently, we consider the modified problem of determining which open-topped box with a specified volume has the smallest surface area.\n\n##### Example $$\\PageIndex{6}$$: Minimizing Surface Area\n\nA rectangular box with a square base, an open top, and a volume of $$216 \\,\\text{in}^3$$ is to be constructed. What should the dimensions of the box be to minimize the surface area of the box? What is the minimum surface area?\n\nSolution\n\nStep 1: Draw a rectangular box and introduce the variable $$x$$ to represent the length of each side of the square base; let $$y$$ represent the height of the box. Let $$S$$ denote the surface area of the open-top box.",
null,
"Figure $$\\PageIndex{8}$$: We want to minimize the surface area of a square-based box with a given volume.\n\nStep 2: We need to minimize the surface area. Therefore, we need to minimize $$S$$.\n\nStep 3: Since the box has an open top, we need only determine the area of the four vertical sides and the base. The area of each of the four vertical sides is $$x⋅y.$$ The area of the base is $$x^2$$. Therefore, the surface area of the box is\n\n$$S=4xy+x^2$$.\n\nStep 4: Since the volume of this box is $$x^2y$$ and the volume is given as $$216\\,\\text{in}^3$$, the constraint equation is\n\n$$x^2y=216$$.\n\nSolving the constraint equation for $$y$$, we have $$y=\\dfrac{216}{x^2}$$. Therefore, we can write the surface area as a function of $$x$$ only:\n\n$S(x)=4x\\left(\\dfrac{216}{x^2}\\right)+x^2.\\nonumber$\n\nTherefore, $$S(x)=\\dfrac{864}{x}+x^2$$.\n\nStep 5: Since we are requiring that $$x^2y=216$$, we cannot have $$x=0$$. Therefore, we need $$x>0$$. On the other hand, $$x$$ is allowed to have any positive value. Note that as $$x$$ becomes large, the height of the box $$y$$ becomes correspondingly small so that $$x^2y=216$$. Similarly, as $$x$$ becomes small, the height of the box becomes correspondingly large. We conclude that the domain is the open, unbounded interval $$(0,∞)$$. Note that, unlike the previous examples, we cannot reduce our problem to looking for an absolute maximum or absolute minimum over a closed, bounded interval. However, in the next step, we discover why this function must have an absolute minimum over the interval $$(0,∞).$$\n\nStep 6: Note that as $$x→0^+,\\, S(x)→∞.$$ Also, as $$x→∞, \\,S(x)→∞$$. Since $$S$$ is a continuous function that approaches infinity at the ends, it must have an absolute minimum at some $$x∈(0,∞)$$. This minimum must occur at a critical point of $$S$$. The derivative is\n\n$S′(x)=−\\dfrac{864}{x^2}+2x.\\nonumber$\n\nTherefore, $$S′(x)=0$$ when $$2x=\\dfrac{864}{x^2}$$. Solving this equation for $$x$$, we obtain $$x^3=432$$, so $$x=\\sqrt{432}=6\\sqrt{2}.$$ Since this is the only critical point of $$S$$, the absolute minimum must occur at $$x=6\\sqrt{2}$$ (see Figure $$\\PageIndex{9}$$).\n\nWhen $$x=6\\sqrt{2}$$, $$y=\\dfrac{216}{(6\\sqrt{2})^2}=3\\sqrt{2}\\,\\text{in.}$$ Therefore, the dimensions of the box should be $$x=6\\sqrt{2}\\,\\text{in.}$$ and $$y=3\\sqrt{2}\\,\\text{in.}$$ With these dimensions, the surface area is\n\n$S(6\\sqrt{2})=\\dfrac{864}{6\\sqrt{2}}+(6\\sqrt{2})^2=108\\sqrt{4}\\,\\text{in}^2\\nonumber$",
null,
"Figure $$\\PageIndex{9}$$: We can use a graph to determine the dimensions of a box of given the volume and the minimum surface area.\n##### Exercise $$\\PageIndex{6}$$\n\nConsider the same open-top box, which is to have volume $$216\\,\\text{in}^3$$. Suppose the cost of the material for the base is $$20¢/\\text{in}^2$$ and the cost of the material for the sides is $$30¢/\\text{in}^2$$ and we are trying to minimize the cost of this box. Write the cost as a function of the side lengths of the base. (Let $$x$$ be the side length of the base and $$y$$ be the height of the box.)\n\nHint\n\nIf the cost of one of the sides is $$30¢/\\text{in}^2,$$ the cost of that side is $$0.30xy$$ dollars.\n\n$$c(x)=\\dfrac{259.2}{x}+0.2x^2$$ dollars\n\n## Key Concepts\n\n• To solve an optimization problem, begin by drawing a picture and introducing variables.\n• Find an equation relating the variables.\n• Find a function of one variable to describe the quantity that is to be minimized or maximized.\n• Look for critical points to locate local extrema.\n\n## Glossary\n\noptimization problems\nproblems that are solved by finding the maximum or minimum value of a function\n\nThis page titled 4.6: Applied Optimization Problems is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8440241,"math_prob":0.9999969,"size":23638,"snap":"2023-40-2023-50","text_gpt3_token_len":7010,"char_repetition_ratio":0.14474909,"word_repetition_ratio":0.073097005,"special_character_ratio":0.3304848,"punctuation_ratio":0.111788616,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000026,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-05T03:50:31Z\",\"WARC-Record-ID\":\"<urn:uuid:214685f9-9289-47de-9a44-99d50c515c46>\",\"Content-Length\":\"163895\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:de839abf-1e90-4e66-9c56-93d8e8ecd739>\",\"WARC-Concurrent-To\":\"<urn:uuid:ba799967-11aa-4d2f-b96d-2021068dc1ee>\",\"WARC-IP-Address\":\"3.162.103.38\",\"WARC-Target-URI\":\"https://math.libretexts.org/Courses/Monroe_Community_College/MTH_210_Calculus_I_(Seeburger)/04%3A_Applications_of_Derivatives/4.06%3A_Applied_Optimization_Problems\",\"WARC-Payload-Digest\":\"sha1:VFRTWWW3U2NX52FB2MGTCCCIU6J5ZNIC\",\"WARC-Block-Digest\":\"sha1:XX26ZZYEU3UPKRGUGVCBF4ZNV2CZJBG7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511717.69_warc_CC-MAIN-20231005012006-20231005042006-00049.warc.gz\"}"} |
https://www.acmicpc.net/problem/5520 | [
"시간 제한메모리 제한제출정답맞힌 사람정답 비율\n1 초 256 MB43715710435.495%\n\n## 문제\n\n|-------| |-------| |-------|\n| | | | | | |\n|---O | |---O | | O |\n| | | | | |\n|-------| |-------| |-------|\nA B C\n\n|-------| |-------| |-------|\n| | | | | |\n| O | | O | | O |\n| | | | | | | | |\n|-------| |-------| |-------|\nD E F\n\n|-------| |-------| |-------|\n| | | | | |\n| O | | O---| | O |\n| | | | | | | |\n|-------| |-------| |-------| (Figure 1)\nG H I\n\nThere are nine clocks in a 3*3 array (figure 1). The goal is to return all the dials to 12 o'clock with as few moves as possible. There are nine different allowed ways to turn the dials on the clocks. Each such way is called a move. Select for each move a number 1 to 9. That number will turn the dials 90' (degrees) clockwise on those clocks which are affected according to figure 2 below.\n\nMove Affected clocks\n\n1 ABDE\n2 ABC\n3 BCEF\n5 BDEFH\n6 CFI\n7 DEGH\n8 GHI\n9 EFHI (Figure 2)\n\n## 입력\n\nRead nine numbers from the standard input. These numbers give the start positions of the dials. 0=12 o'clock, 1=3 o'clock, 2=6 o'clock, 3=9 o'clock. The example in figure 1 gives the following input data file:\n\n## 출력\n\nWrite to the standard output a shortest sequence of moves (numbers) in increasing order, which returns all the dials to 12 o'clock. In case there are many solutions, only one is required.\n\n## 예제 입력 1\n\n3 3 0\n2 2 2\n2 1 2\n\n\n## 예제 출력 1\n\n4 5 8 9\n\n\n## 힌트\n\nEach number represents a time accoring to following table:\n\n0 = 12 o'clock\n1 = 3 o'clock\n2 = 6 o'clock\n3 = 9 o'clock\n3 3 0 3 0 0 3 0 0 0 0 0 0 0 0\n2 2 2 5-> 3 3 3 8-> 3 3 3 4 -> 0 3 3 9-> 0 0 0\n2 1 2 2 2 2 3 3 3 0 3 3 0 0 0 \n\n## 출처\n\n• 빠진 조건을 찾은 사람: cokcjswo"
] | [
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https://manual.dewesoft.com/x/setupmodule/general/math/filters/math-fftfilter | [
"# Frequency domain filter\n\nWhen you press the Setup button on new activated FFT Filter line, the following FFT filter setup window will open:",
null,
"The filter supports multiple input channels.\n\nFor detailed information about basic settings of the input and output channels see -> Setup screen and basic operation.\n\n## FFT filter description\n\nFFT filter is quite different to other types of filters. While IIR and FIR filters are time domain filters, FFT filter calculates the spectrum of the signal with specific number of lines and overlap and then extracts the RMS value of certain range of this signal. Therefore the result is not the full curve, but only one value per frequency spectrum.\n\nThe usage of this filter is to extract low peaks of signals where there are big harmonics nearby where it wouldn’t be possible to choose IIR filter which would extract this low amplitude.\n\nThe example below shows the electromotive winding failure which can be seen as low values at the rotation frequency where the line frequency is very high:",
null,
"We can design our own filter -> see -> Custom FFT Filter.\n\n## FFT filter parameters\n\nTo set the FFT filter, it is recommended to observe the signal in the FFT display and choose the right Number of lines and Window which fits the best and then set the filter parameters.\n\nFor FFT Filter you can set:\n\nFFT calculation parameters\n\nAmplitude extraction parameters\n\nonly for fixed frequency source\n\nonly for tracking frequency source\n\nonly for Tracking Frequency source\n\n### Number of lines\n\nYou can select Number of lines from the list.",
null,
"This defines the resolution of the filter as well as the number of points in the calculation. The resolution needs to be high enough that the wanted harmonic can be clearly extracted, but not too high to have a higher result update.\n\n### Overlap\n\nYou can select Overlap from the list.",
null,
"Overlap defines (same for FFT averaging) how many ‘old’ data is taken for next calculation. This increases the result update rate with the same number of lines.\n\n### Window type\n\nYou can select Window type from the list.",
null,
"The window defines the behavior of the filter in the transition and the stop band (the height of the sidebands and the width of the main band).\n\n### Frequency source\n\nYou can select Frequency source from the list if Tracking filter type is enabled:\n\n• Fixed - Fixed frequency will always take fixed value for center frequency.\n\n• Tracking - Tracking frequency means that the center frequency will depend on a second input channel (for example rotation frequency).",
null,
"#### Delta frequency\n\nIn this field, you can enter Delta frequency in Hz. This value depends on the wanted frequency band. It also depends on the window and number of lines (line resolution). In our example we would choose 0.5 Hz since we don’t want that 50 Hz value will appear in the result.\n\n#### Center frequency\n\nIf we choose Fixed Frequency source, we need to enter Center frequency in Hz. The center frequency is the middle value of frequency for value extraction. In our example above we would take 49 Hz as the center frequency.\n\n#### Filter settings for Tracking frequency source\n\nIf we choose Tracking Frequency source, we need to enter the Frequency channel and Number of harmonics instead of Center frequency:",
null,
"#### Frequency channel\n\nFrequency channel is the channel with the current frequency which needs to be extracted. The unit of this channel must be in Hz.\n\nFrequency channel can be selected from the list:",
null,
"#### Number of harmonics\n\nNumber of harmonics describes how many harmonics need to be extracted from the spectrum. If we enter a value of 5, there will be 5 channels created for each input channel. The first channel will have the center frequency as the frequency channel; second will have twice the frequency of the input and so on.\n\n## Custom FFT filter\n\nIf Custom filter is selected from the Filter type, then we can design our own filter. With this option, we can create any type of filter curve in the frequency domain and calculate RMS value. Sometimes it is not easy to define filter characteristics in the time domain, but we have it defined in the frequency domain. Custom FFT filter is perfect for such case.",
null,
"If the frequency source is external, we can define the channel where the frequency is defined and the filter will change the characteristic to always filter correctly like in the time domain. This is especially useful for example for CA noise calculation on the external clock."
] | [
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"https://manual.dewesoft.com/assets/img/hmfile_hash_e3b7cf88.jpg",
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"https://manual.dewesoft.com/assets/img/dwsmath-jure_sl14_fftfilter-example.jpg",
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"https://manual.dewesoft.com/assets/img/hmfile_hash_0ce09807.jpg",
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"https://manual.dewesoft.com/assets/img/filters_fft_overlap.jpg",
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"https://manual.dewesoft.com/assets/img/filters_fft_window.jpg",
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"https://manual.dewesoft.com/assets/img/hmfile_hash_996aab87.jpg",
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"https://manual.dewesoft.com/assets/img/filters_fft_filtertracking.jpg",
null,
"https://manual.dewesoft.com/assets/img/filters_fft_frequencychannel.jpg",
null,
"https://manual.dewesoft.com/assets/img/filters_fft_folter_custom.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.896186,"math_prob":0.9244113,"size":4368,"snap":"2019-51-2020-05","text_gpt3_token_len":849,"char_repetition_ratio":0.17942254,"word_repetition_ratio":0.005376344,"special_character_ratio":0.1904762,"punctuation_ratio":0.07098381,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95811373,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,5,null,5,null,5,null,5,null,5,null,5,null,5,null,5,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-21T14:12:01Z\",\"WARC-Record-ID\":\"<urn:uuid:82e4d5b2-e019-4688-8009-624532bed581>\",\"Content-Length\":\"23422\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:49b74c94-7f1e-438b-81c1-cfd121bf315e>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a9fb31b-7aa9-427f-b11e-c9797a10e21e>\",\"WARC-IP-Address\":\"104.26.7.33\",\"WARC-Target-URI\":\"https://manual.dewesoft.com/x/setupmodule/general/math/filters/math-fftfilter\",\"WARC-Payload-Digest\":\"sha1:TZSZGRFCRKYZPAVEKQE5K5SFHB4NDC4C\",\"WARC-Block-Digest\":\"sha1:226HPXUG5BK44SJEICZUZ4FQS5ZKMIRE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250604397.40_warc_CC-MAIN-20200121132900-20200121161900-00451.warc.gz\"}"} |
https://engineersedge.com/material_science/deflection-torsion-hollow-shaft.htm | [
"Torsional Deflection of Hollow Cylinder Calculator\n\nStrength / Mechanics of Materials\n\nIn solid mechanics, torsion is the twisting of an object due to an applied torque. It is expressed in newton metres (N·m) or foot-pound force (in·lb). In sections perpendicular to the torque axis, the resultant shear stress in this section is perpendicular to the radius.\n\n Torsional Deflection of Hollow Shaft Variables Applied Torque T (N-mm, in-lb) = Shear Modulus G (Mpa, psi) = Outside Diameter D (mm, in.) = Inside Diameter d (mm, in.) = Unsupported Length L (mm, in) = Results Deflection of shaft (radians) Example calculations are in imperial units (in-lbs, psi, etc.)"
