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https://www.physicsforums.com/search/7767894/ | # Search results
• Users: Zyxer22
• In Introductory Physics Homework Help
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1. ### Accceleration due to gravity of a Black Hole
So, the solution to this problem was posted on my course website, but I'm still having issues. Basically, the answer is the same as mine until: a_{0} = \frac{c^{4}}{4(1+\epsilon)^{2}GM} The (1+ε)2 is simplified as (1+2ε)... ok, if ε is really small, I suppose assuming ε2 = 0 is alright. But...
2. ### Accceleration due to gravity of a Black Hole
If there is anyone who could at least give me a point in the right direction I'd be grateful.
3. ### Accceleration due to gravity of a Black Hole
Homework Statement The radius Rh and mass M of a black hole are related by Rh = \frac{2GM}{c2}, where c is the speed of light. Assume that the gravitational acceleration ag of an object at a distance ro = (1 + ε)Rh from the center of a black hole is given by ag = \frac{GM}{r2}, where ε is a...
4. ### Roational Mechanics of Helicopter Blades
Scratch that... I get the correct answers according to the book, but using the numbers from online I get the wrong answers... makes no sense to me
5. ### Roational Mechanics of Helicopter Blades
Right, rotated about it's end the moment of inertia is I = (m L2) /3 I was assuming that the rotational axis is in the center of the blade though, leading to my equation. Using I = (m L2) /3 actually leads to correct answers. Thanks ^^
6. ### HW problem with force and friction
Solving for T, I got 76.97 N I'm willing to bet the discrepancy is rounding. To be sure, I used mg-Tsin(θ) = N friction = (mg-Tsin(θ))μ Tcos(θ)=friction = (mg-Tsin(θ))μ T(cos(θ) + sin(θ)μ) = mgμ so, T =mgμ/(cos(θ) + sin(θ)μ)
7. ### Uniform Circular motion
The first thing to do is to make a picture. You'll notice that there are only 2 forces acting on the plane. Namely, tension and gravity. You know that the horizontal motion is circular, so that leads to Fnet = mv^2/r You know Fnet in the x direction, and velocity. Find a relationship to...
8. ### Roational Mechanics of Helicopter Blades
This question is still getting the better of me if anyone could help.
9. ### Roational Mechanics of Helicopter Blades
Homework Statement A uniform helicopter rotor blade is 8.82 m long, has a mass of 108 kg, and is attached to the rotor axle by a single bolt. (a) What is the magnitude of the force on the bolt from the axle when the rotor is turning at 302 rev/min? (Hint: For this calculation the blade can...
10. ### Time Dilation - Help needed
The idea of time dilation is that light moves at the same constant rate seen from any perspective and that other moving objects don't. As you move faster you're perception of reality has a decreased rate. Everything else appears slower to you. To find the time distortion you would use Einstein's... | {"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": 0.8513421416282654, "perplexity": 1159.3886300450035}, "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/1679296945289.9/warc/CC-MAIN-20230324211121-20230325001121-00634.warc.gz"} |
http://www.chegg.com/homework-help/questions-and-answers/conduct-experiment-amc-increased-movie-ticket-prices-900-1000-measured-change-ticket-sales-q4185943 | To conduct an experiment, AMC increased movie ticket prices from \$9.00 to \$10.00 and measured the change in ticket sales. Using the data over the following month, they have concluded that the increase was profitable. However, over the subsequent months, they changed their minds and discontinued the experiment. How did the timing affect their conclusion about the profitability of increasing prices?
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To conduct an experiment, AMC increased movie ticket prices from \$9.00 to \$10.00 and mea- sured the change in ticket sales. Using the data over the following month, they concluded that the increase was profitable. However, over the subsequent months, they changed their minds and discontinued the experiment. How did the timing affect their conclusion about the profitability of increasing prices? | {"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": 0.8334000110626221, "perplexity": 2977.349159953918}, "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/1469257827077.13/warc/CC-MAIN-20160723071027-00107-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/98466/how-many-points-are-there-on-an-elliptic-curve-reduced-at-a-bad-prime?sort=oldest | # How many points are there on an elliptic curve reduced at a bad prime?
Given an elliptic curve $E$ defined over $\mathbb{Z}$, and a prime $p$, I know that Hasse's theorem gives, when $p$ is a good prime, a relation between the number of solutions over $\mathbb{F}_{p^n}$ and the number of solutions over $\mathbb{F}_p$ (for this, the coefficients of the equation are reduced mod $p$).
Is there such a relation also at the bad primes?
-
For bad primes it is much easier as there are essentially three cases to consider: additive reduction, split multiplicative reduction and non-split multiplicative reduction. It is a good exercise to try to count the number of points in each case yourself. – Daniel Loughran May 31 '12 at 7:38
If the reduction is additive, there are $p+1$ points including one singular point. If it is split multiplicative it is $p$ and if non-split multiplicative, then it is $p+2$. See Washington "Elliptic curves, Number Theory and Cryptography ", section 2.10 on page 59.
... and see François comment below for $n>1$.
The way to remember it is that when you remove the singularity, the rest has a group structure, and it is isomorphic to the additive group (order $p$) in the case of additive reduction, to the multiplicative group of the base field (order $p-1$) or to the kernel of the norm map from the quadratic extension of the base (order $p+1$) in the two cases of multiplicative reduction. – Chandan Singh Dalawat May 31 '12 at 8:05
So when one considers the reduction of $E$ as a curve defined over $\mathbf{F}_{p^n}$, then the number of points is $p^n+1$, $p^n$ in the case of additive resp. split multiplicative reduction. In the non-split multiplicative case it will depend whether $n$ is odd or even. – François Brunault May 31 '12 at 8:14
The question was specifically when $n>1$, so Francois answered that. As a consequence of all this, the relation $|E(\mathbb{F}_p)| = p+1-a_p$ remains true for primes of bad reduction, where $a_p$ are coefficients of the Hasse-Weil L-function. It fails however when $p$ is replaced by $q=p^n$, except for primes where the reduction is additive. – Anonymous May 31 '12 at 8:55
Sorry I did not see the $n$ in the unedited version. – Chris Wuthrich May 31 '12 at 9:45
Careful. This answer is correct when the elliptic curve is given in Weierstrass form but not in general. For instance, $xy(x-y)=p$ defines an elliptic curve whose reduction modulo $p$ has $3p+1$ points. – Felipe Voloch May 31 '12 at 17:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9488599896430969, "perplexity": 230.7660355690409}, "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/1397609537376.43/warc/CC-MAIN-20140416005217-00034-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/horizontal-spring-with-friction.669380/ | # Horizontal spring with friction
1. Feb 4, 2013
### Ali 2
1. The problem statement, all variables and given/known data
A spring with negligible mass has a constant of 105 N/m. It has been compressed horizantally with a 2 kg mass for a distance of 0.1m. If the mass has moved after release for a distance of 0.25m, what is the coefficient of kinetic friction between mass and horizontal surface.
2. Relevant equations
F=-kx (spring force)
f= C N, f is friction, C is the kinetic friction coefficient, and N is the force exerted on the mass by the horizontal surface
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Feb 4, 2013
### Gee Wiz
i would recommend using work to solve this problem.
Work=force*distance
also work done by a spring is .5*k*x^2
3. Feb 4, 2013
### ap123
Conservation of energy would be the easiest way to solve this.
What happens to the potential energy initially stored in the spring?
4. Feb 4, 2013
### Ali 2
hello
ok the work done by the spring is 0.5*k*x02 - 0.5*k*x12
we know that x0=-0.1m
but what is x1, since the 0.25m given in the question is the distance not the displacement??!?
5. Feb 4, 2013
### Ali 2
ok i see what you mean ap123
the energy stored initially = the work done by the friction force
0.5*k*x0^2= C * mg * distance
6. Feb 4, 2013
### ap123
Yes, you've got it :)
Similar Discussions: Horizontal spring with friction | {"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": 0.9727538228034973, "perplexity": 1729.0294839999833}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886126017.3/warc/CC-MAIN-20170824004740-20170824024740-00073.warc.gz"} |
http://www2.macaulay2.com/Macaulay2/doc/Macaulay2-1.19/share/doc/Macaulay2/TSpreadIdeals/html/_t__Strongly__Stable__Mon.html | # tStronglyStableMon -- give the t-strongly stable set generated by a given monomial
## Synopsis
• Usage:
tStronglyStableMon(u,t)
• Inputs:
• u, a t-spread monomial of a polynomial ring
• t, a positive integer that idenfies the t-spread contest
• Outputs:
• a list, the list of all the t-spread monomials of the t-strongly stable set generated by u
## Description
the function tStronglyStableMon(u,t) gives the list of all the monomials belonging to the t-strongly stable set generated by u, that is, $B_t\{u\}.$
We recall that if $u\in M_{n,d,t}\subset S=K[x_1,\ldots,x_n]$ then $B_t\{u\}$ is the smallest t-strongly stable set of monomials of $M_{n,d,t}$ containing $u.$
Moreover, a subset $N\subset M_{n,d,t}$ is called a t-strongly stable set if taking a t-spread monomial $u\in N$, for all $j\in \mathrm{supp}(u)$ and all $i,\ 1\leq i\leq j$, such that $x_i(u/x_j)$ is a t-spread monomial, then it follows that $x_i(u/x_j)\in N$.
Examples:
i1 : S=QQ[x_1..x_9] o1 = S o1 : PolynomialRing i2 : tStronglyStableMon(x_2*x_5*x_8,2) o2 = {x x x , x x x , x x x , x x x , x x x , x x x , x x x , x x x , x x x , 1 3 5 1 3 6 1 3 7 1 3 8 1 4 6 1 4 7 1 4 8 1 5 7 1 5 8 ------------------------------------------------------------------------ x x x , x x x , x x x , x x x , x x x } 2 4 6 2 4 7 2 4 8 2 5 7 2 5 8 o2 : List i3 : tStronglyStableMon(x_2*x_5*x_8,3) o3 = {x x x , x x x , x x x , x x x } 1 4 7 1 4 8 1 5 8 2 5 8 o3 : List | {"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": 0.913127064704895, "perplexity": 276.87203767428946}, "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-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00698.warc.gz"} |
https://www.intmath.com/blog/mathematics/subtracting-negative-numbers-from-positive-numbers-12499 | Search IntMath
Close
# Subtracting Negative Numbers From Positive Numbers
By Kathleen Cantor, 07 Oct 2020
Addition and subtraction can get complex the more numbers, signs and variables are involved. This is why, before we dive into the subtraction of negative numbers from positive numbers, we will revisit the basics of numbers, addition and subtraction first so that the concepts are built on one another.
## Numbers can be positive or negative
Numbers can be positive or negative. Think of all the numbers that exist; they are all on a spectrum from negative through zero to positive.
Imagine all these numbers on a line; this will make it easier for you to envision each number as positive or negative. For example:
-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
All of the numbers to the right of the zero are positive.
## Denoting positive integers
When denoting positive numbers or integers, we do not need to add the plus "+" sign before the number; it is assumed that the number is positive when there is no negative '-' sign.
This means that the number '3' is actually '+3'. If we want 3 to be noted as negative, we would have to list it as '-3'.
1. Adding a positive number to another positive number
When we are adding two or more positive numbers together, we usually use the plus '+' sign. For example:
5 + 7 = 12
5 and 7 are both positive numbers. 7 is adding to the value of 5 and thus increasing the total value of the sum.
1. Subtracting a positive number from another positive number
When we are subtracting a positive number from another positive number, we usually use the minus '-' sign. For example:
8 - 4 = 4
8 and 4 and both positive numbers (note the lack of the minus sign before either) and thus 4 is taking away value from eight.
1. Adding a negative number to a positive number or vice versa
When the negative sign '-' and the positive sign are added together, the sign with the larger value will be the sign of the sum total. For example:
3 + (-2) = 1 i.e. (+1) because the positive sign has a larger value than the negative sign.
Another example is 2 + (-4) = (-2) because the negative sign has a higher value than that of the positive sign.
or even (-5) + 7 = 2 i.e. (+2) because the positive sign has the larger value.
Remember that a positive number plus a negative number is always a subtraction, and the sign of the total is determined by which of the signs had a larger magnitude. i.e
4 +( -2) = 4 - 2 = 2
and
4 + (-6) = 4 - 6 = (-2)
1. Subtracting a positive number from a negative number
When subtracting a positive number from a negative number, the difference will always have a negative sign. This is because you are actually adding the numbers on the negative side of the number line.
Think of the number line: -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5
Now imagine taking away 2 from 5, what would you do? You would take two steps to the left from five, and that would make your answer 3. The same concept is applied when taking away a positive number from a negative number. For example
(-3) - 2 = (-5) that is, you take two steps to the left of negative three on the number line in order to get your difference.
This gives the same effect as adding two positive numbers, only with a negative sign added before the answer. So, next time you are stuck when subtracting a positive number from a negative number, simply add the two numbers together and put the negative sign '-' before the sum total. For example:
-5 - 7 = -5 - (+7) = (-12) but if you are stuck simply change the question to 5 + 7 = 12 and then add the negative sign '-' to your answer.
1. Subtracting a negative number from another negative number
When subtracting a negative number from another negative number, the minus sign will become positive. This is due to the fact that a double minus signifies a multiplication of signs, that is, -2 - -2 = -2 + 2.
The negative sign for the number two, together with the minus sign that denotes subtraction yield a positive sign; two negatives equal a positive.
Therefore, when subtracting a negative number from another negative number, remember that the sign turns positive.
For example:
-4 -( -5) = 4 +5 = 9
1. Subtracting a negative number from a positive number
When subtracting a negative number from a positive number, the sign will also become a positive sign because, as discussed in step number five, a negative times a negative is a positive.
Knowing that this same principle applies, we can now begin subtracting.
5 - (-9) = 5 + 9 = 14
A negative number subtracted from a positive number always yields a positive result.
## Main Concepts:
The key take away is this:
(+) (+) = + [a positive and a positive will yield a positive]
(+) (-) = - [ a positive and a negative will yield a negative]
(-) (+) = - [ a negative and a positive will also yield a negative]
(-) (-) = + [ a negative and a negative will yield a positive]
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From Math Blogs | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8139176964759827, "perplexity": 521.413878706326}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662562106.58/warc/CC-MAIN-20220523224456-20220524014456-00061.warc.gz"} |
http://www.physicsforums.com/showthread.php?p=3871719 | # dampening of oscillatory motion
by thefancybum
Tags: dampening, motion, oscillatory
P: 1 Is there a quantitative relationship between dampening of a mass on a spring and time. I need some help on how to go about finding a mathematical relationship of dampening of a mass on a spring. thank you | {"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": 0.842214047908783, "perplexity": 506.2434356953337}, "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-2013-48/segments/1386163065934/warc/CC-MAIN-20131204131745-00024-ip-10-33-133-15.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/33778-expected-value-product-two-correlated-variables.html | ## Expected value of the product of two correlated variables
Hello Everyone,
I'm new to this forum and I'm wondering if someone could help me solving the following problem:
Let h be an M x 1 vector whose elements are independent zero-mean and unit-variance circularly symmetric complex Gaussian; v is an M x 1 deterministic complex vector, and c is a real constant. Is there any close form expression for E[hh*v/(c+v*hh*v)]? Here E[.] is the expectation operator and * denotes hermitian conjugate.
Thanks. | {"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": 0.9313445091247559, "perplexity": 334.895339621974}, "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/1518891816462.95/warc/CC-MAIN-20180225130337-20180225150337-00485.warc.gz"} |
https://eng.libretexts.org/Bookshelves/Chemical_Engineering/Map%3A_Fluid_Mechanics_(Bar-Meir)/12%3A_Compressible_Flow_2%E2%80%93Dimensional/12.6%3A_d'Alembert's_Paradox | In ideal inviscid incompressible flows, the movement of body does not encounter any resistance. This result is known as d'Alembert's Paradox, and this paradox is examined here. Supposed that a two–dimensional diamond–shape body is stationed in a supersonic flow as shown in Figure 12.27.
Fig. 12.27 A simplified diamond shape to illustrate the supersonic d'Alembert's Paradox.
Again, it is assumed that the fluid is inviscid. The net force in flow direction, the drag, is
$D = 2 \left( \dfrac{w }{ 2} \, (P_2 - P_4)\right) = w \, (P_2 - P_4) \label{pm:eq:dragG} \tag{38}$
It can be observed that only the area that "seems'' to be by the flow was used in expressing equation (38). The relation between $$P_2$$ and $$P_4$$ is such that the flow depends on the upstream Mach number, $$M_1$$, and the specific heat, $$k$$. Regardless in the equation of the state of the gas, the pressure at zone 2, $$P_2$$, is larger than the pressure at zone 4, $$P_4$$. Thus, there is always drag when the flow is supersonic which depends on the upstream Mach number, $$M_1$$, specific heat, $$k$$, and the "visible'' area of the object. This drag is known in the literature as (shock) wave drag.
### Contributors
• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. | {"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": 0.9370213747024536, "perplexity": 724.5917821355303}, "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-2019-04/segments/1547583684033.26/warc/CC-MAIN-20190119221320-20190120003320-00551.warc.gz"} |
https://socratic.org/questions/what-is-the-domain-and-range-of-y-sqrt-1-x | Algebra
Topics
What is the domain and range of y =- sqrt(1 - x)?
Apr 20, 2018
$\therefore {D}_{f} : x \le 1$
${R}_{f} : y \le 0$
Explanation:
The term inside the Square root must be non-negative for the
function to be defined so ;
Domain of function is ${D}_{f}$ :
${D}_{f} : 1 - x \ge 0$
$\therefore {D}_{f} : x \le 1$
Since the function attains all the negative values and also $0$ .
$\therefore$the range of function is thus ${R}_{f} : y \le 0$
The graph of the function is given below :-
Impact of this question
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https://www.physicsforums.com/threads/oval-shaped-cartesian-curves.127021/ | # Oval Shaped Cartesian Curves
1. Jul 24, 2006
### mubashirmansoor
Hello, are all Oval Shaped Cartesian Curves" +/-(x^2) " or we can have it with other degrees??
2. Jul 24, 2006
### HallsofIvy
Staff Emeritus
I have no idea what you mean. It is true that every ellipse (including the special case of a circle), if that's what you mean by "oval", having axes parallel to the x and y- axes, has equation
$$\frac{(x-x_0)^2}{a^2}+ \frac{(y-y_0)^2}{b^2}= 1$$.
If you allow the axes to be tilted, then is possible to have an equation that has either x2 or y2 but not both (but then, of course, it must also include xy).
If you mean "oval" in the general sense of "any roughly eggshaped closed path" then it may have quite different equations- some involving higher degree functions.
Similar Discussions: Oval Shaped Cartesian Curves | {"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": 0.9217262864112854, "perplexity": 2015.7722713134754}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187822851.65/warc/CC-MAIN-20171018085500-20171018105500-00340.warc.gz"} |
https://mathoverflow.net/questions/33365/wolframs-2-state-3-symbol-turing-machine/33393 | # Wolfram's 2-state 3-symbol Turing machine
A few years ago it was announced that a 2-state symbol Turing machine was proven to be universal. However, Vaughn Pratt disputed the proof, and I gather he still disputes it. Wolfram's prize committee seems to be satisfied.
Is there anyone not on team Wolfram who believes the proof is correct?
-
This is an interesting twist: "Believing in a proof"... – vonjd Jul 26 '10 at 8:05
The question was maybe ill-posed, though I am not an expert. Minsky (1967) required finite input, then repeating infinite was allowed decades later, but Pratt does not like the fact that the (2,3) "universality" allows nonrepeating infinite, though patterned. Without proper definitions, anything is valid. – Junkie Jul 26 '10 at 8:20
According to wolframscience.com/prizes/tm23/solved.html, the proof was supposed to appear in Complex Systems, but that hasn't happened. – Dan Ramras Jul 27 '10 at 0:50
## 3 Answers
See the discussion on FOM mailing list. As far as I remember, according to some members of the prize committee, Wolfram announced it without proper contact with them. There was also discussion about what was Pratt's objection to the proof. See this:
and the other posts in that thread.
-
I've looked through that, but I still can't quite figure out what the (non-Wolfram) consensus is. As Junkie says, it seems it boils down who accepts what definition. – k2forever Jul 26 '10 at 21:10
See Alexander Smith's posts on the FOM. I think he himself agreed at some point that his proof has a problem, don't remember if it was corrected. As far as I remember, no one said that she/he agrees with Wolfram, though I might be wrong. I consider it as a kind of consensus. (IMHO as a non-expert, the question is not completely well defined, and even if it was, it is not as important as it seems.) – Kaveh Jul 27 '10 at 0:00
There is a quite interesting post in Shtetl-Optimized about this topic. Here you can get an idea of many researcher's opinion about after the announcement of the discovery (it does not seem to have changed much since then); if you are looking for the opinion of the community, this is definitely want you should read.
What I get from the discussion: people seem to agree that these Turing Machine are universal (in some interesting but not so trivial sense), and maybe the simplest Turing machine we can hope to find. Yet nobody has found this simple machine very useful for theoretical computer science.
Still it's obviously a cool Turing Machine ;)
-
According to a paper published in Complex Systems in 2010 [1], Smith's paper is still forthcomming. Here is the reference given there to Smith's paper:
[6] A. Smith, “Wolfram’s 2,3 Turing Machine Is Universal,” Complex Systems, to appear. (Aug 12, 2010)
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http://mt-fedfiles.blogspot.com/2012_11_04_archive.html | ## Friday, November 09, 2012
### UHF Changes in Oak Ridge?
A recent posting in the Radio Reference forums has some information on what appears to the an additional P25 UHF trunked site from the National Nuclear Security Administration facility in Oak Ridge, TN. You can see the posting here:
http://forums.radioreference.com/tennessee-radio-discussion-forum/252895-i-found-p25-trunked-system-i-cant-identify.html
The original poster did not list any system information, other than it is reporting as Site 002. The control channel frequencies he was hearing were 408.7125 & 410.1125 MHz. He also reported a voice channel of 408.2250 MHz. My suspicion is that this is an additional site to the already existing Oak Ridge DoE NNSA P25 trunked system:
http://www.radioreference.com/apps/db/?sid=4446
If anyone has any additional information you want to pass along, please do!
### UHF Changes in Pittsburgh?
I recently spotted an entry to the Radio Reference database showing a "new" federal UHF trunked system. You can check out the entry here:
The system is not new, and for you Fed Files readers, it appeared in the March, 2006 column. It belongs to the National Nuclear Security Administration Bechtel Bettis Navy Reactor Facility in West Mifflin, PA a suburb of Pittsburgh. You can see more about this facility at these web sites:
The entry in the Radio Reference database indicates that they may have changed some frequencies in the system since I last monitored it. Originally I had it using these channels:
406.9750 MHz
407.1375 MHz
407.3875 MHz
407.7875 MHz
408.1750 MHz
New new listing has added a frequency of 406.3375 MHz and deleted the 407.1375 MHz channel. Probably some additional re-shuffling of the federal UHF assignments. | {"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": 0.8175559043884277, "perplexity": 4626.854877688841}, "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-2014-52/segments/1418802777438.76/warc/CC-MAIN-20141217075257-00114-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://cris.bgu.ac.il/en/publications/strong-lensing-modeling-in-galaxy-clusters-as-a-promising-method- | # Strong Lensing Modeling in Galaxy Clusters as a Promising Method to Test Cosmography. I. Parametric Dark Energy Models
Juan Magana, Ana Acebrón, Verónica Motta, Tomás Verdugo, Eric Jullo, Marceau Limousin
Research output: Contribution to journalArticlepeer-review
8 Scopus citations
## Abstract
In this paper we probe five cosmological models for which the dark energy equation of state parameter, w(z), is parameterized as a function of redshift using strong lensing data in the galaxy cluster Abell 1689. We constrain the parameters of the w(z) functions by reconstructing the lens model under each one of these cosmologies with strong lensing measurements from two galaxy clusters, Abell 1689 and a mock cluster, Ares, from the Hubble Frontier Fields Comparison Challenge, to validate our methodology. To quantify how the cosmological constraints are biased due to systematic effects in the strong lensing modeling, we carry out three runs considering the following uncertainties for the multiple image positions: 0.″25, 0.″5, and 1.″0. With Ares, we find that larger errors decrease the systematic bias on the estimated cosmological parameters. With real data, our strong-lensing constraints on w(z) are consistent with those derived from other cosmological probes. We confirm that strong lensing cosmography with galaxy clusters is a promising method to constrain w(z) parameterizations. A better understanding of galaxy clusters and their environment is needed, however, to improve the SL modeling and hence to estimate stringent cosmological parameters in alternative cosmologies.
Original language English 122 Astrophysical Journal 865 2 https://doi.org/10.3847/1538-4357/aada7d Published - 1 Oct 2018
## Keywords
• cosmological parameters
• dark energy
• gravitational lensing: strong
## ASJC Scopus subject areas
• Astronomy and Astrophysics
• Space and Planetary Science
## Fingerprint
Dive into the research topics of 'Strong Lensing Modeling in Galaxy Clusters as a Promising Method to Test Cosmography. I. Parametric Dark Energy Models'. Together they form a unique fingerprint. | {"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": 0.8662275075912476, "perplexity": 4953.725955626447}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943471.24/warc/CC-MAIN-20230320083513-20230320113513-00021.warc.gz"} |
http://www.researchgate.net/publication/24015024_HI_Observations_of_the_Supermassive_Binary_Black_Hole_System_in_0402379 | Article
# HI Observations of the Supermassive Binary Black Hole System in 0402+379
The Astrophysical Journal (Impact Factor: 6.73). 02/2009; DOI:10.1088/0004-637X/697/1/37
Source: arXiv
ABSTRACT We have recently discovered a supermassive binary black hole system with a projected separation between the two black holes of 7.3 parsecs in the radio galaxy 0402+379. This is the most compact supermassive binary black hole pair yet imaged by more than two orders of magnitude. We present Global VLBI observations at 1.3464 GHz of this radio galaxy, taken to improve the quality of the HI data. Two absorption lines are found toward the southern jet of the source, one redshifted by 370 +/- 10 km/s and the other blueshifted by 700 +/- 10 km/s with respect to the systemic velocity of the source, which, along with the results obtained for the opacity distribution over the source, suggests the presence of two mass clumps rotating around the central region of the source. We propose a model consisting of a geometrically thick disk, of which we only see a couple of clumps, that reproduces the velocities measured from the HI absorption profiles. These clumps rotate in circular Keplerian orbits around an axis that crosses one of the supermassive black holes of the binary system in 0402+379. We find an upper limit for the inclination angle of the twin jets of the source to the line of sight of 66 degrees, which, according to the proposed model, implies a lower limit on the central mass of ~7 x 10^8 Msun and a lower limit for the scale height of the thick disk of ~12 pc . Comment: 20 pages, 7 figures. Accepted on the Astrophysical Journal
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http://mathhelpforum.com/pre-calculus/9706-rocket-my-pocket-print.html | # rocket in my pocket
• Jan 8th 2007, 09:21 AM
math619
rocket in my pocket
How would you go about solving this?
A rocket is fired into the air and it's height in metres at any given time t seconds can be calculated by h(t)=225+ 49t-0.98t(squared). Find the maximum height of the rocket and the time at which it occurs.
• Jan 8th 2007, 09:33 AM
ThePerfectHacker
Quote:
Originally Posted by math619
How would you go about solving this?
A rocket is fired into the air and it's height in metres at any given time t seconds can be calculated by h(t)=225+ 49t-0.98t(squared). Find the maximum height of the rocket and the time at which it occurs.
The squared coefficient is $a=-.98$
The linear coefficient is $b=49$.
We have a turning point at,
$x=\frac{-b}{2a}=\frac{-49}{2(-.98)}$ | {"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": 3, "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": 0.9084219336509705, "perplexity": 725.5615934105042}, "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/1505818690203.42/warc/CC-MAIN-20170924190521-20170924210521-00008.warc.gz"} |
https://rigtriv.wordpress.com/2009/11/10/segre-classes-of-cones/ | ## Segre Classes of Cones
Last time, we talked about the Normal Cone. We’re going to go back a bit and increase the generality before coming back to it. Let $C$ be a cone over $X$, and let $P=P(C\oplus 1)$ be the projective closure. We define the Segre class of the cone $C$, $s(C)$ in $A_*(X)$ to be $s(C)=q_*\left(\sum_{i\geq 0} c_1(\mathscr{O}(1))^i\cap [P]\right)$, where $q:P\to X$ is the projection.
So…that’s a fairly opaque definition. What does it give us in a case we understand? Let $E$ be a vector bundle, so it’s a special cone. Then $[P(E\oplus 1)]=q^*[X]$, and so the definition can be unraveled (annoying calculation, do it in the privacy of your office) to get $c(E\oplus 1)^{-1}\cap [X]$, but then $c(E\oplus 1)=c(E)$, so for a vector bundle, $s(E)=c(E)^{-1}\cap [X]$.
Additionally, you get geometric multiplicities of the cone out: $s(C)=\sum_{i=1}^t m_i s(C_i)$ where $C_i$ are the cone’s irreducible components.
So this Segre class tells us interesting things, and in the cases we’ve already looked at, gives us familiar things. But what happens if we try to combine cones? Well, if $C$ and $D$ are cones defined by $S^*$ and $T^*$, we get a cone $C\times_X D$ given by $S^*\otimes_{\mathscr{O}_X} T^*$. Now, if $D$ is actually a vector bundle $E$, we define $C\oplus E = C\times_X E$ and get $s(C\oplus E)=c(E)^{-1}\cap s(C)$.
This is actually a general pattern. If we want to do anything interesting with cones, we need to include the special case of vector bundles in order to work it out. For example, a morphism of cones $\phi:C\to C'$ is a morphism of their algebras $\phi^*:S^{*'}\to S^*$, and a short exact sequence is $0\to E\to C\to C'\to 0$, where $E$ is a vector bundle requires that the map on algebras for $C\to C'$ be injective adn the map $S^*\to \mathrm{Sym}(\mathscr{E})$ be surjective, and that there is $\tilde{\mathscr{E}}$ locally free subsheaf of $S^1$ such that $S^{*'}\otimes_{\mathscr{O}_X} \mathrm{Sym}(\tilde{\mathscr{E}})\to S^*$ is an isomorphism. Yeah, it’s kind of a mess.
However, we do get the good result that if $0\to E\to C\to C'\to 0$ is a short exact sequence of cones, we have $s(C')=c(E)\cap s(C)$. In particular, if we have a coherent sheaf $\mathscr{F}$, we get a cone $C(\mathscr{F})=\underline{\mathrm{Spec}}(\mathrm{Sym}(\mathscr{F}))$, and we define $s(\mathscr{F})=s(C(\mathscr{F}))$. Then if we have $0\to \mathscr{F}'\to\mathscr{F}\to\mathscr{E}\to 0$ with $\mathscr{E}$ locally free, we get precisely the above back, and get the same formula. | {"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": 41, "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": 0.9874061942100525, "perplexity": 182.16377615780112}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698540409.8/warc/CC-MAIN-20161202170900-00166-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/188227/how-to-properly-space-unicode-ellipsis | # How to properly space unicode ellipsis?
Please note the discussion at the end of this thread. After discussing the tex codes \ldots et al, a commenter points out that there is a unicode code point for an ellipsis.
However, on my document (which I'm editing in Lyx), the … character prints out in my pdf as something resembling a \cdots code; that is, vertically centered (and in my case, also shifted horizontally to collide with the next word after the whitespace). See the graphic:
In the pdf file itself, I can copy the ellipsis out, and when I paste it into another document, it remains an ellipsis.
This is undesireable behavior; how do I tell TeX to treat the unicode ellipsis correctly as a baseline ellipsis character, and format it accordingly?
ETA: here is an mwe, which isolates the problem:
\documentclass{article}
\usepackage{xeCJK}
\newfontfamily\zhcn [Path=/usr/share/fonts/arphicfonts/]{uming}
\begin{document}
blah … ... blah
\end{document}
On my system, that outputs the ellipsis character as raised and shifted rightwards. If I drop those two lines from the preamble, the text is correctly formatted. But the xeCJK package shouldn't be called unless I specifically flag something to use it, right?
