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http://math.stackexchange.com/questions/289759/is-eigenvector-eigenvalues-of-a-sub-graph-similar-to-the-main-graph/289764 | # Is Eigenvector/eigenvalues of a sub graph similar to the main graph?
In Gephi I visualized a graph to calculate the eigenvalues,then I choose a portion of graph (e.g 6 vertex with their edges) and delete all others. I calculate the eigenvalues again and noticed that I get the same result.By same result I mean same ranking but with different values. The question is if this is happens by chance(I tried three times) or there is a mathematical fact back of it? If yes, what theorem should I search for? or if anybody can explain here and give a proof of it I'll be appreciate it.
-
Yes. The adjacency matrix for the subgraph is a principle submatrix of the adjacency matrix of the entire graph, so the eigenvalues will be related by interlacing inequalities. Here is one paper that discusses these inequalities and some graph-theoretic results derived from them.
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https://brilliant.org/problems/fan-of-this-fan/ | Fan of this fan
When a ceiling fan is switched off, its angular velocity reduces to half after it makes 36 rotations. How many more rotations will it make before coming to rest?
The angular deceleration of the fan is uniform.
× | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9457672238349915, "perplexity": 1233.1773053941706}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676594954.59/warc/CC-MAIN-20180723051723-20180723071723-00280.warc.gz"} |
https://ccssmathanswers.com/eureka-math-geometry-module-2-lesson-29/ | # Eureka Math Geometry Module 2 Lesson 29 Answer Key
## Engage NY Eureka Math Geometry Module 2 Lesson 29 Answer Key
### Eureka Math Geometry Module 2 Lesson 29 Example Answer Key
Example 1.
Scott, whose eye level is 1.5 m above the ground, stands 30 m from a tree. The angle of elevation of a bird at the top of the tree is 36°. How far above ground is the bird?
→ With respect to your diagram, think of the measurement you are looking for.
In our diagram, we are looking for BC.
→ How will you find BC?
I can use the tangent to determine BC in meters.
tan 36 = $$\frac{B C}{A C}$$
tan 36 = $$\frac{B C}{30}$$
30 tan 36 = BC
BC ≈ 21.8
→ Have we found the height at which the bird is off the ground?
No. The full height must be measured from the ground, so the distance from the ground to Scott’s eye level must be added to BC.
The height of the bird off of the ground is 1.5 m + 21.8 m = 23.3 m.
→ So, between the provided measurements, including the angle of elevation, and the use of the tangent ratio, we were able to find the height of the bird.
Example 2
From an angle of depression of 40°, John watches his friend approach his building while standing on the rooftop. The rooftop is 16 m from the ground, and John’s eye level is at about 1.8 m from the rooftop. What is the distance between John’s friend and the building?
Make sure to point out the angle of depression in the diagram below. Emphasize that the 40° angle of depression is the angle between the line of sight and the line horizontal (to the ground) from the eye.
→ Use the diagram to describe the distance we must determine.
We are going to find BC in meters.
→ How will we find BC?
We can use the tangent: tan 40 = $$\frac{A B}{B C}$$
tan 40 = $$\frac{A B}{B C}$$
tan 40 = $$\frac{17.8}{B C}$$
BC =
BC ≈ 21.2
→ Again, with the assistance of a few measurements, including the angle of depression, we were able to determine the distance between John’s friend and the building, which is 21.2 meters.
### Eureka Math Geometry Module 2 Lesson 29 Opening Exercise Answer Key
a. Use a calculator to find the tangent of θ. Enter the values, correct to four decimal places, in the last row of the table.
Note to the teacher: Dividing the values in the first two rows provides a different answer than using the full decimal readout of the calculator.
b. The table from Lesson 29 is provided here for you. In the row labeled $$\frac{\sin \theta}{\cos \theta}$$, divide the sine values by the cosine values. What do you notice?
For each of the listed degree values, dividing the value of sin θ by the corresponding value of cos θ yields a value very close to the value of tan θ. The values divided to obtain $$\frac{\sin \theta}{\cos \theta}$$ were approximations, so the actual
values might be exactly the same.
### Eureka Math Geometry Module 2 Lesson 29 Exercise Answer Key
Exercise 1.
Standing on the gallery of a lighthouse (the deck at the top of a lighthouse), a person spots a ship at an angle of depression of 20°. The lighthouse is 28 m tall and sits on a cliff 45 m tall as measured from sea level. What is the horizontal distance between the lighthouse and the ship? Sketch a diagram to support your answer.
Approximately 201 m
Exercise 2.
A line on the coordinate plane makes an angle of depression of 36°. Find the slope of the line correct to four decimal places.
Choose a segment on the line, and construct legs of a right triangle parallel to the x- and y-axes, as shown. If m is the length of the vertical leg and n is the length of the horizontal leg, then tan 36 = $$\frac{m}{n}$$. The line decreases to the right, so the value of the slope must be negative. Therefore,
slope = –$$\frac{m}{n}$$ = – tan 36 ≈ – 0.7265.
### Eureka Math Geometry Module 2 Lesson 29 Problem Set Answer Key
Question 1.
A line in the coordinate plane has an angle of elevation of 53°. Find the slope of the line correct to four decimal places.
Since parallel lines have the same slope, we can consider the line that passes through the origin with an angle of inclination of 53°. Draw a vertex at (5, 0) on the x-axis, and draw a segment from (5, 0) parallel to the y-axis to the intersection with the line to form a right triangle. If the base of the right triangle is 5-units long, let the height of the triangle be represented by y.
tan 53 = $$\frac{x}{5}$$
5(tan 53) = x
The slope of the line is
slope = tan 53 ≈ 1.3270.
Question 2.
A line in the coordinate plane has an angle of depression of 25°. Find the slope of the line correct to four decimal places.
A line crosses the x-axis, decreasing to the right with an angle of depression of 25°. Using a similar method as in Problem 1, a right triangle is formed. If the leg along the x-axis represents the base of the triangle, let h represent the height of the triangle.
Using tangent:
tan 25 = $$\frac{h}{5}$$
5(tan 25) = h
The slope of the line is negative since the line decreases to the right:
Question 3.
In Problems 1 and 2, why do the lengths of the legs of the right triangles formed not affect the slope of the line?
When using the tangent ratio, the length of one leg of a right triangle can be determined in terms of the other leg. Let x represent the length of the horizontal leg of a slope triangle. The vertical leg of the triangle is then x tan θ, where θ is the measure of the angle of inclination or depression. The slope of the line is
Question 4.
Given the angles of depression below, determine the slope of the line with the Indicated angle correct to four decimal places.
a. 35° angle of depression
tan 35 ≈ 0.7002
slope ≈ – 0.7002
b. 49° angle of depression
tan 49 ≈ 1. 1504
slope ≈ – 1.1504
c. 80° angle of depression
tan 80 ≈ 5. 6713
slope ≈ – 5. 6713
d. 87° angle of depression
tan 87 ≈ 19. 0811
slope ≈ – 19. 0811
e. 89° angle of depression
tan 89 ≈ 57. 2900
slope ≈ – 57.2900
f. 89.9° angle of depression
tan 89.9 ≈ 572.9572
slope ≈ – 572.9572
g. What appears to be happening to the slopes (and tangent values) as the angles of depression get closer to 90°?
As the angles get closer to 90°, their slopes (and tangent values) get much farther from zero.
h. Find the slopes of angles of depression that are even closer to 90° than 89.90. Can the value of the tangent of 90° be defined? Why or why not?
Choices of angles will vary. The closer an angle is in measure to 90°, the greater the tangent value of that angle and the farther the slope of the line determined by that angle is from zero. An angle of depression of 90° would be a vertical line, and vertical lines have 0 run; therefore, the value of the ratio rise: run is undefined. The value of the tangent of 90° would have a similar outcome because the adjacent leg of the “triangle” would have a length of 0, so the ratio $$\frac{\text { opp }}{\text { adj }}=\frac{\text { opp }}{0}$$, which is undefined.
Question 5.
For the indicated angle, express the quotient in terms of sine, cosine, or tangent. Then, write the quotient in simplest terms.
a. $$\frac{4}{2 \sqrt{13}}$$; α
cos α = $$\frac{4}{2 \sqrt{13}}=\frac{2}{\sqrt{13}}=\frac{2 \sqrt{13}}{13}$$
b. $$\frac{6}{4}$$; α
tan α = $$\frac{6}{4}=\frac{3}{2}$$
c. $$\frac{4}{2 \sqrt{13}}$$; β
sin β = $$\frac{4}{2 \sqrt{13}}=\frac{2 \sqrt{13}}{13}$$
d. $$\frac{4}{6}$$; β
tan β = $$\frac{4}{6}=\frac{2}{3}$$
Question 6.
The pitch of a roof on a home Is expressed as a ratio of vertical rise: horizontal run where the run has a length of 12 units. If a given roof design includes an angle of elevation of 22. 5° and the roof spans 36 ft. as shown in the diagram, determine the pitch of the roof. Then, determine the distance along one of the two sloped surfaces of the roof.
The diagram as shown is an isosceles triangle since the base angles have equal measure. The altitude, a, of the triangle is the vertical rise of the roof.
The right triangles formed by drawing the altitude of the given isosceles triangle have a leg of length 18 ft.
tan 225 = $$\frac{a}{18}$$
a = 18 tan 22.5
a ≈ 7.5
Roof pitch:
$$\frac{7.5}{18}=\frac{h}{12}$$
18h = 12 · 7.5
h = 5
cos 22.5 = $$\frac{18}{s}$$
s = $$\frac{18}{\cos 22.5}$$
s ≈ 19.5
The pitch of the roof is 5:12.
The sloped surface of the roof has a distance of approximately 19.5 ft.
Question 7.
An anchor cable supports a vertical utility pole forming a 51° angle with the ground. The cable is attached to the top of the pole. If the distance from the base of the pole to the base of the cable is 5 meters, how tall is the pole?
Let h represent the height of the pole in meters.
Using tangent:
tan 51 = $$\frac{h}{5}$$
5(tan 51) = h
6.17 ≈ h
The height of the utility pole is approximately 6. 17 meters.
Question 8.
A winch is a tool that rotates a cylinder, around which a cable is wound. When the winch rotates in one direction, it draws the cable in. Joey is using a winch and a pulley (as shown in the diagram) to raise a heavy box off the floor and onto a cart. The box is 2 ft. tall, and the winch is 14 ft. horizontally from where cable drops down vertically from the pulley. The angle of elevation to the pulley is 42°. What is the approximate length of cable required to connect the winch and the box?
Let h represent the length of cable in the distance from the winch to the pulley along the hypotenuse of the right triangle shown in feet, and let y represent the distance from the pulley to the floor in feet.
Using tangent:
tan 42 = $$\frac{y}{14}$$
14(tan 42) = y
12.61 ≈ y
Using cosine:
cos 42 = $$\frac{14}{h}$$
h = $$\frac{14}{\cos 42}$$
h ≈ 18.84
Let t represent the total amount of cable from the winch to the box in feet:
t ≈ 12.61 + 18.84 – 2
t ≈ 29.45
The total length of cable from the winch to the box is approximately 29.45 ft.
### Eureka Math Geometry Module 2 Lesson 29 Exit Ticket Answer Key
Question 1.
A line on the coordinate plane makes an angle of depression of 24°. Find the slope of the line correct to four decimal places.
Choose a segment on the line, and construct legs of a right triangle parallel to the x- and y-axes, as shown. If m is the length of the vertical leg and n is the length of the horizontal leg, then tan 24 = $$\frac{m}{n}$$ The line decreases to the right, so the value of the slope must be negative. Therefore,
slope = – $$\frac{m}{n}$$ = – tan 24 ≈ – 0.4452.
Question 2.
Samuel is at the top of a tower and will ride a trolley down a zip line to a lower tower. The total vertical drop of the zip line is 40 ft. The zip line’s angle of elevation from the lower tower is 11. 5°. What is the horizontal distance between the towers?
tan 11.5 = $$\frac{40}{x}$$
x = $$\frac{40}{\tan 11.5}$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8351880311965942, "perplexity": 757.6932416794709}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948965.80/warc/CC-MAIN-20230329085436-20230329115436-00100.warc.gz"} |
http://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-r-section-r-2-fractions-practice-page-r-10/10 | ## Algebra: A Combined Approach (4th Edition)
$\frac{3}{10}$
1. Write numerator and denominator as product of primes: $\frac{12}{40}$ = $\frac{2 * 2 * 3}{2 * 2 * 2 * 5}$ 2. Separate common factors by writing as fraction equal to 1: $\frac{3}{2 * 5}$ * $\frac{4}{4}$ = $\frac{3}{2 * 5}$ = $\frac{3}{10}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9388158917427063, "perplexity": 597.2959804655248}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917119361.6/warc/CC-MAIN-20170423031159-00098-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/107498-evaluate-limits.html | # Math Help - Evaluate the Limits
1. ## Evaluate the Limits
Question: If $f(9) = 9 , f'(9) = 4$ ,evaluate $\lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3 }$
2. Originally Posted by zorro
Question: If $f(9) = 9 , f'(9) = 4$ ,evaluate $\lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3 }$
As both the numerator and the denominator in your function approach 0 when x --> 9 and both are derivable functions around x = 9, you can Use L'Hospital's Rule here. The answer is 4.
Tonio
3. ## Is this correct?
Originally Posted by tonio
As both the numerator and the denominator in your function approach 0 when x --> 9 and both are derivable functions around x = 9, you can Use L'Hospital's Rule here. The answer is 4.
Tonio
Here is what i have done
$\lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{ \sqrt{3} - 3}$
Using L'Hospitals rule
$g'(x)$ = $\frac{d}{dx} \left[ \sqrt{f(x)} - 3 \right]$ = $\frac{f'(x)}{2 \sqrt{f(x)}}$ ...............eq(1)
$h'(x)$ = $\frac{d}{dx} \left[ \sqrt{x} - 3 \right]$ = $\frac{1}{a \sqrt{x}}$ ..................eq(2)
Now taking lim of eq (1) and eq(2) we get
$\lim_{x \to 9} \left[ \frac{g'(x)}{h'(x)} \right]$ = $\lim_{x \to 9} \left[ \frac{f'(x)}{2 \sqrt{f(x)}} . \frac{2 \sqrt{x}}{1} \right]$ = $\frac{4}{2 \sqrt{9}} . \frac{2 \sqrt{9}}{1}$ = $4 ........................Is \ this \ correct \ ? \ ? \ ? \ ? \ ? \ ?$
4. Originally Posted by zorro
Here is what i have done
$\lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{ \sqrt{3} - 3}$
Using L'Hospitals rule
$g'(x)$ = $\frac{d}{dx} \left[ \sqrt{f(x)} - 3 \right]$ = $\frac{f'(x)}{2 \sqrt{f(x)}}$ ...............eq(1)
$h'(x)$ = $\frac{d}{dx} \left[ \sqrt{x} - 3 \right]$ = $\frac{1}{a \sqrt{x}}$ ..................eq(2)
Now taking lim of eq (1) and eq(2) we get
$\lim_{x \to 9} \left[ \frac{g'(x)}{h'(x)} \right]$ = $\lim_{x \to 9} \left[ \frac{f'(x)}{2 \sqrt{f(x)}} . \frac{2 \sqrt{x}}{1} \right]$ = $\frac{4}{2 \sqrt{9}} . \frac{2 \sqrt{9}}{1}$ = $4 ........................Is \ this \ correct \ ? \ ? \ ? \ ? \ ? \ ?$
Yes. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 26, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.957998514175415, "perplexity": 863.4119951095288}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999664205/warc/CC-MAIN-20140305060744-00019-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://puzzling.stackexchange.com/questions/79248/pucks-in-the-arena | # Pucks in the arena
Two identical pucks of radius 10 cm are placed in a round arena of radius 1 m. They are positioned 50 cm away from the center of the arena on opposing sides. Assuming no energy losses during sliding and collisions, can you hit one of the pucks such that the two pucks touch exactly once and then never again? If so, how can you do it?
Notes:
• The collisions conserve energy and momentum.
• The walls of the arena don't move.
• As there is no friction the pucks can't exchange angular momentum.
• Please don't invent additional dimensions. Everybody knows that the universe is two-dimensional.
• Poincaré recurrence is applicable: en.wikipedia.org/wiki/Poincar%C3%A9_recurrence_theorem Feb 4, 2019 at 15:45
• @Johannes This is perfectly true, but would you please place this comment under the answer instead of here? Otherwise it's a too direct hint for everyone familiar with physics. Feb 4, 2019 at 21:21
Without actually solving the problem formally I hypothesize that the answer is
No
My reasoning is that
The dynamics of perfectly elastic collisions is reversible in time. So if there existed a dynamical state where a collision occurs followed by no collisions forever, then that would be equivalent to being able to define a state where no collision occurs forever, but then a collision follows. Which doesn't make logical sense to my way of thinking.
I presume that there is probably a simple geometrical conjecture involving rational angles that could confirm the above. Proof that irrational angles don't produce a solution is as follows:
If both the balls bounces off the wall at an irrational angle (as a fraction of $$360°$$ or $$2\pi$$ radians), then the balls must collide eventually. Just consider stretching their future collision times along a single time axis. A ball $$a$$ or $$b$$ that does not collide with any other will hit the wall at constant time interval $$T_a$$ or $$T_b$$ and each collision with the wall will precess some angle $$\alpha_a$$ or $$\alpha_b$$, where $$–\pi \le \alpha_a \le \pi$$, around arena. So at the $$n$$th collision the ball a will collide with the wall at position $$n\alpha_a \, \text{mod} \, 2\pi$$ at time $$n T_a$$. As angle $$\alpha_a$$ is an irrational fraction of $$2\pi$$, an $$n$$ can be chosen such that the ball is arbitrarily close to the zenith ($$0°$$) position of the arena, and similarly an $$m$$ can be chosen such that the ball will arbitrarily closely revisit that position after a further $$p$$ (or $$2p$$, or $$3p…kp$$) collisions – where $$m$$ and $$k$$ are functions of what you define as arbitrarily close. If both balls are bouncing irrationally then they each get arbitrarily close to the zenith after $$p$$ and $$q$$ collisions respectively at time intervals $$p T_a$$ and $$q T_b$$. Now identify a fraction $$h/k$$ which is arbitrarily close to $$p T_a / q T_b$$ , and after $$kp$$ and $$hq$$ collisions each, the two balls will both be arbitrarily close to the zenith within an arbitrarily short time period and will collide. The same argument still holds if one ball is cycling a rational angle.
So the only remaining possibility is that both balls bounce at rational angle increments after the collision and manage to miss each other, which I still think is unlikely, although it is obviously possible to set two balls bouncing so they never collide...
If such a rational solution exists after one initial collision then:
Both the balls bounce rational angles so each repeats its path exactly after some $$p$$ and $$q$$ collisions. In this case the repeat period for each ball will be $$p T_a$$, and $$q T_b$$ respectively. There is no guarantee that the repeat times are rational or rational ratios of each other, but regardless of that, you can choose some large number of repeat periods $$u$$ and $$v$$ for each ball such that $$u p T_a$$ is arbitrarily equal to $$v q T_b$$. At the next collision after that time, the two balls will strike the walls at positions and relative times arbitrarily close to those that occurred at the time of their first collision with the wall (after they initially struck each other). At that precise moment (when the pattern repeats), trace the ball's positions backwards and you will observe that their time-reversed paths are arbitrarily close to their initial paths after they struck each other. So tracing it backwards they would strike again (or to put it another way – they couldn't get to the arbitrarily identical positions without colliding with each other immediately beforehand).
I think that is sufficient to confirm the hypothesis in all possible cases other than zero initial velocity. I'm sure things could be formalized but I will stick with the hand-wavy "proof".
• We could remove the arena walls and your reasoning still applies. Feb 3, 2019 at 23:20
• I agree with your comment completely - but within a finite arena I believe the "logical sense" line still holds in the sense that "I cant't point out the exact problem, but something smells wrong with the consistency of such a dynamic". That is why I only present it as an informal hypothesis. Feb 3, 2019 at 23:23
• Hmmm, yes, finite arena (with puck of non-zero radius). I think you're right actually... Feb 3, 2019 at 23:27
• rot13(engvbany => crevbqvp, bevtvany nethzrag nccyvrf) Feb 4, 2019 at 0:06
• Rational for both pucks doesn't apply to my original argument as then the contact positions of each puck do not precess around the arena(in the sense that there will be many angles on the wall of the arena that a ball will never get arbitrarily close to). Instead, after some p*q collisions, both balls will be in teh exact same places again and will repeat their dance. But that gives me an idea... Feb 4, 2019 at 1:45
I'm on the same boat with @Penguino, but there's the edge case still to consider:
Can the pucks come to touching distance of each other without actually colliding?
If that were the case, then it could theoretically be possible for the pucks to complete arbitrarily many synchronised loops of some kind without actually colliding, and it would be possible to have a collision in the mix as well.
This corner case seems equivalent to this problem:
What are the constant non-zero speeds (directions given) of these two points, given that they never came closer than ten units of each other?
If there's a solution to this problem, then there could be a solution to the whole, even given @Penguino's excellent argument: after the first collision the pucks must be moving at right angles because of the conservation laws, and they must repeat the position later, (because infinity is quite big), but given an answer, this could be a way to get around @Penguino's argument.
I think the answer must be "nuh-uh, can't do that, $$\text{cos}(\theta)\approx1$$ for small $$\theta$$, and infinite speed ain't cool", but I'm not exactly sober, and I'm way past my bedtime even without that, so I'm afraid I'll have to leave it to you guys. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 40, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8283920884132385, "perplexity": 551.0537843558175}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00203.warc.gz"} |
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## theslytherinhelper 2 years ago Algebra 2, Solving Polynomial Equations Determine the zeros of f(x) = x4 – x3 + 7x2 – 9x – 18. Delete Cancel Submit
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1. theslytherinhelper
• 2 years ago
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I've gotten most of the work down, if you'd like to see it. I'm stuck on the last step :(
2. amistre64
• 2 years ago
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youre not doing ferrari are you?
3. theslytherinhelper
• 2 years ago
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ferrari?
4. amistre64
• 2 years ago
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x4 – x3 + 7x2 – 9x – 18=0 to get rid of the x3 term, let x=(y+1/4) (y+1/4)^4 – (y+1/4)^3 + (y+1/4)^2 – 9(y+1/4) – 18=0 $y^4+\frac58y^2-\frac{69}8y-\frac{5171}{256}=0$ put the linear parts on the other side $y^4+\frac58y^2=\frac{69}8y+\frac{5171}{256}$ complete the square on the left by adding 5y^2/8+25/64 to each side $(y^2+\frac58)^2=\frac58y^2+\frac{69}8y+\frac{5271}{256}$ this is as far as we can go without introducing another variable, like z adding a z into the left side gives us more room to play with, but we have to adjust the right side by the same amount: (a+b+c)^2 = a^2+2ab+b^2+(c^2+2c(a+b)) so lets input the z on the left, and add z^2 + 2z(y^2+5/8) to the right side $(y^2+\frac58+z)^2=\frac58y^2+\frac{69}8y+\frac{5271}{256}+z^2 + 2z(y^2+\frac58)$ lets clean up the right side to see it in its quadraitc form for y $(y^2+\frac58+z)^2=(2z+\frac58)y^2+\frac{69}8y+(z^2+\frac{10}{8}z+\frac{5271}{256})$ solving the right side for yusing the quadratic formula we get: $\Large y=\frac{-\frac{69}{8}\pm\sqrt{(\frac{69}{8})^2-4(2z+\frac58)(z^2+\frac{10}{8}z+\frac{5271}{256})}}{2(2z+\frac58)}$ if we can get the discriminant to equal zero, for some reason, the method will give us roots ... so $(\frac{69}{8})^2-4(2z+\frac58)(z^2+\frac{10}{8}z+\frac{5271}{256})=0$ which simplifies (lol) to:$8z^3+\frac{25}{2}z^2+\frac{5371}{32}z-\frac{11733}{512}=0$which is just Cardanos method now.
5. amistre64
• 2 years ago
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Cardanos method is to reduse it by removing the z^2; let z=(u-25/48) 8(u-25/48)^3+25(u-25/48)^2/2+5371(u-25/48)/32-11733/512 which reduces to:$u^3+\frac{484}{24}u=\frac{2918}{216}$ hence: $u=\sqrt[3]{\sqrt{\frac{(2918/216)^2}{4}+\frac{(484/24)^3}{27}}+\frac{(2918/216)}{2}}\\~~~~~-\sqrt[3]{\sqrt{\frac{(2918/216)^2}{4}+\frac{(484/24)^3}{27}}-\frac{(2918/216)}{2}}$ $u=...$
6. amistre64
• 2 years ago
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lol, im pretty sure i might have mistyped a number or two, but if youve never heard of ferrari then im sure theres a simpler way they want you to look at this
7. amistre64
• 2 years ago
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8. theslytherinhelper
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Thanks sooo much! The last step is simplifying the equation. ^_^
9. amistre64
• 2 years ago
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im not sure of the method you used, so ill need to know a little more about how you approached this
10. amistre64
• 2 years ago
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my u wasnt all the way simplified $u^3+\frac{121}{6}u=\frac{1459}{108}$ might make life on my method here a smidge simpler $u = \sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}-\sqrt[3]{\frac{\sqrt{22361}}{8}-\frac{1459}{216}}=abt~~0.6559$ $z=u-\frac{25}{48}=\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}-\sqrt[3]{\frac{\sqrt{22361}}{8}-\frac{1459}{216}}-\frac{25}{48}$ plugging this into our (y^2...)^2 = y^2+... setup with the zs gives us $(y^2+\frac58+z)^2=(2z+\frac58)y^2+\frac{69}8y+(z^2+\frac{10}{8}z+\frac{5271}{256})$ $(y^2+\frac58+(\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}-\sqrt[3]{\frac{\sqrt{22361}}{8}-\frac{1459}{216}}-\frac{25}{48}))^2\\=(2(\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}-\sqrt[3]{\frac{\sqrt{22361}}{8}-\frac{1459}{216}}-\frac{25}{48})+\frac58)y^2+\frac{69}8y\\+((\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}-\sqrt[3]{\frac{\sqrt{22361}}{8}-\frac{1459}{216}}-\frac{25}{48})^2\\+\frac{10}{8}(\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}-\sqrt[3]{\frac{\sqrt{22361}}{8}-\frac{1459}{216}}-\frac{25}{48})+\frac{5271}{256}$
11. amistre64
• 2 years ago
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im sooo glad for the wolfram :)
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9924290180206299, "perplexity": 3158.2693559892828}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936463181.39/warc/CC-MAIN-20150226074103-00239-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://vectortutorials.in/maths-formulas-for-class-8-pdf/ | # Maths Formulas for Class 8 PDF
Maths formulas for class 8 include the formulas from all the chapters in NCERT class 8 Maths book. Here you will find all the formulas in a PDF file, including Rational Number, Algebraic Expressions, Exponents and Powers, Mensuration, and other chapters.
These formulas for class 8 are beneficial while preparing for your exam. You can get the printout or read the formulas online. Please note that you need to practice more and more to memorize all the maths formulas. You can also review the essential maths formulas for upcoming classes (i.e., Class 9 and Class 10). We have prepared this formula PDF sheet based on the NCERT & CBSE syllabus. We will soon update Hindi medium study materials.
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## Rational Numbers Maths Formulas for Class 8
In class 8, you will learn about another type of number. i.e., Rational Numbers. In this section, we have provided maths formulas for class 8 to study rational numbers. You will also find essential properties of rational numbers and other numbers.
If r = p/q, then r is a rational number.
Where p and q are integers and q≠0. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9500235319137573, "perplexity": 1103.6714936974922}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358323.91/warc/CC-MAIN-20211127223710-20211128013710-00491.warc.gz"} |
http://planetmath.org/SecondFundamentalForm | # second fundamental form
In classical differential geometry the second fundamental form is a symmetric bilinear form defined on a differentiable surface $M$ embedded in $\mathbb{R}^{3}$, which in some sense measures the curvature of $M$ in space.
To construct the second fundamental form requires a small digression. After the digression we will discuss how it relates to the curvature of $M$.
## Construction of the second fundamental form
Consider the tangent planes $\mathrm{T}_{p}M$ of the surface $M$ for each point $p\in M$. There are two unit normals to $\mathrm{T}_{p}M$. Assuming $M$ is orientable, we can choose one of these unit normals (http://planetmath.org/MutualPositionsOfVectors), $n(p)$, so that $n(p)$ varies smoothly with $p$.
Since $n(p)$ is a unit vector in $\mathbb{R}^{3}$, it may be considered as a point on the sphere $S^{2}\subset\mathbb{R}^{3}$. Then we have a map $n\colon M\to S^{2}$. It is called the normal map or Gauss map.
The second fundamental form is the tensor field $\mathcal{II}$ on $M$ defined by
$\mathcal{II}_{p}(\xi,\eta)=-\langle\operatorname{D}n_{p}(\xi),\eta\rangle\,,% \quad\xi,\eta\in\mathrm{T}_{p}M\,,$ (1)
where $\langle,\rangle$ is the dot product of $\mathbb{R}^{3}$, and we consider the tangent planes of surfaces in $\mathbb{R}^{3}$ to be subspaces of $\mathbb{R}^{3}$.
The linear transformation $\operatorname{D}n_{p}$ is in reality the tangent mapping $\operatorname{D}n_{p}\colon\mathrm{T}_{p}M\to\mathrm{T}_{n(p)}S^{2}$, but since $\mathrm{T}_{n(p)}S^{2}=\mathrm{T}_{p}M$ by the definition of $n$, we prefer to think of $\operatorname{D}n_{p}$ as $\operatorname{D}n_{p}\colon\mathrm{T}_{p}M\to\mathrm{T}_{p}M$.
The tangent map $\operatorname{D}n$, is often called the Weingarten map.
###### Proposition 1.
The second fundamental form is a symmetric form.
###### Proof.
This is a computation using a coordinate chart $\sigma$ for $M$. Let $u,v$ be the corresponding names for the coordinates. From the equation
$\left\langle n,\frac{\partial\sigma}{\partial v}\right\rangle=0\,,$
differentiating with respect to $u$ using the product rule gives
$\begin{split}\displaystyle\left\langle n,\frac{\partial^{2}\sigma}{\partial u% \partial v}\right\rangle&\displaystyle=-\left\langle\frac{\partial n}{\partial u% },\frac{\partial\sigma}{\partial v}\right\rangle\\ &\displaystyle=-\left\langle\operatorname{D}n\left(\frac{\partial\sigma}{% \partial u}\right),\frac{\partial\sigma}{\partial v}\right\rangle\\ &\displaystyle=\mathcal{II}\left(\frac{\partial\sigma}{\partial u},\frac{% \partial\sigma}{\partial v}\right)\,.\end{split}$ (2)
(The second equality follows from the definition of the tangent map $\operatorname{D}n$.) Reversing the roles of $u,v$ and repeating the last derivation, we obtain also:
$\left\langle n,\frac{\partial^{2}\sigma}{\partial u\partial v}\right\rangle=% \left\langle n,\frac{\partial^{2}\sigma}{\partial v\partial u}\right\rangle=% \mathcal{II}\left(\frac{\partial\sigma}{\partial v},\frac{\partial\sigma}{% \partial u}\right)\,.$ (3)
Since $\partial\sigma/\partial u$ and $\partial\sigma/\partial v$ form a basis for $\mathrm{T}_{p}M$, combining (2) and (3) proves that $\mathcal{II}$ is symmetric. ∎
In view of Proposition 1, it is customary to regard the second fundamental form as a quadratic form, as it done with the first fundamental form. Thus, the second fundamental form is referred to with the following expression11 Unfortunately the coefficient $M$ here clashes with our use of the letter $M$ for the surface (manifold), but whenever we write $M$, the context should make clear which meaning is intended. The use of the symbols $L,M,N$ for the coefficients of the second fundamental form is standard, but probably was established long before anyone thought about manifolds.:
$L\,du^{2}+2M\,dudv+N\,dv^{2}\,.$
Compare with the tensor notation
$\mathcal{II}=L\,du\otimes du+M\,du\otimes dv+M\,dv\otimes du+Ndv\otimes dv\,.$
Or in matrix form (with respect to the coordinates $u,v$),
$\mathcal{II}=\begin{pmatrix}L&M\\ M&N\end{pmatrix}\,.$
## Curvature of curves on a surface
Let $\gamma$ be a curve lying on the surface $M$, parameterized by arc-length. Recall that the curvature $\kappa(s)$ of $\gamma$ at $s$ is $\gamma^{\prime\prime}(s)$. If we want to measure the curvature of the surface, it is natural to consider the component of $\gamma^{\prime\prime}(s)$ in the normal $n(\gamma(s))$. Precisely, this quantity is
$\langle\gamma^{\prime\prime}(s),n(\gamma(s))\rangle\,,$
and is called the normal curvature of $\gamma$ on $M$.
So to study the curvature of $M$, we ignore the component of the curvature of $\gamma$ in the tangent plane of $M$. Also, physically speaking, the normal curvature is proportional to the acceleration required to keep a moving particle on the surface $M$.