] | [
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https://fr.slideserve.com/roddy/electron-beam-lithography | [
"",
null,
"Download",
null,
"Download Presentation",
null,
"Electron Beam Lithography\n\n# Electron Beam Lithography\n\nDownload Presentation",
null,
"## Electron Beam Lithography\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. Electron Beam Lithography William Whelan-Curtin\n\n2. What is EBL for? LPN • Nanopatterning • High Precision • Reliable • Versatile\n\n3. What is needed? • Very narrow, precisely controllable beam of Electrons • Lots of money, a big complex machine, and a lot of expertise!\n\n4. Outline • System description • Exposure • Examples\n\n5. Electron “Pencil” • System Schematic\n\n6. Electron source I Tungsten /Zirconium Oxide Tip -V • Tungsten wire (Thermionic) • 2300C • Energy Spread 2-3eV • Source size 25um • Thermal (Schottky) field emitter • 1800C • Energy Spread 0.9eV • Source size 20nm • Cold Field Emitter • 20C • Energy Spread 0.22eV • Source size 5nm Suppressor Higher Current density =>EBL Extractor (Anode) +V Unstable current 10-20% => SEM (High Vacuum ~10e-9mB)\n\n7. Electron Lenses • Chromatic dispersion • Monochromatic beam • Aberrations • Use centre of lens F = q · (E+ v B) Electro-magnetic Electro-static\n\n8. Electron Lenses • Magnetic versus Electrostatic Faster deflection Worse Aberrations EM lenses Simpler to implement\n\n9. Beam Blanker • Turns the beam/off quickly • Control current for each pixel • High speed • Electrostatic Extractor +V +V\n\n10. Column • Source • Apertures • Blanking • Collimation • Stigmation control • Deflectors • Focus • Final Lens (VISTEC VB6)\n\n11. Types of Electron Beam Writer • SEM (Electron Beam “Reader”) • Generally magnetic lenses • Up to 30kV • Converted SEM (RAITH) • Addition of (fast) beam blanker • Pattern generator • Deflection needs time to stablise raster scan vector scan\n\n12. Types of Electron Beam Writer • Purpose Built (LEICA/JEOL) • Better control, calibration • Up to 100kV • Higher speeds • Bigger writefields • Secondary deflection system Can also correct for aberrations in the primary lens\n\n13. Types of Electron Beam Writer • Shaped Beam systems • Very complex optics • Higher current, lower resolution • Photomask making • (Not a research tool) GAUSSIAN\n\n14. Patterning LPN • Electron sensitive polymer- the “paper”\n\n15. Resist Overview “Positive” “Negative”\n\n16. Electron-Solid Interactions Primary electrons • Forward scattering • Often • Small angles • High energy (~95% pass through the resist)\n\n17. Electron-Solid Interactions Primary electrons • Back scattering • Rare • Large angles • Still High energy\n\n18. Electron-Solid Interactions Primary electrons • Secondary electrons • Low energy (50eV) • Low penetrating power Responsible for exposure\n\n19. H H H H C C C C H H H H H H H H H H H H H H C C C C C C C C C C C H H H H H O C O C O C C O n H H H H C O C O C O C O H H H H H H H H Electron Sensitive resist • Poly methyl methyl acrylate • Spin Coating • Long polymer chains H H C C H H H H H H C C C H H C C O O H H O O C C H H H H PMMA Substrate\n\n20. H H C C C C H Electron Sensitive resist Secondary Electrons • Bonds broken (induced chain scission) • Dissolved by suitable chemical • Methyl Isobutyl ketone • Isopropanol:Water H H H H H H C C C C C C H H H H H H H H H H H H H H H H H H C C C C C C C C C C H H H H H H C C O O O C O C O C C O n H H H H H H O O C C C O C O C O C O H H H H H H H H H H H H PMMA Substrate\n\n21. Contrast Curve Function of: Voltage Resist Thickness Substrate e.g. 170uAs/cm2 for 500nm thick PMMA on Silicon @ 30kV • Threshold electron density (lower≡faster exposure) • “Clearing Dose” or “Base Dose” • Slope -> Resolution Resist Thickness Dose (electrons/area)\n\n22. Contrast Curve- Experimental • Determine for each new situation • Recheck regularly • Problem diagnosis Dose 50um\n\n23. H H C C C C H Negative Resist Secondary Electrons • Microchem SU8 • Photo acid generator • Baking • Crosslinking • Acid Diffusion H H H H H H C C C C C C H H H H H H H H H H H H H H H H H C C C C C C C C C C H H H H H H C C O O O C O C O C C O n H H H H H H O O C C C O C O C O C O H H H H H H H H H H H H SU8 Substrate\n\n24. Contrast Curve • Threshold electron density • Chemical amplification • Lower clearing dose Resist Thickness e.g. 1uAs/cm2 for 200nm thick SU8 on Silicon @ 30kV Dose (electrons/area)\n\n25. Resolution J. Vac. Sci. Technol. B 12, 1305 (1975) • Most EBL systems -> 1nm spot sizes or less Vb 1nm Rt df df = effective beam diameter (nm) Rt = resist thickness (nm) Vb = acceleration voltage (kV)\n\n26. Resolution EBL advice: Keep resist as thin as possible\n\n27. EBL vs Focused Ion Beam etching • FIB • Energetic Gallium ions • Etching of the material • EBL • Modification • No removal of material Heavy ions Substrate\n\n28. Electron-Solid Interactions Unintended Exposure! Primary electrons • Secondary electrons • Low energy (50eV) • Low penetrating power\n\n29. Proximity Errors • Stray electrons • Bias t dose\n\n30. Correction • Shape Correction • Difficult to generalise\n\n31. Correction • Dose Modulation • Calculate the electron distribution • Reduce in certain areas 2 1\n\n32. Electron Distribution Forward Scattering- α Back Scattering- β\n\n33. Parameters • Depend on voltage/resist/substrate • Determine in each instance • Monte Carlo simulations • Experiment L. Stevens et al., Microelectronic Engineering 5, 141-150 (1986)\n\n34. Correction Programs • Nanopecs, Proxecco • Pattern • Fracture • Calculate electron distributions • Alter pattern • Recalculate • Iterate until convergence is reached\n\n35. Guidelines • PEC Computationally intensive • β<< pattern length scale << β (Homogenous background of scattered electrons) β =3.5um L=2um L=500nm L=50um\n\n36. Laser Stage • Limited Deflection • Expose one “writefield” at a time • Laser interferometer controlled Stage movements • Calibrated • Sub 40nm accuracy • Pattern stitched together • Critical for Photonics\n\n37. Writefield alignment θ ? 100um\n\n38. Writefield alignment θ Xum\n\n39. Examples\n\n40. Liftoff • PMMA • Electron beam evaporation • Acetone\n\n41. Liftoff • Metal thickness <1/3 of resist • Pure PMMA very effective at 30kV (or less)\n\n42. Liftoff • Higher voltages • Better Resolution • Less forward scattering • Bilayer Resist Low molecular Weight PMMA\n\n43. Etch back • Liftoff incomplete • Deposit metal , spin • Expose and develop • Dry etch Pmma Metal Substrate\n\n44. Dry Etching • ZEON ZEP 520A • Xylene • Higher Sensitivity • Tougher\n\n45. Dry Etching • ZEON ZEP 520A • High resolution, good etch resistance • Etch Quality crucial for many applications • Low selectivity etch • Thick Resist (400nm, @30kV)\n\n46. Grayscale Lithography • SU8 Resist • Graded Dose\n\n47. Grayscale Lithography • Dry etch into Silicon • Luneberg Lens Di Falco et al. Optics Express 19, pp. 5156 (2011)\n\n48. The highest Resolution • Polymer resists • Resolution limited by chain length • Hydrogen silsesquioxane • Spin on Dielectric • Negative Resist Exposure/ Curing\n\n49. HSQ 25nm period Grating • Highest resolution available • Good etch resistance 100kV Metal HSQ Substrate Diamond! <15nm dots Lister et al. Microelectronic Engineering 73--74, 319 (2004)\n\n50. RAITH E-Line • Electron beam induced condensation • Gas fed into Chamber • 3D High Resolution Lithography Substrate"
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https://www.physicsforums.com/threads/maximum-minimum-hints-or-pictures.387034/ | [
"# Maximum/Minimum - hints or pictures?\n\n• Carl_M\nIn summary, the conversation discusses three different mathematical problems involving optimization. The first problem involves finding the outside dimensions of a billboard with a minimum total area given specific margins. The second problem involves finding the dimensions of a cylindrical soft drink can that minimize its cost, taking into account different costs for the sides, bottom, and top. The third problem involves determining the speed at which a bale of hay is rising when it is 2m below a loft, given the farmer's walking speed and the placement of a pulley. The conversation also mentions the need for equations and diagrams to solve these problems.\n\n## Homework Statement\n\n1. A billboard is to made with 100 m2 of printed area, with margins of 2m at the top and bottom, and 4m\non each side. Find the outside dimensions of the billboard if its total area is to be a minimum.\n\n2. A cylindrical soft drink can is to have a volume of 500 ml. If the sides and bottom\nare made from aluminum that costs 0.1¢/cm2, while the top is made from a thicker aluminum that costs 0.3\n¢.cm2. Find the dimensions of the can that minimize its cost.\n\n3. A farmer raises a bale of hay to a loft 6m above his shoulder by a 20 m rope using a pulley 1.5 m above\nthe loft. He walks away from the loft at 1.3 m/s. How fast is the bale rising when it is 2m below the loft?\n\nWhat's the third one trying to say?\n\n## Homework Equations\n\nAny pictures or diagrams by any chance?\n\n## The Attempt at a Solution\n\nHave you had a go at this?\n\nFirst of all you need to find equations for your problems, and from there differentiate them to find the maximum/minimum.\n\nIf you have a go at the problems, I'll help you some more.\n\nIn #3,\n\nhttp://img405.imageshack.us/img405/4035/mmmcopy.png [Broken]\n\nIs this what it's saying for the diagram?\n\nLast edited by a moderator:"
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http://mediphacos.com/kggjbzcv/lfspxy.php?id=6567ce-hong-kong-math-curriculum | [
"Evening Tiling Courses, Circuit Court Of Sangamon County Il Case Search, Apartments In Gallatin, Tn Under \\$700, Mint 20 Kde, Cheese Vector Png, Hunting In Tajikistan, Cbsa Interview Questions And Answers, Why Does Portia Send Lucius To The Senate?, Beyerdynamic Dt 1770 Vs 770, \" /> Evening Tiling Courses, Circuit Court Of Sangamon County Il Case Search, Apartments In Gallatin, Tn Under \\$700, Mint 20 Kde, Cheese Vector Png, Hunting In Tajikistan, Cbsa Interview Questions And Answers, Why Does Portia Send Lucius To The Senate?, Beyerdynamic Dt 1770 Vs 770, \" />\n\n## NOTÍCIAS E EVENTOS\n\n### hong kong math curriculum\n\nHong Kong law requires attendance in primary school starting at the age of six. The Junior secondary school covers a broad curriculum that expands upon what students were being taught in primary school. Introducing the rules for finding the area of a rectangle and a parallelogram. Objective: On completion of the lesson the student will be able to compare and order decimals to two decimal places and understand decimal notation to two places. Objective: Students will increase their vocabulary and develop their reading skills by reading, in which the present continuous tense is used. Objective: To use the apostrophe of possession correctly. Objective: On completion of the lesson the student will be able to simplify and evaluate numerical expressions containing patterns, and be able to simplify algebraic expressions that contain like terms. Objective: On completion of the lesson the student will be able to change numbers greater than 1 to scientific notation. Objective: On completion of the lesson the student should be able to: choose the correct unit to measure the mass of small, medium or large objects, and convert measurements from one unit to another. Objective: Students will develop their comprehension skills by reading a short text about Christmas and Chinese New Year. Objective: To form plurals of foreign nouns. Surface area of a triangular/trapezoidal prism. Objective: Students will increase their vocabulary and develop their reading skills by reading a short speech about what a person ‘can’ and ‘can’t’ do. Objective: On completion of the lesson the student will understand simplification of algebraic expressions in step-by-step processing. We offer eLearning classes in both math and English that coincide with all levels of Hong Kong primary and secondary education levels. They will also learn the little words that help to sequence a story. Objective: Students will develop their comprehension skills by reading, and listening to, a rhyme. Objective: On completion of the lesson the student will be able to name and find fractions that represent equal amounts between halves, quarters and eighths – using diagrams and number lines. Objective: Students will increase their vocabulary and develop their reading skills by reading a short text about daily routines. Objective: On completion of the lesson the student will able to use Long Division by repeated subtraction with divisors greater than 50 including division of thousands, and some remainders. Objective: Present phrases that ask for and give opinions. Dividing two and three digit numbers by a single digit number. Objective: Learn about the past continuous tense. Objective: On completion of the lesson the student will be able to divide decimal numbers by one hundred and recognise the pattern formed when decimals are divided by ten, one hundred and one thousand. Objective: To use hyphens, dashes, apostrophes, quotation marks, parentheses and brackets correctly. Objective: On completion of the lesson the student will be able to identify which test to use to show two triangles are congruent. The 14 Best Math Tutors Near Me in Hong Kong, Hong Kong - … Objective: On completion of the lesson the student will be able to: read and interpret problems involving money, interpret the everyday use of decimals, and perform calculations with money. Objective: On completion of the lesson the student will be able to solve and record division problems with whole numbers using a two-digit divisor and show any remainders as a fraction. Objective: On completion of the lesson the student will be able to: compare fractions with the denominators 5, 10, 100, and represent fractions with the denominator 5, 10, 100. Objective: Learn the use of first conditionals. Singapore International School (Hong Kong) is known for our high quality holistic education, academic rigour and bilingual programme. Objective: To construct expressive sentences by using adjectives, adverbs, phrases and clauses. Objective: On completion of the lesson the student will be able to: recognise the need for a formal unit to measure volume, use the abbreviation for cubic centimetre, construct three dimensional objects using cubic centimetre blocks, and use counting to determine vo. Objective: On completion of the lesson the student will be able to identify specific places on a map and use regions on a grid to locate objects or places. Objective: On completion of the lesson the student will be able to find the surface areas of pyramids. Objective: On completion of the lesson the student will be able to apply strategies to solve problems using rectangular prisms. PositionIn a knowledge-based information era driven by technology and creativity, students need knowledge and skills that could help them meet the dynamic challenges in the 21st century, and mathematics constitute an important share. Objective: On completion of the lesson the student will be able to divide numbers by a decimal fraction. Objective: Students will build vocabulary; begin to use conjunctions ‘and’, ‘but’ or ‘because’. Objective: To show a pronoun is used in place of a noun. Objective: On completion of the lesson the student will know how to find the answer to take away number sentences with bigger numbers and will also know how to write these number sentences. Objective: On completion of the lesson the student will be able to reduce a fraction to its lowest equivalent form by dividing the numerator and denominator by a common factor. Objective: On completion of the lesson the student will be able to add and subtract positive and negative numbers in any combination, and understand adding and subtracting positive and negative pronumerals. Objective: On completion of the lesson the student will understand the relationship between the two common units of capacity, the litre and millilitre. Objective: Students will develop their language skills by reading, and listening to, a picture story in which some possessive pronouns are used. Objective: On completion of the lesson the student will be able to use long division by repeated subtraction with divisors less than 55 and where dividends are 3 and 4 digit numbers, and there are some remainders. Objective: On completion of the lesson the student will be able to understand what is meant by the speed of an object, read the instantaneous speed of a vehicle on a speedometer and find the average speed of an object. They will also revise colours and pronouns. Objective: On completion of the lesson the student will be able to solve problems with money. Objective: On completion of the lesson the student will be able to recall the steps in a strategy for solving word problems. We take the best aspects of this consistent and rigorous framework and adapt it to ensure that our primary school curriculum is relevant for children growing up in Hong Kong, whilst also placing a strong emphasis on educating global citizens. Mathematics jobs in Hong Kong. Hong Kong Island: Island Christian Academy: Central & Western: Hong Kong Island: Kellett School (Kowloon Bay) Kowloon City: Kowloon: Kellett School (Pok Fu Lam) Southern: Hong Kong Island: Kent College Hong Kong: Central & Western: Hong Kong Island: Kiangsu-Chekiang College, International (Foundation & Primary) Eastern District: Hong Kong Island Objective: On completion of the lesson the student will know how to solve equations using fractions. Objective: To identify and use contractions. Objective: Students will read and practise the words for basic colours. Using the trigonometric ratios to find an angle in a right-angled triangle. Our Hong Kong primary school curriculum is based on the English National Curriculum and is designed to provide the highest academic quality of education. 2 Heung Yip Road, Wong Chuk Hang, Hong Kong. Objective: On completion of the lesson the student will be capable of subtraction using tens and units and calculating the answer to take away or subtraction number sentences. Objective: On completion of the lesson the student will be able to: name the days of the week in order, recognise them in a written form, be able to answer questions about the days of the week, and have an understanding of a diary. Objective: On completion of the lesson the student will be able to multiply any 4 digit numbers by any 3 digit numbers using long multiplication. Objective: Students will develop their comprehension skills by reading a picture story about food preferences. Directed numbers: addition and subtraction. Objective: On completion of the lesson the student will be able to use the law of multiplication of indices when raising more than one term to the same power. Objective: On completion of the lesson the student will be able to construct and interpret frequency histograms and polygons. Objective: On completion of the lesson the student will be able to expand expressions using a negative multiplier. Objective: On completion of the lesson the student will be able to associate the numerals 3, 6 and 9 with 15, 30 and 45 minutes, and use the terms ‘quarter to’ and ‘quarter past’. They will learn some irregular verbs in the present perfect tense. They will revise adjectives that describe personality. Education in Hong Kong is largely modelled on that of the United Kingdom, particularly the English system.It is overseen by the Education Bureau and the Social Welfare Department.. Mathematics and its applications pervade all aspects of life in the modern world. Objective: To use commas: to list, to make meaning clear, and when writing the date. Objective: Learn the correct structure for forming yes/no questions. Objective: On completion of the lesson the student will be able to specify multiples and factors of whole numbers, and calculate the product of squared numbers. Objective: On completion of the lesson the student will know a way to find volume and capacity. Objective: To identify and use conjunctions correctly. Objective: On completion of the lesson the student will be able to use the calculator to find values for the sine, cosine and tangent ratios of acute angles. Objective: Learn the uses of ‘will’ and ‘going to’ to talk about the future. Objective: Students will increase their vocabulary, develop their reading skills and practise using the simple past tense of the verb ‘to be’. Objective: On completion of the lesson the student will understand the concepts of mass. Objective: On completion of the lesson the student will be able to apply strategies to solve problems using rectangular prisms and larger unit. Prof. Jimmy Chi-Hung FUNG George Stephenson Medal Award (Second Best Paper Overall) in ICE Publishing Awards 2020 for his paper \"Downwind flow behaviours of cuboid-shaped obstacles: modelling and experiments\" (2020) The IB is a framework more than a curriculum and is less perscriptive than other \"curricula\", ... mathematics, science and technology, arts, and personal, social and physical education. Multiplying 2-digit numbers by multiple of 10, Multiplying 3 and 4-digit numbers by multiples of 100, Recognise nets for prisms, pyramids, cubes and cones. Objective: On completion of the lesson the student will be able to find the size of an unknown angle of a triangle using the cosine rule given the lengths of the 3 sides. Simplifying Algebraic expressions: combining addition and subtraction. Objective: On completion of the lesson the student will be able to write about equal groups and rows and will understand how to count them. Objective: On completion of the lesson the student will be able to use long division with repeated subtraction by multiples of 1,2,3 where the divisors are less than 20 and there are remainders. Objective: To identify, and make use of, onomatopoeia, assonance, and alliteration. Objective: On completion of the lesson the student will be able to use the student’s knowledge of place value to solve addition problems with 3, 4 and 5 digit numbers and word problems. Objective: On completion of the lesson the student will be able to use standard deviation as a measure of deviation from a mean. To understand an adverb tells something about (modifies) a verb. Objective: On completion of the lesson the student will be able to calculate the area in square centimetres of surfaces or objects and record their results correctly. Objective: On completion of this lesson the student will be able to distinguish between mutually exclusive and non mutually exclusive events and be able to find the probabilities of both. Objective: Students will increase their vocabulary and develop their reading skills by reading a text in which ‘There is’ and ‘There are’ in relation to zoo animals is used. Objective: On completion of the lesson the student will be able to write the numeral and word for the numbers 11 to 20, count up to 20 and match a group of objects to the correct number. Objective: On completion of this lesson the student will be able to use Pythagoras’ Theorem to calculate the length of the hypotenuse. It ushers in a new era for curriculum development to keep abreast of the macro and dynamic changes in various aspects in the local, regional and global landscapes in maintaining the competitiveness of Hong Kong. Displayed here are Job Ads that match your query. They will increase their vocabulary for family relationships and occupations, and they will learn to use ‘this’ and ‘that’. Copyright © 2018. Objective: Students will develop their comprehension skills by reading a short text in which quarter past, quarter to, 5 past, 5 to, 10 past and 10 to the hour are used. Math 37500 (2) General Education electives and/or course meeting multidisciplinary requirement (1) Free Elective Study Abroad course replacement options MA 2504 (3) General Education, Multidisciplinary course, free electives Information about university City University of Hong Kong … Objective: Students will develop their comprehension skills by reading a text about Bruce Lee. ‘and the area of the curved surface ‘ p r l’. It is highly selective, causing many students to go abroad. Objective: On completion of the lesson the student will be able to use the formal units millimetre, centimetre, metre and kilometre to measure and convert. Objective: To identify and eliminate common errors. Objective: On completion of the lesson the student will be able to convert units of time and read and interpret simple timetables, timelines and calendars. Objective: On completion of the lesson the student will be able to calculate the gradient of a line given any two points on the line and also be capable of checking whether 3 or more points lie on the same line and what an unknown point will make to parallel lines. The student will be able to label the side lengths in relation to a given angle e.g. Objective: Students will develop their comprehension skills by reading a text about animals. Objective: Students will develop their language skills by reading, and listening to, a picture story. Objective: Students will develop their comprehension skills by reading a story about the weather. Objective: On completion of the lesson the student will be able to: identify a rhombus, learn how to find the formula for the area of a rhombus, and use it in solving problems.",
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https://books.google.co.uk/books?id=B574tQbP6WcC&source=gbs_navlinks_s | [
"Vector Analysis for Computer Graphics\n\nSpringer Science & Business Media, Jun 18, 2007 - Computers - 259 pages\nIn my last book, Geometry for Computer Graphics, I employed a mixture of algebra and vector analysis to prove many of the equations used in computer graphics. At the time, I did not make any distinction between the two methodologies, but slowly it dawned upon me that I had had to discover, for the first time, how to use vector analysis and associated strategies for solving geometric problems. I suppose that mathematicians are taught this as part of their formal mathematical training, but then, I am not a mathematician! After some deliberation, I decided to write a book that would introduce the beginner to the world of vectors and their application to the geometric problems encountered in computer graphics. I accepted the fact that there would be some duplication of formulas between this and my last book; however, this time I would concentrate on explaining how problems are solved. The book contains eleven chapters: The first chapter distinguishes between scalar and vector quantities, which is reasonably straightforward. The second chapter introduces vector repres- tation, starting with Cartesian coordinates and concluding with the role of direction cosines in changes in axial systems. The third chapter explores how the line equation has a natural vector interpretation and how vector analysis is used to resolve a variety of line-related, geometric problems. Chapter 4 repeats Chapter 3 in the context of the plane.\n\nWhat people are saying -Write a review\n\nWe haven't found any reviews in the usual places.\n\nContents\n\n Scalars and Vectors 1 Vector Representation 11 Straight Lines 61 The Plane 101 Reflections 123 Rotating Vectors 179 Vector Differentiation 201\n Projections 213 Rendering 225 Motion 241 Appendix B 249 Index 256 Copyright"
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http://gskmaturemo.lookstorebr4.xyz/thermodynamic-equations-sheet-physics.html | [
"# Thermodynamic equations sheet physics\n\nSheet thermodynamic\n\n## Thermodynamic equations sheet physics\n\n67 × 10- equations 27 kg Neutron = 0 C = 1. THERMODYNAMICS 73 THERMODYNAMICS physics PROPERTIES OF SINGLE- COMPONENT SYSTEMS Nomenclature 1. The net energy crossing the system boundary is equal to the change sheet in. In particle physics it reflects the changes in the underlying force laws ( codified in a quantum field theory) as the energy scale at which physical processes occur sheet varies, energy/ momentum resolution. Carnot used the phrase motive power for work. GRE PHYSICS STUDY GUIDE by the Department of Physics and Astronomy. These scales are related by the equations K = ° C + 273.\n\nor Maxwell’ s equations for electromagnetic fields. equations of a thermodynamic. Learn for free sheet about math biology, medicine, history, , economics, physics, finance, chemistry, art, sheet computer programming more. Mcat Physics Equations Sheet The physics Gold Standard Prep. sheet Heat And Thermodynamics Physics In 24 Hrs.\n\n1 General De nitions Equations First Law of Thermodynamics: Heat Qis a form of energy, energy is conserved. Revision Notes on Thermodynamics. Fuel Economy engines need only modest amounts of propellant. thermodynamic processes equations of state, kinetic theory, ideal gases ensembles. Thermodynamic equations sheet physics.\n\n{ S 0 Third Law of Thermodynamics: For a system with a nondegenerate ground state S! Which really cuts into the payload mass. Thermodynamics is concerned with macroscopic behavior rather than microscopic behavior of the system. Statistical Physics Spring. Thermo Final Cheat Sheet Docsity.\n\nHere’ s a list of the most important equations ones you need to do the calculations necessary for solving thermodynamics problems. Thermodynamics is filled with equations and formulas. 022 × 10 23 atoms in one atomic mass unit thermodynamic e is the elementary charge: 1. 602 19 × 10- 19 C Potential Energy, velocity of electron: PE = eV = ½mv 2 1V = 1J/ physics C 1N/ C = 1V/ m 1J = 1 N· m = 1 C· V. Thermodynamics: physics - It is the branch of physics which deals with process involving heat work internal energy.\n\nMuscle physics engines have the thrust to perform Hohmann trajectories but the low specific impulse means they have to carry sheet outrageous amounts of propellant. 11 × 10- 31 kg Proton = 1. tists do not study relatively particle physics, but thermodynamics is an integral . A series of thermodynamic processes is shown in the pV- diagram. The behaviour of a complex thermodynamic system,.\n\n{ dU= sheet dQ dW Second Law of Thermodynamics: The entropy Sof a system must increase. Intensive properties are independent of mass. Thermodynamic Equations Formulas; Important Thermodynamic Equations thermodynamic Formulas. PHYSICS FORMULAS 2426 Electron = - 1. How to Calculate Displacement in a Physics Problem. 67 × 10- 27 kg 6.\n\nWork equations done by physics a gas. sheet We are living in a computer programmed reality sheet the only clue we have to it is when some variable is changed some alteration in our reality occurs. For Dummies Cheat Sheet. physics Physical Chemistry Symbols Gallery Free sheet Symbol thermodynamic Design. One of the fundamental thermodynamic equations is the description of thermodynamic work in analogy to mechanical work , weight lifted through an elevation against sheet gravity as defined in 1824 by French physicist Sadi thermodynamic Carnot. Cambridge Handbook Of Physcis Formulas Index Page 6. Thermodynamic equations sheet physics. sheet to be used in the above equations for.\nIt shows you how to solve problems associated with PV diagrams , heat, internal energy work. 0 S= k Bln 0, where 0 is the degeneracy of the. Course Page For physics Chemistry 2290. Important Thermodynamic Equations and Formulas. philosophy of physics: Thermodynamics. Combustion equations: Air- fuel ratio: Hydrocarbon fuel combustion reaction: Compressibility calculations: Compressibility equations factor Z: Pv = ZRT.\n\nconservation of energy in a thermodynamic system. In theoretical sheet physics, the renormalization group ( thermodynamic RG) refers to a mathematical apparatus equations that allows systematic investigation of the changes of a physical system as viewed at different scales. equations such equation is the van der. AP Physics B - Thermodynamics 602 19 × 10- 19 C = 9. 602 19 × 10- 19 C = 1.\n\nsheet Khan Academy is a nonprofit with the mission of providing a free world- class education for anyone anywhere.\n\n## Sheet physics\n\nThermodynamics - 19. Welcome to the homepage for Junior Honours Thermal Physics Part 1: Thermodynamics. Here you will find lecture notes, tutorial sheets, hand- in questions etc. The recommended text on which the course is based is \" Thermal Physics\" by Finn. There are several copies of the second edition in the library. Frequently used equations in physics.\n\n``thermodynamic equations sheet physics``\n\nAppropriate for secondary school students and higher. Mostly algebra based, some trig, some calculus, some fancy calculus."