• Could we have a minimal working example? – Bernard Jul 6 '14 at 15:16
• For its working (I don't know exactly why), xeCJK assigns … (U+2026) to character class 3; it also defines interchar tokens between classes 0 (default) and 3, with the result that the character … is taken from the main CJK font. – egreg Jul 6 '14 at 22:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.9683868885040283, "perplexity": 2660.233473058338}, "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-2021-17/segments/1618038476606.60/warc/CC-MAIN-20210418103545-20210418133545-00322.warc.gz"} |
https://cms.math.ca/cjm/kw/%24p%24-adic | location: Publications → journals
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Search: All articles in the CJM digital archive with keyword $p$-adic
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1. CJM Online first
Smith, Jerrod Manford
Relative discrete series representations for two quotients of $p$-adic $\mathbf{GL}_n$ We provide an explicit construction of representations in the discrete spectrum of two $p$-adic symmetric spaces. We consider $\mathbf{GL}_n(F) \times \mathbf{GL}_n(F) \backslash \mathbf{GL}_{2n}(F)$ and $\mathbf{GL}_n(F) \backslash \mathbf{GL}_n(E)$, where $E$ is a quadratic Galois extension of a nonarchimedean local field $F$ of characteristic zero and odd residual characteristic. The proof of the main result involves an application of a symmetric space version of Casselman's Criterion for square integrability due to Kato and Takano. Keywords:$p$-adic symmetric space, relative discrete series, Casselmanâs CriterionCategories:22E50, 22E35
2. CJM 2016 (vol 68 pp. 1227)
Brasca, Riccardo
Eigenvarieties for Cuspforms over PEL Type Shimura Varieties with Dense Ordinary locus Let $p \gt 2$ be a prime and let $X$ be a compactified PEL Shimura variety of type (A) or (C) such that $p$ is an unramified prime for the PEL datum and such that the ordinary locus is dense in the reduction of $X$. Using the geometric approach of Andreatta, Iovita, Pilloni, and Stevens we define the notion of families of overconvergent locally analytic $p$-adic modular forms of Iwahoric level for $X$. We show that the system of eigenvalues of any finite slope cuspidal eigenform of Iwahoric level can be deformed to a family of systems of eigenvalues living over an open subset of the weight space. To prove these results, we actually construct eigenvarieties of the expected dimension that parameterize finite slope systems of eigenvalues appearing in the space of families of cuspidal forms. Keywords:$p$-adic modular forms, eigenvarieties, PEL-type Shimura varietiesCategories:11F55, 11F33
3. CJM 2016 (vol 68 pp. 571)
Gras, Georges
Les $\theta$-régulateurs locaux d'un nombre algébrique : Conjectures $p$-adiques Let $K/\mathbb{Q}$ be Galois and let $\eta\in K^\times$ be such that $\operatorname{Reg}_\infty (\eta) \ne 0$. We define the local $\theta$-regulators $\Delta_p^\theta(\eta) \in \mathbb{F}_p$ for the $\mathbb{Q}_p\,$-irreducible characters $\theta$ of $G=\operatorname{Gal}(K/\mathbb{Q})$. A linear representation ${\mathcal L}^\theta\simeq \delta \, V_\theta$ is associated with $\Delta_p^\theta (\eta)$ whose nullity is equivalent to $\delta \geq 1$. Each $\Delta_p^\theta (\eta)$ yields $\operatorname{Reg}_p^\theta (\eta)$ modulo $p$ in the factorization $\prod_{\theta}(\operatorname{Reg}_p^\theta (\eta))^{\varphi(1)}$ of $\operatorname{Reg}_p^G (\eta) := \frac{ \operatorname{Reg}_p(\eta)}{p^{[K : \mathbb{Q}\,]} }$ (normalized $p$-adic regulator). From $\operatorname{Prob}\big (\Delta_p^\theta(\eta) = 0 \ \& \ {\mathcal L}^\theta \simeq \delta \, V_\theta\big ) \leq p^{- f \delta^2}$ ($f \geq 1$ is a residue degree) and the Borel-Cantelli heuristic, we conjecture that, for $p$ large enough, $\operatorname{Reg}_p^G (\eta)$ is a $p$-adic unit or that $p^{\varphi(1)} \parallel \operatorname{Reg}_p^G (\eta)$ (a single $\theta$ with $f=\delta=1$); this obstruction may be lifted assuming the existence of a binomial probability law confirmed through numerical studies (groups $C_3$, $C_5$, $D_6$). This conjecture would imply that, for all $p$ large enough, Fermat quotients, normalized $p$-adic regulators are $p$-adic units and that number fields are $p$-rational. We recall some deep cohomological results that may strengthen such conjectures. Keywords:$p$-adic regulators, Leopoldt-Jaulent conjecture, Frobenius group determinants, characters, Fermat quotient, Abelian $p$-ramification, probabilistic number theoryCategories:11F85, 11R04, 20C15, 11C20, 11R37, 11R27, 11Y40
4. CJM 2015 (vol 67 pp. 1046)
Dubickas, Arturas; Sha, Min; Shparlinski, Igor
Explicit Form of Cassels' $p$-adic Embedding Theorem for Number Fields In this paper, we mainly give a general explicit form of Cassels' $p$-adic embedding theorem for number fields. We also give its refined form in the case of cyclotomic fields. As a byproduct, given an irreducible polynomial $f$ over $\mathbb{Z}$, we give a general unconditional upper bound for the smallest prime number $p$ such that $f$ has a simple root modulo $p$. Keywords:number field, $p$-adic embedding, height, polynomial, cyclotomic fieldCategories:11R04, 11S85, 11G50, 11R09, 11R18
5. CJM 2009 (vol 62 pp. 34)
Campbell, Peter S.; Nevins, Monica
Branching Rules for Ramified Principal Series Representations of $\mathrm{GL}(3)$ over a $p$-adic Field We decompose the restriction of ramified principal series representations of the $p$-adic group $\mathrm{GL}(3,\mathrm{k})$ to its maximal compact subgroup $K=\mathrm{GL}(3,R)$. Its decomposition is dependent on the degree of ramification of the inducing characters and can be characterized in terms of filtrations of the Iwahori subgroup in $K$. We establish several irreducibility results and illustrate the decomposition with some examples. Keywords:principal series representations, branching rules, maximal compact subgroups, representations of $p$-adic groupsCategories:20G25, 20G05
6. CJM 2008 (vol 60 pp. 1067)
Kariyama, Kazutoshi
On Types for Unramified $p$-Adic Unitary Groups Let $F$ be a non-archimedean local field of residue characteristic neither 2 nor 3 equipped with a galois involution with fixed field $F_0$, and let $G$ be a symplectic group over $F$ or an unramified unitary group over $F_0$. Following the methods of Bushnell--Kutzko for $\GL(N,F)$, we define an analogue of a simple type attached to a certain skew simple stratum, and realize a type in $G$. In particular, we obtain an irreducible supercuspidal representation of $G$ like $\GL(N,F)$. Keywords:$p$-adic unitary group, type, supercuspidal representation, Hecke algebraCategories:22E50, 22D99
7. CJM 2006 (vol 58 pp. 897)
Courtès, François
Distributions invariantes sur les groupes réductifs quasi-déployés Soit $F$ un corps local non archim\'edien, et $G$ le groupe des $F$-points d'un groupe r\'eductif connexe quasi-d\'eploy\'e d\'efini sur $F$. Dans cet article, on s'int\'eresse aux distributions sur $G$ invariantes par conjugaison, et \a l'espace de leurs restrictions \a l'alg\ebre de Hecke $\mathcal{H}$ des fonctions sur $G$ \a support compact biinvariantes par un sous-groupe d'Iwahori $I$ donn\'e. On montre tout d'abord que les valeurs d'une telle distribution sur $\mathcal{H}$ sont enti\erement d\'etermin\'ees par sa restriction au sous-espace de dimension finie des \'el\'ements de $\mathcal{H}$ \a support dans la r\'eunion des sous-groupes parahoriques de $G$ contenant $I$. On utilise ensuite cette propri\'et\'e pour montrer, moyennant certaines conditions sur $G$, que cet espace est engendr\'e d'une part par certaines int\'egrales orbitales semi-simples, d'autre part par les int\'egrales orbitales unipotentes, en montrant tout d'abord des r\'esultats analogues sur les groupes finis. Keywords:reductive $p$-adic groups, orbital integrals, invariant distributionsCategories:22E35, 20G40
8. CJM 2005 (vol 57 pp. 648)
Nevins, Monica
Branching Rules for Principal Series Representations of $SL(2)$ over a $p$-adic Field We explicitly describe the decomposition into irreducibles of the restriction of the principal series representations of $SL(2,k)$, for $k$ a $p$-adic field, to each of its two maximal compact subgroups (up to conjugacy). We identify these irreducible subrepresentations in the Kirillov-type classification of Shalika. We go on to explicitly describe the decomposition of the reducible principal series of $SL(2,k)$ in terms of the restrictions of its irreducible constituents to a maximal compact subgroup. Keywords:representations of $p$-adic groups, $p$-adic integers, orbit method, $K$-typesCategories:20G25, 22E35, 20H25
9. CJM 2004 (vol 56 pp. 897)
Borwein, Jonathan M.; Borwein, David; Galway, William F.
Finding and Excluding $b$-ary Machin-Type Individual Digit Formulae Constants with formulae of the form treated by D.~Bailey, P.~Borwein, and S.~Plouffe (\emph{BBP formulae} to a given base $b$) have interesting computational properties, such as allowing single digits in their base $b$ expansion to be independently computed, and there are hints that they should be \emph{normal} numbers, {\em i.e.,} that their base $b$ digits are randomly distributed. We study a formally limited subset of BBP formulae, which we call \emph{Machin-type BBP formulae}, for which it is relatively easy to determine whether or not a given constant $\kappa$ has a Machin-type BBP formula. In particular, given $b \in \mathbb{N}$, $b>2$, $b$ not a proper power, a $b$-ary Machin-type BBP arctangent formula for $\kappa$ is a formula of the form $\kappa = \sum_{m} a_m \arctan(-b^{-m})$, $a_m \in \mathbb{Q}$, while when $b=2$, we also allow terms of the form $a_m \arctan(1/(1-2^m))$. Of particular interest, we show that $\pi$ has no Machin-type BBP arctangent formula when $b \neq 2$. To the best of our knowledge, when there is no Machin-type BBP formula for a constant then no BBP formula of any form is known for that constant. Keywords:BBP formulae, Machin-type formulae, arctangents,, logarithms, normality, Mersenne primes, Bang's theorem,, Zsigmondy's theorem, primitive prime factors, $p$-adic analysisCategories:11Y99, 11A51, 11Y50, 11K36, 33B10
10. CJM 2003 (vol 55 pp. 711)
Broughan, Kevin A.
Adic Topologies for the Rational Integers A topology on $\mathbb{Z}$, which gives a nice proof that the set of prime integers is infinite, is characterised and examined. It is found to be homeomorphic to $\mathbb{Q}$, with a compact completion homeomorphic to the Cantor set. It has a natural place in a family of topologies on $\mathbb{Z}$, which includes the $p$-adics, and one in which the set of rational primes $\mathbb{P}$ is dense. Examples from number theory are given, including the primes and squares, Fermat numbers, Fibonacci numbers and $k$-free numbers. Keywords:$p$-adic, metrizable, quasi-valuation, topological ring,, completion, inverse limit, diophantine equation, prime integers,, Fermat numbers, Fibonacci numbersCategories:11B05, 11B25, 11B50, 13J10, 13B35
top of page | contact us | privacy | site map | | {"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": 0.9695264101028442, "perplexity": 1424.670630990974}, "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/1531676591543.63/warc/CC-MAIN-20180720061052-20180720081052-00550.warc.gz"} |
https://www.physicsforums.com/threads/older-nuclear-structure-models.748060/ | # Older nuclear structure models.
1. Apr 10, 2014
### carllacan
Hi people.
and I can't get my head around it.
How is the Uncertainty Relation uncompatible with the idea of a proton-electron nucleus?
And can you explain why this model predicted those values for the nuclear spin?
Thank you.
2. Apr 10, 2014
### Bill_K
In this model the deuteron would consist of two protons and one electron, i.e. three spin-1/2 particles. To get the resulting overall spin, you would have to combine three spin-1/2's with whatever orbital angular momentum there was. However you do it, the total spin comes out half-integer. | {"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": 0.9262887239456177, "perplexity": 1707.3121482466622}, "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/1521257647649.70/warc/CC-MAIN-20180321121805-20180321141805-00070.warc.gz"} |
https://www.physicsforums.com/threads/proof-by-contradiction.161973/ | • Start date
• #1
2,255
1
How widely used is this method? Is it the most popular proof technique? All it takes it to find a condradiction or counter example in order to prove something which is often much simpler than listing all the cases (not to mention if there are infinte of them to start with) or directly proving something.
i.e To prove A=>B
Proof by contradiction works by showing ~B=>~A
So if A is true than B must be true which is what we set out to prove.
When doing undergrad abstract algebra, I am finding that I am using this method a lot. Is this good? Wasn't this Godfrey Hardy's favouriate proof technique?
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• #2
123
0
counter examples and proof by contradiction are not the same thing.
You can make an conjecture and use a counter example to show it is false.
conjecture
Two odd whole numbers add to give an odd whole number
counter example 5 + 3 = 8 which is even.
This does not prove that Two odd whole numbers add to give an even whole number for all whole numbers
showing ~B=>~A to prove A=>B is not a proof by contradiction and is often a more complicated proof.
A n, m are two odd whole numbers or two even whole numbers
B n+m is even
~A n, m are two whole numbers one or which is even
~B n+m is odd
n + m is odd
therefore n+m=2p+1 for some whole number p
therefore n=2p-m+1
m is either odd or even
m odd m=2q+1 for some whole number q
since n>0 and m>0 then n+m>m therefore 2p+1>2q+1 therefore p>q therefore p-q>0
n=2p-2q-1+1 = 2(p-q) therefore n is even
m even m=2q for some whole number q
n=2p-2q+1 = 2(p-q)+1 n is odd
so if m is odd n is even and if m is even n is odd and by symmetry
if n is odd m is even and if n is even m is odd.
hence ~B=>~A and so A=>B not a contradiction in sight
A n, m are two odd whole numbers
B n+m is even
~B n+m is odd
Assume A=> ~B
m is odd therefore m=2p+1
n is odd therefore n=2q+1
n+m is odd therefore n+m = 2r+1
then 2p+1 + 2q +1 = 2r+1
if and only if 2(p+q)+1 = 2r
if and only if 2(p+q-r)=1
if and only if p+q+r=1/2 contradicting p,q,r are whole numbers
and so our assumption is wrong and A=>B
Of course in this case the best way is to prove A=>B directly
Where you might be getting confused about counter example and proof by contradiction is the $$\sqrt{2}$$ is irrational proof.
A n is a positive whole number
B $$\sqrt{n}$$ is rational
you disprove A=>B using a counter example by showing $$\sqrt{2}$$ is irrational
you show $$\sqrt{2}$$ is irrational by using proof by contradition
start with the opposite assumption that $$\sqrt{2}$$ is rational and arrive at a contradiction.
• #3
370
0
~B=>~A is called the contrapostive, which is equivalent to A=>B.
Here's the first proof that popped into my head for proof by contradiction:
If every sequence x_n that converges to A has the property that f(x_n) converges to L then the limit x->A L.
You can't prove this by direct proof because there are soooooooooo many sequences converging to a. So a proof by contradiction is used.
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• #4
90
0
In ~B -> ~A, you've given the form for a contrapositive proof.
For an assertion of the form, A -> B, a proof by contradition would look like:
(A and ~B) -> some contradiction, e.g., ~A or B or perhaps you can pull an R out of your hat and arrive at (R and ~R).
Sometimes a proof is started as a proof by contradiction,
and in the end you arrive at say ~A (an apparent "contradiction").
But, if in the course of the proof A is never explicitly used, then the proof is actually a proof by contraposition.
You might keep in mind that virtually all assertions come with some form of quantification. Often times left implicit, but there never the less. To say that this can complicate things is an understatement.
PS.
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• #5
uart
2,776
9
More generally proof by contradiction doesn't need to be set out in the form of "A implies B" etc, but rather it just requires that we start with a conjecture "C" that we wish to prove and then show that if C is false then that leads to a contradiction.
If the conjecture "C" that we begin with is that "A implies B" then assuming C to be false is equivalent to assumming that "A does not imply B" rather than assumming that "A implies Not_B".
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• #6
123
0
fopc - not sure what you mean by the PS.
The first sentence of what?
The link is not mine in the sense that I wrote the page to which the link goes I just found it by searching for proof by contradiction.
• #7
HallsofIvy
Homework Helper
41,833
963
"Proof of the contrapositive" is a special case of "proof by contradiction"- you assume the conclusion is true and then prove that the hypothesis is false- that's the contradiction.
But proof by contradiction is more general- you assume the conclusion is false and use that along with the hypotheses to prove "statement A", which may have no obvious relation to the hypotheses- then turn around and prove "statement B" which also may have not obvious relation to the hypotheses, but contradicts "statement A".
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• #8
123
0
uart - take
A n=2
B $$\sqrt{n}$$ is irrational
conjecture C A=>B assuming C to be false is equivalent to assuming that A does not imply B, ie n=2 does not imply $$\sqrt{n}$$ is irrational does not get anywhere as it gives no indication as to what n=2 could imply. n=2 does not imply $$\sqrt{n}$$ is irrational does not imply that n=2 implies that $$\sqrt{n}$$ is rational
But I am also incorrect in saying A=>B is false is equivalent to A=>~B
revising my boolean logic
A=>B is equivalent to ~A V B (V OR) ~ NOT
so ~(A=>B) is equivalent to ~(~A V B) is equivalent to A & ~B (& AND)
so assuming C (above) to be false gives
n=2 AND $$\sqrt{n}$$ is rational (given that all real numbers are either irrational or rational) and hence the proof by contradiction can proceed.
Sorry for the basic error but it is a while since I did predicate calculus
• #9
90
0
"The first sentence of what?"
I don't think there should be any confusion about what sentence I was referring to. But OK, I was referring to the first sentence on the page of the link that you provided.
"The link is not mine in the sense that I wrote the page to which the link goes I just found it by searching for proof by contradiction."
I didn't say you wrote the page. My reference to "your link" only meant the link that you provided to this thread. In that sense you are responsible for the link; but it doesn't mean that you wrote it.
By the way, your words, "is worth a look", strongly suggests (to me at least) that you actually bothered to review the page. Otherwise, why provide the link and say its "worth a look", if you haven't "looked at it"
yourself? My suggestion that you read (again?) the first sentence was to remind you that your statement, "to prove A=>B start with the opposite conjecture A=>~B and arrive at a contradition", has some serious problems. I meant no offense by it.
Anyway, have a nice day.
On another issue:
1.) A->B <->
2.) ~B -> ~A <->
3.) (A and ~B) -> any contradiction.
If you choose to prove 1 directly, obviously there is no contradiction.
If you choose to prove 2 directly, there is no contradiction (this is considered a direct contrapositive proof).
You're simply trying to establish that if B is false, then A is necessarily false.
In proving 3 directly, obviously there will be a contradition (this is considered a proof by contradiction).
If you choose to prove 2 by contradiction, then you are proving (~B and A) -> any contradiction. But as you can see, this is just 3, and clearly 3 can be derived from either 1 or 2. So the status of 3 does not have to have anything whatsoever to do with the contapositive.
Contrapositive proof does not involve producing a contradiction.
Of course, if you say proving 2 by contradiction amounts to a proof by contradiction of the contrapositive, then I think terminology becomes clouded. But I would argue that proving 2 by contradiction is just proving
1 by contradiciton, and now the word contrapositive no longer enters into the discussion.
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• #10
123
0
fopc - No offense taken at all. I genuinely did not follow what you were saying.
I took "The first sentence will suffice" to mean "Nothing after the first sentence is required" rather than "It is sufficient to read just the first sentence of the link to see that the statement ......had serious errors
I did realise my mistake after further thought, and revision reading as you will see by my later post
I was trying to clear up some confusion in the OP about counter examples, proof by contradiction and the way it works and unfortunately not quite succeeded. However if pivoxa15 now reads the whole thread (s)he should be able to sort out the differences between counter examples, direct proof, contrapositive proofs and proof by contradiction.
Unfortunately in sorting this out nobody has answered the original questions directly.
• #11
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I think I see what the general version of proof by contradiction is.
To prove A=>B do
~B=>C where C is wrong when assuming A. The only way for ~B=>C to be correct is for ~B to be false so B is correct (when we assumed A in the statement ~B=>C). Hence when assuming A to be correct, we proved that B must be correct. Proof complete.
C could be ~A hence a special case as you people pointed out.
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That looks kind of tough to follow. It works (ie I've seen it in my abstract algebra textbook), but you don't actually have to start off with a statement like A=>B. You can use it to prove any general kind of statement A simply by finding a contradiction when assuming ~A.
At its absolute most basic, it will look just like HallsofIvy and others have described.
HallsofIvy said:
But proof by contradiction is more general- you assume the conclusion is false and use that along with the hypotheses to prove "statement A", which may have no obvious relation to the hypotheses- then turn around and prove "statement B" which also may have not obvious relation to the hypotheses, but contradicts "statement A".
Here's a step by step summary. I want to prove some statement A.
I assume ~A.
I find some statement B which is implied by ~A and any other hypotheses.
I also find some statment ~B which is implied by ~A and any other hypotheses.
I now have a contradiction (B and ~B).
At this point we discharge the assumption of ~A. Because I have a contradiction, I can apply negation introduction to get ~~A. And by negation elimination I have A. Proof complete.
As far as taking undergrad math courses, I've found that I use mathematical induction and direct proof the most. Proof by contradiction popped up every once in a while when I wanted to prove something existed but there was no obvious way to do it with the direct proof method.
Proof by contradiction is a staple in most philosophy courses though. They like to call it reductio ad absurdum.
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When doing algebra with rings, I found that if I don't see a direct proof immediately, I try proof by contradiction and it tends to fall out. Is this method the most popular in mathematics? And especially in algebra?
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752 | {"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": 0.8597918748855591, "perplexity": 936.0870387499034}, "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-2021-17/segments/1618039604430.92/warc/CC-MAIN-20210422191215-20210422221215-00268.warc.gz"} |
http://mathhelpforum.com/calculus/47215-limits.html | # Math Help - limits
1. ## limits
could someone please explain to me how to do these?
3x^2 + 2x + 1
limx --------------
x^2 + x - 3
x^2 + 1
limx ---------
x^2 - 1
thank you
2. Originally Posted by hollies_yardbirds_kinks
could someone please explain to me how to do these?
3x^2 + 2x + 1
limx --------------
x^2 + x - 3
x^2 + 1
limx ---------
x^2 - 1
thank you
where do we approach x?
3. Hello !
Originally Posted by hollies_yardbirds_kinks
could someone please explain to me how to do these?
3x^2 + 2x + 1
limx --------------
x^2 + x - 3
x^2 + 1
limx ---------
x^2 - 1
thank you
When you have a division of polynomials (and limit to infinity), there are a few possible ways to solve the limit :
- l'hospital's rule
- find the greatest power of x in the numerator and the denominator and factor in the numerator and denominator. For example, if you have x²+1, factor by x² : $x^2+1=x^2 \left(1+\frac 1{x^2}\right)$. $x^2-1=x^2 \left(1-\frac 1{x^2}\right)$
So you can simplify by x². What's left is in general a finite limit
- know some rules : if the degree of the polynomial in the numerator is greater than the one of the denominator, the limit is $\infty$ (the sign depends on the leading coefficient)
if the degree of the polynomial in the numerator is smaller than the one of the denominator, the limit is 0.
if they're the same, the limit is the quotient of the leading coefficients.
These rules derive from the point just above
Sidenote : Leading coefficient : Definition | {"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": 3, "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": 0.9618507027626038, "perplexity": 679.7905525890924}, "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-2014-52/segments/1418802768982.112/warc/CC-MAIN-20141217075248-00071-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/404035/does-every-krull-ring-have-a-height-1-prime-ideal | # Does every Krull ring have a height 1 prime ideal?
Let $A$ be a Krull ring. According to Theorem 12.3 in Matsumura's Commutative Ring Theory, the family of localizations of $A$ at height 1 prime ideals of $A$ forms a defining family of $A$.
Question: Why such family exists? In other words, why does a Krull ring have at least one height 1 prime ideal?
Remark: By definition a Krull ring is the intersection of DVRs and each DVR has dimension 1, hence its maximal ideal has height 1. However, if we contract this maximal ideal to $A$, it is not necessary that the prime ideal we get will have height 1.
Edit: I realized that the definition of a Krull ring given in Wikipedia is quite different from the one given in Matsumura. In fact, the Wikipedia definition trivially answers my question. Matsumura's definition is: an integral domain is called Krull if it is the intersection of a family of DVRs and every non-zero element in the domain is non-zero in only a finite number of corresponding discrete valuations. How to obtain that such a ring contains a height 1 ideal is not obvious to me.
• @2015 The OP said "non-zero in only a finite number of corresponding discrete valuations" which is correct. You think in terms of DVRs, which is also correct, but please let the OP to state his question as they wish. – user26857 Sep 19 '17 at 9:00
Let $R=\cap_{\lambda\in\Lambda}R_{\lambda}$ with $R_{\lambda}$ DVRs (as in Matsumura's definition of Krull domains). Assume that the intersection is irredundant, that is, if $\Lambda'\subsetneq\Lambda$ then $\cap_{\lambda\in\Lambda}R_{\lambda}\subsetneq\cap_{\lambda'\in\Lambda}R_{\lambda'}$.
Let's prove that $m_{\lambda}\cap R$ is a prime ideal of height one, for all $\lambda\in\Lambda$. First note that $m_{\lambda}\cap R\neq (0)$. If the height of some $m_{\alpha}\cap R$ is at least $2$, then there exists a nonzero prime $p\subsetneq m_{\alpha}\cap R$. From Kaplansky, Commutative Rings, Theorem 110, there exists $m_{\alpha'}\cap R\subseteq p$ (obviously $\alpha'\neq\alpha$). Let $x\in\cap_{\lambda\ne\alpha}R_{\lambda}$, $x\notin R$ (so $x\notin R_{\alpha}$), and $y\in m_{\alpha'}\cap R$, $y\neq 0$. One can choose $m,n$ positive integers such that $z=x^my^n$ is a unit in $R_{\alpha}$. Since $x\in R_{\alpha'}$ and $y\in m_{\alpha'}\cap R$ we get $z\in m_{\alpha'}$. Obviously $z\in R_{\lambda}$ for all $\lambda\ne \alpha, \alpha'$, so $z\in R$, $z$ is invertible in $R_{\alpha}$ and not invertible in $R_{\alpha'}$, a contradiction with $m_{\alpha'}\cap R\subseteq m_{\alpha}\cap R$.
(This argument is adapted from Kaplansky's proof of Theorem 114. Furthermore, using again Theorem 110 one can see that $m_{\lambda}\cap R$ are the only height one prime ideals of $R$.)
• How can we make a given intersection irredundant (in the above sense) ? Is it always possible? – user276115 Sep 26 '17 at 8:04
Any finite-dimensional ring (indeed, any ring with a prime of finite, nonzero height) has prime ideals of height 1, basically by definition of the dimension: if $P$ has finite height $n$, there's a maximal chain $P_0 \subseteq P_1 \subseteq \dotsb \subseteq P_n = P$, and $P_1$ can't contain any more than one prime ideal, since otherwise we could make a longer chain. Indeed, so does any Noetherian ring, since it satisfies the descending chain condition on prime ideals.
If you're really interested in the non-Noetherian case, things get confusing -- for example, there are domains where every nonzero prime has infinite height. If $R$ has this property, though, then all localizations of $R$ have this property (the primes of any localization of $R$ corresponding to a downward-closed subset of the primes of $R$). In particular, no localization of $R$ can be a DVR, since DVRs have finite dimension, so $R$ can't be a Krull ring, even by Matsumura's definition. This should prove the equivalence of the two definitions, too.
• Thanks for your answer +1. A first question: lets work with Matsumura's definition. Thus $A=\cap_{\lambda \in \Lambda} R_{\lambda}$, where $R_{\lambda}$ is a DVR of the field of fractions $K$ of $A$. Why must we have that some localization of $A$ is DVR? – Manos May 30 '13 at 20:55
• Each $R_\lambda$ is contained in $K$ and contains $A$, so it's a localization of $A$ -- not necessarily at a prime ideal, but just in the sense of inverting some elements of $A$. For any multiplicative set $S$, the primes of $S^{-1}A$ correspond to the primes of $A$ that don't meet $S$. – Paul VanKoughnett May 30 '13 at 22:38
The short version is: $R$ is a Krull ring if and only if $\mathrm{Div}(R)$ (= the semigroup of divisorial fractional ideals, its neutral element is the submodule $R \subseteq Q(R)$) is (as a partially ordered group) isomorphic via some iso $\phi$ to a direct sum $\bigoplus_{i \in I}{\mathbb{Z}}$ equipped with the canonical partial order. If $R$ is a Krull ring but not a field, then $\mathrm{Div}(R) \neq 0$ and the canonical basis vectors $e_i, i \in I$ form a positive basis (Here is the existence statement!). Their preimages $\mathfrak{p}_i , i \in I$ under $\phi$ form a positive basis of $\mathrm{Div}(R)$ and they are in fact precisely the prime ideals of height one in $R$. | {"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": 0.9783062934875488, "perplexity": 133.39141842637494}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525312.3/warc/CC-MAIN-20190717141631-20190717163631-00098.warc.gz"} |
http://mathoverflow.net/questions/91944/flattening-a-corner-in-a-convex-d-polytope-into-d-1-dimensions-without-ove/91946 | # Flattening a corner in a convex $d$-polytope (into $d-1$ dimensions, without overlap)?
I'm interested in the following question, which seems to be assumed all over the place (at least for 3 dimensions) in convex geometry, and which I cannot find a proof of.
Suppose we have a corner of a convex polytope in $\mathbb{R}^d$. How do we show that we can 'flatten' the surrounding facets into $d-1$ dimensions without overlap?
What do I mean by that? Well, for the three-dimensional case, it means that if you sum the angles around a given vertex of a convex polyhedron, you get a sum of less than $2 \pi$. Here's another way to say that:
(0) Given a collection of 2-dim cones $C_i$ with angles $a_i$, if the sum of the $a_i$ is greater than $2 \pi$, then $C_i$ cannot be facets of a 3-dim cone.
(I'm defining a cone to be the convex hull of a collection of rays; so, cones are assumed to be convex.)
And in general dimensions:
(1) Given a collection of $(d-1)$-dim cones $C_i$ with total $(d-1)$-angle measures $a_i$, if the sum of the $a_i$ is greater than the total angle surrounding a point in $\mathbb{R}^{(d-1)}$, then the $C_i$ cannot be facets of a d-dim cone.
This fact can be restated in a lot of other ways. Maybe one of these is easier to prove?
(2) If a collection of $(d-1)$-dim cones $C_i$ with total $(d-1)$-angle sum greater than the total angle surrounding a point in $\mathbb{R}^{(d-1)}$ is configured in $\mathbb{R}^d$ with all cone points set at the origin and every $(d-2)$-face identified with a $(d-2)$-face of some other cone (i.e., the cones are glued to make a simplicial complex), and if this configuration lies on one side of a hyperplane, then the configuration is not convex.
(3) The facets of any d-cone can be isometrically mapped (unfolded!) into a $(d-1)$-hyperplane, retaining the coincidence of the cone point and without overlap.
(4) The convex spherical polygon of largest perimeter is a great circle, i.e. any spherical polygon with perimeter larger than $2 \pi$ is not convex.
It seems to me like there should be a straightforward, convex-geometry proof of this fact, but I can't find it. If you know another way to prove it (say, using ideas from curvature?) I'd be very interested in that too!
-
I suspect [Miller-Pak 2003] should be relevant: math.ucla.edu/~pak/papers/FoldLI.pdf – Allen Knutson Mar 22 '12 at 23:55
Allen -- thanks! This paper, in fact, proves much stronger results than the one I'm asking about. On the other hand, the tools it invokes go well beyond convex geometry, so I'm still curious about whether there's a simple proof of this fact. – Emily Peters Mar 23 '12 at 15:32
One way of doing this (in $d=3,$ say) requires the "Archimedes axiom": if one convex body $K$ contains another body $L,$ then the perimeter of $K$ is greater than that of $L$ (the nicest proof uses Crofton's formula, which says that the perimeter is proportional to the measure of the set of lines which intersect the set). Then, you set $K$ to the hemisphere, and $L$ to your spherical convex polygon, and you are done. (Archimedes actually introduced this as an axiom in the Euclidean case when computing the perimeter of the circle, since he needed to know that the inscribed polygons provided a lower bound). | {"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": 0.8029806017875671, "perplexity": 294.46420662443865}, "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/1448398447758.91/warc/CC-MAIN-20151124205407-00032-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://www.contrib.andrew.cmu.edu/~ryanod/index.php?p=439&replytocom=2160 | # §2.5: Highlight: Arrow’s Theorem
When there are just $2$ candidates, the majority function possesses all of the mathematical properties that seem desirable in a voting rule (e.g., May’s Theorem and Theorem 32). Unfortunately, as soon as there are $3$ (or more) candidates the problem of social choice becomes much more difficult. For example, suppose we have candidates $a$, $b$, and $c$, and each of $n$ voters has a ranking of them. How should we aggregate these preferences to produce a winning candidate?
In his 1785 Essay on the Application of Analysis to the Probability of Majority Decisions [dC85], Condorcet suggested using the voters’ preferences to conduct the three possible pairwise elections, $a$ vs. $b$, $b$ vs. $c$, and $c$ vs. $a$. This calls for the use of a $2$-candidate voting rule $f : \{-1,1\}^n \to \{-1,1\}$; Condorcet suggested $f = \mathrm{Maj}_n$ but we might consider any such rule. Thus a “$3$-candidate Condorcet election” using $f$ is conducted as follows:
election \ voter #1 #2 #3 societal
aggregation
$a$ ($+1$) vs. $b$ ($-1$) $+1$ $+1$ $-1$ $= x$ $f(x)$
$b$ ($+1$) vs. $c$ ($-1$) $+1$ $-1$ $+1$ $= y$ $f(y)$
$c$ ($+1$) vs. $a$ ($-1$) $-1$ $-1$ $+1$ $= z$ $f(z)$
In the above example, voter $\#1$ ranked the candidates $a > b > c$, voter $\#2$ ranked them $a > c > b$, voter $\#3$ ranked them $b > c > a$, etc. Note that the $i$th voter has one of $3! = 6$ possible rankings, and these translate into a triple of bits $(x_i, y_i, z_i)$ from the following set: $\Bigl\{(+1, +1, -1), (+1, -1, -1), (-1,+1,-1), (-1,+1,+1), (+1,-1,+1), (-1, -1, +1)\Bigr\}.$ These are precisely the triples satisfying the not-all-equal predicate $\mathrm{NAE}_3$.
In the example above, if $n = 3$ and $f = \mathrm{Maj}_3$ then the societal outcome would be $(+1,+1,-1)$, meaning that society elects $a$ over $b$, $b$ over $c$, and $a$ over $c$. In this case it is only natural to declare $a$ the overall winner.
Definition 54 In an election employing Condorcet’s method with $f : \{-1,1\}^n \to \{-1,1\}$, we say that a candidate is a Condorcet winner if it wins all of the pairwise elections in which it participates.
Unfortunately, as Condorcet himself noted, there may not be a Condorcet winner. In the example above, if voter $1$’s ranking was instead $c > a > b$ (corresponding to $(+1, -1, +1)$), we would obtain the “paradoxical” outcome $(+1, +1, +1)$: society prefers $a$ over $b$, $b$ over $c$, and $c$ over $a$! This lack of a Condorcet winner is termed Condorcet’s paradox; it occurs when the outcome $(f(x), f(y), f(z))$ is one of the two “all-equal” triples $\{(-1, -1, -1), (+1,+1,+1)\}$. One might wonder if the Condorcet paradox can be avoided by using a voting rule $f : \{-1,1\}^n \to \{-1,1\}$ other than majority. However in 1950 Arrow [Arr50] famously showed that the only means of avoidance is an unappealing one:
Arrow’s Theorem Suppose $f : \{-1,1\}^n \to \{-1,1\}$ is a unanimous voting rule used in a $3$-candidate Condorcet election. If there is always a Condorcet winner, then $f$ must be a dictatorship.
(In fact, Arrow’s Theorem is slightly stronger than this, as will be discussed in an exercise.)
To prove Arrow’s Theorem, let’s take our cue from the title of Condorcet’s work and compute the probability of a Condorcet winner. We will do this under the “impartial culture assumption” for $3$-candidate elections: each voter independently chooses one of the $6$ possible rankings uniformly at random.
Theorem 55 Consider a $3$-candidate Condorcet election using $f : \{-1,1\}^n \to \{-1,1\}$. Under the impartial culture assumption, the probability of a Condorcet winner is precisely $\tfrac34 – \tfrac34 \mathbf{Stab}_{-1/3}[f]$.