We now come to the motivation for defining the second fundamental form:
###### Proposition 2.
Let $\gamma$ be a curve on $M$, parameterized by arc-length, and $\gamma(s)=p$. Then
$\langle\gamma^{\prime\prime}(s),n(p)\rangle=\mathcal{II}\bigl{(}\gamma^{\prime% }(s),\gamma^{\prime}(s)\bigr{)}\,.$
###### Proof.
From the equation
$\langle n(\gamma(s)),\gamma^{\prime}(s)\rangle=0\,,$
differentiate with respect to $s$:
$\displaystyle\langle n(\gamma(s)),\gamma^{\prime\prime}(s)\rangle$ $\displaystyle=-\left\langle\frac{d}{ds}n(\gamma(s)),\gamma^{\prime}(s)\right\rangle$ $\displaystyle=-\bigl{\langle}\operatorname{D}n(\gamma^{\prime}(s)),\gamma^{% \prime}(s)\bigr{\rangle}$ $\displaystyle=\mathcal{II}\bigl{(}\gamma^{\prime}(s),\gamma^{\prime}(s)\bigr{)% }\,.\qed$
It is now time to mention an important consequence of Proposition 1: the fact that $\mathcal{II}$ is symmetric means that $-\operatorname{D}n$ is self-adjoint with respect to the inner product $\mathcal{I}$ (the first fundamental form). So, if $-\operatorname{D}n$ is expressed as a matrix with orthonormal coordinates (with respect to $\mathcal{I}$), then the matrix is symmetric. (The minus sign in front of $\operatorname{D}n$ is to make the formulas work out nicely.)
Certain theorems in linear algebra tell us that, $-\operatorname{D}n_{p}$ being self-adjoint, it has an orthonormal basis of eigenvectors $e_{1},e_{2}$ with corresponding eigenvalues $\kappa_{1}\leq\kappa_{2}$. These eigenvalues are called the principal curvatures of $M$ at $p$. The eigenvectors $e_{1},e_{2}$ are the principal directions. The terminology is justified by the following theorem:
###### Theorem 1 (Euler’s Theorem).
The normal curvature of a curve $\gamma$ has the form
$\langle\gamma^{\prime\prime}(s),n(p)\rangle=\kappa_{1}\,\cos^{2}\theta+\kappa_% {2}\,\sin^{2}\theta\,,\quad p=\gamma(s)\,.$
It follows that the minimum possible normal curvature is $\kappa_{1}$, and the maximum possible is $\kappa_{2}$.
###### Proof.
Since $e_{1},e_{2}$ form an orthonormal basis for $\mathrm{T}_{p}M$, we may write
$\gamma^{\prime}(s)=\cos\theta\,e_{1}+\sin\theta\,e_{2}$
for some angle $\theta$. Then
$\displaystyle\langle\gamma^{\prime\prime}(s),n(p)\rangle$ $\displaystyle=\mathcal{II}\bigl{(}\gamma^{\prime}(s),\gamma^{\prime}(s)\bigr{)}$ $\displaystyle=\bigl{\langle}-\operatorname{D}n_{p}(\gamma^{\prime}(s)),\gamma^% {\prime}(s)\bigr{\rangle}$ $\displaystyle=\langle\kappa_{1}\cos\theta\,e_{1}+\kappa_{2}\sin\theta\,e_{2},% \cos\theta\,e_{1}+\sin\theta\,e_{2}\rangle$ $\displaystyle=\kappa_{1}\,\cos^{2}\theta+\kappa_{2}\,\sin^{2}\theta\,.\qed$
## Matrix representations of second fundamental form and Weingarten map
At this point, we should find the explicit prescriptions for calculating the second fundamental form and the Weingarten map.
Let $\sigma$ be a coordinate chart for $M$, and $u,v$ be the names of the coordinates. For a test vector $\xi\in\mathrm{T}_{p}M$, we write $\xi_{u}$ and $\xi_{v}$ for the $u,v$ coordinates of $\xi$.
We compute the matrix $W$ for $-\operatorname{D}n$ in $u,v$-coordinates. We have
$\displaystyle\begin{pmatrix}\xi_{u}&\xi_{v}\end{pmatrix}\begin{pmatrix}L&M\\ M&N\end{pmatrix}\begin{pmatrix}\xi_{u}\\ \xi_{v}\end{pmatrix}$ $\displaystyle=\mathcal{II}(\xi,\xi)=\langle-\operatorname{D}n(\xi),\xi\rangle$ $\displaystyle=\left(Q\begin{pmatrix}\xi_{u}\\ \xi_{v}\end{pmatrix}\right)^{\mathrm{T}}QW\begin{pmatrix}\xi_{u}\\ \xi_{v}\end{pmatrix}$ $\displaystyle=\begin{pmatrix}\xi_{u}&\xi_{v}\end{pmatrix}\,(Q^{\mathrm{T}}Q)\,% W\begin{pmatrix}\xi_{u}\\ \xi_{v}\end{pmatrix}\,,$
where $Q$ is the matrix that changes from $u,v$-coordinates to orthonormal coordinates for $\mathrm{T}_{p}M$ — this is necessary to compute the inner product. But
$Q^{\mathrm{T}}Q=\begin{pmatrix}E&F\\ F&G\end{pmatrix}=\mathcal{I}\quad\text{(the first fundamental form),}$
because $Q$ is the matrix with columns $\partial\sigma/\partial u$ and $\partial\sigma/\partial v$ expressed in orthonormal coordinates.
(More to be written…)
## References
Title second fundamental form Canonical name SecondFundamentalForm Date of creation 2013-03-22 15:29:02 Last modified on 2013-03-22 15:29:02 Owner stevecheng (10074) Last modified by stevecheng (10074) Numerical id 5 Author stevecheng (10074) Entry type Definition Classification msc 53A05 Related topic FirstFundamentalForm Related topic ShapeOperator Related topic NormalSection Related topic NormalCurvatures Defines normal curvature Defines principal direction Defines principal curvature Defines Weingarten matrix Defines Weingarten map Defines Gauss map | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 123, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9919683933258057, "perplexity": 144.793325317795}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647706.69/warc/CC-MAIN-20180321215410-20180321235410-00045.warc.gz"} |
http://physics.stackexchange.com/tags/symmetry-breaking/new | # Tag Info
2
To grasp the relevant physics at a sloppy level, perhaps you simply need a few examples. You know a concept is commonly constructed by the manner you refer to it together with other concepts. Symmetry breaking usually results in ground state degeneracy and long range order. Order parameter field aids you in identifying degenerate sectors with the symmetries ...
2
There is at least one philosopher before Plato and he is Anaximander. There are many passages in his works that relate to the concept of symmetry: The basic elements of nature (water, air, fire, earth) which the first Greek philosophers believed that constituted the universe represent in fact the primordial forces of previous thought. Their collision ...
1
To start with, special relativity is necessary when describing elementary particles and hadrons composed by them. Special relativity defines the invariant mass of a particle or an ensemble of particles as the measure of the four momentum vector carried by the particle/ensemble. In the case of one elementary particle, an electron for example, it is ...
0
99.9% of the mass of a hadron or a meson comes from confinement in QCD. Confinement is a special feature of QCD due to its non abelian symmetry which leads to a negative beta function. It is confinement that also leads to a breaking of the chiral symmetry at about 200 MeV or the radius of a hadron (about 1 femto meter).
4
The dilaton $\sigma$ is the Goldstone boson of scale invariance. Scale transformations $x\rightarrow x/\lambda$ are generated non linearly, e.g. $$\sigma(x)\rightarrow \sigma(\lambda x)+f \log\lambda\,,\qquad \lambda>0$$ where $f$ is the dilaton decay constant (see below). An effective field theory for this Goldstone boson can be easily written with ...
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http://math.stackexchange.com/questions/107258/exponential-distribution-as-limiting-distribution | Exponential distribution as limiting distribution
I wonder if there are well-known and studied cases involving Exponential distribution as limiting distribution. I also wonder if this would contradict Central Limit theorem.
-
For example, consider a uniform (on a unit interval) sample of size $n$. Then $$\lim_{n \to \infty} \mathbb{P}\left( \min(u_1,u_2,\ldots,u_n) \leqslant \frac{x}{n}\right) = \lim_{n \to \infty} 1- \left(1-\frac{x}{n}\right)^n = 1 - \exp(-x)$$ The latter is exactly the CDF of the exponential random variable with unit mean. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9848366975784302, "perplexity": 242.24651285882132}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042990112.92/warc/CC-MAIN-20150728002310-00061-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.bartleby.com/solution-answer/chapter-8-problem-37e-chemistry-in-focus-7th-edition/9781337399692/francium-223-decays-to-polonium-215-via-three-decay-steps-propose-a-possible-three-step-sequence/40b3424d-90e6-11e9-8385-02ee952b546e | # Francium-223 decays to polonium-215 via three decay steps. Propose a possible three-step sequence for the overall process.
### Chemistry In Focus
7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692
Chapter
Section
### Chemistry In Focus
7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692
Chapter 8, Problem 37E
Textbook Problem
1 views
## Francium-223 decays to polonium-215 via three decay steps. Propose a possible three-step sequence for the overall process.
Interpretation Introduction
Interpretation:
The three-step sequence for the given decay process is to be represented.
Concept introduction:
There is a difference between a chemical reaction and a nuclear reaction. A chemical reaction involves atoms, molecules and electrons whereas a nuclear reaction involves the nucleus of an atom.
### Explanation of Solution
In a nuclear equation, each element is represented by its appropriate chemical symbol. There are two numbers written along with the chemical symbol. One is the mass number, which is written on the upper side and the other is the atomic number, which is written on the lower side. The mass number is denoted by A and the atomic number is denoted by Z.
In a balanced nuclear equation, the sum of atomic number and mass number is the same on both sides of the nuclear equation.
The three-step sequence for the overall process is represented as follows:
87223Fr88223Ra+&
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https://www.arxiv-vanity.com/papers/1708.08444/ | # Boundedness of singular integrals on C1,α intrinsic graphs in the Heisenberg group
Vasileios Chousionis, Katrin Fässler, and Tuomas Orponen Department of Mathematics, University of Connecticut, USA Department of Mathematics, University of Fribourg, Switzerland Department of Mathematics and Statistics, University of Helsinki, Finland
###### Abstract.
We study singular integral operators induced by -dimensional Calderón-Zygmund kernels in the Heisenberg group. We show that if such an operator is bounded on vertical planes, with uniform constants, then it is also bounded on all intrinsic graphs of compactly supported functions over vertical planes.
In particular, the result applies to the operator induced by the kernel
K(z)=∇H∥z∥−2,z∈H∖{0},
the horizontal gradient of the fundamental solution of the sub-Laplacian. The boundedness of is connected with the question of removability for Lipschitz harmonic functions. As a corollary of our result, we infer that the intrinsic graphs mentioned above are non-removable. Apart from subsets of vertical planes, these are the first known examples of non-removable sets with positive and locally finite -dimensional measure.
###### Key words and phrases:
Singular integrals, Heisenberg group, Removable sets for harmonic functions
###### 2010 Mathematics Subject Classification:
42B20 (Primary) 31C05, 35R03, 32U30, 28A78 (Secondary)
V.C. is supported by the Simons Foundation via the project ‘Analysis and dynamics in Carnot groups’, Collaboration grant no. 521845. K.F. is supported by Swiss National Science Foundation via the project ‘Intrinsic rectifiability and mapping theory on the Heisenberg group’, grant no. 161299. T.O. is supported by the Academy of Finland via the project ‘Restricted families of projections and connections to Kakeya type problems’, grant no. 274512.
## 1. Introduction
The purpose of this paper is to study the boundedness of certain -dimensional singular integrals on intrinsic graphs in the first Heisenberg group , a -dimensional manifold with a -dimensional metric structure. All the formal definitions will be deferred to Section 2, so this introduction will be brief, informal and not entirely rigorous.
We study singular integral operators (SIOs) of convolution type. In , this refers to objects of the following form:
Tμf(p)=∫K(q−1⋅p)f(q)dμ(p), (1.1)
where is a kernel, and is a locally finite Borel measure. Specifically, we are interested in the boundedness of the operator on certain -regular surfaces , where is -dimensional Hausdorff measure restricted to . The relevant surfaces are the intrinsic Lipschitz graphs, introduced by Franchi, Serapioni and Serra Cassano [24] in 2006. These are the Heisenberg counterparts of (co-dimension ) Lipschitz graphs in . In the Euclidean environment, the boundedness of SIOs on Lipschitz graphs, and beyond, is a classical topic, developed by Calderón [6], Coifman-McIntosh-Meyer [15], David [18], David-Semmes [17], and many others.
If is a vertical plane (a plane in containing the vertical axis), then the boundedness of on is essentially a Euclidean problem. In fact, as long as , the group operation behaves like addition in . Also, is simply a constant multiple of -dimensional Lebesgue measure. So, can be identified with a convolution type SIO in .111One should keep in mind, however, that if is a Calderón-Zygmund kernel in , then the restriction of to satisfies the standard growth and Hölder continuity estimates with respect to a non-Euclidean metric on . The boundedness question for such operators is classical, see Stein’s book [34], and for instance Fourier-analytic tools are applicable.
The main result of the paper, see Theorem 1.1 below, asserts that solving the Euclidean problem automatically yields information on the non-Euclidean problem. Before making that statement more rigorous, however, we ask: what are the natural SIOs in , in the context of the -dimensional surfaces ? In , a prototypical singular integral is the -dimensional Riesz transform, whose kernel is the gradient of the fundamental solution of the Laplacian,
KRd(x)=∇|x|−(d−2).
The boundedness of the associated singular integral operator is connected with the problem of removability for Lipschitz harmonic functions. A closed set is removable for Lipschitz harmonic functions, or just removable, if whenever is open, and is Lipschitz and harmonic in , then is harmonic in . In brief, the connection between and removability is the following: if is bounded on for some closed -regular set , then , or positive measure closed subsets of , are not removable for Lipschitz harmonic functions, see Theorem 4.4 in [28]. The importance of the Riesz transform in the study of removability is highlighted in the seminal papers by David and Mattila [16], and Nazarov, Tolsa and Volberg [30, 31]. Using, among other things, techniques from non-homogeneous harmonic analysis, they characterise removable sets as the purely -unrectifiable sets in , that is, the sets which intersect every hypersurface in a set of vanishing -dimensional Hausdorff measure.
In , the counterparts of harmonic functions are solutions to the sub-Laplace equation , see Section 2.1, or [4]. With this notion of harmonicity, the problem of removability in makes sense, and has been studied in [12, 10]. Also, as in , removability is connected with the boundedness of a certain singular integral , now with kernel
KH(z)=∇H∥z∥−2,
where denotes the Korányi distance. In contrast to the Euclidean case, this kernel is not antisymmetric in the sense . Nevertheless, it is known that the associated SIO is bounded on vertical planes, see Remark 3.15 in [12]. This is due to the fact that is horizontally antisymmetric: for . On vertical planes , this amount of antisymmetry suffices to guarantee boundedness on by classical results, see for instance Theorem 4 on p. 623 in [34].
We now introduce the main theorem. We propose in Conjecture 2.11 that convolution type SIOs with Calderón-Zygmund kernels which are uniformly -bounded on vertical planes, are also bounded on intrinsic Lipschitz graphs . This would prove that such sets are non-removable – a fact which, before the current paper, was only known for the vertical planes . In this paper, we verify Conjecture 2.11 for the intrinsic graphs of compactly supported intrinsically -functions, defined on vertical planes , see Definitions 2.12 and 2.16. This class contains all compactly supported Euclidean -functions, with the identification , see Remark 2.21.
###### Theorem 1.1.
Let , and assume that has compact support. Then, any convolution type SIO with a Calderón-Zygmund kernel which is uniformly -bounded on vertical planes, is bounded on for any -Ahlfors-David regular measure supported on the intrinsic graph of . In particular, this is true for the -dimensional Hausdorff measure on .
In particular, the result applies to the operator , as its -boundedness on vertical planes is known. The formal connection between the boundedness of the singular integral , and removability, is explained in the following result:
###### Theorem 1.2.
Assume that is a non-trivial positive Radon measure on , satisfying the growth condition for and , and such that the support has locally finite -dimensional Hausdorff measure. If is bounded on , then is not removable for Lipschitz harmonic functions.
We prove Theorem 1.2 in Section 5; the argument is nearly the same as the one used by Mattila and Paramonov [28] in the Euclidean case. There are a few subtle differences, however, so we provide all the details. The proof also requires an auxiliary result of some independent interest, on slicing a set in by horizontal lines, see Lemma 5.3.
With Theorems 1.1 and 1.2 in hand, the following corollaries are rather immediate:
###### Corollary 1.3.
Let . Assume that is compactly supported. If is a closed subset of the intrinsic graph of with positive -dimensional Hausdorff measure, then is not removable.
###### Corollary 1.4.
Let . Assume that is open, and is Euclidean on . If is a closed subset of the intrinsic graph of over with positive -dimensional Hausdorff measure, then is not removable.
The structure of the paper is the following. In Section 2, we introduce all the relevant concepts, from singular integrals to (intrinsic) functions, and prove some simple lemmas. In Section 3, we prove Theorem 1.1. In Section 4, we study how well the (intrinsic) graphs of functions are approximated by vertical planes; this analysis is required in the proof of Theorem 1.1. Finally, in Section 5, we study the connection with the removability problem, and prove Theorem 1.2 and Corollaries 1.3 and 1.4.
## 2. Definitions and preliminaries
### 2.1. The Heisenberg group and general notation
The first Heisenberg group is endowed with the group law
z1⋅z2=(x1+x2,y1+y2,t1+t2+12[x1y2−x2y1]), (2.1)
for . The neutral element in these coordinates is given by and the inverse of is denoted by and given by . The Korányi distance is defined as
d(z1,z2):=∥(z2)−1⋅z1∥,z1,z2∈H, (2.2)
where
∥(x,y,t)∥:=4√(x2+y2)2+16t2,for (x,y,t)∈H1. (2.3)
A frame for the left invariant vector fields is given by
X=∂x−y2∂t,Y=∂y+x2∂t,andT=∂t.
The horizontal gradient of a function on an open set is
∇Hu=(Xu)X+(Yu)Y.
and the sub-Laplacian of is
ΔHu=X2u+Y2u. (2.4)
We consider the horizontal gradient as a mapping with values in and write . For a thorough introduction to the Heisenberg group, we refer the reader to Chapter 2 of the monograph [7].
#### 2.1.1. Notation
We usually denote points of by or ; in coordinates, we often write with . Points on vertical subgroups (see Section 2.3.1) are typically denoted by .
Unless otherwise specified, all metric concepts in the paper, such as the diameter and distance of sets, are defined using the metric given in (2.2). The notation refers to Euclidean norm, and refers to the quantity defined in (2.3). A closed ball in of radius and centre is denoted by .
For , we use the notation to signify that there exists a constant , depending only on the parameter "", such that . If no "" is specified, the constant is absolute. We abbreviate the two-sided inequality by .
The notation stands for the -dimensional Hausdorff measure (with respect to the metric ), and Lebesgue measure on is denoted by ; this notation is also used to denote Lebesgue measure on the subgroups under the identification .
### 2.2. Kernels and singular integral operators in H
An aim of the paper is to study the boundedness of singular integral operators (SIOs) on fairly smooth -Ahlfors-David regular surfaces in (see Section 2.3 for a more precise description of our surfaces). But what are these SIOs – and what are their kernels? In this paper, a kernel is any continuous function . Motivated by similar considerations in Euclidean spaces, it seems reasonable to impose the following growth and Hölder continuity estimates:
|K(z)|≲1∥z∥3and|K(z1)−K(z2)|≲∥z−12⋅z1∥β∥z1∥3+β, (2.5)
for some , and for all and with . We call such kernels -dimensional Calderón-Zygmund (CZ) kernels in . The conditions above, and the lemma below, imply that our terminology is consistent with standard terminology, see for instance p. 293 in Stein’s book [34].
###### Lemma 2.1.
Assume that a kernel satisfies the second (Hölder continuity) estimate in (2.5) for some . Then,
|K(q−1⋅p1)−K(q−1⋅p2)|+|K(p−11⋅q)−K(p−12⋅q)|≲∥p−12⋅p1∥β/2∥q−1⋅p1∥3+β/2 (2.6)
for with .
###### Proof.
Write and . Then by left-invariance of , so the first summand in (2.6) has the correct bound by (2.5), even with replaced by . Hence, to find a bound for the second summand, we only need to prove that
|K(z−11)−K(z−12)|≲∥z−12⋅z1∥β/2∥z1∥3+β/2.
We may moreover assume that for a suitable large constant . We would like to apply (2.5) as follows,
|K(z−11)−K(z−12)|≲∥z2⋅z−11∥β∥z1∥3+β, (2.7)
but we first need to make sure that . Write and , and observe that
d(z−11,z−12)=∥z2⋅z−11∥ =∥(x2−x1,y2−y1,t2−t1−12[x2y1−y2x1]∥ ≲∥(x2−x1,y2−y1,t2−t1+12[x2y1−y2x1]∥ +√|x2y1−x1y2| =d(z1,z2)+√|(x2−x1)y1−(y2−y1)x1| ≲d(z1,z2)+√d(z1,z2)√∥z1∥. (2.8)
It follows from (2.8) that , if the constant was chosen large enough. Hence, the estimate (2.7) is legitimate, and we may further use (2.8) obtain
|K(z−11)−K(z−12)|≲∥z−12⋅z1∥β∥z1∥3+β+∥z−12⋅z1∥β/2∥z1∥β/2∥z1∥3+β≲∥z−12⋅z1∥β/2∥z1∥3+β/2,
as claimed. ∎
We now recall some basic notions about SIOs. Fix a -dimensional Calderón-Zygmund kernel , and a complex Radon measure . For , we define
Tϵν(p):=∫∥q−1⋅p∥>ϵK(q−1⋅p)dν(q),p∈H,
whenever the integral on the right hand side is absolutely convergent; this is, for instance, the case if has finite total variation. Next, fix a positive Radon measure on satisfying the growth condition
μ(B(p,r))≤Cr3,p∈H,r>0, (2.9)
where is a constant. Given a complex function and , we define
Tμ,ϵf(p):=Tϵ(fdμ)(p),p∈H.
It easily follows from the growth conditions on and , and Cauchy-Schwarz inequality, that the expression on the right makes sense for all .
###### Definition 2.2.
Given a -dimensional CZ kernel , and a measure satisfying (2.9), we say that the SIO associated to is bounded on , if the operators
f↦Tμ,ϵf
are bounded on with constants independent of .
For the rest of the paper, we are mainly concerned with measures satisfying the -sided inequality for all and , and for some fixed constants . Such measures are called -Ahlfors-David regular, or -ADR in short.
###### Remark 2.3.
Given a -dimensional kernel , the kernel , defined by , is the kernel of the formal adjoint of since
∫(Tμ,ϵf)gdμ =∫(∫∥q−1⋅p∥>ϵK(q−1⋅p)f(q)dμ(q))g(p)dμ(p) =∫(∫∥p−1⋅q∥>ϵK∗(p−1⋅q)g(p)dμ(p))f(q)dμ(q)=∫(T∗μ,ϵg)fdμ.
It is easy to check that satisfies the growth condition in (2.5). Moreover, satisfies the Hölder continuity requirement in (2.5) with exponent ; this is a corollary of Lemma 2.1.
#### 2.2.1. Two examples
In this short section, we give two examples of concrete -dimensional CZ kernels.
###### Example 2.4 (The 3-dimensional H-Riesz kernel).
Consider the kernel
K(z)=∇H∥z∥−2,z∈H∖{0}. (2.10)
Note that agrees (up to a multiplicative constant) with the fundamental solution of the sub-Laplacian , as proved by Folland [21], see also [4, Example 5.4.7]. We call the -dimensional -Riesz kernel; it gives rise to a SIO , which we call the -dimensional -Riesz transform. Studying the -boundedness of on subsets of is connected with the removability of these sets for Lipschitz harmonic functions on , see Theorem 5.1 for the precise statement.
The neat formula (2.10) can be expanded to the following rather unwieldy expression:
K(x,y,t)=(−2x|(x,y)|2+8yt∥(x,y,t)∥6,−2y|(x,y)|2−8xt∥(x,y,t)∥6)=:(K1(x,y,t),K2(x,y,t)). (2.11)
From the formula above, one sees that is not antisymmetric in the usual sense ; for instance, . However, both components of are horizontally antisymmetric, as in the definition below.
###### Definition 2.5 (Horizontal antisymmetry).
A kernel is called horizontally antisymmetric, if
K(x,y,t)=−K(−x,−y,t),(x,y,t)∈H∖{0}.
It is clear from the formula (2.11) that is horizontally antisymmetric.
###### Example 2.6 (The 3-dimensional quasi H-Riesz kernel).
Consider
Ω(x,y,t):=(x∥(x,y,t)∥4,y∥(x,y,t)∥4,t∥(x,y,t)∥5),(x,y,t)≠(0,0,0). (2.12)
It is easy to see that is a -dimensional CZ kernel which is antisymmetric in the sense that for all . We will call the -dimensional quasi -Riesz kernel; it defines the -dimensional quasi -Riesz transform . The kernel , which resembles in form the Euclidean Riesz kernels, was introduced in [11]. It was proved there that if is a -ADR measure and is bounded in then can be approximated at almost every point and at arbitrary small scales by homogeneous subgroups. It is unknown if the -Riesz transform has the same property.
#### 2.2.2. Cancellation conditions
Fix a -dimensional CZ kernel . Without additional assumptions, the SIO associated with is generally not bounded on , even when is nice, such as the -dimensional Hausdorff measure on a vertical plane (see Section 2.3.1 for a definition of these planes), which is a constant multiple of Lebesgue measure on . So, for positive results, one needs to impose cancellation conditions. The horizontal antisymmetry or antisymmetry would be such conditions, but neither of them holds both for the -Riesz kernel and the quasi -Riesz kernel simultaneously. Here is a more general cancellation condition, which encompasses antisymmetric and horizontally antisymmetric kernels:
###### Definition 2.7 (Ab).
A kernel satisfies the annular boundedness condition (AB for short) if the following holds. For every every -radial function satisfying , there exists a constant such that
∣∣∣∫W[ψR(w)−ψr(w)]K(w)dL2(w)∣∣∣≤Aψ (2.13)
for all , and for all vertical planes . Above,
ψr(z):=(ψ∘δ1/r)(z),
where is the (-homogeneous) dilatation for .
It turns out that the AB condition for -dimensional CZ kernels is equivalent to the following condition:
###### Definition 2.8 (Ubvp).
Given a kernel with , we say that it is uniformly bounded on vertical planes (UBVP in short), if the SIO associated to is bounded on for every vertical plane (in the sense of Definition 2.2), with constants independent of .
The measure is -ADR, so it makes sense to discuss boundedness of on . The following lemma is an analog of [34, Proposition 2, p. 291].
###### Lemma 2.9.
Assume that a kernel with satisfies the UBVP condition. Then also satisfies the AB condition.
###### Proof.
Fix a vertical plane . The group operation "" restricted to coincides with usual (Euclidean) addition in the plane : if , then . Also, for some positive constant . Hence, for ,
TH3|W,ϵf(w)=c∫WK(w−v)Bϵ(w−v)f(v)dL2(v)=c(KBϵ)∗f(w),f∈C∞0(W),
where is the indicator function of , the notation "" means Euclidean convolution, and stands for smooth and compactly supported functions on . Since is bounded on , it follows that the Fourier transform of is a bounded function on , independently of :
|ˆKBϵ(ξ)|≤A,ξ∈W,ϵ>0.
For the proof see e.g. [25, 2.5.10]. Now, fix a function as in Definition 2.7. Fix also , and a vertical plane . Note that vanishes in the ball . Hence, if , we have for , and hence
∣∣∣∫[ψR(w)−ψr(w)]K(w)dL2(w)∣∣∣ =∣∣∣∫[ψR(w)−ψr(w)]K(w)Bϵ(w)dL2(w)∣∣∣ ≲A∫∣∣ˆψR(ξ)−ˆψr(ξ)∣∣dL2(ξ),
using Plancherel before passing to the second line. Moreover,
∫|ˆψR(ξ)|dL2(ξ)=∫|ˆψ(δR(ξ))|R3dL2(ξ)=∫|ˆψ(ξ)|dL2(ξ)≲1,
and the same holds with "" in place of "". This completes the proof. ∎
Now, recall the main result, Theorem 1.1. With the terminology above, it states that if a -dimensional CZ kernel satisfies the UBVP, then the associated SIO is bounded on certain spaces, which we will define momentarily (see Section 2.3). The strategy of proof is to infer, from the lemma above, that the kernel satisfies the AB condition, and proceed from there. In particular, it remains to prove the following version Theorem 1.1:
###### Theorem 2.10.
Let , and assume that has compact support. Assume that a -dimensional CZ kernel satisfies the AB condition. Then, the associated SIO is bounded on for any -Ahlfors-David regular measure supported on the intrinsic graph of .
As simple corollaries, the -Riesz transform and the quasi -Riesz transform , recall Section 2.2.1, are bounded on the intrinsic graphs mentioned in Theorem 2.10. Since the associated kernels and are -dimensional CZ kernels, it suffices to verify that they satisfy the AB condition. But this is a consequence of either horizontal antisymmetry (in the case of ) or antisymmetry (in the case of ). In fact, the key cancellation condition (2.13) even holds in the stronger form
∫[ψR(w)−ψr(w)]K(w)dL2(w)=0
for all functions as in Definition 2.7, for all , and all vertical planes .
### 2.3. Intrinsic graphs and the boundedness of SIOs
For which -ADR measures are the SIOs associated to -dimensional kernels satisfying the UBVP condition bounded on ? The following seems like a natural conjecture:
###### Conjecture 2.11.
Let be a vertical subgroup with complementary subgroup , and let be an intrinsic Lipschitz function (see Definition 2.12). If is a convolution type SIO with a -dimensional CZ kernel which satisfies the UBVP condition, then it is bounded on for all -ADR measures supported on the intrinsic graph . In particular, this is true for (since is -ADR by Theorem 3.9 in [22]).
Recall that the main theorem of the paper, Theorem 1.1, states that the conjecture holds for , which are compactly supported and intrinsically -smooth for some , see Definition 2.16.
#### 2.3.1. Intrinsic Lipschitz graphs
In our terminology, the vertical subgroups in are all the nontrivial homogeneous normal subgroups of , except for the center of the group. Recall that homogeneous subgroups are subgroups of which are preserved under dilations of the Heisenberg group, see [33]. With the choice of coordinates as in (2.1), the vertical subgroups coincide therefore with the -dimensional subspaces of that contain the -axis. To every vertical subgroup we associate a complementary horizontal subgroup . In our coordinates this is simply the -dimensional subspace in which is perpendicular to . Every point can be written as with a uniquely determined vertical component and horizontal component . This gives rise to the Heisenberg projections
πW:H→W,πW(p)=pW
and
πV:H→V,πV(p)=pV.
###### Definition 2.12.
An intrinsic graph is a set of the form
Γ(ϕ)={w⋅ϕ(w):w∈W},
where is a vertical subgroup with complementary horizontal subgroup , and is any function. We often use the notation for the graph map . To define intrinsic Lipschitz graphs, fix a parameter , and consider the set (cone)
Cγ={z∈H:∥πW(z)∥≤γ∥πV(z)∥}.
We say that is an intrinsic -Lipschitz function, and an intrinsic -Lipschitz graph, if
(z⋅Cγ)∩Γ(ϕ)={z},for z∈Γ(ϕ) and 0<γ<1L.
The function is said to be intrinsic Lipschitz if it is intrinsic -Lipschitz for some constant .
###### Remark 2.13.
Every vertical subgroup can be parametrised as
W={(−w1sinθ,w1cosθ,w2):(w1,w2)∈R2}
with an angle uniquely determined by . The complementary horizontal subgroup is then given by
V={(vcosθ,vsinθ,0):v∈R}.
We often denote points on in coordinates by "", and points on by real numbers "". Then, expressions such as and should be interpreted as elements in , namely the products of the corresponding elements on and .
Intrinsic Lipschitz graphs were introduced by Franchi, Serapioni and Serra Cassano in [24], motivated by the study of locally finite perimeter sets and rectifiability in the Heisenberg group [23]. While intrinsic Lipschitz functions continue to be studied as a class of mappings which are interesting in their own right, they have also recently found a prominent application in [29]. Various properties of intrinsic Lipschitz functions are discussed in detail in [33]. For instance, it is known that an intrinsic Lipschitz function has a well-defined intrinsic gradient , which we will use to perform integration on intrinsic Lipschitz graphs.
#### 2.3.2. Intrinsic differentiability
To define the intrinsic gradient, we recall that the notion of intrinsic graph is left invariant. Indeed, given a function with intrinsic graph , for every , the set is again the intrinsic graph of a function , which we denote by , so that . For instance if is the -plane, the -axis, and , then we can compute explicitly
ϕp0(y,t)=ϕ(y−y0,t−t0+12x0y0−yx0)+x0. (2.14)
We also recall that in our context an intrinsic linear map is a function whose intrinsic graph is a vertical subgroup.
###### Definition 2.14.
A function with is intrinsically differentiable at if there exists an intrinsic linear map such that
∥(Gw)−1⋅ψ(w)∥=o(∥w∥),as w→0. (2.15)
The map is called the intrinsic differential of at 0 and denoted by .