] | [
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https://visualfractions.com/calculator/square-root/379/ | [
"# Square Root of 379\n\nIn this article we're going to calculate the square root of 379 and explore what the square root is and answer some of the common questions you might. We'll also look at the different methods for calculating the square root of 379 (both with and without a computer/calculator).\n\n## Square Root of 379 Definition\n\nIn mathematical form we can show the square root of 379 using the radical sign, like this: √379. This is usually referred to as the square root of 379 in radical form.\n\nSo what is the square root? In this case, the square root of 379 is the quantity (which we will call q) that when multiplied by itself, will equal 379.\n\n√379 = q × q = q2\n\n## Is 379 a Perfect Square?\n\nIn math, we refer to 379 being a perfect square if the square root of 379 is a whole number.\n\nIn this case, as we will see in the calculations below, we can see that 379 is not a perfect square.\n\nTo find out more about perfect squares, you can read about them and look at a list of 1000 of them in our What is a Perfect Square? article.\n\n## Is The Square Root of 379 Rational or Irrational?\n\nA common question is to ask whether the square root of 379 is rational or irrational. Rational numbers can be written as a fraction and irrational numbers cannot.\n\nA quick way to check this is to see if 379 is a perfect square. If it is, then it is a rational number. If it's not a perfect square then it's an irrational number.\n\nWe already know if 379 is a perfect square so we also can see that √379 is an irrational number.\n\n## Can the Square Root of 379 Be Simplified?\n\n379 can be simplified only if you can make 379 inside the radical symbol smaller. This is a process that is called simplifying the surd. In this example square root of 379 cannot be simplified.\n\n## How to Calculate The Square Root of 379 with a Calculator\n\nIf you have a calculator then the simplest way to calculate the square root of 379 is to use that calculator. On most calculators you can do this by typing in 379 and then pressing the √x key. You should get the following result:\n\n√379 ≈ 19.4679\n\n## How to Calculate the Square Root of 379 with a Computer\n\nOn a computer you can also calculate the square root of 379 using Excel, Numbers, or Google Sheets and the SQRT function, like so:\n\nSQRT(379) ≈ 19.467922333932\n\n## What is the Square Root of 379 Rounded?\n\nSometimes you might need to round the square root of 379 down to a certain number of decimal places. Here are the solutions to that, if needed.\n\n10th: √379 ≈ 19.5\n\n100th: √379 ≈ 19.47\n\n1000th: √379 ≈ 19.468\n\n## What is the Square Root of 379 as a Fraction?\n\nWe covered earlier in this article that only a rational number can be written as a fraction, and irrational numbers cannot.\n\nLike we said above, since the square root of 379 is an irrational number, we cannot make it into an exact fraction. However, we can make it into an approximate fraction using the square root of 379 rounded to the nearest hundredth.\n\n√379\n\n≈ 19.5/1\n\n≈ 1947/100\n\n≈ 19 47/100\n\n## What is the Square Root of 379 Written with an Exponent?\n\nAll square root calculations can be converted to a number (called the base) with a fractional exponent. Let's see how to do that with the square root of 379:\n\n√b = b½\n\n√379 = 379½\n\n## How to Find the Square Root of 379 Using Long Division\n\nFinally, we can use the long division method to calculate the square root of 379. This is very useful for long division test problems and was how mathematicians would calculate the square root of a number before calculators and computers were invented.\n\n### Step 1\n\nSet up 379 in pairs of two digits from right to left and attach one set of 00 because we want one decimal:\n\n3\n79\n00\n\n### Step 2\n\nStarting with the first set: the largest perfect square less than or equal to 3 is 1, and the square root of 1 is 1 . Therefore, put 1 on top and 1 at the bottom like this:\n\n 1 3 79 00 1\n\n### Step 3\n\nCalculate 3 minus 1 and put the difference below. Then move down the next set of numbers.\n\n 1 3 79 00 1 2 79\n\n### Step 4\n\nDouble the number in green on top: 1 × 2 = 2. Then, use 2 and the bottom number to make this problem:\n\n2? × ? ≤ 279\n\nThe question marks are \"blank\" and the same \"blank\". With trial and error, we found the largest number \"blank\" can be is 9. Replace the question marks in the problem with 9 to get:\n\n29 × 9 = 261\n\nNow, enter 9 on top, and 261 at the bottom:\n\n 1 9 3 79 00 1 2 79 2 61\n\n### Step 5\n\nCalculate 279 minus 261 and put the difference below. Then move down the next set of numbers.\n\n 1 9 3 79 00 1 2 79 2 61 0 18 00\n\n### Step 6\n\nDouble the number in green on top: 19 × 2 = 38. Then, use 38 and the bottom number to make this problem:\n\n38? × ? ≤ 1800\n\nThe question marks are \"blank\" and the same \"blank\". With trial and error, we found the largest number \"blank\" can be is 4.\n\nNow, enter 4 on top:\n\n 1 9 4 3 79 00 1 2 79 2 61 0 18 00\n\nThat's it! The answer shown at the top in green. The square root of 379 with one digit decimal accuracy is 19.5. Notice that the last two steps actually repeat the previous two. To add decimal places to your answe you can simply add more sets of 00 and repeat the last two steps."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8951418,"math_prob":0.9944457,"size":5151,"snap":"2022-05-2022-21","text_gpt3_token_len":1426,"char_repetition_ratio":0.18884787,"word_repetition_ratio":0.17775665,"special_character_ratio":0.31547272,"punctuation_ratio":0.0959276,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99930906,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-18T17:27:46Z\",\"WARC-Record-ID\":\"<urn:uuid:deaeb0fb-592e-4da7-a638-f89f10a0a854>\",\"Content-Length\":\"25704\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a0a46478-9743-41d7-b229-6e4465c10aa3>\",\"WARC-Concurrent-To\":\"<urn:uuid:3a047fed-073c-4d79-a580-b8338e1df9ae>\",\"WARC-IP-Address\":\"172.67.147.3\",\"WARC-Target-URI\":\"https://visualfractions.com/calculator/square-root/379/\",\"WARC-Payload-Digest\":\"sha1:BSF46JAALCCD2RG3DKDE42PQ43CHYXIC\",\"WARC-Block-Digest\":\"sha1:I34ZFL6IOXMMLGTVMK6UI5KNZYCLDFV4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662522284.20_warc_CC-MAIN-20220518151003-20220518181003-00713.warc.gz\"}"} |
https://lists.archlinux.org/pipermail/arch-commits/2019-January/574509.html | [
"[arch-commits] Commit in maxima-ecl/repos/community-x86_64 (8 files)\n\nAntonio Rojas arojas at archlinux.org\nSat Jan 26 08:34:33 UTC 2019\n\n``` Date: Saturday, January 26, 2019 @ 08:34:32\nAuthor: arojas\nRevision: 427942\n\narchrelease: copy trunk to community-x86_64\n\nmaxima-ecl/repos/community-x86_64/PKGBUILD\n(from rev 427941, maxima-ecl/trunk/PKGBUILD)\nmaxima-ecl/repos/community-x86_64/build-fasl.patch\n(from rev 427941, maxima-ecl/trunk/build-fasl.patch)\nmaxima-ecl/repos/community-x86_64/matrixexp.patch\n(from rev 427941, maxima-ecl/trunk/matrixexp.patch)\nmaxima-ecl/repos/community-x86_64/stack.patch\n(from rev 427941, maxima-ecl/trunk/stack.patch)\nDeleted:\nmaxima-ecl/repos/community-x86_64/PKGBUILD\nmaxima-ecl/repos/community-x86_64/build-fasl.patch\nmaxima-ecl/repos/community-x86_64/matrixexp.patch\nmaxima-ecl/repos/community-x86_64/stack.patch\n\n------------------+\nPKGBUILD | 108 ++++++++++++++++-----------------\nbuild-fasl.patch | 46 +++++++-------\nmatrixexp.patch | 26 ++++----\nstack.patch | 170 ++++++++++++++++++++++++++---------------------------\n4 files changed, 175 insertions(+), 175 deletions(-)\n\nDeleted: PKGBUILD\n===================================================================\n--- PKGBUILD\t2019-01-26 08:34:04 UTC (rev 427941)\n+++ PKGBUILD\t2019-01-26 08:34:32 UTC (rev 427942)\n@@ -1,54 +0,0 @@\n-# Maintainer: Antonio Rojas <arojas at archlinux.org>\n-# Contributor: Ronald van Haren <ronald.archlinux.org>\n-# Contributor: Damir Perisa <damir at archlinux.org>\n-# Modified to compile against ecl by: maribu\n-\n-pkgname=maxima-ecl\n-_pkgname=maxima\n-pkgver=5.42.1\n-_eclver=16.1.3\n-pkgrel=5\n-pkgdesc=\"A sophisticated computer algebra system (compiled against ecl)\"\n-arch=(x86_64)\n-url=\"https://maxima.sourceforge.net\"\n-depends=(ecl=\\$_eclver texinfo shared-mime-info)\n-makedepends=(python emacs)\n-optdepends=('gnuplot: plotting capabilities' 'rlwrap: readline support via /usr/bin/rmaxima' 'tk: graphical xmaxima interface')\n-conflicts=(maxima)\n-provides=(maxima)\n-options=(!zipman) # don't zip info pages or they won't work inside maxima\n- build-fasl.patch matrixexp.patch stack.patch)\n-sha256sums=('8f555aec33bc61b5a3ee0fe2e9d6c1179db67a2ff7e0eceb6bb614058eeb40cd'\n- '90ced3b33361fa24c2b417e0aeba8956892f0965b4a22d57d0c04115f2a3274b'\n- 'ef1bc6a15fc982ff8c6aa1800bbbd3284d9e060ca27abf9d8c1266632c0c2619'\n- '60ed7d96da06361a2f2f2e9df36aecae0384fe454bf3f963a2cab6033e1bd7a5')\n-\n-prepare() {\n- cd \\$_pkgname-\\$pkgver\n-\n-# build maxima ecl library\n- patch -p1 -i ../build-fasl.patch\n-# fix matrix exponentiation\n- patch -p1 -i ../matrixexp.patch\n-# fix segfaults in sagemath\n- patch -p1 -i ../stack.patch\n-}\n-\n-build() {\n- cd \\$_pkgname-\\$pkgver\n-\n- ./configure --prefix=/usr --mandir=/usr/share/man --infodir=/usr/share/info \\\n-\t--libexecdir=/usr/lib --enable-ecl --with-default-lisp=ecl\n- make\n-}\n-\n-package() {\n- cd \\$_pkgname-\\$pkgver\n- make DESTDIR=\"\\$pkgdir\" emacsdir=/usr/share/emacs/site-lisp/maxima install\n-\n- _ecldir=\"/usr/lib/ecl-\\$_eclver\"\n- mkdir -p \"\\$pkgdir/\\$_ecldir\"\n- install src/binary-ecl/maxima.fas \"\\$pkgdir/\\$_ecldir\"\n-}\n\nCopied: maxima-ecl/repos/community-x86_64/PKGBUILD (from rev 427941, maxima-ecl/trunk/PKGBUILD)\n===================================================================\n--- PKGBUILD\t (rev 0)\n+++ PKGBUILD\t2019-01-26 08:34:32 UTC (rev 427942)\n@@ -0,0 +1,54 @@\n+# Maintainer: Antonio Rojas <arojas at archlinux.org>\n+# Contributor: Ronald van Haren <ronald.archlinux.org>\n+# Contributor: Damir Perisa <damir at archlinux.org>\n+# Modified to compile against ecl by: maribu\n+\n+pkgname=maxima-ecl\n+_pkgname=maxima\n+pkgver=5.42.2\n+_eclver=16.1.3\n+pkgrel=1\n+pkgdesc=\"A sophisticated computer algebra system (compiled against ecl)\"\n+arch=(x86_64)\n+url=\"https://maxima.sourceforge.net\"\n+depends=(ecl=\\$_eclver texinfo shared-mime-info)\n+makedepends=(python emacs)\n+optdepends=('gnuplot: plotting capabilities' 'rlwrap: readline support via /usr/bin/rmaxima' 'tk: graphical xmaxima interface')\n+conflicts=(maxima)\n+provides=(maxima)\n+options=(!zipman) # don't zip info pages or they won't work inside maxima\n+ build-fasl.patch matrixexp.patch stack.patch)\n+sha256sums=('167e11d6513a65c829a35f24d4ba539bcd0a82fc3dc7a6721e4f9f118c67b64d'\n+ '90ced3b33361fa24c2b417e0aeba8956892f0965b4a22d57d0c04115f2a3274b'\n+ 'ef1bc6a15fc982ff8c6aa1800bbbd3284d9e060ca27abf9d8c1266632c0c2619'\n+ '60ed7d96da06361a2f2f2e9df36aecae0384fe454bf3f963a2cab6033e1bd7a5')\n+\n+prepare() {\n+ cd \\$_pkgname-\\$pkgver\n+\n+# build maxima ecl library\n+ patch -p1 -i ../build-fasl.patch\n+# fix matrix exponentiation\n+ patch -p1 -i ../matrixexp.patch\n+# fix segfaults in sagemath\n+ patch -p1 -i ../stack.patch\n+}\n+\n+build() {\n+ cd \\$_pkgname-\\$pkgver\n+\n+ ./configure --prefix=/usr --mandir=/usr/share/man --infodir=/usr/share/info \\\n+\t--libexecdir=/usr/lib --enable-ecl --with-default-lisp=ecl\n+ make\n+}\n+\n+package() {\n+ cd \\$_pkgname-\\$pkgver\n+ make DESTDIR=\"\\$pkgdir\" emacsdir=/usr/share/emacs/site-lisp/maxima install\n+\n+ _ecldir=\"/usr/lib/ecl-\\$_eclver\"\n+ mkdir -p \"\\$pkgdir/\\$_ecldir\"\n+ install src/binary-ecl/maxima.fas \"\\$pkgdir/\\$_ecldir\"\n+}\n\nDeleted: build-fasl.patch\n===================================================================\n--- build-fasl.patch\t2019-01-26 08:34:04 UTC (rev 427941)\n+++ build-fasl.patch\t2019-01-26 08:34:32 UTC (rev 427942)\n@@ -1,23 +0,0 @@\n-Build a fasl library for ecl in addition to an executable program.\n-\n-References:\n-* http://trac.sagemath.org/ticket/16178\n-* https://github.com/cschwan/sage-on-gentoo/issues/226\n-* https://bugs.gentoo.org/show_bug.cgi?id=499634\n-\n-Index: maxima-5.29.1/src/maxima.system\n-===================================================================\n---- maxima-5.29.1.orig/src/maxima.system\n-+++ maxima-5.29.1/src/maxima.system\n-@@ -75,6 +75,11 @@\n- \t\t\t ;; Convert dir/foo.fas to dir/foo.o\n- \t\t\t (make-pathname :type \"o\" :defaults p))\n- \t\t\t files)))\n-+\t(c::build-fasl \"binary-ecl/maxima\" :lisp-files obj\n-+\t\t\t :ld-flags\n-+\t\t\t (let ((x (symbol-value (find-symbol \"*AUTOCONF-LD-FLAGS*\"\n-+\t\t\t\t\t\t\t (find-package \"MAXIMA\")))))\n-+\t\t\t (if (and x (not (string= x \"\"))) (list x))))\n- \t(c::build-program \"binary-ecl/maxima\" :lisp-files obj\n- \t\t\t :ld-flags\n- \t\t\t (let ((x (symbol-value (find-symbol \"*AUTOCONF-LD-FLAGS*\"\n\nCopied: maxima-ecl/repos/community-x86_64/build-fasl.patch (from rev 427941, maxima-ecl/trunk/build-fasl.patch)\n===================================================================\n--- build-fasl.patch\t (rev 0)\n+++ build-fasl.patch\t2019-01-26 08:34:32 UTC (rev 427942)\n@@ -0,0 +1,23 @@\n+Build a fasl library for ecl in addition to an executable program.