Proof: Let ${\boldsymbol{x}}, \boldsymbol{y}, \boldsymbol{z} \in \{-1,1\}^n$ be the votes for the elections $a$ vs. $b$, $b$ vs. $c$, and $c$ vs. $a$, respectively. Under impartial culture, the bit triples $({\boldsymbol{x}}_i, \boldsymbol{y}_i, \boldsymbol{z}_i)$ are independent and each is drawn uniformly from the $6$ triples satisfying the not-all-equal predicate $\mathrm{NAE}_3 : \{-1,1\}^3 \to \{0,1\}$. There is a Condorcet winner if and only if $\mathrm{NAE}_3(f({\boldsymbol{x}}),f(\boldsymbol{y}),f(\boldsymbol{z})) = 1$. Hence $$\label{eqn:prob-cond} \mathop{\bf Pr}[\exists \text{ Condorcet winner}] = \mathop{\bf E}[\mathrm{NAE}_3(f({\boldsymbol{x}}),f(\boldsymbol{y}),f(\boldsymbol{z}))].$$ The multilinear (Fourier) expansion of $\mathrm{NAE}_3$ is $\mathrm{NAE}_3(w_1, w_2, w_3) = \tfrac34 - \tfrac14w_1 w_2 -\tfrac14 w_1 w_3 - \tfrac14 w_2w_3;$ thus $\eqref{eqn:prob-cond} = \tfrac34 - \tfrac14\mathop{\bf E}[f({\boldsymbol{x}})f(\boldsymbol{y})] – \tfrac14\mathop{\bf E}[f({\boldsymbol{x}})f(\boldsymbol{z})]- \tfrac14\mathop{\bf E}[f(\boldsymbol{y})f(\boldsymbol{z})].$ In the joint distribution of ${\boldsymbol{x}}, \boldsymbol{y}$ the $n$ bit pairs $({\boldsymbol{x}}_i, \boldsymbol{y}_i)$ are independent. Further, by inspection we see that $\mathop{\bf E}[{\boldsymbol{x}}_i] = \mathop{\bf E}[{\boldsymbol{y}}_i] = 0$ and that $\mathop{\bf E}[{\boldsymbol{x}}_i \boldsymbol{y}_i] = (2/6)(+1) + (4/6)(-1) = -1/3$. Hence $\mathop{\bf E}[f({\boldsymbol{x}})f(\boldsymbol{y})]$ is precisely $\mathbf{Stab}_{-1/3}[f]$. Similarly $\mathop{\bf E}[f({\boldsymbol{x}})f(\boldsymbol{z})] = \mathop{\bf E}[f(\boldsymbol{y})f(\boldsymbol{z})] = \mathbf{Stab}_{-1/3}[f]$ and the proof is complete. $\Box$
Arrow’s Theorem is now an easy corollary:
Proof of Arrow’s Theorem: By assumption, the probability of a Condorcet winner is $1$; hence $1 = \tfrac34 - \tfrac34 \mathbf{Stab}_{-1/3}[f] = \frac34 – \frac34 \sum_{k=0}^n(-1/3)^{k} \mathbf{W}^{k}[f].$ Since $(-1/3)^k \geq -1/3$ for all $k$, the equality above can only occur if all of $f$’s Fourier weight is on degree $1$; i.e., $\mathbf{W}^{1}[f] = 1$. By Exercise 1.19(a) this implies that $f$ is either a dictator or a negated-dictator. Since $f$ is unanimous, it must in fact be a dictator. $\Box$
An advantage of this analytic proof of Arrow’s Theorem is that we can deduce several more interesting results about the probability of a Condorcet winner. For example, combining Theorem 55 with Theorem 44 we get Guilbaud’s formula:
Guilbaud’s Formula In a $3$-candidate Condorcet election using $\mathrm{Maj}_n$, the probability of a Condorcet winner tends to $\tfrac34 - \tfrac{3}{2\pi} \arcsin(-1/3) = \tfrac{3}{\pi}\arccos(1/\sqrt{3}) \approx 91.2\%.$ as $n \to \infty$.
This is already a fairly high probability. Unfortunately, if we want to improve on it while still using a reasonably fair election scheme, we can only set our hopes higher by a sliver:
Theorem 56 In a $3$-candidate Condorcet election using an $f : \{-1,1\}^n \to \{-1,1\}$ with all $\widehat{f}(i)$ equal, the probability of a Condorcet winner is at most $\frac79 + \frac{4}{9\pi} + o_n(1) \approx 91.9\%$.
The condition in Theorem 56 seems like it would be satisfied by most reasonably fair voting rules $f : \{-1,1\}^n \to \{-1,1\}$ (e.g., it is satisfied if $f$ is transitive-symmetric or is monotone with all influences equal). In fact, we will show that Theorem 56‘s hypothesis can be relaxed in Chapter 5; we will further show in Chapter 13 that $\frac79 + \frac{4}{9\pi}$ can be improved to the tight value $\tfrac34 – \tfrac{3}{2\pi} \arcsin(-1/3)$ of majority. To return to Theorem 56, it is an immediate consequence of the following two results, the first being an exercise (based on Theorem 32) and the second being an easy corollary of Theorem 55.
Proposition 57 Suppose $f : \{-1,1\}^n \to \{-1,1\}$ has all $\widehat{f}(i)$ equal. Then $\mathbf{W}^{1}[f] \leq 2/\pi + o_n(1)$.
Corollary 58 In a $3$-candidate Condorcet election using $f : \{-1,1\}^n \to \{-1,1\}$, the probability of a Condorcet winner is at most $\frac79 + \frac29 \mathbf{W}^{1}[f]$.
Proof: From Theorem 55, the probability is \begin{align*} \tfrac34 – \tfrac34 \mathbf{Stab}_{-1/3}[f] &= \tfrac34 – \tfrac34(\mathbf{W}^{0}[f] – \tfrac13 \mathbf{W}^{1}[f] + \tfrac19 \mathbf{W}^{2}[f] – \tfrac{1}{27} \mathbf{W}^{3}[f] + \cdots)\\ &\leq \tfrac34 + \tfrac14 \mathbf{W}^{1}[f] + \tfrac{1}{36} \mathbf{W}^{3}[f] + \tfrac{1}{324} \mathbf{W}^{5}[f] + \cdots\\ &\leq \tfrac34 + \tfrac14 \mathbf{W}^{1}[f] + \tfrac{1}{36} (\mathbf{W}^{3}[f] + \mathbf{W}^{5}[f] + \cdots)\\ &\leq \tfrac34 + \tfrac14 \mathbf{W}^{1}[f] + \tfrac{1}{36} (1 – \mathbf{W}^{1}[f]) \quad=\quad \tfrac79 + \tfrac29\mathbf{W}^{1}[f]. \quad \Box \end{align*}
Finally, using Corollary 58 we can prove a “robust” version of Arrow’s Theorem, showing that a Condorcet election is almost paradox-free only if it is almost a dictatorship.
Corollary 59 Suppose that in a $3$-candidate Condorcet election using $f : \{-1,1\}^n \to \{-1,1\}$, the probability of a Condorcet winner is $1 – \epsilon$. Then $f$ is $O(\epsilon)$-close to $\pm \chi_i$ for some $i \in [n]$.
Proof: From Corollary 58 we obtain that $\mathbf{W}^{1}[f] \geq 1 – \tfrac92 \epsilon$. The conclusion now follows from the FKN Theorem. $\Box$
Friedgut–Kalai–Naor (FKN) Theorem Suppose $f : \{-1,1\}^n \to \{-1,1\}$ has $\mathbf{W}^{1}[f] \geq 1-\delta$. Then $f$ is $O(\delta)$-close to $\pm \chi_i$ for some $i \in [n]$.
We will see the proof of the FKN Theorem in later chapter. We’ll also show in a later chapter that the $O(\delta)$ closeness can be improved to $\delta/4 + O(\delta^2 \log(2/\delta))$.
### 9 comments to §2.5: Highlight: Arrow’s Theorem
• Awesome! Here is a typo I found: “In the joint distribution of x,y the n bits pairs (xi,yi) are independent.”
• Thanks!
• I have one other question. Taking elections that respect IIA and then viewing them as boolean functions in this way seems to be a great insight. Are there versions where we quantitatively relax IIA? (A survey-esque post of Nisan in the same vein also seemed to mention only strict IIA.)
• Hi David — not that I personally know of, but I wouldn’t doubt it; there’s tons of literature on this with which I’m not so familiar.
• just after definition 54 you say “if voter 2’s ranking was instead b>c>a (corresponding to (−1,+1,+1)), we would obtain the “paradoxical” outcome (−1,−1,−1)”. It seems wrong. Would it be right to say “if voter 1’s ranking was instead b>a>c (corresponding to (−1,+1,-1)), we would obtain the “paradoxical” outcome (−1,−1,−1)”?
At the next line there is also a typo: “two “all-equal” triples {(−1,−1,1),(+1,+1,+1)}” instead of “two “all-equal” triples {(−1,−1,-1),(+1,+1,+1)}”
• Fixed both, thanks!
• John Engbers
Just after definition 54, you wrote:
we would obtain the “paradoxical” outcome (−1,−1,−1): society prefers a over b, b over c, and c over a
This should be:
…society prefers b over a, c over b, and a over c
• John Engbers
In the first line to the proof of Theorem 2.56, there should be a space between vs. and $c$ in `$b$ vs.$c$’
• Thanks! I think I fixed these. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9722561836242676, "perplexity": 784.9352889794686}, "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/1518891814566.44/warc/CC-MAIN-20180223094934-20180223114934-00182.warc.gz"} |
http://mathhelpforum.com/geometry/174353-question-coordinate-geometry-circles.html | # Thread: Question on coordinate geometry of circles
1. ## Question on coordinate geometry of circles
Hi, I'm currently doing AS-level Maths in the UK, and I'm struggling with this question on coordinate geometry of circles:
A circle passes through the points (0, -3) and (0, 3) and has radius 5. Draw a sketch to show the two possible positions of the circles and find their equations.
Any help would be greatly appreciated!
2. Originally Posted by sakuraxkisu
Hi, I'm currently doing AS-level Maths in the UK, and I'm struggling with this question on coordinate geometry of circles:
A circle passes through the points (0, -3) and (0, 3) and has radius 5. Draw a sketch to show the two possible positions of the circles and find their equations.
Any help would be greatly appreciated!
Things we know. The equation of a circle is given by the formula
$(x-h)^2+(y-k)^2=r^2$ Since we know r we get
$(x-h)^2+(y-k)^2=25$
Since we know two points that lie on the circle we can plug them in for (x,y) and this will give a system of equations in two unknowns that you can solve for h and
k.
Can you finish from here?
3. Thank you! I was on the right tracks, with the substitution of (0, -3) and (0, 3) into the equation, but thank you for clarifying | {"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": 2, "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": 0.9089049696922302, "perplexity": 195.36136530814872}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541692.55/warc/CC-MAIN-20161202170901-00161-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://www.cs.mcgill.ca/~rwest/link-suggestion/wpcd_2008-09_augmented/wp/m/Mechanical_work.htm | # Mechanical work
In physics, mechanical work is the amount of energy transferred by a force. Like energy, it is a scalar quantity, with SI units of joules. The term work was first coined in the 1830s by the French mathematician Gaspard-Gustave Coriolis.
According to the work-energy theorem if an external force acts upon an object, causing its kinetic energy to change from Ek1 to Ek2, then the mechanical work (W) is given by:
$W = \Delta E_k = E_{k2} - E_{k1} = \frac{1}{2}m \Delta (v^2) \,\!$
where m is the mass of the object and v is the object's speed.
The mechanical work applied to an object can be calculated from the scalar multiplication of the applied force (F) and the displacement (d) of the object. This is given by:
$W = \mathbf{F}\cdot \mathbf{d}=Fd \cos\theta \,\!$
## Introduction
A baseball pitcher does positive work on the ball by transferring energy into it. The catcher does negative work on it.
Work can be zero even when there is a force. The centripetal force in circular motion, for example, does zero work because the kinetic energy of the moving object doesn't change. Likewise, when a book sits on a table, the table does no work on the book, because no energy is transferred into or out of the book.
Heat conduction is not considered to be a form of work, since there is no macroscopically measurable force, only microscopic forces occurring in atomic collisions.
## Units
The SI unit of work is the joule (J), which is defined as the work done by a force of one newton acting over a distance of one meter. This definition is based on Sadi Carnot's 1824 definition of work as "weight lifted through a height", which is based on the fact that early steam engines were principally used to lift buckets of water, through a gravitational height, out of flooded ore mines. The dimensionally equivalent newton-meter (N·m) is sometimes used instead; however, it is also sometimes reserved for torque to distinguish its units from work or energy.
Non-SI units of work include the erg, the foot-pound, the foot-poundal, and the liter-atmosphere.
## Mathematical calculation
### Force and displacement
Force and displacement are both vector quantities and they are combined using the dot product to evaluate the mechanical work, a scalar quantity:
$W = \bold{F} \cdot \bold{d} = F d \cos\phi$ (1)
where φ is the angle between the force and the displacement vector.
In order for this formula to be valid, the force and angle must remain constant. The object's path must always remain on a single, straight line, though it may change directions while moving along the line.
In situations where the force changes over time, or the path deviates from a straight line, equation (1) is not generally applicable although it is possible to divide the motion into small steps, such that the force and motion are well approximated as being constant for each step, and then to express the overall work as the sum over these steps.
The general definition of mechanical work is given by the following line integral:
$W_C := \int_{C} \bold{F} \cdot \mathrm{d}\bold{s}$ (2)
where:
C is the path or curve traversed by the object;
F is the force vector;
s is the position vector.
The expression δW=F·ds is an inexact differential which means that the calculation of WC is path-dependent and cannot be differentiated to give F·ds.
Equation (2) explains how a non-zero force can do zero work. The simplest case is where the force is always perpendicular to the direction of motion, making the integrand always zero. This is what happens during circular motion. However, even if the integrand sometimes takes nonzero values, it can still integrate to zero if it is sometimes negative and sometimes positive.
The possibility of a nonzero force doing zero work illustrates the difference between work and a related quantity, impulse, which is the integral of force over time. Impulse measures change in a body's momentum, a vector quantity sensitive to direction, whereas work considers only the magnitude of the velocity. For instance, as an object in uniform circular motion traverses half of a revolution, its centripetal force does no work, but it transfers a nonzero impulse.
### Mechanical energy
The mechanical energy of a body is that part of its total energy which is subject to change by mechanical work. It includes kinetic energy and potential energy. Some notable forms of energy that it does not include are thermal energy (which can be increased by frictional work, but not easily decreased) and rest energy (which is constant as long as the rest mass remains the same).
If an external force F acts upon a body, causing its kinetic energy to change from Ek1 to Ek2, then:
$W = \Delta E_k = E_{k2} - E_{k1} = \Delta E_k = \frac{1}{2} mv_2 ^2 - \frac{1}{2} mv_1 ^2 = \frac{1}{2} m \Delta (v^2)$
Thus we have derived the result, that the mechanical work done by an external force acting upon a body is proportional to the difference in the squares of the speeds. (It should be observed that the last term in the equation above is Δv2 rather than v)2.)
The principle of conservation of mechanical energy states that, if a system is subject only to conservative forces (e.g. only to a gravitational force), or if the sum of the work of all the other forces is zero, its total mechanical energy remains constant.
For instance, if an object with constant mass is in free fall, the total energy of position 1 will equal that of position 2.
$(E_k + E_p)_1 = (E_k + E_p)_2 \,\!$
where
The external work will usually be done by the friction force between the system on the motion or the internal-non conservative force in the system or loss of energy due to heat. | {"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": 6, "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": 0.9467547535896301, "perplexity": 343.1380449262141}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187826840.85/warc/CC-MAIN-20171023221059-20171024001059-00596.warc.gz"} |
https://mathoverflow.net/questions/335863/is-there-a-name-for-a-stable-physical-measure | # Is there a name for a “stable” physical measure?
Let $$M$$ be a Riemannian manifold with volume measure $$\lambda$$, let $$f \colon M \to M$$ be a continuous map, and let $$\mu$$ be a probability measure on $$M$$ with compact support.
Definition. The measure $$\mu$$ is a physical measure of $$f$$ if there is a set $$V \subset M$$ with [PropertyX] such that for all $$x \in V$$, the sequence $$\frac{1}{n}\sum_{i=0}^{n-1} \delta_{f^i(x)}$$ converges weakly to $$\mu$$ as $$n \to \infty$$.
Different definitions give different versions of Property X; three that I've seen are:
• $$V$$ is a $$\lambda$$-positive measure set;
• $$V$$ includes $$\lambda$$-almost all points in some neighbourhood of $$\mathrm{supp}\,\mu$$;
• $$V$$ is a neighbourhood $$\mathrm{supp}\,\mu$$.
Example. Take $$M=\mathbb{S}^1$$ with $$\lambda=\mathrm{Lebesgue}$$, and let $$f \colon M \to M$$ be a homeomorphism with a unique fixed point $$p$$. (E.g. $$f(x)=x+\varepsilon(1-\cos(2\pi x))$$, with $$p=0$$.) Since all trajectories of $$f$$ converge to $$p$$, we clearly have that $$\delta_p$$ is a physical measure (under any version of Property X).
However, in the above example, the fixed point $$p$$ is not stable (not even stable in the sense of Lyapunov), and therefore it seems intuitively strange to me to consider $$\delta_p$$ a "physical" measure. I feel like there should be a stronger stability requirement, such as what I will now suggest.
Definition. Given a topological space $$T$$, we say that a sequence of sets $$S_n \subset T$$ converges to a point $$x^\ast \in T$$ if for every neighbourhood $$U$$ of $$x^\ast$$ there exists $$N \in \mathbb{N}$$ such that for all $$n \geq N$$, $$S_n \subset U$$.
Definition. The measure $$\mu$$ is a stable physical measure of $$f$$ if there is a neighbourhood $$V$$ of $$\mathrm{supp}\,\mu$$ such that the sequence of sets $$S_n = \left\{ \frac{1}{n}\sum_{i=0}^{n-1} \delta_{f^i(x)} : x \in V \right\}$$ converges to $$\mu$$ with respect to the topology of weak convergence.
Remark. The sequence $$S_n$$ converges to $$\mu$$ with respect to the topology of weak convergence if and only if for every bounded continuous function $$g \colon M \to \mathbb{R}$$, $$\frac{1}{n}\sum_{i=0}^{n-1}g(f^i(x)) \to \mu(g)$$ uniformly across $$x \in V$$ as $$n \to \infty$$.
Is there an existing term for what I have called a "stable physical measure", or for something conceptually similar to this? Are there any references that simply define the term "physical measure" along similar lines to how I have defined a "stable physical measure"?
Further thoughts. I suppose the issue I'm raising is somewhat philosophical -- actually there's a very natural sense in which it is reasonable to consider $$\delta_p$$ in my example as being "stable": For all $$\varepsilon>0$$ there exists $$\delta>0$$ such that given any $$\delta$$-pseudo-orbit $$(x_n)$$ of $$f$$, for sufficiently large $$n$$ we have $$\frac{1}{n}\sum_{i=0}^{n-1}\mathbf{1}_{B_\varepsilon(p)}(x_i)>1-\varepsilon$$.
I wonder whether some analogous statement holds for more general physical measures -- i.e. whether for a large class of "physical measures" under the definition where $$V$$ is simply a $$\lambda$$-positive set, "typical" pseudo-orbits starting in $$V$$ will eventually have their empirical measures close to $$\mu$$.
I suppose my intuition for being "physical" was about the physical realisticness of being able to model a time-series $$(X_n)_{n=0,\ldots,N}$$ recorded at a "random" time from a process that has been running since "indefinitely long into the past" as a stochastic process whose law is the image measure of $$\mu$$ under $$x \mapsto (x,f(x),\ldots,f^N(x))$$. The conventional definition of a "physical" measure (with $$V$$ simply being $$\lambda$$-positive) seems to describe exactly this property if one ignores the potential for small perturbations. That's why I wondered whether there is a version that describes this property and also takes into account the potential for small perturbations. Perhaps the best way to achieve this is simply to specify that the measure $$\mu$$ is not only physical but also has the property that every trajectory in $$\mathrm{supp}\,\mu$$ is stable in the sense of Lyapunov.
• I've never seen this definition before. It looks a bit too strong to me, though; not because of your addition, but because you require $V$ to be a neighborhood of the support. Once you go beyond delta measures, the most important examples of physical measures are the SRB measures, and for uniformly hyperbolic systems these certainly do not satisfy even your first definition if you require $V$ to be a neighborhood of the support; there are large sets of "irregular points" whose empirical measures do not converge to $\mu$. – Vaughn Climenhaga Jul 11 '19 at 3:00
I realize this doesn't directly answer the "reference request" part of the question, but I believe that if you require $$V$$ to be full (Lebesgue) measure in a neighborhood of the support of $$\mu$$, then this definition is too strong to be satisfied by any of the usual examples that one would want, going beyond delta measures to SRB measures where the expanding direction is nontrivial.
Here's why: Suppose $$\Lambda$$ is a topologically mixing Axiom A attractor for a diffeo $$f$$, and $$\mu$$ is the SRB measure. Let $$\xi(x) = \log \det Df|_{E^u(x)}$$, so that $$\mu$$ satisfies the following lower Gibbs bound: $$\mu \{ y : d(f^k y, f^k x) \leq \epsilon\} \geq C(\epsilon) e^{-S_n\xi(x)}.$$ Theorem 1 (part 3) of Lai-Sang Young's 1990 TAMS paper "Large deviation results for dynamical systems" gives, for each continuous $$\phi\colon \Lambda\to \mathbb{R}$$, the estimate $$\varliminf_{n\to\infty} \frac 1n \log \mu \{ x : \frac 1n S_n\phi(x) > c \} \geq \sup \{ h_\nu(f) - \int \xi\,d\nu : \nu \text{ is f-invariant and } \int\phi\,d\nu > c\}.$$ Write the supremum as $$\alpha(\phi,c)$$, so that for each $$\beta < \alpha(\phi,c)$$ we have some constant $$K$$ such that $$\mu \{x : \frac 1n S_n\phi(x) > c \} \geq K e^{\beta n}.$$ Choose any continuous $$\phi$$ such that there is an $$f$$-invariant $$\nu$$ with $$\int\phi\,d\nu > \int\phi\,d\mu$$ (such a $$\phi$$ exists since $$f$$ is not uniquely ergodic), and choose $$c$$ between these two integrals. Then taking $$U = \{ m : \int \phi \,dm < c\}$$, we see that $$U$$ is an open set containing $$\mu$$ for which we have $$\mu \{ x : \frac 1n \sum_{k=0}^{n-1} \delta_{f^k x} \notin U \} \geq K(\beta) e^{\beta n}$$ whenever $$\beta < \alpha(\phi,c)$$; here $$\alpha>-\infty$$ since we can take $$\nu=m$$ in the supremum. I expect (though I didn't check details) that a similar bound would hold replacing $$\mu$$ by $$\lambda$$ restricted to a neighborhood. In particular if $$V$$ contains a full measure set in a neighborhood of the support of $$\mu$$, then the sets $$S_n$$ that you define would never be contained in $$U$$, so the sequence of sets would not converge to $$\mu$$.
This doesn't rule out obtaining this behavior for a positive measure set of $$V$$, and indeed you could probably get it for sets $$V$$ whose complements intersect the neighborhood of the support with arbitrarily small Lebesgue measure. But at this point I've strayed quite far from your actual request for a reference... I just had the thought about large deviations results creating problems for some possible definitions, and then got carried away writing down the details. | {"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": 106, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.985756516456604, "perplexity": 118.88640818833544}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107905965.68/warc/CC-MAIN-20201029214439-20201030004439-00185.warc.gz"} |
https://epjc.epj.org/articles/epjc/abs/2002/11/100520327/100520327.html | 2018 Impact factor 4.843
Particles and Fields
Eur. Phys. J. C 25, 327-332 (2002)
DOI: 10.1007/s10052-002-0986-y
## Superstring on pp-wave orbifold from large- N quiver gauge theory
N. Kim1, A. Pankiewicz1, S.-J. Rey2, 3 and S. Theisen1
1 Max-Planck-Institut für Gravitationsphysik, Albert-Einstein-Institut, Am Mühlenberg 1, 14476 Golm, Germany
2 School of Physics and Center for Theoretical Physics, Seoul National University, Seoul 151-747, Korea
3 Isaac Newton Institute for Mathematical Sciences, 20 Clarkson Road, Cambridge CB3 0EH, UK
(Received: 14 March 2002 / Published online: 5 July 2002 )
Abstract
We extend the proposal of Berenstein, Maldacena and Nastase to the Type IIB superstring propagating on a pp-wave over the orbifold. We show that first-quantized free string theory is described correctly by the large- N, fixed gauge coupling limit of [U(N)]k quiver gauge theory. We propose a precise map between gauge theory operators and string states for both untwisted and twisted sectors. We also compute leading-order perturbative correction to the anomalous dimensions of these operators. The result is in agreement with the value deduced from the string energy spectrum, thus substantiating our proposed operator-state map.
© Società Italiana di Fisica, Springer-Verlag 2002 | {"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": 0.8806453347206116, "perplexity": 3842.5990730293784}, "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/1568514573124.40/warc/CC-MAIN-20190917223332-20190918005332-00326.warc.gz"} |
https://atcoder.jp/contests/agc013/tasks/agc013_b | Contest Duration: - (local time) (150 minutes) Back to Home
B - Hamiltonish Path /
Time Limit: 2 sec / Memory Limit: 256 MB
問題文
N 頂点 M 辺の、連結な単純無向グラフが与えられます。 頂点には 1 から N までの番号がついており、辺には 1 から M までの番号がついています。 辺 i は、頂点 A_i と頂点 B_i を結んでいます。 次の条件を満たすパスを 1 つ見つけて、出力してください。
• 2 個以上の頂点を通る
• 同じ頂点を 2 度以上通らない
• パスの少なくとも一方の端点と直接辺で結ばれている頂点は、必ずパスに含まれる
ただし、この問題の制約の下で、このようなパスが必ず存在することが証明できます。 また、あり得る答えのうちどれを出力しても構いません。
制約
• 2 \leq N \leq 10^5
• 1 \leq M \leq 10^5
• 1 \leq A_i < B_i \leq N
• 与えられるグラフは連結かつ単純（どの 2 頂点を直接結ぶ辺も高々 1 つ）である
入力
N M
A_1 B_1
A_2 B_2
:
A_M B_M
入力例 1
5 6
1 3
1 4
2 3
1 5
3 5
2 4
出力例 1
4
2 3 1 4
入力例 2
7 8
1 2
2 3
3 4
4 5
5 6
6 7
3 5
2 6
出力例 2
7
1 2 3 4 5 6 7
入力例 3
12 18
3 5
4 12
9 11
1 10
2 5
6 10
8 11
1 3
4 10
2 4
3 7
2 10
3 12
3 9
1 7
2 3
2 11
10 11
出力例 3
8
12 4 2 5 3 9 11 8
Score : 500 points
Problem Statement
You are given a connected undirected simple graph, which has N vertices and M edges. The vertices are numbered 1 through N, and the edges are numbered 1 through M. Edge i connects vertices A_i and B_i. Your task is to find a path that satisfies the following conditions:
• The path traverses two or more vertices.
• The path does not traverse the same vertex more than once.
• A vertex directly connected to at least one of the endpoints of the path, is always contained in the path.
It can be proved that such a path always exists. Also, if there are more than one solution, any of them will be accepted.
Constraints
• 2 \leq N \leq 10^5
• 1 \leq M \leq 10^5
• 1 \leq A_i < B_i \leq N
• The given graph is connected and simple (that is, for every pair of vertices, there is at most one edge that directly connects them).
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
A_2 B_2
:
A_M B_M
Output
Find one path that satisfies the conditions, and print it in the following format. In the first line, print the count of the vertices contained in the path. In the second line, print a space-separated list of the indices of the vertices, in order of appearance in the path.
Sample Input 1
5 6
1 3
1 4
2 3
1 5
3 5
2 4
Sample Output 1
4
2 3 1 4
There are two vertices directly connected to vertex 2: vertices 3 and 4. There are also two vertices directly connected to vertex 4: vertices 1 and 2. Hence, the path 2314 satisfies the conditions.
Sample Input 2
7 8
1 2
2 3
3 4
4 5
5 6
6 7
3 5
2 6
Sample Output 2
7
1 2 3 4 5 6 7 | {"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": 0.8730716109275818, "perplexity": 772.1297961821274}, "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-2020-50/segments/1606141681209.60/warc/CC-MAIN-20201201170219-20201201200219-00304.warc.gz"} |
https://chat.stackexchange.com/transcript/36?m=7091064 | 00:00 - 16:0016:00 - 00:00
user40730
4:09 PM
short question: a convergent sequence is bounded. so does every bounded sequence have a bounded subsequence? does it apply for all subsequences?
4:45 PM
@KaliMa Not sure why yellow is the area you want - why not the red part as well?
a<=b ?
i dont know if my bounds are right because when i try in wolfram it stops well below X
if you go into wolfram and type "plot -100<=-ab<=100 and a<=b" it seems to graph and stop before 100 on the axes
why is this?
is it supposed to go on forever and taper off at 100?
or is the graph just ending early
The graphs for $ab=X$ etc should go on - not sure why W|A has stopped them there
@p05 you mean convergent?
4:51 PM
do they taper off to points?
user40730
@AlexeiAverchenko you mean: every convergent sequence has bounded subsequences?
no...
@KaliMa the curve $ab=X$ approaches the horizontal and vertical axes as asymptotes - the further along you go, the closer the curve becomes to the axis, but never actually touches it
user40730
@AlexeiAverchenko well maybe this makes it more clear: I have a convergent sequence in $\mathbb R$. so this sequence is bounded. Now I wanna consider its subsequences. are they bounded?
but i mean at the X threshold, does the whole thing intersect at X to a point, or stop off early/truncate and become a hard line like in the graph
4:55 PM
yeah
because they are convergent
i assume it gets cut off hard
@KaliMa sorry - not sure what you mean by "at the X threshold"
user40730
@AlexeiAverchenko but without having that its subsequences are convergent you can't say it?
sure i can
when a or b = X
4:57 PM
any subset of a bounded set is bounded
the cutoff point that should have nothing past it
user40730
yeah and the same for the sequences...
@KaliMa sorry - that is not making any sense to me
user40730
okay
user40730
thx
4:58 PM
np
back later folks
5:47 PM
hey
6:28 PM
If you have $x \in A$. And then it turns out A is the empty set, can you say this is a contradiction as this is an impossibility?
6:59 PM
@sonicboom It just means x can not have any values because there are no members in the empty set, right?
Well the question seems to me to be 'Can we have a point x in the empty set'. Which would be a contradiction as it's wouldn't be empty then!
But x is a symbol used to represent members of the set.
I should have said 'Can we have an element x in the empty set'.
I think I should've mentioned that $x \in X$ and also $x \in A$. So when A turns out to be the empty set it is a contradiction as we can't have x in X and x in the empty set.
7:14 PM
Yes, 'Can we have an element in the empty set'. Is a contradiction.
But x is a symbol used to represent members in the set.
It is the members that are in the set, not x.
Yes an x is equal to a member of the set X. The set X being the entire space. So we can't have a member of the set X in the empty set.
putnam = halfway done
second half in T minus 35 min
@anon how was first half?
There was an easy one I got right. Another one I BS'd on. Other four I got nada.
Guys, I have a simple question. Say I have an equation like a^x = a^3. I know it's common sense that allows me to say that x = 3, but how would you prove that?
7:26 PM
@ZERO lake logs of both sides
in the positive reals? logarithm with base a. or, show x->a^x is an increasing function and hence injective.
nada = nil = zilch = $\varnothing$ = 0
ahh
in India, nada means the thread which is strung through the pyajamas to tighten them. :P
There was a hard one that basically says the affine group of $\Bbb F_p^n$ has an element for which the cyclic group generated by it acts transitively.
@JayeshBadwaik I was going to ask what it meant in Indian :)
7:28 PM
@OldJohn Indian!! Actually, what I said is in Hindi.
(I use "Indian" loosely !! _ know there are loads of languages)
:-) I know. Actually, it reminded me of a joke by Russell Peters with the same idea.
@JayeshBadwaik I have a book somewhere entitled "Sanskrit in 30 Days" - I bought it just for the insane title :)
@OldJohn, could you show me?
7:30 PM
@ZERO show what?
the book?
I would imagine that it would look the same even if you took the logs of both sides
@OldJohn Taking into account 21.12.2012, it would be hard to learn Sanskrit even using that book
@OldJohn Hahaha. Sanskrit in thirty days! I don't know Sanksrit even an iota BTW. My brother took it for six years though.
@Ilya No... Not again!! This is driving me crazy!! (Everyone is crazy about it)
ah - if you take logs, you get $x\log a = 3\log a$ (assuming something about $a$) then divide by $\log a$
You need $\log a \ne 0$ - but you can probably get that or look at it as a special case
@OldJohn Is this the book?