More generally, a function is intrinsically differentiable at a point if is intrinsic differentiable at for . The intrinsic differential of at is given by
dϕw0:=dψ0.
Recall that can be identified with through our choice of coordinates, see Remark 2.13. Under this identification the restriction of the Korányi distance to agrees with the Euclidean distance so that (2.15) reads
|ψ(w)−Gw|=o(∥w∥),as w→0.
With the parametrisation from Remark 2.13, every intrinsic linear map has the form for a constant .
###### Definition 2.15.
Assume that is intrinsically differentiable at a point . Then its intrinsic gradient at is the unique number such that
dϕw0(w1,w2)=∇ϕϕ(w0)w1,for all (w1,w2)∈W.
The intrinsic gradient is simply a number determined by the "angle" between and the vertical plane .
#### 2.3.3. C1,α-intrinsic Lipschitz functions and graphs
The goal of the paper is to prove that certain SIOs are bounded in on intrinsic graphs , where is a compactly supported function satisfying a (Heisenberg analogue of) -regularity. The most obvious definition of would be to require the intrinsic gradient to be locally -Hölder function in the metric space , but this condition is not left-invariant: the parametrisation of the left-translated graph , for , would not necessarily be locally -Hölder continuous with the same exponent , see Example 4.5. So, instead, we define an "intrinsic" notion of , which is (a) left-invariant in the sense above, and (b) is well-suited for the application we have in mind, and (c) is often easy to verify, see Remark 2.21 below.
###### Definition 2.16.
We say that a function is an intrinsic function if exists at every point , and is continuous. We further define the subclasses , , as follows: , if , and there exists a constant such that
|∇ϕ(p−10)ϕ(p−10)(w)−∇ϕ(p−10)ϕ(p−10)(0)|≤H∥w∥α, (2.16)
for all , and all . For notational convenience, we also define . Intrinsic graphs of (or ) functions will be called intrinsic (or ) graphs.
Several remarks are now in order.
###### Remark 2.17.
(a) It is well-known, see for instance Proposition 4.4 in [14] or Lemma 4.6 in [8], that if is intrinsic Lipschitz, then .
(b) Conversely, if with , then is intrinsic Lipschitz. This is well-known and follows from existing results, but it was difficult to find a reference to this particular statement; hence we include the argument in Lemma 2.22 below.
###### Remark 2.18.
Note that if has compact support, then , and hence is intrinsic Lipschitz by (b) above.
###### Remark 2.19.
If is the -plane, the condition (2.16) for can be written in coordinates as follows:
|∇ϕϕ(y+y0,t+t0+ϕ(y0,t0)y)−∇ϕϕ(y0,t0)|≤H∥(y,t)∥α. (2.17)
To see this, apply the representation (2.14) with replaced by .
###### Remark 2.20.
It is known by Theorem 4.95 in [33] that the intrinsic graph of an intrinsic function is an -regular surface; in particular, satisfies an area formula, see Section 2.3.4 for more details.
The definitions of (intrinsic) and are quite different from their standard Euclidean counterparts, which we denote by and (a function belongs to if its partial derivatives exist and are -Hölder continuous with respect to the Euclidean metric). However, at least for compactly supported functions, sufficient regularity in the Euclidean sense also implies regularity in the intrinsic Heisenberg sense, as the following remark shows.
###### Remark 2.21.
Assume that is the -plane, and identify with . Then, any compactly supported -function is in the class . Indeed, if then has the following expression:
∇ϕϕ=ϕy+ϕϕt, (2.18)
see [8, (4.4)]. Since are bounded, is and are Euclidean -Hölder, we infer that is Euclidean -Hölder continuous. Since also is bounded, we obtain
≲min{1,|(y,t+ϕ(y0,t0)y)|α} ≲min{1,|(y,t)|α}≲∥(y,t)∥α,
which by (2.17) verifies that .
###### Lemma 2.22.
Assume that with . Then, is intrinsic Lipschitz.
###### Proof.
For simplicity, we assume that is the -plane. Write . By Proposition 4.56(iii) in [33], it suffices to verify that
|ϕ(p−1)(y,t)|=|ϕ(p−1)(y,t)−ϕ(p−1)(0,0)|≲L∥(y,t)∥,(y,t)∈W,p∈Γ(ϕ).
Write . Then, we estimate as follows:
|ϕ(p−1)(y,t)| ≤|ϕ(p−1)(y,t)−∇ϕϕ(w)y|+|∇ϕϕ(w)y| ≲L∥(y,t)∥+L∥(y,t)∥,
as claimed. The estimate leading to the last line follows from Proposition 2.23 (with ) below. ∎
###### Proposition 2.23.
Fix , and assume that is the -plane. Assume that with . Then, for ,
|ϕ(p−1)(y,t)−∇ϕϕ(w)y|≲∥(y,t)∥1+α,(y,t)∈W,
where the implicit constants only depend on , and, if , also on the Hölder continuity constant "" in the definition of .
We postpone the proof to Section 4.
###### Remark 2.24.
If , Proposition 2.23 above shows that functions in with are uniformly intrinsically differentiable, see [2, Definition 3.16].
The next lemma verifies that being an intrinsic -graph is a left-invariant concept.
Let . Let | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9916762709617615, "perplexity": 706.5197988291009}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662534693.28/warc/CC-MAIN-20220520223029-20220521013029-00460.warc.gz"} |
https://usa.cheenta.com/usajmo-2012-questions/ | Categories
# USAJMO 2012 questions
1. Given a triangle ABC, let P and Q be the points on the segments AB and AC, respectively such that AP = AQ. Let S and R be distinct points on segment BC such that S lies between B and R, ∠BPS = ∠PRS, and ∠CQR = ∠QSR. Prove that P, Q, R and S are concyclic (in other words these four points lie on a circle).
2. Find all integers such that among any n positive real numbers with there exist three that are the side lengths of an acute triangle.
3. Let a, b, c be positive real numbers. Prove that .
4. Let be an irrational number with , and draw a circle in the plane whose circumference has length 1. Given any integer , define a sequence of points as follows. First select any point on the circle, and for define as the point on the circle for which the length of the arc is , when travelling counterclockwise around the circle from to . Suppose that and are the nearest adjacent points on either side of . Prove that .
5. For distinct positive integers a, b < 2012, define f(a, b) to be the number of integers k with (1le k < 2012) such that the remainder when ak divided by 2012 is greater than that of bk divided by 2012. Let S be the minimum value of f(a, b), where a and b range over all pairs of distinct positive integers less than 2012. Determine S.
6. Let P be a point in the plane of triangle ABC, and be a line passing through P. Let A’, B’, C’ be the points where reflections of the lines PA, PB, PC with respect to intersect lines BC, AC, AB, respectively. Prove that A’, B’ and C’ are collinear.
## By Ashani Dasgupta
Ph.D. in Mathematics from University of Wisconsin Milwaukee (USA)
Founder - Faculty at Cheenta | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9379130005836487, "perplexity": 370.9952715418367}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141194171.48/warc/CC-MAIN-20201127191451-20201127221451-00525.warc.gz"} |
https://www.physicsforums.com/threads/what-does-the-navier-stokes-equation-look-like-after-time-discretization.895587/ | # What does the Navier-Stokes equation look like after time discretization?
Tags:
1. Dec 2, 2016
### Kukkat
Hi,
I know the general form of the Navier Stokes Equation as follows.
I am following a software paper of "Gerris flow solver written by Prof. S.Popinet"
and he mentions after time discretization he ends with the following equation:
where n-1 is the previous time step, n+1 is the next time step and n+0.5 is mid time for the present time step.
Solving equation implicitly/ explicitly in time means solving for next time data however in the equation there are rather two unknowns un+0.5 and
un+1.
Not sure why he uses different terms at different time intervals. Density at n+0.5, velocity at n, n-1, n+0.5 etc..
Can anyone point me or explain me how he arrives at this specific sort of discretized equation.
2. Dec 2, 2016
### eys_physics
The link to the paper doesn't work.
3. Dec 2, 2016
### Staff: Mentor
I'm not familiar with this particular finite difference scheme, but presumably un+0.5 is already know when you are calculating un+1
4. Dec 2, 2016
### Kukkat
http://www.sciencedirect.com/science/article/pii/S002199910900240X
@Chestermiller guess thats true. It is being solved by the time step projection method which means an intermediate velocity is computed and later updated to a divergence free velocity by solving the laplace of pressure term as in the mentioned paper.
The notation is a bit strange for me as he uses density at time n+0.5 without solving any advection equation. From what I see density terms at n and n+0.5 should be the same.
Draft saved Draft deleted
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http://cpr-astrophsr.blogspot.com/2013/01/13016766-boaz-katz-et-al.html | ## An exact integral relation between the Ni56 mass and the bolometric light curve of a type Ia supernova [PDF]
Boaz Katz, Doron Kushnir, Subo Dong
An exact relation between the Ni56 mass and the bolometric light curve of a type Ia supernova can be derived as follows, using the following excellent approximations: 1. the emission is powered solely by Ni56-> Co56 ->Fe56; 2. each mass element propagates at a non-relativistic velocity which is constant in time (free coasting); and 3. the internal energy is dominated by radiation. Under these approximations, the energy E(t) carried by radiation in the ejecta satisfies: dE/dt=-E(t)/t-L(t)+Q(t), where Q(t) is the deposition of energy by the decay which is precisely known and L(t) is the bolometric luminosity. By multiplying this equation by time and integrating over time we find: E(t)*t=\int_0^t Q(t')t'dt' -\int_0^t L(t')t'dt'. At late time, t>> t_peak, the energy inside the ejecta decreases rapidly due to its escape, and thus we have \int_0^t Q(t')t'dt'=\int_0^t L(t')t'dt'. This relation is correct regardless of the opacities, density distribution or Ni56 deposition distribution in the ejecta and is very different from "Arnett's rule", L_peak ~ Q(t_peak). By comparing \int_0^t Q(t')t'dt' with \int_0^t L(t')t'dt' at t~40 day after the explosion, the mass of Ni56 can be found directly from UV, optical and infrared observations with modest corrections due to the unobserved gamma-rays and due to the small residual energy in the ejecta, E(t)*t>0.
View original: http://arxiv.org/abs/1301.6766 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9928796887397766, "perplexity": 2995.192776329188}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267158279.11/warc/CC-MAIN-20180922084059-20180922104459-00317.warc.gz"} |
https://papers.nips.cc/paper/7372-posterior-concentration-for-sparse-deep-learning | # NIPS Proceedingsβ
## Posterior Concentration for Sparse Deep Learning
[PDF] [BibTeX] [Supplemental] [Reviews]
### Abstract
We introduce Spike-and-Slab Deep Learning (SS-DL), a fully Bayesian alternative to dropout for improving generalizability of deep ReLU networks. This new type of regularization enables provable recovery of smooth input-output maps with {\sl unknown} levels of smoothness. Indeed, we show that the posterior distribution concentrates at the near minimax rate for alpha-Holder smooth maps, performing as well as if we knew the smoothness level alpha ahead of time. Our result sheds light on architecture design for deep neural networks, namely the choice of depth, width and sparsity level. These network attributes typically depend on unknown smoothness in order to be optimal. We obviate this constraint with the fully Bayes construction. As an aside, we show that SS-DL does not overfit in the sense that the posterior concentrates on smaller networks with fewer (up to the optimal number of) nodes and links. Our results provide new theoretical justifications for deep ReLU networks from a Bayesian point of view. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8134456872940063, "perplexity": 997.5896366082922}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370500426.22/warc/CC-MAIN-20200331084941-20200331114941-00497.warc.gz"} |
http://math.stackexchange.com/questions/166679/vertex-cover-upper-bound | # Vertex Cover - upper bound
A few definitions:
• $\mathsf{VC} = \{ (G,k) \mid \text{There exists a vertex cover of size$k$in$G$}\}$
• $\mathsf{VC_{LOG}} = \{ G \mid \text{There exists a vertex cover of size$\leq \log |V|$in$G$} \}$
Does $\mathsf{VC_{LOG}} \in \mathsf P$? Does $\mathsf{VC_{LOG}} \in \mathsf{NP}\setminus \mathsf P$?
Is there something wrong with the following reduction that shows that $\mathsf{VC \leq VC_{LOG}}$?
Let $(G,k) \$ where $G = (V,E)$. If $k < \log(|V|)$, return $G$. Else, construct a new graph $G_{k}$, where $G_k = (V_k,E)$ with $$V_k = V \cup \{ v_i \mid 1 \leq i \leq 2^k -|V| \}$$
The reason is I want to add enough useless vertices such that $k$ would be $\log |V_k|$ which would translate elements in $\mathsf{VC}$ to elements in $\mathsf{VC_{LOG}}$.
-
A reduction $f$ must map all pairs $(G,k)$ to graphs $f\bigl((G,k)\bigr)$ such that $(G,k) \in \mathsf{VC} \iff f\bigl((G,k)\bigr) \in \mathsf{VC_{LOG}}$. You can't just map those in $\mathsf{VC}$. – martini Jul 4 '12 at 19:26
@martini You're right, I edited the question, though I think the reduction is still valid. – Shmoopy Jul 4 '12 at 19:44
Your reduction is not valid, because you miss the if and only if part. You need an algorithm that transforms all pairs $(G,k)$ into graphs $G_k'$ such that $(G,k) \in VC$ if and only if $G_k' \in VC_{LOG}$. The point of a reduction is to be able to solve $VC$ using an oracle/algorithm that can solve $VC_{LOG}$ in polynomial time. However, with your reduction it is not possible, and also you may need to add a huge (exponential) number of vertices.
Moreover, vertex cover is fixed-parameter tractable (for each unprocessed vertex you take it or take all of its neighbours) and this yields polynomial-time algorithm for $VC_{LOG}$ (e.g. check wikipedia or try to google it). Besides, $VC_{LOG} \in (NP\,\backslash\,P)$ would imply $P \neq NP$ which is a very hard problem. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8469487428665161, "perplexity": 358.57150549020133}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443737911339.44/warc/CC-MAIN-20151001221831-00108-ip-10-137-6-227.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/42736/showing-equivalence-of-two-definitions-of-the-blow-up-of-a-variety | Showing equivalence of two definitions of the Blow-Up of a Variety
Let $X=\mathop{\mathrm{Spec}}(A)$ be an affine variety over some algebraically closed field $\Bbbk$ and $I\subseteq A$ an ideal of $A$. There are two ways to define the blow-up $\tilde X$ of $X$ along $I$, namely
1. Set $\tilde X := \mathop{\mathrm{Proj}}(A[IT])$ where $A[IT]$ is the graded ring $\bigoplus_{d\ge 0} I^dT^d \subseteq A[T]$ and $I^0:= A$.
2. Let $I=(f_0,\ldots,f_r)$ be a set of generators for $I$ and define a rational map $\varphi: X \to \mathbb{P}_\Bbbk^r$ by $\varphi(P):=[f_0(P):\ldots:f_r(P)]$. It is defined over $U:=X\setminus Z(I)$. Then, we define $\tilde X := \Gamma_\varphi$ to be the graph of $\varphi$, which is the closure of the graph of $\varphi|_U$.
I would like to show that both definitions are equivalent; let me give you my approach (which is basically just a more general variant of Example II.7.12.1 in Hartshorne):
Define a map $\pi: A[y_0,\ldots,y_r] \to A[IT]$ by $y_i\mapsto f_iT$. It induces an embedding of $\tilde X$ into $\mathbb{P}_\Bbbk^r \times X$ whose image should be $\Gamma_\varphi$. On any open subset $D(f_iT)$, we should now be able to prove that the kernel of the induced map
$A\left[\frac{y_0}{y_i},\ldots,\frac{y_r}{y_i}\right]\to \left(A[IT]_{f_iT}\right)_0$
is equal to $\left(y_kf_j-y_jf_k\,\vert\,0\le k<j\le r\right)$. However, I don't seem to be able to verify this. If anyone could show me how to proceed from here or even give a completely different approach, I would be very grateful.
-
By the way, it is obvious that $f_ky_j-f_jy_k\in\ker(\pi)$ for all $j$ and $k$. Hence, the inclusion $\tilde X \subseteq \Gamma_\varphi$ is obvious. – Jesko Hüttenhain Jun 2 '11 at 12:44
Also, since $\dim(A[IT])>\dim(A)$, we can immediately see that $\dim(\tilde X)\ge\dim(\Gamma_\varphi)$. Hence by my previous comment, $\dim(\tilde X)=\dim(\Gamma_\varphi)$. This is me trying to upgrade the inclusion to an equality by some topological argument, but I am not really there yet. – Jesko Hüttenhain Jun 2 '11 at 21:04
You try to show that the kernel of $\pi:A[y_0, \ldots, y_r] \rightarrow A[IT]$ by $y_i \mapsto f_iT$ is generated by the $y_if_j - y_jf_i$ for $1\le i < j \le r$. It is not a surprise that you don't succeed, because this is not true for general ideals $I$. Consider for example $$I = (x^2,xy,y^2)$$ in $k[x,y]$. The kernel of $\pi: k[x,y,a,b,c] \rightarrow k[x,y][IT]$ under $a \mapsto x^2T$, $b\mapsto xyT$ and $c\mapsto y^2T$ contains $ac-b^2$ which does not lie in the ideal $(axy - bx^2, ay^2-cx^2,by^2-cxy)$. The statement holds if (iff?) $f_0,\ldots, f_r$ is a regular sequence according to Fulton's "Introduction to Intersection Theory in Algebraic Geometry", if I remember correctly (I don't have a copy at hand).
However, it holds the following: $$\ker \pi = (y_1 - f_1T, \ldots, y_r-f_rT) \cap A[y_1, \ldots, y_r],$$ where $(y_1 - f_1T, \ldots, y_r-f_rT)$ is an ideal in $A[y_0,\ldots,y_r,T]$. So I think one could build a proof like this:
1) $\mathrm{Proj}(A[IT]) = \mathrm{Proj}(A[y_0, \ldots,y_r]/\ker \pi)$ should be clear.
2) Furthermore, $\mathrm{Proj}(A[y_0, \ldots,y_r]/\ker \pi)$ is the closure of the projection of $$\mathrm{Proj}(A[y_0,\ldots,y_r,T]/(y_1 - f_1T, \ldots, y_r-f_rT) )$$ to the hyperplane $\{t=0\}$. This is the geometrical interpretation of elimination according to e.g. Cox, Little, O'Shea "Ideals, Varieties and Algorithms". Remember that $\ker \pi$ is obtained by elimination of $T$ from $(y_1 - f_1T, \ldots, y_r-f_rT)$. The closure of the projection should be the closure of $(y_0 : \ldots:y_r)=(f_0:\ldots:f_r)$, hence the closure of the graph of $\varphi|_U$, i.e. $\Gamma_\varphi$.
Please ask me to clarify if something is unclear.
-
I think my comments have, by now, answered the question: Since $X$ is irreducible, so is $U$. Let $U' := \mathbb{P}_\Bbbk^r \times U$. Then, $U'\cap\Gamma_\varphi \cong U$ is irreducible. Since $\tilde X \cap U'$ is a closed subset of $U'\cap\Gamma_\varphi$ and of the same dimension, we must have $U'\cap\Gamma_\varphi = U'\cap\tilde X$. By passing to the closure, we obtain the desired result.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9825268387794495, "perplexity": 62.34603225071819}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461861727712.42/warc/CC-MAIN-20160428164207-00193-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://slideplayer.com/slide/8519055/ | # Energy in the Atmosphere
## Presentation on theme: "Energy in the Atmosphere"— Presentation transcript:
Energy in the Atmosphere
Section 1 Pages 42-45
1-Electromagnetic waves
Form of energy that can travel through space (ex: from the sun) Classified by wavelength Type of energy transfer Radiation
Direct transfer of energy by electromagnetic waves
2- Radiation Direct transfer of energy by electromagnetic waves
3- Most of the energy from the sun reaches the Earth in the form of visible light and infrared radiation, and a small amount of ultraviolet radiation.
4- Infrared Radiation Form of energy with wavelengths that are longer than red light. Not visible but can be felt as heat.
Form of radiation that has wavelengths that are shorter than violet light. Not visible but causes sunburn, skin cancer and eye damage.
6- What happens to the sun’s energy when it reaches the Earth’s atmosphere?
It is either absorbed or reflected
7- What absorbs the sun’s energy?
Water vapor & carbon dioxide absorb infrared radiation Ozone Layer absorbs most of the ultraviolet radiation Clouds, dust and gases also absorb some of the sun’s energy
8- What reflects the sun’s energy?
Clouds Dust Gases
9- Define scattering. Reflection of light in all directions
Why is the sky blue? When the white light (from the sun) hits the gas molecules, the light is scattered in mostly shorter wavelengths (blue and violet), which is what we see.
10- “Energy in the Atmosphere”
11- Explain what happens to the energy that reaches the Earth’s surface.
Energy is absorbed by Earth’s surfaces (at different rates) The energy is turned into heat It is radiated back into the atmosphere as infrared energy (heat) The energy gets absorbed and held by greenhouse gases (water vapor & carbon dioxide)
12- Greenhouse effect The process by which gases hold heat in the air.
Two main greenhouse gases are water vapor and carbon dioxide. How is the greenhouse effect related to global warming? The amount of CO2 is increasing (more gases that hold heat well) causing the temperature of Earth’s atmosphere to also rise. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.926181435585022, "perplexity": 1688.0482981266928}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988724.75/warc/CC-MAIN-20210505234449-20210506024449-00412.warc.gz"} |
http://www.lsc-group.phys.uwm.edu/~ballen/grasp-distribution/GRASP/doc/html/node104.html | Next: Chirp generation for test Up: GRASP Routines: Waveforms from Previous: The waveform Contents
## A note on the allowed frequency range
The frequency range of a signal calculated from black hole perturbation theory is bounded from above by the innermost stable circular orbit. This corresponds, by virtue of equation (), to an orbital velocity of . In practice there is also a lower bound. Since zero frequency would correspond to an infinite orbital radius we have to introduce a cutoff. In the present data files . Figure shows the allowed frequency range as a function of the total mass .
Since in equation () diverges at calculating at the upper limit is numerically difficult. However this is not a problem in the time domain, since the system spends very little time near . An value of inaccurate by several percent will typically not change the signal by even one wave cycle.
Next: Chirp generation for test Up: GRASP Routines: Waveforms from Previous: The waveform Contents
Bruce Allen 2000-11-19 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9602769613265991, "perplexity": 1142.5609534838852}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187821088.8/warc/CC-MAIN-20171017110249-20171017130249-00083.warc.gz"} |
http://math.stackexchange.com/questions/150615/references-for-kolmogorovs-strong-law-of-a-large-numbers | References for Kolmogorov's strong law of a large numbers
On the Wikipedia law of large numbers site, they mention "Kolmogorov's strong law of large numbers", which works even if the random variables are not identically distributed.
Where can I find this theorem shown and proven? I know that a reference is provided on the Wikipedia site, but that book is out of availability. Are there any other references out there?
(Interestingly, Allan Gut's book "Probability: A Graduate Course", has a theorem by the name of "Kolmogorov's strong law", but in his book, the random variables have to be identically distributed. Any ideas why this is?)
-
The result that they cite as "Kolmogorov's Strong Law" is not what I always refer to as Kolmogorov's Strong Law (I suspect that what I refer to as Kolmogorov's Strong Law is the same thing that Allan Gut does). The result given on Wikipedia requires a finite second moment and that $\sum \frac{\mbox{Var} X_k}{k^2} < \infty$, but in exchange you can drop the requirement of being identically distributed. The proof of this version I think is actually pretty easy, the sketch being: because the series converges we can apply the Khintchine-Kolmogorov convergence theorem so that $\sum \frac{(X_k - \mu_k)}{k}$ converges almost surely, and the result follows after an application of Kronecker's Lemma. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9071730971336365, "perplexity": 322.7346930718461}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510275111.45/warc/CC-MAIN-20140728011755-00269-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://www.cuemath.com/functions/applying-modulus/ | # Applying modulus
Go back to 'Functions'
The following figure shows the graph of the function $$f\left( x \right) = {x^2} - 5x+ 4$$:
Using this graph, can we plot the graph of $$y = g\left( x \right) = \left|{f\left( x \right)} \right|$$? The answer is simple: wherever f is positive, we leave the curve untouched; wherever f is negative, we reflect that part in the x-axis, because the effect of the modulus operation is to give us the positive magnitude. In this case, the graph of $$y = \left| {f\left( x \right)} \right|$$ will be as follows:
Thus, the part of the curve below the x-axis in the original graph gets reflect in the x-axis.
Let’s see another example of this transformation. The following is the graph of $$y= f\left( x \right)$$, where f is some arbitrary function:
And, the following is the graph of $$y = \left| {f\left( x \right)} \right|$$:
Functions
Functions
grade 10 | Questions Set 2
Functions
Functions
grade 10 | Questions Set 1 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.914527177810669, "perplexity": 528.3505804696741}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987829507.97/warc/CC-MAIN-20191023071040-20191023094540-00293.warc.gz"} |
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https://www.physicsforums.com/threads/friction-equlibrium.108572/ | # Friction/ Equlibrium
1. Jan 29, 2006
### wayneo
A mass m,is attached to 2 equal peices of string each of length, whose ends are attached to rings around a rod. If the static coefficient of friction between the rings and the rod is u, find the largest distance d between the rings such that the mass is at equilibrium.
2. Jan 29, 2006
### Hootenanny
Staff Emeritus
Can you show some of your working / thoughts?
3. Jan 29, 2006
### wayneo
Ive got the answer as d= 2uL/square root of (1 + u squared) but Im a bit stuck with the working out
4. Jan 29, 2006
### Hootenanny
Staff Emeritus
Have you drawn a diagram showing all the forces acting?
5. Jan 29, 2006
### wayneo
no but there arent any forces given as it is in equilibrium
6. Jan 29, 2006
### wayneo
nemore ideas?
7. Jan 29, 2006
### wayneo
any1 ??????
8. Jan 29, 2006
### wayneo
hard huh ?
9. Jan 29, 2006
### Hootenanny
Staff Emeritus
There are forces but they are all balanced. Try calculating the tension in the strings, then resolving the tension horizontally parallel to the rod.
Last edited: Jan 29, 2006
10. Jan 29, 2006
### sporkstorms
I think what you're trying to say is that there is no net force. There are, most certainly, forces acting in the system.
Listen to Hootenanny, s/he speaks words of wisdom.
Your goal is to find out where those rings will be when the net force is zero. In order to do so, you really need to make a diagram.
11. Jan 29, 2006
### Hootenanny
Staff Emeritus
Thank-you sporkstorms | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8232771754264832, "perplexity": 3520.2263573571395}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721067.83/warc/CC-MAIN-20161020183841-00360-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://rupress.org/jgp/article/106/2/149/26446/Permeation-properties-of-a-Ca-2-blockable | A Ca(2+)-blockable monovalent cation channel is present in the apical membrane of the ectoderm of the gastrulating chick embryo. We used the patch clamp technique to study several single-channel permeation properties of this channel. In symmetrical conditions without Ca2+, the Na+ current carried by the channel rectifies inwardly. The channel has an apparent dissociation constant for extracellular Na+ of 115 mM at 0 mV and a low density of negative surface charge (-0.03 e/nm2) at its extracellular entrance. The minimal pore diameter is approximately 5.8 A, as calculated from the relative permeabilities of 10 small organic cations. Extracellular application of six large organic cations decreased the inward Na+ current in a voltage-dependent manner, which strongly suggests an intrachannel block. The presence of at least two ion binding sites inside the pore is inferred from the Na+ dependence of the block by the organic cations. This hypothesis is strengthened by the fact that the extracellular Ca2+ block is also modified by the Na+ concentration. In particular, the rise of the unblocking rate with increased Na+ concentrations clearly suggests the presence of an interaction between Ca2+ and Na+ inside the channel. A low probability of double occupancy at physiological ionic conditions is implied from the absence of an anomalous mole fraction effect with mixtures of extracellular Li+ and K+. Finally, the absence of inward current at very strong hyperpolarizations and in the presence of 10 mM extracellular Ca2+ demonstrates the absence of significant Ca2+ current through this channel. It is argued that this embryonic epithelial Ca(2+)-blockable monovalent cation channel is related to both L-type Ca2+ channel and cyclic nucleotide-gated channels.
This content is only available as a PDF. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9019735455513, "perplexity": 3216.7460048319463}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780060201.9/warc/CC-MAIN-20210928032425-20210928062425-00153.warc.gz"} |
https://gis.meta.stackexchange.com/questions/4545/is-this-the-right-community-for-this-type-of-question-latex-support | # Is this the right community for this type of question + latex support
I posted this question on robotics.stackexchange because the lack of latex support for gis.stackexchange made me wonder whether or not it was the appropriate community to ask such questions. So I actually have two questions: do you think that this question is more appropriate where it was posted, or would gis.stackexchange be a better option? Second, will there be any latex support (well, maybe there already is, but it just is not accessible through the usual syntax?)? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9494816064834595, "perplexity": 416.11726193224274}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989874.84/warc/CC-MAIN-20210518222121-20210519012121-00384.warc.gz"} |
http://math.stackexchange.com/questions/255274/are-normal-subgroups-transitive | # Are normal subgroups transitive?
Suppose $G$ is a group and $K\lhd H\lhd G$ are normal subgroups of $G$. Is $K$ a normal subgroup of $G$, i.e. $K\lhd G$? If not, what extra conditions on $G$ or $H$ make this possible?
Applying the definitions, we know $\{ghg^{-1}|h\in H\}=H$ and $\{hkh^{-1}|k\in K\}=K$, and want $\{gkg^{-1}|k\in K\}=K$. Clearly, the best avenue for a counterexample is if $gkg^{-1}\not\in K$ for some $k\in K$ and $g\in G-H$.
If no such element exists, $\{gkg^{-1}|k\in K\}\subseteq K$ implies $\{gkg^{-1}|k\in K\}=K$ because if $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}|k\in K\}$.
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$K$ characteristic in $H$ and $H$ normal in $G$ then $K$ is normal in $G$ – jim Dec 10 '12 at 7:48
Take a look at $D_8$, the dihedral group with 8 elements. – Hans Giebenrath Dec 10 '12 at 7:51
@HansGiebenrath I'm not seeing a counterexample in $D_4$. It has $C_4=\{1,r,r^2,r^3\}$ as a normal subgroup, but the only subgroup of $C_4$ is $C_2=\{1,r^2\}$, which is normal in $D_4$. – Mario Carneiro Dec 10 '12 at 8:02
@PatrickDaSilva: They do, $\langle r^2,s\rangle$ is normal in $D_8$ and contains $\langle s \rangle$, which is not normal in $D_8$. – Hans Giebenrath Dec 10 '12 at 9:08
@Hans : I guess I am tired for saying false things. Sorry to have doubted you. – Patrick Da Silva Dec 10 '12 at 9:16
Using some suggestions from the other commenters:
The alternating group, $A_4$, has the set $H=\{I,(12)(34),(13)(24),(14)(23)\}\cong V_4$ as a subgroup. If $f\in S_4\supseteq A_4$ is a permutation, then $f^{-1}[(12)(34)]f$ has the effect of swapping $f(1)$ with $f(2)$ and $f(3)$ with $f(4)$. One of these is $1$, and depending on which it is paired with, the conjugated element may be any of $H-\{I\}$, since the other two are also swapped. Thus $H\lhd S_4$ is normal, so $H\lhd A_4$ as well. Similarly, $H$ has three nontrivial subgroups, and taking $K=\{I,(12)(34)\}\cong C_2$, this is normal because $V_4$ is abelian. But $K\not\lhd A_4$, since $$[(123)][(12)(34)][(132)]=(13)(24)\in H-K.$$
Moreover, this is a minimal counterexample, since $|A_4|=12=2\cdot 2\cdot 3$ is the next smallest number which factors into three integers, which is required for $K\lhd H\lhd G$ but $\{I\}\subset K\subset H\subset G$ so that $[G\,:\,H]>1$, $[H\,:\,K]>1$, $|K|>1$ and $$|G|=[G\,:\,H]\cdot[H\,:\,K]\cdot|K|.$$ The smallest integer satisfying this requirement is 8, but the only non-abelian groups with $|G|=8$ are the dihedral group $D_4$ and the quaternion group $Q_8$, and neither of these have counterexamples. (Note that if $G$ is abelian, then all subgroups are normal.) Thus $A_4$ is a minimal counterexample. (Edit: Oops, $D_4$ has a counterexample, as mentioned in the comments: $\langle s\rangle\lhd\langle r^2,s\rangle\lhd \langle r,s\rangle=D_4$, but $\langle s\rangle\not\lhd D_4$.)
However, if $H\lhd G$ and $K$ is a characteristic subgroup of $H$, then $K$ is normal in $G$. This is because the group action $f$ defines an automorphism on $G$, $\varphi(g)=f^{-1}gf$, and because $H$ is normal, $\varphi(H)=H$ so that $\varphi|_H$ is an automorphism on $H$. Thus $\varphi(K)=K$ since $K$ is characteristic on $H$ and so $\{f^{-1}kf\,|\,k\in K\}=K\Rightarrow K\lhd G$.