\n+\n+References:\n+* http://trac.sagemath.org/ticket/16178\n+* https://github.com/cschwan/sage-on-gentoo/issues/226\n+* https://bugs.gentoo.org/show_bug.cgi?id=499634\n+\n+Index: maxima-5.29.1/src/maxima.system\n+===================================================================\n+--- maxima-5.29.1.orig/src/maxima.system\n++++ maxima-5.29.1/src/maxima.system\n+@@ -75,6 +75,11 @@\n+ \t\t\t ;; Convert dir/foo.fas to dir/foo.o\n+ \t\t\t (make-pathname :type \"o\" :defaults p))\n+ \t\t\t files)))\n++\t(c::build-fasl \"binary-ecl/maxima\" :lisp-files obj\n++\t\t\t :ld-flags\n++\t\t\t (let ((x (symbol-value (find-symbol \"*AUTOCONF-LD-FLAGS*\"\n++\t\t\t\t\t\t\t (find-package \"MAXIMA\")))))\n++\t\t\t (if (and x (not (string= x \"\"))) (list x))))\n+ \t(c::build-program \"binary-ecl/maxima\" :lisp-files obj\n+ \t\t\t :ld-flags\n+ \t\t\t (let ((x (symbol-value (find-symbol \"*AUTOCONF-LD-FLAGS*\"\n\nDeleted: matrixexp.patch\n===================================================================\n--- matrixexp.patch\t2019-01-26 08:34:04 UTC (rev 427941)\n+++ matrixexp.patch\t2019-01-26 08:34:32 UTC (rev 427942)\n@@ -1,13 +0,0 @@\n---- a/share/linearalgebra/matrixexp.lisp\n-+++ b/share/linearalgebra/matrixexp.lisp\n-@@ -138,8 +138,8 @@\n- \t (print `(ratvars = ,\\$ratvars gcd = '\\$gcd algebraic = ,\\$algebraic))\n- \t (print `(ratfac = ,\\$ratfac))\n- \t (merror \"Unable to find the spectrum\")))\n--\n-- (setq res (\\$fullratsimp (ncpower (sub (mult z (\\$ident n)) mat) -1) z))\n-+\n-+ (setq res (\\$fullratsimp (\\$invert_by_lu (sub (mult z (\\$ident n)) mat) '\\$crering) z))\n- (setq m (length sp))\n- (dotimes (i m)\n- (setq zi (nth i sp))\n\nCopied: maxima-ecl/repos/community-x86_64/matrixexp.patch (from rev 427941, maxima-ecl/trunk/matrixexp.patch)\n===================================================================\n--- matrixexp.patch\t (rev 0)\n+++ matrixexp.patch\t2019-01-26 08:34:32 UTC (rev 427942)\n@@ -0,0 +1,13 @@\n+--- a/share/linearalgebra/matrixexp.lisp\n++++ b/share/linearalgebra/matrixexp.lisp\n+@@ -138,8 +138,8 @@\n+ \t (print `(ratvars = ,\\$ratvars gcd = '\\$gcd algebraic = ,\\$algebraic))\n+ \t (print `(ratfac = ,\\$ratfac))\n+ \t (merror \"Unable to find the spectrum\")))\n+-\n+- (setq res (\\$fullratsimp (ncpower (sub (mult z (\\$ident n)) mat) -1) z))\n++\n++ (setq res (\\$fullratsimp (\\$invert_by_lu (sub (mult z (\\$ident n)) mat) '\\$crering) z))\n+ (setq m (length sp))\n+ (dotimes (i m)\n+ (setq zi (nth i sp))\n\nDeleted: stack.patch\n===================================================================\n--- stack.patch\t2019-01-26 08:34:04 UTC (rev 427941)\n+++ stack.patch\t2019-01-26 08:34:32 UTC (rev 427942)\n@@ -1,85 +0,0 @@\n-diff --git a/src/hayat.lisp b/src/hayat.lisp\n-index 07699d6..ab8984d 100644\n---- a/src/hayat.lisp\n-+++ b/src/hayat.lisp\n-@@ -2189,6 +2189,23 @@\n- (or (alike1 (exp-pt (get-datum (datum-var (car l)))) (exp-pt (car l)))\n- \t (return () ))))\n-\n-+;; SUBTREE-SEARCH\n-+;;\n-+;; Search for subtrees, ST, of TREE that contain an element equal to BRANCH\n-+;; under TEST as an immediate child and return them as a list.\n-+;;\n-+;; Examples:\n-+;; (SUBTREE-SEARCH 2 '(1 2 3)) => '((1 2 3))\n-+;; (SUBTREE-SEARCH 2 '(1 2 2 3)) => '((1 2 2 3))\n-+;; (SUBTREE-SEARCH 2 '(1 (2) 3)) => '((2))\n-+;; (SUBTREE-SEARCH 4 '(1 (2) 3)) => NIL\n-+;; (SUBTREE-SEARCH 2 '(1 (2) 3 (2))) => '((2) (2))\n-+\n-+(defun subtree-search (branch tree &optional (test 'equalp))\n-+ (unless (atom tree)\n-+ (if (find branch tree :test test) (list tree)\n-+ (mapcan (lambda (child) (subtree-search branch child test)) tree))))\n-+\n- (defun taylor2 (e)\n- (let ((last-exp e))\t ;; lexp-non0 should be bound here when needed\n- (cond ((assolike e tlist) (var-expand e 1 () ))\n-@@ -2232,8 +2249,31 @@\n- \t\t ((null l) t)\n- \t\t (or (free e (car l)) (return ()))))\n- \t (newsym e))\n--\t(t (let ((exact-poly () ))\t; Taylor series aren't exact\n--\t (taylor2 (diff-expand e tlist)))))))\n-+\t(t\n-+ ;; When all else fails, call diff-expand to try to expand e around the\n-+ ;; point as a Taylor series by taking repeated derivatives. This might\n-+ ;; fail, unfortunately: If a required derivative doesn't exist, then\n-+ ;; DIFF-EXPAND will return a form of the form \"f'(x)\" with the\n-+ ;; variable, rather than the expansion point in it.\n-+ ;;\n-+ ;; Sometimes this works - in particular, if there is a genuine pole at\n-+ ;; the point, we end up passing a sum of terms like x^(-k) to a\n-+ ;; recursive invocation and all is good. Unfortunately, it can also\n-+ ;; fail. For example, if e is abs(sin(x)) and we try to expand to first\n-+ ;; order, the expression \"1/1*(cos(x)*sin(x)/abs(sin(x)))*x^1+0\" is\n-+ ;; returned. If we call taylor2 on that, we will end up recursing and\n-+ ;; blowing the stack. To avoid doing so, error out if EXPANSION\n-+ ;; contains E as a subtree. However, don't error if it occurs as an\n-+ ;; argument to %DERIVATIVE (in which case, we might well be fine). This\n-+ ;; happens from things like taylor(log(f(x)), x, x0, 1).\n-+\n-+ (let* ((exact-poly nil) ; (Taylor series aren't exact)\n-+ (expansion (diff-expand e tlist)))\n-+ (when (find-if (lambda (subtree)\n-+ (not (eq (\\$op subtree) '%derivative)))\n-+ (subtree-search e expansion))\n-+ (exp-pt-err))\n-+ (taylor2 expansion))))))\n-\n- (defun compatvarlist (a b c d)\n- (cond ((null a) t)\n-@@ -2968,7 +3008,21 @@\n- (and (or (member '\\$inf pt-list :test #'eq) (member '\\$minf pt-list :test #'eq))\n- \t (unfam-sing-err)))\n-\n--(defun diff-expand (exp l)\t\t;l is tlist\n-+;; DIFF-EXPAND\n-+;;\n-+;; Expand EXP in the variables as specified in L, which is a list of tlists. If\n-+;; L is a singleton, this just works by the classic Taylor expansion:\n-+;;\n-+;; f(x) = f(c) + f'(c) + f''(c)/2 + ... + f^(k)(c)/k!\n-+;;\n-+;; If L specifies multiple expansions, DIFF-EXPAND works recursively by\n-+;; expanding one variable at a time. The derivatives are computed using SDIFF.\n-+;;\n-+;; In some cases, f'(c) or some higher derivative might be an expression of the\n-+;; form 1/0. Instead of returning an error, DIFF-EXPAND uses f'(x)\n-+;; instead. (Note: This seems bogus to me (RJS), but I'm just describing how\n-+;; EVAL-DERIV works)\n-+(defun diff-expand (exp l)\n- (check-inf-sing (mapcar (function caddr) l))\n- (cond ((not l) exp)\n- \t(t\n\nCopied: maxima-ecl/repos/community-x86_64/stack.patch (from rev 427941, maxima-ecl/trunk/stack.patch)\n===================================================================\n--- stack.patch\t (rev 0)\n+++ stack.patch\t2019-01-26 08:34:32 UTC (rev 427942)\n@@ -0,0 +1,85 @@\n+diff --git a/src/hayat.lisp b/src/hayat.lisp\n+index 07699d6..ab8984d 100644\n+--- a/src/hayat.lisp\n++++ b/src/hayat.lisp\n+@@ -2189,6 +2189,23 @@\n+ (or (alike1 (exp-pt (get-datum (datum-var (car l)))) (exp-pt (car l)))\n+ \t (return () ))))\n+\n++;; SUBTREE-SEARCH\n++;;\n++;; Search for subtrees, ST, of TREE that contain an element equal to BRANCH\n++;; under TEST as an immediate child and return them as a list.\n++;;\n++;; Examples:\n++;; (SUBTREE-SEARCH 2 '(1 2 3)) => '((1 2 3))\n++;; (SUBTREE-SEARCH 2 '(1 2 2 3)) => '((1 2 2 3))\n++;; (SUBTREE-SEARCH 2 '(1 (2) 3)) => '((2))\n++;; (SUBTREE-SEARCH 4 '(1 (2) 3)) => NIL\n++;; (SUBTREE-SEARCH 2 '(1 (2) 3 (2))) => '((2) (2))\n++\n++(defun subtree-search (branch tree &optional (test 'equalp))\n++ (unless (atom tree)\n++ (if (find branch tree :test test) (list tree)\n++ (mapcan (lambda (child) (subtree-search branch child test)) tree))))\n++\n+ (defun taylor2 (e)\n+ (let ((last-exp e))\t ;; lexp-non0 should be bound here when needed\n+ (cond ((assolike e tlist) (var-expand e 1 () ))\n+@@ -2232,8 +2249,31 @@\n+ \t\t ((null l) t)\n+ \t\t (or (free e (car l)) (return ()))))\n+ \t (newsym e))\n+-\t(t (let ((exact-poly () ))\t; Taylor series aren't exact\n+-\t (taylor2 (diff-expand e tlist)))))))\n++\t(t\n++ ;; When all else fails, call diff-expand to try to expand e around the\n++ ;; point as a Taylor series by taking repeated derivatives. This might\n++ ;; fail, unfortunately: If a required derivative doesn't exist, then\n++ ;; DIFF-EXPAND will return a form of the form \"f'(x)\" with the\n++ ;; variable, rather than the expansion point in it.\n++ ;;\n++ ;; Sometimes this works - in particular, if there is a genuine pole at\n++ ;; the point, we end up passing a sum of terms like x^(-k) to a\n++ ;; recursive invocation and all is good. Unfortunately, it can also\n++ ;; fail. For example, if e is abs(sin(x)) and we try to expand to first\n++ ;; order, the expression \"1/1*(cos(x)*sin(x)/abs(sin(x)))*x^1+0\" is\n++ ;; returned. If we call taylor2 on that, we will end up recursing and\n++ ;; blowing the stack. To avoid doing so, error out if EXPANSION\n++ ;; contains E as a subtree. However, don't error if it occurs as an\n++ ;; argument to %DERIVATIVE (in which case, we might well be fine). This\n++ ;; happens from things like taylor(log(f(x)), x, x0, 1).\n++\n++ (let* ((exact-poly nil) ; (Taylor series aren't exact)\n++ (expansion (diff-expand e tlist)))\n++ (when (find-if (lambda (subtree)\n++ (not (eq (\\$op subtree) '%derivative)))\n++ (subtree-search e expansion))\n++ (exp-pt-err))\n++ (taylor2 expansion))))))\n+\n+ (defun compatvarlist (a b c d)\n+ (cond ((null a) t)\n+@@ -2968,7 +3008,21 @@\n+ (and (or (member '\\$inf pt-list :test #'eq) (member '\\$minf pt-list :test #'eq))\n+ \t (unfam-sing-err)))\n+\n+-(defun diff-expand (exp l)\t\t;l is tlist\n++;; DIFF-EXPAND\n++;;\n++;; Expand EXP in the variables as specified in L, which is a list of tlists. If\n++;; L is a singleton, this just works by the classic Taylor expansion:\n++;;\n++;; f(x) = f(c) + f'(c) + f''(c)/2 + ... + f^(k)(c)/k!\n++;;\n++;; If L specifies multiple expansions, DIFF-EXPAND works recursively by\n++;; expanding one variable at a time. The derivatives are computed using SDIFF.\n++;;\n++;; In some cases, f'(c) or some higher derivative might be an expression of the\n++;; form 1/0. Instead of returning an error, DIFF-EXPAND uses f'(x)\n++;; instead. (Note: This seems bogus to me (RJS), but I'm just describing how\n++;; EVAL-DERIV works)\n++(defun diff-expand (exp l)\n+ (check-inf-sing (mapcar (function caddr) l))\n+ (cond ((not l) exp)\n+ \t(t\n```"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5296693,"math_prob":0.7903099,"size":16475,"snap":"2019-26-2019-30","text_gpt3_token_len":5490,"char_repetition_ratio":0.13241455,"word_repetition_ratio":0.36056337,"special_character_ratio":0.42312595,"punctuation_ratio":0.19153225,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9539968,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-25T06:40:45Z\",\"WARC-Record-ID\":\"<urn:uuid:d132fd5d-87b0-49ad-819b-918925c6fda0>\",\"Content-Length\":\"22302\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:98e2acea-5341-4411-809f-f46fe98d217f>\",\"WARC-Concurrent-To\":\"<urn:uuid:4d708bf7-5aac-4ddb-9c01-529466906ea0>\",\"WARC-IP-Address\":\"5.9.250.164\",\"WARC-Target-URI\":\"https://lists.archlinux.org/pipermail/arch-commits/2019-January/574509.html\",\"WARC-Payload-Digest\":\"sha1:N6CCJIQERR3PNLKSQH6HZBP5CB7LLJML\",\"WARC-Block-Digest\":\"sha1:VDHGIPE2IGTU7HRVWWFQAYXCGURCGEHI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999800.5_warc_CC-MAIN-20190625051950-20190625073950-00387.warc.gz\"}"} |
https://issuu.com/ijtsrd.com/docs/331_rp-75_formulation_of_two_specia/2 | [
"",
null,
"International Journal of Trend in Scientific Research and Development (IJTSRD) @ www.ijtsrd.com eISSN: 2456-6470 Thus, Here\n\nis a solution of the quadratic congruence: is the modular inverse of\n\nSuch congruence has exactly two solutions, if it is solvable. Solvability test is found in literature in terms of Legendre’s symbol . If\n\nis a quadratic residue of p, then the congruence is solvable. Thus the solvability condition is:\n\nis a quadratic residue of p.\n\nThus, the congruence under consideration is reduced to a standard quadratic congruence of prime modulus. It can be solved easily by author’s previous formulation of standard quadratic congruence of prime modulus. Thus, the congruence\n\nis solvable if\n\nis a quadratic residue of p.\n\nThe equivalent quadratic congruence is Case-II: Consider the congruence be a solution of the said congruence. Then,\n\nLet Then, Multiplying (B) by\n\nThus, This type:\n\nis a solution of the cubic congruence: shows\n\nthat\n\nthe\n\nsaid\n\ncongruence\n\ncan\n\nbe\n\nreduced\n\nto\n\na\n\ncubic\n\ncongruence\n\nSuch cubic congruence of prime modulus have two types of solutions. Some has unique Solution if have exactly three solutions, if\n\nof\n\nthe\n\nand others\n\n.\n\nLet us consider that Then, Let Then, from (C): It is true by Fermat’s theorem. Hence,\n\nis the unique solution of\n\nThus, the congruence is solvable for all values of a. solutions . These three solutions are the members of the residues .\n\n, if Then the cubic congruence has exactly three\n\nof p such that\n\nis modular inverse of a\n\nILLUSTRATIONS Consider Here, Then the said congruence can be written as It is of the type\n\nhence, it can be reduced to:\n\n.\n\ngiving solutions: Therefore, solutions of the given congruence are Consider\n\n@ IJTSRD | Unique Paper ID - IJTSRD23417 | Volume – 3 | Issue – 3 | Mar-Apr 2019\n\nPage: 1532\n\n# RP 75 Formulation of two Special Classes of Standard Congruence of Prime Modulus of Higher Degree\n\nby Prof. B M Roy \"RP-75: Formulation of two Special Classes of Standard Congruence of Prime Modulus of Higher Degree\" Published in Interna...\n\n# RP 75 Formulation of two Special Classes of Standard Congruence of Prime Modulus of Higher Degree\n\nby Prof. B M Roy \"RP-75: Formulation of two Special Classes of Standard Congruence of Prime Modulus of Higher Degree\" Published in Interna..."