7:35 PM
Ack, I feel like a total noob. What are the '$' for and '\'? @aDangerousIdea Yes! @ZERO LaTeX markup afk - must eat later afk - exam 7:36 PM good luck and off they went 7:54 PM to see the wizard of Oz 8:43 PM sheesh - quiet here tonight 9:01 PM @Ilya ho-ho! What a person! :) user19161 @OldJohn Nice new picture. Very sexy. Does anybody know about Kolmogorov Complexity? @WillHunting thought it about time I showed my face user19161 @OldJohn Yes, considering you are almost at 6k! @WillHunting creeping closer :) and creeping up on the generalist badge, I think user19161 9:12 PM Wow @amwhy you are getting closer to 10k! =) user19161 @anon Good luck! I am with you in spirit. Hm, how to show that$\int\limits_{0}^{\infty} f(x) dg(x) = \langle f(x), g'(x) \rangle$for Lebesgue-Stiltjes integral? 10:17 PM @Nimza What are the hypothesis on$f$and$g$? @PeterTamaroff$f \in \mathcal{D}(0,\infty)$,$g$is arbitrary nondecreasing If$g$is differentiable, then $$\int_0^\infty f dg =\int_0^\infty f g' dx=\langle f,g'\rangle$$ @Nimza What is$\mathcal D(A)$? @PeterTamaroff test functions,$C^{\infty}(0,\infty)$with compact support @Nimza What is a test function? @PeterTamaroff base functions on which distributions are defined 10:20 PM @Nimza Oh, OK. Distributions are "über function" right? @PeterTamaroff generalized functions, functionals - do you mean that by über function? I just bought "Integral, measure, derivative" and "Elementary complex and real analysis" by Shilov. @PeterTamaroff I didn't read that @Nimza über means "over" 10:23 PM @PeterTamaroff aha, but what does mean "over-function"? @Nimza I usually use it to mean "super" or "supra".... like übermensch @PeterTamaroff :) @OldJohn that looks likea mug shot =P "I just bought "Integral, measure, derivative" and "Elementary complex and real analysis" by Shilov." @OldJohn Do you know those? (I also bought Arkham City and a XBOX360 controller for windows, but that's classified procrastrination) @PeterTamaroff It is ... a shot of a mug, yes :) Not sure I know Shilov's book - but I have a few books with titles very much like that :) @OldJohn I like unification! 10:29 PM Has anyone seen the comment following my comment here? @PeterTamaroff me too @OldJohn HJAAHAHAH yes, I upvoted it Barcelona won 5 to 1 And Real Madrid is playing right now @PeterTamaroff It is actually a continuation of an email conversation will WJ - I imagined that the rest of MSE would be baffled by it :) @OldJohn Hahhaah =P Gotta go. Bye byes,. user19161 @PeterTamaroff You might like his Elementary Functional Analysis too. user19161 @old You choosing to use the side view makes it look very artistic. I wonder if you are looking into the past or the future, or both. user19161 10:37 PM @gus Your school is starting soon, yay! @WillHunting At the time I was actually looking at a camel I was about to ride ... nothing so symbolic as past and future :))) user19161 @OldJohn Perhaps I should take one facing the left, then we can face opposite directions! @WillHunting facing apart or facing together? :) user19161 @OldJohn That's exactly what I was thinking. Great minds think alike! @WillHunting :) 10:45 PM @OldJohn Is that you in your avatar? user19161 @Argon Hello Aaron! @WillHunting Hi Jasper! How are you? @Argon yep :) You aren't that old! 60 is old enough :) 10:50 PM Until 120, as they say! that photo was 2 weeks ago on holiday in the middle east - in case anyone was thinking it was taken 20 years ago :) HAHAHAHA! user19161 @Argon Bad, as usual. @WillHunting "Bad, as usual"? That's not good! user19161 @Argon Haha, don't you know some of my secrets already? 10:58 PM @WillHunting Yes, but still! user19161 @Argon I hope some kind of miracle will happen soon. There can be miracles when you believe. -Mariah Carey user19161 Haha, the voters clearly did not understand my answer. math.stackexchange.com/questions/248817/… user19161 Of course I omitted some details, but I think they are not aware that that is a correct answer. @WillHunting Yep - +1 @WillHunting am I right here? 11:12 PM @WillHunting, I am wanting to give a Part 3 to the Hint 2 found here, but I can't figure out how to integrate the integrals with the quadratics in the denominator. @WillHunting, any tips? user19161 @OldJohn Yes. Of course, it need not be injective if you allow a point to be represented twice! @WillHunting Yeah - I am not totally sure of the exact conditions required - but I thought I needed to correct the impression given in the comments :) user19161 @Limitless Hmm I can't remember my antiderivatives, but maybe the integrals involving D and E can't be computed in that manner of splitting. One may need to split it up otherwise. Would be good if you try to work it out yourself too. @WillHunting, I have tried everything from Weierstrass substitution to finding if it can match the form$\frac{1}{a^2+u^2}$. It just won't play nicely. user19161 @Limitless You mean the integral involving D and E? 11:22 PM @WillHunting, indeed. Partial fraction decomposition, then integrate each fraction. Nope user19161 @Limitless Ah I used to do plenety of these integrals in high school. Let me find something to help you. The roots of the denominator are in$\mathbb{C}$. I don't know complex analysis, @Argon. Do you know partial fraction decomposition? @Limitless It needs no CA 11:23 PM @Argon, CA? @Limitless Complex analysis @Argon, how to do you use partial fraction decomposition when the denominator does not factorize over$\mathbb{R}$??? Let's see Broaden my horizons, @WillHunting! Broaden them, please. :P Multiply out denominators to get user19161 11:25 PM @limitless What is D and E specifically? @WillHunting, I don't know. I was trying to write this without giving that so that the asker could fill in the details. I'm not trying to post a full solution. $$6x^5+x^2+x+2 = (x^2 + 2x + 1)(2x^2 - x + 4)(x+1)\left(\frac{A}{(x+1)^3}+\frac{B}{(x+1)^2}+\frac{C}{x+1}+\frac{Dx+E}{2x^2-x+4}+F\right)$$ user19161 One can use the trick that the antiderivative of f'(x)/f(x) is ln |f(x)|. @Argon,$2x^2-x+4$has roots$x_{+,-}=\frac{1}{4}\pm \frac{i\sqrt{31}}{4}$. I don't think partial fraction decomposition works here. @WillHunting, this applies to the first integral, yes? user19161 One can also try to complete the square in the denominator of the fraction. 11:28 PM @Limitless Expand, then then group the powers It should still work does it work if you just multiply out all the brackets and equate coefficients? Hmm, @WillHunting, it would appear that doesn't work on the first integral involving$D$: If you let$f(x)=2x^2-x+4$, then$f'(x)=4x-1\ne x$. @OldJohn Yep, this is what I mean It should work @Argon, I'm not trying to post a full solution. I don't understand what you're saying given that fact. I agree - if you have the right expressions, then it has to work :) user19161 11:30 PM So for example, if I have$\frac{2x+3}{x^2+x+1}$, I can split it up into$\frac{2x+1}{x^2+x+1}$and$\frac{2}{x^2+x+1}$. @Limitless You simply clear denominators and then group together all the terms of equal powers! @Argon, I'm trying to tell you: I'm not trying to solve for the variables$A,B,C,D,E,$and$F$. I'm trying to give a hint as to how to evaluate the following integrals:$\int \frac{1}{2x^2-x+4}dx$and$\int \frac{x}{2x^2-x+4}dx$. I am sorry this was not clear. user19161 After that in the second fraction above I can complete the square as$(x+\frac{1}{2})^2+\frac{3}{4}$. @Limitless for those integrals, the standard method is to complete the square in the denominator and then use a trig substitution hmm 11:34 PM @OldJohn, how do I do that? I'm sorry that I don't know how. I feel like I should know this, but it's not like completing the square in Algebra II.$\int \frac{1}{2x^2-x+4}dx = \int\frac{1/2}{x^2-1/2x+2}dx = \int\frac{1/2}{(x-1/4)^2+15/16}dx$then put ... user19161 @limitless See my steps above and this article. en.wikipedia.org/wiki/List_of_integrals_of_rational_functions @OldJohn, okay. I see what you're doing there. @Charlie hmmm.........................................$x- 1/4 = \sqrt{15}/16\tan\theta$11:37 PM @Argon hmmmm........................................... @Limitless I see. For #2, logs will work, no? @Charlie hhhhhhhhhhhhhmmmmmmmmmmmmmmmmmmmmm..................... user19161 Also, don't attempt the integral questions until you are really sure you can compute them! @WillHunting, why? It's always a great learning process. @Argon hahahahahahhah user19161 @Limitless Hmm OK. I mean you should post only when you are sure you can do it yourself, but it's alright. 11:39 PM @WillHunting, Calculus is one of those subjects where I'm never entirely confident in what I'm doing since I just started learning it. Things are harder than they seem. user19161 @Limitless Also, I got your email. @Limitless I have been doing calculus for 42 years and still feel like that sometimes @WillHunting, wonderful. Thanks for the help, guys. I'm really trying to get something productive done this weekend. I can't get anything done on the weekdays. @Limitless can anybody? :P @Charlie, true story. The school I attend is worthless sometimes. 11:45 PM every day seems like a weekend when you have retired :) 6 @Limitless correction to my earlier substitution: should probably be$x- 1/4 = \sqrt{15}/4\tan\theta$@OldJohn, I think you messed up here:$\int \frac{\frac{1}{2}}{x^2-1/2x+2}dx=\int\frac{1/2}{(x-1/4)^2+15/16}dx. I am getting \begin{align} \int \frac{1}{2x^2-x+4}dx&=\int \frac{\frac{1}{2}}{x^2-\frac{1}{2}x+2}dx\\ &=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^2+\frac{31}{16}}dx. \end{align} @Limitless yeah -31/16$- sorry it is late here, and I am tired :) so the substitution will be$x- 1/4 = \sqrt{31}/4\tan\theta$you should end up with$\sec^2\theta d\theta$cancelling out$\tan^2\theta+1$on the bottom of the fraction @OldJohn Interesting. We can sub that in and simply change$dx$to$d\theta$? And yeah, I noticed that. :) no -$dx$becomes$\sec^2\theta d\theta$within a rational multiple to be precise:$\sqrt{31}/4 \sec^2\theta d\theta$@OldJohn Ah. So you take$x=\sqrt{31}/4 \tan \theta+1/4$, and$dx=(\sqrt{31}/4 \tan \theta+1/4)'d\theta$as per usual. Makes much more sense. 11:58 PM I prefer to work out$dx/d\theta$from the substitution$x- 1/4 = \sqrt{31}/4\tan\theta$you get$dx/d\theta = \sqrt{31}/4\sec^2\theta\$
@OldJohn Typo
00:00 - 16:0016:00 - 00:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8098887205123901, "perplexity": 1418.8670899831895}, "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-2020-24/segments/1590347391309.4/warc/CC-MAIN-20200526191453-20200526221453-00099.warc.gz"} |
https://math.stackexchange.com/questions/649557/finding-the-boundary-conditions-for-a-laplaces-equation-in-polar-coordinates | # Finding the Boundary Conditions for a Laplace's Equation in Polar Coordinates
I have solved Laplace's equation in Polar Coordinates for the scalar electric potential in a circle of radius R and have the solution $$\phi(r,\varphi) = \phi_{0} + \sum^{\infty}_{k=1}r^{k_{\varphi}}\left(A_{k}\sin(k\varphi)+B_k\cos(k\varphi)\right)$$
Where $k_{\varphi}$ is a constant times k.
Now the boundary condition that I have is $\vec{n}\cdot\vec{J}=0$ where n is the normal to the surface, and $\vec{J}$ is the current density. This can be written as $$\vec{n}\cdot \vec{\nabla}\phi=-\vec{n}\cdot\frac{\partial\vec{A}}{\partial t}$$ Since the system is quasi-static the magnetic vector potential, A, can be separated into a spacial and time component written as $\vec{A}(\vec{s},t)= \vec{A}(s)f(t)$ for an arbitrary function of time. The magnetic vector potential is known and is: $$\vec{A}= \frac x{r^2}\hat{\varphi}$$ Where x is just a constant.
My attempt was to write the Boundary condition as $$\vec{n}\cdot \left( \begin{matrix} \frac{\partial\phi}{\partial r} \\ \frac{\partial\phi}{r\partial \varphi } \end{matrix} \right) = =\vec{n}\cdot \left( \begin{matrix} 0\\ \frac{x}{r^2} \end{matrix} \right)\cdot \dot{f}(t)$$ What I am not sure about is what is the normal vector to a circle of radius R? or if my approach is correct?
Thank you.
• The normal vector to the circle is $(R cos \phi , R sin \phi)$ Jan 24 '14 at 4:56
• @Semsem From the context, $\vec n$ is a unit normal vector, which is $(\cos \varphi, \sin\varphi)$. Jan 24 '14 at 5:03
• yes it is ok !! Jan 24 '14 at 5:06
Taking full gradient and then multiplying by $\vec n$ is doing too much. The outward normal derivative at the boundary of the circle is simply the partial derivative with respect to $r$. So, $$\frac{\partial \phi}{\partial r} = -\vec{n}\cdot\frac{\partial\vec{A}}{\partial t} \tag{1}$$ Next, your formula for $\vec A$ says that this vector is a multiple of $\hat \varphi$, which I understand to be a unit tangent vector. If so, then the time derivative of $\vec A$ is tangent to the circle, too. Consequently, the right hand side of (1) is zero.
So, it appears that $\phi$ must be a harmonic function with zero normal derivative; the only such functions are the constant ones.
An aside: using both $\phi$ and $\varphi$ in the same computation is usually a bad idea. | {"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": 0.9680585265159607, "perplexity": 70.48930683728864}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304915.53/warc/CC-MAIN-20220126041016-20220126071016-00661.warc.gz"} |
http://physics.hmc.edu/course/p080/ | ## Physics 80 — Energy and the Environment
An area of physics is studied, together with its applications and social impact. Possible areas include energy and the environment, climate change, and sustainability. Active participation and group activities are stressed. | {"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": 0.8194904327392578, "perplexity": 3150.7962055556322}, "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-2020-05/segments/1579251788528.85/warc/CC-MAIN-20200129041149-20200129071149-00148.warc.gz"} |
http://math.stackexchange.com/questions/61774/how-do-i-solve-this-equation-involving-a-logarithm/61805 | # How do I solve this equation involving a logarithm?
I'm running in circles and I don't understand how to do this.
$$x\log(x) = 100$$
Where the $\log$ is in base $10$, I understand that $\log(y)=x$ is $10^x = y$. So is it the same for $x\log(x) = 100$? Would it be $10^{100}=x\cdot x$? It doesn't come out right when I do it, and it's clear that I have holes in my knowledge on logarithms, could someone please tell me my flaws and explain this to me?
-
Wait, suddenly I'm not so sure it is something for [algebra-precalculus] anymore. – Asaf Karagila Sep 4 '11 at 9:24
You will need the services of the Lambert function $W(x)$ to solve this equation. Briefly, the Lambert function is the inverse of the function $xe^x$: if $x=ye^y$, then $y=W(x)$.
To turn your equation into a form where the Lambert function's appearance becomes transparent, let's first turn everything into natural logarithms:
$$x\ln\,x=100\ln\,10$$
and then we make the left side a "little" complicated:
$$(\ln\,x)e^{\ln\,x}=100\ln\,10$$
We now recognize the Lambert form, and thus perform the inversion:
$$\ln\,x=W(100\ln\,10)$$
from which
$$x=e^{W(100\ln\,10)}\approx 56.961248432\dots$$
-
WOW. This is a mouth full. I did not think it would lead to so much math. – Doug Sep 4 '11 at 9:48
@J. M.: Do you think the [special-functions] tag should be added to the question? Or even create one for the Lambert function. – Américo Tavares Sep 4 '11 at 12:42
@Américo: I'm not quite sure if there should be a special-functions tag, much less a more specialized lambert-w tag. On the other hand, we now have quite a pile of questions where the Lambert function shows up in the solution... tell you what, let's create a tag for Lambert questions if at least five people upvote your comment. How's that sound? – Ｊ. Ｍ. Sep 4 '11 at 12:47
It sounds good. – Américo Tavares Sep 4 '11 at 12:49
Here is the link for a meta thread about adding a tag for the Lambert W function: meta.math.stackexchange.com/q/2908/622 – Asaf Karagila Sep 4 '11 at 15:52
If $x \log_b (y) = z$ then taking anti-logarithms you get $y^x = b^z$.
So in this case with $y=x$ and $b=10$ you get $x^x = 10^{100}$.
You will not find it easy to solve this explicitly for $x$; try reading about the Lambert W function or use numerical methods to get something just over 56.96.
-
Why is x the exponent for y? I don't find that really intuitive, could you please explain? – Doug Sep 4 '11 at 9:16
You forgot some braces there, I hope you did not mind. – Asaf Karagila Sep 4 '11 at 9:16
@Asaf: thank you – Henry Sep 4 '11 at 9:42
@Doug: it is a basic property of logarithms: see the power formula here – Henry Sep 4 '11 at 9:44
@Asaf: To the meta-cave! – The Chaz 2.0 Sep 4 '11 at 15:54
There is no way to solve $x\log_{10}x=100$ exactly using the methods of school algebra. There is a way, called Newton's Method, to get a solution to as many decimals as you want, using Calculus. Newton's Method is in a thousand intro Calculus textbooks, also a thousand websites. If you haven't done Calculus yet, you have something to look forward to.
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http://math.stackexchange.com/questions/152257/analyzing-the-complex-function-ez-1-z?answertab=votes | # Analyzing the complex function $(e^z-1)/z$
I'm trying to write the function $f(z)= (e^z-1)/z$ when $z \neq 0$ and 1 when $z=0$ in the form of $u(x,y)+iv(x,y)$ where $z=x+iy$. How should I proceed?
I have already shown that it is differentiable (continuously) for all $z \in \mathbb C$ in an earlier problem, but is this needed to solve this?
Also, another problem asks to show that $u(x,y)$ is $C^\infty$; this is simply a result of the fact that $(e^z)'=e^z$ correct? Thank you.
-
Pity. $$\frac z {e^z-1}$$ is far mor interesting. – Pedro Tamaroff May 31 '12 at 23:33
Welcome to m.se, Mr President. – Gerry Myerson Jun 1 '12 at 0:27
## 1 Answer
$$\frac{e^z-1}{z}=\frac{e^xe^{iy}-1}{x+iy}=\frac{e^x\cos y-1+ie^x\sin y}{x+iy}$$ and now do the usual to divide two complex numbers.
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Now, the real test... Can our helpful members refrain from doing more until James returns and gives this a try? – GEdgar Jun 1 '12 at 0:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9138320088386536, "perplexity": 757.905590443868}, "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-2014-52/segments/1418802768044.102/warc/CC-MAIN-20141217075248-00033-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://normaldeviate.wordpress.com/2013/07/ | ## Monthly Archives: July 2013
### The Steep Price of Sparsity
The Steep Price of Sparsity
We all love sparse estimators these days. I am referring to things like the lasso and related variable selection methods.
But there is a dark side to sparsity. It’s what Hannes Leeb and Benedikt Potscher call the “return of the Hodges’ estimator.” Basically, any estimator that is capable of producing sparse estimators has infinitely bad minimax risk.
1. Hodges Estimator
Let’s start by recalling Hodges famous example.
Suppose that ${X_1,\ldots,X_n \sim N(\theta,1)}$. Define ${\hat\theta}$ as follows:
${\hat\theta =\overline{X}}$ if ${|\overline{X}| \geq n^{-1/4}}$
${\hat\theta=0}$ if ${|\overline{X}| < n^{-1/4}}$.
If ${\theta\neq 0}$, then ${\hat\theta}$ will equal ${\overline{X}}$ for all large ${n}$. But if ${\theta=0}$, then eventually ${\hat\theta =0}$. The estimator discovers that ${\theta}$ is 0.
This seems like a good thing. This is what we want whenever we do model selection. We want to discover that some coefficients are 0. That’s the nature of using sparse methods.
But there is a price to pay for sparsity. The Hodges estimator has the unfortunate property that the maximum risk is terrible. Indeed,
$\displaystyle \sup_\theta \mathbb{E}_\theta [n (\hat\theta-\theta)^2] \rightarrow \infty.$
Contrast this will the sample mean:
$\displaystyle \sup_\theta \mathbb{E}_\theta [n (\overline{X}-\theta)^2] =1.$
The reason for the poor behavior of the Hodges estimator — or any sparse estimator — is that there is a zone near 0 where the estimator will be very unstable. The estimator has to decide when to switch from ${\overline{X}}$ to 0 creating a zone of instability.
I plotted the risk function of the Hodges estimator here. The risk of the mle is flat. The large peaks in the risk function of the Hodges estimator are very clear (and very disturbing).
2. The Steep Price of Sparsity
Leeb and Potscher (2008) proved that this poor behavior holds for all sparse estimators and all loss functions. Suppose that
$\displaystyle Y_i = x_i^T \beta + \epsilon_i,\ \ \ i=1,\ldots, n.$
here, ${\beta\in \mathbb{R}^k}$. Let ${s(\beta)}$ be the support of ${\beta}$: ${s_j(\beta)=1}$ of ${\beta_j \neq 0}$ and ${s_j(\beta)=0}$ of ${\beta_j = 0}$. Say that ${\hat\beta}$ is sparsistent (a term invented by Pradeep Ravikumar) if, for each ${\beta}$,
$\displaystyle P^n_\beta(s(\hat\beta)=s(\beta)) \rightarrow 1$
as ${n\rightarrow \infty}$.
Leeb and Potscher (2008) showed that if ${\hat\beta}$ is sparsistent, then
$\displaystyle \sup_\beta E[ n\, ||\hat\beta-\beta||^2]\rightarrow \infty$
as ${n\rightarrow \infty}$. More generally, for any non-negative loss function ${\ell(\hat\beta-\beta)}$, we have
$\displaystyle \sup_\beta E[ \ell(n^{1/2}(\hat\beta -\beta))]\rightarrow \sup_\beta \ell(\beta).$
3. How Should We Interpret This Result?
One might object that the maximum risk is too conservative and includes extreme cases. But in this case, that is not true. The high values of the risk occur in a small neighborhood of 0. (Recall the picture of the risk of the Hodges estimator.) This is far from pathological.
Another objection is that the proof assumes ${n}$ grows while ${k}$ stays fixed. But in many applications that we are interested in, ${k}$ grows and is even possibly larger than ${n}$. This is a valid objection. On the other hand, if there is unfavorable behavior in the ideal case of fixed ${k}$, we should not be sanguine about the high-dimensional case.
I am not suggesting we should give up variable selection. I use variable selection all the time. But we should keep in mind that there is a steep price to pay for sparsistency.
References
Leeb, Hannes and Potscher, Benedikt M. (2008). Sparse estimators and the oracle property, or the return of Hodges’ estimator. Journal of Econometrics, 142, 201-211.
Leeb, Hannes and Potscher, Benedikt M. (2005). Model selection and inference: Facts and fiction. Econometric Theory, 21, 21-59.
### THE FIVE: Jeff Leek’s Challenge
Jeff Leek, over at Simply Statistics asks an interesting question: What are the 5 most influential statistics papers of 2000-2010?
I found this to be incredibly difficult to answer. Eventually, I came up with this list:
Donoho, David (2006). Compressed sensing. IEEE Transactions on Information Theory. 52, 1289-1306.
Greenshtein, Eitan and Ritov, Ya’Acov. (2004). Persistence in high-dimensional linear predictor selection and the virtue of overparametrization. Bernoulli, 10, 971-988.
Meinshausen, Nicolai and Buhlmann, Peter. (2006). High-dimensional graphs and variable selection with the lasso. The Annals of Statistics, 34, 1436-1462.
Efron, Bradley and Hastie, Trevor and Johnstone, Iain and Tibshirani, Robert. (2004). Least angle regression. The Annals of statistics, 32, 407-499.
Hofmann, Thomas and Scholkopf, Bernhard and Smola, Alexander J. (2008). Kernel methods in machine learning. The Annals of Statistics. 1171–1220.
These are all very good papers. These papers had a big impact on me. More precisely, they are representative of ideas that had an impact on me. It’s more like there are clusters of papers and these are prototypes from those clusters. I am not really happy with my list. I feel like I must be forgetting some really important papers. Perhaps I am just getting old and forgetful. Or maybe our field is not driven by specific papers.
What five would you select? (Please post them at Jeff’s blog too.)
### LOST CAUSES IN STATISTICS II: Noninformative Priors
LOST CAUSES IN STATISTICS II: Noninformative Priors
I thought I would post at a higher frequency in the summer. But I have been working hard to finish some papers which has kept me quite busy. So, apologies for the paucity of posts.
Today I’ll discuss another lost cause: noninformative priors.
I like to say that noninformative priors are the perpetual motion machines of statistics. Everyone wants one but they don’t exist.
By definition, a prior represents information. So it should come as no surprise that a prior cannot represent lack of information.
The first “noninformative prior” was of course the flat prior. The major flaw with this prior is lack of invariance: if it is flat in one parameterization it will not be flat in most other parameterizations. Flat prior have lots of other problems too. See my earlier post here.
The most famous noninformative prior (I’ll stop putting quotes around the phrase from now on) is Jeffreys prior which is proportional to the square root of the determinant of the Fisher information matrix. While this prior is invariant, it can still have undesirable properties. In particular, while it may seem noninformative for a parameter ${\theta}$ it can end up being highly informative for functions of ${\theta}$. For example, suppose that ${Y}$ is multivariate Normal with mean vector ${\theta}$ and identity covariance. The Jeffreys prior is the flat prior ${\pi(\theta) \propto 1}$. Now suppose that we want to infer ${\psi = \sum_j \theta_j^2}$. The resulting posterior for ${\psi}$ is a disaster. The coverage of the Bayesian ${1-\alpha}$ posterior interval can be close to 0.
This is a general problem with noninformative priors. If ${\pi(\theta)}$ is somehow noninformative for ${\theta}$, it may still be highly informative for sub-parameters, that is for functions ${\psi = g(\theta)}$ where ${\theta\in \mathbb{R}^d}$ and ${\psi: \mathbb{R}^d \rightarrow \mathbb{R}}$.
Jim Berger and Jose Bernardo wrote a series of interesting papers about priors that were targeted to be noninformative for particular functions of ${\theta}$. These are often called reference priors. But what if you are interested in many functions of ${\theta}$. Should you use a different prior for each function of interest?
A more fundamental question is: what does it mean for a prior to be noninformative? Of course, people have argued about this for many, many years. One definition, which has the virtue of being somewhat precise, is that a prior is noninformative if the ${1-\alpha}$ posterior regions have frequentist coverage equal (approximately) to ${1-\alpha}$. These are sometimes called “matching priors.”
In general, it is hard to construct matching priors especially in high-dimensional complex models. But matching priors raise a fundamental question: if your goal is to match frequentist coverage, why bother with Bayes at all? Just use a frequentist confidence interval.
These days I think that most people agree that the virtue of Bayesian methods is that it gives you a way to include prior information in a systematic way. There is no reason to formulate a “noninformative prior.”
On the other hand, in practice, we often deal with very complex, high-dimensional models. Can we really formulate a meaningful informative prior in such problems? And if we do, will anyone care about our inferences?
In 1996, I wrote a review paper with Rob Kass on noninformative priors (Kass and Wasserman 1996). We emphasized that a better term might be “default prior” since that seems more honest and promises less. One of our conclusions was:
“We conclude that the problems raised by the research on priors chosen by formal rules are serious and may not be dismissed lightly: When sample sizes are small (relative the number of parameters being estimated), it is dangerous to put faith in any default solution; but when asymptotics take over, Jeffreys’s rules and their variants remain reasonable choices.”
Looking at this almost twenty years later, the one thing that has changed is the “the number of parameters being estimated” which these days is often very, very large.
My conclusion: noninformative priors are a lost cause.
Reference
Kass, Robert E and Wasserman, Larry. (1996). The selection of prior distributions by formal rules. Journal of the American Statistical Association, 91, 1343-1370. | {"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": 55, "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": 0.8103100061416626, "perplexity": 830.4658743216253}, "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/1656104209449.64/warc/CC-MAIN-20220703013155-20220703043155-00643.warc.gz"} |
https://www.physicsforums.com/threads/physics-help-unsure-where-i-went-wrong.307362/ | # Physics help- unsure where I went wrong
1. Apr 14, 2009
### cpat
1. A thin, uniform, metal bar, 2.5 m long and weighing 90 N, is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.3 m below the ceiling by a small 4-kg ball, initially traveling horizontally at 14 m/s. The ball rebounds in the opposite direction with a speed of 7 m/s.
Part A. Find the angular speed of the bar just after the collision.
I tried to solve using conservation of angular momentum (it must be a calculation error but I keep getting the same answer):
m1v_0d = -m1vd +(1/3)m2L^2(omega)
(4.0kg)(14m/s)(1.3m) = -(3kg)(7m/s)(1.3m)+(1/3)(90N/9.82m/s)(2.5m)^2(omega)
72.8 = 57.398 + (-27.3)
1001.1 = 57.398
2. A Ball Rolling Uphill. A bowling ball rolls without slipping up a ramp that slopes upward at an angle beta to the horizontal. Treat the ball as a uniform, solid sphere, ignoring the finger holes.
Part A. What is the acceleration of the center of mass of the ball?
I tried drawing it out, f = mew N, N = wsin(beta)
- (WsinF+ muWcos(beta) = macm
I entered acm = -g(sinF+mucos (beta)
It was incorrect, so I thought I over-did it, and tried just -g(sin(beta)), but that was also incorrect, I'm not sure where to go from here.
Part B. What minimum coefficient of static friction is needed to prevent slipping?
I tried mu = tan (beta)
2. Apr 14, 2009
### LowlyPion
OK
You have
|r X P| = r*m*vi = r*m*(-vf) + I*ω
So r*m*(vi + vf) = 1.3*4*(14 + 7) = 1.3*4*21 = 109.2 = 1/3*90/9.8*L2
So then ...
ω = 109.2*9.8*3/(90*2.5*2.5) = ... ?
3. Apr 14, 2009
### cpat
thanks!- I've got a calculator now (much easier! Mine stopped working last week so I've been borrowing from a friend), 5.70752.
Any tips on the next question?
4. Apr 14, 2009
### LowlyPion
For 2) consider the torque about the center of the solid sphere. That might yield "a" angular acceleration which you can then relate to linear acceleration along the incline.
Similar Discussions: Physics help- unsure where I went wrong | {"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": 0.8545710444450378, "perplexity": 2134.2382891688517}, "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/1518891817523.0/warc/CC-MAIN-20180225225657-20180226005657-00147.warc.gz"} |
https://www.ocean-sci.net/6/361/2010/ | Journal cover Journal topic
Ocean Science An interactive open-access journal of the European Geosciences Union
Journal topic
• IF 2.539
• IF 5-year
3.129
• CiteScore
2.78
• SNIP 1.217
• IPP 2.62
• SJR 1.370
• Scimago H
index 48
• h5-index 32
# Abstracted/indexed
Abstracted/indexed
Ocean Sci., 6, 361–378, 2010
https://doi.org/10.5194/os-6-361-2010
Special issue: Thermophysical properties of seawater
Ocean Sci., 6, 361–378, 2010
https://doi.org/10.5194/os-6-361-2010
18 Mar 2010
18 Mar 2010
# A model for predicting changes in the electrical conductivity, practical salinity, and absolute salinity of seawater due to variations in relative chemical composition
R. Pawlowicz R. Pawlowicz
• Dept. of Earth and Ocean Sciences, University of British Columbia, Canada
Abstract. Salinity determination in seawater has been carried out for almost 30 years using the Practical Salinity Scale 1978. However, the numerical value of so-called practical salinity, computed from electrical conductivity, differs slightly from the true or absolute salinity, defined as the mass of dissolved solids per unit mass of seawater. The difference arises because more recent knowledge about the composition of seawater is not reflected in the definition of practical salinity, which was chosen to maintain historical continuity with previous measures, and because of spatial and temporal variations in the relative composition of seawater. Accounting for these spatial variations in density calculations requires the calculation of a correction factor δSA, which is known to range from 0 to 0.03 g kg−1 in the world oceans. Here a mathematical model relating compositional perturbations to δSA is developed, by combining a chemical model for the composition of seawater with a mathematical model for predicting the conductivity of multi-component aqueous solutions. Model calculations for this estimate of δSA, denoted δSRsoln, generally agree with estimates of δSA based on fits to direct density measurements, denoted δSRdens, and show that biogeochemical perturbations affect conductivity only weakly. However, small systematic differences between model and density-based estimates remain. These may arise for several reasons, including uncertainty about the biogeochemical processes involved in the increase in Total Alkalinity in the North Pacific, uncertainty in the carbon content of IAPSO standard seawater, and uncertainty about the haline contraction coefficient for the constituents involved in biogeochemical processes. This model may then be important in constraining these processes, as well as in future efforts to improve parameterizations for δSA.
Special issue | {"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": 0.8798269033432007, "perplexity": 3645.0712945861414}, "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/1568514573570.6/warc/CC-MAIN-20190919183843-20190919205843-00411.warc.gz"} |
https://arxiv.org/abs/1602.03076 | math.CV
(what is this?)
# Title: Hole probability for zeroes of Gaussian Taylor series with finite radii of convergence
Abstract: We study a family of random Taylor series $$F(z) = \sum_{n\ge 0} \zeta_n a_n z^n$$ with radius of convergence almost surely $1$ and independent identically distributed complex Gaussian coefficients $(\zeta_n)$; these Taylor series are distinguished by the invariance of their zero sets with respect to isometries of the unit disk. We find reasonably tight upper and lower bounds on the probability that $F$ does not vanish in the disk $\{|z|\le r\}$ as $r\uparrow 1$. Our bounds take different forms according to whether the non-random coefficients $(a_n)$ grow, decay or remain of the same order. The results apply more generally to a class of Gaussian Taylor series whose coefficients $(a_n)$ display power-law behavior.
Comments: 42 pages, new aproach taken in Section 7 Subjects: Complex Variables (math.CV); Probability (math.PR) Cite as: arXiv:1602.03076 [math.CV] (or arXiv:1602.03076v2 [math.CV] for this version)
## Submission history
From: Jeremiah Buckley [view email]
[v1] Tue, 9 Feb 2016 17:11:37 GMT (30kb)
[v2] Wed, 15 Mar 2017 10:11:08 GMT (31kb) | {"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": 0.8449584245681763, "perplexity": 1238.8739978850583}, "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/1512948514238.17/warc/CC-MAIN-20171212002021-20171212022021-00439.warc.gz"} |
https://www.math-stockholm.se/en/kalender/mlseminarium/erik-lindell-postdoc-day-seminar-3-abelian-cycles-in-the-homology-of-the-torelli-group-1.1155586 | Abstract: The mapping class group of a compact and orientable surface of genus g has an important subgroup called the Torelli group, which is the kernel of the action on the homology of the surface. In this talk we will discuss the stable rational homology of the Torelli group of a surface with a boundary component, about which very little is known in general. These homology groups are representations of the arithmetic group $$\mathrm{Sp}_{2g}(\mathbb{Z})$$ and we study them using an $$\mathrm{Sp}_{2g}(\mathbb{Z})$$-equivariant map induced on homology by the so-called Johnson homomorphism. The image of this map is a finite dimensional and algebraic representation of $$\mathrm{Sp}_{2g}(\mathbb{Z})$$. By considering a type of homology classes called abelian cycles, which are easy to write down for Torelli groups and for which we can derive an explicit formula for the map in question, we may use classical representation theory of symplectic groups to describe a large part of the image. | {"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": 0.9538005590438843, "perplexity": 99.31984423373986}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296944606.5/warc/CC-MAIN-20230323003026-20230323033026-00437.warc.gz"} |
http://patch.typepad.com/minor_details/2008/03/latex-table-col.html | ## March 12, 2008
Thank you, this was very helpful!
Glad to have helped! Thanks for the comment.
Thank you! Your method made the tables in my thesis much better.
Hello and Thanks you very much for this small and really effective tutorial.
However I have a small question, what is if i want a border line on the left/right of the table?
I tried changing the header to
\begin{tabular}{|l p{.5cm} p{4cm} p{.5cm} l|},
but I got some blank on the borders,
any idea how to fix it?
I would appreciate.
thanks once more
Piccolo
Piccolo, the gaps that you see in the vertical borders are by products of the \smallskip element and the double \\ delimiters, which add a blank line. What I would recommend doing is removing those and adding a horizontal line instead.
Thanks,
It helps.
Regards
There is a much cleaner method of doing this, and it took me forever to find this too: you can call ragged right in the column deceleration:
\begin{tabular}{| >{\raggedright}p{.5cm} | >{raggedright}p{4cm} | >{\centering}p{.5cm} | }
This will work with a variety of commands, in this example the left two columns will typeset with ragged right, and the right column will center. This is for every column in the table. Then you can ditch the parbox.
wow. many thanks, you just saved me the exact same headache... | {"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": 0.9417803287506104, "perplexity": 1832.4852476162614}, "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-2014-35/segments/1408500822053.47/warc/CC-MAIN-20140820021342-00185-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://byjus.com/maths/maxima-and-minima/ | # Local Maxima And Minima
Maxima and Minima are one of the most common concepts in differential calculus. A branch of Mathematics called “Calculus of Variations” deals with the maxima and the minima of the functional. The calculus of variations is concerned with the variations in the functional, in which small change in the function leads to the change in the functional value.
The first variation is defined as the linear part of the change in the functional, and the second part of the variation is defined in the quadratic part. Functional is expressed as the definite integrals which involve the functions and their derivatives.
The functions that maximize or minimize the functional are can be found using the Euler – Lagrange of the calculus of variations. These two Latin maxima and minima words basically mean the maximum and minimum value of a function respectively, which is quite evident. The maxima and minima are collectively called “Extrema”. Here, we assume our function to be continuous for its entire domain. Before knowing how to find maxima and minima, we should first learn about derivatives. Assuming that you all know how to find derivatives, let us go ahead and learn about some curves. What are the curves?
Also, learn:
## What are the Curves?
Figure 1: Curves
Figure 2: Value of a Function
A curve is defined as one-dimensional continuum. In figure 1, that curve is graph of a function
$$\begin{array}{l} f ~in~ x \end{array}$$
.
$$\begin{array}{l} f(x) \end{array}$$
represents the value of function at
$$\begin{array}{l} x \end{array}$$
. The value of
$$\begin{array}{l} f \end{array}$$
when
$$\begin{array}{l} x = a \end{array}$$
, will be
$$\begin{array}{l} f(a) \end{array}$$
. Similarly, for
$$\begin{array}{l} B,~ C ~and ~D\end{array}$$
. You can refer fig. 2 to understand this. From the figure it is quite clear that the value of the given function has its maximum value at x=b, i.e.
$$\begin{array}{l} f(b) \end{array}$$
.