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Great! You understand the theory verywell. +1! And sorry for the false comments. – Patrick Da Silva Dec 10 '12 at 9:17
look at $S_4$ and its following subgroup $A = \langle (12)(34) \rangle$ and $B=\{(12)(34),(13)(42),(23)(41),e \}$ try to show $A$ is normal in $B$ and $B$ is normal in $S_4$ but $A$ is not normal in $G$
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Good eye! I like this example. How about you show it? The proof is not long. No calculations required. I challenge you (unless you were leaving the exercise to the OP). – Patrick Da Silva Dec 10 '12 at 7:59
@PatrickDaSilva I've fleshed out this argument in an answer below, but this leads to an interesting line of investigation: What is the smallest counterexample? As you point out, $|G|\geq 8$, but $D_8$ fails, and the quaternion group fails too since $\{1,-1\}\lhd Q_8$ is the only subgroup of order 2. – Mario Carneiro Dec 10 '12 at 9:09
@PatrickDaSilva $A_4$ is normal in $S_4$ and the sylow 2 subgroup of of $A_4$ is a unique group of order 4 hence the group of $B$ given above will be normal in $S_4$ and the normality of $A$ in $B$ is becuase of the fact $[B:A]=2$ – jim Dec 10 '12 at 10:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9370303750038147, "perplexity": 151.4648379818624}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246637364.20/warc/CC-MAIN-20150417045717-00305-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/calculating-the-curvature-drift-in-a-plasma.948076/ | I Calculating the curvature drift in a plasma
Tags:
1. May 24, 2018
TimeRip496
How do i get this equation,
$$\frac{db}{dl}=(\hat{B}.∇)\hat{b}$$
This equation is a vector whose direction is towards the centre of the circle which most closely approximates the magnetic field-line at a given point, and whose magnitude is the inverse of the radius of this circle.
However I have no idea how to obtain the above and the source basically state it as by definition. Where is this equation obtained from?
Source: https://ocw.mit.edu/courses/nuclear...a-physics-i-fall-2003/lecture-notes/chap2.pdf page 8
2. May 29, 2018
PF_Help_Bot
Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8857934474945068, "perplexity": 442.23534839236135}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376827992.73/warc/CC-MAIN-20181216191351-20181216213351-00381.warc.gz"} |
https://www.ctanujit.in/2013/06/iitjee-previous-years-solved-papers.html | ## Pages
### IITJEE Previous Years Solved Papers
IIT-JEE Previous Years Solved Paper On Probability
Year-wise Solution of IIT-JEE paper on Probability
Q. [2010] A signal which can be green or red with probability 4/5 and 1/5 respectively,is received by station A and then transmitted to station B. The probability of each station receiving the signal correctly is 3/4 .If the signal received at station B is green, then the probability that the original signal was green is:
(a) 3/5 (b) 6/7 (c) 20/23 (d) 9/20
Sol:- (c)
Let us consider three events:
A₁ : The original signal is green.
A₂ : The original signal is red.
A : Signal received at station B is green.
P(A₁) = 4/5, P(A₂) =1/5,
P(A|A₁)=(3/4)² + (1/4)² =5/8,
P(A|A₂)=(3/4)(1/4)+(1/4)(3/4)=3/8
Now apply Bayes Theorem to find P(A₁|A). Ans is=20/23 .
Q.[2008] An experiment has 10 equally likely outcomes. Let A and B be two non-empty events of the experiment. If A consists of 4 outcomes,the number of outcomes that B must have so that A and B are independent,is:
(a) 2,4 or 8 (b) 3,6 or 9 (c) 4 or 8 (d) 5 or 10
Sol:- (d)
Given n(A)=4 ; P(A)=4/10
Let number of outcomes for B be n(B) ; P(B)=n(B)/10
Since A and B are independent, P(A∩B)=P(A)P(B)=(4/10)(n(B)/10)
⇒n(A∩B)=2(n(B))/5
Since n(A∩B) is an integer so n(B)=5 or 10.
Q.[2004] If three distinct numbers are chosen randomly from the first 100 natural numbers,then the probability that all three of them are divisible by both 2 and 3 is:
(a) 4/25 (b) 4/35 (c) 4/33 (d) 4/1155
Sol:- (d)
A natural number is divisible by 2 or 3, if it is divisible by 6.
So, numbers divisible by 6 are 6,12,18, . . . . ,96
So there are total 16 numbers out of which 3 distinct number can be chosen in 16×15×14=n(A)
Total no of ways=100×99×98=n(S)
Required probability=n(A)/n(S)=4/1155 .
Q.[2003] Two selected numbers are selected randomly from the set S={1,2,3,4,5,6} without replacement one by one. The probability that minimum of the two numbers is less than 4 is:
(a) 1/15 (b) 14/15 (c) 1/5 (d) 4/5
Sol:- (d)
Total no of ways= 6C2=15.
Two numbers can be={(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)}
Number of favourable ways=12.
Required probability=12/15=4/5.
Q.[1998] A fair coin is tossed repeatedly. If tail appears on the first four tosses,then the probability of head appearing in the fifth toss equals
(a) 1/2 (b) 1/32 (c) 31/32 (d) 1/5
Sol:- (a)
Since the tosses are independent events. So, probability that a head appears on the 5th toss is not dependent on the earlier tosses. So, the required probability is 1/2.
Note:- Each set on different topics will contain 5 distinct Solved problems from IIT-JEE papers. Keep visiting. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8545892834663391, "perplexity": 2188.542951284}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303845.33/warc/CC-MAIN-20220122103819-20220122133819-00539.warc.gz"} |
https://discourse.acados.org/t/matrix-should-be-binary-error/776 | # Matrix should be binary error
I am using acados to solve a decentralized NMPC problem with a flock of drones. More specifically, I am implementing tube MPC. In this formulation, one has to implement the following initial state constraint:
Since the minimum robust invariant set is quite complex for our system, when we write this constraint using acados in Matlab, we get this error:
“ocp_create: error setting constr_Jbx_0, matrix should be binary”
Is there a way to define Jbx_0 as a non-binary matrix in acados framework?
If you set the matrix Jbx_0, you are formulating box constraints on x at shooting node 0 in acados.
i.e. lbx_0 \leq Jbx_0 * x_0 \leq ubx_0. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9178535342216492, "perplexity": 3170.2275683004073}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00009.warc.gz"} |
https://www.physicsforums.com/threads/boundary-value-problem-and-ode.605001/ | # Homework Help: Boundary Value problem and ODE
1. May 10, 2012
### fionamb83
Hi, I'm not sure if this is on the right thread but here goes. It's a perturbation type problem.
Consider the boundry value problem
$$\epsilon y'' + y' + y = 0$$
Show that if $$\epsilon = 0$$ the first order constant coefficient equation has
the solution
$$y_{outer} (x) = e^{1-x}$$
I have done this fine.
Find a suitable rescaling $$X = x/\delta$$ so that the highest derivative is important and balances another term and find the solutions $y_{inner}(X)$ of
this equation, containing one free parameter, satisfying the boundary
condition at $$x = X = 0$$
So I am at the rescaling part and solved the differential equation (after neglecting the δy part of the full equation)
$$\frac{d^2y}{dX^2} + \frac{dy}{dX} = 0$$
yielding $$y = Ae^{-X} + B$$
imposing the boundry condition $$x = X = 0$$
gives $$A = B$$
so is $$y_{inner} = Ae^{-X}$$ ??
I think I covered that the highest derivative is important (Although again I was unsure about the wording)
When I continue on I think I either have this part wrong or the matching is wrong as I am not getting the right answer. I am supposed to be doing intermediate scaling. If anyone has any comments about this that would also be much appreciated.
P.S. This is posted on a different thread. A mistake on my part. Not sure how to delete it so please could someone tell me!
2. May 10, 2012
### HallsofIvy
Why do you call this a "boundary value problem" when there is no boundary condition given?
If a condition is that y(0)= e, yes, that would be correct.
Yes, that's good.
??? at X= 0, y(0)= A+ B but if you are given no specific value of y there, it gives you nothing. If you are taking y(0)= 0, then you would have A= -B, not A= B.
Why? If you do have A= B, then it would be y= A(e^{-X}+ 1). If A= -B, it would be y= A(e^{-X}- 1)
Now, as X goes to infinity (x is not infinitesmal) this would give y= -A and you want to match that to y= e^{1- x} which is e at x= 0. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9452365040779114, "perplexity": 744.9412906831883}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676589179.32/warc/CC-MAIN-20180716041348-20180716061348-00136.warc.gz"} |
https://www.physicsforums.com/threads/difference-between-power-sets-and-sample-space.418822/ | # Difference between Power Sets and Sample Space
1. Jul 27, 2010
### thrillhouse86
Hey All,
In my probability theory class we have just started learning about how a probability space is defined by a sample space (which contains all possible events), events and a measure.
We briefly went over the idea of the Power Set, which seems to be the set of all subsets in your sample space. My question is what is the difference between the Power Set and the Sample Space ? aren't they both just a collection of all possible outcomes ?
Thanks
2. Jul 28, 2010
### ZQrn
The powerset is the sample space of the experiment:
'Of all subsets of X, what is the chance that this subset is one specific subset Y. provided an even distribution'
Well, then your chance happens to be 1/|P(X)|..
This is of course nothing special. This is just because the powerset is the defined as the set of all subsets.
If you ask, 'what is the chance that you have one specific subset which meets property Q'
Then your chance is also simply 1/|{x sub X : Q(x)}| ...
The powerset can be defined with the identity property (the property which is always true), which is x = x basically. So we get P(x) = {x sub X : x = x}.
Edit: not sure you know 'set builder notation', but if Phi(x) and Theta(x) are formulae in x, such as x > 3, or x ^ 2 = 3 et cetera. See {Phi(x) : Theta(x)} as the set of all x for which both formulae are true.
3. Jul 28, 2010
### mathman
Simple example:
Sample space {1,2,3}
Power set {(),(1),(2),(3),(1,2),(1,3),(2,3),(1,2,3)}
4. Jul 28, 2010
### ZQrn
That's not an example of a sample space and a powerset, that's A: a set, and B: the power set of that set.
If A is a sample space, then B is the powerset of a sample space.
A sample space is with respect to some experiment.
5. Jul 29, 2010
### mathman
When I learned probability theory, it was taught on an abstract basis. A sample space was defined on the basis of Kolmogoroff axioms and did not require a particular experiment. In essence a sample space was a measure space with total measure = 1.
6. Jul 29, 2010
### ZQrn
Yes, all and well, the first set was a possible sample space. But so was the second.
You just gave a sample space, and the powerset of that space as an illustration that powerspaces and sample sets are some-how different concepts, which they are, but not via this logic, because a powerset is of course always different to its set.
7. Jul 30, 2010
### g_edgar
Perhaps he is talking about the set of all events, and asking why that is not just the power set ...
8. Aug 1, 2010 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.933139979839325, "perplexity": 667.2199113967948}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738661640.68/warc/CC-MAIN-20160924173741-00256-ip-10-143-35-109.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/908033/uniqueness-of-the-direct-product-decomposition-of-finite-groups | # Uniqueness of the direct product decomposition of finite groups
A group $G$ is indecomposable if: $G = H \times K \Rightarrow \{ H,K \} = \{1, G \}$.
Then, a finite group $G$ decomposes into a direct product of indecomposable groups: $G = \prod_i G_i$.
Question: Is this decomposition unique (up to permutation and isomorphism)?
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What does $\;\{H,K\}=\{1,G\}\;$ mean? – Timbuc Aug 24 '14 at 19:58
@Timbuc: it's an equality of sets. This means "$H=1$ and $K=G$" or "$H=G$ and $K=1$" (with $1$ the trivial group). – Sébastien Palcoux Aug 24 '14 at 19:59
@SébastienPalcoux What does $[H,K]$ mean, then? – Thomas Andrews Aug 24 '14 at 20:00
@IttayWeiss, doesn't that theorem apply only to abelian groups? – Timbuc Aug 24 '14 at 20:03
@Timbuc you are right. I was sure OP was talking about abelian groups. – Ittay Weiss Aug 24 '14 at 20:11
## 2 Answers
Yes, the decomposition is unique for finite groups. This is a consequence of the Remak-Krull-Schmidt Theorem, which applies to groups that satisfy both the minimum and maximum conditions on normal subgroups. Since finite groups certainly have these properties, R-K-S applies here.
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This decomposition is not unique in the class of all groups, although, as the other answer points out, it is true fir finite groups.
One reason that the result fails in general is that groups are not cancellable in general. A group $H$ is cancellable if the following holds. $$H\times Q\cong H\times P\Rightarrow P\cong Q$$ Finite groups are cancellable, but in general groups are not (see this question - $\mathbb{Z}$ is a counter-example). If your result held then the decompositions of $P$ and $Q$ would have to be the same, and hence $P$ and $Q$ they would have to be isomorphic, so all groups would be cancellable.
As finite groups are cancellable, this proof only works for general groups and not for finite groups.
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Your answer is a long comment, because my question is precisely for finite groups. Thank you for the link! – Sébastien Palcoux Aug 24 '14 at 21:21
You start your answer by "No, this decomposition is not unique" and you finish by "but I suspect that it still does not hold". It would be more correct to start by "I suspect this decomposition is not unique" or "This decomposition is not unique in general and I suspect the same for the finite groups". – Sébastien Palcoux Aug 24 '14 at 21:22
@Sebastian Sorry, my mistake, I didn't see the the word "finite". I've edited my answer to fit it in. – user1729 Aug 25 '14 at 8:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8484696745872498, "perplexity": 539.5694099120274}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443737940789.96/warc/CC-MAIN-20151001221900-00225-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://support.google.com/docs/answer/3094097?hl=en&ref_topic=3105600 | # POISSON.DIST
Returns the value of the Poisson distribution function (or Poisson cumulative distribution function) for a specified value and mean.
### Sample Usage
POISSON.DIST(2.4,1,FALSE)
POISSON.DIST(A2,A3,TRUE)
### Syntax
POISSON.DIST(x, mean, cumulative)
• x - The input to the Poisson distribution function.
• mean - The mean (mu) of the Poisson distribution function.
• cumulative - Whether to use the Poisson cumulative distribution function rather than the distribution function..
### Notes
• The Poisson distribution function is typically used to calculate the number of 'arrivals' or 'events' over a period of time, such as the number of network packets or login attempts given some mean.
• If cumulative is TRUE then POISSON.DIST returns the probability of x or fewer events, otherwise the probability of exactly x events.
• You can use POISSON or POISSON.DIST to perform this function.
WEIBULL: Returns the value of the Weibull distribution function (or Weibull cumulative distribution function) for a specified shape and scale.
NORMSINV: Returns the value of the inverse standard normal distribution function for a specified value.
NORMSDIST: Returns the value of the standard normal cumulative distribution function for a specified value.
NORMINV: Returns the value of the inverse normal distribution function for a specified value, mean, and standard deviation.
NORMDIST: The NORMDIST function returns the value of the normal distribution function (or normal cumulative distribution function) for a specified value, mean, and standard deviation.
NEGBINOMDIST: Calculates the probability of drawing a certain number of failures before a certain number of successes given a probability of success in independent trials.
LOGNORMDIST: Returns the value of the log-normal cumulative distribution with given mean and standard deviation at a specified value.
LOGINV: Returns the value of the inverse log-normal cumulative distribution with given mean and standard deviation at a specified value.
EXPONDIST: Returns the value of the exponential distribution function with a specified lambda at a specified value.
BINOMDIST: Calculates the probability of drawing a certain number of successes (or a maximum number of successes) in a certain number of tries given a population of a certain size containing a certain number of successes, with replacement of draws. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9153625965118408, "perplexity": 448.2492682937321}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627999298.86/warc/CC-MAIN-20190624084256-20190624110256-00363.warc.gz"} |
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1. Free, publicly-accessible full text available April 1, 2023
2. Abstract The ProtoDUNE-SP detector is a single-phase liquid argon time projection chamber (LArTPC) that was constructed and operated in the CERN North Area at the end of the H4 beamline. This detector is a prototype for the first far detector module of the Deep Underground Neutrino Experiment (DUNE), which will be constructed at the Sandford Underground Research Facility (SURF) in Lead, South Dakota, U.S.A. The ProtoDUNE-SP detector incorporates full-size components as designed for DUNE and has an active volume of 7 × 6 × 7.2 m 3 . The H4 beam delivers incident particles with well-measured momenta and high-purity particle identification. ProtoDUNE-SP's successful operationmore »
Free, publicly-accessible full text available January 1, 2023
3. Free, publicly-accessible full text available October 1, 2022
4. null (Ed.)
Abstract The Deep Underground Neutrino Experiment (DUNE), a 40-kton underground liquid argon time projection chamber experiment, will be sensitive to the electron-neutrino flavor component of the burst of neutrinos expected from the next Galactic core-collapse supernova. Such an observation will bring unique insight into the astrophysics of core collapse as well as into the properties of neutrinos. The general capabilities of DUNE for neutrino detection in the relevant few- to few-tens-of-MeV neutrino energy range will be described. As an example, DUNE’s ability to constrain the $$\nu _e$$ ν e spectral parameters of the neutrino burst will be considered.
5. Abstract The Deep Underground Neutrino Experiment (DUNE) will be a powerful tool for a variety of physics topics. The high-intensity proton beams provide a large neutrino flux, sampled by a near detector system consisting of a combination of capable precision detectors, and by the massive far detector system located deep underground. This configuration sets up DUNE as a machine for discovery, as it enables opportunities not only to perform precision neutrino measurements that may uncover deviations from the present three-flavor mixing paradigm, but also to discover new particles and unveil new interactions and symmetries beyond those predicted in the Standardmore »
6. Abstract The sensitivity of the Deep Underground Neutrino Experiment (DUNE) to neutrino oscillation is determined, based on a full simulation, reconstruction, and event selection of the far detector and a full simulation and parameterized analysis of the near detector. Detailed uncertainties due to the flux prediction, neutrino interaction model, and detector effects are included. DUNE will resolve the neutrino mass ordering to a precision of 5 $$\sigma$$ σ , for all $$\delta _{\mathrm{CP}}$$ δ CP values, after 2 years of running with the nominal detector design and beam configuration. It has the potential to observe charge-parity violation in themore » | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8493826389312744, "perplexity": 3014.2942446953857}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335276.85/warc/CC-MAIN-20220928180732-20220928210732-00431.warc.gz"} |
https://portonmath.wordpress.com/2010/12/17/filters-error-corrected/ | The error was in Appendix B in the proof of the theorem stating $(t;x)\not\in S$ (I messed $t$ and $\{t\}$.) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 3, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9958491325378418, "perplexity": 138.89279465345146}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257646914.32/warc/CC-MAIN-20180319120712-20180319140712-00049.warc.gz"} |
https://brilliant.org/problems/work-energy-power/ | # Work, energy, power
A pump, taking water from a large reservoir,is used to spray a jet of water with speed 20 m/s and radius 0.05 metres, from a nozzle level with the surface of the reservoir.Calculate the power of the pump.
× | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8429034948348999, "perplexity": 2486.6881398025102}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084892059.90/warc/CC-MAIN-20180123171440-20180123191440-00412.warc.gz"} |
http://repository.bilkent.edu.tr/browse?type=subject&value=Expectation-maximization%20approaches | Now showing items 1-1 of 1
• #### A complexity-reduced ML parametric signal reconstruction method
(2011)
The problem of component estimation from a multicomponent signal in additive white Gaussian noise is considered. A parametric ML approach, where all components are represented as a multiplication of a polynomial amplitude ... | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8091437816619873, "perplexity": 2002.8783504021776}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703644033.96/warc/CC-MAIN-20210125185643-20210125215643-00091.warc.gz"} |
https://www.dsprelated.com/freebooks/mdft/Radix_2_FFT.html | When is a power of , say where is an integer, then the above DIT decomposition can be performed times, until each DFT is length . A length DFT requires no multiplies. The overall result is called a radix 2 FFT. A different radix 2 FFT is derived by performing decimation in frequency.
A split radix FFT is theoretically more efficient than a pure radix 2 algorithm [73,31] because it minimizes real arithmetic operations. The term split radix'' refers to a DIT decomposition that combines portions of one radix 2 and two radix 4 FFTs [22].A.3On modern general-purpose processors, however, computation time is often not minimized by minimizing the arithmetic operation count (see §A.7 below).
#### Radix 2 FFT Complexity is N Log N
Putting together the length DFT from the length- DFTs in a radix-2 FFT, the only multiplies needed are those used to combine two small DFTs to make a DFT twice as long, as in Eq.(A.1). Since there are approximately (complex) multiplies needed for each stage of the DIT decomposition, and only stages of DIT (where denotes the log-base-2 of ), we see that the total number of multiplies for a length DFT is reduced from to , where means on the order of ''. More precisely, a complexity of means that given any implementation of a length- radix-2 FFT, there exist a constant and integer such that the computational complexity satisfies
for all . In summary, the complexity of the radix-2 FFT is said to be N log N'', or .
Next Section:
Fixed-Point FFTs and NFFTs
Previous Section:
Decimation in Time | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9057120084762573, "perplexity": 2090.2659617748227}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487621699.22/warc/CC-MAIN-20210616001810-20210616031810-00091.warc.gz"} |
http://mathoverflow.net/questions/8816/result-of-repeated-applications-of-the-binomial-distribution?sort=newest | # Result of repeated applications of the binomial distribution?
What is the result of multiplying several (or perhaps an infinite number) of binomial distributions together?
To clarify, an example.
Suppose that a bunch of people are playing a game with k (to start) weighted coins, such that heads appears with probability p < 1. When the players play a round, they flip all their coins. For each heads, they get a coin to flip in the next round. This is repeated every round until they have a round with no heads.
How would I calculate the probability distribution of the number or coins a player will have after n rounds? Especially if n is extremely high and p extremely close to 1?
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This looks like a branching process problem in disguise. If you want to calculate the exact distribution for a single player, then my first offhand impression is that one would get some kind of recurrence relation which should be amenable to direct attack or else to some kind of "generating function" machine? Does that sound plausible, familiar or useful to you? or are the expressions you get still intractable for your purposes? – Yemon Choi Dec 14 '09 at 2:28
To clarify my previous comment: nothing personal, but I'd prefer to have some indication of how far you've got or how much you've worked at this problem before I sit down and start scribbling. It might also save people time in their answers if they knew whether or not to start from scratch. – Yemon Choi Dec 14 '09 at 2:31
Sorry about that. I just want to know if there is a simple and straight-forward answer that I just couldn't find in a literature search. I'm actually trying to answer a more generalized version of this. Anyone who tries to sit down and solve it from scratch will probably encounter some pretty harsh difficulties, I think. All I've managed to do is find a recurrence and a very loose bound, and I want to know asymptotic behavior. Darsh's answer below looks very much like what I want. If not, I'll edit the question with more details, but for now I'm trying to do this mostly on my own. – DoubleJay Dec 14 '09 at 3:43
Here's how I interpret your example: there are a bunch of coins (k initially), each being flipped every round until it comes up tails, at which point the coin is "out," And you want to know, after n rounds, the probability that exactly j coins are still active, for j = 0, ..., k. (The existence of multiple players seems irrelevant.)
In that case, this is pretty elementary: after n rounds, the probability of each individual coin being active is p^n, so you have a binomial distribution with parameter p^n, k trials. Since you want to send p to 1 and n to infinity, note that replacing p by its square root and doubling n gives you the same distribution.
Your problem has a surprisingly fascinating generalization, which I believe is called the Galton-Watson process. Its solution has a very elegant representation in terms of generating functions, but I think there are very few examples in which the probabilities are simple to compute in general. Your instance is one of those. (The generalization: at each round, you have a certain number of individuals, each of which turns (probabilistically, independently) into a finite number of identical individuals. If the individuals are coins and each coin turns into one coin with probability p and zero coins with probability 1-p, and you begin with k coins, then we recover your example.)
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Do you want to get a quick but approximate answer or rather the exact law (or may be just expectation and variation)?
This is indeed the clasical Galton-Watson tree (http://en.wikipedia.org/wiki/Galton%E2%80%93Watson_process). A very good reference (and quite elementary!) is: "Athreya, Ney" - Branching processes. Each coin is an invidual which either survies to the next round with probability $p$ or dies with probability $1-p$.
1. (Approximate answer). You start with $n$ inviduals of which each has chance $p^k$ that it survives $k$ rounds. If $n$ is large, $p^k$ is small we can use the law of rare events. It says that approximately the number of the inviduals surviving $k$ rounds is the Poisson r.v. with parameter $\lambda := n p^k$. The error of this approximation is upperbounded by $\lambda^2/n$.
2. (Exact solution). Let $\mathbb{P}(X=1) = p^k = 1- \mathbb{P}(X=0)$. $X$ denote if an individual survied $k$ rounds (1) or not (0). Its generating function is $F_X(s) = (1-p^k) + p^k s$. If you start with $n$ individuals and denote the number of them surviving $k$ rounds by $Z$ then its generating function is $F_Z(s) = F(s)^n = \sum_{l=0}^n \binom{n}{l}(1-p^k)^{n-l}p^{kl}s^l$. Therefore
$\mathbb{P}(Z = l) = \binom{n}{l}(1-p^k)^{n-l}p^{kl}s^l$
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8322996497154236, "perplexity": 251.87596177671793}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645367702.92/warc/CC-MAIN-20150827031607-00004-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://infoscience.epfl.ch/record/216370 | Infoscience
Journal article
# The stability of a rising droplet: an inertialess non-modal growth mechanism
Prior modal stability analysis (Kojima et al., Phys. Fluids, vol. 27, 1984, pp. 19-32) predicted that a rising or sedimenting droplet in a viscous fluid is stable in the presence of surface tension no matter how small, in contrast to experimental and numerical results. By performing a non-modal stability analysis, we demonstrate the potential for transient growth of the interfacial energy of a rising droplet in the limit of inertialess Stokes equations. The predicted critical capillary numbers for transient growth agree well with those for unstable shape evolution of droplets found in the direct numerical simulations of Koh & Leal (Phys. Fluids, vol. 1, 1989, pp. 1309-1313). Boundary integral simulations are used to delineate the critical amplitude of the most destabilizing perturbations. The critical amplitude is negatively correlated with the linear optimal energy growth, implying that the transient growth is responsible for reducing the necessary perturbation amplitude required to escape the basin of attraction of the spherical solution. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9388546943664551, "perplexity": 948.7826799723791}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818689897.78/warc/CC-MAIN-20170924062956-20170924082956-00654.warc.gz"} |
http://math.stackexchange.com/questions/263151/show-that-a-finite-dimensional-banach-space-has-a-bijective-compact-operator | # Show that a finite-dimensional Banach space has a bijective compact operator
It is clear that if $T: X \rightarrow X$ is a bijective compact operator, where $X$ is a Banach space, then $\dim(\text{Range}(T)) = \dim(X)$, which implies that $\dim(X)$ must be $< \infty$.
How do I prove the converse: If $\dim(X) < \infty$, then there exists a bijective compact operator $T: X \rightarrow X$?
Thank you!
-
There's some sort of typo in the first sentence of your question. – Christopher A. Wong Dec 21 '12 at 10:55
@ChristopherA.Wong : i think now its correct . – Theorem Dec 21 '12 at 11:00
The first statement, namely proving that the existence of a bijective compact operator $T: X \rightarrow X$ implies that $\dim(X) < \infty$, is actually a non-trivial result. It is a more difficult problem than the one that the OP is posing. Every operator on a finite-dimensional Banach space is compact. Hence, the identity operator on a finite-dimensional Banach space is a bijective compact operator. – Haskell Curry Dec 22 '12 at 0:41
In fact, a finite-dimensional Banach space $X$ is nothing other than a finite-dimensional Euclidean space. This follows from the fact that all norms on a finite-dimensional vector space (over $\mathbb{R}$ or $\mathbb{C}$) are equivalent. Hence, every bijective operator on $X$ can be represented as an invertible matrix with respect to a fixed finite basis of $X$. As such, there are $2^{\aleph_{0}}$-many bijective compact operators on a finite-dimensional Banach space, with the identity operator being one of them. – Haskell Curry Dec 22 '12 at 0:54
add comment
## 2 Answers
I suppose the theorem you want to prove is this one:
Let $(X,\Vert \cdot \Vert)$ be a Banach space. There exists a linear continuous operator $T \colon X \to X$ compact if and only if $\dim X <+\infty$.
One way (if) is clear: indeed, if $\dim X<+\infty$ then every operator $T \colon X \to X$ is compact (since its range is finite dimensional: this is a well-known sufficient condition for compactness). For example, take identity of $X$: it is bijective (obviusly!) and compact.
Now, the other way (only if): suppose $T\colon X \to X$ is bijective and compact. There exists $T^{-1}$ and, moreover, it is continuous: so $TT^{-1}=\text{id}_X$ is compact, since $\mathcal K(X)$ is a closed ideal in $\mathcal L(X)$. In particular, the closure of the unit ball of $X$ is compact, hence the space $X$ is finite dimensional.
Hope this helps.
-
But if $T$ is compact then is it required that rank of $T$ is always finite dimensional ? – Theorem Dec 21 '12 at 11:10
No. The implication you ask for should be the easy one. – flavio Dec 21 '12 at 11:17
I perfectly agree with you, jiku1797. Anyway, I've added some details. – Romeo Dec 21 '12 at 11:23
@Romeo: My comment was meant to answer Theorem's misunderstanding regarding compact operators. Your comment is completely fine! – flavio Dec 21 '12 at 11:29
@Romeo : Thanks! – Theorem Dec 21 '12 at 11:40
show 1 more comment
Theorem, it is not true that $\dim(\text{Range}(T)) = \dim(X)$ implies that $\dim(X) < \infty$. Another way of reasoning is as follows. Let $T: X \rightarrow X$ be a bijective compact operator. Then by the Bounded Inverse Theorem, $T^{-1}$ exists and is continuous. Hence, $T$ is also a homeomorphism. Let $B_{X}$ be the closed unit ball of $X$. Then $T[B_{X}]$ is closed and has compact closure, which implies that $T[B_{X}]$ is compact (a closed subset of a compact space is also compact). However, as $T$ is a homeomorphism, $B_{X}$ must then be compact. Hence, as a consequence of Riesz's Lemma, $\dim(X) < \infty$.
Of course, Romeo's answer is very slick in the sense that reasoning with the ideal $\mathcal{K}(X)$ saves us a lot of work.
-
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http://mathhelpforum.com/number-theory/13002-solutions-2x-3mod4.html | # Math Help - Solutions to 2x = 3mod4
1. ## Solutions to 2x = 3mod4
How would i start by finding all the solutions to
2x=3(mod4).....the = is congruence. Thank you this site is awesome
2. Originally Posted by smoothi963
How would i start by finding all the solutions to
2x=3(mod4).....the = is congruence. Thank you this site is awesome
There are none.
gcd(2,4)=2 but 2 does not divide 3 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8397516012191772, "perplexity": 1922.8792665510061}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448399326483.8/warc/CC-MAIN-20151124210846-00225-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://ccrma.stanford.edu/~jos/pasp/Traveling_Wave_Solution.html | Next | Prev | Up | Top | Index | JOS Index | JOS Pubs | JOS Home | Search
### Traveling-Wave Solution
It can be readily checked (see §C.3 for details) that the lossless 1D wave equation
(where all terms are defined in Eq. (6.1)) is solved by any string shape which travels to the left or right with speed
If we denote right-going traveling waves in general by and left-going traveling waves by , where and are arbitrary twice-differentiable functions, then the general class of solutions to the lossless, one-dimensional, second-order wave equation can be expressed as
(7.2)
Note that we have and (derived in §C.3.1) showing that the wave equation is satisfied for all traveling wave shapes and . However, the derivation of the wave equation itself assumes the string slope is much less than at all times and positions (see §B.6). An important point to note is that a function of two variables is replaced by two functions of a single (time) variable. This leads to great reductions in computational complexity, as we will see. The traveling-wave solution of the wave equation was first published by d'Alembert in 1747 [100]7.1
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http://mymathforum.com/calculus/342117-pointwise-total-convergence-sum-power-sequence.html | My Math Forum Pointwise, total convergence, and sum of a (power?) sequence
Calculus Calculus Math Forum
October 3rd, 2017, 09:13 AM #1 Newbie Joined: Oct 2017 From: UK Posts: 2 Thanks: 0 Pointwise, total convergence, and sum of a (power?) sequence Hi guys... I seriously cannot help myself solving this problem: $\displaystyle \sum_1^\infty n^{2}e^{-nx^2-nx}$ (I'm not that good writing formulas but "n=1" and "+inf" are the sub and the sup) So I was asked to evaluate the pointwise and the total convergence of this sequence and find the sum. I have been thinking about to reduce it to a power sequence the fact is that I truly cannot think a way to do it.. Can you help me by resolving it and writing the steps? Thanks a lot! Best regards, An engineering student.
October 3rd, 2017, 12:28 PM #2 Global Moderator Joined: May 2007 Posts: 6,754 Thanks: 695 I am not sure what you mean by evaluate. For x=0, divergent. Otherwise convergent. Power series doesn't exist (as far as I can tell).