] | [
null,
"https://image.isu.pub/190613065959-17df329ce284c30b9b2b181d4f307d4d/jpg/page_2_thumb_large.jpg",
null
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https://www.jstage.jst.go.jp/article/trbane/2017/62/2017_37/_article/-char/ja | [
"ばね論文集\nOnline ISSN : 1348-1479\nPrint ISSN : 0385-6917\nISSN-L : 0385-6917\n\nジャーナル フリー\n\n2017 年 2017 巻 62 号 p. 37-47",
null,
"J-STAGE : trbane 2018 63 pp.103-104\n\nIn this study, 3D finite element analyses have been made on the stress intensity factors for semi-elliptical surface cracks on the wire surface of a tension coil spring in order to investigate the effects of spring parameters on the correction factors, Fi (i=I, II and III) for the three modes of the stress intensity factors. The parameters selected for the present analyses are crack inclination angle θ, crack position on wire surface, normalized half crack surface length c/r, crack aspect ratio a/c and spring index s=R/r where c is the half crack surface length, r the wire radius, a the crack depth and R the coil radius of the spring. The mode I correction factor, or the normalized mode I stress intensity factor FI was found dominant over the correction factors for the other two modes in all the cases investigated. The most influential parameter for FI was the crack aspect ratio a/c, and the largest FI value was calculated as approximately 2.75 for a/c=4 near wire surface, i.e., φ=0°. The most influential parameter for the other correction factors FII and FIII was the crack inclination angle θ. The correction factors FII and FIII were revealed to take on smaller maximum values than FI: i.e., the maximum absolute value of FII was found approximately 1.3 for θ=0° and 90°near wire surface (φ=0° and 180°), and that of FIII approximately 0.59 for θ=0° and 90°at the deepest point of the crack (φ=90°).",
null,
"著者関連情報"
] | [
null,
"https://www.jstage.jst.go.jp/web/images/articlelink-correction-icn.png",
null,
"https://www.jstage.jst.go.jp/web/images/data-services/J-GLOBAL.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8957287,"math_prob":0.85592526,"size":1446,"snap":"2021-31-2021-39","text_gpt3_token_len":346,"char_repetition_ratio":0.14771152,"word_repetition_ratio":0.01632653,"special_character_ratio":0.23720609,"punctuation_ratio":0.096666664,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9750326,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,8,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-18T22:52:39Z\",\"WARC-Record-ID\":\"<urn:uuid:0c0d3aef-02c8-4e86-a047-dc2d6e299ec4>\",\"Content-Length\":\"70321\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6cfdbcd5-c8cb-4d32-a6b9-24ff48fc0dc5>\",\"WARC-Concurrent-To\":\"<urn:uuid:4624dd49-9b33-4c8b-86a0-fd0461dbdf90>\",\"WARC-IP-Address\":\"160.74.72.32\",\"WARC-Target-URI\":\"https://www.jstage.jst.go.jp/article/trbane/2017/62/2017_37/_article/-char/ja\",\"WARC-Payload-Digest\":\"sha1:MIH4GCOZXIXR3EHSFKIULUOEANT4P55Z\",\"WARC-Block-Digest\":\"sha1:H2GT5PGPILVX2HSL5ZLR7WL6VBVMYRRV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056578.5_warc_CC-MAIN-20210918214805-20210919004805-00506.warc.gz\"}"} |
https://ask.sagemath.org/questions/9273/revisions/ | [
"# Pneumatic Air Gun Calculations\n\nimport matplotlib.pyplot as plt\n\nVm=var('Vm')\n\nPatm=var('Patm')\n\nVc=var('Vc')\n\nAb=var('Ab')\n\nLb=var('Lb')\n\nk=var('k')\n\ncb=var('cb')\n\neng=var('eng')\n\neng2=var('eng2')\n\nmd=var('md')\n\nPc=var('Pc')\n\n# Definitions\n\nk = 1.4 Patm = 101235 cb = 0.261208\n\n# m/s^2\n\nVm= 91.44 md=2 Pc=10 eng=0.618081\n\n# Main Formula\n\nsolve(eng=(k-1)mdVm^2/(2PatmVc*(Pc-Pc^(1/k)),Vc)\n\nCan anyone tell me why this is giving a syntax error and pointing to my variable definition Vm? It's really bugging me!",
null,
"2 No.2 Revision",
null,
"DSM",
null,
"5242 ●21 ●68 ●112\n\n# Pneumatic #Pneumatic Air Gun Calculations\n\n Calculations import matplotlib.pyplot as plt\nplt #Variables Outlined Variables Outlined Muzzle Velocity Vm=var('Vm') Atmospheric Pressure Patm=var('Patm') Gas #Muzzle Velocity Vm=var('Vm') #Atmospheric Pressure Patm=var('Patm') #Gas Chamber Volume Vc=var('Vc') Barrel Area[Internal] Ab=var('Ab') Barrel Length Lb=var('Lb') Specific Volume Vc=var('Vc') #Barrel Area[Internal] Ab=var('Ab') #Barrel Length Lb=var('Lb') #Specific Heat of Air k=var('k') CB Air k=var('k') #CB Ratio[see table] cb=var('cb') formula variable eng=var('eng') 2nd table] cb=var('cb') #formula variable eng=var('eng') #2nd Formula Variable eng2=var('eng2') Mass Variable eng2=var('eng2') #Mass of the Projectile md=var('md') Operating Projectile md=var('md') #Operating Pressure in atmatm Pc=var('Pc') Pc=var('Pc') Definitions #Definitions k = 1.4 Patm = 101235 cb = 0.261208 m/s^2 0.261208 #m/s^2 Vm= 91.44 md=2 Pc=10 eng=0.618081eng=0.618081 Main Formula solve(eng=(k-1)mdVm^2/(2PatmVc*(Pc-Pc^(1/k)),Vc)#Main Formula solve(eng=(k-1)*md*Vm^2/(2*Patm*Vc*(Pc-Pc^(1/k)),Vc) \n\nCan anyone tell me why this is giving a syntax error and pointing to my variable definition Vm? It's really bugging me!",
null,
"3 retagged",
null,
"FrédéricC",
null,
"4385 ●3 ●37 ●93\n\n### What's Wrong With This?\n\n#Pneumatic Air Gun Calculations\nimport matplotlib.pyplot as plt\n\n#Variables Outlined\n\n#Muzzle Velocity\nVm=var('Vm')\n#Atmospheric Pressure\nPatm=var('Patm')\n#Gas Chamber Volume\nVc=var('Vc')\n#Barrel Area[Internal]\nAb=var('Ab')\n#Barrel Length\nLb=var('Lb')\n#Specific Heat of Air\nk=var('k')\n#CB Ratio[see table]\ncb=var('cb')\n#formula variable\neng=var('eng')\n#2nd Formula Variable\neng2=var('eng2')\n#Mass of the Projectile\nmd=var('md')\n#Operating Pressure in atm\nPc=var('Pc')\n\n#Definitions\nk = 1.4\nPatm = 101235\ncb = 0.261208\n#m/s^2\nVm= 91.44\nmd=2\nPc=10\neng=0.618081\n\n#Main Formula\nsolve(eng=(k-1)*md*Vm^2/(2*Patm*Vc*(Pc-Pc^(1/k)),Vc)\n\n\nCan anyone tell me why this is giving a syntax error and pointing to my variable definition Vm? It's really bugging me!"
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https://homework.cpm.org/category/CCI_CT/textbook/int1/chapter/8/lesson/8.1.4/problem/8-65 | [
"",
null,
"",
null,
"### Home > INT1 > Chapter 8 > Lesson 8.1.4 > Problem8-65\n\n8-65.\n\nTickets for a concert have been in incredibly high demand, and as the date for the concert draws closer, the price of tickets increases exponentially. The cost of a pair of concert tickets was $$150$ yesterday, and today it is$$162$. As you complete the parts below, assume that each day’s percent increase from the day before is the same.\n\n1. What is the daily percent rate of increase? What is the multiplier?\n\nWhat percent of $$150$ is$$162$?\n\n2. What will be the cost of a pair of concert tickets one week from now?\n\nConstruct a table of values or an equation describing the relation\nbetween the number of days since yesterday and the price of a ticket.\n\n3. What was the cost of a pair of tickets two weeks ago?\n\nIn order to go backwards, make the exponent negative.\nRemember exponential equations look like this, y = a bx where a is initial value, b multiplier and x is time.\n\n4. What will be the cost of a pair of tickets two weeks from now? Does your answer make sense? Why or why not?\n\nTime in the equation is in days, the problem gives you weeks. What do you do?"
] | [
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null,
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",
null
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https://www.clutchprep.com/chemistry/practice-problems/85161/a-binary-compound-between-an-unknown-element-e-and-hydrogen-contains-91-27-e-and | [
"# Problem: A binary compound between an unknown element E and hydrogen contains 91.27% E and 8.73% H by mass. If the formula of the compound is E3H8. Calculate the atomic mass of E.\n\n🤓 Based on our data, we think this question is relevant for Professor Rubin's class at UCSC.\n\n###### FREE Expert Solution\n\nSince the given percentages are % by mass, let’s assume a 100 g of sample of E3H8.",
null,
"Assume Mass of each component (% to g):",
null,
"###### Problem Details\n\nA binary compound between an unknown element E and hydrogen contains 91.27% E and 8.73% H by mass. If the formula of the compound is E3H8. Calculate the atomic mass of E."
] | [
null,
"https://lightcat-files.s3.amazonaws.com/problem_images/a75fbedad868a1ce-1558567684520.jpg",
null,
"https://cdn.clutchprep.com/assets/button-view-text-solution.png",
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https://quantumcomputing.stackexchange.com/questions/12387/how-to-decompose-unitary-quantum-gate-in-current-simulator-or-emulator/12388 | [
"# How to decompose unitary quantum gate in current simulator or emulator?\n\nI have a question about how to decompose a unitary quantum gate in a currently existing simulator or emulator. I have read some papers about SK algorithm and other algorithms which aim to decompose unitary quantum gates. Is there any specific method to decompose a quantum gate in a currently existing simulator? Some papers about decomposition methods mention Trotter-Suzuki decomposition but I don't exactly know if this is true? To be more specific, what is the decomposition algorithm in Qiskit or Project Q?\n\nIn Qiskit, the iso() function allows you to add a gate, defined by means of a unitary, to your quantum circuit:\n\nhttps://qiskit.org/documentation/stubs/qiskit.circuit.QuantumCircuit.iso.html\n\nThe decomposition used in the iso() function was introduced by Iten et al. in https://arxiv.org/abs/1501.06911.\n\nOf course this is related to the circuit, not to the backend (which can be real or simulated).\n\n• So all the simulator or emulator decompose a unitary gate in that way? I mean, the decomposion strategy is same in existing simulators or emulators or it is depend on specific simulators or emulators? Thx – Henry_Fordham Jun 9 at 12:14\n• In Qiskit, you may either build the circuit from the scratch by adding gates, or by providing the unitary matrix of the circuit and decompose it using iso(). As I wrote in my answer, decomposition is independent on whether you want to run the circuit on a real backend or in a simulated one. Once you have the decomposition, then you choose the backend. – Michele Amoretti Jun 9 at 12:18\n• So after decompose it into CNOT and single qubit gate as the paper mentioned. How to decompose the single qubit gate into universal quantum gates set? like Clifford +T, Thx – Henry_Fordham Jun 9 at 14:01\n• Usually, you choose the universal quantum gate set depending on the backend you are using. For IBM Q backends, the most general single qubit gate is $U_3$ (qiskit.org/textbook/ch-states/…). In Qiskit, to compile your circuit using specific gates (e.g., CNOT and $U_3$), you need to use the transpile() function: qiskit.org/documentation/stubs/qiskit.compiler.transpile.html – Michele Amoretti Jun 9 at 14:16\n• So, the universal quantum gates {Clifford +T} is not commonly used in any backend or simulated one anymore, right? I read the reference you sent, one U3 gate seems can represent any single qubit gate, However, in SK algorithm, by running the C++ code, the H and T gate grow exponentially. so if the basic gate set canbe {U3 and CNOT}, why we need {Clifford+T},like in SK-algorithm or some of its optimization. – Henry_Fordham Jun 9 at 15:07"
] | [
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https://holmwildboken.web.app/1099.html | [
"# Nnnelectric field physics pdf books\n\nFind the speed of the proton after it has moved a distance of 2. Eleven science questions for the new century, a 2002 report of the national research council. It builds on the foundations laid in books 14, which covered many parts of mathematics and physics, and on some parts of book 5, which refers to the structure and. Field computation for accelerator magnets analytical and numerical methodsfor electromagnetic design and optimization 2010 isbn 9783527407699 stock, r.\n\nFree college physics textbook available for download openstax. The expression for the magnitude of the electric field between two uniform metal plates is. The physics of the universe is a free pdf ebook from nasa. So thats this charge here, and lets say its charge is q. This field guide serves as a reference guide to the primary results, explanations. Cut out a piece of length about 20 cm with the ball at one end and flatten the cut end. Free electricity magnetism books download ebooks online. Encyclopedia of applied high energy and particle physics 2009. What is the electric field at the midpoint between the plates. Basic books in science learning development institute. These lecture notes are not meant to replace the course textbook.\n\nCoulombs law, electric field, gauss law, electrical potential michel van biezen calculate the magnitude of the electric field at the center of a square. Ncert physics books free pdf download for class 11 and class 12. Unlike gravitational fields, which can only exert attractive forces, electric fields can attract or. If you read this book youll actually learn quantum field theory, which is about. Calculate the electric field produced at the point a in terms of k, q and d.\n\nPhys102 general physics ii electric fields electric fields topics electric field force per unit charge electric field lines and electric flux electric field from more than 1 charge electric dipoles motion of point charges in an electric field examples of finding electric fields from continuous charges the electric field. A positive test charge, q, at a certain point in an electric field is acted on by a force, f, due to the electric field. Free physics books download ebooks online textbooks. Once the electric field strength is known, the force on a charge is found using f qe.\n\nSince the charges of the proton and electron are exactly equal in magnitude and opposite in sign, atoms are electrically neutral. The equation further allows us to derive the electric field equation in case of a unit charge. Please provide steps a uniform electric field has a magnitude of 2. Electric charge and electric field christopher chui 4 coulombs law coulomb 17361806 investigated electric forces using a torsion balance coulombs law. In physics, a field is a dynamical quantity which takes a value at every point in.\n\nElectric field, an electric property associated with each point in space when charge is present in any form. Lets imagine that instead of having two charges, we just have one charge by itself, sitting in a vacuum, sitting in space. In physics, a field is a physical quantity, represented by a number or tensor, that has a value for each point in spacetime. Find the magnitude and direction of the electric field at the five points indicated with open circles. This is strongly reminiscent of introductory classical mechanics courses that. Quantum field theory is the basic mathematical language that is used to describe and analyze the physics of elementary particles. Calculate the electric field produced at the point a if q 1010 and d 3 cm. This book covers the electricity and magnetism topics from physics with calculus at the university level. Unlike gravitational fields, which can only exert attractive forces, electric fields can attract or repel objects that are charged. Electric field physics problems point charges, tension.\n\nThe electric field is a vector field around a charged particle. Basic books in science about this book this book, like the others in the series1, is written in simple english the language most widely used in science and technology. The field strength is highest at the centre and decreases as the distance from the centre increases. Take a large bottle that can hold this rod and a cork which will fit in the opening of the bottle. Start studying physics study guide electrostatics and electric field. When solving electric field problems, you need to find the magnitude and the direction of the electric field. Ncert physics books for class 11 and class 12 are published by the officials of ncert national council of educational research and training, new delhi. Sep 20, 2016 this physics video tutorial explains the concept behind electric field, electric force, charge, and distance. The magnitude of the field depends on the distance \\r \\ to the center of the metal ball and the field is directed away from the center of the ball. The units of the magnetic field are nn n t tesla mcam cm ss. Quantum field theory ucsb physics uc santa barbara. Kinematics and dynamics, units and vectors, motion in one dimension, motion in two and three dimensions, forces, work and energy, linear momentum and collisions rotations, vibrations and waves, rolling motion, oscillatory motion, electric fields, electric charge and coulombs law, gauss law. All quantities of charge must be a whole number multiple of e 1. The electric field around a point charge is called a radial field.\n\nElectric charge and electric field physics libretexts. The whole point of a uniform electric field is that the field does not change anywhere between the plates at the midpoint, as anywhere, it is 90 nc. Of the physics teachers responding, only 3% and 5%, respectively, reported that they worked not very well or not well at all, and 64% of physics teachers reported that the books worked quite well for their classes. Given a point charge, or a particle of infinitesimal size, that contains a certain charge, electric field lines emanate from equally in.\n\nLearn vocabulary, terms, and more with flashcards, games, and other study tools. Energy physics with an important minority of theorists. Electric field analysis is both a studentfriendly textbook and a valuable tool for engineers and physicists engaged in the design work of highvoltage insulation systems. May 17, 2014 electricity and magnetism benjamin crowell book 4 in the light. Electric potential in a uniform electric field physics. Alevel physics advancing physicselectric fieldworked. Download course materials electricity and magnetism. The openstax name, openstax logo, openstax book covers, openstax cnx name, openstax cnx. The interagency working group on the physics of the universe iwg presents its conclusions on the actions necessary to implement the recommendations of connecting quarks with the cosmos. This website is created solely for jee aspirants to download pdf, ebooks, study materials for free. Equivalent to advanced placement if this is too advanced for you, you can try the more basic version here. Electric fields and charge 3 electron cloud is equal to the number of protons in the nucleus. And i want to know, if i were to place another charge close to this q, within its sphere.\n\nThis is reflected in the above formula, which shows that e electric is proportional to 1 r 2 \\displaystyle \\frac 1r2. Coulombs law, electric field, gauss law, electrical potential michel van biezen calculate the magnitude of the electric field at. Thanks for contributing an answer to physics stack exchange. Another unit based on the cgs metric system is the gauss, where 1 g 10 t.\n\nThe magnitude and direction of the electric field are expressed by the value of e, called electric field strength or electric field intensity or simply the electric field. Electric fields are defined using positive test charges i. Shown are snapshots during the dft iterations leading to the evaporation of mg and melting of the top row. Electric charge,electrical conductor insulator,coulombs law,electrostatic induction,electroscope,electric fields i,electric fields ii,faraday cage,cathode ray. In an inhomogeneous electric field the magnitude and direction at each location varies. Phy2061 enriched physics 2 lecture notes magnetic fields d. When the potentials of the nodes are obtained, a numerical derivative evaluation technique is used to calculate the electric field intensity. The quantum, or smallest unit, of charge is that carried by an electron. Coulombs law, superposition, energy of a system of charges, basic field concept, flux, gausss law, fields and potentials around conductors, the electrostatic uniqueness theorem,rc circuits, thevenin equivalence, forces and fields in special relativity. You can print it, but the pdf version has many advantages, like.\n\nIt builds on the foundations laid in books 14, which covered many parts of mathematics and physics, and on some parts of book 5. Griffiths, but the beginning of the book explains both electromagnetism and. Take a thin aluminium curtain rod with ball ends fitted for hanging the curtain. January 19, 2020 november 14, 2010 by mini physics this topic covers electric fields of a level physics. Since the electric field is in only one direction, we can write this equation in terms of the magnitudes, f qe.\n\nThe text begins by introducing the physical and mathematical fundamentals of electric fields, presenting problems from power and. These notes are only meant to be a study aid and a supplement to your own notes. Principles with applications giancoli and college physics serway et al. E that is, when viewed far away, the field is just that due to a point charge.\n\nReview problems for introductory physics 2 february 6, 2014 robert g. The numerical estimate of the electric field is calculated from the potential using the function gradient e gradientv,hx. For electricity, this becomes there is no special name for its unit, nor does it reduce to anything simpler. The pdf version can be more convenient than paper books, with web links and a. This chapter begins the study of electromagnetic phenomena at a fundamental level. This book extensively discusses the basic electromagnetic principles and laws involved in electrostatics, steady magnetic fields, timevarying magnetic fields, and uniform plane waves. The electric field is introduced as the mediator of electrostatic interactions. In classical field theory, the strength of the field at a point is the normalized value of the field. You cannot just look for one and forget about the other.\n\nPhysics students can make a simple electroscope as follows fig. Find materials for this course in the pages linked along the left. It represents the force that other charged particles would feel if placed near the particle creating the electric field. This physics video tutorial explains the concept behind electric field, electric force, charge, and distance. Figure 2 shows the potential and electric field calculated by the relaxation. Phys 201 lecture 2 electric fields open yale courses. The next several chapters will cover static electricity, moving electricity, and magnetismcollectively known as electromagnetism. Given a point charge, or a particle of infinitesimal size, that contains a certain charge, electric field lines emanate from equally in all radial directions. Use these results and symmetry to find the electric field at as many.\n\nBrown physics textbooks introductory physics i and ii a lecture note style textbook series intended to support the teaching of introductory physics, with calculus, at a level suitable for duke undergraduates. But avoid asking for help, clarification, or responding to other answers. In physics, a field is a physical quantity, represented by a number or tensor, that has a value for. Use these results and symmetry to find the electric field at as many points as possible without additional calculation.\n\n1042 772 86 1453 1598 1116 1277 1178 233 396 633 1507 72 251 1112 1040 1558 1136 801 594 19 884 1327 450 1352 656 146 674 328 955 1288 1029 747 896 58 889 157 20 363 206 642 1200 475 1277 586"
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https://stats.stackexchange.com/questions/473337/why-are-the-ends-of-the-prediction-interval-wider-in-the-regression | [
"# Why are the ends of the prediction interval wider in the regression? [duplicate]\n\nUsually, the prediction interval has this shape in the image.",
null,
"I don't know why the end of the interval is wider than the center.\n\n• It's because you have no information about the variance of values greater than $20$ and smaller than $0$. – Joshua Jun 21 '20 at 23:02\n• I posted an image as an example. But in the other occasion, even when data are plenty around (for example) -40 and 60, the ends of the shape are still wider. Can you tell me more specific, please? – Hello_World Jun 21 '20 at 23:06\n• Maybe there are less observations or more scattered observations in areas where the interval becomes wider? – Joshua Jun 21 '20 at 23:12\n• There are math answers. But the intuitive answer is this. Take that graph and, understanding that you may be erroneous in estimating both the slope and the average value at the center, draw a bunch of lines that fall within the realm of those possibilities. What does that bundle of lines look like? – Russ Lenth Jun 22 '20 at 2:02\n\nWhen performing a linear regression, there are 2 types of uncertainty in the prediction.\n\nFirst is the prediction of the overall mean of the estimate (ie the center of the fit). The second is the uncertainly in the estimate calculating the slope.\nThus when you combine both uncertainties of the prediction there is a spread between the high and low estimates. Then as further away from the center, uncertainty of the slope becomes a large and more noticeable factor, thus the limits widen.\n\n• +1. In addition, there is (an estimate of) residual variance. – Stephan Kolassa Jun 22 '20 at 10:34\n\nIts very easy to determine the prediction interval for the data.\n\n$$\\operatorname{Var}(y) = \\operatorname{Var}(\\beta_0 + \\beta_1 x) + \\operatorname{Var}(\\varepsilon) = \\sigma^2_{\\beta_0} + \\sigma^2_{\\beta_1}x^2 + 2x \\operatorname{Cov}(\\beta_0, \\beta_1)+ \\sigma^2_{\\epsilon}$$\n\nAs you can see, this is a quadratic function in x, which means for larger values of $$x$$ (well...larger as compared to the sample mean of $$x$$), the variance of the prediction will be larger.\n\nThis reference, for example, clearly gives the formula of a prediction interval for a simple linear regression model, which contains the expression:\n\n$$\\sqrt{({1/n + (x_p - x_m)^2}/{(n-1){s_x}^2}}$$\n\nSo, as the prediction for the explanatory variable $$x_p$$ becomes more removed from its mean $$x_m$$, the interval widens.\n\nIn practice, I would be cautious about applying the prediction interval far from the mean, particularly a point estimate not included in the observed range of the data, as the structural equation itself may no longer be accurate/valid."