Interval of a function plays a very important role to find extreme values of a function. If the interval for which the function
$$\begin{array}{l} f \end{array}$$
is defined in
$$\begin{array}{l} R \end{array}$$
, then we can’t talk about maxima and minima of
$$\begin{array}{l} f \end{array}$$
. We can understand it logically that though
$$\begin{array}{l} f(b) \end{array}$$
appears to have the maximum value, we can’t be sure it has the largest value till we have seen the graph for its entire domain.
## Local Maxima and Minima
We may not be able to tell whether
$$\begin{array}{l} f(b) \end{array}$$
is the maximum value of
$$\begin{array}{l} f \end{array}$$
, but we can give some credit to point . We can do this by declaring
$$\begin{array}{l} B \end{array}$$
as the local maximum for function
$$\begin{array}{l} f \end{array}$$
. These are also called relative maxima and minima. These local maxima and minima are defined as:
• If
$$\begin{array}{l} f(a) \leq f(x) \end{array}$$
for all
$$\begin{array}{l} x \end{array}$$
in
$$\begin{array}{l} P’s\end{array}$$
neighborhood (within the distance nearby
$$\begin{array}{l} P \end{array}$$
, where
$$\begin{array}{l} x = a \end{array}$$
),
$$\begin{array}{l} f \end{array}$$
is said to have a local minimum at
$$\begin{array}{l} x = a \end{array}$$
.
• If
$$\begin{array}{l} f(a) \geq f(x) \end{array}$$
for all in
$$\begin{array}{l} P’s \end{array}$$
neighborhood (within the distance nearby
$$\begin{array}{l} P \end{array}$$
, where
$$\begin{array}{l} x = a \end{array}$$
),
$$\begin{array}{l} f \end{array}$$
is said to have a local maximum at
$$\begin{array}{l} x = a \end{array}$$
.
In the above example,
$$\begin{array}{l} B ~and~ D \end{array}$$
are local maxima and
$$\begin{array}{l} A ~and~ C\end{array}$$
are local minima. Local maxima and minima are together referred to as Local extreme.
Let us now take a point
$$\begin{array}{l} P \end{array}$$
, where
$$\begin{array}{l} x = a\end{array}$$
and try to analyze the nature of the derivatives. There are total of four possibilities:
• If
$$\begin{array}{l} f'(a) = 0 \end{array}$$
, the tangent drawn is parallel to
$$\begin{array}{l} x -axis \end{array}$$
, i.e. slope is zero. There are three possible cases:
• The value of
$$\begin{array}{l} f \end{array}$$
, when compared to the value of
$$\begin{array}{l} f \end{array}$$
at
$$\begin{array}{l} P \end{array}$$
, increases if you move towards right or left of
$$\begin{array}{l} P \end{array}$$
(Local minima: look like valleys)
• The value of
$$\begin{array}{l} f \end{array}$$
, when compared to the value of
$$\begin{array}{l} f \end{array}$$
at
$$\begin{array}{l} P \end{array}$$
, decreases if you move towards right or left of
$$\begin{array}{l} P \end{array}$$
(Local maxima: look like hills)
• The value of
$$\begin{array}{l} f \end{array}$$
, when compared to the value of
$$\begin{array}{l} f \end{array}$$
at
$$\begin{array}{l} P \end{array}$$
, increases and decreases as you move towards left and right respectively of
$$\begin{array}{l} P \end{array}$$
(Neither: looks like a flat land)
• If, the tangent is drawn at a negative slope. The value of f'(a), at p, increases if you move towards left of and decreases if you move towards the right of . So, in this case, also, we can’t find any local extrema.
• If, the tangent is drawn at a positive slope. The value of f'(a), at P, increases if you move towards the right of and decreases if you move towards left of . So, in this case, we can’t find any local extrema.
• $$\begin{array}{l} f’ \end{array}$$
doesn’t exist at point
$$\begin{array}{l} P \end{array}$$
, i.e. the function is not differentiable at
$$\begin{array}{l} P \end{array}$$
. This normally happens when the graph of
$$\begin{array}{l} f \end{array}$$
has a sharp corner somewhere. All the three cases discussed in the previous point also hold true for this point.
To remember this, you can refer the Table 1.
Table 1: Various possibilities of derivatives of a function
Nature of f'(a) Nature of Slope Example Local Extremum f'(a) > 0 Positive Neither f'(a) < 0 Negative Neither f'(a) = 0 Zero Local Minimum Local Maximum Neither Not Defined Not Defined Local Minimum Local Maximum Neither
### What is Critical Point?
In mathematics, a Critical point of a differential function of a real or complex variable is any value in its domain where its derivative is 0. We can hence infer from here that every local extremum is a critical point but every critical point need not be a local extremum. So, if we have a function which is continuous, it must have maxima and minima or local extrema. This means that every such function will have critical points. In case the given function is monotonic, the maximum and minimum values lie at the endpoints of the domain of the definition of that particular function.
Maxima and minima are hence very important concepts in the calculus of variations, which helps to find the extreme values of a function. You can use these two values and where they occur for a function using the first derivative method or the second derivative method.
To know and learn new concepts every day, download BYJU’S – The Learning App. | {"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": 0.8912473917007446, "perplexity": 378.5638581129413}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663012542.85/warc/CC-MAIN-20220528031224-20220528061224-00761.warc.gz"} |
https://wikieducator.org/VirtualMV/Research/Academic_writing | ## Introduction
### Overview
By the end of this section you will be able to: Understand what is meant by academic writing | {"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": 0.9279474020004272, "perplexity": 745.862660188355}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104189587.61/warc/CC-MAIN-20220702162147-20220702192147-00327.warc.gz"} |
http://www.physicsimplified.com/2013/09/einsteins-postulates-of-special-theory.html | ## Pages
### S6. Einstein's Postulates of Special theory of Relativity
There was no absolute standard of rest. There exists a relativity principle that applies to all of physics, but it's not Galilean relativity. Laws of mechanics must be revised. Perhaps laws of electrodynamics can survive intact.”
In the year 1905, Albert Einstein published the theory of special relativity.
Einstein explained that when two objects are moving at a constant speed, relative motion between the two objects was essential, instead of appealing to the ether as an absolute frame of reference that defined what was going on. For instance, if you and some astronaut, Amber, are moving in different spaceships and want to compare your observations, all that matters is how fast you and Amber are moving with respect to each other and nothing more.
What Einstein does is to extend the concept of the old Galilean relativity to include light too. He asserts that when this be done the laws of mechanics has to be modified. We must note here, that this holds only as long as the system being considered is moving with relative velocity. We thus have the special case (hence the name Special Relativity) .
The motion it explains is only if you’re travelling in a straight line at a constant speed. As soon as you accelerate or curve — or do anything that changes the nature of the motion in any way — special relativity ceases to apply. That’s where Einstein’s general theory of relativity comes in, because it can explain the general case of any sort of motion.
My apologies that, I presented the notes on basics of General Theory of Relativity first.
Anyway, Einstein’s theory has two key postulates:
The principle of relativity:
The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems in uniform translatory motion relative to each other.
The principle of the speed of light:
· As measured in any inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body.
The genius of Einstein’s discoveries is that he looked at the experiments and assumed the findings were true. This was the exact opposite of what other physicists seemed to be doing. Instead of assuming the theory was correct and that the experiments failed, he assumed that the experiments were correct and the theory had failed.
The belief in ether had caused a mess of things. In the latter part of the 19th century, physicists were searching for the mysterious thing called ether — the medium they believed existed for light waves to wave through. However, Einstein just removed the ether entirely and assumed that the laws of physics, including the speed of light, worked the same regardless of how you were moving and that was the genius of the Einstein.
In the next notes we will be discussion on the concept of the LorentzTransformation. | {"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": 0.923826277256012, "perplexity": 381.22871034500474}, "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-2021-43/segments/1634323588341.58/warc/CC-MAIN-20211028131628-20211028161628-00399.warc.gz"} |
https://searxiv.org/search?author=Xinzhong%20Er | ### Results for "Xinzhong Er"
total 250took 0.12s
Gravitational lensing properties of isothermal universal halo profileDec 17 2012N-body simulations predict that dark matter halos with different mass scales are described by a universal model, the Navarro-Frenk-White (NFW) density profiles. As a consequence of baryonic cooling effects, the halos will become more concentrated, and ... More
Measurement of differential magnificationAug 11 2014In gravitational lensing, the magnification effect changes the luminosity and size of a background galaxy. If the image sizes are not small compared to the scale over which the magnification and shear vary, higher-order distortions occur which are termed ... More
Estimate of dark halo ellipticity by lensing flexionSep 16 2010Aims. The predictions of the ellipticity of the dark matter halos from models of structure formation are notoriously difficult to test with observations. A direct measurement would give important constraints on the formation of galaxies, and its effect ... More
Effects of plasma on gravitational lensingOct 22 2013Nov 06 2013We study gravitational lensing when plasma surrounds the lens. An extra deflection angle is induced by the plasma in addition to the deflection generated by gravity. An inhomogeneous plasma distribution generates a greater effect than a homogeneous one, ... More
Increasing the Lensing Figure of Merit through Higher Order Convergence MomentsFeb 08 2018The unprecedented quality, the increased dataset, and the wide area of ongoing and near future weak lensing surveys allows to move beyond the standard two points statistics thus making worthwhile to investigate higher order probes. As an interesting step ... More
Estimation of halo ellipticity using spin-3 flexionSep 14 2012Estimating the ellipticity of dark matter haloes at the galaxy or galaxy-cluster scale can provide important constraints on the formation of galaxies or clusters, as well as on the nature of dark matter. We show in this paper that the spin-3 gravitational ... More
Weak lensing goes bananas: What flexion really measuresSep 07 2007In weak gravitational lensing, the image distortion caused by shear measures the projected tidal gravitational field of the deflecting mass distribution. To lowest order, the shear is proportional to the mean image ellipticity. If the image sizes are ... More
Mass reconstruction by gravitational shear and flexionAug 18 2010Galaxy clusters are considered as excellent probes for cosmology. For that purpose, their mass needs to be measured and their structural properties needs to be understood. We propose a method for galaxy cluster mass reconstruction which combines information ... More
Measurement of halo properties with weak lensing shear and flexionDec 20 2011We constrain properties of cluster haloes by performing likelihood analysis using lensing shear and flexion data. We test our analysis using two mock cluster haloes: an isothermal ellipsoid (SIE) model and a more realistic elliptical Navarro-Frenk-White ... More
Biases in physical parameter estimates through differential lensing magnificationMar 21 2013We study the lensing magnification effect on background galaxies. Differential magnification due to different magnifications of different source regions of a galaxy will change the lensed composite spectra. The derived properties of the background galaxies ... More
An exact analytic spectrum of relic gravitational waves in an accelerating universeApr 21 2006Nov 28 2008An exact analytic calculation is presented for the spectrum of relic gravitational waves in the scenario of accelerating Universe $\Omega_{\Lambda}+\Omega_m = 1$. The spectrum formula contains explicitly the parameters of acceleration, inflation, reheating, ... More
Gravitational lensing effects on sub-millimetre galaxy countsJan 03 2013We study the effects on the number counts of sub-millimetre galaxies due to gravitational lensing. We explore the effects on the magnification cross section due to halo density profiles, ellipticity and cosmological parameter (the power-spectrum normalisation ... More
Estimate of halo ellipticity as a function of radius with flexionsJul 07 2011The cold dark matter theory predicts triaxial dark matter haloes. The radial distribution of halo ellipticity depends on baryonic processes and the nature of dark matter particles (collisionless or collisional). Here we show that we can use lensing flexion ... More
The Radiative Stress TensorMar 31 2000We use the transfer equation in relativistic form to develop an expansion of the one-photon distribution for a medium with constant photon mean free path, $\epsilon$. The resulting radiative viscosity tensor may be expressed either as a simple integral ... More
Cosmic tidal reconstructionNov 15 2015May 09 2016The gravitational coupling of a long-wavelength tidal field with small-scale density fluctuations leads to anisotropic distortions of the locally measured small-scale matter correlation function. Since the local correlation function is known to be statistically ... More
Causal Relativistic Fluid DynamicsJul 01 2011We derive causal relativistic fluid dynamical equations from the relaxation model of kinetic theory as in a procedure previously applied in the case of non-relativistic rarefied gases. By treating space and time on an equal footing and avoiding the iterative ... More
Improving lensing cluster mass estimate with flexionJul 20 2016Gravitational lensing has long been considered as a valuable tool to determine the total mass of galaxy clusters. The shear profile as inferred from the statistics of ellipticity of background galaxies allows to probe the cluster intermediate and outer ... More
Large Size Scatter of Passively Evolving Lensed Galaxies at z~2 in CLASHDec 30 2012In a systematic search over 11 cluster fields from Cluster Lensing And Supernova survey with Hubble (CLASH) we identify ten passively evolving massive galaxies at redshift z~2.We derive the stellar properties of these galaxies using HST WFC3/ACS multiband ... More
Calibration of colour gradient bias in shear measurement using HST/CANDELS dataAug 21 2017Apr 19 2018Accurate shape measurements are essential to infer cosmological parameters from large area weak gravitational lensing studies. The compact diffraction-limited point-spread function (PSF) in space-based observations is greatly beneficial, but its chromaticity ... More
A Continuum Description of Rarefied Gas Dynamics (III)--- The Structures of Shock WavesMay 20 2001We use the one-dimensional steady version of the equations derived in paper I to compute the structure of shock waves. The agreement with experiment is good, especially when we retain the experimental value of the Prandtl number adopted in II.
A Continuum Description of Rarefied Gas Dynamics (I)--- Derivation From Kinetic TheoryMay 20 2001We describe an asymptotic procedure for deriving continuum equations from the kinetic theory of a simple gas. As in the works of Hilbert, of Chapman and of Enskog, we expand in the mean flight time of the constituent particles of the gas, but we do not ... More
An Efficient Watermarking Algorithm to Improve Payload and Robustness without Affecting Image Perceptual QualityApr 26 2010Capacity, Robustness, & Perceptual quality of watermark data are very important issues to be considered. A lot of research is going on to increase these parameters for watermarking of the digital images, as there is always a tradeoff among them. . In ... More
Effect of Crosstalk on Permutation in Optical Multistage Interconnection NetworksApr 26 2010Optical MINs hold great promise and have advantages over their electronic networks.they also hold their own challenges. More research has been done on Electronic Multistage Interconnection Networks, (EMINs) but these days optical communication is a good ... More
Mobile Zigbee Sensor NetworksApr 26 2010OPNET Modeler accelerates network R&D and improves product quality through high-fidelity modeling and scalable simulation. It provides a virtual environment for designing protocols and devices, and for testing and demonstrating designs in realistic scenarios ... More
The Advantage of Rightmost Ordering for gamma5 in Dimensional RegularizationMay 10 2009We propose a gamma5 scheme in dimensional regularization by analytically continuing the dimension after all the gamma5 matrices have been moved to the rightmost position. All Feynman amplitudes corresponding to diagrams with no fermion loops regulated ... More
An Immersive Telepresence System using RGB-D Sensors and Head Mounted DisplayNov 21 2015We present a tele-immersive system that enables people to interact with each other in a virtual world using body gestures in addition to verbal communication. Beyond the obvious applications, including general online conversations and gaming, we hypothesize ... More
Gauge Invariant Treatment of $γ_{5}$ in the Scheme of 't Hooft and VeltmanMay 11 2009Nov 29 2010We propose moving all the $\gamma_{5}$ matrices to the rightmost position before continuing the dimension, and show that this simple prescription will enable the dimension regularization scheme proposed by 't Hooft and Veltman to be consistent with gauge ... More
Maintaining Gauge Symmetry in Renormalizing Chiral Gauge TheoriesDec 16 2010It is known that the $\gamma_{5}$ scheme of Breitenlohner and Maison (BM) in dimensional regularization requires finite counter-term renormalization to restore gauge symmetry and implementing this finite renormalization in practical calculation is a daunting ... More
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https://openstax.org/books/chemistry/pages/chapter-6 | Chemistry
# Chapter 6
ChemistryChapter 6
1.
The spectrum consists of colored lines, at least one of which (probably the brightest) is red.
3.
3.15 m
5.
3.233 $××$ 10−19 J; 2.018 eV
7.
ν = 4.568 $××$ 1014 s; λ = 656.3 nm; Energy mol−1 = 1.823 $××$ 105 J mol−1; red
9.
(a) λ = 8.69 $××$ 10−7 m; E = 2.29 $××$ 10−19 J; (b) λ = 4.59 $××$ 10−7 m; E = 4.33 $××$ 10−19 J; The color of (a) is red; (b) is blue.
11.
E = 9.502 $××$ 10−15 J; ν = 1.434 $××$ 1019 s−1
13.
Red: 660 nm; 4.54 $××$ 1014 Hz; 3.01 $××$ 10−19 J. Green: 520 nm; 5.77 $××$ 1014 Hz; 3.82 $××$ 10−19 J. Blue: 440 nm; 6.81 $××$ 1014 Hz; 4.51 $××$ 10−19 J. Somewhat different numbers are also possible.
15.
5.49 $××$ 1014 s−1; no
17.
Quantized energy means that the electrons can possess only certain discrete energy values; values between those quantized values are not permitted.
19.
$E=E2−E5=2.179×10−18(1n22−1n52)J=2.179×10−18(122−152)=4.576×10−19J=4.576×10−19J1.602×10−19JeV−1=2.856eVE=E2−E5=2.179×10−18(1n22−1n52)J=2.179×10−18(122−152)=4.576×10−19J=4.576×10−19J1.602×10−19JeV−1=2.856eV$
21.
−8.716 $××$ 10−18 J
23.
−3.405 $××$ 10−20 J
25.
33.9 Å
27.
1.471 $××$ 10−17 J
29.
Both involve a relatively heavy nucleus with electrons moving around it, although strictly speaking, the Bohr model works only for one-electron atoms or ions. According to classical mechanics, the Rutherford model predicts a miniature “solar system” with electrons moving about the nucleus in circular or elliptical orbits that are confined to planes. If the requirements of classical electromagnetic theory that electrons in such orbits would emit electromagnetic radiation are ignored, such atoms would be stable, having constant energy and angular momentum, but would not emit any visible light (contrary to observation). If classical electromagnetic theory is applied, then the Rutherford atom would emit electromagnetic radiation of continually increasing frequency (contrary to the observed discrete spectra), thereby losing energy until the atom collapsed in an absurdly short time (contrary to the observed long-term stability of atoms). The Bohr model retains the classical mechanics view of circular orbits confined to planes having constant energy and angular momentum, but restricts these to quantized values dependent on a single quantum number, n. The orbiting electron in Bohr’s model is assumed not to emit any electromagnetic radiation while moving about the nucleus in its stationary orbits, but the atom can emit or absorb electromagnetic radiation when the electron changes from one orbit to another. Because of the quantized orbits, such “quantum jumps” will produce discrete spectra, in agreement with observations.
31.
Both models have a central positively charged nucleus with electrons moving about the nucleus in accordance with the Coulomb electrostatic potential. The Bohr model assumes that the electrons move in circular orbits that have quantized energies, angular momentum, and radii that are specified by a single quantum number, n = 1, 2, 3, …, but this quantization is an ad hoc assumption made by Bohr to incorporate quantization into an essentially classical mechanics description of the atom. Bohr also assumed that electrons orbiting the nucleus normally do not emit or absorb electromagnetic radiation, but do so when the electron switches to a different orbit. In the quantum mechanical model, the electrons do not move in precise orbits (such orbits violate the Heisenberg uncertainty principle) and, instead, a probabilistic interpretation of the electron’s position at any given instant is used, with a mathematical function ψ called a wavefunction that can be used to determine the electron’s spatial probability distribution. These wavefunctions, or orbitals, are three-dimensional stationary waves that can be specified by three quantum numbers that arise naturally from their underlying mathematics (no ad hoc assumptions required): the principal quantum number, n (the same one used by Bohr), which specifies shells such that orbitals having the same n all have the same energy and approximately the same spatial extent; the angular momentum quantum number l, which is a measure of the orbital’s angular momentum and corresponds to the orbitals’ general shapes, as well as specifying subshells such that orbitals having the same l (and n) all have the same energy; and the orientation quantum number m, which is a measure of the z component of the angular momentum and corresponds to the orientations of the orbitals. The Bohr model gives the same expression for the energy as the quantum mechanical expression and, hence, both properly account for hydrogen’s discrete spectrum (an example of getting the right answers for the wrong reasons, something that many chemistry students can sympathize with), but gives the wrong expression for the angular momentum (Bohr orbits necessarily all have non-zero angular momentum, but some quantum orbitals [s orbitals] can have zero angular momentum).
33.
n determines the general range for the value of energy and the probable distances that the electron can be from the nucleus. l determines the shape of the orbital. m1 determines the orientation of the orbitals of the same l value with respect to one another. ms determines the spin of an electron.
35.
(a) 2p; (b) 4d; (c) 6s
37.
(a) 3d; (b) 1s; (c) 4f
39.
41.
(a) x. 2, y. 2, z. 2; (b) x. 1, y. 3, z. 0; (c) x. 4 0 0 $12,12,$ y. 2 1 0 $12,12,$ z. 3 2 0 $12;12;$ (d) x. 1, y. 2, z. 3; (e) x. l = 0, ml = 0, y. l = 1, ml = –1, 0, or +1, z. l = 2, ml = –2, –1, 0, +1, +2
43.
12
45.
n l ml s
4 0 0 $+12+12$
4 0 0 $−12−12$
4 1 −1 $+12+12$
4 1 0 $+12+12$
4 1 +1 $+12+12$
4 1 −1 $−12−12$
47.
For example, Na+: 1s22s22p6; Ca2+: 1s22s22p6; Sn2+: 1s22s22p63s23p63d104s24p64d105s2; F: 1s22s22p6; O2–: 1s22s22p6; Cl: 1s22s22p63s23p6.
49.
(a) 1s22s22p3; (b) 1s22s22p63s23p2; (c) 1s22s22p63s23p64s23d6; (d) 1s22s22p63s23p64s23d104p65s24d105p4; (e) 1s22s22p63s23p64s23d104p65s24d105p66s24f9
51.
The charge on the ion.
53.
(a)
(b)
(c)
(d)
(e)
55.
Zr
57.
Rb+, Se2−
59.
Although both (b) and (c) are correct, (e) encompasses both and is the best answer.
61.
K
63.
1s22s22p63s23p63d104s24p64d105s25p66s24f145d10
65.
Co has 27 protons, 27 electrons, and 33 neutrons: 1s22s22p63s23p64s23d7.
I has 53 protons, 53 electrons, and 78 neutrons: 1s22s22p63s23p63d104s24p64d105s25p5.
67.
Cl
69.
O
71.
Rb < Li < N < F
73.
15 (5A)
75.
Mg < Ca < Rb < Cs
77.
Si4+ < Al3+ < Ca2+ < K+
79.
Se, As
81.
Mg2+ < K+ < Br < As3–
83.
O, IE1
85.
Ra | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 29, "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": 0.8906996846199036, "perplexity": 1668.986245833523}, "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/1606141672314.55/warc/CC-MAIN-20201201074047-20201201104047-00532.warc.gz"} |
https://books.compclassnotes.com/rothphys110-2e/2021/06/29/section-9-2-v2/ | # Chapter 9: Force
## 9.2 Newton’s laws
In the late 1600s and early 1700s, Isaac Newton published the Principia Mathematica in which he built upon discoveries of scientists (at the time called “natural philosophers”) before him, laying the foundation of classical mechanics. In the Principia, he described three laws of motion:
1. An object’s momentum will only change if the object is acted on by some force. This is known as the principle of inertia.
2. The total force acting on an object is equal to the rate of change of the object’s momentum: $$\vec{F}_\textit{net} = \frac{\Delta\vec{p}}{\Delta t} \tag{9.1}$$
3. For every force acting on an object, there is a reaction force equal in magnitude and opposite in direction acting on another object.
We have already discussed the first law in the context of energy; if you understand inertia, you can explain many phenomena you observe. The second law is a mathematical statement which we can use to quantitatively solve problems. The third law, as we’ll see later, is a result of conservation of momentum.
Let’s consider situations where an object’s mass remains constant (which is often the case), and apply Newton’s second law:
\begin{align} \vec{F}_\textit{net} &= \frac{\Delta\vec{p}}{\Delta t} \\ &= \frac{\vec{p}_f – \vec{p}_i}{\Delta t} \\ &= \frac{m\vec{v}_f – m\vec{v}_i}{\Delta t} \end{align}
Since the mass is not changing, we can factor it out:
\begin{align*} \vec{F}_\textit{net} &= \frac{m\left(\vec{v}_f – \vec{v}_i\right)}{\Delta t} \\ &= m\frac{\Delta\vec{v}}{\Delta t} \end{align*}
The rate of change of velocity is acceleration:
$\vec{F}_\textit{net} = m\vec{a} \tag{9.2}$
This is a specific case of Newton’s second law; it only applies when the mass of a system is not changing.
Remember that force, momentum and acceleration are all vector quantities. Breaking them up into components gives us:
\begin{align} F_\textit{net,x} &= \frac{\Delta p_x}{\Delta t} & F_\textit{net,y} &= \frac{\Delta p_y}{\Delta t} \tag{9.3} \\ F_\textit{net,x} &= ma_x & F_\textit{net,y} &= ma_y \tag{9.4} \end{align}
The unit of force is named the newton (N) in reference to Isaac Newton. Using Newton’s second law, we can break the newton down into base SI units: N = kg·m/s2. One newton is approximately the force required to push down one key on a computer keyboard.
#### Example 9.1
A 1380 kg car is moving due East with an initial speed of 27 m/s. After 8 s, the car has slowed down to 17 m/s. Find the net force acting on the car.
Let’s call due East the $$+x$$ direction. There are no $$y$$ components of the car’s motion. Since this is a one-dimensional problem, we simply use + and – signs to indicate direction instead of using vector notation.
\begin{align*} F_\textit{net,x} &= \frac{\Delta p_x}{\Delta t} \\ &= m\frac{\Delta v}{\Delta t} \\ &= m\frac{v_f – v_i}{\Delta t} \\ &= (1380\ \textrm{kg})\left(\frac{17 – 27\ \textrm{m/s}}{8\ \textrm{s}}\right) \\ &= -1725\ \textrm{N} \end{align*}
The negative value for the net force indicates that it is applied in the opposite direction as the car’s motion.
#### Example 9.2
##### Newton’s third law, examined
Two carts are on a nearly-frictionless track. Cart $$A$$ has a mass of 0.5 kg and is traveling towards cart $$B$$ with a speed of 2.0 m/s. Cart $$B$$ is initially at rest, and has a mass of 0.25 kg. The collision is perfectly inelastic. Determine the final velocity of the two carts, and the force that each cart exerts on the other.
We’ll start by analyzing conservation of momentum (revisit chapter 7 for a refresher) to determine the final velocity of the two carts. Let’s say cart $$A$$ is traveling in the $$+x$$ direction with speed $$v_i$$.
\begin{align*} p_{ix} &= p_{fx} \\ p_a &= p_f \\ m_av_i &= (m_a + m_b)v_f \\ \hookrightarrow v_f &= \frac{m_av_i}{m_a + m_b} \\ &= \frac{(0.5\ \textrm{kg})(2.0\ \textrm{m/s})}{(0.5 + 0.25)\ \textrm{kg}} \\ &= 1.3\ \textrm{m/s} \end{align*}
Now we’ll apply Newton’s second law to each cart individually. Let’s say the collision lasts for a time of 0.2 s. This is the period of time during which the carts’ velocities are changing, and the period of time that separates “initial” from “final” in our momentum analysis above. Note that $$v_{i,b} = 0$$ and $$v_f = 1.3\ \textrm{mps}$$ for both carts.
\begin{align*} \text{Cart A} & & \text{Cart B} & \\ F_\textit{on a} &= m_a\frac{\Delta p_a}{\Delta t} & F_\textit{on b} &= m_b\frac{\Delta p_b}{\Delta t} \\ &= m_a\frac{v_f – v_i}{\Delta t} & &= m_b\frac{v_f – 0}{\Delta t} \\ &= \left(0.5\ \textrm{kg}\right)\left(\frac{(1.3 – 2.0)\ \textrm{m/s}}{0.2\ \textrm{s}}\right) & &= \left(0.25\ \textrm{kg}\right)\left(\frac{1.3\ \textrm{m/s}}{0.2\ \textrm{s}}\right) \\ &= -1.7\ \textrm{N} & &= 1.7\ \textrm{N} \end{align*}
So, the force that cart $$B$$ exerted on cart $$A$$ has the exact same magnitude as the force that $$A$$ exerted on , but in the opposite direction—this is Newton’s third law!
As seen in the previous example, Newton’s third law is a direct consequence of conservation of momentum. During the collision, cart $$A$$ transferred some of its momentum to cart $$B$$; since momentum is conserved, cart $$B$$ gained the exact same amount of momentum that cart $$A$$ lost.
Speaking in terms of force, we say that cart $$A$$ exerted a force on cart $$B$$, which caused cart $$B$$’s momentum to change. Through this interaction, cart $$A$$’s momentum also changed. In order for this to happen, cart $$B$$ needed to exert a force on cart $$A$$. Since momentum is conserved, cart $$B$$ must have exerted the exact same amount of force on cart $$A$$ that cart $$A$$ exerted on cart $$B$$.
Remember: all forces describe interactions between two objects; all forces come in pairs. | {"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": 2, "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": 1.0000048875808716, "perplexity": 1114.9809305867086}, "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-20230329085436-20230329115436-00247.warc.gz"} |
https://www.semanticscholar.org/paper/Strong-Multiplicity-One-for-the-Selberg-Class-Soundararajan/4822267e1fc69d1fce329daf4dd5af34a532e845 | # Strong Multiplicity One for the Selberg Class
@article{Soundararajan2004StrongMO,
title={Strong Multiplicity One for the Selberg Class},
author={Kannan Soundararajan},
year={2004},
volume={47},
pages={468 - 474}
}
• K. Soundararajan
• Published 18 October 2002
• Mathematics
Abstract We investigate the problem of determining elements in the Selberg class by means of their Dirichlet series coefficients at primes.
### Strong multiplicity one for the Selberg class
We study the problem of determining elements of the Selberg class by information on the coefficents of the Dirichlet series at the squares of primes, or information about the zeroes of the functions.
### Distinguishing L-functions by joint universality
• J. Steuding
• Mathematics
Lithuanian Mathematical Journal
• 2021
In this note, we present results for distinguishing L -functions by their multisets of zeros and unique factorizations in an axiomatic setting; our tools stem from universality theory.
### Multiplicity one for $L$-functions and applications
• Mathematics
• 2013
We give conditions for when two Euler products are the same given that they satisfy a functional equation and their coefficients satisfy a partial Ramanujan bound and do not differ by too much.
### On a generalization of the Euler totient function
For a general polynomial Euler product F(s) we define the associated Euler totient function φ(n, F) and study its asymptotic properties. We prove that for F(s) belonging to certain subclass of the
### A refinement of strong multiplicity one for spectra of hyperbolic manifolds
Let $\calM_1$ and $\calM_2$ denote two compact hyperbolic manifolds. Assume that the multiplicities of eigenvalues of the Laplacian acting on $L^2(\calM_1)$ and $L^2(\calM_2)$ (respectively,
### Analytic monoids and factorization problems
The main goal of this paper is to initiate study of analytic monoids as a general framework for quantitative theory of factorization. So far the latter subject was developed either in concrete
### N T ] 1 9 Ju l 2 02 1 lADIC IMAGES OF GALOIS FOR ELLIPTIC CURVES OVER Q
• Mathematics
• 2021
We discuss the l-adic case of Mazur’s “Program B” over Q, the problem of classifying the possible images of l-adic Galois representations attached to elliptic curves E over Q, equivalently,
### $\ell$ -adic images of Galois for elliptic curves over $\mathbb {Q}$ (and an appendix with John Voight)
• Mathematics
Forum of Mathematics, Sigma
• 2022
Abstract We discuss the $\ell$ -adic case of Mazur’s ‘Program B’ over $\mathbb {Q}$ : the problem of classifying the possible images of $\ell$ -adic Galois representations attached to
### A database of genus-2 curves over the rational numbers
• Mathematics
LMS Journal of Computation and Mathematics
• 2016
## References
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### On the structure of the Selberg class, I: 0≤d≤1
• Mathematics
• 1999
The Selberg class S is a rather general class of Dirichlet series with functional equation and Euler product and can be regarded as an axiomatic model for the global L-functions arising from number
• Mathematics
• 1996
• Mathematics
• 1974
### Old and new conjectures about a class of Dirichlet series
• Collected papers II, Springer, Berlin
• 1991 | {"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": 0.8773142695426941, "perplexity": 1350.3309779305494}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337421.33/warc/CC-MAIN-20221003133425-20221003163425-00535.warc.gz"} |
http://math.stackexchange.com/questions/601483/point-are-zariski-dense-in-projective-space | # point are Zariski-dense in projective space
Statement i read a few times :
if $k$ is a field, then the rational points are Zariski-dense in the projective space $\mathbb{P}^n$. Does anybody could provide a proof of this fact or a reference? (i think it should be only true if $k$ is infinite).
-
What are rational points in $\mathbb{P}^n?$ Aren't they all rational? – Igor Rivin Dec 10 '13 at 16:14
I assume that it means that you fix an algebraically closed field extension $K/k$, the Zariski topology on $\mathbb{P}^n(K)$ is the topology whose closed sets are locus of systems of homogeneous equations in $k[X_0,...,X_n]$. This give you the Zariski $k$-toplogy, for which line in $k^{n+1}$ should be dense... If i dont miss something. – Louis La Brocante Dec 10 '13 at 16:23
A $k$-point is a proper closed subset, so if $k$ is finite, the set of $k$-points cannot be Zariski-dense. – user64687 Dec 10 '13 at 16:29
yes i see the point with finite field, if $Z_1$,...,$Z_n$ are the finite elements, they are all solution of 1 polynomial equation, thus there is an open subset not containing all those. – Louis La Brocante Dec 10 '13 at 16:40
@IgorRivin: in this context "rational points" is being used a shorthand for "$k$-rational points", which is a fancy way of saying "points with coordinates in $k$". The reason for this seemingly superfluous terminology is that $\mathbf{P}^n_k$, considered as a scheme, contains lots of other points besides these. – user64687 Dec 10 '13 at 16:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9037015438079834, "perplexity": 298.61695280364864}, "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-2016-07/segments/1454701159155.63/warc/CC-MAIN-20160205193919-00217-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://proofwiki.org/wiki/Definition:Real-Valued_Function | # Definition:Real-Valued Function
## Definition
Let $f: S \to T$ be a function.
Let $S_1 \subseteq S$ such that $f \left({S_1}\right) \subseteq \R$.
Then $f$ is said to be real-valued on $S_1$.
That is, $f$ is defined as real-valued on $S_1$ if and only if the image of $S_1$ under $f$ lies entirely within the set of real numbers $\R$.
A real-valued function is a function $f: S \to \R$ whose codomain is the set of real numbers $\R$.