October 3rd, 2017, 01:03 PM #3 Newbie Joined: Oct 2017 From: UK Posts: 2 Thanks: 0 Yes, I probably make some mistakes while writing the first task of the exercise, ... maybe I got some problems with the translation since I am not English, sorry... but by the way I solved the first task.. what I cannot do now is to calculate the sum, that is the most difficult part...any ideas sir ?
October 5th, 2017, 01:34 PM #4 Global Moderator Joined: May 2007 Posts: 6,754 Thanks: 695 Good news! Sum is relatively easy. First note - divergence is for x in the closed interval from -1 to 0, not just 0. Sum: Recognize that $\displaystyle e^{-x^2-x}$ is an unnecessary complication. The problem is then $\displaystyle g(u)=\sum_{n=0}^\infty n^2u^n$, where $\displaystyle u=e^{-x^2-x}$. Starting at n=0 doesn't change anything. Step 1: $\displaystyle \sum_{n=0}^\infty n^2u^n=\sum_{n=0}^\infty n(n-1)u^n+\sum_{n=0}^\infty nu^n=g_1(u)+g_2(u)$ Step 2: Look at $\displaystyle h(u)= \sum_{n=0}^\infty u^n=\frac{1}{1-u}$, where |u|<1. Then $\displaystyle h'(u)=\sum_{n=0}^\infty nu^{n-1}=\frac{1}{(1-u)^2}$ and $\displaystyle h''(u)=\sum_{n=0}^\infty n(n-1)u^{n-2}=\frac{2}{(1-u)^3}$ Put all this together to get $\displaystyle g_1=u^2h''(u)=\frac{2u^2}{(1-u)^3}$ and $\displaystyle g_2=uh'(u)=\frac{u}{(1-u)^2}$ so that $\displaystyle g(u)=\frac{u^2+u}{(1-u)^3}$. Thanks from greg1313
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Contact - Home - Forums - Cryptocurrency Forum - Top | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8525646328926086, "perplexity": 1733.1611432911552}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232254882.18/warc/CC-MAIN-20190519121502-20190519143502-00254.warc.gz"} |
https://eguruchela.com/math/Calculator/sector-area | # Sector Area Calculator
Calculate the area of sector for given value which can be in radian or in degree.
Enter SA (Sector Angle) in radians / degrees
Area of Sector (If angle given in radians) Area of Sector (If angle given in degrees)
one degree is equal to 0.0174533 radian (approx), simlarlly one radian is equal to 57.2958 degree (approx).
The fraction is determined by the ratio of the arc length to the entire circumference.
$$\text {The area of the circle } = \pi \times radius^{2}$$ $$\text {and the circumference circle} = 2 \pi \times radius$$ Now we can use of the angle is clearly delineated, therefore we can replace once around (2π) with the subtended angle and we get the formula we needed for the sector as follows: $$Area = (\frac {Angle}{2}) \times radius^{2}$$ Let's take the example : radius is 7 cm and angle is 60o, now we can convert the 60o into radian value as follow:
Since 360o in radian = 2π, therefore 60o in radian = 60 * (2/360) = 1/3π = 0.3333π.
= 0.33333 X 3.141592654 = 1.047619
We can enter the value in both form as degree (60) or as radian (1.047619).
The area of sector can be calculated using following formula $$\text {Area of sector = } \pi r^{2} \cdot \frac {\theta^{o}}{360}$$ where π is 3.141592654, r is radius of thr circle and θ is angle in degree | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9950233101844788, "perplexity": 1620.0098161646881}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303779.65/warc/CC-MAIN-20220122073422-20220122103422-00161.warc.gz"} |
http://mathhelpforum.com/algebra/230520-rational-roots-problem.html | 1. ## Rational roots problem
A box manufacturer makes boxes of volume 500cm3 from pieces of tin 20cm on a side by cutting squares from each corner and folding up the edges. Find the length of each side of a square.
From the above from, I found some steps in solving the above problem. However, I followed through, gotten the answer but couldn't make sense of it. The followings are how I did it:
First : since we know that it cut out squares from the corners, the side of the squares is x. Hence, the length would be 20-2x.
The formula to solve for x would be x(20-x)(20-x). Factoring everything, I will have the answer 5 and 1.9 or 13.09(13.09 is not in the answer sheet)
My questions are, isn't the box suppose to be a square? Why does it have different measurement of the sides 5 and 1.9? Secondly the formula for volume is l*b*h. With sides at 5 and 1.9, I wonder how do i get the volume of 500?
2. ## Re: Rational roots problem
I'm not sure what you mean by "isn't the box supposed to be a square". The box is three dimensional so can't be a square which is two dimensional. Do you mean a cube? It could theoretically be but then 20- 2x would have to be the same as x: 20- 2x= x is the same as 3x= 20 or x= 20/3. Then the volume of the box would be (20/3)^3= 8000/27= 297.29... cubic inches, not 500.
Solving a general cubic can be very difficult but you titled this 'rational roots'. If this equation has a rational root, it must be an integer divisor of the constant term, -50. Trying x= 5 we have 5^3- 20(5^2)+ 100(5)- 125= 125- 500+ 500- 125= 0. Yes, x= 5 is a root. The other two roots are, as you say, irrational numbers around 1.9 and 13.
However, the roots are NOT all lengths of sides. x is a side of the square cut off and the height of the box. Its base is a square with sides of length 20- 2x. Taking x= 5, 20-2x= 20- 10= 10 so the box has a "10 by 10 base" with height 5 so volume 5(10)(10)= 500 as desired. Taking x= 1.9, 20- 2x= 20- 3.8= 16.2. The volume would be 1.9(16.2)(16.2)= 498.636, approximately 500 because we rounded the irrational sides to one decimal place. Taking x= 13 would give base length 20- 26= -6 which is impossible.
There are, in fact, two solutions to this problem: cut 5 cm from each corner to get a 5 by 10 by 10 cm box with volume 500 cc or cut $\frac{15- 5\sqrt{5}}{2}$ cm (approximately 1.9 cm) from each side to get a $\frac{15- 5\sqrt{5}}{2}$ by $5\sqrt{5}+5$ by $5\sqrt{5}+ 5$ (approximately 1.9 by 16.2 by 16.2).
3. ## Re: Rational roots problem
Hey hallsofivy, thank you for your reply. But I do not get it on how the height and the base could be of different measurements since a 20x20 sheet is shaved off uniformly in squares. Shouldn't the measurements of height width and base be the same? In addition, you were explaining that height is affected, but the manufacture were cutting off from the sides, which formed a shape of a plus sign. The centre of the plus sign is the base while folded up the edges, they form the height of the box, in that case height isn't affected. Did I get the question wrongly?
4. ## Re: Rational roots problem
Hello, xwy!
A box manufacturer makes boxes of volume 500cm3 from square pieces of tin 20cm on a side
by cutting squares from each corner and folding up the edges. Find the length of each side of a square.
From the above from, I found some steps in solving the above problem.
However, I followed through, gotten the answer but couldn't make sense of it.
The followings are how I did it:
First: since we know that it cut out squares from the corners, the side of the squares is x.
Hence, the length would be 20 - 2x.
The equation solve for x would be: . $x(20 - 2x)(20 - 2x) \:=\:500$
Factoring everything, I will have the answer 5 and 1.9 or 13.09(13.09 is not in the answer sheet)
My questions are, isn't the box suppose to be a square?
Why does it have different measurement of the sides 5 and 1.9?
Secondly the formula for volume is l*w*h.
With sides at 5 and 1.9, I wonder how do i get the volume of 500?
You forgot what $x$ represents.
$x$ = side of the corner squares to be removed.
Solving your cubic equation, we get three roots:
. . [tex]x \:=\:5,
**
5. ## Re: Rational roots problem
Hello, xwy!
A box manufacturer makes boxes of volume 500cm3 from square pieces of tin 20cm on a side
by cutting squares from each corner and folding up the edges. Find the length of each side of a square.
From the above from, I found some steps in solving the above problem.
However, I followed through, gotten the answer but couldn't make sense of it.
The followings are how I did it:
First: since we know that it cut out squares from the corners, the side of the squares is x.
Hence, the length would be 20 - 2x.
The equation solve for x would be: . $x(20 - 2x)(20 - 2x) \:=\:500$
Factoring everything, I will have the answer 5 and 1.9 or 13.09(13.09 is not in the answer sheet)
My questions are, isn't the box suppose to be a square?
Why does it have different measurement of the sides 5 and 1.9?
Secondly the formula for volume is l*w*h.
With sides at 5 and 1.9, I wonder how do i get the volume of 500?
You forgot what $x$ represents.
$x$ = side of the corner squares to be removed.
Solving your cubic equation, we get three roots:
. . $x \:=\:5,\;1.909830056,\;13.09016994$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8073504567146301, "perplexity": 905.7896849031695}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280850.30/warc/CC-MAIN-20170116095120-00098-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://proofwiki.org/wiki/Ring_of_Integers_Modulo_Prime_is_Field/Proof_1 | # Ring of Integers Modulo Prime is Field/Proof 1
## Theorem
Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$.
Then:
$m$ is prime
$\struct {\Z_m, +, \times}$ is a field.
## Proof
### Prime Modulus
$\struct {\Z_m, +, \times}$ is a commutative ring with unity by definition.
From Reduced Residue System under Multiplication forms Abelian Group, $\struct {\Z'_m, \times}$ is an abelian group.
$\Z'_m$ consists of all the elements of $\Z_m$ coprime to $m$.
Now when $m$ is prime, we have, from Reduced Residue System Modulo Prime:
$\Z'_m = \set {\eqclass 1 m, \eqclass 2 m, \ldots, \eqclass {m - 1} m}$
That is:
$\Z'_m = \Z_m \setminus \set {\eqclass 0 m}$
where $\setminus$ denotes set difference.
Hence the result.
$\Box$
### Composite Modulus
Now suppose $m \in \Z: m \ge 2$ is composite.
From Ring of Integers Modulo Composite is not Integral Domain, $\struct {\Z_m, +, \times}$ is not an integral domain.
From Field is Integral Domain $\struct {\Z_m, +, \times}$ is not a field.
$\blacksquare$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9958868026733398, "perplexity": 707.1588267970747}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655890092.28/warc/CC-MAIN-20200706011013-20200706041013-00574.warc.gz"} |
https://math.stackexchange.com/users/119075/volperossa?tab=topactivity | volperossa
### Questions (6)
7 Orthogonal Projection onto the ${L}_{2}$ Unit Ball 1 divergence (or net flow) on graphs and incidence matrix: difference with the classic divergence operator 1 gradient norm of a simple function 0 question about absolute value and inequality 0 verifing a limit with definition; finding delta when the interval is not circular …
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### Tags (22)
0 inequality × 3 0 norm 0 laplacian 0 optimization 0 limits 0 partial-derivative 0 linear-algebra 0 radicals 0 matlab 0 subgradient
### Bookmarks (19)
74 Understanding of the theorem that all norms are equivalent in finite dimensional vector spaces 54 how does expectation maximization work? 32 Divergence as transpose of gradient? 28 Why are additional constraint and penalty term equivalent in ridge regression? 25 how does the dot product determine similarity? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8524496555328369, "perplexity": 2047.0587725559535}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704804187.81/warc/CC-MAIN-20210126233034-20210127023034-00677.warc.gz"} |
https://www.physicsforums.com/threads/electron-confined-in-a-one-dimensional-box.870281/ | # Electron confined in a one dimensional box
1. May 3, 2016
### tboyers
1. The problem statement, all variables and given/known data
An electron confined in a one-dimensional box is observed, at different times, to have energies of 27 eV , 48 eV , and 75 eV .
What is the length of the box? Hint: Assume that the quantum numbers of these energy levels are less than 10.
2. Relevant equations
E=h^2n^2/(8mL^2)
3. The attempt at a solution
I tried using that equation to solve for length, but I dont know what energy levels these are at, so i cant seem to solve it.
2. May 3, 2016
### drvrm
energy of the states are proportional to n^2/L^2 and you have energies for three states so can you find out L?
Draft saved Draft deleted
Similar Discussions: Electron confined in a one dimensional box | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8771367073059082, "perplexity": 746.5726486542445}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948513512.31/warc/CC-MAIN-20171211125234-20171211145234-00101.warc.gz"} |
https://web.mit.edu/16.unified/www/SPRING/thermodynamics/notes/node44.html | # 6.2 The Thermodynamic Temperature Scale
The considerations of Carnot cycles in this section have not mentioned the working medium. They are thus not limited to an ideal gas and hold for Carnot cycles with any medium. Earlier we derived the Carnot efficiency with an ideal gas as a medium and the temperature definition used in the ideal gas equation was not essential to the thermodynamic arguments. More specifically, we can define a thermodynamic temperature scale that is independent of the working medium. To see this, consider the situation shown below in Figure 6.2, which has three reversible cycles. There is a high temperature heat reservoir at and a low temperature heat reservoir at . For any two temperatures , , the ratio of the magnitudes of the heat absorbed and rejected in a Carnot cycle has the same value for all systems.
We choose the cycles so is the same for A and C. Also is the same for B and C. For a Carnot cycle
Also
But
Hence
We thus conclude that has the form , and similarly . The ratio of the heat exchanged is therefore
In general,
so that the ratio of the heat exchanged is a function of the temperature. We could choose any function that is monotonic, and one choice is the simplest: . This is the thermodynamic scale of temperature, . The temperature defined in this manner is the same as that for the ideal gas; the thermodynamic temperature scale and the ideal gas scale are equivalent.
Douglas Quattrochi 2006-08-06 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9721073508262634, "perplexity": 286.8180031618076}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030333541.98/warc/CC-MAIN-20220924213650-20220925003650-00046.warc.gz"} |
https://www.prepanywhere.com/prep/textbooks/calculus-and-vectors-nelson/chapters/chapter-4-curve-sketching/materials/4-5-an-algorithm-for-curve-sketching | 4.5 An Algorithm for Curve Sketching
Chapter
Chapter 4
Section
4.5
Solutions 23 Videos
If a polynomial function of degree three has a local minimum, explain how the functions values behave as x\to \infty and x\to -\infty. Consider all cases.
1.19mins
Q1
How many local maximum and local minimum values are possible for a polynomial function of degree three, four, or n? Explain.
0.47mins
Q2
Determine whether each function has vertical asymptotes. If it does, state the equations of the asymptotes.
y = \displaystyle{\frac{x}{x^2 +4x + 3}}
0.19mins
Q3a
Determine whether each function has vertical asymptotes. If it does, state the equations of the asymptotes.
y = \displaystyle{\frac{5x -4}{x^2-6x + 12}}
0.42mins
Q3b
Determine whether each function has vertical asymptotes. If it does, state the equations of the asymptotes.
y = \displaystyle{\frac{3x + 2}{x^2-6x + 9}}
0.17mins
Q3c
Use the algorithm for curve sketching to sketch the following:
\displaystyle y = x^3 - 9x^2 + 15x +30
3.54mins
Q4a
Use the algorithm for curve sketching to sketch the following:
\displaystyle f(x) = -4x^3 + 18x^2 + 3
2.08mins
Q4b
Use the algorithm for curve sketching to sketch the following:
\displaystyle y = 3 + \frac{1 }{(x + 2)^2}
2.47mins
Q4c
Use the algorithm for curve sketching to sketch the following:
\displaystyle f(x) = x^4 -4x^3 - 9x^2 + 48x
7.26mins
Q4d
Use the algorithm for curve sketching to sketch the following:
\displaystyle y = \frac{2x}{x^2 - 25}
5.32mins
Q4e
Use the algorithm for curve sketching to sketch the following:
\displaystyle f(x) =\frac{1}{x^2 -4x}
7.12mins
Q4f
Use the algorithm for curve sketching to sketch the following:
\displaystyle y = \frac{6x^2 -2}{x^3}
6.00mins
Q4g
Use the algorithm for curve sketching to sketch the following:
\displaystyle y = \frac{x + 3}{x^2 -4}
8.49mins
Q4h
Use the algorithm for curve sketching to sketch the following:
\displaystyle y = \frac{x^2 -3x +6}{x -1}
5.43mins
Q4i
Use the algorithm for curve sketching to sketch the following:
\displaystyle y = (x - 4)^{\displaystyle{\frac{2}{3}}}
2.18mins
Q4j
Determine the constants a, b, c, and d so that the curve defined by y = ax^3 + bx^2 +cx +d has a local maximum at the point (2, 4) and a point of inflection at the origin.
4.44mins
Q6
Given the following results of the analysis of a function, sketch a possible graph for the function:
f(0) = 0, the horizontal asymptote is y =2, the vertical asymptote is x = 3 and f'(x) < 0 and f''(x) < 0 for x < 3 and f''(x) > 0 for x > 3.
1.25mins
Q7a
Give the following results of the analysis of a function, sketch a possible graph for the function:
f(0) = 6, f(-2) = 0 the horizontal asymptote is y = 7, the vertical asymptote is x = -4, and f'(x) > 0 and f''(x) > 0 for x < -4; f'(x) > 0 and f''(x)< 0 for x > -4.
3.28mins
Q7b
Sketch the graph of f(x) = \displaystyle{\frac{k -x}{k^2 + x^2}}, where k is any positive constant.
14.12mins
Q8
Sketch the cute defined by g(x) = x^{{\frac{1}{3}}}(x + 3)^{{\frac{2}{3}}}.
9.29mins
Q9
Find the horizontal asymptotes.
f(x) = \displaystyle{\frac{x}{\sqrt{x^2 + 1}}}
1.21mins
Q10a
Find the horizontal asymptotes.
g(t) = \sqrt{t^2+4t} - \sqrt{t^2 + t}
3.02mins
Q10b
Show that, for any cubic function of the form y = ax^3 + bx^2 + cx + d, there is a single point of inflection, and the slope of the cute at that point is c-\displaystyle{\frac{b^2}{3a}}.
1.39mins
Q11
Lectures 2 Videos
Full Graphing ex1
Sketch f(x) = x^4 -8x^3
Sketch \displaystyle f(x) =x^{\frac{2}{3}}(6-x)^{\frac{1}{3}} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9340656399726868, "perplexity": 1034.2576663220502}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141171126.6/warc/CC-MAIN-20201124053841-20201124083841-00384.warc.gz"} |
https://scicomp.stackexchange.com/questions/14760/how-to-impose-an-integral-conservation-in-solving-odes-boundary-value-problems | # How to impose an integral conservation in solving ODEs boundary value problems (BVP)?
I have a system of coupled ODEs that I want to solve. The functions are A(x), B(x), C(x). It is a boundary values problem. I am using Matlab bvp4c.
So far I am not satisfied with my solutions. For the boundaries that are of interest for me, the solver fail (Matlab return "a singular Jacobian encountered"). For some other boundaries, the result depend of the initial guess. So I think that my system is ill defined. I am thinking of adding one constraint but I don't find how to implement it
How to enforce $\int_{a}^{b}d x(A(x)+B(x)+C(x))=N$ ?
then $I(a) = 0$, $I(b) = N$, $\dot{I}(x) = A(x) + B(x) + C(x)$, and you have another boundary value problem that you can solve in MATLAB using bvp4c.
• Is the condition $I(a)=0$ necessary? Since $I(x)=\int_{a}^{x} (A(s)+B(s)+C(s)) ds$, then $I(a)=0$ comes by definition so it should not be required. I am asking this because I have now a system with too many boundary conditions since I added only one equation of first order but with two additional constraints ($I(a)$ and $I(b)$) – David Dec 23 '14 at 14:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9360867738723755, "perplexity": 260.4488967418391}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178378043.81/warc/CC-MAIN-20210307170119-20210307200119-00576.warc.gz"} |
http://tex.stackexchange.com/questions/16748/how-can-i-prevent-citations-with-the-second-author-as-firstname-lastname?answertab=active | # How can I prevent citations with the second author as Firstname Lastname
I am using biblatex with the following command:
\usepackage[backend=biber,style=authoryear-comp,
bibstyle=authoryear,citestyle=authoryear-comp]{biblatex}
Now when I do a \textcite I sometimes get the following output:
Smit and John Doe, 2011
And in the bibliography
Smit, Peter and John Doe (May 2011). “Using
I find this ugly. How can I get the citation to be:
Smit and Doe, 2011
And the bibliography to be
Smit, Peter and Doe, John (May 2011). “Using
-
It is possible that \textcite will result in citations like Smit and John Doe, 2011 (including Does first name) because biblatex since v1.4 will fully disambiguate individual author names and name lists. Just see my example.
The Month appearing in addition to the year is actually a built-in feature of biblatex. Just use the date field instead of year, and conform to the required format. (Details may be found in section 2.3.8 of the documentation.)
For the reordering of names, I prefer the form \DeclareNameAlias{sortname}{last-first}. See Guidelines for customizing biblatex styles ("Bibliography -- order of first names and last names") for details.
\documentclass{article}
\usepackage[style=authoryear-comp,backend=biber]{biblatex}
\DeclareNameAlias{sortname}{last-first}
\begin{filecontents}{\jobname.bib}
@misc{S11x,
author = {Smit, Peter and Doe, Jane},
date = {2011},
title = {Me and Jane},
}
@misc{S11y,
author = {Smit, Peter and Doe, John},
date = {2011},
title = {Me and John},
}
@misc{S11z,
author = {Smit, Peter and McGee, Bobby},
date = {2011-05},
title = {Me and Bobby McGee},
}
\end{filecontents}
\nocite{*}
\begin{document}
\textcite{S11x}
\textcite{S11y}
\textcite{S11z}
\printbibliography
\end{document}
EDIT: Name disambiguation only works with biber as backend. Replace biber with bibtex8 in my example and see what happens. ;-)
-
Thanks a lot! This was indeed my problem. I had a certain bib entry with Doe, J and an entry with Doe, John. Fixing that solved my problem. About the month, I don't mind that one being there at all. – Peter Smit Apr 27 '11 at 16:45
You don't need to use multiple bibstyle and citestyle commands if you are using the same format for both. So your command to load biblatex should be:
\usepackage[backend=biber,style=authoryear-comp]{biblatex}
(Since the authoryear-comp and authoryear bbx files are the same, there's no need to have separate bibstyle and citestyle specifications in this case.)
Your citations should show up as you request with the standard authoryear-comp format. If only certain items show up incorrectly, then this is likely a problem with how the .bib record itself is entered. If after comparing a .bib file record that works with one that doesn't, you are still having problems with this, you should edit your question showing the actual record from your .bib file that causes the problem.
With respect to the ordering of names, add the following to your preamble:
\DeclareNameAlias{last-first/first-last}{last-first}
With respect to having the Month appear in the date field in the bibliography, this is quite unorthodox. Are you sure you want to do this? I'm sure it's possible, but it might require some work.
-
Thanks for the tips. Actually, when you add the month field to the bib file, it will be included automatically. For the citation, I tried it now with an empty tex file, and it seems the same commands give me now only last names. Are there any packages that could interfere with biblatex? – Peter Smit Apr 27 '11 at 16:41
A, no packages, but lockstep's answer was spot on. Thanks for the help anyway! – Peter Smit Apr 27 '11 at 16:46
@Peter: You should still accept @Alan's advice with regard to using multiple style options. – lockstep Apr 27 '11 at 18:09
I switched to BibLaTeX just today and these information are priceless – especially the last \DeclareNameAlias. Thank you. – Harold Cavendish Oct 15 '11 at 20:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8624991178512573, "perplexity": 3673.6619377157344}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00657-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.enotes.com/homework-help/sketch-the-graph-of-a-function-that-satisfies-all-2231389 | # Math
Sketch the graph of a function that satisfies all of the given conditions:
• f'(0) = f'(4)=0
• f'(x) =1 if x<-1
• f'(x)>0 if 0<x<2
• f'(x)<0 if -1<x<0 or 2 <x <4 or x>4
• lim x--2- f'(x) = infinity
• lim x--2+ f'(x) = negative infinity
• f''(x)>0 if -1<x<2 or 2<x<4
• f''(x)<0 if x>4
We are asked to sketch the graph of a function subject to the following constraints:
f'(0)=f'(4)=0f'(x)=1 for x<-1f'(x)>0 for 0<x<2f'(x)<0 for -1<x<0; 2<x<4; x>4`lim_(x->2^(-1))=oo`
`lim_(x->2^(+)=-oo`
f''(x)>0 -1<x<2; 2<x<4f''(x)<0 x>4
(See attachment for sketch)
If we assume the function to be continuous we have a local...
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We are asked to sketch the graph of a function subject to the following constraints:
f'(0)=f'(4)=0
f'(x)=1 for x<-1
f'(x)>0 for 0<x<2
f'(x)<0 for -1<x<0; 2<x<4; x>4
`lim_(x->2^(-1))=oo`
`lim_(x->2^(+)=-oo`
f''(x)>0 -1<x<2; 2<x<4
f''(x)<0 x>4
(See attachment for sketch)
If we assume the function to be continuous we have a local minimum at x=-1 and a local maximum at x=2 as the derivative changes sign at a critical point. (x=-1 is a critical point as the first derivative is zero; x=2 is a critical point as the first derivative does not exist there.)**If we do not assume continuity the function could have a vertical asymptote at x=2.
At x=4 the first derivative is zero but it does not change sign on either side; the second derivative changes sign at x=4 thus there is an inflection point there.
Since the first derivative is constant on x<-2 you have a straight line with slope 1. At x=0 we have a local minimum as discussed above (also by the second derivative test.) The graph is concave up on the interval (-1,2).
At x=2 the first derivative changes sign (goes from positive to negative) while the second derivative remains positive (the graph remains concave up). Either the graph has a cusp at x=2 or the graph has a vertical asymptote at x=2 (your choice).
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http://tex.stackexchange.com/questions/91254/can-i-redefine-begin-so-that-it-does-not-complain-about-undefined-environment | # Can I redefine \begin so that it does not complain about undefined environment
I'd like to be able to do the following kind of thing:
\documentclass{standalone}
\begin{document}
Foo
\begin{MyUndefinedEnv}
Bar
\end{MyUndefinedEnv}
\end{document}
without LaTeX complaining about undefined environment. Is it possible without breaking existing environments?
-
Of course, I can patch the definition of \begin to remove the \@latex@error (p.204 of source2e.pdf). As a \csname is invoked it should not be a problem if the control sequence is not defined. But I'd like to hear it is safe from gurus :). – cjorssen Jan 10 '13 at 22:53
Why do you want to use undefined environments? Just curious... – marczellm Jan 10 '13 at 22:56
@marczellm Coding environments is a lot more fun than writing a whole lecture about the phase plane. So every time I need a new environment, I spend time coding it rather than doing the right thing :). – cjorssen Jan 10 '13 at 22:59
@cjorssen: Hmmm, I think you are missing the whole point of procrastinating.. :-) – Peter Grill Jan 10 '13 at 23:02
Yes, you can:
\documentclass{article}
\makeatletter
\let\xbegin\begin % store original \begin
\let\xifundefined\@ifundefined % store original \@ifundefined
\def\begin{%
% "inactivate" \@ifundefined, but only once, hence reverting
% it to original definition immediately
\def\@ifundefined##1##2##3{\global\let\@ifundefined\xifundefined##3}%
\xbegin}
\makeatother
\begin{document}
\begin{ttgg}
Hello!
\end{ttgg}
\begin{center}
Hello!
\end{center}
\end{document}
-
Interesting, thanks. – cjorssen Jan 10 '13 at 23:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8351452946662903, "perplexity": 1431.325741998983}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257826773.17/warc/CC-MAIN-20160723071026-00009-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://amathew.wordpress.com/tag/ufds/ | Earlier I went over the definition and first properties of a discrete valuation ring. Today, it’s time to say how we can tell a ring is a DVR–it turns out to be not too bad, which is nice because the properties we need in this criterion are often easier to work with than the existence of some discrete valuation.
Today’s result is:
Theorem 1 If the domain ${R}$ is Noetherian, integrally closed, and has a unique nonzero prime ideal ${\mathfrak{m}}$, then ${R}$ is a DVR. Conversely, any DVR has those properties. (more…) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 3, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8341214060783386, "perplexity": 214.00795885626908}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347436466.95/warc/CC-MAIN-20200603210112-20200604000112-00020.warc.gz"} |
http://mathoverflow.net/questions/23026/how-can-i-understand-the-groupoid-quotient-of-a-group-action-as-some-sort-of/23040 | # How can I understand the “groupoid” quotient of a group action as some sort of “product”?
Recall the notion of groupoid (Wikipedia, nLab). An important construction of groupoids is as "action groupoids" for group actions. Namely, let $X$ be a groupoid and $G$ a group, and suppose that $G$ acts on $X$ by groupoid automorphisms. Then we can form a new groupoid $X//G$, which has as objects the objects of $X$, but the morphisms include, in addition to the original morphisms of $X$, a morphism $x \overset g \to gx$ for each $g\in G$ and $x\in X$. The composition of morphisms is well-defined if the action is by groupoid automorphisms. (When $X$ is a set, then $X//G$ is equivalent to the skeletal groupoid whose objects are the elements of the "coarse" quotient $X/G$, and with ${\rm Aut}(\bar x) = {\rm Stab}_G(x)$.)
(Probably there is a fancier construction, in which the conditions on the word "group action" be relaxed to an "action" up to specified natural isomorphism, and then $G$ could act on $X$ by autoequivalences, rather than autoisomorphisms, but this generalization won't concern me.)
Let $1$ denote the one-point set, thought of as a groupoid with only identity morphisms. Then any group $G$ acts uniquely on $1$, and so we have the groupoid $1//G$. In general, although $X\times 1 \cong X$, we do not have $X \times (1//G) \cong X//G$ for arbitrary $G$-actions on $X$ unless the action is trivial. (Here $\times$ denotes the groupoid product, which is just what you think it is.) However, the construction provides natural bijections between the objects of $X//G$ and the objects of $X \times (1//G)$, and between the morphisms of $X//G$ and the morphisms of $X \times (1//G)$.
Question: Is there some sort of "semidirect" or "crossed" product of groupoids, which presumably depends on extra data, so that we do have $X//G \cong X \rtimes (1//G)$? By which I mean, what is the correct notion of "action" of a groupoid $Y$ on a groupoid $X$ and what is the corresponding correct notion of $X \rtimes Y$?
I see that the page semidirect product in nLab defines $X \rtimes G$ as something closely related to $X//G$. But clearly this ought to be called $X\rtimes (1//G)$, but then I do not know what the right definition for $X\rtimes Y$ is, hence the question. And really I'd like to know about a "double crossed product" $X\bowtie Y$.
My motivation for this question is from my answer to Do rational numbers admit a categorification which respects the following “duality”?.
-
By the way, Theo, the nLab has a forum where you will be much more likely to get an answer quickly to all of these questions. math.ntnu.no/~stacey/Vanilla/nForum – Harry Gindi Apr 29 '10 at 21:20
@HG: Thanks! I didn't know about that forum. – Theo Johnson-Freyd Apr 30 '10 at 3:58
Thanks for asking this question! I was working on it after your answer to my question, but wasn't getting very far. – Steven Gubkin Apr 30 '10 at 16:02
Let $X$ and $Y$ be groupoids. An action of $Y$ on $X$ is a functor $\rho: Y \to B\operatorname{Aut}(X)$, where $B\operatorname{Aut}(X)$ is the one-object 2-groupoid such that $\operatorname{Hom}(\ast, \ast)$ is the 2-group of autoequivalences of $X$.
We define $X \rtimes Y$ as follows. Its objects are simply $\operatorname{Ob}(X) \times \operatorname{Ob}(Y)$. An element of $(X \rtimes Y)((x_1, y_1), (x_2, y_2))$ consists of a pair $(f, g)$, where $g \in Y(y_1, y_2)$, and $f \in X(x_1, \rho(g)x_2)$. Given $(f, g) \in (X \rtimes Y)((x_1, y_1), (x_2, y_2))$, and $(f', g') \in (X \rtimes Y)((x_2, y_2), (x_3, y_3))$, we define $(f', g') \circ (f, g) \in \operatorname{Hom}((x_1, y_1), (x_3, y_3))$ as $(\rho(g)f' \circ f, g' \circ g)$. It is straightforward to check that in the case that $Y$ is $1 // G$, $X \rtimes Y \cong X // G$.
-
Also, this is a special case of the Grothendieck construction (ncatlab.org/nlab/show/Grothendieck%20construction) for pseudofunctors into Cat, where the domain is a groupoid $Y$ and the pseudofunctor happens to take all the objects of $Y$ to the object $X\in Cat$. – Mike Shulman Apr 30 '10 at 14:29
You can also view the groupoid $X$ together with a $Y$-action as a groupoid in the category $Y-Set \simeq Set^{Y^{op}},$ hence a presheaf of groupoids on $Y,$ and perform the Grothendieck construction to this. – David Carchedi Feb 14 '13 at 21:05
The notion of semidirect product $\Gamma \rtimes G$ where $G$ is a group acting on a groupoid $\Gamma$ is set up in Chapter11, Section 11.4, of my book "Topology and Groupoids".
It is used there in connection with studying orbit groupoids, and their relevance to the fundamental groupoid of an orbit space by a group action.
One nice point is that this semidirect product includes the case $\Gamma$ is a discrete groupoid, i.e. essentially a set, when you get what is commonly called the action groupoid. In this case the morphism $p: \Gamma \rtimes G \to G$ is known as a covering morphism of groupoids, and all covering morphisms of $G$ arise in this way.