] | [
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"https://i.stack.imgur.com/segMg.png",
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https://openwetware.org/wiki/Beauchamp:NED | [
"# Beauchamp:NED\n\n## Estimating amounts of McGurk Fusion across Stimuli\n\n### System setup\n\n1. Install GNU R\n3. Extract the zip file to your Desktop/ or other preferred location\n\n### Data setup\n\nThe model code assumes the data are stored in a matrix format with rows as subjects and each column a separate stimulus. The first row is used as labels for the columns. Each cell stores the proportion of times the subject reported a fused perception for the given stimulus. If there are 20 subjects and 14 stimuli, the file will have 21 rows (1st row is header row) and 14 columns. See data.csv in the code pack for a sample of the data (full data available on request) used in Magnotti & Beauchamp.\n\nIf you are unfamiliar with R, the best approach is to run through all the model building steps using the included data.csv file, and then try with your own data.\n\n### Program setup\n\nWe need to ensure R can find the data and code files\n\n1. Launch R\n\n2. Open the file fit_model.R: File-> Open Document\n\n3. We need to make 2 changes before running the code. See the comments in the code file for additional direction\n\nSet the path to be the location of the downloaded files. If you extracted the code pack to your desktop, the path may already be correct\n\n``` setwd('~/Desktop/mcg_code_pack/')\n```\n\nSet the filename of the data to be fit. If you are using the example data, the filename is already correct\n\n``` mcg_data = read.csv(file='data.csv', row.names=NULL)\n```\n\n4. Run the setup code to make sure there are no errors\n\n1. Highlight lines 1 through 9 using the mouse (click and drag to highlight)\n2. Execute the code by using the R menu: Edit -> Execute\n\n### Fitting the model\n\nHighlight and execute each of the following lines in turn ``` ```\n\n``` cl = makeCluster(detectCores()) `````` # Fit the model # For 20 subjects with 14 stimuli, this takes about 12s per repetition mcg_model = optim.mcg(mcg_data, n.sim=5, n.iter=8) ```\n\n### Model Parameters\n\nYou can obtain the model parameters for each subject using the getElement function applied to each subject ``` ```\n\n``` disparities = mcg_model\\$disparities `````` thresholds = clip(sapply(mcg_model\\$subjs, getElement, 'threshold'), range=THRESHOLD_RANGE) `````` noises = clip(sapply(mcg_model\\$subjs, getElement, 'sd'), range=SENSORY_NOISE_RANGE) ```\n\n### Model Fit Statistics\n\nYou can obtain the model fit (RMSE) for each subject using the getElement function applied to each subject ``` ```\n\n``` rmses = sapply(mcg_model\\$subjs, getElement, 'rmse') ```\n\n### Predicted Susceptibility Values\n\nTo obtain the actual and predicted fusion rates, use the predict function on each subejct\n\n``` ```\n\n``` # mean illusion percept pF.mean = colMeans(sapply(mcg_model\\$subjs, getElement, 'y')) `````` #mean predicted percept fitted = t(sapply(mcg_model\\$subjs, predict, disparities)) pF.mean.pred = rowMeans(fitted) ```\n\n### Save model output to a csv file\n\nYou can save all the model results to a csv file for further processing/graphing\n\n``` ```\n\n``` write.csv(cbind(noises, thresholds, rmse, pF.mean, pF.mean.hat), file='model_results.csv') ```\n\nWe mention here some other useful functions for those comfortable analyzing data with the R language. These functions assume you have run all the code in the previous section. Intrepid users are encouraged to let the source be their guide.\n\nIf you have previously saved a model fit with save(...) you can load the model using load(...) ``` load(file='mcg_model.RDATA') ``` Once loaded, the variable will be in the current scope. Note that there is no explicit assignment with the load() function\n\n## Contact Information\n\nIf you run into trouble with any step, please contact me: john dot magnotti at gmail dot com. If the model fitting fails to converge for your dataset, you may need to send me at least a portion of the data so I can replicate the error."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.76024973,"math_prob":0.7279335,"size":4054,"snap":"2020-45-2020-50","text_gpt3_token_len":962,"char_repetition_ratio":0.119753085,"word_repetition_ratio":0.025316456,"special_character_ratio":0.22076961,"punctuation_ratio":0.1154345,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9796789,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-26T12:24:26Z\",\"WARC-Record-ID\":\"<urn:uuid:205a5f5c-81cc-4c5b-95ee-b85eac6ffb6a>\",\"Content-Length\":\"22086\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7503a5d6-a68d-4e03-9bca-4d07175a7aca>\",\"WARC-Concurrent-To\":\"<urn:uuid:a9b17b44-cb0c-4f5c-b9f6-04474bb25da4>\",\"WARC-IP-Address\":\"138.68.62.156\",\"WARC-Target-URI\":\"https://openwetware.org/wiki/Beauchamp:NED\",\"WARC-Payload-Digest\":\"sha1:JHIM3BMHTNL445LXIULVJYIP2VGLNPSF\",\"WARC-Block-Digest\":\"sha1:GUBTTODBZM422PUKEK5WYJ75OM4DBRJR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141188146.22_warc_CC-MAIN-20201126113736-20201126143736-00076.warc.gz\"}"} |
https://github.com/christianversloot/machine-learning-articles/blob/main/how-to-use-categorical-multiclass-hinge-with-keras.md | [
"{{ message }}\nSwitch branches/tags\nNothing to show",
null,
"Christian Versloot\n0 contributors\n\n### Users who have contributed to this file\n\n514 lines (376 sloc) 27 KB\ntitle date categories tags\nHow to use categorical / multiclass hinge with TensorFlow 2 and Keras?\n2019-10-17\n buffer deep-learning frameworks\n categorical-hinge-loss deep-learning hinge-loss keras loss-function machine-learning mlxtend\n\nRecently, I've been looking into loss functions - and specifically these questions: What is their purpose? How does the concept of loss work? And more practically, how can loss functions be implemented with the TensorFlow 2 based Keras framework for deep learning?\n\nThis resulted in blog posts that e.g. covered huber loss and hinge & squared hinge loss. Today, in this tutorial, we'll extend the latter to multiclass classification: we cover categorical hinge loss, or multiclass hinge loss. How can categorical hinge / multiclass hinge be implemented with TF2 based Keras? That's what well find out today.\n\nAfter reading this tutorial, you will understand...\n\n• What it means to go from binary hinge loss to multiclass hinge loss.\n• How categorical (multiclass) hinge loss works.\n• How `tensorflow.keras.losses.CategoricalHinge` can be used in your TensorFlow 2 based Keras model.\n\nLet's go! 😎\n\nUpdate 10/Feb/2021: ensure that article is up to date. Code examples now reflect TensorFlow 2 ecosystem and have been upgraded from TensorFlow/Keras 1.x.\n\n[toc]\n\n## Code example: multiclass hinge loss with TensorFlow 2 based Keras\n\nThis code example demonstrates quickly how to use categorical (multiclass) hinge loss with TensorFlow 2 based Keras. You can use this in your model straight away. If you want to understand the background details for multiclass hinge, make sure to read the rest of this tutorial as well 🚀\n\n``````loss_function_used = 'categorical_hinge'\n``````\n\n## From binary hinge to multiclass hinge\n\nIn that previous blog, we looked at hinge loss and squared hinge loss - which actually helped us to generate a decision boundary between two classes and hence a classifier, but yep - two classes only.\n\nHinge loss and squared hinge loss can be used for binary classification problems.\n\nUnfortunately, many of today's problems aren't binary, but rather, multiclass: the number of possible target classes is [latex]> 2[/latex].\n\nAnd hinge and squared hinge do not accommodate for this.\n\nBut categorical hinge loss, or multiclass hinge loss, does - and it is available in Keras!\n\n## How does multiclass hinge work?\n\nMulticlass hinge was introduced by researchers Weston and Watkins (Wikipedia, 2011):",
null,
"What this means in plain English is this:\n\nFor a prediction [latex]y[/latex], take all [latex]y[/latex] values unequal to [latex]t[/latex], and compute the individual losses. Eventually, sum them together to find the multiclass hinge loss.\n\nThe name categorical hinge loss, which is also used in place of multiclass hinge loss, already implies what's happening here:\n\nWe first convert our regular targets into categorical data. That is, if we have three possible target classes {0, 1, 2}, an arbitrary target (e.g. 2) would be converted into categorical format (in that case, [latex][0, 0, 1][/latex]).\n\nNext, for any sample, our DL model generates a multiclass probability distribution over all possible target classes. That is, for the total probability of 100% (or, statistically, [latex]1[/latex]) it generates the probability that any of the possible categorical classes is the actual target class (in the scenario above, e.g. [latex][0.25, 0.25, 0.50][/latex] - which would mean class two, but with some uncertainty.\n\nComputing the loss - the difference between actual target and predicted targets - is then equal to computing the hinge loss for taking the prediction for all the computed classes, except for the target class, since loss is always 0 there. The hinge loss computation itself is similar to the traditional hinge loss.\n\nCategorical hinge loss can be optimized as well and hence used for generating decision boundaries in multiclass machine learning problems. Let's now see how we can implement it with TensorFlow 2 based Keras.\n\n## Today's dataset: extending the binary case\n\n...which requires defining a dataset first :-)\n\nIn our post covering traditional hinge loss, we generated data ourselves because this increases simplicity.\n\nWe'll do so as well in today's blog. Specifically, we create a dataset with three separable clusters that looks as follows:",
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"How? Let's find out.\n\nFirst, open some folder and create a Python file where you'll write your code - e.g. `multiclass-hinge.py`.\n\nNext, open a development environment as well as the file, and you can start coding 😊\n\n### Importing software dependencies\n\n``````'''\nKeras model discussing Categorical (multiclass) Hinge loss.\n'''\nimport tensorflow\nfrom tensorflow.keras.models import Sequential\nfrom tensorflow.keras.layers import Dense\nfrom tensorflow.keras.utils import to_categorical\nimport matplotlib.pyplot as plt\nimport numpy as np\nfrom sklearn.datasets import make_blobs\nfrom mlxtend.plotting import plot_decision_regions\n``````\n\nWe need TensorFlow 2 (`pip install tensorflow`) since we build the model by means of its APIs and functionalities. From its `tensorflow.keras` representation of Keras, we need:\n\n• The Sequential API, which allows us to stack neural network layers;\n• The densely-connected layer type, since we'll build our network by means of such layers.\n\nWe also need Matplotlib for generating visualizations of our dataset, Numpy for basic number processing, Scikit-learn for generating the dataset and Mlxtend for visualizing the decision boundary of our model.\n\n### Model & data configuration\n\nWe next add some configuration options:\n\n``````# Configuration options\nnum_samples_total = 3000\ntraining_split = 1000\nnum_classes = 3\nloss_function_used = 'categorical_hinge'\nlearning_rate_used = 0.03\nnum_epochs = 30\nbatch_size = 5\nvalidation_split = 0.2 # 20%\n``````\n\nThe three clusters contain 3000 samples in total divided over three classes or clusters, as we saw in the image above. The `training_split` value is 1000, which means that 1000 samples are split off the training set to serve as testing data.\n\nNext, we specify the hyper parameters. Obviously, we'll use categorical hinge loss. We set the learning rate to 0.03 since traditional hinge required a more aggressive value contrary to 0.001, which is default in Keras. We use the Adam optimizer and configure it to use this learning rate, which is very common today since Adam is the de facto standard optimizer used in DL projects.\n\nAs an additional metric, we specify accuracy, as we have done before in many of our blog posts. Accuracy is more intuitively understandable to humans.\n\nThe model will train for 30 epochs with a batch size of 5 samples per forward pass, and 20% of the training data (2000 samples, hence 400 samples) will be used for validating each epoch as validation data.\n\n### Generating a dataset\n\nNext, we can generate the data:\n\n``````# Generate data\nX, targets = make_blobs(n_samples = num_samples_total, centers = [(0,0), (15,15), (0,15)], n_features = num_classes, center_box=(0, 1), cluster_std = 1.5)\ncategorical_targets = to_categorical(targets)\nX_training = X[training_split:, :]\nX_testing = X[:training_split, :]\nTargets_training = categorical_targets[training_split:]\nTargets_testing = categorical_targets[:training_split].astype(np.integer)\n\n# Set shape based on data\nfeature_vector_length = len(X_training)\ninput_shape = (feature_vector_length,)\n``````\n\nWe use Scikit-learns `make_blobs` function to generate data. It simply does as it suggests: it generates blobs of data, or clusters of data, where you specify them to be. Specifically, it generates `num_samples_total` (3000, see model configuration section) in our case, splits them across three clusters centered at [latex]{ (0, 0), (15, 15), (0,15) }[/latex]. The standard deviation in a cluster is approximately 1.5 to ensure that they are actually separable.\n\nNext, we must convert our target values (which are one of [latex]{ 0, 1, 2 }[/latex]) into categorical format since our categorical hinge loss requires categorical format (and hence no integer targets such as [latex]2[/latex], but categorical vectors like [latex][0, 0, 1][/latex].\n\nSubsequently, we can split our feature vectors and target vectors according to the `training_split` we configured in our model configuration. Note that we add `.astype(np.integer`) to the testing targets. We do this because when visualizing categorical data, the Mlxtend library requires the vector contents to be integers (instead of floating point numbers).\n\nFinally, we set the `input_shape` based on the length of our feature vector, which originates from the training data.\n\n### Visualizing our dataset\n\nWe can finally visualize the data we generated:\n\n``````# Generate scatter plot for training data\nplt.scatter(X_training[:,0], X_training[:,1])\nplt.title('Three clusters ')\nplt.xlabel('X1')\nplt.ylabel('X2')\nplt.show()\n``````\n\n...which, as illustrated before, looks like this:",
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"As illustrated before, this is what is generated 😎\n\nWe can work with this!\n\n## Creating the multiclass hinge Keras model\n\n### What you'll need to run this model\n\nIf you wish to run this model on your machine, you'll need to install some dependencies to make the code work. First of all, you need Keras, the deep learning framework with which this model is built. It's the most essential dependency and can be installed by installing TensorFlow 2.x today, e.g. 2.4.0. It is then available as `tensorflow.keras`.\n\nAdditionally, you'll need the de facto standard Python libraries Matplotlib, Numpy and Scikit-learn - they can be installed with `pip` quite easily.\n\nAnother package, which can also be installed with `pip`, is Sebastian Raschka's Mlxtend. We use it to visualize the decision boundary of our model.Creating the model architecture\n\nWe will create a very simple model today, a four-layered (two hidden layers, one input layer and one output layer) MLP:\n\n``````# Create the model\nmodel = Sequential()\n``````\n\nMore specifically, we use the Keras Sequential API which allows us to stack multiple layers on top of each other. We subsequently `add` the Dense or densely-connected layers; the first having four neurons, the second two, and the last `num_classes`, or three in our case. The hidden layers activate by means of the ReLU activation function and hence are initialized with He uniform init. The last layer activates with tanh.\n\n### Model configuration & training\n\nNext, we configure the model and start the training process:\n\n``````# Configure the model and start training\nhistory = model.fit(X_training, Targets_training, epochs=num_epochs, batch_size=batch_size, verbose=1, validation_split=validation_split)\n``````\n\nIt's as simple as calling `model.compile` with the settings that we configured under model configuration, followed by `model.fit` which fits the training data to the model architecture specified above. The training history is saved in the `history` object which we can use for visualization purposes.\n\nNext, we must add some more code for testing the model's ability to generalize to data it hasn't seen before.\n\n## Model performance\n\n### Generalization power with testing set\n\nIn order to test model performance, we add some code that evaluates the model with the testing set:\n\n``````# Test the model after training\ntest_results = model.evaluate(X_testing, Targets_testing, verbose=1)\nprint(f'Test results - Loss: {test_results} - Accuracy: {test_results*100}%')\n``````\n\nWhat it will do is this: it takes the testing data (both features and targets) and feeds them through the model, comparing predicted target with the actual prediction. Since the model has never seen the data before, it tells us something about the degree of overfitting that occurred during training. When the model performs well during validation but also during testing, it's useful to practice.\n\n### Visualizing the decision boundary\n\nVisualizing the decision boundaries of the model (remember, we have a three-class classification problem!) is the next step.\n\nI must admit, I had a little help from dr. Sebastian Raschka here, the creator of Mlxtend (also see rasbt/mlxtend#607). As noted before, we had to convert our targets into categorical format, or e.g. [latex]target = 2[/latex] into [latex]target = [0, 0, 1][/latex]. Mlxtend does not natively support this, but fortunately, Raschka helped out by creating a helper class that embeds the model yet converts the way it makes predictions (back into non-categorical format). This looks as follows:\n\n``````'''\nThe Onehot2Int class is used to adapt the model so that it generates non-categorical data.\nThis is required by the `plot_decision_regions` function.\nThe code is courtesy of dr. Sebastian Raschka at https://github.com/rasbt/mlxtend/issues/607.