That is, $f$ is real-valued if and only if it is real-valued over its entire domain. | {"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": 0.9847038984298706, "perplexity": 57.71358185789573}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669847.1/warc/CC-MAIN-20191118205402-20191118233402-00466.warc.gz"} |
https://epjc.epj.org/articles/epjc/abs/2003/03/100520073/100520073.html | 2018 Impact factor 4.843
Particles and Fields
Eur. Phys. J. C 27, 73-85 (2003)
DOI: 10.1140/epjc/s2002-01100-8
## Unintegrated gluon in the photon and heavy quark production
L. Motyka1, 2 and N. Tîmneanu1
1 High Energy Physics, Uppsala University, Box 535, 751 21 Uppsala, Sweden
2 Institute of Physics, Jagellonian University, Reymonta 4, 30-059 Kraków, Poland
(Received: 4 September 2002 / Revised version: 6 November 2002 / Published online: 20 December 2002 )
Abstract
The unintegrated gluon density in the photon is determined, using the Kimber-Martin-Ryskin prescription. In addition, a model of the unintegrated gluon is proposed, based on the saturation model extended to the large- x region. These gluon densities are applied to obtain cross sections for charm and bottom production in and collisions using the kt factorization approach. We investigate both direct and resolved photon contributions and make a comparison with the results from the collinear approach and the experimental data. An enhancement of the cross section due to inclusion of non-zero transverse momenta of the gluons is found. The charm production cross section is consistent with the data. The data exceed our conservative estimate for bottom production in collisions, but theoretical uncertainties are too large to claim a significant inconsistency. A substantial discrepancy between theory and the experiment is found for , which is not cured by the kt factorization approach.
© Società Italiana di Fisica, Springer-Verlag 2003 | {"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": 0.9495213627815247, "perplexity": 2041.4447858883557}, "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-2020-05/segments/1579251778168.77/warc/CC-MAIN-20200128091916-20200128121916-00188.warc.gz"} |
http://tex.stackexchange.com/questions/83000/texshop-gives-null-does-not-exist-error-when-compiling-postscript?answertab=votes | # TeXShop gives “(null) does not exist” error when compiling postscript [closed]
Mac: 10.7.3
TeXShop version: 2.47
When I try to compile .tex into a Post Script file, TeXShop gives me an error as follows:
(null) does not exist. Perhaps TeXLive was not installed or was removed during a system upgrade. If so, go to the TeXShop web site and follow the instructions to (re)install TeXLive. Another possibility is that a tool path is incorrectly configured in TeXShop preferences. This can happen if you are using the fink teTeX distribution.
I can compile PDFs fine though. I don't remember updating TeXShop recently, so have no idea what's going on. Though I almost never compile to PS, so not sure if this problem is new either. Found a post suggesting I delete the preference file, but that doesn't work. If not a solution to this problem, does anybody know which script it is looking for and how to run it from command line?
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## closed as too localized by Alan Munn, Joseph Wright♦Dec 1 '12 at 21:12
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.
Welcome to TeX.sx! – lockstep Nov 16 '12 at 16:29
Open the TeXShop preferences and click on the Engine tab, and then in the part that says "TeX + dvips + distiller" click on the two default buttons (one for TeX and one for LaTeX). Then try again. Do you still get the error? – Alan Munn Nov 16 '12 at 19:42
@Lena Please let us know what Alan's suggestion reveals. If there is no more info then the question is likely to be closed. – Joseph Wright Nov 19 '12 at 8:46
I really wish this hadn't been closed.... – jvriesem Jun 24 at 17:14 | {"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": 0.8723268508911133, "perplexity": 2117.3436216174787}, "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-2015-35/segments/1440646312602.99/warc/CC-MAIN-20150827033152-00271-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/241897/maple-how-do-i-show-vectors-with-many-elements | # Maple: How do I show vectors with many elements? [closed]
As the header says: How do I show vectors in their full form, when they have many elements? Instead of just seeing some vector data that Maple shows, when the vector is too long:
I recall seeing a command before that let you change this setting, so Maple shows vectors with up to as many elements as you decide. Can anyone help my memory?
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## closed as off-topic by froggie, Lucian, user86418, Lost1, ShuchangFeb 28 '14 at 0:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is not about mathematics, within the scope defined in the help center." – Lucian, user86418, Lost1, Shuchang
If this question can be reworded to fit the rules in the help center, please edit the question. | {"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": 0.9048965573310852, "perplexity": 1725.4438495911918}, "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-2015-22/segments/1432207929956.54/warc/CC-MAIN-20150521113209-00166-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://www.math.sinica.edu.tw/www/seminar/abstract_view.jsp?seminar_id=2006 | Seminar on PDE 主講者: 鄺國權 教授 (國立成功大學) 講題: Monotone quantities along inverse curvature flows and Minkowski formulas 時間: 2014-11-28 (Fri.) 15:30 - 16:30 地點: 數學所 722 研討室 (台大院區) Abstract: Monotone quantities along hypersurfaces evolving under the inverse mean curvature flow have many applications in geometry and relativity. In this talk, I will discuss a family of new monotone increasing quantities along inverse curvature flows in the Euclidean space. I will also discuss a related Minkowski type formula and a geometric inequality for closed k-convex hypersurfaces. Part of it is joint work with Pengzi Miao. || Close window || | {"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": 0.9609880447387695, "perplexity": 3244.2011628662017}, "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/1568514574182.31/warc/CC-MAIN-20190921022342-20190921044342-00298.warc.gz"} |
https://www.physicsforums.com/threads/ode-again.85932/ | # Ode Again!
1. Aug 22, 2005
### heman
Here is another problem which is driving me insane!I just need a hint on how to proceed!
To Solve:-
p^7 + p^3 - p^2 + 1 =0
here p = dy/dx
2. Aug 22, 2005
### HallsofIvy
Staff Emeritus
Since that is a linear homogeneous d. e., it characteristic equation is just what you give: p7+ p3- p2+ 1= 0. There are no general ways of solving 7th degree polynomial equations so the best you could hope for is some simple equation. Because the leading and ending coefficients are both 1, the only possible rational solutions are p= 1 and p= -1- and neither of those satisfy the equation!
My suggestion- go insane!
3. Aug 22, 2005
### heman
Halls,You triggered in the right direction!
I asked my tutor and he said that there will be atleast 1 real root
=> p=k
or y=kx + c
From here i am thinking ahead!
4. Aug 22, 2005
### saltydog
I have a question:
So we look for solutions of the form y=mx+b, plug it in, get a polynomial in m and solve for the roots. So there's one real root. So we find it numerically (I ain't proud), and then we get the solution:
$$y(x)=m_rx+b$$
So, how do we know if there are other solutions?
5. Aug 22, 2005
### heman
The same thing i asked him!
He said our aim is to get just 1 family of curves of solution!
The other six can be complex,real who knows!
6. Aug 22, 2005
### HallsofIvy
Staff Emeritus
heman, I interpreted "p^7" where p= dy/dx to mean the seventh derivative (operator notation). saltydog tells me that he interpreted it as the seventh power of dy/dx so that this is not a linear equation at all. Which is it?
(Yes, the polynomial p^7 + p^3 - p^2 + 1 =0, by DesCartes' "rule of signs", has one negative real root and either 0 or 2 positive real roots. That is, the equation may have (1) 1 negative real root and 3 pair of conjugate complex roots or (2) 1 negative real root, 2 positive real roots, and 2 pair of conjugate comples roots. However, the real roots are not rational.)
7. Aug 22, 2005
### heman
Ohhhh....I am the source of miscommunication. :surprised ...Halls,it's 7th power,i should have had clarified more to avoid this..!
(1) 1 negative real root and 3 pair of conjugate complex roots
I understand that it can have 3 pair of conjugate complex roots-but how do you say that real root will be negative?
(2) 1 negative real root, 2 positive real roots, and 2 pair of conjugate comples roots
How are you able to tell the signs??
I have taken the Complex Analysis Course last semester but i haven't come across any such theorem which tells about the sign! | {"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": 0.8615970015525818, "perplexity": 1228.3717737473432}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542323.80/warc/CC-MAIN-20161202170902-00367-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://cs.stackexchange.com/questions/45499/complexity-of-given-a-graph-g-with-vertex-v-is-there-a-maximum-clique-cont | # Complexity of "given a graph $G$ with vertex $v$, is there a maximum clique containing $v$"?
The usual way of translating the maximum clique problem into a decision problem is to ask "does there exist a clique of size $\ge k$ in $G$?" Clearly this problem is in NP (and is NP-hard).
Another possible way of making it into a decision problem would be:
Instance: Graph $G=(V,E)$ and vertex $v\in V$.
Question: Does there exist a clique $C\subseteq V$ in $G$ that contains $v$ and is of maximum cardinality among cliques?
If P = NP, this problem is polynomial-time solvable. The problem is also contained in $\Sigma_2^p$ (does there exist a clique such that all other cliques are not bigger?). Has its precise complexity been studied?
(Very similar questions can of course also be asked about vertex cover, dominating set, independent set and the like).
I do not recall having seen this precise problem before, but it does have a Cook reduction (polynomial-time Turing reduction) to $k$-Clique:
Let $Y$ be an oracle for $k$-Clique. On input $(G,v)$:
1. Determine the maximum clique size $\omega(G)$ via at most $log_{2}(n)$ calls to $Y$ using binary search.
2. Create $G'$ by removing all vertices $u \neq v$ where $uv \notin E(G)$ (i.e. $V(G') = N[v]$).
3. Determine $\omega(G')$ using $Y$ and binary search.
4. If $\omega(G') = \omega(G)$ answer YES, otherwise answer NO.
This puts it in $\Delta_{2}^{P} = P^{NP} = P^{SAT}$, a refinement of your observation.
As there are $O(\log{}n)$ calls to the oracle, we can squash containment a little further; the problem is in $\Delta_{2}^{P}[O(\log{}n)] = P^{NP}[O(\log{}n)] = P^{SAT}[O(\log{}n)]$.
[Thanks here to Ricky Demer] If $P=NP$, then $P^{NP} = P^{P} = P$, so you are correct about the polynomial time solvability in this case (contrary to my original, now edited, musings).
So perhaps the problem is $P^{NP}[O(\log{}n)]$-complete, it shouldn't be (as Dominik observed in his answer) $P^{NP}$-complete. I have tried to use the results of Theorem 3.5 from Krentel's paper to construct a reduction, but unless $P^{NP}$ is low for itself (which I don't think it is, but don't know), this has been without luck. (If it is low for itself, then I have a marvellous proof, sadly there are no margins here to put it in.)
• This is very interesting, thank you. There seem to be few $\Delta_2^p$-complete problems around, but I'll have a look into the richer $\text{FP}^{\text{NP}}$-complete function-problem landscape. Aug 24, 2015 at 10:12
• @DominikPeters, following your answer, and the paper you linked, I refined the observations a little. From Thm 3.5, there's a Church reduction using $P^{NP}$ oracles from the clique problem, but this doesn't really help unless, as I say in the answer, $P^{NP}$ is low for itself. $NP$ is low for $P^{NP}$, but that seems to be the limit I can find. Aug 28, 2015 at 6:55
• I don't really understand this answer but it's probably offtopic, so I posted a new question: cs.stackexchange.com/questions/45635/… Aug 28, 2015 at 7:38
• @Luke: Presumably since $NP \subseteq P^{NP}$, we have $\Sigma_2^p = NP^{NP} \subseteq (P^{NP})^{P^{NP}}$ so that we don't have $P^{NP} =(P^{NP})^{P^{NP}}$ unless the polynomial hierarchy collapses. Thanks for all your help! For the actual application I have in mind, I don't care too much where exactly between $NP$ and $P^{NP}[\log n]$ this problem lies, but I am still curious. Aug 28, 2015 at 21:57
• The problem is in fact complete for $P^{NP}[\log n]$, as can be seen by reducing from a similar problem about being a member of a min vertex cover, proved complete in this paper: sciencedirect.com/science/article/pii/S0304397505005785 Feb 27, 2017 at 22:28
For what it's worth, here is a reduction showing NP-hardness. As regards $\Delta_2^p$-completeness, I'm not too optimistic, since one complete problem for this class is "given a formula, does its lexicographically maximal satisfying assignment set the last variable to true?" which seems harder than this problem.
We reduce from 3SAT. Let a formula $\phi'$ with $m-1$ clauses be given; add a fresh variable $x$ and add the one-literal clause $x$; this forces $x$ to be true. Thus we obtain a formula $\phi$ with $m$ clauses which is satisfiable if and only if $\phi'$ is.
Next we construct the graph $G$, which up to an extra component is identical to the graph used in Karp's original reduction showing that CLIQUE is NP-hard. Take a vertex $\ell$ for each literal occurrence in $\phi$. Include an edge between literal occurrences $\ell_1$ and $\ell_2$ if and only if (a) the occurrences are in different clauses, and (b) $\ell_1$ and $\ell_2$ are not contradictory (i.e. are negations of each other). Further, add a disjoint copy of the complete graph $K_m$ of $m$ vertices. This completes the description of $G$. As special vertex $v$ use the single literal occurrence of $x$.
Suppose the formula is satisfiable. Note first that there cannot be a clique among the literals of size larger than $m$ since every clique can contain at most 1 vertex per clause. Of course, among $K_m$, there is also no clique larger than $m$. Now take a satisfying assignment of $\phi$, and pick one true literal per clause. Since these literals cannot be contradictory, this forms a clique of size $m$, and includes the vertex $v$; it is of maximum cardinality $m$.
Suppose there is a maximum-cardinality clique in $G$ containing $v$. Then from the presence of $K_m$ in $G$, this clique must have size at least $m$. From what we said before, it can't have size larger than $m$, so it is of size exactly $m$. The clique thus chooses one literal per clause, and we can set them all true in a contradiction-free assignment. Hence $\phi$ is satisfiable.
• Nice reduction! We might be able to put something together from that paper as well. Aug 28, 2015 at 6:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8887017965316772, "perplexity": 337.1191459163191}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104215790.65/warc/CC-MAIN-20220703043548-20220703073548-00599.warc.gz"} |
http://zbmath.org/?q=an:0611.10001 | # zbMATH — the first resource for mathematics
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Pi and the AGM. A study in analytic number theory and computational complexity. (English) Zbl 0611.10001
Canadian Mathematical Society Series of Monographs and Advanced Texts. A Wiley-Interscience Publication. New York etc.: John Wiley & Sons. xv, 414 pp. £48.00 (1987).
The central theme of this book is the efficient calculation of mathematical constants.
A brief sketch of the contents is as follows. In chapters 1 and 2 the arithmetic-geometric mean is defined and its connection with elliptic integrals and theta functions is shown. In chapter 3 Jacobi’s triple product is introduced and applied to theta functions and in other ways. Chapter 4 gives higher order transformations and modular functions, and chapter 5 uses the previous material to obtain algebraic approximations to $\pi$. In chapter 6 the complexity of computational methods is discussed, and in chapter 7 the complexity of algorithms applied to particular functions is dealt with. Chapter 8 introduces general means, chapter 9 gives various applications of theta functions, and chapter 10 gives methods for accelerating the convergence of classical methods of calculation of various functions, especially exp and log. Chapter 11 gives a history of the calculation of $\pi$, and a discussion of transcendence and irrationality. An extensive bibliography follows. Many results in the text are given as exercises for the reader to prove.
Aside from the main course of the book there are interesting digressions into, for example, results on representation as sums of squares, series that enumerate partitions, and lattice sums that arise from chemistry.
This is a delightful book in the classical tradition, full of beautiful formulae, and ably complemented by the excellence of the typography and layout.
Reviewer: H.J.Godwin
##### MSC:
11-02 Research monographs (number theory) 11Y60 Evaluation of constants 11F03 Modular and automorphic functions 68Q25 Analysis of algorithms and problem complexity 33E05 Elliptic functions and integrals 65B99 Acceleration of convergence (numerical analysis) 65D20 Computation of special functions, construction of tables 11J81 Transcendence (general theory) 33B10 Exponential and trigonometric functions | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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": 0.8363505601882935, "perplexity": 3228.8671845124923}, "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/1398223202457.0/warc/CC-MAIN-20140423032002-00300-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/schwarzschild-orbits-in-cartesian-coordinates.126996/ | # Schwarzschild Orbits in Cartesian coordinates
## Is Carl going to find the Schwarzschild orbits in Cartesian DE form?
50.0%
2. ### Nope.
50.0%
1. Jul 24, 2006
### CarlB
My Java applet gravity simulator http://www.gaugegravity.com/testapplet/SweetGravity.html
draws beautiful orbits, however the GR simulation is very badly broken as one can tell when comparing it with Newton at long distances. The source code is at:
http://www.gaugegravity.com/testapplet/SwGrav_Top.java
Test bodies are required to all lie in a plane. The Newtonian physics is produced by writing the x and y accelerations in terms of the phase space parameters of the test body, that is, in terms of its (x,y) position and its (Vx,Vy) velocity. With Newton, the acceleration has no dependence on velocity of course.
The GR "physics" is produced by a complicated calculation that follows the artificial potential that is discussed with respect to the Schwarzschild orbits in MTW and in various places all over the web. These methods use energy and angular momentum as constants of the motion to describe the orbits but they are less than optimal when used in an integration form.
I've seen a computer method for computing GR orbits that should work better, but I am hesitating to implement it because it is based on polar coordinates. This means that the calculation has to use trig functions to convert back and forth and these are slow.
So I'm wondering how hard it is to put the Schwarzschild orbits into phase space form in Cartesian coordinates. Here's the basic plan:
(1) Write the Schwarzschild metric in Cartesian coordinates.
(2) Write the proper length of a path as an integral over coordinate time.
(3) Vary the path and use the Euler-Lagarange equation to determine a pair of 2nd order differential equations that the orbits solve.
(4) Find $$d^2x/dt^2$$ and $$d^2y/dt^2$$ from the two 2nd order DEs.
Now for the poll: Do you think that this can be done reasonably easily? Part of the reason I am working on it is that I want to get a better idea on how these gravity theories differ one from another. I will post calculations as I make them, if convenient.
Carl
Last edited: Jul 24, 2006
2. Jul 24, 2006
### CarlB
Step (1), write the Schwarzschild metric in Cartesian coordinates.
$$x = r \cos(\phi)$$
$$y = r \sin(\phi)$$
$$dx = \cos(\phi) dr - r\sin(\phi) d\phi$$
$$dy = \sin(\phi) dr + r\cos(\phi) d\phi$$
$$dr = \cos(\phi) dx + \sin(\phi) dy$$
$$rd\phi = -\sin(\phi) dx + \cos(\phi) dy$$
$$dr^2 = \cos^2(\phi) dx^2 + \sin^2(\phi) dy^2 + \sin(2\phi) dx\;dy$$
$$r^2d\phi^2 = \sin^2(\phi) dx^2 + \cos^2(\phi) dy^2 - \sin(2\phi) dx\;dy$$
$$dr^2 = (x^2dx^2 + y^2dy^2 + 2xy dx\;dy)/r^2$$
$$r^2d\phi^2 = (y^2dx^2 + x^2dy^2 - 2xy dx\;dy)/r^2$$
$$ds^2 = -\frac{(r-2)}{r}dt^2 + \frac{r}{r-2}dr^2 + r^2d\phi^2$$
$$ds^2 = -\frac{(r-2)}{r}dt^2 + \frac{1}{r(r-2)}(x^2dx^2 + y^2dy^2 + 2xy dx\;dy) + \frac{1}{r^2}(y^2dx^2 + x^2dy^2 - 2xy dx\;dy)$$
$$ds^2 = -\frac{(r-2)}{r}dt^2 + \frac{1}{r^2(r-2)}((rx^2+ ry^2 -2y^2)dx^2 + (ry^2+rx^2-2x^2)dy^2 + (2rxy -2rxy +4rxy) dx\;dy)$$
$$ds^2 = -\frac{(r-2)}{r}dt^2 + \frac{1}{r^2(r-2)}((r^3 -2y^2)dx^2 + (r^3-2x^2)dy^2 + (4rxy) dx\;dy)$$
$$ds^2 = -\frac{(r-2)}{r}dt^2 + \frac{(r^3-2y^2)}{(r^3-2r^2)}dx^2 + \frac{(r^3-2x^2)}{(r^3-2r^2)}dy^2 +\frac{4rxy}{(r^3-2r^2)}dx\;dy$$
Last edited: Jul 24, 2006
3. Jul 24, 2006
### Jorrie
Interesting excercise Carl, but how do you think having the orbital equation in Cartesian form (as against polar form) will help you get a "better idea on how these gravity theories differ one from another."?
4. Jul 24, 2006
### pervect
Staff Emeritus
There isn't any such thing as "Cartesian coordinates" for a black hole. Cartesian coordinates work only for flat space-times. The space-time around a black hole is not flat. In particular, r is not a radial coordinate.
The closest you can come to "cartesian" coordiantes are isotropic coordinates.
Howver, from experience, I can tell you it will be far far easier to get your program working for Schwarzschild coordinates than isotropic cooridinates.
As I mentioned before, you can just use the geodesic equations to express your set of equations as a second order system that look pretty much like the second order set you have already for Newton's equations.
This is the "geodesic" equation
http://en.wikipedia.org/wiki/Solving_the_geodesic_equations
(Wiki somewhat confusingly uses "t" instead of "tau" for the variables.
The necessary Christoffel symbols are simplest for the schwarzschild metric, they are
Code (Text):
r m
CC [r r] = - -----------
(r - 2 m) r
theta
CC [theta r] = 1/r
phi
CC [phi r] = 1/r
t m
CC [t r] = -----------
(r - 2 m) r
theta
CC [r theta] = 1/r
r
CC [theta theta] = -r + 2 m
phi cos(theta)
CC [phi theta] = ----------
sin(theta)
phi
CC [r phi] = 1/r
phi cos(theta)
CC [theta phi] = ----------
sin(theta)
r 2
CC [phi phi] = -(r - 2 m) sin(theta)
theta
CC [phi phi] = -sin(theta) cos(theta)
t m
CC [r t] = -----------
(r - 2 m) r
r (r - 2 m) m
CC [t t] = -----------
3
r
As I also mentioned, you should be able to obtain the same set of geodesic equations from the conserved quantites, as well, by replacing "conserve quantity = constant" with "d/dtau (conserved quantity) = 0"
Last edited: Jul 24, 2006
5. Jul 24, 2006
### pervect
Staff Emeritus
Let's try to make things really, really, really simple.
theta will always be 90 degrees. Therfore d theta / dtau will be zero.
We then have for our variables
r(tau), t(tau), phi(tau).
The second order geodesic equations of motion (the geodesic equations) will therfore be
$$\frac{d^2 r}{d tau^2} + \Gamma^r{}_{rr} \left( \frac{dr}{dtau} \right) ^2 + \Gamma^r{}_{tt} \left( \frac{dt}{d tau} \right)^2 + \Gamma^r{}_{\phi\phi} \left( \frac{d \phi}{d tau} \right)^2 = 0$$
$$\frac{d^2 t}{d tau^2} + \left( \Gamma^t{}_{tr} + \Gamma^t{}_{rt} \right) \left( \frac{dr}{d tau} \right) \left( \frac{dt}{dtau} \right) = 0$$
$$\frac{d^2 \phi}{d tau^2} + \left( \Gamma^{\phi}{}_{r\phi} + \Gamma^{\phi}{}_{\phi r} \right) \left( \frac{dr}{d tau} \right) \left( \frac{d \phi} {d tau} \right) = 0$$
See the previous post for the values for the Christoffel symbols.
Note for instance that since r^2 ${d \phi}/ {d tau}$ = constant, when differentiated, gives by the chain rule
$$2 r \frac{dr}{d\tau} \frac{d \phi}{d tau} + r^2 \frac{d^2 \phi} {d tau^2} = 0$$
and that this is equivalent to the last equation above when you substitute in the correct value for the Christoffel symbol.
The conseved energy when differentiated will give one of the other geodesic equations (the one for time).
The last geodesic equation is a consequence of the others plus the equation for the metric.
Last edited: Jul 24, 2006
6. Jul 24, 2006
### Jorrie
Cartesian coordinates
Are you not a bit harsh on CarlB? I understood that he simply wants to express Schwarzschild coordinates in terms of x,y,z rather than in terms of r,phi,theta. Nothing fundamentally wrong with that, I think! Maybe his (Cartesian) choice of words was wrong, but let’s not get too semantic!
Nevertheless, I agree that we should rather stick to the tried and tested procedure of treating black holes.
7. Jul 24, 2006
### CarlB
pervect, one of the things I'm trying to get away from is differential equations written in proper time. It makes for a more complicated simulation and also makes it a lot easier for the simulation to fail in various ways. So I'm continuing with the process of writing the equations in Cartesian coordinates. I will allow you to continue calling them isotropic coordinates if that's what you want to do.
Carl
8. Jul 24, 2006
### pervect
Staff Emeritus
Carl (and Jorrie, too) I think you've totally missed the point.
The 'r' in the Scwarzschild coordinate system is not the same 'r' in the isotropic coordinate system.
In fact, one can see that r = r'(1+2m/r')^2 from the webpage, where r' is the isotropic radial coordinate and r is the Schwarzschild r coordinate.
Neither of these 'r' coordinates can be interpreted as 'radial distance'. The isotropic coordinates, though, define a set of coordinates where light cones will apear to be circles.
Thus assuming that r^2 = x^2 + y^2 does not make any physical sense. This is assumed implicitly when one writes x = r cos(phi) y = r sin(phi), one assumes that x^2 + y^2 = r^2. One can think of it as a purely formal mathematical expression, I suppose.
As such a formal exercise, unless Carl has made an algebraic error he's got a new and totally non-standard metric that doesn't even have a name in terms of new and toatally non-standard coordiantes that don't have names either (except what Carl has christened them). They definitely shouldn't be confused with Cartesian coordinates, which is what Carl has chosen to call his variables.
One can easily see that Carl's metric is not the same as the isotropic metric either by comparing the line elements.
Aside from the fact that it is non-standard, and thus will be hard to check solutions for, even deriving the correct solutions is going to be a giant pain. Re-writing a simple metric in a more complicated form is not going to get anywyere as far as generating a solution for the metric.
If the problem is with the factor of 'tau' in the standard equations, just eliminate that tau. This shouldn't be too hard since
(1-2m/r) dt/dtau = E
The most annoying thing is going to be this
If we write r as a function of t : r = rr(t) then
r(tau) = rr(t(tau))
and dr/dtau = drr/dt dt/dtau
but by the chain rule
$$\frac{d^2 r} {d tau^2} = \frac{d^2 rr}{d t^2} \left( \frac{dt}{d tau} \right) ^2 + \frac{dr}{dt} \frac{d^2 t}{d tau^2}$$
Still, a little bit of algebra should eliminate tau as a variable if this is really what is desired.
Solving for phi shouldn't be terriby difficult, since
r^2 dphi/dtau =L
therfore
r^2 dphi/dt * dt/dtau = L
And we've already noted that dt/dtau is easy to eliminate
9. Jul 24, 2006
### pervect
Staff Emeritus
Unless I've made a mistake (quite possible, even with computer assistance) the equations with proper time eliminated for the Schwarzschld metric are just
r(t), phi(t)
d phi/dt = K*(1-2*m/r) / r^2
K is a constant of motion, equal to "L/E" in the old notation. Knowing dphi/dt and r at one point will determine the value of K
The second order equation for r(t) is
$$\frac{d^2 r}{dt^2} = \frac{2\,m}{r^2 \left( 1 - \frac{2m}{r} \right) } \left( \frac{dr}{dt} \right) ^2 - \left( 1-{\frac {2 m}{r}} \right) \frac{m}{r^2}+ \left( 1-{\frac {2m}{r}} \right) ^{3}\frac{{K}^{2}}{r^3}$$
It looks sensible at first glance, dr/dt is forced to zero, however I have not tested it.
Compare this to the newtonian equations in polar coordinates (r, phi)
r^2 * d phi/dt = K
where K = L/m
d^2 r / dt^2 = -m/r^2 + r (dphi/dt)^2 = -m/r^2 + K^2 / r^3
we can also see that the Schwarzschild equations reduce to the Newtonian equations when dr/dt << 1 and r >> 2m as they should, another sign that they are probably right.
Last edited: Jul 24, 2006
10. Jul 24, 2006
### pervect
Staff Emeritus
The Christoffel symbols for the Painleve metric are:
Code (Text):
1/2 1/2
T 2 (M/r) M
CC [T T] = ---------------
2
r
r (r - 2 M) M
CC [T T] = -----------
3
r
T M
CC [r T] = ----
2
r
1/2 1/2
r 2 (M/r) M
CC [r T] = - ---------------
2
r
T M
CC [T r] = ----
2
r
1/2 1/2
r 2 (M/r) M
CC [T r] = - ---------------
2
r
1/2
T 2 M
CC [r r] = -------------
1/2 2
2 (M/r) r
r M
CC [r r] = - ----
2
r
theta
CC [theta r] = 1/r
phi
CC [phi r] = 1/r
theta
CC [r theta] = 1/r
T 1/2 1/2
CC [theta theta] = -2 (M/r) r
r
CC [theta theta] = -r + 2 M
phi cos(theta)
CC [phi theta] = ----------
sin(theta)
phi
CC [r phi] = 1/r
phi cos(theta)
CC [theta phi] = ----------
sin(theta)
T 1/2 1/2 2
CC [phi phi] = -2 (M/r) r sin(theta)
r 2
CC [phi phi] = -(r - 2 M) sin(theta)
theta
CC [phi phi] = -sin(theta) cos(theta)
a similar analysis using the geodesic equations and eliminating tau should be possible.
11. Jul 24, 2006
### CarlB
It is my intention to work only in the Schwarzschild metric, but with the polar coordinates replaced by Cartesian coordinates. I realize that you think this is unphysical, but here in the United States computers are built with screens that have rectangular arrays of pixels, and so any computation I make must be eventually so converted. Now that you've explained what they are, I don't mean to use "isotropic" coordinates at all. It's just extra confusion you've brought in, not me. I thought it was you calling my coordinates "isotropic". Recall, it was I who "allowed" you to call them that if you like.
Nor am I the only person in the world who calls these things I'm using "Cartesian". The United States is a very primitive place, filled with horses, saloons and cowboys. Other obscure, amateurish authors here use the same confusing and incorrect terminology. For example see:
Numerical Relativity for Inspiraling Binaries in Co-Rotating Coordinates: Test Bed for Lapse and Shift Equations
Kip S. Thorne, 1998
http://arxiv.org/abs/gr-qc/9808024
By the way, I should mention that while I only took one graduate level General Relativity class at the University of California, Irvine, I was the top student, the only student in the class receiving an "A+". I don't think that taking one class makes you an expert in a subject, but you can be sure that I understand the principles of general relativity at least a little better than some.
But I see that it is not only on this side of the Atlantic that one finds the use of Cartesian coordinates in describing black holes:
Shortcut Method of Solution of Geodesic Equations for Schwarzschild Black Hole
Jean-Alain Marck,
D´epartement d’Astrophysique Relativiste et de Cosmologie
Observatoire de Paris - Section de Meudon
F-92195 Meudon Cedex, France
Classical and Quantum Gravity 13 (1996) 393-402
$$r^2 = x^2 + y^2 + z^2\;\;\;\;\;\;\;(6)$$
http://arxiv.org/abs/gr-qc/9505010
[/edit]
Carl
Last edited: Jul 24, 2006
12. Jul 24, 2006
### CarlB
This is exactly what I need and much thanks. To convert it into Cartesian form, I need only split the radial acceleration into x and y accelerations. I should have a new simulation running later today, but Seattle is sooooooo hot (about 90 F) that everyone has left for the beach and very little work is getting done. Maybe I'll wait until it begins to cool down.
I was in mathematics grad school at the University of Washington 10 years ago and got a student priced copy of Mathematica. I played around with it, but never worked at it long enough to get around to messing with the symbolic calculus. Since then, most of my work has been in Clifford algebra and for that I wrote Java programs. But after this fiasco I'm convinced that I need to get some sort of a symbolic math package.
Carl
Last edited: Jul 24, 2006
13. Jul 24, 2006
### pervect
Staff Emeritus
Symbolic math really rocks. Couldn't live without it.
I've also added horizon crossing solutions to the Painleve thread (but only in terms of conseved quantites) I haven't eliminated dt/dtau, it appears to be a royal pain to do so. dt/dtau will range in values from 1 to E for the horizon crossing solutions.
I note that non-horizon crossing alternate solutions exist, but I haven't explored them very far.
I've also computed d^2 r / dtau^2 only in the special case where dr/dtau = 0.
Last edited: Jul 24, 2006
14. Jul 25, 2006
### CarlB
Unfortunately, I couldn't get it to work. This could always be a programming error of course. It acts like there is too much energy in the angular kinetic energy term, that is the K^2 term. In fact, the simulation doesn't match Newton at large r because of this term. To get it to work, I have to divide by an extra factor of r.
Take a look at the units for large r, in the first and third terms of the above and see if there is a transcription error, most likely in the last term of the above, but at least conceivably in the formula for $$K = \dot{\phi} r^2 / (1-2m/r)$$ which can be used to simplify the third term.
Having a differential equation in r and a constant of the motion in phi is unsatisfying to me. I realize that I can put the Newtonian equation into the same form, but I am one of those crazy guys who thinks that the equations of motion are closer to the physics than the constants of the motion. Yes, I believe that the constants of the motion are nothing more than convenient methods we use for simplifying the original physical problem, which to me are always defined in terms of positions and velocities. So it looks like I will continue my process of deriving the Newton style equations of motion for the Schwarzschild metric.
Carl
15. Jul 25, 2006
### CarlB
Step (2), write the proper length of a path as an integral over coordinate time.
Here's where I begin:
Note that in the above I've swapped over to the other coast metric in order to make s time like instead of spacelike. Sorry for starting the problem with the spacelike signature, it's a habit from Euclidean relativity where s is interpreted as a little curled up spatial dimension. Dividing by (dt)^2 gives:
$$\left(\frac{ds}{dt}\right)^2 = \frac{(r-2)}{r} - \frac{(r^3-2y^2)}{(r^3-2r^2)}\dot{x}^2 - \frac{(r^3-2x^2)}{(r^3-2r^2)}\dot{y}^2 -\frac{4rxy}{(r^3-2r^2)}\dot{x}\dot{y}$$
where the dots indicate derivatives with respect to coordinate time t rather than the more usual proper time. Thus the proper time along the path is:
$$S = \int_{t_1}^{t_2} \frac{ds}{dt}\; dt$$
$$= \int_{t_1}^{t_2} \sqrt{\frac{(r-2)}{r} - \frac{(r^3-2y^2)}{(r^3-2r^2)}\dot{x}^2 - \frac{(r^3-2x^2)}{(r^3-2r^2)}\dot{y}^2 -\frac{4rxy}{(r^3-2r^2)}\dot{x}\dot{y} }\;\; dt$$
Last edited: Jul 25, 2006
16. Jul 25, 2006
### CarlB
The next step is to vary the path by the Euler-Lagrange equations. Discretion being the better part of valor, I will wait until I have a symbolic mathematics engine before I attempt that slope. Basically, I need to calculate the partial derivatives with respect to the coordinates and velocities. The equations of motion are then the time derivatives of the partials with respect to velocities set equal to the partials with respect to positions. That is, for example,
$$\frac{d}{dt} \left(\frac{\partial ds/dt}{\partial x}\right) = \frac{\partial ds/dt}{\partial \dot{x}}$$
Carl
Okay, now I've got Maxima running on my laptop. Pretty cool. And cheap.