I feel the use of covering morphisms of groupoids makes for a nice exposition, base point free, of the theory of covering spaces. Such an idea was pointed out for the simplicial case in the 1967 book on simplicial theory by Gabriel and Zisman, was used in the first 1968 edition of my book, and is used in Peter May's 1999 book "A concise course in algebrac topology".
Update: I should also add that the notion of action of a groupoid on a groupoid is given, following Ehresmann, in my paper
[11] Groupoids as coefficients'', Proc. London Math Soc. (3) 25 (1972) 413-426.
available here. The aim was cohomology with coefficients in a groupoid. One of the methods exploited is fibrations of grouopids.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9589172601699829, "perplexity": 161.39097069910582}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860121534.33/warc/CC-MAIN-20160428161521-00131-ip-10-239-7-51.ec2.internal.warc.gz"} |
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# How many numbers of two digits are divisible by $3$ ?
Last updated date: 25th Mar 2023
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Answer
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Hint: Select the lowest and highest two digit terms divisible by $3$. To see whether the series is an A.P or not. Then if it is in A.P solve it by taking the ${{n}^{th}}$ term of the A.P. Find the value of $n$. You will get the answer.
We use ten digits in the way we count. Sometimes we use one digit to represent a number and sometimes we use more. In this lesson, you'll learn how many digits are in any given numeral, and recognize a digit when you see it. You will also begin learning about place value.
This is the numeral: $51$.These are the digits in that numeral: $5$ and $1$. The numeral has two digits because two symbols, or digits, make up the numeral $51$. What about this numeral ? $452$. This numeral has three digits: $4,5$ and $2$.
That $7$ is a special kind of symbol. It's called a digit. We have ten digits we use to make up all numerals. A numeral is a number written down. These digits are $0,1,2,3,4,5,6,7,8$ and $9$. That's it! You can make any numeral you want out of those ten digits.
Just like a red light means 'stop', a $7$ means 'seven'. It's a single symbol that represents a numeral. Yes, just like this is a symbol to stop.
We know, first two digit number divisible by $3$ is $12$ and the last two digit number divisible by $3$ is $99$. Thus, we get $12,15,......,99$.
So the lowest two digit number divisible by $3$ is $12$.
Highest two digit number divisible by $3$ is $99$.
So we can see the difference between the numbers that are divisible by 3, is $3$.
So the above series is in A.P.
We have to find it in terms of $n$.
And here$a=12,d=3,{{a}_{n}}=99$
Thus, the ${{n}^{th}}$term of A.P is :
${{a}_{n}}=a+(n-1)d$
Where,
$a=$First-term
$d=$ Common difference
$n=$ number of terms
${{a}_{n}}={{n}^{th}}$term
So now applying the formula for 99, we get,
$99=12+(n-1)3$
Simplifying further we get,
\begin{align} & 99-12=(n-1)3 \\ & 87=3n-3 \\ & 90=3n \\ & n=30 \\ \end{align}
So we get $n=30$.
Therefore, the number of two digits divisible by $3$ are $30$.
Note: Read the question properly. Also, we should know the lowest and highest two digit terms divisible by $3$. So the concepts related to A.P should be clear. Here we have used the concept of A.P that is we have used ${{n}^{th}}$ term of A.P which is ${{a}_{n}}=a+(n-1)d$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9682738780975342, "perplexity": 556.3578902584226}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949644.27/warc/CC-MAIN-20230331144941-20230331174941-00547.warc.gz"} |
https://rakhesh.com/tag/docx/ | Word 2010 – The xxxx.docx cannot be opened because there are problems with the contents
Got the following error for a Word document at work.
Obviously your mileage may vary in terms of the fix but here’s what I did so there’s a starting point.
Since this is a docx file I extracted it using 7-Zip. Went through the XML files in it but they seemed fine. Next I extracted another working docx file and replaced the “[Content_Types].xml” file of the broken one with that of the working one. Zipped it all back into a docx file, double clicked, and I got a different error now but the document opened fine. It complained about comments or something missing, but all that was expected as obviously I had replaced a master file with another one. The fact that it opened fine (more or less) confirmed that this file must be the culprit.
Next I tried removing bits and pieces from the broken “[Content_Types].xml” file but that didn’t help. Finally I compared the two side by side, starting with the stuff I hadn’t removed. I noticed that the broken file had an entry like this:
The same one in the working file was different:
So I replaced the line in the broken file with the one in the working file, zipped it all back, double clicked and voila! it opens fine now. :)
From this MIME types document it seems like “application/vnd.ms-word.document.macroEnabled.12” is a “.docm” file so at this point my guess is that the user copy pasted something from another document and that possible corrupted his destination document? I don’t know.
How to remove complex scripts from Word DOCX documents
Recently came across a Word document where some parts of the document seemed to ignore the general rules. The document was in English, and its language was set to English (U.S.) but certain parts were set to Arabic (Saudi) and none of the usual methods of selecting the text and marking it as English (U.S.) was helping. Very weird.
After a lot of fiddling around I also noticed that if I change the style of a paragraph containing such text, the adjoining text changes but this particular one stays as it. I am able to change the font and size directly by applying them, but changes via styles seem to get ignored.
Then I realized that although this text was in English, since it was marked as Arabic (Saudi) they were being treated as “complex scripts” in the style definitions and hence had separate rules. I guess that at some point someone had marked this text as of being Arabic (Saudi) and continued typing in English, or perhaps the original text was Arabic but someone had changed the font to an English one like “Times New Roman” and typed in English, so even though the text was appearing as English in fact Word was treating it as Arabic written in English (I guess). Anyways, point was Word was treating these blocks as complex scripts (as opposed to Latin for other parts) and so the usual formatting rules didn’t apply to them. Moreover I could change the language from Arabic (Saudi) to Arabic (UAE), for instance, so that seemed to support my theory that it was letting me changing the language to other complex scripts – just not from complex to Latin and vice versa.
This being a DOCX file, it is really just a zip file. So I unzipped it using 7-Zip. Went to the word\styles.xml file (which I came across through trial and error actually, I went through pretty much all the XML files there) in the extracted folder and found the following:
Since I didn’t want the document to have any Arabic at all, I simply changed the “ar-SA” to “en-US”. Saved the XML file, went back to the extracted folder, and zipped all its contents up again. Renamed this from .zip to .docx and opened the document, and bingo! now all that complex stuff weirdness was gone! :)
(A word to note about zipping back the folder. The format is ZIP. And also, don’t zip the top level folder as then your zip file will be the top level folder followed by all the sub-folders. No, what we want is that the zip file is all the sub-folders directly). | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8597467541694641, "perplexity": 1653.0456806477468}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376828018.77/warc/CC-MAIN-20181216234902-20181217020902-00156.warc.gz"} |
https://www.physicsforums.com/threads/dynamics-problem.774817/ | # Dynamics problem
1. Oct 7, 2014
### phyziks4lyfe
• Warning! Posting template must be used for homework questions.
this is a part of a much bigger problem, I have a 600lb motorcycle going from point A to point B, which are 500 ft apart, i know the force exerted by the road on the motorcycle when the throttle is pressed is F(t)=300(1-.1t).
I need time in order velocity at B ,
I'm really lost on where to start on this, i was thinking f=ma could give me acceleration, but then i dont know what to do with that.
2. Oct 7, 2014
### paisiello2
Are you trying to determine the velocity at point B?
If you know the acceleration then by definition you should be able to get the velocity.
3. Oct 7, 2014
### phyziks4lyfe
so f=ma will give me a acceleration as a function of t, and yes i need velocity at B, but then won't i only have velocity as a function of t if by definition?.
4. Oct 7, 2014
### RUber
Right, but if you have velocity, you can find position. Find how long it takes to get to point B, then use that t for your velocity.
5. Oct 7, 2014
### SteamKing
Staff Emeritus
Gee, if only there were a way that acceleration could be used to get velocity, and that velocity could then be used to find distance. It would seem to be a good reason to study rectilinear motion in physics, or something. Maybe somebody wrote a textbook on how to do this stuff. IDK, that would probably take a lot of work to figure out.
6. Oct 7, 2014
### phyziks4lyfe
so would i use Vi^2=Vf^2+2A(500) and substitute in V and A as their function of t equations? because then i get t= 10.05 does that seem right?
7. Oct 7, 2014
### RUber
What units are you using for m, F, and distance? I hope F is in ft/lbs.
8. Oct 7, 2014
### phyziks4lyfe
m = slugs, F lbs, distance = ft
9. Oct 7, 2014
### haruspex
No, that formula only works for constant acceleration. You need to use the general relationships between acceleration, velocity, and distance - namely, integral formulas.
Draft saved Draft deleted
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# proper tail recursion proposal, take 2
My original concern is resolved. Not to be nitpicky, but I just plain
don't understand this sentence:
A Scheme implementation is
properly tail-recursive if it supports an unbounded number of active
tail calls (a call is {\em active} if the called procedure may still
return).
In this, are you saying that if F calls G normally and G calls H as a tail
call then F, G, and H are still active? The parenthetical remark would seem
to describe G as inactive, since it can no longer return--H will return
directly to F, and and G is effectively inactive. But if I'm to believe there
is an unbounded number of active tail calls permitted, I presumably must model
G as still "active" even when removed from the "stack" or "chain" of calls.
To clarify, the wording that makes sense to me (but seems in conflict with
your chosen terms above) would be:
A Scheme implementation is {\em properly tail recursive} if making a tail
call does not increase the number of {\em active} functions (i.e., functions
that are still waiting to return).
However, this is not a specific proposal for change. In truth, I kinda liked
the old wording which talked about not using additional space, etc. It seemed
somehow more precise. I guess I can infer that if it has enough space for
many that it must have enough space for 1, but... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9355376958847046, "perplexity": 3895.1584526478146}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578678807.71/warc/CC-MAIN-20190425014322-20190425040322-00150.warc.gz"} |
http://mathhelpforum.com/calculus/173499-monotonic-decreasing-print.html | # monotonic decreasing
• March 5th 2011, 06:28 AM
transgalactic
monotonic decreasing
$\sum_{K=1}^{INFINITY}\frac{k^{2}}{e^{k^{3}}}$
the dividing of a_n+1 /a_n is not conlusive
?
• March 5th 2011, 06:33 AM
Plato
Quote:
Originally Posted by transgalactic
$\sum_{K=1}^{INFINITY}\frac{k^{2}}{e^{k^{3}}}$ the dividing of a_n+1 /a_n is not conlusive?
What exactly is the question? The title is almost meaningless.
• March 5th 2011, 08:41 AM
transgalactic
i need to prove that a_n is monotonickly decreasing
• March 5th 2011, 08:45 AM
Plato
Quote:
Originally Posted by transgalactic
i need to prove that a_n is monotonickly decreasing
Can you show that $\dfrac{x^2}{e^{x^2}}$ is decreasing for $x\ge 1~?$
• March 5th 2011, 10:01 AM
Sambit
hint: the denominator increases more rapidly than the numerator.
• March 5th 2011, 10:43 PM
transgalactic
i can show a_n+1 -a_n <0
or
a_n+1 /a_n=q
and show that q<1
both i have written but i dont know how to conclude from the expression
• March 6th 2011, 04:42 AM
HallsofIvy
Quote:
Originally Posted by transgalactic
i can show a_n+1 -a_n <0
or
a_n+1 /a_n=q
and show that q<1
both i have written but i dont know how to conclude from the expression
Now, you have me confused! $a_{n+1}- a_n< 0$ is the same as $a_{n+1}< a_n$ and that is the definition of "decreasing".
Or are you saying that you know you have to prove one of those but don't know how?
I would look at $\frac{a_{k+1}}{a_k}= \frac{(k+1)^2}{e^{(k+1)^3}}\frac{e^{k^3}}{k^2}$ $= \frac{(k+1)^2}{k^2}e^{k^3-((k+1)^3}= \frac{(k+1)^2}{k^2}e^{-3k^2- 3k- 1}$.
Now, in the limit, as k goes to infinity, that fraction goes to 1 so this "eventually" decreases to 0.
If you really need to prove that it is decreasing for all k greater than or equal to 1, I recommend Plato's suggestion. Take the dervative of $x^2e^{-x^3}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9619415998458862, "perplexity": 2157.6160238844927}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931009004.88/warc/CC-MAIN-20141125155649-00153-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://www.rdocumentation.org/packages/quadprog/versions/1.5-7/topics/solve.QP.compact | # solve.QP.compact
0th
Percentile
##### Solve a Quadratic Programming Problem
This routine implements the dual method of Goldfarb and Idnani (1982, 1983) for solving quadratic programming problems of the form $\min(-d^T b + 1/2 b^T D b)$ with the constraints $A^T b >= b_0$.
Keywords
optimize
##### Usage
solve.QP.compact(Dmat, dvec, Amat, Aind, bvec, meq=0, factorized=FALSE)
##### Arguments
Dmat
matrix appearing in the quadratic function to be minimized.
dvec
vector appearing in the quadratic function to be minimized.
Amat
matrix containing the non-zero elements of the matrix $A$ that defines the constraints. If $m_i$ denotes the number of non-zero elements in the $i$-th column of $A$ then the first $m_i$ entries of the $i$-th column of Amat hold these non-zero elements. (If $maxmi$ denotes the maximum of all $m_i$, then each column of Amat may have arbitrary elements from row $m_i+1$ to row $maxmi$ in the $i$-th column.)
Aind
matrix of integers. The first element of each column gives the number of non-zero elements in the corresponding column of the matrix $A$. The following entries in each column contain the indexes of the rows in which these non-zero elements are.
bvec
vector holding the values of $b_0$ (defaults to zero).
meq
the first meq constraints are treated as equality constraints, all further as inequality constraints (defaults to 0).
factorized
logical flag: if TRUE, then we are passing $R^{-1}$ (where $D = R^T R$) instead of the matrix $D$ in the argument Dmat.
##### Value
a list with the following components:
solution
vector containing the solution of the quadratic programming problem.
value
scalar, the value of the quadratic function at the solution
unconstrained.solution
vector containing the unconstrained minimizer of the quadratic function.
iterations
vector of length 2, the first component contains the number of iterations the algorithm needed, the second indicates how often constraints became inactive after becoming active first.
Lagrangian
vector with the Lagragian at the solution.
iact
vector with the indices of the active constraints at the solution.
##### References
D. Goldfarb and A. Idnani (1982). Dual and Primal-Dual Methods for Solving Strictly Convex Quadratic Programs. In J. P. Hennart (ed.), Numerical Analysis, Springer-Verlag, Berlin, pages 226--239.
D. Goldfarb and A. Idnani (1983). A numerically stable dual method for solving strictly convex quadratic programs. Mathematical Programming, 27, 1--33.
solve.QP
##### Aliases
• solve.QP.compact
##### Examples
# NOT RUN {
##
## Assume we want to minimize: -(0 5 0) %*% b + 1/2 b^T b
## under the constraints: A^T b >= b0
## with b0 = (-8,2,0)^T
## and (-4 2 0)
## A = (-3 1 -2)
## ( 0 0 1)
## we can use solve.QP.compact as follows:
##
Dmat <- matrix(0,3,3)
diag(Dmat) <- 1
dvec <- c(0,5,0)
Aind <- rbind(c(2,2,2),c(1,1,2),c(2,2,3))
Amat <- rbind(c(-4,2,-2),c(-3,1,1))
bvec <- c(-8,2,0)
solve.QP.compact(Dmat,dvec,Amat,Aind,bvec=bvec)
# } | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8662521839141846, "perplexity": 1534.3753061541015}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514576047.85/warc/CC-MAIN-20190923043830-20190923065830-00499.warc.gz"} |
http://mathhelpforum.com/pre-calculus/71028-find-expressions-quadratic-functions.html | # Math Help - Find expressions for the quadratic functions
1. ## Find expressions for the quadratic functions
Find expressions for the quadratic functions whose graphs are shown:
2. Originally Posted by question111
Find expressions for the quadratic functions whose graphs are shown:
there are several ways to approach this. here's one
note that the vertex is on the x-axis, that is, we have a double root for the vertex
thus, our quadratic is of the form $y = a(x - 3)^2$
we are given that the point (4,2) is on the graph, we can use this to find $a$, by plugging in $x = 4$ and $y = 2$ into the form above
3. Originally Posted by question111
Find expressions for the quadratic functions whose graphs are shown:
Your image is so small that it is very hard to see the co-ordinate of vertex and a point in parabola. Please write these coordinates, then it will be solved.
4. Originally Posted by Shyam
Your image is so small that it is very hard to see the co-ordinate of vertex and a point in parabola. Please write these coordinates, then it will be solved.
if you click on the image, it takes you to a larger one
5. So a would be 2, and when I plug that in, I am finally getting a graph like it's shown in the book.
Thank you very much.
6. Originally Posted by question111
So a would be 2, and when I plug that in, I am finally getting a graph like it's shown in the book.
Thank you very much.
7. ## and so what is your expression
8. Originally Posted by lovegreen
i do not understand your question. we found that a = 2. just plug that into the expression given in post #2 to get the quadratic
9. ## Thanks!
I see it is y = 2(x squared) -12x + 18
10. Originally Posted by lovegreen
I see it is y = 2(x squared) -12x + 18
Why not just leave it as $y= 2(x- 3)^2$? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8100169897079468, "perplexity": 416.8357852257707}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860113010.58/warc/CC-MAIN-20160428161513-00054-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://homework.cpm.org/category/CCI_CT/textbook/calc/chapter/4/lesson/4.1.2/problem/4-19 | ### Home > CALC > Chapter 4 > Lesson 4.1.2 > Problem4-19
4-19.
For each $f(x)$, find its general antiderivative, $F(x)$.
Just as $f^\prime(x)$ is the slope function of $f(x)$, $f(x)$ is the slope function of $F(x)$.
1. $f(x) =-2$
If the slope function, $f(x)$, is horizontal, then the original function, $F(x)$, is _______________.
If $f(x)$ is horizontal, then $F(x)$ will be linear: $F(x) = -2x + C$
1. $f ( x ) = \frac { 3 } { 2 } x ^ { - 1 / 2 }$
Check your work by finding the derivative of $F(x)$. Are your $+$ and $−$ signs correct?
1. f(x) = −3x2 + 6x
For all of these, don't forget the $+C$.
1. f(x) = 2(x + 3)
If the slope function, $f(x)$, is linear, then the original function, $F(x)$, will be _____________________________. | {"extraction_info": {"found_math": true, "script_math_tex": 19, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9889230132102966, "perplexity": 987.5272558751781}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662636717.74/warc/CC-MAIN-20220527050925-20220527080925-00416.warc.gz"} |
https://arxiv.org/abs/1708.02570 | math.CO
(what is this?)
# Title: Decomposition spaces and restriction species
Abstract: We show that Schmitt's restriction species (such as graphs, matroids, posets, etc.) naturally induce decomposition spaces (a.k.a. unital 2-Segal spaces), and that their associated coalgebras are an instance of the general construction of incidence coalgebras of decomposition spaces. We introduce the notion of directed restriction species that subsume Schmitt's restriction species and also induce decomposition spaces. Whereas ordinary restriction species are presheaves on the category of finite sets and injections, directed restriction species are presheaves on the category of finite posets and convex maps. We also introduce the notion of monoidal (directed) restriction species, which induce monoidal decomposition spaces and hence bialgebras, most often Hopf algebras. Examples of this notion include rooted forests, directed graphs, posets, double posets, and many related structures. A prominent instance of a resulting incidence bialgebra is the Butcher-Connes-Kreimer Hopf algebra of rooted trees. Both ordinary and directed restriction species are shown to be examples of a construction of decomposition spaces from certain cocartesian fibrations over the category of finite ordinals that are also cartesian over convex maps. The proofs rely on some beautiful simplicial combinatorics, where the notion of convexity plays a key role. The methods developed are of independent interest as techniques for constructing decomposition spaces.
Comments: 44 pages. This is the last of six papers that formerly constituted the long manuscript arXiv:1404.3202 Subjects: Combinatorics (math.CO); Algebraic Topology (math.AT); Category Theory (math.CT) MSC classes: 18G30, 16T10, 06A07, 18-XX, 55Pxx Cite as: arXiv:1708.02570 [math.CO] (or arXiv:1708.02570v1 [math.CO] for this version)
## Submission history
From: Joachim Kock [view email]
[v1] Tue, 8 Aug 2017 17:36:37 GMT (51kb) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8115167021751404, "perplexity": 2221.065032006196}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583510969.31/warc/CC-MAIN-20181017023458-20181017044958-00542.warc.gz"} |
https://plainmath.net/94313/if-lambda-int-0-1-dx-1-x-3-then-eval | Deborah Proctor
2022-10-18
If:
$\lambda ={\int }_{0}^{1}\frac{dx}{1+{x}^{3}};$
Then evaluate:
$p=\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{\prod _{r=1}^{n}\left({n}^{3}+{r}^{3}\right)}{{n}^{3n}}\right)}^{1/n};$
Do you have a similar question?
Expert
First, to evaluate $\lambda$, use partial fractions:
$\frac{1}{1+{x}^{3}}=\frac{1}{3}\left(\frac{1}{1+x}-\frac{x-2}{{x}^{2}-x+1}\right)$
The second piece in the parentheses is more amenable to integration when expressed as follows:
$\frac{x-2}{{x}^{2}-x+1}=\frac{x-1/2}{\left(x-1/2{\right)}^{2}+3/4}-\frac{3/2}{\left(x-1/2{\right)}^{2}+3/4}$
Now, let
$A={\int }_{0}^{1}\frac{dx}{1+x}=\mathrm{log}2$
$B={\int }_{0}^{1}dx\phantom{\rule{mediummathspace}{0ex}}\frac{x-1/2}{\left(x-1/2{\right)}^{2}+3/4}={\left[\mathrm{log}\left[{\left(x-\frac{1}{2}\right)}^{2}+\frac{3}{4}\right]\right]}_{0}^{1}=0$
$C={\int }_{0}^{1}dx\phantom{\rule{mediummathspace}{0ex}}\frac{3/2}{\left(x-1/2{\right)}^{2}+3/4}=\sqrt{3}{\left[\mathrm{arctan}\left[\frac{2}{\sqrt{3}}\left(x-\frac{1}{2}\right)\right]\right]}_{0}^{1}=\frac{\pi }{\sqrt{3}}$
Therefore
$\lambda =\frac{1}{3}\left(A-B+C\right)=\frac{1}{3}\mathrm{log}2+\frac{\pi }{3\sqrt{3}}$
Now for the limit, which is evaluated by taking logs of both sides:
$\mathrm{log}p=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{r=0}^{n}\mathrm{log}\left(1+\frac{{r}^{3}}{{n}^{3}}\right)$
This is a Riemann sum, in the sense that
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{r=0}^{n}f\left(\frac{r}{n}\right)={\int }_{0}^{1}dx\phantom{\rule{mediummathspace}{0ex}}f\left(x\right)$
Therefore
$\begin{array}{rl}\mathrm{log}p& ={\int }_{0}^{1}dx\phantom{\rule{mediummathspace}{0ex}}\mathrm{log}\left(1+{x}^{3}\right)\\ & =\left[x\mathrm{log}\left(1+{x}^{3}\right){\right]}_{0}^{1}-{\int }_{0}^{1}dx\phantom{\rule{mediummathspace}{0ex}}\frac{3{x}^{3}}{1+{x}^{3}}\\ & =\mathrm{log}2-3{\int }_{0}^{1}dx\phantom{\rule{mediummathspace}{0ex}}\left(1-\frac{1}{1+{x}^{3}}\right)\\ & =\mathrm{log}2-3+3\lambda \\ & =2\mathrm{log}2-3+\frac{\pi }{\sqrt{3}}\\ \therefore p& =4{e}^{-\left(3-\frac{\pi }{\sqrt{3}}\right)}\end{array}$
Still Have Questions?
Free Math Solver | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 38, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9726088047027588, "perplexity": 1209.7221997801282}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711637.64/warc/CC-MAIN-20221210005738-20221210035738-00120.warc.gz"} |
https://clay6.com/qa/37354/if-omega-is-a-complex-root-of-unity-one-root-of-the-equation-beginx-1-omega?show=37356 | # If $\omega$ is a complex root of unity, one root of the equation $\begin{vmatrix}x+1 &\omega&\omega^2\\\omega&x+\omega^2&1\\\omega^2&1&x+\omega\end{vmatrix}=0$ is
$\begin{array}{1 1}(A)0 \\ (B) 1 \\ (C) \omega \\(D) \omega^2 \end{array}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8697501420974731, "perplexity": 56.74749064705245}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400204410.37/warc/CC-MAIN-20200922063158-20200922093158-00145.warc.gz"} |
https://www.physicsforums.com/threads/more-friction.101082/ | # More Friction
1. Nov 22, 2005
### Izekid
Here it comes, I have a frictiongrade who is 0,45 A car who is driving in 18m/s and mass of the car 1200kg
For the Frictionpower it is : 0,45 * (mg) = 5,5 kN
And how do I calculate the meters the car travels until it stops?
2. Nov 22, 2005
### skeeter
initial kinetic energy - work done by friction = 0
(1/2)m(vo)^2 - f(delta x) = 0
solve for delta x
3. Nov 23, 2005
### andrevdh
Use the work-kinetic energy theorem. That is the work done by all of the forces acting on an object (or the resultant of all of these) is equal to it's change in kinetic energy:
$$W_{F_R}=\Delta K$$
Similar Discussions: More Friction | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9101459980010986, "perplexity": 2305.034235067991}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948514113.3/warc/CC-MAIN-20171211222541-20171212002541-00075.warc.gz"} |
http://inverseprobability.com/publications/pena-fbd-tech04.html | # Optimising Kernel Parameters and Regularisation Coefficients for Non-linear Discriminant Analysis
Tonatiuh Peña-CentenoNeil D. Lawrence
, 2004.
#### Abstract
In this paper we consider a Bayesian interpretation of Fisher’s discriminant. By relating Rayleigh’s coefficient to a likelihood function and through the choice of a suitable prior we use Bayes’ rule to infer a posterior distribution over projections. Through the use of a Gaussian process prior we show the equivalence of our model to a regularised kernel Fisher’s discriminant. A key advantage of our approach is the facility to determine kernel parameters and the regularisation coefficient through optimisation of the marginalised likelihood of the data. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8938260078430176, "perplexity": 793.7189479138873}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588113.25/warc/CC-MAIN-20211027084718-20211027114718-00071.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-1-foundations-for-algebra-1-2-order-of-operations-and-evaluating-expressions-practice-and-problem-solving-exercises-page-14/49 | ## Algebra 1
$68$
We start with the given expression: $2p^2+(2q)^2$ We plug in values for $p$ and $q$: $2(4)^2+(2(3))^2$ The order of operations states that first we perform operations inside grouping symbols, such as parentheses, brackets, and fraction bars. Then, we simplify powers. Then, we multiply and divide from left to right. Finally, we add and subtract from left to right. We use these rules to simplify the expression. First, we simplify inside parentheses by multiplying: $2(4)^2+(6)^2$ Next, we simplify exponents: $2(16)+36$ Next, we multiply: $32+36$ Finally, we add: $68$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.978915810585022, "perplexity": 455.15706880149514}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989018.90/warc/CC-MAIN-20210509213453-20210510003453-00276.warc.gz"} |
https://www.physicsforums.com/threads/age-of-rocks-calculation-i-only-need-little-help.598069/ | # Age of rocks calculation - I only need little help
1. Apr 19, 2012
### wannie112
How many half lives have passed since a particular rock was formed if a sample of the rock contains 31 times as much lead-207 than uranium-235?
I know that I need to do a table and calculate
I know how to do this question but the thing I need help on is how do I know the number of atoms in order to do a table and calculate. Like I have the ratio of parents:daughter isotope which is 1:31 right? but how do I do this? Just tell me how and I will do it myself. Actually I don't know physics, I'm studying it all by myself and I got this question from my friend's homework and I'm stuck on it, there are 3 questions like it and I need help on how to turn this ratio into actual numbers in order to calculate how many half lives have passed. If I don't need to turn it into numbers then please tell me how do I solve this question? Thank you :)
EDIT: okay so thats what I did is this correct?
1 - 31 half lives=5
2 - 30 half lives=4
4 - 28 half lives=3
8 - 24 half lives=2
16 - 16 half lives=1
32 - 0 half lives=0
so 5 half lives have passed
Correct??
Last edited: Apr 19, 2012
2. Apr 19, 2012
### ehild
5 half lives is correct.
ehild
Last edited: Apr 19, 2012
3. Apr 19, 2012
### wannie112
Hmm? You didn't really explain anything...how did you get to 32? I got to 5.
4. Apr 19, 2012
### ehild
I meant 5, sorry... As the original amount decreased to 1/32-th.
hild
5. Apr 19, 2012
### wannie112
oo thank you :)..and my way is correct do I have to do any formulas?
Thank you very much ^^
6. Apr 19, 2012
### ehild
The daughter atoms present + number of un-decayed U235 atoms=N(0):
31N + N =N(0)
The number of un-decayed isotopes present at time t is
N(t)=N(0) (1/2)t/halflife,
where N(0) is the number initially.
N=N(0)/32, 32 =25, ....
ehild
7. Apr 19, 2012
### wannie112
I see :D Now I understand everything, thank you!
Similar Discussions: Age of rocks calculation - I only need little help | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.865264356136322, "perplexity": 1616.716623203536}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812556.20/warc/CC-MAIN-20180219072328-20180219092328-00261.warc.gz"} |
https://www.stevenabbott.co.uk/practical-adhesion/particle-flow.php | ## Particle Flow
### Quick Start
If you have found it hard to get particles to flow, then the two apps below help explain why. It's all rather complicated, but the apps help clarify things for those who have this specific interest!
Getting particles to flow through a hopper can be difficult. A relatively small outlet at the bottom would be more elegant to control, and a relatively shallow angle in the hopper would mean that for a given height it could hold more material. Yet we are often forced to have large outlets (diameter D in the diagram), otherwise the material jams into an arch, blocking the flow. Note that this is different from the observation that an orifice which is less than ~5x the particle size will tend to block as the 5 particles jam together. There is also a need to have steep walls. Without steep walls the powder only flows in the middle ("funnel flow") rather than across the full width of the hopper ("mass flow").
To further complicate matters, although conical hoppers are generally more convenient, "plain-" or "wedge-"shaped hoppers tend to jam less easily so are sometimes necessary, even though this means that the contents pour out in a wide line rather than a round heap.
Clearly particle-particle and particle-wall adhesion are both important. But a powder is so complex that in reality measurements are made of friction coefficients as these can be measured relatively easily. As is well-known, a friction coefficient η can be expressed in terms of the angle φ needed for something to slide down a plane (η=tan(φ)) and so the powder handling world tends to talk of friction in terms of "wall angles". So one condition for mass flow is that the hopper angle θ (or, rather, 90-θ because θ is defined so that θ=0 for a vertical wall) should be slightly greater than the measured "wall friction angle" φ.
There is a further complication. The particles themselves stick together or, equivalently, have a coefficient of friction. If there is some significant degree of attraction between the particles (especially if they are moist - see the capillary adhesion calculations) then this coefficient of friction will be higher than that with a surface. This in turn means that the equivalent angle (in terms of sliding down a plane) is higher. It seems odd to talk of an "internal angle of friction", δ, but mathematically it is helpful because the wall friction angle φ is itself so important. For very free-flowing powders, δ and φ are close. Tests give one more parameter τ which is the yield shear stress, or "cohesion coefficient".
Unfortunately, life is even more complicated than that. The compaction of a powder under pressure (e.g. under its own weight in a hopper) has an effect on a powder's compressive strength - the stress needed for it to break under pressure. To understand hopper flows you need to know the size of the compressive strength at all relevant degrees of compaction. One way to do this is to compress some powder in a cell with a stress σ1, then to remove the walls around the sample so it is free-standing, and then push on the surface of the column of powder to find the stress σc at which it fails - which is the compressive strength. Such an experiment is easy to understand but hard to do because for each value of σ1 you have to start off with a fresh sample of powder.
So the powder community uses a more indirect method. They measure the shear stress necessary to break the powder which has been compacted under a known load. This can be done in a Jenike box which is able to slide along its middle, or in a ring shear cell, a rather more convenient device. Unfortunately, shear testers only measure shear and we want compression! Fortunately via a series of tests done on a single powder sample but under many different compaction loads, and tested at loads up to those compaction loads (it's a lot of data!), it is possible to extract σ1 and σc via "Mohr Circle" techniques which need not detain us here.
Whether done directly (which is unlikely) or indirectly (more usual), the result is a graph of σc versus σ1. The graph here uses a 3-parameter non-linear version to describe the curve where E and F are in kPa and q is a power dependency. These values are used in the calculation below of the minimum diameter Dmin to avoid blocking.
### Particle Failure Stresses
Modulus E kPa
Force F kPa
Power q
σ1 Max kPa
FF 25%
FF 75%
We can gain a key bit of information from this curve. From the slope of the curve a Flow Factor (FF) can be calculated. The larger it is, the easier the flow of the powder. Values above 4 are "easy flowing" above 10 the powder is "free flowing". For a non-linear curve FF varies, of course, so it is calculated at 25% and 75% of the maximum σ1, with the value at higher compaction generally being higher (better flow), which is non-intuitive.