\nThanks!\n'''\n# No hot encoding version\nclass Onehot2Int(object):\n\ndef __init__(self, model):\nself.model = model\n\ndef predict(self, X):\ny_pred = self.model.predict(X)\nreturn np.argmax(y_pred, axis=1)\n\n# fit keras_model\nkeras_model_no_ohe = Onehot2Int(model)\n\n# Plot decision boundary\nplot_decision_regions(X_testing, np.argmax(Targets_testing, axis=1), clf=keras_model_no_ohe, legend=3)\nplt.show()\n'''\nFinish plotting the decision boundary.\n'''\n``````\n\n### Visualizing the training process\n\nFinally, we can visualize the training process itself by adding some extra code - which essentially plots the Keras `history` object with Matplotlib:\n\n``````# Visualize training process\nplt.plot(history.history['loss'], label='Categorical Hinge loss (training data)')\nplt.plot(history.history['val_loss'], label='Categorical Hinge loss (validation data)')\nplt.title('Categorical Hinge loss for circles')\nplt.ylabel('Categorical Hinge loss value')\nplt.yscale('log')\nplt.xlabel('No. epoch')\nplt.legend(loc=\"upper left\")\nplt.show()\n``````\n\n### How does the model perform?\n\nNow that we've completed our code, we can actually run the model!\n\nOpen up a terminal where you have access to the software dependencies required to run the code, `cd` to the directory where your file is located, and execute e.g. `python multiclass-hinge.py`.\n\nAfter the visualization of your dataset (with the three clusters), you'll see the training process run and complete - as well as model evaluation with the testing set:\n\n``````Epoch 1/30\n2019-10-16 19:39:12.492536: I tensorflow/stream_executor/platform/default/dso_loader.cc:44] Successfully opened dynamic library cublas64_100.dll\n1600/1600 [==============================] - 1s 906us/step - loss: 0.5006 - accuracy: 0.6950 - val_loss: 0.3591 - val_accuracy: 0.6600\nEpoch 2/30\n1600/1600 [==============================] - 1s 603us/step - loss: 0.3397 - accuracy: 0.6681 - val_loss: 0.3528 - val_accuracy: 0.6500\nEpoch 3/30\n1600/1600 [==============================] - 1s 615us/step - loss: 0.3398 - accuracy: 0.6681 - val_loss: 0.3721 - val_accuracy: 0.7425\nEpoch 4/30\n1600/1600 [==============================] - 1s 617us/step - loss: 0.3379 - accuracy: 0.8119 - val_loss: 0.3512 - val_accuracy: 0.8500\nEpoch 5/30\n1600/1600 [==============================] - 1s 625us/step - loss: 0.3368 - accuracy: 0.8869 - val_loss: 0.3515 - val_accuracy: 0.8600\nEpoch 6/30\n1600/1600 [==============================] - 1s 608us/step - loss: 0.3358 - accuracy: 0.8906 - val_loss: 0.3506 - val_accuracy: 0.9325\nEpoch 7/30\n1600/1600 [==============================] - 1s 606us/step - loss: 0.3367 - accuracy: 0.9344 - val_loss: 0.3532 - val_accuracy: 0.9375\nEpoch 8/30\n1600/1600 [==============================] - 1s 606us/step - loss: 0.3365 - accuracy: 0.9375 - val_loss: 0.3530 - val_accuracy: 0.9425\nEpoch 9/30\n1600/1600 [==============================] - 1s 625us/step - loss: 0.3364 - accuracy: 0.9419 - val_loss: 0.3528 - val_accuracy: 0.9475\nEpoch 10/30\n1600/1600 [==============================] - 1s 627us/step - loss: 0.3364 - accuracy: 0.9450 - val_loss: 0.3527 - val_accuracy: 0.9500\nEpoch 11/30\n1600/1600 [==============================] - 1s 606us/step - loss: 0.3363 - accuracy: 0.9506 - val_loss: 0.3525 - val_accuracy: 0.9525\nEpoch 12/30\n1600/1600 [==============================] - 1s 642us/step - loss: 0.3366 - accuracy: 0.9425 - val_loss: 0.3589 - val_accuracy: 0.6475\nEpoch 13/30\n1600/1600 [==============================] - 1s 704us/step - loss: 0.3526 - accuracy: 0.8606 - val_loss: 0.3506 - val_accuracy: 0.9850\nEpoch 14/30\n1600/1600 [==============================] - 1s 699us/step - loss: 0.3364 - accuracy: 0.9925 - val_loss: 0.3502 - val_accuracy: 0.9875\nEpoch 15/30\n1600/1600 [==============================] - 1s 627us/step - loss: 0.3363 - accuracy: 0.9944 - val_loss: 0.3502 - val_accuracy: 0.9875\nEpoch 16/30\n1600/1600 [==============================] - 1s 670us/step - loss: 0.3363 - accuracy: 0.9937 - val_loss: 0.3502 - val_accuracy: 0.9875\nEpoch 17/30\n1600/1600 [==============================] - 1s 637us/step - loss: 0.3362 - accuracy: 0.9694 - val_loss: 0.3530 - val_accuracy: 0.9400\nEpoch 18/30\n1600/1600 [==============================] - 1s 637us/step - loss: 0.3456 - accuracy: 0.9744 - val_loss: 0.3537 - val_accuracy: 0.9825\nEpoch 19/30\n1600/1600 [==============================] - 1s 635us/step - loss: 0.3347 - accuracy: 0.9975 - val_loss: 0.3501 - val_accuracy: 0.9950\nEpoch 20/30\n1600/1600 [==============================] - 1s 644us/step - loss: 0.3344 - accuracy: 0.9994 - val_loss: 0.3501 - val_accuracy: 0.9950\nEpoch 21/30\n1600/1600 [==============================] - 1s 655us/step - loss: 0.3344 - accuracy: 0.9994 - val_loss: 0.3501 - val_accuracy: 0.9950\nEpoch 22/30\n1600/1600 [==============================] - 1s 636us/step - loss: 0.3344 - accuracy: 0.9994 - val_loss: 0.3501 - val_accuracy: 0.9950\nEpoch 23/30\n1600/1600 [==============================] - 1s 648us/step - loss: 0.3344 - accuracy: 0.9994 - val_loss: 0.3501 - val_accuracy: 0.9950\nEpoch 24/30\n1600/1600 [==============================] - 1s 655us/step - loss: 0.3344 - accuracy: 0.9994 - val_loss: 0.3501 - val_accuracy: 0.9950\nEpoch 25/30\n1600/1600 [==============================] - 1s 656us/step - loss: 0.3344 - accuracy: 0.9994 - val_loss: 0.3501 - val_accuracy: 0.9950\nEpoch 26/30\n1600/1600 [==============================] - 1s 641us/step - loss: 0.3344 - accuracy: 0.9994 - val_loss: 0.3501 - val_accuracy: 0.9950\nEpoch 27/30\n1600/1600 [==============================] - 1s 644us/step - loss: 0.3344 - accuracy: 0.9994 - val_loss: 0.3500 - val_accuracy: 0.9950\nEpoch 28/30\n1600/1600 [==============================] - 1s 666us/step - loss: 0.3344 - accuracy: 0.9994 - val_loss: 0.3500 - val_accuracy: 0.9950\nEpoch 29/30\n1600/1600 [==============================] - 1s 645us/step - loss: 0.3344 - accuracy: 0.9994 - val_loss: 0.3500 - val_accuracy: 0.9950\nEpoch 30/30\n1600/1600 [==============================] - 1s 669us/step - loss: 0.3344 - accuracy: 0.9994 - val_loss: 0.3500 - val_accuracy: 0.9950\n1000/1000 [==============================] - 0s 46us/step\nTest results - Loss: 0.3260095896720886 - Accuracy: 99.80000257492065%\n``````\n\nIn my case, it was able to achieve very high accuracy - 99.5% on the testing set and 99.8% on the training set! Indeed, the decision boundaries allow us to classify the majority of samples correctly:",
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"...and the training process looks like this:",
null,
"Just after the first epoch, model performance pretty much maxed out.\n\n...which is not unsurprising given the fact that our datasets are quite separable by nature, or perhaps, by design 😉 The relative ease with which the datasets are separable allows us to focus on the topic of this blog post, which was the categorical hinge loss.\n\nAll in all, we've got a working model using categorical hinge in Keras!\n\n## All code merged together\n\nWhen merging all code together, we get this:\n\n``````'''\nKeras model discussing Categorical (multiclass) Hinge loss.\n'''\nimport tensorflow\nfrom tensorflow.keras.models import Sequential\nfrom tensorflow.keras.layers import Dense\nfrom tensorflow.keras.utils import to_categorical\nimport matplotlib.pyplot as plt\nimport numpy as np\nfrom sklearn.datasets import make_blobs\nfrom mlxtend.plotting import plot_decision_regions\n\n# Configuration options\nnum_samples_total = 3000\ntraining_split = 1000\nnum_classes = 3\nfeature_vector_length = len(X_training)\ninput_shape = (feature_vector_length,)\nloss_function_used = 'categorical_hinge'\nlearning_rate_used = 0.03\nnum_epochs = 30\nbatch_size = 5\nvalidation_split = 0.2 # 20%\n\n# Generate data\nX, targets = make_blobs(n_samples = num_samples_total, centers = [(0,0), (15,15), (0,15)], n_features = num_classes, center_box=(0, 1), cluster_std = 1.5)\ncategorical_targets = to_categorical(targets)\nX_training = X[training_split:, :]\nX_testing = X[:training_split, :]\nTargets_training = categorical_targets[training_split:]\nTargets_testing = categorical_targets[:training_split].astype(np.integer)\n\n# Generate scatter plot for training data\nplt.scatter(X_training[:,0], X_training[:,1])\nplt.title('Three clusters ')\nplt.xlabel('X1')\nplt.ylabel('X2')\nplt.show()\n\n# Create the model\nmodel = Sequential()\n\n# Configure the model and start training\nhistory = model.fit(X_training, Targets_training, epochs=num_epochs, batch_size=batch_size, verbose=1, validation_split=validation_split)\n\n# Test the model after training\ntest_results = model.evaluate(X_testing, Targets_testing, verbose=1)\nprint(f'Test results - Loss: {test_results} - Accuracy: {test_results*100}%')\n\n'''\nThe Onehot2Int class is used to adapt the model so that it generates non-categorical data.\nThis is required by the `plot_decision_regions` function.\nThe code is courtesy of dr. Sebastian Raschka at https://github.com/rasbt/mlxtend/issues/607.\nThanks!\n'''\n# No hot encoding version\nclass Onehot2Int(object):\n\ndef __init__(self, model):\nself.model = model\n\ndef predict(self, X):\ny_pred = self.model.predict(X)\nreturn np.argmax(y_pred, axis=1)\n\n# fit keras_model\nkeras_model_no_ohe = Onehot2Int(model)\n\n# Plot decision boundary\nplot_decision_regions(X_testing, np.argmax(Targets_testing, axis=1), clf=keras_model_no_ohe, legend=3)\nplt.show()\n'''\nFinish plotting the decision boundary.\n'''\n\n# Visualize training process\nplt.plot(history.history['loss'], label='Categorical Hinge loss (training data)')\nplt.plot(history.history['val_loss'], label='Categorical Hinge loss (validation data)')\nplt.title('Categorical Hinge loss for circles')\nplt.ylabel('Categorical Hinge loss value')\nplt.yscale('log')\nplt.xlabel('No. epoch')\nplt.legend(loc=\"upper left\")\nplt.show()\n``````\n\n## Summary\n\nIn this blog post, we've seen how categorical hinge extends binary (normal) hinge loss and squared hinge loss to multiclass classification problems. We considered the loss mathematically, but also built up an example with Keras that allows us to use categorical hinge with a real dataset, generating visualizations of the training process and decision boundaries as well. This concludes today's post.\n\nI hope you've learnt something here. If you did, I'd appreciate it if you let me know! 😊 You can do so by leaving a comment below 👇 Thanks a lot - and happy engineering! 😎"
] | [
null,
"https://camo.githubusercontent.com/9fdd1b6e34d92258d8cae7392fd7e1c0b9910ff0b91399ec8d20732c2519fce5/68747470733a2f2f322e67726176617461722e636f6d2f6176617461722f31646235633639353433653335656266393366353261653935313965666631393f643d68747470732533412532462532466769746875622e6769746875626173736574732e636f6d253246696d6167657325324667726176617461727325324667726176617461722d757365722d3432302e706e6726723d6726733d313430",
null,
"https://github.com/christianversloot/machine-learning-articles/raw/main/images/image-2-1024x170.png",
null,
"https://github.com/christianversloot/machine-learning-articles/raw/main/images/mh_3.png",
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"https://github.com/christianversloot/machine-learning-articles/raw/main/images/mh_3.png",
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"https://github.com/christianversloot/machine-learning-articles/raw/main/images/mh_boundary-1024x587.png",
null,
"https://github.com/christianversloot/machine-learning-articles/raw/main/images/mh_loss-1024x564.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73270345,"math_prob":0.8803403,"size":24123,"snap":"2022-27-2022-33","text_gpt3_token_len":6278,"char_repetition_ratio":0.19428666,"word_repetition_ratio":0.22497676,"special_character_ratio":0.32301122,"punctuation_ratio":0.19005044,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97152036,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,2,null,2,null,2,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-28T12:32:16Z\",\"WARC-Record-ID\":\"<urn:uuid:d61bef3d-fa5c-4540-8143-d6e8ea8005a0>\",\"Content-Length\":\"217450\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:339d3a42-4921-4f9c-978c-6fd2ce9f189c>\",\"WARC-Concurrent-To\":\"<urn:uuid:da3a1ccb-3759-45c7-8657-b04916bf894d>\",\"WARC-IP-Address\":\"140.82.112.3\",\"WARC-Target-URI\":\"https://github.com/christianversloot/machine-learning-articles/blob/main/how-to-use-categorical-multiclass-hinge-with-keras.md\",\"WARC-Payload-Digest\":\"sha1:AJZ7KKBYNHH62QWFSOQG2BCF3JGY2FKV\",\"WARC-Block-Digest\":\"sha1:SKDEFRGAZYIGQNCOHTB5EX72EMCKANSJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103516990.28_warc_CC-MAIN-20220628111602-20220628141602-00774.warc.gz\"}"} |
https://www.easycalculation.com/physics/astrodynamics/relativistic-pressure-density.php | [
"# Relativistic Pressure Calculator\n\nIn astrodynamics, measure the relativistic pressure based on Neutron mass and density of the gas. Physically, the gas pressure is a measure of force trade inside the gas.\n\n## Calculate the Relativistic Pressure based on Density\n\nm2kg/s\nm/s\nρ\nweight\nN/m2\n\nIn astrodynamics, measure the relativistic pressure based on Neutron mass and density of the gas. Physically, the gas pressure is a measure of force trade inside the gas.\n\nCode to add this calci to your website",
null,
"",
null,
"#### Formula:\n\nP = [ ((h/2π)c (3π2)1/3) / 4] * (ρ / mn)4/3 Where, P = Relativistic Pressure h = Planck Constant c = Speed of Light ρ = Density mn = Neutron Mass"
] | [
null,
"https://www.easycalculation.com/images/embed-plus.gif",
null,
"https://www.easycalculation.com/images/embed-minus.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91052943,"math_prob":0.9944318,"size":341,"snap":"2022-40-2023-06","text_gpt3_token_len":74,"char_repetition_ratio":0.18397626,"word_repetition_ratio":0.9230769,"special_character_ratio":0.18475074,"punctuation_ratio":0.125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9931804,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-28T12:54:50Z\",\"WARC-Record-ID\":\"<urn:uuid:33739389-aea1-4374-9f4d-fee94aabec41>\",\"Content-Length\":\"25370\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f6fc1cbe-27cd-4e53-84dc-a5dbf27df0ff>\",\"WARC-Concurrent-To\":\"<urn:uuid:f767cf58-7e09-4b5a-8b81-4c3344f78efc>\",\"WARC-IP-Address\":\"66.228.40.80\",\"WARC-Target-URI\":\"https://www.easycalculation.com/physics/astrodynamics/relativistic-pressure-density.php\",\"WARC-Payload-Digest\":\"sha1:TDC5BUHSBOLPZOT22WTL5JFTSOKIW6IC\",\"WARC-Block-Digest\":\"sha1:QO4VZ7U7EXVFIT3IA4EYIQ4LJMK5FCE4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499634.11_warc_CC-MAIN-20230128121809-20230128151809-00796.warc.gz\"}"} |
https://engineering.stackexchange.com/questions/3348/calculating-pitch-yaw-and-roll-from-mag-acc-and-gyro-data/3350 | [
"Calculating pitch, yaw, and roll from mag, acc, and gyro data\n\nI have an Arduino board with a 9 degree of freedom sensor, from which I must determine the pitch, yaw, and roll of the board.\n\nHere is an example of one set of data from the 9-DOF sensor:\n\nAccelerometer (m/s)\n\n• $\\text{Acc}_{X}$ = -5,85\n• $\\text{Acc}_{Y}$ = 1,46\n• $\\text{Acc}_{Z}$ = 17,98\n\nGyroscope (RPM)\n\n• $\\text{Gyr}_{X}$ = 35,14\n• $\\text{Gyr}_{Y}$ = -40,22\n• $\\text{Gyr}_{Z}$ = -9,86\n\nMagnetometer (Gauss)\n\n• $\\text{Mag}_{X}$ = 0,18\n• $\\text{Mag}_{Y}$ = -0,04\n• $\\text{Mag}_{Z}$ = -0,15\n\nHow can I calculate pitch, yaw, and roll from these data?\n\n• Basic principle: from the detection of gravity in your accelerometer you know which way is down; from the detection of the earth's magnetic field in your magnetometer you know which way is North. Based on this and assuming no other significant accelerations or strong magnetic fields you can determine your own attitude.\n– welf\nJun 27 '15 at 0:00\n• Gyroscope data provides a rate of rotation, but not an absolute position. It can be integrated to estimate change from a known attitude, but this is typically noisy and prone to drift if it is not used in conjunction with the other sensors.\n– welf\nJun 27 '15 at 0:05\n• also refer to kalman filters, as the static numbers need to be processed quite a lot, to give reliable estimates of roll-pitch and yaw. Also note that, the position of the sensor is important (you need to take it into account). Jun 28 '15 at 17:24\n\nPitch, roll and yaw are defined as the rotation around X, Y and Z axis. Below as a picture to illustrate the definition.",
null,