Last edited: Jul 25, 2006
17. Jul 26, 2006
### pervect
Staff Emeritus
If you use Maxima to calculate the Christoffel symbols, watch out for the reversed ordering of the indices. I fell afoul of that issue in GrTensor. Maxima does it to, I believe (I've only used that briefly).
If you have mathematica I think grtensor version M will work with it. Grtensor is better than Maxima in my experience, and Maxima is much better than nothing at all.
I've recalculated the formula for the Schwarzschild metric, and gotten a different answer. The main difference is that I removed some manual copying steps, collecting the values of the Christoffel symbols automatically instead of transcribing them by hand. The new formula seems to conserve energy and momentum in the numerical test case I ran, though some variance can be noted near the horizon. This variance is improved by tightning up the tolearances of the integration routine, though, so I think it's a numerical issue.
The new formula are:
$$\frac{d^2 r}{d t^2} = \frac {3 m{{\it vr}}^{2}}{ \left( r-2\,m \right) r} + \left( r-2\,m \right) \left( {{\it vphi}}^{2}-{\frac {m}{{r}^{3}}} \right)$$
$$\frac{d^2 phi}{d t^2} = -\frac {2 {\it vr}\,{\it vphi}\, \left( r -3\,m \right) }{ \left( r-2\,m \right) r}$$
vr = dr/dt and vphi = dphi/dt
Last edited: Jul 26, 2006
18. Jul 26, 2006
### pervect
Staff Emeritus
Here's what the solution looks like at the code level, minus some manual simplifications of the formula.
manual simplifications are checked by computing simplify(manual - auto) and making sure it is zero.
CC(up,dn,dn) calculates the Christoffel symbols - it's a macro that uses grtensor's built-in routine and changes the order to the standard order.
>grcalc(CC(up,dn,dn));
> grdisplay(_);
> tmp := grarray(CC(up,dn,dn));
> vlist := [rdot,0,phidot,tdot];
> eq1 := 0;
> eq2 := 0;
> eq3 := 0;
> eq4 := 0;
> # sum geodesic equations
> for j from 1 by 1 to 4 do
> for k from 1 by 1 to 4 do
> eq1 := eq1 + tmp[1,j,k]*vlist[j]*vlist[k];
> eq2 := eq2 + tmp[2,j,k]*vlist[j]*vlist[k];
> eq3 := eq3 + tmp[3,j,k]*vlist[j]*vlist[k];
> eq4 := eq4 + tmp[4,j,k]*vlist[j]*vlist[k];
> end do;
> end do;
>
> eq1 := subs(sin(theta)=1,eq1);
> eq2 := subs(cos(theta)=0,eq2);
> eq3 := subs(sin(theta)=1,eq3);
> eq4 := subs(sin(theta)=1,eq4);
> # we have t(tau), r(tau), phi(tau)
> # tdot = dt/dtau, rdot = dr/dtau, phidot = dphi/dtau
> # we want vr(t) = dr/dt, vphi(t) = dphi/dt
> # vr * tdot = rdot; vphi * tdot = phidot
> # rdotdot = ar*tdot^2+vr*tdotdot
> # but tdotdot = -eq4, thus
> # equation for r, ar is d^2 r / dt^2
> eq1a := ar*tdot^2 -vr*eq4 + eq1;
> eq1b := subs(rdot = vr*tdot, phidot = vphi*tdot, eq1a);
> eq1c:= simplify(solve(eq1b,ar));
> collect(eq1c,vr);
> # phi equation
> eq3a := aphi*tdot^2 -vphi*eq4 + eq3;
> eq3b := subs(rdot = vr*tdot, phidot = vphi*tdot, eq3a);
> eq3c := simplify(solve(eq3b,aphi));
> collect(eq3c,vr);
19. Jul 26, 2006
### pervect
Staff Emeritus
The standard textbook approach is to write the Lagrangian as
$$L = g_{\mu \nu} \frac{d x^\mu}{d \lambda} \frac{d x^\nu}{d \lambda} d \lambda$$
(Summation is implied).
Applyin the Euler-Lagrange equations to this Lagrangian will yield the geodesic equations.
Of course this approach yields equations parameterized in terms of proper time - i.e. the standard geodesic equations.
AFAIK the easiest way to eliminate proper time is to use the approach I outlined earlier. It is not immediately obvious why one doesn't need to solve for "tdot", but it seems to factor itself out automatically, so one need not solve for it.
20. Jul 26, 2006
### pervect
Staff Emeritus
Finally, the painleve metric equations with proper time eliminated. They also seemed to conserve energy and angular momentum in the one test case I ran
$$\frac{d^2 r}{d t^2} = \sqrt {{\frac {2 M}{r}}}{\it vr}\, \left( {\frac {{{\it vr }}^{2}}{2 r}}+3\,{\frac {M}{{r}^{2}}}-r{{\it vphi}}^{2} \right) +3\,{ \frac {M{{\it vr}}^{2}}{{r}^{2}}}+ \left( r-2\,M \right) \left( {{ \it vphi}}^{2}-{\frac {M}{{r}^{3}}} \right)$$
$$\frac{d^2 phi}{d t^2} = {\it vphi}\, \left( \sqrt {{\frac {2 M}{r}}} \left( {\frac {{{\it vr}}^{2}}{2 r}}-r{{\it vphi}}^{2}+{\frac {M}{{r}^{2}}} \right) -2\,{\frac {{\it vr}\, \left( r-M \right) }{{r}^{2}}} \right)$$
One can see that while the Painleve equations are simple in the newtonian limit when vr=vphi is almost zero they are not as simple as the Schwarzschild equations in the fully relativistic case.
This could be anticipated by the "count the number of Christoffel symbols" method.
Last edited: Jul 26, 2006
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http://mathhelpforum.com/algebra/187510-finding-stopping-distance-using-d-0-20v-0-15v-2-a.html | # Thread: Finding the stopping distance using d=0.20v+0.15v^2.
1. ## Finding the stopping distance using d=0.20v+0.15v^2.
The stopping distance, d in meters for a car traveling at v km/h is given by the formula d=0.20v+0.15v^2. Find the increase in stopping distance when v is increase by x km/h.
I do not get this at all much help appreciated
2. ## Re: Need help math question
Replace v with v+x.
what do you get?
3. ## Re: Need help math question
Originally Posted by BobNeedsMuchHelp
The stopping distance, d in meters for a car traveling at v km/h is given by the formula d=0.20v+0.15v^2. Find the increase in stopping distance when v is increase by x km/h.
$\displaystyle d_{new}=0.20(v+x)+0.15(v+x)^2$
increase = (d new) - (d old) | {"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": 0.952485978603363, "perplexity": 3158.728208977443}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2018-26/segments/1529267863100.8/warc/CC-MAIN-20180619154023-20180619174023-00031.warc.gz"} |
http://math.stackexchange.com/questions/95028/vector-dimension-of-a-set-of-functions | # Vector dimension of a set of functions
Let $F$ be a field and $S$ an infinite set. Set $V=\{f:S \rightarrow F\}$ endowed with the vector space structure that results from the pointwise operations of $F$.
It is easy to prove that $|S| \leq \dim V$, since the functions $\delta_s$ defined by $\delta_s(s)=1$ and $\delta_s(t)=0$ for all $t \neq s$ are linearly independent.
Is it true that $|S| < \dim V$? If so how could it be proved? I tried some kind of Cantor-like diagonal argument, but nothing worked so far.
-
Case 1 when $|F|\leq\aleph_0$: Let $W$ be the set of functions $S\to F$ with finite support. Then $W$ is clearly a vector space with $\dim W=|S|$. We also have $|W|=|S|$, because there are $|F|\times|S|=|S|$ scalar multiples of base vectors and the number of finite sums of such vectors is $\sum_{i<\omega}|S|^i = \sum_{i<\omega}|S| = \aleph_0\times|S|=|S|$. On the other hand $V$ contains at least $2^{|S|}$ elements. Since $W$ and $V$ have different cardinality, they cannot be isomorphic. In particular, $V$ must have dimension $>|S|$.
Case 2 for arbitrary $F$: Let $K$ be the smallest subfield of $F$, that is, $\mathbb{Q}$ if $F$ has characteristic $0$ and $\mathbb{F}_p$ for characteristic $p > 0$. In either case, $K$ is small enough to match the condition in case 1.
Let $\mathfrak{B}$ be a basis for $\{f:S\to K\}$. By case 1, $|\mathfrak{B}|>|S|$. But each member of $\mathfrak{B}$ is also a member of $V$, and $\mathfrak{B}$ is still linearly independent over $F$. Namely, suppose that $f_1\mathbf{b}_1+\cdots+f_n\mathbf{b}_n$ is an $F$-linear relation among vectors $\mathbf{b}\in\mathfrak{B}$. Then the $f_i$'s are a solution to a homogenous system of $|S|$ linear equations with coefficients in $K$, and this system has only the trivial solution in $K^n$. Then $n$ of the equations must be linearly independent over $K$, which means that their determinant is nonzero whether evaluated in $K$ or in $F$. Therefore the $f_i$'s must all be zero, too.
So the dimension of $V$ is at least $|\mathfrak{B}|>|S|$.
-
If $\kappa=|S|$ is a regular cardinal, $\dim V>\kappa$.
For $A\subseteq S$ let $$f_A:S\to F:s\mapsto\begin{cases}1,&\text{if }s\in A\\0,&\text{if }s\notin A\;.\end{cases}$$
Suppose that $\mathscr{A}$ be a collection of subsets of $S$, each of cardinality $\kappa$, such that $|A_0\cap A_1|<\kappa$ whenever $A_0,A_1\in\mathscr{A}$ and $A_0\ne A_1$; I’ll call such a family almost disjoint. I claim that $\{f_A:A\in\mathscr{A}\,\}$ is linearly independent.
Suppose that $A_0,\dots,A_n$ are distinct members of $\mathscr{A}$ and that $c_0f_{A_0}+\cdots+c_nf_{A_n}=0$, where $c_0,\dots,c_n\in F$ are not all $0$. Without loss of generality assume that $c_0\ne 0$. Since $|A_0|=\kappa$, but $|A_0\cap A_k|<\kappa$ for $k=1,\dots,n$, $|A_0\setminus (A_1\cup\dots\cup A_n)|=\kappa$. In particular, $A_0\setminus (A_1\cup\dots\cup A_n)\ne\varnothing$, so we can choose $s\in A_0\setminus (A_1\cup\dots\cup A_n)$. Then $$c_0f_{A_0}(s)+c_1f_{A_1}(s)+\cdots+c_nf_{A_n}(s)=c_0\ne 0\;;$$ this gives us the desired contradiction, showing that $\{f_A:A\in\mathscr{A}\,\}$ is indeed linearly independent.
Thus, to show that $\dim V>\kappa$, it suffices to show that the almost disjoint family $\mathscr{A}$ can be chosen so that $|\mathscr{A}\,|>\kappa$. Clearly $\mathscr{A}$ can be chosen to have cardinality at least $\kappa$, since $S$ can be partitioned into $\kappa$ subsets of cardinality $\kappa$. Let $\mathscr{A}_0$ be such a family. Suppose that $\alpha<\kappa$, and for every $\beta<\alpha$ we’ve constructed an almost disjoint family $\mathscr{A}_\beta$ in such a way that $\mathscr{A}_\beta\subseteq\mathscr{A}_\gamma$ whenever $\beta<\gamma<\alpha$. Let $\mathscr{A}_\alpha^*=\bigcup_{\beta<\alpha}\mathscr{A}_\beta$; clearly $|\mathscr{A}_\alpha^*|=\kappa$, so let $\mathscr{A}_\alpha^*=\{A_\xi:\xi<\kappa\}$. Suppose that $\eta<\kappa$. For each $\xi<\eta$ we have $|A_\eta\cap A_\xi|<\kappa$, and $\kappa$ is regular, so $$\left|\bigcup_{\xi<\eta}(A_\eta\cap A_\xi)\right|<\kappa\;,$$ and we can choose $$s_\eta\in A_\eta\setminus\bigcup_{\xi<\eta}A_\xi\;.$$ Now let $A=\{s_\eta:\eta<\kappa\}$; clearly $|A_\eta|=\kappa$. Moreover, $A\cap A_\eta\subseteq\{s_\xi:\xi\le\eta\}$ for each $\eta<\kappa$, so $|A\cap A_\eta|<\kappa$ for each $\eta<\kappa$, and $\mathscr{A}_\alpha^*\cup\{A\}$ is almost disjoint. Thus, we may set $\mathscr{A}_\alpha=\mathscr{A}_\alpha^*\cup\{A\}$, and the construction goes through to $\kappa^+$. Finally, let $$\mathscr{A}=\bigcup_{\alpha<\kappa^+}\mathscr{A}_\alpha\;;$$ clearly $|\mathscr{A}|=\kappa^+>\kappa$, and $\mathscr{A}$ is almost disjoint since any two members of it already belong to one of the almost disjoint families $\mathscr{A}_\alpha$ with $\alpha<\kappa^+$.
-
Here's my two cents. The space $F^{\oplus S}$ of the set of all functions $S\to F$ with finite support in $S$ is a vector space of dimension $\#(S)$. So, if $F^S$ had dimension $\#(S)$ then it would be in bijective correspondence with $F^S$. In most tame cases this is impossible. For example, if $F$ is countable then $F^{\oplus S}$ is equipotent with $S$ while $F^S$ is certainly larger, cardinal wise, $S$ since $2^S$ naturally (set wise) embeds into $F^S$ (note $F$ must have at least two elements) and so Cantor's theorem applies.
EDIT: Of course, for the general case you could just reduce to the prime subfield.
EDIT EDIT:I see Henning just said this.
EDIT EDIT EDIT: Of course, this is is how one generally proves that for infinitey dimensional spaces the dual space is strictly larger, in dimension, than the original space! Indeed, if $F[S]$ is some infinite dimensional vector space, then $\text{Hom}_F(F[S],F)\cong F^S$ whereas $F[S]\cong F^{\oplus S}$.
-
Indeed, the dimension of the dual space was the original context for the text I reused in my answer. – Henning Makholm Dec 29 '11 at 22:12
Indeed, the vector space $\mathbf R^\mathbf N$ has no countable Hamel basis. Fix any countable set of sequences $\{ a_i \}$ where $a_i \in \mathbf R^\mathbf N$. Our goal is to produce an element of sequence $b$ not in the span of this set, using a Cantor-style diagonalisation argument.
Let $\mathcal U$ be the set of finite subsets of $\mathbf N$. Since $\mathcal U$ is countable, we enumerate the members of $\mathcal U$ as $\{ U_1, U_2, \ldots, U_n, \ldots \}$. For each $U_n$, let $I_n$ be a set of $|U_n|+1$ “private” coordinates (so that all $I_n$'s are disjoint). Now, for each $i \in U_n$, the vector $(a_{i,j})_{j \in I_n}$ is of length $|U_n|+1$. Since $|U_n|$ vectors cannot span the whole of $\mathbf R^{|U_n|+1}$, there exists a vector $x_n$ in $\mathbf R^{|U_n|+1}$ that is not in the span of the vectors $$\left\{ (a_{i,j})_{j \in I_n} \ : \ i \in U_n \right\}.$$ Finally define $b$ so that $b|_{I_n}$ is equal $x_n$ above.
Now to complete the argument, just note that the construction ensures that for each finite $U \subseteq \mathbf N$, $b$ is not in the span of the vectors $\{ a_i \}_{i \in U}$. That is, $b$ is not in the span of $\{ a_i \}$.
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I do not address the question in full generality; moreover, Henning's answer seems to subsume this. But I will just leave the answer as it is just in case any one is interested. :-) – Srivatsan Dec 29 '11 at 22:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9940517544746399, "perplexity": 57.85724046736947}, "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/1469257824756.90/warc/CC-MAIN-20160723071024-00322-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/128142-solving-matrix-equations.html | 1. ## Solving matrix equations
P is a 2 times 2 matrix.Determine p if
Attached Thumbnails
2. Have you even tried to do this yourself? It's very straight forward.
$\displaystyle 3\begin{bmatrix}1 & 0 \\ 2 & 4\end{bmatrix}- 2P= \begin{bmatrix}-2 & -3 \\ 4 & -5\end{bmatrix]$.
Since P must be a 2x2 matrix in order to be able to subtract 2P, write it as [tex]\begin{bmatrix}a & b \\ c & d\end{bmatrix}:
$\displaystyle 3\begin{bmatrix}1 & 0 \\ 2 & 4\end{bmatrix}- 2\begin{bmatrix}a & b \\ c & d\end= \begin{bmatrix}-2 & -3 \\ 4 & -5\end{bmatrix]$.
Do the indicated calculations and you will have 4 separate linear equations for a, b, c, and d. For example, one equation is 1- 2a= -2. That's easy to solve.
3. Originally Posted by mastermin346
P is a 2 times 2 matrix.Determine p if
let $\displaystyle \begin{bmatrix} 1 & 0\\ 2 & 4 \end{bmatrix}$ be A
and let
$\displaystyle \begin{bmatrix} -2 & -3\\ 4 & -5 \end{bmatrix}$ be C.
So we have $\displaystyle 3A-2P=C$
to find P, rearange it
$\displaystyle -P=\frac{C-3A}{2}$
$\displaystyle P=-\frac{C-3A}{2}$
Can you do it now? | {"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": 0.9868342280387878, "perplexity": 1501.6970804517073}, "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/1524125946011.29/warc/CC-MAIN-20180423125457-20180423145457-00340.warc.gz"} |
https://tex.stackexchange.com/questions/4624/a-symbol-for-the-quotient-of-two-objects/4625 | # A symbol for the quotient of two objects
One often needs a symbol to denote the quotient of two (algebraic) objects (e.g. quotient by a subgroup, subring, submodule etc.). In simple cases people use A/B. But when both A and B are complicated to write, this doesn't look good. e.g. \mathcal{O}_{(V',0)}/\mathcal{O}_{(V,0)}
For some reason, people do not use just \frac{A}{B}. Is there some way to achieve the following:
$A$ raised a bit, then \Big/ then $B$ a bit lowered.
• Mar 1, 2015 at 13:49
## 7 Answers
How about using \left, \right, and \raisebox?
\newcommand{\bigslant}[2]{{\raisebox{.2em}{$#1$}\left/\raisebox{-.2em}{$#2$}\right.}}
$$\bigslant{\mathcal{O}_{(V',0)}}{\mathcal{O}_{(V,0)}}$$
which gives
The only problem is that I need to re-enter mathmode inside the \raiseboxes. Anyone know how to avoid that?
• Would it not be a better solution to make the slash a mid, as opposed to a left delimiter, as in: \left.\raisebox{.2em}{$#1$}\mid/\raisebox{-.2em}{$#2$}\right. Note that this requires the amsmath package. Jun 14, 2011 at 18:32
• Sorry, that \mid should have been \middle. Jun 14, 2011 at 19:08
• Yes, I agree, using middle is a better idea. Jun 26, 2011 at 6:11
• The problem with this solution, is that the parameters are interpreted as inline command. E.g. if I write \bigslant{\bigcup_{n \in \mathbb N} S_n}{\sim} the indices are written aside the \bigcup symbol, instead of under it. Do you know how to avoid it? Mar 19, 2012 at 12:54
• Nevermind, I found a solution. I use \bigcup\limits_{n \in \mathbb N}. Mar 19, 2012 at 13:02
If you want to invoke the Someone Else's Problem principle, there's also the faktor package, which ostensibly was designed to do what you want, and which implements using the AMS symbol \diagup. But IMHO the slash is a bit small.
For in-line expressions, you can also consider the nicefrac package, which makes both the "denominator" and "numerator" small.
• Oct 27, 2010 at 16:16
• The current thing to use instead of nicefrac is apparently xfrac. Oct 27, 2010 at 20:44
• Thanks for mentioning my package. I'll have a look on what can be done against the too small slash. Jul 31, 2011 at 4:36
As already mentioned, there are two packages to solve this problem:
• xfrac - typeset fractions in the form n/d generally
• faktor - especially to typeset factor structures
Here is a comparison between the \sfrac{n}{d} and \faktor{n}{d} commands which also demonstrates how they behave in comparison to normal text:
\documentclass[12pt,preview, border={2pt,2pt,2pt,2pt}]{standalone}
\usepackage[english]{babel}
\usepackage{xfrac}
\usepackage{amsmath}
\usepackage[amsmath,thmmarks,standard]{ntheorem}
\usepackage{faktor}
\begin{document}
\begin{alignat*}{4}
\text{This } &\sfrac{\mathcal{O}_{(V',0)}}{\mathcal{O}_{(V,0)}} &\text{ and this } &\sfrac{\mathcal{S}^n}{\equiv_m} &\text{ and this } &\sfrac{A}{B} &&\text{ is \texttt{sfrac}.}\\
\text{This } &\faktor{\mathcal{O}_{(V',0)}}{\mathcal{O}_{(V,0)}} &\text{ and this } &\faktor{\mathcal{S}^n}{\equiv_m} &\text{ and this } &\faktor{A}{B} &&\text{ is \texttt{faktor}.}
\end{alignat*}
\end{document}
I would try with $^{a}/_{b}$, or, based in your example
$^{\mathcal{O}_{(V',0)}}/_{\mathcal{O}_{(V,0)}}$
Also it is possible to do $^{a}\Big/_{b}$.
• Welcome to TeX.SX! You can have a look at our starter guide to familiarize yourself further with our format. Nov 3, 2014 at 23:36
Something like:
\documentclass{article}
\def\quotient#1#2{%
\raise1ex\hbox{$#1$}\Big/\lower1ex\hbox{$#2$}%
}
\begin{document}
$\quotient{\mathcal{O}_{(V',0)}}{\mathcal{O}_{(V,0)}}$
\end{document}
Update, here is a real plain example:
\def\quotient#1#2{%
\raise1ex\hbox{$#1$}\Big/\lower1ex\hbox{$#2$}%
}
$$\quotient{{\cal O}_{(V',0)}}{{\cal O}_{(V,0)}}$$
\bye
• That is essentially the same as Yossi's answer, albeit more plainish. Oct 27, 2010 at 14:29
• $$...$$, really? You should have \m@th inside the inline math in your \hboxes.
– TH.
Oct 27, 2010 at 14:29
• When I wrote it first it was a plain TeX file, but since there is no \mathcal in plain I quickly converted the document to LaTeX just to try it. Oct 27, 2010 at 14:31
• In plain TeX you can use ${\cal O}_{(V',0)}$ Oct 28, 2010 at 5:55
I like Eduardo's answer, but with properly sized numerators/denominators.
(Note the double set of braces in the definition for proper spacing. The outer pair will disappear on use, and without the second set, it will place the quotient smack up against whatever's to the left, trying to make it a superscript.)
\documentclass{article}
\usepackage{amsmath}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%% For quotient groups / modding equiv relations
%%%% Use: \quot{A}{B} --> A/B
\newcommand*\quot[2]{{^{\textstyle #1}\big/_{\textstyle #2}}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\begin{align*}
\text{Reference size: }A
\quad ^{A}\big/_{B}
&\quad \quot{A}{B}
\quad ^{\textstyle A}\Big/_{\textstyle B}\\
\text{Reference size: }E
\quad ^{E}\big/_{\sim}
&\quad \quot{E}{\sim}
\quad ^{\textstyle E}\Big/_{\textstyle \sim}
\end{align*}
\end{document}
I give a very small amelioration which takes care of the different math styles.
% Source : http://tex.stackexchange.com/questions/4624/a-symbol-for-the-quotient-of-two-objects
\documentclass[10pt,a4paper]{article}
\usepackage[utf8x]{inputenc}
\usepackage{ucs}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\newcommand\quotient[2]{
\mathchoice
{% \displaystyle
\text{\raise1ex\hbox{$#1$}\Big/\lower1ex\hbox{$#2$}}%
}
{% \textstyle
#1\,/\,#2
}
{% \scriptstyle
#1\,/\,#2
}
{% \scriptscriptstyle
#1\,/\,#2
}
}
\newcommand{\setA}{{\cal O}_{(V',0)}}
\newcommand{\setB}{{\cal O}_{(V,0)}}
\begin{document}
One formula in one text : $\frac{4}{5} = \quotient{\setA}{\setB}$
and one formula alone...
$\frac{4}{5} = \quotient{\setA}{\setB}$
What about quotient of quotients ?
$\quotient{\left( \quotient{\setA}{\setB} \right)}{\left( \quotient{\setA}{\setB} \right)}$
Better like this ?
$\quotient{\textstyle \left( \quotient{\setA}{\setB} \right)}{\textstyle \left( \quotient{\setA}{\setB} \right)}$
\end{document} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9427560567855835, "perplexity": 3363.604448349262}, "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-2022-21/segments/1652662532032.9/warc/CC-MAIN-20220520124557-20220520154557-00765.warc.gz"} |
https://www.h2knowledgecentre.com/content/journal1071 | 1900
Thermal Efficiency of On-site, Small-scale Hydrogen Production Technologies using Liquid Hydrocarbon Fuels in Comparison to Electrolysis a Case Study in Norway
Abstract
The main goal of this study was to assess the energy efficiency of a small-scale, on-site hydrogen production and dispensing plant for transport applications. The selected location was the city of Narvik, in northern Norway, where the hydrogen demand is expected to be 100 kg/day. The investigated technologies for on-site hydrogen generation, starting from common liquid fossil fuels, such as heavy naphtha and diesel, were based on steam reforming and partial oxidation. Water electrolysis derived by renewable energy was also included in the comparison. The overall thermal efficiency of the hydrogen station was computed including compression and miscellaneous power consumption.
Related subjects:
Countries: | {"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": 0.9771960973739624, "perplexity": 4755.256646823127}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00102.warc.gz"} |
http://www.maplesoft.com/support/help/Maple/view.aspx?path=FunctionAdvisor | provide information on mathematical functions in general
Parameters
topics - literal name; 'topics'; specify that the FunctionAdvisor command return the topics for which information is available quiet - (optional) literal name; 'quiet'; specify that only the computational result in Maple syntax is returned Topic - (optional) name; FunctionAdvisor topic function - name; mathematical function or function class. For some topics, you can specify multiple mathematical functions opts - (optional) topic-specific options
Description
• The FunctionAdvisor() command returns basic instructions for the use of the FunctionAdvisor function.
• The FunctionAdvisor(function) command returns a summary of information related to the function function.
• The FunctionAdvisor(Topic, function) command returns information related to the topic Topic for the function function.
• The requirement concerning mathematical functions is not just computational. Typically, you need supporting information on definitions, identities, possible simplifications, integral forms, different types of series expansions, and mathematical properties in general. This information is in handbooks of mathematical functions like the one by Abramowitz and Stegun. You can now access this information directly from Maple, using the routines in the MathematicalFunctions package and the FunctionAdvisor command. This command is particularly useful when studying, teaching, and solving problems where mathematical function properties are relevant.
• Using the FunctionAdvisor command, you can access mathematical language information easily that is both readable and directly usable in Maple mathematical computations. The FunctionAdvisor command provides information on the following topics.
The FunctionAdvisor command provides information on the following mathematical functions.
Like the conversion facility for mathematical functions, the FunctionAdvisor command also works with the concept of function classes and considers assumptions on the function parameters, if any. The FunctionAdvisor command provides information on the following function classes.
• The FunctionAdvisor command can be considered to be between a help and a computational special function facility. Due to the wide range of information this command can handle and in order to facilitate its use, it includes two distinctive features:
– If you call the FunctionAdvisor command without arguments, it returns information that you can follow until the appropriate information displays.
– If you call the FunctionAdvisor command with a topic or function misspelled, but a match exists, it returns the information with a warning message.
You do not have to remember the exact Maple name of each mathematical function or the FunctionAdvisor topic. However, to avoid these messages and all FunctionAdvisor verbosity, specify the optional argument quiet when calling the FunctionAdvisor command from another routine.
Examples
The following example uses the FunctionAdvisor command with no arguments specified.
> $\mathrm{FunctionAdvisor}\left(\right)$
The usage is as follows: > FunctionAdvisor( topic, function, ... ); where 'topic' indicates the subject on which advice is required, 'function' is the name of a Maple function, and '...' represents possible additional input depending on the 'topic' chosen. To list the possible topics: > FunctionAdvisor( topics ); A short form usage, > FunctionAdvisor( function ); with just the name of the function is also available and displays a summary of information about the function.
> $\mathrm{FunctionAdvisor}\left(\mathrm{topics}\right)$
The topics on which information is available are:
$\left[{\mathrm{DE}}{,}{\mathrm{analytic_extension}}{,}{\mathrm{asymptotic_expansion}}{,}{\mathrm{branch_cuts}}{,}{\mathrm{branch_points}}{,}{\mathrm{calling_sequence}}{,}{\mathrm{class_members}}{,}{\mathrm{classify_function}}{,}{\mathrm{definition}}{,}{\mathrm{describe}}{,}{\mathrm{differentiation_rule}}{,}{\mathrm{function_classes}}{,}{\mathrm{identities}}{,}{\mathrm{integral_form}}{,}{\mathrm{known_functions}}{,}{\mathrm{periodicity}}{,}{\mathrm{plot}}{,}{\mathrm{relate}}{,}{\mathrm{required_assumptions}}{,}{\mathrm{series}}{,}{\mathrm{singularities}}{,}{\mathrm{special_values}}{,}{\mathrm{specialize}}{,}{\mathrm{sum_form}}{,}{\mathrm{symmetries}}{,}{\mathrm{synonyms}}{,}{\mathrm{table}}\right]$ (1)
To avoid all FunctionAdvisor verbosity, specify the optional argument quiet.
> $\mathrm{FunctionAdvisor}\left(\mathrm{function_classes},\mathrm{quiet}\right)$
$\left[{\mathrm{trig}}{,}{\mathrm{trigh}}{,}{\mathrm{arctrig}}{,}{\mathrm{arctrigh}}{,}{\mathrm{elementary}}{,}{\mathrm{GAMMA_related}}{,}{\mathrm{Psi_related}}{,}{\mathrm{Kelvin}}{,}{\mathrm{Airy}}{,}{\mathrm{Hankel}}{,}{\mathrm{Bessel_related}}{,}{\mathrm{0F1}}{,}{\mathrm{orthogonal_polynomials}}{,}{\mathrm{Ei_related}}{,}{\mathrm{erf_related}}{,}{\mathrm{Kummer}}{,}{\mathrm{Whittaker}}{,}{\mathrm{Cylinder}}{,}{\mathrm{1F1}}{,}{\mathrm{Elliptic_related}}{,}{\mathrm{Legendre}}{,}{\mathrm{Chebyshev}}{,}{\mathrm{2F1}}{,}{\mathrm{Lommel}}{,}{\mathrm{Struve_related}}{,}{\mathrm{hypergeometric}}{,}{\mathrm{Jacobi_related}}{,}{\mathrm{InverseJacobi_related}}{,}{\mathrm{Elliptic_doubly_periodic}}{,}{\mathrm{Weierstrass_related}}{,}{\mathrm{Zeta_related}}{,}{\mathrm{complex_components}}{,}{\mathrm{piecewise_related}}{,}{\mathrm{Other}}{,}{\mathrm{Bell}}{,}{\mathrm{Heun}}{,}{\mathrm{trigall}}{,}{\mathrm{arctrigall}}{,}{\mathrm{integral_transforms}}\right]$ (2)
The type of information that the FunctionAdvisor command returns general information, for example, "the Maple names for the Bessel functions",
> $\mathrm{FunctionAdvisor}&Ap$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "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": 0.8746899962425232, "perplexity": 1579.1008379903274}, "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-2017-30/segments/1500549424960.67/warc/CC-MAIN-20170725022300-20170725042300-00399.warc.gz"} |
http://math.stackexchange.com/questions/158791/recurrence-telescoping-tn-tn-1-1-n-and-tn-tn-1-log-n | # Recurrence telescoping $T(n) = T(n-1) + 1/n$ and $T(n) = T(n-1) + \log n$
I am trying to solve the following recurrence relations using telescoping. How would I go about doing it?
1. $T(n) = T(n-1) + 1/n$
2. $T(n) = T(n-1) + \log n$
thanks
-
In the first case you have $T(n) - T(n-1) = \frac{1}{n}$. If you sum with $n$ going from $2$ to $k$ you get:
\begin{align*} \big(T(k) - T(k-1)\big) &+ \big(T(k-1) + T((k-2)\big) + \big(T(k-2) + T((k-3)\big) + \cdots + \\ &+ \big(T(4) - T(3)\big) +\big(T(3) - T(2)\big) + \big(T(2) - T(1)\big) = T(k) - T(1) = \sum_{m=2}^k \frac{1}{m}\end{align*}
Notice how every term except $T(k)$ and $T(1)$ cancels out (I added "backwards" to make this more evident). This is the telescoping part. Then we get $T(n) = T(1) - 1 + H_n$, where $H_n$ is the $n$-th harmonic number $\displaystyle \sum_{m=1}^n \frac{1}{m}$ (I substracted $1$ because the first term $\displaystyle \frac{1}{1}$ doesn't appear in $\displaystyle \sum_{m=2}^n \frac{1}{m}$).
Likewise, in the second case you get: $\displaystyle T(n) = T(1) + \sum_{m=2}^n \log m = T(1) + \sum_{m=1}^n \log m$ (this last equality is given by $\log 1 = 0$; it doesn't make any difference but to me it looks nicer). Or you can use Cameron's hint: $T(n) = T(1) + \log n!$, by applying the product rule for logarithms.
The exact solution depends on the value of $T(1)$ in both cases.
-
Hint: a recurrence of the form $T(n) = T(n-1) + f(n)$ will have solutions $T(n) = C + \sum_{k=1}^n f(k)$.
- | {"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": 0.9921268820762634, "perplexity": 535.7466589023668}, "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-2016-07/segments/1454701151789.24/warc/CC-MAIN-20160205193911-00320-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://cs.stackexchange.com/questions/41681/maximize-distance-between-k-nodes-in-a-graph | Maximize distance between k nodes in a graph
I have an undirected unweighted graph $G$ and I want to select $k$ nodes from $G$ such that they are pairwise as far as possible from each other, in terms of geodesic distance. In other words they have to be spread around the graph as possible.
Let $d(u,v)$ be the length of a shortest path between $u$ and $v$ in $G$. Now, for a set of vertices $X \subseteq V(G)$, define $$d(X) = \sum_{\{u,v\} \subseteq X}d(u,v).$$
Let the problem SCATTERED SET be the problem which on input $G,k$ asks to find a set of $k$ vertices $X$ maximizing $d(X)$.
Is there an efficient algorithm solving SCATTERED SET?
• @D.W. The objective would be to maximize the distance between all the chosen nodes. So in your example 5,5,5 would be better since it would be 15. Another way to look at it is that I need to maximize the number of intermediate nodes in the graph one would need to traverse in order to eventually visit all the chosen nodes. Not so sure about clustering, do you have any particular approach in mind?
– jbx
Apr 22 '15 at 5:51
• This is somewhat similar to the geodetic number problem. A set of vertices $X$ of a graph $G$ is geodetic if every vertex of $G$ lies on a shortest path between two (not necessarily) distinct vertices in $X$. Given a graph $G$ and an integer $k$, the problem is to decide if $G$ has a geodetic set of cardinality at most $k$. The problem is NP-complete even for chordal bipartite graphs.