Now we have all but one the key parameters. We need to add the density of the powder in the hopper. Of course this density depends somewhat on the powder's self-compaction (see below) but is generally taken as a constant in these estimations, presumably measured under a "representitive" degree of compaction. Now, given all these parameters, how does one calculate a minimum outlet size for a hopper for a given wall angle θ? There is a vast literature of different methods, each claiming to be more elegant and accurate than the others. The results, after a few decades of debate, seem to be that they all more-or-less give the same answer, and the answer is pessimistic by a factor of 2; in other words, if they say that the minimum outlet diameter Dmin should be X, in reality the flow is OK (no arching) for an outlet of X/2. Out of interest I've provided two answers, Mróz & Szymański (M&Z) and Arnold & McLean (A&M). The formulae for each are complex and I've used the two Drescher papers1 (and the non-linear IYLs based on E, F, q) to help me through the algebraic maze.
### Particle Flow Minimum Diameter
θ Wall Angle
φ Wall Friction Angle
δ Internal Friction Angle
ρ g/cc
DM&Z m
DA&M m
Plane
Conical
### Compaction
All powders can compact on tapping, vibrating or pushing in some controlled manner. It is useful to be able to compare the compaction behaviour of different powders. One quick number is Carr's compressibility index
C=100(ρ_t-ρ_b)/ρ_t = 100(V_b-V_t)/V_0
i.e. the % compression between the bulk density ρb and the tapped density ρt, or, equivalently, the respective Volumes. The transition from freely-flowing to non-freely-flowing is ~C=20%. An even simpler ratio is the Hausner Ratio H
H=ρ_t/ρ_b = V_b/V_t
However, a single number gives no idea of the real behaviour. So it is usual to look at Kawakita or Heckel plots. It is arguable that they are much the same thing, though different communities tend to prefer one over the other.
Kawakita
For an applied pressure P the observed compaction
C=(V_0-V)/V_0
is given by
P/C=1/(ab)+P/a
where a and b are fitted constants. a is nominally equal to the original porosity ε0 but the meaning of b (dimensions 1/stress) is unclear.
Heckel
Given a starting porosity ε0 and measured porosity ε at pressure P,
ln(1/ε)=ln(1/ε_0)+KP
where K is a fitting parameter and 1/K is equal to the yield pressure Py or to 3σ0, the yield strength. Instead of porosity ε it is common to use 1-D, where D is the relative density = ρ/ρsolid compared to the pure solid.
Kawakita and Heckel
From my observations of the literature, it seems to make sense to always plot both Kawakita and Heckel. They seem to reveal different/complementary aspects of a powder's behaviour and using just one seems to miss out on insights from the other. My intention at this point is to add some curves from some high-quality data in the literature obtained on some representative powders. But first I must get the authors' permission! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8651155829429626, "perplexity": 1265.302452910571}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027314752.21/warc/CC-MAIN-20190819134354-20190819160354-00322.warc.gz"} |
https://www.physicsforums.com/threads/prove-series-converges.321452/ | # Prove series converges
• Start date
• #1
726
1
## Homework Statement
If a_1 + a_2 + ... is an infinite series converging to A, and b1, b2, ... is an infinite sequence that is bounded and monotonic, prove that (a_1)(b_1) + (a_2)(b_2) + ... converges
## The Attempt at a Solution
I don't really know where to start...all I can say is that if a_1 + a_2 + ... converges, then a_n approaches 0 as n goes to infinity, and so (a_n)(b_n) also has a limit of 0, since b_n converges to some finite value.
Related Calculus and Beyond Homework Help News on Phys.org
• #2
Office_Shredder
Staff Emeritus
Gold Member
3,750
99
You know that bn is bounded and monotonic. Have you tried using that? Given a general number b, you know
$$\sum a_n b$$
converges right? Now try to compare that to the sequence
$$\sum a_n b_n$$ choosing b such that |bn| < b for all n
• #3
tiny-tim
Homework Helper
25,832
249
Hi JG89!
… since b_n converges to some finite value.
Yes … concentrate on that value (call it b) …
then use deltas and epsilons.
• #4
726
1
I have an intuitive idea of what's going on, but I'm having a hard time fleshing it out into an epsilon argument.
First off, since a_1 + a_2 + ... converges then b(a_1 + a_2 + ...) converges. I know that for large enough n, |(a_n)(b_n)| gets 'really' close to |(a_n)b|, since b is the limit of b_n. Since the b_n are monotonic, then as n increases the |(a_n)(b_n)| gets even closer to |(a_n)b|. I can in fact make this difference as small as I please, provided n is taken large enough and so it seems that after a certain n, the difference between (a_n)(b_n) and (a_n)b will become "negligible" and since (a_n)b is Cauchy, then (a_n)(b_n) is also Cauchy.
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785 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9620288610458374, "perplexity": 1845.112951842908}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875141396.22/warc/CC-MAIN-20200216182139-20200216212139-00465.warc.gz"} |
https://www.stat.berkeley.edu/~aldous/RWG/Book_Ralph/Ch6.S7.html | # 6.7 Distributional aspects
In many examples one can apply the following result to show that hitting time distributions become exponential as the size of state space increases.
###### Corollary 6.32
Let $i,j$ be arbitrary states in a sequence of reversible Markov chains.
(i) If $E_{\pi}T_{j}/\tau_{2}\rightarrow\infty$ then
$P_{\pi}\left(\frac{T_{j}}{E_{\pi}T_{j}}>x\right)\rightarrow e^{-x},\ 0
(ii) If $E_{i}T_{j}/\tau_{1}\rightarrow\infty$ and $E_{i}T_{j}\geq(1-o(1))E_{\pi}T_{j}$ then $E_{i}T_{j}/E_{\pi}T_{j}\rightarrow 1$ and
$P_{i}\left(\frac{T_{j}}{E_{i}T_{j}}>x\right)\rightarrow e^{-x},\ 0
Proof. In continuous time, assertion (i) is immediate from Chapter 3 Proposition yyy. The result in discrete time now holds by continuization: if $T_{j}$ is the hitting time in discrete time and $T^{\prime}_{j}$ in continuous time, then $E_{\pi}T^{\prime}_{j}=E_{\pi}T_{j}$ and $T^{\prime}_{j}-T_{j}$ is order $\sqrt{E_{\pi}T_{j}}$. For (ii) we have (cf. Chapter 4 section yyy) $T_{j}\leq U_{i}+T^{*}_{j}$ where $T_{j}$ is the hitting time started at $i$, $T^{*}_{j}$ is the hitting time started from stationarity, and $E_{i}U_{i}\leq\tau^{(2)}_{1}$. So $ET_{j}\leq ET^{*}+O(\tau_{1})$, and the hypotheses of (ii) force $ET_{j}/ET^{*}_{j}\rightarrow 1$ and force the limit distribution of $T_{j}/ET_{j}$ to be the same as the limit distribution of $T^{*}_{j}/ET^{*}_{j}$, which is the exponential distribution by (i) and the relation $\tau_{2}\leq\tau_{1}$. $\Box$
In the complete graph example, $C$ has mean $\sim n\log n$ and s.d. $\Theta(n)$, so that $C/EC\rightarrow 1$ in distribution, although the convergence is slow. The next result shows this “concentration” result holds whenever the mean cover time is essentially larger than the maximal mean hitting time.
###### Theorem 6.33 ([23])
For states $i$ in a sequence of (not necessarily reversible) Markov chains,
$\mbox{if }E_{i}C/\tau^{*}\rightarrow\infty\mbox{ then }P_{i}\left(\left|\frac{% C}{E_{i}C}-1\right|>\varepsilon\right)\rightarrow 0,\ \varepsilon>0.$
The proof is too long to reproduce. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 29, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9820449352264404, "perplexity": 343.7394351821899}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948599549.81/warc/CC-MAIN-20171218005540-20171218031540-00751.warc.gz"} |
http://mathhelpforum.com/geometry/19326-triangle-angle-problem.html | # Math Help - Triangle Angle Problem
1. ## Triangle Angle Problem
Hi everyone, I think that I have got the first part (a) of this question correct but would appreciate someone checking and pointing me in the right direction as I have a feeling that I am missing something as the question seems too simple to be true.
In triangle ABC, angle A=90 and sec B=2
a/ Find cos B - I have got:
sec B=2
1/cos B=2
1=2cos B
cosB=1/2
b/ Find angles B and C
I've got:
cosB=1/2
B=60,300
can't use B=300 as angles in a triangle = 180 so B=60
180-A-B=C so 180-90-60=C=30
2. Originally Posted by Tom G
Hi everyone, I think that I have got the first part (a) of this question correct but would appreciate someone checking and pointing me in the right direction as I have a feeling that I am missing something as the question seems too simple to be true.
In triangle ABC, angle A=90 and sec B=2
a/ Find cos B - I have got:
sec B=2
1/cos B=2
1=2cos B
cosB=1/2
b/ Find angles B and C
I've got:
cosB=1/2
B=60,300
can't use B=300 as angles in a triangle = 180 so B=60
180-A-B=C so 180-90-60=C=30
Too simple, perhaps, but also correct.
-Dan
3. Originally Posted by Tom G
Hi everyone, I think that I have got the first part (a) of this question correct but would appreciate someone checking and pointing me in the right direction as I have a feeling that I am missing something as the question seems too simple to be true.
In triangle ABC, angle A=90 and sec B=2
a/ Find cos B - I have got:
sec B=2
1/cos B=2
1=2cos B
cosB=1/2
b/ Find angles B and C
I've got:
cosB=1/2
B=60,300
can't use B=300 as angles in a triangle = 180 so B=60
180-A-B=C so 180-90-60=C=30
Yes, seems fine
4. I've just looked at the next section of the question and am stuck (I have a feeling its really simple but my brain just doesn't seem to be working today).
c/ Find tan B
I've got tan B = Sin B / Cos B
I've already worked out cos B, but could someone tell me how I find Sin B? Thanks
5. Originally Posted by Tom G
I've just looked at the next section of the question and am stuck (I have a feeling its really simple but my brain just doesn't seem to be working today).
c/ Find tan B
I've got tan B = Sin B / Cos B
I've already worked out cos B, but could someone tell me how I find Sin B? Thanks
You've just worked B out to be 30 degrees didn't you?
Now use your special triangles to aid you.
AND Sin B = cos (90 - B)
6. Originally Posted by janvdl
You've just worked B out to be 30 degrees didn't you?
Now use your special triangles to aid you.
AND Sin B = cos (90 - B)
Actually, the more direct one to use would be either
$cos(B) = \pm \sqrt{1 - sin^2(B)}$<-- Choose the "+" since cosine is positive for an acute angle.
or
$cos(B) = sin(90 - B)$
-Dan
7. Hi, I'm still struggling with this one.
I have now got;
sin B = +Squareroot(1-cos^2 B)
but am now confused as I know cos B x cos B = cos^2 B ... so does this mean that cos 60 x cos 60 = 1/2 x 1/2 = 1/4
Then 1-1/4 = 3/4 so sin B = \pm \sqrt{3/4} = 0.866025........ and so B=59.99999 ( I already have worked out that B=60)
I really have no clue what is going on and am going around in circles - would really appreciate someone showing me how its done - Thanks.
8. Originally Posted by Tom G
Hi, I'm still struggling with this one.
I have now got;
sin B = +Squareroot(1-cos^2 B)
but am now confused as I know cos B x cos B = cos^2 B ... so does this mean that cos 60 x cos 60 = 1/2 x 1/2 = 1/4
Then 1-1/4 = 3/4 so sin B = \pm \sqrt{3/4} = 0.866025........ and so B=59.99999 ( I already have worked out that B=60)
I really have no clue what is going on and am going around in circles - would really appreciate someone showing me how its done - Thanks.
Sin B = 0.866025 (which is actually $\frac{ \sqrt{3} }{2}$ )
And $Cos B = \frac{1}{2}$
And $Tan B = \frac{Sin B}{Cos B}$
Take it from here. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8966770768165588, "perplexity": 764.6245086949444}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375095671.53/warc/CC-MAIN-20150627031815-00181-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/109052/description-of-regular-covering-maps-between-surfaces/109093 | # Description of regular covering maps between surfaces.
This is an improved and hopefully a more precise version of the question Covering spaces of surfaces.
Question: Given a regular covering map $\pi:\Sigma_g\to\Sigma_h$, where $\Sigma_n$ denotes a surface of genus $n$, is it possible to describe the covering map?
One example of such a description is the following. In the decomposition of $\Sigma_h$ into the connected sum of tori $T_1$#...#$T_h$, one torus, say $T_1$, is covered $k$-times by another torus $T_1'$, which appears in a similar decomposition of $\Sigma_g$; every other torus $T_i$ in the decomposition of $\Sigma_h$ is covered by $k$ different tori $T_{i_1}'$,...,$T_{i_k}'$ (each covering $T_i$ identically) in the decomposition of $\Sigma_g$.
Any other explicit description would likely also be useful.
A weaker version of the question would be the following: given a regular covering $\rho:\Sigma_l\to \Sigma_h$, is there another regular covering $\rho':\Sigma_g\to \Sigma_l$, such that the composition $\pi=\rho\rho'$ has a description as above (for example)?
-
1. What do you mean by "describe"? If you mean "realize surface in 3-space so that the covering group action extends", then the answer is "of course not". I do not see a real question here. 2. Your weaker question has obvious negative answer: just take any $\rho$ with nonabelian finite covering group. – Misha Oct 7 '12 at 12:17
I am interested in any kind of description, really, the simpler the better. In particular, is the description stated in the example above valid in general? – George Oct 7 '12 at 12:54
Cut your surface $\Sigma_h$ into a $2h$-sided polygon $P$ glued in the traditional pattern $a_1b_1a_1^{-1}b_1^{-1}\ldots$. Letting the degree of the covering by $D$ (which could be infinity), take $D$ copies of $P$. Glue them up appropriately to form $\Sigma_g$. Map each copy to $P$. The covering map has been described. – Lee Mosher Oct 7 '12 at 12:58
Not every torus is necessarily covered by another torus. If the circles that separate the toruses in the connect-sum decomposition pull back to longer circles in the cover, rather than just disjoint unions of a bunch of copies of themselves, then the covering of the torus, which you get by gluing discs onto those circles, will be a ramified covering (ramified at the center of the disc), which necessarily increases the genus of the torus. – Will Sawin Oct 7 '12 at 16:08
George: It is a good exercise to see that in the example of a covering (call it $q$) that you gave, the covering group is abelian, see Will's comments. Therefore, no covering $\rho$, with nonabelian deck-group, appears in $q=\rho\circ \rho'$. – Misha Oct 7 '12 at 21:37
## 1 Answer
I agree with the commenters that the question is a bit vague, but the most concrete description I know of is the permutation representation of the fundamental group (permuting the "sheets" of the covering. This tends to give a rather concrete "cut and paste" description of the cover (and in addition, I believe that this description, going back to Hurrwitz, might have been historically the first way to look at covering maps). For more along these lines, check out "On extension of coverings" by M. Droste and I. Rivin, and references therein.
-
Igor, I think you just mean that the deck-group $G$ of a regular covering $M\to N$ admits a fundamental domain $F$ in $M$ (translates $g(F), g\in G$, are the "sheets" of the covering). This is valid for arbitrary (not necessarily finite) regular coverings of arbitrary smooth/PL/topological manifolds. I guess, in the surface case you can also say that permutation representation is transitive and is given by a collection of permutations s.t. product of their commutators is $1$. Is there more to it? – Misha Oct 7 '12 at 21:33
@Misha: there is a lot more to it: in particular, the inverse problem (trying to reconstruct the covering from the monodromy at a collection of points) is the "Riemann-Hilbert problem" in one of its guises, and is still open in important special cases. – Igor Rivin Oct 7 '12 at 23:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8352978825569153, "perplexity": 293.61188040433575}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207925696.30/warc/CC-MAIN-20150521113205-00074-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://studyforce.com/finding/ | # Resolving Vectors in Quadrant I
Any vector can be replaced by two vectors which, acting together, exactly duplicate the effect of the original vector. These replacement vectors are called the components of the vector, and are usually chosen perpendicular (at right angles) to each other. Another name for these perpendicular component vectors is rectangular components. The process of of breaking a 2D vector into its vertical (y) and horizontal (x) components is known as resolving a vector. The first video shows how a vector in the first quadrant is broken down into its rectangular components.
Another way to represent the answer in the video is by writing it in rectangular form:
v = ai + bj where a = vx and b = vy
v = 25.3i + 49.3i
Note that i is horizontal component vector of the vector v, and j is vertical component vector. The vector sum ai + bj is called a linear combination of the vector i and j. The magnitude of v = ai + bj is given by:
null
The process above can also be reversed, where the rectangular components are added up to obtain the resultant vector. This is typically done using the tangent ratio and the Pythagorean theorem. The technique is shown below:
Challenge Question:
Find the resultant of two perpendicular vectors whose magnitudes are 485 and 627. Also find the angle that it makes with the 627-magnitude vector. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9802674651145935, "perplexity": 671.1289107250515}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104054564.59/warc/CC-MAIN-20220702101738-20220702131738-00560.warc.gz"} |
https://www.transtutors.com/questions/show-transcribed-image-text-evaluate-the-following-improper-integrals-using-only-the-1293674.htm | # Show transcribed image text Evaluate the following improper integrals using only the definition. Be
Show transcribed image text Evaluate the following improper integrals using only the definition. Be sure to use L'Hospital's rule, when appropriate, to compute limits.
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http://guitarpenguin.is-programmer.com/tag/latex | Consider the fact that I have used Tikz (Tex) to finished drawing some nice graphs. In another words, I have some source files in .tex format. The question is, how do I show these diagrams drawn using Tikz on the html file? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9996768236160278, "perplexity": 927.1484580454995}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945660.53/warc/CC-MAIN-20180422212935-20180422232935-00157.warc.gz"} |
https://www.physicsoverflow.org/27067/quivers-representations-in-susy-gauge-theories | # Quivers Representations in SUSY gauge theories
+ 4 like - 0 dislike
153 views
I would like to hear some reasons and ideas on how quivers are useful in SUSY gauge theories. There is a nice answer about the case of D-branes here but it is not clear on their appearance in gauge theory independently of the D-branes. More specifically I have heard that quivers can describe BPS states. Is this correct? And why so?
This post imported from StackExchange Physics at 2015-02-07 11:48 (UTC), posted by SE-user user39726
SUSY quivers are dissected here arxiv.org/abs/hep-th/0201205
This post imported from StackExchange Physics at 2015-02-07 11:48 (UTC), posted by SE-user Autolatry
Indeed, quivers first appeared in the context of D-branes at (conifold) singularities (there are various nice expositions in Klebanov-Witten theory reviews) where the D-branes "conspire" to give a $\mathcal{N}=1$ SYM theory. Additionally, gauge theories are strongly encoded inside the physics of D-branes, so I am not sure in what way you can "separate" these notions. Usually, quivers are used to describe the physics of BPS bound states of $\mathcal{N}=2$ susy and sugra. I will say a few words on this as an example of quivers in gauge theories. So let us consider $\mathcal{N}=2$ theory in four dimensions. As you will probably know this theory has a moduli space with a Coulomb and a Higgs branch. Let us consider a point $u$ in the Coulomb branch $\mathcal{C}$ of the moduli space. There we have a gauged $U(1)^r$ symmetry group together with a lattice $\Gamma$ from which the various BPS states take their charges $(p,q)$. From Seiberg-Witten theory we know how to consider the above on an elliptic curve $\Sigma_u$ that varies along $\mathcal{C}$. It is very well known that the homology classes of 1-cycles along the tori we are considering can be identified with $\Gamma$. This is all standard Seiberg-Witten stuff. Seiberg-Witten is of course solved in the IR. To study the BPS states at some specific point $u \in \mathcal{C}$ we need to introduce the quiver. These theories also have a central charge $Z$. Now, we take half the plain of the plane on which the central charge $Z$ takes values and we name it $H$. On this plane there exist a set of $2r+f$ (where $f$ is the number of flavors of the theory) states which are customary to denote as $\gamma_i$ (that we can naively consider them as particles). It turns out that such a basis, if it exists, it is the only possible one, it is unique. Using this basis, the set $\{ \gamma_i \}$ we can construct a quiver. For every $\gamma_i$ we draw a node and for every pair we draw arrows that connect them. Then we can use quiver quantum mechanics to find the BPS bound states of the BPS "particles" $\gamma_i$. So the moral/summary is the following: In the $\mathcal{N}=2$ theory consider a point $u$ of the Coulomb branch and use its data to form (if possible) a basis $\{ \gamma_i \}$ of the hypers. Then put a node on each one, and arrows between them. Then use quiver quantum mechanics to find the BPS bound states.
Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor) Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8825696706771851, "perplexity": 515.3516294066059}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550249530087.75/warc/CC-MAIN-20190223183059-20190223205059-00077.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/28239-galois-theory.html | 1. ## galois theory
Thanks Edgar
Let K be a field and s an indeterminate. Then K(s) is a field extension of K(s^n). Prove that [K(s) : K(s^n)] = n.
Hence show that the minimum polynomial of s over K(s^n) is t^n - s^n.
Hint: first show that s satisifies a polynomial of degree n over K(s^n); this gives <=. Then show that {1,s,...,s^(n-1)} is linearly independent over K(s^n); this gives you >=.
2. Originally Posted by edgar davids
Let $\displaystyle f(x) = x^n - s^n \in K(s^n)$ then clearly $\displaystyle f(s) = 0$ so if $\displaystyle p(x)$ is the minimal polynomial for $\displaystyle s$ over $\displaystyle K(s^n)$ we have that $\displaystyle p(x) | f(x) \implies k= \deg p(x) \leq \deg f(x) = n \implies k\leq n$. Now that means that $\displaystyle \{ 1, s,s^2, ... ,s^{k-1} \}$ is a basis for $\displaystyle K(s)$ over $\displaystyle K(s^n)$. If we can show that $\displaystyle \{1,s,s^2 , ... s^{n-1} \}$ is linearly independent then it would means $\displaystyle k\geq n$ because the dimension of a linearly independent set cannot the dimension of a basis. To show that $\displaystyle \{1,s,s^2,...,s^{n-1} \}$ is linearly independent over $\displaystyle K(s^n)$ it is equivalent to showing that no element in this set can be expressed as a linear combination of the other elements. Take for example, $\displaystyle 1$, we cannot express $\displaystyle 1$ as a linear combination $\displaystyle a_1 s+a_2s^n + ... + a_{n-1}s^{n-1}$ where $\displaystyle a_i \in K(s^n)$ because $\displaystyle a_i = b^{(i)}_0 + b^{(i)}_1 s^n + b^{(i)}_2 s^{2n} + ...$ so it is impossible to get intermediately exponents between $\displaystyle 1$ and $\displaystyle n$. Thus, $\displaystyle k=n$ and this means $\displaystyle p(x) = x^n - s^n$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.979146420955658, "perplexity": 173.479434505971}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945448.38/warc/CC-MAIN-20180421203546-20180421223546-00486.warc.gz"} |
http://projecteuclid.org/euclid.aoms/1177730349 | ## The Annals of Mathematical Statistics
### On Small-Sample Estimation
George W. Brown
#### Abstract
This paper discusses some of the concepts underlying small sample estimation and reexamines, in particular, the current notions on "unbiased" estimation. Alternatives to the usual unbiased property are examined with respect to invariance under simultaneous one-to-one transformation of parameter and estimate; one of these alternatives, closely related to the maximum likelihood method, seems to be new. The property of being unbiased in the likelihood sense is essentially equivalent to the statement that the estimate is a maximum likelihood estimate based on some distribution derived by integration from the original sampling distribution, by virtue of a "hereditary" property of maximum likelihood estimation. An exposition of maximum likelihood estimation is given in terms of optimum pairwise selection with equal weights, providing a type of rationale for small sample estimation by maximum likelihood.
#### Article information
Source
Ann. Math. Statist. Volume 18, Number 4 (1947), 582-585.
Dates
First available in Project Euclid: 28 April 2007
Permanent link to this document
http://projecteuclid.org/euclid.aoms/1177730349
Digital Object Identifier
doi:10.1214/aoms/1177730349
Mathematical Reviews number (MathSciNet)
MR22337
Zentralblatt MATH identifier
0029.40701
JSTOR | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8402521014213562, "perplexity": 1160.2069649322361}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982292181.27/warc/CC-MAIN-20160823195812-00034-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-do-you-solve-x-2-7x-44-0 | Algebra
Topics
# How do you solve x^2 + 7x - 44 = 0?
Apr 8, 2016
$x = - 11 , 4$
#### Explanation:
Begin with writing the equation into brackets.
$\left(x + a\right) \left(x + b\right) = 0$
You want to find $a$ and $b$ such that $a + b = 7$ and $a \cdot b = - 44$.
Through a process of trial and error, this gives $a = - 4$ and $b = 11$.
$\left(x - 4\right) \left(x + 11\right) = 0$
Anything multiplied by $0$ is $0$, meaning at least one of the brackets has to be equal to $0$. This actually gives two answers for $x$. Set each one equal to zero and then solve.
$x - 4 = 0$
$x = 4$
or
$x + 11 = 0$
$x = - 11$
##### Impact of this question
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http://math.stackexchange.com/questions/105222/finding-poles-and-zeros-in-the-z-domain | # Finding Poles and Zeros in the Z-Domain
I have the transfer function$$H(z) = \frac{z-.75}{.1 z+.15}$$
how do I find the Poles and Zeros?
-
Poles are the values of $z$ where the denominator becomes zero; zeros are the values of $z$ where the numerator becomes, well, zero. I presume you know how to find the zeros of a linear function? – Guess who it is. Feb 3 '12 at 6:28
yes I do. So finding poles and zeros in the z domain is exactly like in the s-domain? – skipfer0712 Feb 3 '12 at 6:29
Well, that is a function of $z$ in there, no? – Guess who it is. Feb 3 '12 at 6:34
I remember that one is not supposed to tag a question merely as homework, right? Perhaps complex-analysis can be applied? – awllower Feb 3 '12 at 9:43
## 2 Answers
First of all, I'd advice to get rid of those awful decimal numbers and use fractions: $$H(z):=\frac{z-\frac{3}{4}}{\frac{1}{10}z+\frac{3}{20}}=\frac{\frac{1}{4}}{\frac{1}{20}}\frac{4z-3}{2z+3}=5\,\frac{4z-3}{2z+3}$$ From here, we clearly see the only zero of the function (i.e., exactly where its numerator vanishes) is $\,z=3/4\,$ , and its pole (i.e., exactly where the denominator vanishes) is $\,z=-3/2\,$, both simple.
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roots of numerator are zeros and roots of denonumerator are poles
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https://homework.cpm.org/category/CCI_CT/textbook/pc/chapter/2/lesson/2.2.2/problem/2-56 | ### Home > PC > Chapter 2 > Lesson 2.2.2 > Problem2-56
2-56.
Write the following expression using summation notation.
$\frac { 1 } { 4 } + \frac { 4 } { 9 } + \frac { 9 } { 16 } + \ldots + \frac { 81 } { 100 } + \frac { 100 } { 121 }$
Write an expression for the sequence $1, 4, 9, ..., 81, 100$.
Write an expression for the sequence $4, 9, 16, ..., 100, 121$.
Use your sequences to determine the number of terms in the expression.
$\displaystyle\sum\limits_{k=1}^{?}\frac{k^2}{(?)^2}$ | {"extraction_info": {"found_math": true, "script_math_tex": 4, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.89547199010849, "perplexity": 1663.7666328953778}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662647086.91/warc/CC-MAIN-20220527112418-20220527142418-00632.warc.gz"} |
http://mathoverflow.net/questions/57649/applications-of-koszuls-formula-other-than-the-fundamental-lemma-of-riemannian | # Applications of Koszul's formula other than the fundamental lemma of Riemannian geometry
I'm wondering what else one can do with Koszul's formula
$$2\langle\nabla_XA,B\rangle = X\langle A,B\rangle-B\langle X,A\rangle + A\langle X,B\rangle - \langle A,[X,B]\rangle + \langle[B,X],A\rangle - \langle B,[A,X]\rangle$$
beyond proving existence and uniqueness of the Levi-Civita connection. I haven't yet seen anybody using it for anything else, which would be quite curious.
Here's a pretty and simple example. I don't know if it is known...
Let $\nabla^{LC}$ be the Levi-Civita connection and $\nabla$ be some other metric connection with torsion $T$. Then
$$2\langle\nabla_XA,B\rangle - 2\langle\nabla_X^{LC}A,B\rangle= \langle T(A,X),B\rangle -\langle T(A,B),X\rangle - \langle T(B,X),A\rangle$$
An application of this would be to compute the LC-connection for the metric $\langle K\cdot,K\cdot\rangle$ in terms of the endomorphism $K$ and the LC-connection for $\langle \cdot,\cdot\rangle$. The computation starts with the connection $K^{-1}\nabla^{LC} K$, which is metric for $\langle K\cdot,K\cdot\rangle$...
I can't guarantee correct letters and signs :-)
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Riemannian geometry IS the study of the Levi-Civita connection and the Koszul formula is an explicit expression for the connection. So ultimately anything you do in Riemannian geometry is bound to be using it at some level. If, on the other hand, you mean what other formulae you can get by formally manipulating the Koszul formula, then there are plenty, but I would not necessarily call them "applications". Killing's equation comes to mind, as well as the O'Neill formulae for submersions, the basic formulae for homogeneous geometry,... Pick any book on differential geometry: e.g., do Carmo. – José Figueroa-O'Farrill Mar 7 '11 at 11:18
Indeed. Except when the glory breaks down and coordinates get introduced (or used implicitly). (If not, wait till play with Laplacians begins and frames introduced...) BTW, one application of my application would be the coordinate representation of the LC connection. PLEASE NOTE: I'm far away from the library and made it there only a few times this century - and it was open a few of these occasions. But I've seen and possibly remember quite a few of these books, e.g. none on Ricci flow without ungeometricly resorting to coordinate stuff when things get most basic – Martin Gisser Mar 7 '11 at 15:09
So let me get this straight: you make a distinction between the Koszul formula and the formula for the Christoffel symbols? I don't think there is any conceptual difference between the two. I suppose it's a question of how much pain are you willing to endure in order to keep everything coordinate-free. Analysts as a general rule seem to have a low pain threshold :) – José Figueroa-O'Farrill Mar 7 '11 at 17:40
Hm, yes I'm actually trying to avoid pain. I might perhaps have an infinitude of indices (symbolic or not) or try to keep track of the tensor product rule (after all, it's tensor calculus) and not get lost (or waste sparse IQ) in the debauch of indices. Why else Koszul connections? Of course Christoffel stuff and Koszul covariant derivatives (plus Cartan's calculus, Penrose spaghetti, etc. pp.) are all about the same thing, but at least there should be some motivation to stay in one concept (plus, avoid coordinates (it could be a ringed space) (or the coordinates being unknown or irrelevant)). – Martin Gisser Mar 7 '11 at 18:05
Quite a borderline comment but I find it funny. There is (at least) one thing I don't know how to prove without coordinates. Whena metric evolves by the Ricci flow, the evolution of the curvature operator is given by : $\partial_t R=\Delta R + R^2+R^#$, where $R^#$ can be defined in a coordinate free way using the Lie algebra structure of $\Lambda^2 TM$. A substantial part of the proof can be done without coordinates (as in Toppong lectures) but I never saw anyone that goes from the expression ypu found in Topping book to the Lie algebraic expression without coordinates. – Thomas Richard Mar 10 '11 at 7:51
Koszul's formula simply expresses the Levi-Civita connection explicitly in terms of the Riemannian metric. It is quite useful any time you want to eliminate the connection from a formula and write the formula in terms of the metric only. José cited some nice examples. I haven't checked, but I bet the book by Cheeger and Ebin discusses these examples and maybe more quite explicitly.
But it is no different from writing a formula in local co-ordinates and replacing all appearances of Christoffel symbols by their formula in terms of partial derivatives of the Riemannian metric. This is often an equally useful thing to do when, for example, you want to apply PDE techniques or theorems that are stated in terms of co-ordinates to a problem in Riemannian geometry.
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Deane, 1) my example even involves 2 connections... 2a) I find it disadvantageous to have to learn at least 3 calculi for the whole picture. 2b) I found me developing my own calculus to sanely get into Hilbert space via Stokes theorem (for doing geometric PDE geometrically, like Bochner theorems) Indeed José has an excellent answer. There's sure much more. Alas I had to suffer through O'Neill's formulae etc. in Christoffelian (by a famous German differential ring theorist, no less). – Martin Gisser Mar 8 '11 at 3:05
Martin, I also prefer to understand everything from a single approach. And to some extent I have succeeded in developing a single way to do everything I need to do in Riemannian geometry. But it appears that each of us ends up developing our own preferred approach and even notation. But even then my approach does not accomplish absolutely everything I want, just most of it. Depending on the context, I still sometimes find either the differential form or local co-ordinate approach the easiest way to do what I want. So I find it quite handy to be able to use them, too. – Deane Yang Mar 8 '11 at 3:22
P.S.: Said German differential ring theorist's lectures had at least a complete proof of the fundamental lemma of Riem. Geom., completely proving tensoriality on the r.h.s. P.P.S.: My answer-comment to Jose is hidden (anyhow syntacticly mangled) - Killing fields indeed serving an example of Koszul vs. not-Koszul. P.P.P.S.: Something learned today. Thanks, sirs. – Martin Gisser Mar 8 '11 at 3:24
What's the "fundamental lemma of Riemannian geometry"? – Deane Yang Mar 8 '11 at 4:09
Haha, good question. My few books call it the fundamental theorem: Existence and uniqueness of the Levi-Civita connection. -- I have currently no idea why I called it lemma. – Martin Gisser Mar 8 '11 at 11:44
I've meanwhile found out that my innocent example ...