"In a previous project I used a ADXL345 Accelerometer from Analog Devices to calculate roll and pitch. Below are the equations used to calculated roll and pitch. I have made some of source code available for public use.\n\naccelerationX = (signed int)(((signed int)rawData_X) * 3.9);\naccelerationY = (signed int)(((signed int)rawData_Y) * 3.9);\naccelerationZ = (signed int)(((signed int)rawData_Z) * 3.9);\npitch = 180 * atan (accelerationX/sqrt(accelerationY*accelerationY + accelerationZ*accelerationZ))/M_PI;\nroll = 180 * atan (accelerationY/sqrt(accelerationX*accelerationX + accelerationZ*accelerationZ))/M_PI;\n\nComplete source code can be found here.\n\nBase on the above definitions\n\nyaw = 180 * atan (accelerationZ/sqrt(accelerationX*accelerationX + accelerationZ*accelerationZ))/M_PI;\n\nNote: M_PI = 3.14159265358979323846 it is constant defined in math.h\n\nBelow are some references including Arduino base source code that might help you.\n\nReferences:\n\n• Nice answer, it would be worth adding that the position and orientation of the sensor in the vehicle would be important, and that the data must be processed further, to give reliable results. (filtered or fused with more reliable low-frequency data, such as GPS) Jun 28 '15 at 17:26\n• (@Zubair) \"yaw = 180 * atan (accelerationZ/sqrt(accelerationXaccelerationX + accelerationZaccelerationZ))/M_PI;\" what is that 'M_PI'??\n– Wasabi\nApr 22 '16 at 14:54\n• @Wasabi M_PI=3.14159265358979323846. It is constant defined in the math.h library. Apr 23 '16 at 0:59\n• why sometimes i find pitch and roll both using pythagoras and sometimes only one uses it Aug 9 '20 at 16:23\n• @Mr-Programs, It has been many years since I worked on the problem. I suggest posting a fresh question and linking this response. Aug 10 '20 at 2:50\n\nSo my longer answer below assumes that the board will undergo acceleration and during this time you still need to be able to measure your pitch, roll and yaw within a short amount of time. If the board will be stationary for all measurements then Mahendra Gunawardena's answer will work perfectly for you. If this is going into a device like a segway or model plane or multirotor or anything that moves around, you may want to keep reading. This post is about how to use all three sensors though a method called sensor fusion. Sensor fusion allows you to get the strengths of each sensor and minimize the effects of each sensors weaknesses.\n\nSensor characteristics and background\n\nFirst understand that an accelerometer measure all forces being applied to it, not just the force of gravity. So in a perfect world with the accelerometer in a stationary position without any vibrations you could perfectly determine which way is up using some basic trigonometry as shown by Mahendra Gunawardena's answer. However since an accelerometer will pick up all forces, any vibrations will result in noise. It should also be noted that if the board is accelerating you can not just use simple trigonometry as the force the accelerometer is reporting is not only the earths force of gravity but also the force that is causing you to accelerate.\n\nA magnetometer is more straightforward then an accelerometer. Movement will not cause problems with it but things like iron and other magnets will end up effecting your output. If the sources causing this interference are constant its not to hard to deal with but if these sources are not constant it will create tons of noise that is problematic to remove.\n\nOf the three sensors, the gyroscope is arguable the most reliable and they are normally very very good at measuring rotational speed. It is not affected by things like iron sources and accelerations have basically no impact on their ability to measure rotational speed. They do a very good job of reporting the speed at which the device is turning at, however since you are looking for an absolute angle you have to integrate the speed to get position. Doing this will add the error of the last measurement to the error of the new measurements since integration is basically a sum of values over a range, even if the error for one measurement is only 0.01 degrees per second off, in 100 measurements, your position can be off by 1 degree, by 1000 measurements, you can by off by 10 degrees. If you are taking hundreds of measurements a second, you can see this causes problems. This is commonly called gyro drift.\n\nSensor fusion\n\nNow the beauty of having all of these sensors work together is that you can use the information from the accelerometer and magnetometer to cancel out gyro drift. This ends up allowing you to giving you the accuracy and speed of the gyro without the fatal flaw of gyro drift.\n\nCombining the data from these three sensors can be done in more then one way, I'll talk about using a complementary filter because its far simpler then a kalman filter and kalman filters will eat up much more resources on embedded systems. Often times a complementary filter is good enough, simpler to implement(assuming your not using a pre-built library) and lets you process the data faster.\n\nNow onto the process. The first steps you need to do is to integrate the gyroscope output to convert the angular speed into angular position. You will also most likely have to apply a low pass filter on the accelerometer and magnetometer to deal with noise in the output. A simple FIR filter like the one shown below works here. With some trigonometry you can find the pitch and roll with the accelerometer and the yaw with the magnetometer.\n\nfilteredData = (1-weight)*filteredData + weight*newData\n\nThe weight is just a constant that can be adjusted depending on how much noise you have to deal with, the higher the noise is the smaller the weight value will be. Now combining the data from the sensors can be done by the following line of code.\n\nfusedData = (1-weight)*gyroData + weight*accelMagData\n\nIt should be noted that the data is a vector of the pitch, roll and yaw. You can just use three variables to do this as well instead of arrays if you want. For this calculation the gyro provides a position in degrees in pitch, roll and yaw, the magnetometer provides an angle for yaw while the accelerometer provides its own numbers for pitch and roll.\n\nFrom the accelerator sensor data, you can only calculate pitch and roll. The bellow document from Freescale explains with plenty of information what you need:\n\nAN3461 - Tilt Sensing Using a Three-Axis Accelerometer\n\nBased on the sayings of the document,\n\n$$\\tan \\phi_{xyz} = \\frac{G_{py}}{G_{pz}}$$\n\n$$\\tan \\theta_{xyz} = \\frac{-G_{px}}{G_{py}\\sin \\phi + G_{pz}\\cos \\phi} = \\frac{-G_{px}}{\\sqrt{G_{py}^2 + G_{pz}^2}}$$\n\nwhich equates to:\n\nroll = atan2(accelerationY, accelerationZ)\n\npitch = atan2(-accelerationX, sqrt(accelerationY*accelerationY + accelerationZ*accelerationZ))\n\nOf course, the result is this only when the rotations are occurring on a specific order (Rxyz):\n\n1. Roll around the x-axis by angle $\\phi$\n2. Pitch around the y-axis by angle $\\theta$\n3. Yaw around z-axis by angle $\\psi$\n\nDepending on the rotations order, you get different equations. For the $R_{xyz}$ rotation order, you can not find the angle $\\psi$ for the Yaw around z-axis.\n\n• He, welcome on the Engineering SE! This site supports Latex, look how beautiful your answer became now. :-) Jun 10 '18 at 14:11\n• i still don't understand why they have to use Pythagoras for second angle here Aug 9 '20 at 16:35"
] | [
null,
"https://i.stack.imgur.com/8IuOw.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.94435745,"math_prob":0.96441036,"size":4571,"snap":"2022-05-2022-21","text_gpt3_token_len":915,"char_repetition_ratio":0.12130501,"word_repetition_ratio":0.0025348542,"special_character_ratio":0.19273682,"punctuation_ratio":0.06117647,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9908346,"pos_list":[0,1,2],"im_url_duplicate_count":[null,10,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-21T09:21:05Z\",\"WARC-Record-ID\":\"<urn:uuid:bf8b0d2a-cde2-4a82-8372-d9e2f29de8f5>\",\"Content-Length\":\"177951\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f9c6592b-d9de-4c4d-aba9-4d976d801239>\",\"WARC-Concurrent-To\":\"<urn:uuid:fec1cf59-4928-48e4-ba97-eaaf9e80c5b3>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://engineering.stackexchange.com/questions/3348/calculating-pitch-yaw-and-roll-from-mag-acc-and-gyro-data/3350\",\"WARC-Payload-Digest\":\"sha1:OJBQJNB7ZB4CSEN62X7E77O6GV2W4VI3\",\"WARC-Block-Digest\":\"sha1:ISHZOCCWF75OFMCB7MCDFJZPVMBYU6KI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320302740.94_warc_CC-MAIN-20220121071203-20220121101203-00409.warc.gz\"}"} |
https://giotto-ai.github.io/gtda-docs/0.2.1/modules/generated/mapper/utils/gtda.mapper.utils.pipeline.transformer_from_callable_on_rows.html | [
"transformer_from_callable_on_rows¶\n\ngtda.mapper.utils.pipeline.transformer_from_callable_on_rows(func, validate=True)[source]\n\nConstruct a transformer from a callable acting on 1D arrays.\n\nGiven a callable which can act on 1D arrays, this function returns a fit-transformer which applies the callable to slices of 2D arrays along axis 1. When possible, the array output by the transformer’s fit_transform is two-dimensional.\n\nParameters\n• func (callable) – A callable object.\n\n• validate (bool, optional, default: True) – Whether the output transformer should implement input validation.\n\nReturns\n\nfunction_transformer – Output fit-transformer.\n\nReturn type\n\nsklearn.preprocessing.FunctionTransformer object\n\nExamples\n\n>>> import numpy as np\n>>> from gtda.mapper import transformer_from_callable_on_rows\n>>> function_transformer = transformer_from_callable_on_rows(np.sum)\n>>> X = np.array([[0, 1], [2, 3]])\n>>> print(function_transformer.fit_transform(X))\n[,\n]"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5601719,"math_prob":0.90583265,"size":961,"snap":"2022-05-2022-21","text_gpt3_token_len":217,"char_repetition_ratio":0.20794149,"word_repetition_ratio":0.0,"special_character_ratio":0.23204994,"punctuation_ratio":0.18309858,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9690088,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-21T08:42:27Z\",\"WARC-Record-ID\":\"<urn:uuid:cc04e811-0879-49b8-84ca-742d9b42ab80>\",\"Content-Length\":\"53005\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e98ddda-decf-4130-b79f-5c08be377e8f>\",\"WARC-Concurrent-To\":\"<urn:uuid:36b64e30-a5f0-4ad4-a70b-1a7f0626e929>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://giotto-ai.github.io/gtda-docs/0.2.1/modules/generated/mapper/utils/gtda.mapper.utils.pipeline.transformer_from_callable_on_rows.html\",\"WARC-Payload-Digest\":\"sha1:W3G3S4PC2WMSUQMDM2SEVXM5IWVB3NHP\",\"WARC-Block-Digest\":\"sha1:HGIZXZNQLZ55WEW7H7U4SO4EQZAMUPYH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320302740.94_warc_CC-MAIN-20220121071203-20220121101203-00380.warc.gz\"}"} |
http://cdbug.org/standard-error/linear-regression-prediction-standard-error.php | [
"Home > Standard Error > Linear Regression Prediction Standard Error\n\n# Linear Regression Prediction Standard Error\n\n## Contents\n\nThe least-squares estimates b0 and b1 are usually computed by statistical software. The estimated coefficient b1 is the slope of the regression line, i.e., the predicted change in Y per unit of change in X. These authors apparently have a very similar textbook specifically for regression that sounds like it has content that is identical to the above book but only the content related to regression The estimate for the response is identical to the estimate for the mean of the response: = b0 + b1x*. http://cdbug.org/standard-error/linear-model-prediction-standard-error.php\n\nOften X is a variable which logically can never go to zero, or even close to it, given the way it is defined. In linear regression, one wishes to test the significance of the parameter included. How to decipher Powershell syntax for text formatting? However, more data will not systematically reduce the standard error of the regression. http://onlinestatbook.com/2/regression/accuracy.html\n\n## Standard Error Of Prediction\n\nJoin for free An error occurred while rendering template. temperature What to look for in regression output What's a good value for R-squared? In a simple regression model, the standard error of the mean depends on the value of X, and it is larger for values of X that are farther from its own There are various formulas for it, but the one that is most intuitive is expressed in terms of the standardized values of the variables.\n\nFrom your table, it looks like you have 21 data points and are fitting 14 terms. The numerator is the sum of squared differences between the actual scores and the predicted scores. A scatterplot of the two variables indicates a linear relationship: Using the MINITAB \"REGRESS\" command with \"sugar\" as an explanatory variable and \"rating\" as the dependent variable gives the following result: Standard Error Of Estimate Calculator In a multiple regression model with k independent variables plus an intercept, the number of degrees of freedom for error is n-(k+1), and the formulas for the standard error of the\n\nThe test statistic t is equal to b1/sb1, the slope parameter estimate divided by its standard deviation. Standard Error Of Estimate Formula You can choose your own, or just report the standard error along with the point forecast. The standard error of the mean is usually a lot smaller than the standard error of the regression except when the sample size is very small and/or you are trying to https://www.researchgate.net/post/What_is_standard_error_of_prediction_from_linear_regression_with_known_SE_for_y-values All rights reserved.About us · Contact us · Careers · Developers · News · Help Center · Privacy · Terms · Copyright | Advertising · Recruiting We use cookies to give you the best possible experience on ResearchGate.\n\nIn a simple regression model, the percentage of variance \"explained\" by the model, which is called R-squared, is the square of the correlation between Y and X. Linear Regression Standard Error The terms in these equations that involve the variance or standard deviation of X merely serve to scale the units of the coefficients and standard errors in an appropriate way. Formulas for R-squared and standard error of the regression The fraction of the variance of Y that is \"explained\" by the simple regression model, i.e., the percentage by which the The estimated standard error of a prediction error is based on a sigma, but not of the population of y, but instead on the residuals, or for weighted least squares (WLS)\n\n## Standard Error Of Estimate Formula\n\nblog comments powered by Disqus Who We Are Minitab is the leading provider of software and services for quality improvement and statistics education. http://stats.stackexchange.com/questions/66946/how-are-the-standard-errors-computed-for-the-fitted-values-from-a-logistic-regre Is there a way to view total rocket mass in KSP? Standard Error Of Prediction more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed Standard Error Of The Regression You interpret S the same way for multiple regression as for simple regression.\n\nThe model is probably overfit, which would produce an R-square that is too high. http://cdbug.org/standard-error/linear-regression-standard-error.php However... 5. The standard error of the estimate is a measure of the accuracy of predictions. Red balls and Rings more hot questions default about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture Standard Error Of Regression Coefficient\n\nThanks for the question! In multiple regression output, just look in the Summary of Model table that also contains R-squared. The standard error of the forecast for Y at a given value of X is the square root of the sum of squares of the standard error of the regression and http://cdbug.org/standard-error/linear-regression-standard-error-vs-standard-deviation.php Why do people move their cameras in a square motion?\n\nAssume the data in Table 1 are the data from a population of five X, Y pairs. Standard Error Of Prediction In R Want to make things right, don't know with whom Why does Luke ignore Yoda's advice? Further, this demonstrates an analysis of this process. ----- Note that confidence bounds on b would make a wedge-shaped appearing figure within the predicted y bounds shown.\n\n## The numerator is the sum of squared differences between the actual scores and the predicted scores.\n\nThe confidence interval for the predicted value is given by + t*s, where is the fitted value corresponding to x*. The population standard deviation is STDEV.P.) Note that the standard error of the model is not the square root of the average value of the squared errors within the historical sample I am aware or robust 'sandwich' errors, eg, but those are for you betas, not for predicted y. –gung Jul 31 '14 at 4:27 2 Check out the car package. Standard Error Of Estimate Excel The critical value that should be used depends on the number of degrees of freedom for error (the number data points minus number of parameters estimated, which is n-1 for this\n\nYou don′t need to memorize all these equations, but there is one important thing to note: the standard errors of the coefficients are directly proportional to the standard error of the Obs Sugars Rating Fit StDev Fit Residual St Resid 1 6.0 68.40 44.88 1.07 23.52 2.58R 2 8.0 33.98 40.08 1.08 -6.09 -0.67 3 5.0 59.43 47.28 1.14 12.15 1.33 4 Your cache administrator is webmaster. news I think it should answer your questions.\n\nLane PrerequisitesMeasures of Variability, Introduction to Simple Linear Regression, Partitioning Sums of Squares Learning Objectives Make judgments about the size of the standard error of the estimate from a scatter plot Your article is informative, but my regression line does not go through the origin, the dependent variable is normally-distributed (by the Shapiro-Wilks test) and its variance is constant (rvariance,mean = +0.251, What does a profile's Decay Rate actually do? Why is JK Rowling considered 'bad at math'?\n\nRegressions differing in accuracy of prediction. The correlation between Y and X , denoted by rXY, is equal to the average product of their standardized values, i.e., the average of {the number of standard deviations by which The usual default value for the confidence level is 95%, for which the critical t-value is T.INV.2T(0.05, n - 2). Smaller values are better because it indicates that the observations are closer to the fitted line.\n\nThe standardized version of X will be denoted here by X*, and its value in period t is defined in Excel notation as: ... Each of the two model parameters, the slope and intercept, has its own standard error, which is the estimated standard deviation of the error in estimating it. (In general, the term Related 16How to understand output from R's polr function (ordered logistic regression)?8How do I run Ordinal Logistic Regression analysis in R with both numerical / categorical values?5How to evaluate fit of standard error of regression Hot Network Questions Publishing a mathematical research article on research which is already done?\n\nIf I denote the covariance matrix as $\\Sigma$ and and write the coefficients for my linear combination in a vector as $C$ then the standard error is just $\\sqrt{C' \\Sigma C}$ Technical questions like the one you've just found usually get answered within 48 hours on ResearchGate. The variance ² may be estimated by s² = , also known as the mean-squared error (or MSE). Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the\n\nis a privately owned company headquartered in State College, Pennsylvania, with subsidiaries in the United Kingdom, France, and Australia. Intuitively, that shifts the data far from pop=1029 without altering the regression line and therefore should result in a much wider prediction interval. The values fit by the equation b0 + b1xi are denoted i, and the residuals ei are equal to yi - i, the difference between the observed and fitted values. Smaller is better, other things being equal: we want the model to explain as much of the variation as possible.\n\nIn the regression output for Minitab statistical software, you can find S in the Summary of Model section, right next to R-squared."
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