1.) ... amounts to a rediscovery of Schouten's contorsion tensor. (See e.g. this note).
This concept is important in torsion gravity (which isn't as exotic as it may sound - except for some charlatanry derived from it...) My example equation seems to express the equivalence principle (cf. link eq. (255)). See also Rodrigues & Oliveira: The many faces of Maxwell, Dirac and Einstein equations.
As I've already hinted here, contorsion stuff (or what I would term the contorsion operator) can also occur in computations with torsion-free connections.
2.) ... leads to a generalization of the fundamental "lemma", a.k.a. Levi-Civita theorem: For any given vector-valued 2-form $T$ there exists a unique metric connection with torsion $T$.
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https://www.physicsforums.com/threads/simple-projectile-question.228734/ | # Simple Projectile Question
1. Apr 13, 2008
### watsup91749
1. The problem statement, all variables and given/known data
Which one of the following statements is true concerning the motion of an ideal projectile launched at an angle of 45 degrees to the horizontal?
ok so, it seems simple enough, this was a multiple choice questions and i narrowed it down to 2 solutions:
1) the speed at the top of the trajectory is zero.
2) the vertical speed decreases on the way up and increases on the way down.
I got stumped here; both of them seem to be true, so when i looked in the answers it said that 2 was right because "the acceleration of the projectile is always down..[the rest of the answers] are wrong. I dont get this, anyone explain this better?
2. Apr 13, 2008
### Nabeshin
Speed is simply the absolute value of velocity, which is a VECTOR quantity. This is to say it has direction, and components, which is referred two in part 2. Consider that all components figure in to the total speed of the projectile. If a projectile's speed was zero at the top, why does it continue going? :)
3. Apr 14, 2008
### tiny-tim
Hi watsup91749!
The horizontal speed is always the same.
In other words, v.costheta is constant.
It starts at vi.cos45º (vi is initial speed), and it stays at that value, however large v is.
"the acceleration of the projectile is always down.." means that the horizontal acceleration is zero!
So, using Newtons's second law, the change in horizontal momentum must be zero.
In other words, the horizontal speed is contant. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9610740542411804, "perplexity": 596.1202819729097}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171163.39/warc/CC-MAIN-20170219104611-00108-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://math.stackexchange.com/users/53783/rustyn?tab=activity | # Rustyn
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I like math, art, music, yoga, running, other things too.
# 1,483 Actions
Jan13 comment $\delta$ = min {1, $\epsilon$} works for proving $\lim_{x->0}$ $x^3$ = 0? Nevermind Did, ok? I'll delete whatever I said. Dec20 awarded Constituent Dec20 awarded Yearling Dec17 revised let $(a_n)$ be a sequence of real numbers such that $|a_{n+1}-a_n|\leq \frac {n^2}{2^n}$ for all $n\in \mathbb N$. Then I messed up Dec17 revised let $(a_n)$ be a sequence of real numbers such that $|a_{n+1}-a_n|\leq \frac {n^2}{2^n}$ for all $n\in \mathbb N$. Then added 77 characters in body Dec17 answered let $(a_n)$ be a sequence of real numbers such that $|a_{n+1}-a_n|\leq \frac {n^2}{2^n}$ for all $n\in \mathbb N$. Then Dec17 comment Number of functions from domain to codomain @Mark sure no problem. Dec17 comment Number of functions from domain to codomain @Mark There are $b^a$ number of functions. Dec15 awarded Nice Answer Dec11 answered How is the area of this triangle calculated Dec11 comment What does non-zero integer mean? do you know what zero is? do you know what non is? Dec11 comment What is a counting number? hahhahhahhahaha Dec9 awarded Caucus Dec4 reviewed Approve A difficult trigonometry problem Dec4 accepted No-where dense sets in the reals Dec4 comment No-where dense sets in the reals @bof That makes sense. Yeah, I didn't think very deeply about the question... Dec4 comment No-where dense sets in the reals @AsafKaragila Yes!!! those are the things that I mean by that. I guess weird to me is just simply things I have less familiarity with. I've played with vitali sets, and cantor space a lot-- fat cantor sets yeah, those are nice. But bernstein sets! I need to play with those some more... I'm a newcomer to desc. set theory so weird means unfamiliar :) Dec4 comment No-where dense sets in the reals However!!!!! Great answer. Dec4 comment No-where dense sets in the reals Guess what I'm looking for isn't there... I'm trying to think of some way something weird could happen--but alas too much is known about $\mathbb{R}$ at this point.... Dec4 comment No-where dense sets in the reals I somehow find these examples unsatisfying.. However, intuitively--these types of examples seem to be, (right now), as the only ones--so I guess there probably isn't a "satisfying" example anyway... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8446815013885498, "perplexity": 1967.6484817906717}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422115858727.26/warc/CC-MAIN-20150124161058-00223-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/questions/51302/hamiltonian-and-non-conservative-force | Hamiltonian and non conservative force
I have to find the Hamiltonian of a charged particle in a uniform magnetic field; the potential vector is $\vec {A}= B/2 (-y, x, 0)$.
I know that $$H=\sum_i p_i \dot q_i -L$$ where $p_i$ is conjugated momentum, $\dot q_i$ is the velocity and $L$ is the Lagrangian.
The result that I should obtain is $$H=\frac{1}{2} m(\dot x^2+ \dot y^2)= \frac {1}{2m}(p_x^2+ p_y^2)+ \frac{1}{2}\omega^2 (x^2+y^2)+ \omega (p_x y- p_y x)$$, where $\omega= \frac{eB}{2mc}$.
I obtain this result only if I don't consider in the Hamiltonian the potential magnetic, and if I substitute only at the last step the values of velocities in function of conjugata momenta.
Considering that the magnetic field isn't a field of conservative forces, I ask you:
if I have a system in a non conservative field, is it correct to not consider the potential of the non conservative force when I'm writing the Hamiltonian? are my steps correct? thank you.
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1) We start writing the Lagrangian
$$L=\frac{1}{2}mv^2+q\vec{v}\cdot\vec{A}=\frac{1}{2}m(v_x^2+v_y^2+v_z^2)+q\frac{B}{2}\left(-yv_x+xv_y \right)$$
2) We find the momenta
$$p_{x}=\frac{\partial L}{\partial v_{x}}=mv_{x}-\frac{qBy}{2}$$
$$p_{y}=\frac{\partial L}{\partial v_{y}}=mv_{y}+\frac{qBx}{2}$$
$$p_{z}=\frac{\partial L}{\partial v_{z}}=mv_{z}$$
3) We express the Hamiltonian, performing a Legrende transformation between velocities and momenta as usual, and solving for the velocities as a function of the coordinates and momenta
$$v_{x}=\frac{1}{m}\left(p_x+\frac{qBy}{2}\right)$$
$$v_{y}=\frac{1}{m}\left( p_y-\frac{qBx}{2} \right)$$
$$v_{z}=\frac{p_{z}}{m}$$
So
$$H=\sum_{i} p_{i}v_{i}-L\bigg|_{v=v(p,q)}=\frac{p_{x}}{m}\left(p_{x}+\frac{qBy}{2}\right)+\frac{p_{y}}{m}\left(p_{y}-\frac{qBx}{2} \right)+\frac{p_z^2}{m}-\frac{1}{2m}\left[\left(p_{x}+\frac{qBy}{2}\right)^2+\left(p_{y}-\frac{qBx}{2} \right)^2+p_z^2\right]-q\frac{B}{2}\left[-\frac{y}{m}\left(p_{x}+\frac{qBy}{2}\right)+\frac{x}{m}\left(p_{y}-\frac{qBx}{2} \right) \right]$$
Expanding the squares
$$H=\frac{p_x^2}{m}+\frac{p_xqBy}{2m}+\frac{p_y^2}{m}-\frac{p_yqBx}{2m}+\frac{p_z^2}{m} -\frac{1}{2m}\left(p_x^2+\frac{q^2B^2y^2}{4}+p_xqBy+p_y^2+\frac{q^2B^2x^2}{4}-p_yqBx+\frac{p_{z}^2}{m}\right)-\frac{qB}{2m}\left( -yp_x -\frac{qBy^2}{2}+xp_y-\frac{qBx^2}{2}\right)$$
Taking common factors
$$H=\frac{p_x^2}{2m}+\frac{p_y^2}{2m}+\frac{p^2_{z}}{2m}+p_{x}\left( \frac{qBy}{2m}-\frac{qBy}{2m}+y\right)+p_y\left( -\frac{qBx}{2m}+\frac{qBx}{2m}-x\right) -\frac{q^2B^2y^2}{2·4m}-\frac{q^2B^2x^2}{2·4m}+\frac{qB^2y^2}{4m}+\frac{qB^2x^2}{4m}$$
Finally
$$H=\frac{p_x^2}{2m}+\frac{p_y^2}{2m}+\frac{p^2_{z}}{2m}+\frac{q^2B^2}{4m}(x^2+y^2)+\frac{qB}{2m}(p_xy-p_yx)$$
In the textbook they problably decided to obviate the $p_z$ because, given that $A_z=0$ the Lagrangian does not depende on $v_z$ so $p_z$ is a constant.
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thanks for your answer! :) – sunrise Jan 17 '13 at 9:25
can I follow these steps for every case, either I have a velocity-dependent potential either I have a non generalized potential? – sunrise Jan 17 '13 at 9:28
In this case you have a velocity dependent potential via the term $q\vec{v}\cdot\vec{A}$ and you can use it following the usual steps. What do you mean by a non generalized potential? if you mean a force non derivable from a potential, please take a look at this question and specifically at the first answer physics.stackexchange.com/questions/41034/… – Jorge Jan 17 '13 at 9:34
For "generalized potential" I mean a "velocity-dependent potential" and for "not generalized potential" a potential dependent only on coordinates.. – sunrise Jan 17 '13 at 13:34
Then the answer is yes, this is the very general prescription to translate a problem of Lagrangian mechanics into a problem of Hamiltonian mechanics. 1)Lagrangian 2) Hamiltonian $\sum v_ip_i -L$ (as a function coordinates and momenta, not velocities!) 3) Hamilton's equations 4) Profit! – Jorge Jan 17 '13 at 13:42
The potential of a charged particle in an electromagnetic field is:
$$U(r,v,t)=q\phi -q\mathbf{v}\cdot A$$
Being $\phi$ the electric potential, $v$ the speed of the particle, $q$ the charge of the particle, and $A$ the vector potential of the magnetic field. Make sure su haven't made any mistakes calculating the lagrange equations (when you derive by $d/dt$, remember $\mathbf{v}$ is a function of time.
The Lagrangian will be:
$$L=\frac{1}{2}mv^2-q\phi +q\mathbf{v}\cdot A$$
Again, make sure you haven't made arithmetic mistakes.
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thank you, but the particle is in a magnetic field and not in an electromagnetic field... – sunrise Jan 15 '13 at 18:33
@sunrise Then the only thing you have to do is remove the electric field term of the potential: $q\phi$, and there you go. – MyUserIsThis Jan 15 '13 at 18:39
sure! but my problem is about the Hamiltonian not about Lagrangian.. :( – sunrise Jan 15 '13 at 18:52
$H=m\dot x ^2+m\dot y^2-1/2m\dot x^2-1/2m\dot y^2-qvA\sin\alpha$, being $\alpha$ the angle bewtween $v$ and $A$, get that angle and operate. – MyUserIsThis Jan 15 '13 at 19:04
the book tells me that the result is $H=\frac{1}{2}m(\dot x^2+ \dot y^2)$ and I have no information about any angle.. :( – sunrise Jan 15 '13 at 19:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9754171967506409, "perplexity": 442.0373680258486}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440646242843.97/warc/CC-MAIN-20150827033042-00044-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/321082/reflectivity-with-complex-refraction-indices | # Reflectivity with complex refraction indices
So the general equation for the reflectivity at the interface between two materials is given by: $$R=\left(\frac{n_1-n_2}{n_1+n_2}\right)^2$$ in case of air/glass $n$ is real, but for, say, semiconductors or metals, where radiation is absorbed, $n$ is a complex number, with $\underline{n}=n_r-ik$. $k$ is described as the extinction coefficient and is related to the absorption coefficient with $\alpha=\frac{4\pi k}{\lambda}$, $\lambda$ being the wavelength.
I am looking to derive a formula for the reflectivity which only includes the real and imaginary parts of the complex refractive index. As far as I can tell, the equation above gives the reflectivity as long as the norm of the index is known, that is $$n_1=\sqrt{n_{r_1}^2+k_1^2} \\ n_2=\sqrt{n_{r_2}^2+k_2^2}$$ in the above formula for the reflectivity, I replaced the norms of the complex numbers and not the numbers themselves,obviously. So doing that, I get a fraction where square root terms appear. On the other hand Wikipedia writes(https://en.wikipedia.org/wiki/Refractive_index) $$R=\left|\frac{n_1-n_2}{n_1+n_2}\right|^2$$which also makes sense and leads to $$R=\frac{(n_{r_1}-n_{r_2})^2+(k_1-k_2)^2}{(n_{r_1}+n_{r_2})^2+(k_1+k_2)^2}$$ Which formula is right?
• As for me, the quantity $n = \sqrt{n_r^2 + k^2}$ does not make any sense. You don't want to mix the real and imaginary part of the refractive index as they describe different phenomena. – Ilya Mar 24 '17 at 16:45
• $n$ is the norm of $\underline{n}$, $\left|\underline{n}\right|=n=\sqrt{n_r^2+k^2}$ , $n_r$ being the real and $k$ being the imaginary part. If $k=0$, there is no point writing an index $\small r$, since it's $n=n_r$ – pitfermi Mar 24 '17 at 16:48
• Yes, I see that, but what is the physical meaning of this norm? The ratio of $n_r$ for two materials refers to the ratio of the wave speeds in them. $k$ is related to the decay rate of the amplitude. What about the norm? – Ilya Mar 24 '17 at 16:59
• If I knew i wouldn't have asked in the first place, but mathematically, the complex refraction index is automatically assigned a norm. why do you assume that it's false to build a norm out of 2 constants which describe a different phenomena? both are constants. there is no problem with units etc. – pitfermi Mar 24 '17 at 17:10
The first equation (Fresnel reflectivity) is derived assuming you have a lossless system, meaning $n$ is always real. When you introduce absorption you get the second formula as all the $k$ go to zero. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9296121597290039, "perplexity": 196.5044034552596}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232258620.81/warc/CC-MAIN-20190526004917-20190526030917-00487.warc.gz"} |
https://www.taylorfrancis.com/chapters/mono/10.1201/b16977-57/section-55-inverse-matrix-alan-jeffrey?context=ubx&refId=63e383d1-5f7c-4d3e-ad1f-89a21d76962a | ## ABSTRACT
Matrix division is not defined, but provided a square matrix A has a non-zero associated determinant jA j, a multiplicative inverse A1 exists with the property that
A1AAA1I :
Thus A1 and A are mutually inverse, in the sense that A1 is the multiplicative inverse of A , and A is the multiplicative inverse of A1. To discover how to compute A1 from A , we need to make use of the Laplace expansion rule for determinants and a result mentioned only in Problem 25 of Problems
52 that the sum of the products of the elements in any row (or column) of a determinant with the corresponding cofactors of a different row (or column)
is zero. It then follows that if C is the matrix of cofactors of A ,
ACTCTA
½ A ½ 0 0 0 0 ½ A ½ 0 0 0 0 ½ A ½ 0 0 0 0 ½ A ½
2 66664
3 77775 ½ A ½ I ,
and so, provided j A j"/0,
A
CT
½ A ½
CT
½ A ½
AI :
Thus, the inverse A1 of A is given by
A1 CT
½ A ½ , provided ½ A ½"0:
The transpose of the matrix of cofactors CT is important in its own right and it is called the adjoint matrix and written adj A . Thus, provided jA j"/0, | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8900142908096313, "perplexity": 815.0658578923046}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948708.2/warc/CC-MAIN-20230327220742-20230328010742-00258.warc.gz"} |
http://math.stackexchange.com/questions/216518/convergence-of-sum-n-1a-n-given-convergence-of-sum-a-n | # Convergence of $\sum (n - 1)a_n$ given convergence of $\sum a_n$
I haven't studied basic calculus/analysis for a while so I apologise if this is obvious. In the course of solving another problem I have found that a sum of the form $$\sum_{n=1}^{\infty}(n - 1)a_n$$ converges. I also know that the sum $\sum_{n=1}^{\infty}a_n$ converges, and also that each term $a_n$ is non-negative. Can I then conclude that $\sum_{n=1}^{\infty}na_n$ converges? I have a feeling this is true but I'm not really sure why.
-
If $a_n \rightarrow a$ and $b_n \rightarrow b$, then $(a+b)_n \rightarrow a + b.$ Your partial sums can certainly be written in this form. – mjqxxxx Oct 18 '12 at 20:46
This is however not exactly matching the title of the question. Knowing just that $\sum a_n$ converges does not allow to conclude that $\sum(n-1)a_n$ converges (e.g. if $a_n=\frac1{(n-1)^2}$ for $n>1$). – Hagen von Eitzen Oct 18 '12 at 20:49
For $n>1$ you have $0\le a_n\le(n-1)a_n$, and $\sum_{n\ge 1}(n-1)a_n$ converges, so by the comparison test $\sum_{n\ge 1}a_n$ converges. The convergence is absolute, since these are non-negative series, so
$$\sum_{n\ge 1}na_n=\sum_{n\ge 1}(n-1)a_n+\sum_{n\ge 1}a_n$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9944079518318176, "perplexity": 103.8340428517679}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645281325.84/warc/CC-MAIN-20150827031441-00317-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://cstheory.stackexchange.com/questions/47936/generating-k-random-bits-from-a-pdf-with-entropy-hp-k/47956 | # Generating $k$ random bits from a pdf with entropy $H(p) = k$
All the sources online say that, intuitively, a distribution with entropy $$k$$ has $$k$$ bits of pure randomness in it.
So can we formalize this as follows? Suppose I can only sample from my distribution, is there an algorithm or procedure to generate a uniform $$k$$ bit binary string?
• Google “randomness extractor”. Note that you need to bound the min-entropy of the source distribution, it is not possible to do this with only a bound on the Shannon entropy. Nov 28, 2020 at 8:46
• @EmilJerabek can you do it with Shannon entropy if you don't care about efficiency? Nov 28, 2020 at 20:09
• I think there’s known examples where one cannot extract from Shannon entropy (in Vadhan’s survey): a source on $n$ bits that is all-zero with .99 probability and otherwise uniform has $\Omega(n)$ Shannon entropy, but even seeded extraction with logarithmic random bits is not possible. The reason is that the output of any such extractor on this source is concentrated on a polynomial number of strings, but no such output distribution can be close to uniform on $\Omega(n)$ bits.
– J.G
Nov 28, 2020 at 22:01
• Ah, Yuval Peres's answer is what I was thinking of. @J.G. Do you know how this answer squares with the example you mentioned? Seems like it's something about order of quantifiers, but I'm not entirely sure. Dec 2, 2020 at 19:02
• @JoshuaGrochow I think the answer shows that Shannon entropy is the right measure given a "large" number of i.i.d. samples. I thought we get just one sample from a source (+ some uniform bits on the side, i.e. seeded extractors) or some constant number of i.i.d. samples (i.e. two-source extractors)--these need min-entropy guarantees. In the example, you expect to get a nonzero string every 100 independent samples on average (where the number 100 crucially depends on the source itself) which is (almost) uniform on $n$ bits, so you get $\Omega(n)$ uniform bits per sample asymptotically I think.
– J.G
Dec 2, 2020 at 22:31
The relevance of Shannon entropy is to repeated sampling: Given $$n$$ independent samples from a source with binary Shannon Entropy $$k$$, you can extract $$nk(1+o(1)$$ i.i.d. uniform bits as $$n$$ tends to infinity with probability tending to 1. This follows e.g. from the Keane-Smorodinsky [1] finitary isomorphism theorem. See also [2]-[5] below.
[1] M. Keane and M. Smorodinsky (1979), Bernoulli schemes of the same entropy are finitarily isomorphic. Annals of Math. 109, 397–406.
[2] P. Elias (1972), The efficient construction of an unbiased random sequence. Ann. Math. Statist. 43, 865–870.
[3] D. E. Knuth and A. C. Yao (1976), The complexity of nonuniform random number generation. Algorithms and complexity (Proc. Sympos., Carnegie-Mellon Univ., Pittsburgh, Pa., 1976), 357–428. Academic Press, New York.
[4] Y. Peres (1992), Iterating von Neumann’s procedure for extracting random bits. Ann. Stat. 20, 590–597.
[5] Harvey, Nate, Alexander E. Holroyd, Yuval Peres, and Dan Romik. "Universal finitary codes with exponential tails." Proceedings of the London Mathematical Society 94, no. 2 (2007): 475-496.
• Sorry, what do you mean by 'extract'? Dec 4, 2020 at 1:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8479264974594116, "perplexity": 366.03216544472144}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662531762.30/warc/CC-MAIN-20220520061824-20220520091824-00463.warc.gz"} |
https://www.physicsforums.com/threads/dont-believe-that-pi-is-a-real-number.742002/ | # Don't believe that pi is a real number
1. Mar 6, 2014
### robertjford80
Wiki defines a real number as
But then it turns out that that definition is not rigorous:
Since we don't have a rigorous definition of a real number it is hard to justify that pi is a real number. Let's discuss what it means to 'represent a quantity'. Say I were to ask you what is the quantity of x, ultimately, you would have to give me an answer which is a complete sentence with a period at the end, in other words, you're explanation of the quantity would have to end. You would then indicate where on the number line the quantity exists. You can't indicate where pi exists on the number line. If you say that it is between 3.14 and 3.15 then you haven't told me where it exists exactly you have only told where it exists roughly. This could go on ad infinitum. Hence, you can't "represent the quantity" of pi.
Admittedly, though in order to give a real rigorous proof that pi is not real, we would need a rigorous definition of real number, something much more precise then represent the quantity and we don't have that yet.
2. Mar 6, 2014
### Blackhawk4560
Pi is real, just irrational as well- because you can graph it, it's real- like try graphing i (or the square root of -1), that's when numbers get imaginary-
Oh imaginary numbers.... Gotta love math sometimes :D
3. Mar 6, 2014
### jgens
This is nonsense. We have perfectly rigorous definitions of real numbers. Axiomatically they are the unique Dedekind-complete ordered field and constructions of the real numbers via Dedekind cuts or Cauchy sequences have been known for over a century. A rigorous definition for π is also easy to give. For example:
$$\pi = 2\int_{-1}^1 \sqrt{1-x^2} \; \mathrm{d}x$$
So I am confused where you are getting all these wild ideas from since everything I just mentioned is very standard stuff.
4. Mar 6, 2014
### robertjford80
I need to know what you mean by graphing. What does it mean to graph a number?
5. Mar 6, 2014
### Blackhawk4560
I should clarify,
Put it on a number line-
*beautiful demonstration*
|------------------|------------------|
-5 0 5
.i, or sqrRt(-1)
*note, i is on its own, because i is an introvert- no really though I just doesn't follow logic, therefore it's not on the line
6. Mar 6, 2014
### robertjford80
I would be interested to know if you could give me a perfectly rigorous definition of perfectly rigorous?
If you cannot, then I'll give you mine and you can analyze my definition. A perfectly rigorous definition starts with a group of indefinable words. The axioms tell us the true and false combination of these indefinables. Every definition is built up from these indefinables. A perfectly rigorous (aka decidedable) sentence is then justified when it can be broken down into a group of sentences which are all composed of indefinables and each sentence has already been declared true but the axioms. For example, say you start with the indefinable words:
x, y, z, R, S, T, &
Then you have a new sentence: gh FG ui
You break that sentence down into:
xRy & zSx & yTx
Those three sentences are already declared true by the axioms. So the new sentence gh FG ui is a perfectly rigorous definition. I seriously doubt there is such a perfect definition as to what Dedkind cuts are. I could be wrong and I'm willing to listen.
7. Mar 6, 2014
### robertjford80
You can't put pi on a number line. If you put it between 3.14 and 3.15, that's wrong. So then you try to be more precise and you put it between 3.1415 and 3.1416 that's still wrong. No matter how much you increase your precision, you're still wrong.
8. Mar 6, 2014
### Blackhawk4560
You're never gonna be perfect, but with 5 trillion digits, I think you are going to be fine in a practical sense- at that point you start to run out of molecules in the paper to be so precise!
9. Mar 6, 2014
### micromass
Staff Emeritus
Weird way of formulating things, but yes, you can give a definition of what it means to be perfectly rigorous. Just take any logic book and work through it, you'll see soon enough what mathematicians call perfectly rigorous.
Basically, you are given some collection of well-formed formula's, called axioms. And you are also given inference rules which allow you to go from one well-formed formula to another. A perfectly rigorous proof is now a list of well-formed formulas. Every well-formed formula follows from a previous one by either an axiom or an inference rule.
Giving definitions of real numbers and giving definitions of pi is then perfectly possible.
Also, claiming that pi is not a real number counts as crackpottery and is not allowed on this forum.
10. Mar 6, 2014
### robertjford80
Applied math is about what's true for getting something practical accomplished. We want to know what's true in the abstract. You have yet to prove that pi can be represented on a line.
11. Mar 6, 2014
### Blackhawk4560
Who knows? EVENTUALLY we may hit an end... But at 5 trillion digits I wouldn't hold your breath... Regardless of its precise location on a number line, it's irrational, but real | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8431560397148132, "perplexity": 724.0056131350389}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501173405.40/warc/CC-MAIN-20170219104613-00200-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://astrobites.org/2011/02/07/magnetic-fields-and-gravitational-collapse-2/ | # Magnetic Fields and Gravitational Collapse
A professor once told me that the key to doing astrophysics is to understand what physics is important in a given problem. Can radiative transfer and cooling be ignored? Are relativistic effects relevant? These and many other questions are often best answered by considering the typical time/size/energy scales of a problem, and then deciding what physical processes are important at those scales.
Tangled magnetic field lines in the authors' simulation of a collapsed gas cloud.
It has become apparent in recent years that in the gravitational collapse of gas clouds (a process which can lead, for example, to star formation), magnetic fields may play a crucial role. Even with an extremely small initial field (or “seed”), the compression of the gas and small-scale dynamo action can lead to a greatly amplified field, by many orders of magnitude. One major way that the field gets enhanced is simply through the collapse of the gas; since astrophysical gas tends to carry the magnetic field lines along with it—the so-called “frozen flux approximation”, valid in the case of low resistivity and high ionization— so that the collapsed cloud has a much higher magnetic pressure than the initial, diffuse cloud. Another important mechanism of magnetic field amplification is dynamo effects, where, on smaller scales, astrophysical turbulence can fold and tangle the field lines (see Brandenburg & Subramanian 2005 for an extensive review of this phenomenon). Also, see the picture for an idea of how tangled field lines can get in a collapsed gas cloud.
In order to get a good handle on the role of magnetic fields in gas collapse scenarios, it is often necessary to run computer simulations. The reason for this is that the combination of self-gravity, turbulence, and magnetic fields creates a 3-D, highly non-linear system that is often difficult to capture in a purely analytic calculation. In light of this, the authors use magnetohydrodynamic (MHD) simulations to explore the dynamics of magnetic fields in collapsing gas clouds.
## Gravitational Collapse: Important Scales
Let us back up a little bit. At the beginning of this astrobite, I briefly mentioned the importance of getting a handle on relevant scales of the problem. For a collapsing gas cloud, one relevant time-scale is the “free-fall time”, or the characteristic time that it takes something to collapse under its own self-gravity, assuming no other forces resist the collapse. The free-fall time $t_{ff}$ is proportional to $rho^{-1/2}$ where $rho$ denotes the density of the object (interestingly enough, $t_{ff}$ is independent of the actual size of the object). Another time-scale which is often important, especially in the context of star formation, is the Kelvin-Helmholtz timescale, which gives an idea of how long a cloud of a given luminosity would need to collapse, if it lost all of its gravitational potential energy by radiating it away. An important mass scale is set by the Jeans mass, which has been touched on in past astrobites. This is the mass of a cloud beyond which gravitational attraction will overcome the stabilizing effect of the internal gas pressure, and the cloud will collapse. The related Jeans length $lambda_J$ is the size of such a cloud (so clouds of the same density that have size $lambda > lambda_J$ will collapse).
## Simulating Gas Collapse
We’ll now return to the idea of simulating magnetized gas clouds, and discuss the methods and results of the paper. One crucial consideration in astrophysical simulations is the question of resolution; is the size of a given cell or particle in your simulation small enough to adequately resolve the important scales of a problem. For instance, from the above discussion it is clear that if we did a simulation on a Cartesian grid where the grid spacing was greater than a Jeans length, then we would not be able to adequately resolve gravitational collapse. Alternatively, having a tiny grid spacing is computationally prohibitive. What, then, is the optimum grid spacing when trying to resolve gravitational collapse in an MHD simulation? The authors of this paper attack this problem by running the same simulation several times, each time with finer grid spacing (the authors use a grid-based code with adaptive mesh refinement (AMR), as opposed to particle-based approaches such as those discussed by Nathan in an earlier astrobite). Then they carry out a Fourier analysis on each simulation. The benefit of a Fourier analysis is that, after averaging over all directions, it shows the relevant scales of whatever is being analyzed—in this case, the magnetic field (please see figure).
This figure, taken from the paper, shows the Fourier power spectra of the different simulations. The different lines show the number of cells per Jeans length, where the y-axis shows the magnetic field "power spectrum", and the x-axis is the wavenumber (proportional to the inverse of length, so smaller k means larger physical size). Only the high-resolution simulations with 32 or more cells per Jeans length clearly show the peak of the spectrum, with the peak shifting further to smaller scales as the resolution is increased. The authors take this as evidence that MHD simulations need a minimum of about 30 cells per Jeans length to properly resolve the magnetic processes at play.
What they find is that, in agreement with past literature, during the collapse, gravitational potential energy is converted into turbulent gas motion, which, consequently, acts to amplify the magnetic field through the mechanisms discussed above. Furthermore, they find that the scale at which this gravitational energy–> turbulence shift happens is on a scale that is roughly comparable to the Jeans length. In addition, they find that MHD simulations, as a rule of thumb, should have a resolution of about 30 grid cells per Jeans length in order to properly capture the magnetic field amplification by small-scale dynamo processes (see figure). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 6, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9096840023994446, "perplexity": 654.786442101853}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370500331.13/warc/CC-MAIN-20200331053639-20200331083639-00517.warc.gz"} |
http://tex.stackexchange.com/questions/36406/an-enumerate-environment-in-which-the-label-is-omitted-when-item-number-is-just | # An enumerate environment in which the label is omitted when item number is just one and displayed as usual when item number is more than one
How to define an enumerate environment in which the label is omitted when item number is just one and displayed as usual when item number is more than one?
-
You must write the contents in a file (or a macro), and then test the number, and then typeset it. – Leo Liu Nov 28 '11 at 14:42
@Leo Thanks. I have seen the manual of thmtools in which (page 3) a package unique is mentioned. But I cannot find it in CTAN. The package might give any help? – Tau spit Nov 28 '11 at 15:02
Instead of "item number" did you mean "the number of items", as it has a different meaning. The solution by @egreg below solves the problem with this rephrasing, not your original wording. – Peter Grill Nov 28 '11 at 16:57
You can typeset the environment inside a box just to get the number of items and then retypeset it:
\usepackage{environ}
\makeatletter
\NewEnviron{fenumerate}
{\setbox0=\vbox{\enumerate\BODY\endenumerate\expandafter}%
\expandafter\def\expandafter\@tempfenum\expandafter{\the\value{\@enumctr}}%
\begin{enumerate}
\ifnum\@tempfenum=\@ne\expandafter\def\csname label\@enumctr\endcsname{}\fi
\BODY
\end{enumerate}}
\makeatother
First of all, the infrastructure set up by environ gathers the environment's text; then it typesets it in a temporary box and at the end the value of the relevant counter (enumi, enumii, enumiii or enumiv) is stored in \@tempfenum, which is used for the "real" typesetting.
Usage:
\begin{fenumerate}
\item this won't get a number
\end{fenumerate}
\begin{fenumerate}
\item this will have a number
\item since it's followed by another item
\end{fenumerate}
-
@GonzaloMedina Yes, of course! – egreg Nov 28 '11 at 17:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9645005464553833, "perplexity": 1824.9869598609275}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246657041.90/warc/CC-MAIN-20150417045737-00070-ip-10-235-10-82.ec2.internal.warc.gz"} |
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