– Juho
Apr 22 '15 at 6:34
• The objective can be rewritten as to maximizing the average distance. Apr 22 '15 at 7:23
• somewhat similar to finding hubs/ centers in a graph
– vzn
Jul 1 '15 at 21:43
I don't know if there is a polynomial-time algorithm (it feels like it might be NP-hard), but here are some plausible algorithmic approaches you might could consider, if you need to solve it in practice:
Heuristics
One well-explored algorithm is Furthest Point First (FPF). At each iteration, it chooses a point that is furthest from the set of points selected so far. Iterate $k$ times. As this is a greedy strategy, there is no reason to expect this to give an optimal answer or even close to optimal, and it was designed to optimize a slightly different objective function... but in some contexts it gives a reasonable approximation, so it could be worth a try.
FPF comes out of the literature on graph-based clustering and was introduced in the following research paper:
Teofilo F. Gonzalez. Clustering to minimize the maximum intercluster distance. Theoretical Computer Science, vol 38, pp.293-306, 1985.
You could try exploring the literature on graph-based clustering to see if anyone has studied your specific problem.
Exact algorithms
If you have this problem in practice and need an exact optimum solution, you could try to solve this using an ILP solver.
Here's how. Introduce 0-or-1 variables $x_i$, where $x_i$ indicates whether the $i$th vertex was selected, and 0-or-1 variables $y_{i,j}$, with the intended meaning that $y_{i,j}=1$ only if $x_i=1$ and $x_j=1$. Now maximize the objective function $\sum_{i,j} d(i,j) y_{i,j}$, subject to the constraints $\sum x_i \le k$ and $x_i \ge y_{i,j}$ and $x_j \ge y_{i,j}$. Now solve this ILP with an off-the-shelf ILP solver. As ILP is NP-hard, there's no guarantee this will be efficient, but it might work on some problem instances.
Another approach is to use weighted MAX-SAT. In particular, introduce boolean variables $x_i$, where $x_i$ is true if the $i$th vertex was selected, and variables $y_{i,j}$. The formula is $\phi \land \land_{i,j} y_{i,j}$, where $\phi$ must be true (its clauses have weight $W$ for some very large $W$) and each clause $y_{i,j}$ is given weight $d(i,j)$. Define the formula $\phi$ to be true if at most $k$ of the $x_i$'s are true (see here for details on how to do that) and if $y_{i,j}=x_i \land x_j$ for all $i,j$. Now the solution to this weighted MAX-SAT problem is the solution to the original problem, so you could try throwing a weighted MAX-SAT solver at the problem. The same caveats apply.
No, the problem is NP-complete.
Let $(G,k)$ be an instance of INDEPENDENT SET. Construct $G'$ by adding a universal vertex $u$ to $G$. The crucial observation is that the distance between two vertices in $G$ is 1 in $G'$ if and only if they are adjacent in $G$, and 2 otherwise.
Now, the optimal solution of SCATTERED SET on input $(G',k)$ is $2\binom{k}{2}$ if and only if $G$ has an independent set of size $k$. | {"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": 0.8410205245018005, "perplexity": 151.64531265048893}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057787.63/warc/CC-MAIN-20210925232725-20210926022725-00514.warc.gz"} |
https://cstheory.stackexchange.com/questions/36630/notion-similar-to-k-wise-independence | # Notion similar to k-wise independence
I want to construct a family of functions $H:\{0,1\}^n \rightarrow \{0,1\}$ with a property that is similar to k-wise independence. Specifically, I want $H$ to satisfy the following property. Let $k$ be some large natural number that is fixed. There exists a $d$ such that for every $n^k$ tuple of $n$ bit strings $x_1,...,x_{n^k}$,
$Pr_{h \leftarrow H}[ \cap_{j=2}^{n^k} h(x_j)=0 \, \cap \, h(x_1)=1] \geq 1/n^d$
The additional (and crucial constraint) is that I want to be able to compute this family of functions using (a family of) circuits of size at most $n^3$ ($k$ is much bigger than 3). Note that if we don't have this constraint, then it is possible to do this (using family of circuits of size at least $n^e$).
Is it even possible for such a family to exist?
We'll imagine picking a hash function from $H$ as follows: first, pick $w_1^0,\ldots,w_n^0,w_1^1,\ldots,w_n^1$ uniformly and independently from integer weights in $[1,4n]$. Then pick a threshold $T$ in $[1,4n^2]$ also uniformly and independently at random. Let $w(x) = \sum_i x_iw_i^1 + (1-x_i)w_i^0$. Now define $h(x)$ to be one if $w(x) \le T$ and zero otherwise.
We can then lower-bound the probability you're interested in by \begin{align*} P_{h\gets H}[ h(x_2) = \cdots = h(x_{n^k}) = 0, h(x_1) = 1 ] \\ &\ge P_{h \gets H}[ \forall i > 1 : w(x_1) < w(x_i), T = w(x_1) ] \\ &= P_{h \gets H}[ \forall i > 1 : w(x_1) < w(x_i) ] \cdot P_{h\gets H}[ w(x_1) = T \mid \forall i > 1 : w(x_1) < w(x_i) ] \end{align*} which is to say that $x_1$ is the unique minimum-weight element, and $T$ equals $w(x_1)$. We can easily compute the right factor using the uniformity and independence of $T$: $$P_{h\gets H}[ w(x_1) = T \mid \forall i > 1: w(x_1) < w(x_i) ] = \frac{1}{4n^2}$$ Bounding the other factor will involve the isolation lemma. The lemma tells us that $$P_{h \gets H}[\exists j \forall i\ne j : w(x_j) < w(x_i)] \ge 1-\frac{2n}{4n} = \frac{1}{2}$$ By a union bound, the left hand side is at most $$\sum_{j=1}^{n^k} P_{h \gets H}[\forall i\ne j: w(x_j) < w(x_i)]$$ From here, observe that the distribution of $w(x)$ doesn't depend on $x$. Hence, the terms in the above sum are all identical (they don't depend on $j$). Combining with the previous inequality and rearranging, we thus have $$P_{h \gets H}[\forall i > 1 : w(x_1) < w(x_i)] \ge \frac{1}{2n^k}$$ Plugging this into our earlier work, we conclude $$P_{h\gets H}[ h(x_2) = \cdots = h(x_{n^k}) = 0, h(x_1) = 1 ] \ge \frac{1}{8n^{k+2}}$$ Lastly, each $h$ can be computed very easily: a fixed $h$ needs $O(n\log(n))$ bits to store the $w_i^b$ and $T$, and then some circuitry to sum the appropriate weights and compare against $T$. | {"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": 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": 0.9923492670059204, "perplexity": 216.63800273379084}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575541317967.94/warc/CC-MAIN-20191216041840-20191216065840-00528.warc.gz"} |
https://support.numxl.com/hc/en-us/articles/115000142626-Understanding-Exponential-Weighted-Volatility-EWMA- | # Understanding Exponential Weighted Volatility (EWMA)
After receiving several inquiries about the exponential weighted moving average (EWMA) function in NumXL, we decided to dedicate this issue to exploring this simple function in greater depth.
The main objective of EWMA is to estimate the next-day (or period) volatility of a time series and closely track the volatility as it changes.
## Background
Define $\sigma_n$ as the volatility of a market variable on day n, as estimated at the end of day n-1. The variance rate is The square of volatility,$\sigma_n^2$, on day n.
Suppose the value of the market variable at the end of day i is $S_i$. The continuously compounded rate of return during day i (between end of prior day (i.e. i-1) and end of day i) is expressed as:
$$r_t = \ln{\frac{S_i}{S_{i-1}}}$$
Next, using the standard approach to estimate $\sigma_n$ from historical data, we’ll use the most recent m-observations to compute an unbiased estimator of the variance:
$$\sigma_n^2=\frac{\sum_{i=1}^m (r_{n-i}-\bar r)^2}{m-1}$$
Where $\bar r$ is the mean of $r_{i}$:
$$\bar r = \frac{\sum_{i=1}^m r_{n-i}}{m}$$
Next, let’s assume $\bar r = 0$ and use the maximum likelihood estimate of the variance rate:
$$\sigma_n^2=\frac{\sum_{i=1}^m r_{n-i}^2}{m}$$
So far, we have applied equal weights to all $r_i^2$, so the definition above is often referred to as the equally-weighted volatility estimate.
Earlier, we stated our objective was to estimate the current level of volatility $\sigma_n$, so it makes sense to give higher weights to recent data than to older ones. To do so, let’s express the weighted variance estimate as follows:
$$\sigma_n^2=\sum_{i=1}^m \alpha_i \times r_{n-i}^2$$
Where:
• $\alpha_i$ is the amount of weight given to an observation i-days ago.
• $\alpha_i\geq 0$
• $\sum_{i=1}^m\alpha_i=1$
So, to give higher weight to recent observations, $\alpha_i \geq \alpha_{i+1}$
### Long-run average variance
A possible extension of the idea above is to assume there is a long-run average variance $V_L$, and that it should be given some weight:
$$\sigma_n^2=\gamma V_L+\sum_{i=1}^m \alpha_i \times r_{n-i}^2$$
Where:
• $\gamma+\sum_{i=1}^m\alpha_i=1$
• $V_L > 0$
The model above is known as the ARCH (m) model, proposed by Engle in 1994.
$$\sigma_n^2=\omega+\sum_{i=1}^m \alpha_i \times r_{n-i}^2$$
## EWMA
EWMA is a special case of the equation above. In this case, we make it so that the weights of variable $\alpha_i$ decrease exponentially as we move back through time.
$$\alpha_{i+1}=\lambda \alpha_i = \lambda^2 \alpha_{i-1} = \cdots = \lambda^{n+1}\alpha_{i-n}$$
Unlike the earlier presentation, the EWMA includes all prior observations, but with exponentially declining weights throughout time.
Next, we apply the sum of weights such that they equal the unity constraint:
$$\sum_{i=1}^\infty \alpha_i = \alpha_1 \sum_{i=1}^\infty \lambda^i=1$$
For $\left | \lambda \right | < 1$, the value of $\alpha_1=1-\lambda$
Now we plug those terms back into the equation. For the $\sigma_{n-1}^2$ estimate:
$$\sigma_{n-1}^2=\sum_{i=1}^{n-1}\alpha_i r_{n-i-1}^2=\alpha_1 r_{n-2}^2+\lambda\alpha_1 r_{n-3}^2+\cdots+\lambda^{n-3}\alpha_1 r_1^2$$ $$\sigma_{n-1}^2=(1-\lambda)(r_{n-2}^2+\lambda r_{n-3}^2+\cdots+\lambda^{n-3} r_1^2)$$
And the $\sigma_n^2$ estimate can be expressed as follows:
$$\sigma_n^2=(1-\lambda)(r_{n-1}^2+\lambda r_{n-2}^2+\cdots+\lambda^{n-2} r_1^2)$$ $$\sigma_n^2=(1-\lambda)r_{n-1}^2+\lambda(1-\lambda)(r_{n-2}^2+\lambda r_{n-3}^2+\cdots+\lambda^{n-3} r_1^2)$$ $$\sigma_n^2=(1-\lambda)r_{n-1}^2+\lambda\sigma_{n-1}^2$$
Now, to understand the equation better:
$$\sigma_n^2=(1-\lambda)r_{n-1}^2+\lambda\sigma_{n-1}^2$$ $$\sigma_n^2=(1-\lambda)r_{n-1}^2+\lambda ((1-\lambda)r_{n-2}^2+\lambda\sigma_{n-2}^2)$$ $$\cdots$$ $$\sigma_n^2=(1-\lambda)(r_{n-1}^2+\lambda r_{n-2}^2+\lambda^2 r_{n-3}^2+\cdots+\lambda^{k+1} r_{n-k}^2)+\lambda^{k+2}\sigma_{n-k}$$
For a larger data set, the $\lambda^{k+2}\sigma_{n-k}$ is sufficiently small to be ignored from the equation.
The EWMA approach has one attractive feature: it requires relatively little stored data. To update our estimate at any point, we only need a prior estimate of the variance rate and the most recent observation value.
A secondary objective of EWMA is to track changes in the volatility. For small values, recent observations affect the estimate promptly. For $\lambda$values closer to one, the estimate changes slowly based on recent changes in the returns of the underlying variable.
The RiskMetrics database (produced by JP Morgan and made public available) uses the EWMA with $\lambda=0.94$ for updating daily volatility.
IMPORTANT: The EWMA formula does not assume a long run average variance level. Thus, the concept of volatility mean reversion is not captured by the EWMA. The ARCH/GARCH models are better suited for this purpose.
### Lambda
A secondary objective of EWMA is to track changes in the volatility, so for small$\lambda$ values, recent observation affect the estimate promptly, and for $\lambda$ values closer to one, the estimate changes slowly to recent changes in the returns of the underlying variable.
The RiskMetrics database (produced by JP Morgan) and made public available in 1994, uses the EWMA model with $\lambda=0.94$ for updating daily volatility estimate. The company found that across a range of market variables, this value of $\lambda$ gives forecast of the variance that come closest to realized variance rate. The realized variance rates on a particular day was calculated as an equally-weighted average of $r_i^2$ on the subsequent 25 days.
$$\sigma_n^2=\frac{\sum_{i=0}^{24} r_{n-i}^2}{25}$$
Similarly, to compute the optimal value of lambda for our data set, we need to calculate the realized volatility at each point. There are several methods, so pick one. Next, calculate the sum of squared errors (SSE) between EWMA estimate and realized volatility. Finally, minimize the SSE by varying the lambda value.
Sounds simple? It is. The biggest challenge is to agree on an algorithm to compute realized volatility. For instance, the folks at RiskMetrics chose the subsequent 25-day to compute realized variance rate. In your case, you may choose an algorithm that utilizes Daily Volume, HI/LO and/or OPEN-CLOSE prices. | {"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": 0.9796021580696106, "perplexity": 928.4187535005733}, "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-2021-17/segments/1618038088245.37/warc/CC-MAIN-20210416161217-20210416191217-00415.warc.gz"} |
http://math.stackexchange.com/questions/7122/determining-orthonormal-matrix-of-rank-n-with-special-first-row/7124 | # determining orthonormal matrix of rank N with special first row
Is there a more efficient algorithm besides Gram-Schmidt that would produce an orthonormal matrix of rank N, with first row equal to [1 1 1 1 1 ... 1] / sqrt(N)?
e.g. for N = 3, the matrix \begin{align} \mathsf A_3 &= \begin{bmatrix} 1/\sqrt 3 & 1/\sqrt 3 & 1/\sqrt 3 \\ 2/\sqrt 6 & -1/\sqrt 6 & -1/\sqrt 6 \\ 0 & 1/\sqrt 2 & -1/\sqrt 2 \end{bmatrix}\end{align} suffices, but I'm not sure how to generalize.
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please help me with my math latex, I tried adapting from other posts but can't seem to get the matrix rows to show up. – Jason S Oct 18 '10 at 14:05
ah-- thanks! so you have to escape the \\\\ symbols. – Jason S Oct 18 '10 at 14:10
If you'd settle for unitary matrices, you could use a Fourier matrix! – Mariano Suárez-Alvarez Oct 18 '10 at 14:29
yeah, I had thought of that (see my answer) unfortunately I'm looking for real-valued matrices. – Jason S Oct 18 '10 at 17:40
There's a reflection swapping $v=(\sqrt{n},0,\ldots,0)$ with $w=(1,1,\ldots,1)$ which will do what you want. It will fix $v+w$ and all vectors orthogonal to $v$ and $w$. It will negate $v-w$. It will be a symmetric matrix; its first row and column is all $1/\sqrt{n}$. The remaining entries $a_{i,j}$ for $i$, $j\ge2$ will depend only on whether $i=j$ or not. This should be enough information to reconstruct it.
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+1, thanks. (p.s. this isn't meant as a criticism, but I am amused by the difference between programmer + mathematician mindsets: "solved" in math circles seems to be proof of existence, "solved" in programmer circles is a description of an algorithm) – Jason S Oct 18 '10 at 15:02
I'm familiar w/ Householder reflections: is there an easy way to get the normal vector n from the two vectors "v" and "w" as you described? – Jason S Oct 18 '10 at 15:04
As I said, the vector that gets negated is $v-w$. – Robin Chapman Oct 18 '10 at 17:53
ah -- great, thanks, I didn't pick up on that one. – Jason S Oct 18 '10 at 20:52
Hmmm. I don't follow why "its first row and column is all 1/sqrt(n)" is consistent with it being a reflection swapping v and w. – Jason S Oct 18 '10 at 20:59
Given the special form, you can probably construct a rotation matrix that would turn one of the standard basis vectors there.
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oh, that makes sense. Any further suggestions how to do that? – Jason S Oct 18 '10 at 14:09
haven't you just rephrased the question? – Mariano Suárez-Alvarez Oct 18 '10 at 14:30
You should be able to use K-1 rows of Normalized Hadamard matrices of size K, where K is a power of 2 as building blocks. (How to normalize see this: http://math.stackexchange.com/questions/4797/what-is-sum-of-rows-of-hadamard-matrix/4798#4798) You should be able to construct these rows recursively as K is a power of 2.
Something like
[1 1 .. 1 ]
[ H_1 | 0 |0 0 ]
[ 0 | H_2|0 0 ]
[ 0 | 0 | H_4]
etc
At the end O(LogN) rows will remain to be filled, which I suppose you can use Gram-Schmidt on.
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I took @Robin Chapman's answer and ran with it:
The matrix A that is the Householder reflection of v-w where v = [sqrt(N), 0, 0, 0, ... ] and w = [1 1 1 1 1 ... 1] can be defined by:
A(1,i) = A(i,1) = 1/sqrt(N)
A(i,i) = 1 - K for i >= 2
A(i,j) = -K for i,j >= 2, i != j
K = 1/sqrt(N)/(sqrt(N)-1)
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hmm, I think one possibility for orthonormal vectors for N odd is
[ 1 1 1 1 1 ... 1 ]
[ 1, cos phi, cos 2*phi, cos 3*phi, ... cos (N-1)*phi ]
[ 0, sin phi, sin 2*phi, sin 3*phi, ... sin (N-1)*phi ]
[ 1, cos 2*phi, cos 4*phi, cos 6*phi, ... cos 2*(N-1)*phi ]
[ 0, sin 2*phi, sin 4*phi, sin 6*phi, ... sin 2*(N-1)*phi ]
...
[ 1, cos k*phi, cos 2*k*phi, cos 3*k*phi, ... cos k*(N-1)*phi ]
[ 0, sin k*phi, sin 2*k*phi, sin 3*k*phi, ... sin k*(N-1)*phi ]
where k = (N-1)/2
but I'm not sure how to extend to N even.
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Take the transpose of this pattern $$\left( \begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 \end{array} \right).$$ and divide what will now be the rows by a suitable square root, different for each row.
Here is the transpose that you want, $$\left( \begin{array}{rrrrrrrrrr} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ -1 & -1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ -1 & -1 & -1 & 3 & 0 & 0 & 0 & 0 & 0 & 0 \\ -1 & -1 & -1 & -1 & 4 & 0 & 0 & 0 & 0 & 0 \\ -1 & -1 & -1 & -1 & -1 & 5 & 0 & 0 & 0 & 0 \\ -1 & -1 & -1 & -1 & -1 & -1 & 6 & 0 & 0 & 0 \\ -1 & -1 & -1 & -1 & -1 & -1 & -1 & 7 & 0 & 0 \\ -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & 8 & 0 \\ -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & 9 \end{array} \right).$$
That worked.
For row 1, divide by $\sqrt n.$ For row $i$ with $2 \leq i \leq n,$ divide by $\sqrt{i^2 - i},$ which is double a triangular number. Note that the diagonal entry in row $i$ is $i-1,$ just one of those things. So, except for the first row, if you call the diagonal entry $d,$ divide by $\sqrt{d^2 + d}.$ Which works out the same.
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http://sites.millersville.edu/bikenaga/basic-algebra/inverse-functions/inverse-functions.html | Inverse Functions
Functions and are inverses if
for all and . If f has an inverse, it is often denoted . However, does not mean " "!
Example. and are inverses, since
if x is a real number.
Notice that but the inverse is not !
Example. Functions which are inverses "undo" one another. Thus, if f and are inverses and f takes 4 to 17, then must take 17 to 4.
In symbols,
Example. In some cases, it's possible to find the inverse of a function algebraically. I'll find the inverse of .
First, I'll write it as .
Next, switch x's and y's:
(This means you should replace each "x" with a "y" and replace each "y" with an "x".)
Now solve for y in terms of x:
Since I was able to solve for y in terms of x, the result is the inverse function: .
The procedure I used tells something about the relation between the graphs of a function and its inverse. Since the inverse is obtained by swapping x's and y's, the graph of is a mirror image of the graph of f across the line .
In the picture below, I've shown the graphs of , , and :
Example. Find the inverse of .
Let . Swap x's and y's to obtain
Solve for y:
Therefore, .
Example. Find the inverse of .
Let . Swap x's and y's to obtain
Solve for y:
Hence, .
Not every function has an inverse. For example, consider . Now , so should take 4 back to 2. But as well, so apparently should take 4 to -2. can't do both, so there is no inverse! The problem is that you can't undo the effect of the squaring function in a unique way.
On the other hand, if I restrict to , then it has an inverse function: .
A function f is one-to-one or injective if different inputs go to different outputs:
Example. is not one-to-one, because different inputs can produce the same output. For example,
On the other hand, is one-to-one. For suppose the inputs a and b produce the same output: . Then
Then
That is, the inputs a and b were the same to begin with.
A graph of a function represents a one-to-one function if every horizontal line hits the graph at most once.
A one-to-one function has an inverse: Since a given output could have only come from one input, you can undo the effect of the function.
Contact information | {"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": 0.9495655298233032, "perplexity": 466.8132825677426}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084887065.16/warc/CC-MAIN-20180118032119-20180118052119-00231.warc.gz"} |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/The_Four_Laws_of_Thermodynamics/First_Law_of_Thermodynamics | # 1st Law of Thermodynamics
To understand and perform any sort of thermodynamic calculation, we must first understand the fundamental laws and concepts of thermodynamics. For example, work and heat are interrelated concepts. Heat is the transfer of thermal energy between two bodies that are at different temperatures and is not equal to thermal energy. Work is the force used to transfer energy between a system and its surroundings and is needed to create heat and the transfer of thermal energy. Both work and heat together allow systems to exchange energy. The relationship between the two concepts can be analyzed through the topic of Thermodynamics, which is the scientific study of the interaction of heat and other types of energy.
## Introduction
To understand the relationship between work and heat, we need to understand a third, linking factor: the change in internal energy. Energy cannot be created nor destroyed, but it can be converted or transferred. Internal energy refers to all the energy within a given system, including the kinetic energy of molecules and the energy stored in all of the chemical bonds between molecules. With the interactions of heat, work and internal energy, there are energy transfers and conversions every time a change is made upon a system. However, no net energy is created or lost during these transfers.
Law of Thermodynamics
The First Law of Thermodynamics states that energy can be converted from one form to another with the interaction of heat, work and internal energy, but it cannot be created nor destroyed, under any circumstances. Mathematically, this is represented as
$\Delta U=q + w \label{1}$
with
• $$ΔU$$ is the total change in internal energy of a system,
• $$q$$ is the heat exchanged between a system and its surroundings, and
• $$w$$ is the work done by or on the system.
Work is also equal to the negative external pressure on the system multiplied by the change in volume:
$w=-p \Delta V \label{2}$
where $$P$$ is the external pressure on the system, and $$ΔV$$ is the change in volume. This is specifically called "pressure-volume" work.
The internal energy of a system would decrease if the system gives off heat or does work. Therefore, internal energy of a system increases when the heat increases (this would be done by adding heat into a system). The internal energy would also increase if work were done onto a system. Any work or heat that goes into or out of a system changes the internal energy. However, since energy is never created nor destroyed (thus, the first law of thermodynamics), the change in internal energy always equals zero. If energy is lost by the system, then it is absorbed by the surroundings. If energy is absorbed into a system, then that energy was released by the surroundings:
$\Delta U_{system} = -\Delta U_{surroundings}$
where ΔUsystem is the total internal energy in a system, and ΔUsurroundingsis the total energy of the surroundings.
Table 1
Process Sign of heat (q) Sign of Work (w)
Work done by the system N/A -
Work done onto the system N/A +
Heat released from the system- exothermic (absorbed by surroundings) - N/A
The above figure is a visual example of the First Law of Thermodynamics. The blue cubes represent the system and the yellow circles represent the surroundings around the system. If energy is lost by the cube system then it is gained by the surroundings. Energy is never created nor destroyed. Since the area of the clue cube decreased the visual area of the yellow circle increased. This symbolizes how energy lost by a system is gained by the surroundings. The affects of different surroundings and changes on a system help determine the increase or decrease of internal energy, heat and work.
Table 2
The Process Internal Energy Change Heat Transfer of Thermal Energy (q) Work (w=-PΔV) Example
q=0 Adiabatic + 0 + Isolated system in which heat does not enter or leave similar to styrofoam
ΔV=0 Constant Volume + + 0 A hard, pressure isolated system like a bomb calorimeter
Constant Pressure + or - enthalpy (ΔH) -PΔV Most processes occur are constant external pressure
ΔT=0 Isothermal 0 + - There is no change in temperature like in a temperature bath
Example $$\PageIndex{1}$$
A gas in a system has constant pressure. The surroundings around the system lose 62 J of heat and does 474 J of work onto the system. What is the internal energy of the system?
Solution
To find internal energy, ΔU, we must consider the relationship between the system and the surroundings. Since the First Law of Thermodynamics states that energy is not created nor destroyed we know that anything lost by the surroundings is gained by the system. The surrounding area loses heat and does work onto the system. Therefore, q and w are positive in the equation ΔU=q+w because the system gains heat and gets work done on itself.
\begin{align} ΔU &= (62\,J) + (474\,J) \\[4pt] &= 536\,J \end{align}
Example $$\PageIndex{2}$$
A system has constant volume (ΔV=0) and the heat around the system increases by 45 J.
1. What is the sign for heat (q) for the system?
2. What is ΔU equal to?
3. What is the value of internal energy of the system in Joules?
Solution
Since the system has constant volume (ΔV=0) the term -PΔV=0 and work is equal to zero. Thus, in the equation ΔU=q+w w=0 and ΔU=q. The internal energy is equal to the heat of the system. The surrounding heat increases, so the heat of the system decreases because heat is not created nor destroyed. Therefore, heat is taken away from the system making it exothermic and negative. The value of Internal Energy will be the negative value of the heat absorbed by the surroundings.
1. negative (q<0)
2. ΔU=q + (-PΔV) = q+ 0 = q
3. ΔU = -45J | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9066231846809387, "perplexity": 420.9987333623941}, "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/1631780057421.82/warc/CC-MAIN-20210923104706-20210923134706-00135.warc.gz"} |
https://www.isa-afp.org/entries/Wetzels_Problem.html | # Wetzel's Problem and the Continuum Hypothesis
February 18, 2022
### Abstract
Let $F$ be a set of analytic functions on the complex plane such that, for each $z\in\mathbb{C}$, the set $\{f(z) \mid f\in F\}$ is countable; must then $F$ itself be countable? The answer is yes if the Continuum Hypothesis is false, i.e., if the cardinality of $\mathbb{R}$ exceeds $\aleph_1$. But if CH is true then such an $F$, of cardinality $\aleph_1$, can be constructed by transfinite recursion. The formal proof illustrates reasoning about complex analysis (analytic and homomorphic functions) and set theory (transfinite cardinalities) in a single setting. The mathematical text comes from Proofs from THE BOOK by Aigner and Ziegler. | {"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": 0.9656897783279419, "perplexity": 299.5983366831464}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334596.27/warc/CC-MAIN-20220925193816-20220925223816-00075.warc.gz"} |
https://www.jiskha.com/questions/368138/I-have-to-make-this-half-circle-from-a-piece-of-white-paper-It-has-to-be-circle | English creative project
I have to make this half circle from a piece of white paper.
It has to be circle shaped like a beach ball but made out of paper and only half of a circle.
Does anyone know how to make this with a piece of paper?
Could someone tell me a video too
Thanks
1. 0
1. posted by Writeacher
2. Half of a sphere (3D)
posted by Amy~
3. Read through all the links in the first search results above.
posted by Writeacher
4. Another thought --
Use papier mache and a volleyball or other ball for a form. Papier mache is made from paper, but not "a piece of paper."
posted by Writeacher
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http://mathhelpforum.com/calculus/179352-fourier-series.html | 1. ## Fourier Series
I want to find the fourier series for the following function:
f(t)= cost (t) t>=0 and = 0 cos(t)<0
Now since the function is 2pi periodic and even the sin coeff will be 0 and I need only to calculate the fourier cosine.
But what should I use for limits?Becuase the coefficient,defined by the integral between the function and cos(kt), will be equal to cos(kt)sin(t),which is 0 if the integral is defined from 0 to pi os 2pi=>the fourier cosine to be equal to 0.Any help?
2. Originally Posted by StefanM
I want to find the fourier series for the following function:
f(t)= cost (t) t>=0 and = 0 cos(t)<0
This is pretty unclear: did you mean f(t) = cos(t) when t >= 0 , and f(t) = 0 when t < 0, or you did mean f(t) = cos(t) whenever cos(t) >= 0 ,and
f (t) = 0 otherwise? I think it should be the second one since then the function is clearly periodic, but you better make it clearer.
Tonio
Now since the function is 2pi periodic and even the sin coeff will be 0 and I need only to calculate the fourier cosine.
But what should I use for limits?Becuase the coefficient,defined by the integral between the function and cos(kt), will be equal to cos(kt)sin(t),which is 0 if the integral is defined from 0 to pi os 2pi=>the fourier cosine to be equal to 0.Any help?
.
3. the second part.....f(t) = cos(t) whenever cos(t) >= 0 ,and f (t) = 0 otherwise
4. Originally Posted by StefanM
the second part.....f(t) = cos(t) whenever cos(t) >= 0 ,and f (t) = 0 otherwise
Ok, but then why do you think the limits of the integrals defining the coefficients are between 0 and 2*pi? If you sketch the function f(t) you can see it is
periodic 2*pi, but different from zero ONLY in (-pi/2 , pi/2) , so these must be the integrals' limits times 1/pi each.
Unfortunately we have no latex anymore so it's impossilbe for me to write down my mind, but if you do it you'll get the integral of cos(t)cos(nt) , which equals
sin(n-1)x/2(n-1) + sin(n+1)x/2(n+1) , and then you have to distinguish between n = 2 (mod 4) and n = 0 (mod 4) , which gives you
2/(n^2-1) and -2/(n^2-1), resp. (and all the time divided by pi)...
Tonio | {"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": 0.9836071133613586, "perplexity": 974.7680011976595}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187820927.48/warc/CC-MAIN-20171017052945-20171017072945-00645.warc.gz"} |
http://caelinux.org/wiki/index.php?title=Doc:FEM_Learning:Finite_Element_Method&diff=7147 | # Difference between revisions of "Doc:FEM Learning:Finite Element Method"
2D FEM solution for a magnetostatic configuration (lines denote the direction of calculated flux density and colour - its magnitude)
2D mesh for the image above (mesh is denser around the object of interest)
The finite element method (FEM) is used for finding approximate solutions of partial differential equations (PDE) as well as of integral equations such as the heat transport equation. The solution approach is based either on eliminating the differential equation completely (steady state problems), or rendering the PDE into an approximating system of ordinary differential equations, which are then solved using standard techniques such as finite differences, Runge-Kutta, etc.
In solving partial differential equations, the primary challenge is to create an equation that approximates the equation to be studied, but is numerically stable, meaning that errors in the input data and intermediate calculations do not accumulate and cause the resulting output to be meaningless. There are many ways of doing this, all with advantages and disadvantages. The Finite Element Method is a good choice for solving partial differential equations over complex domains (like cars and oil pipelines), when the domain changes (as during a solid state reaction with a moving boundary), when the desired precision varies over the entire domain, or when the solution lacks smoothness. For instance, in simulating the weather pattern on Earth, it is more important to have accurate predictions over land than over the wide-open sea, a demand that is achievable using the finite element method.
## History
The finite-element method originated from the needs for solving complex elasticity, structural analysis problems in civil engineering and aeronautical engineering. Its development can be traced back to the work by Alexander Hrennikoff (1941) and Richard Courant (1942). While the approaches used by these pioneers are dramatically different, they share one essential characteristic: mesh discretization of a continuous domain into a set of discrete sub-domains. Hrennikoff's work discretizes the domain by using a lattice analogy while Courant's approach divides the domain into finite triangular subregions for solution of second order elliptic partial differential equations (PDEs) that arise from the problem of torsion of a cylinder. Courant's contribution was evolutionary, drawing on a large body of earlier results for PDEs developed by Rayleigh, Ritz, and Galerkin. Development of the finite element method began in earnest in the middle to late 1950s for airframe and structural analysis and gathered momentum at the University of Stuttgart through the work of John Argyris and at Berkeley through the work of Ray W. Clough in the 1960s for use in civil engineering.[1] The method was provided with a rigorous mathematical foundation in 1973 with the publication of Strang and Fix's An Analysis of The Finite Element Method, and has since been generalized into a branch of applied mathematics for numerical modeling of physical systems in a wide variety of engineering disciplines, e.g., electromagnetism and fluid dynamics.
The development of the finite element method in structural mechanics is often based on an energy principle, e.g., the virtual work principle or the minimum total potential energy principle, which provides a general, intuitive and physical basis that has a great appeal to structural engineers.
## Technical discussion
We will illustrate the finite element method using two sample problems from which the general method can be extrapolated. It is assumed that the reader is familiar with calculus and linear algebra.
P1 is a one-dimensional problem
$\Large \mbox{P1}:\begin{cases}u''=f \mbox{ in }(0,1),\\u(0)=u(1)=0,\end{cases}$
where $f$ is given, $u$ is an unknown function of $x$, and $u$ is the second derivative of $u$ with respect to $x$.
The two-dimensional sample problem is the Dirichlet problem
$\Large \mbox{P2}:\begin{cases} u_{xx}+u_{yy}=f & \mbox{ in } \Omega, \\ u=0 & \mbox{ on } \partial \Omega, \end{cases}$
where $\Omega$ is a connected open region in the $(x,y)$ plane whose boundary $\partial \Omega$ is "nice" (e.g., a smooth manifold or a polygon), and $u_{xx}$ and $u_{yy}$ denote the second derivatives with respect to $x$ and $y$, respectively.
The problem P1 can be solved "directly" by computing antiderivatives. However, this method of solving the boundary value problem works only when there is only one spatial dimension and does not generalize to higher-dimensional problems or to problems like $u + u = f$. For this reason, we will develop the finite element method for P1 and outline its generalization to P2.
Our explanation will proceed in two steps, which mirror two essential steps one must take to solve a boundary value problem (BVP) using the FEM. In the first step, one rephrases the original BVP in its weak, or variational form. Little to no computation is usually required for this step, the transformation is done by hand on paper. The second step is the discretization, where the weak form is discretized in a finite dimensional space. After this second step, we have concrete formulae for a large but finite dimensional linear problem whose solution will approximately solve the original BVP. This finite dimensional problem is then implemented on a computer.
## Variational formulation
The first step is to convert P1 and P2 into their variational equivalents. If $u$ solves P1, then for any smooth function $v$ that satisfies the displacement boundary conditions, i.e. $v = 0$ at $x = 0$ and $x = 1$,we have
(1)$\int_0^1 f(x)v(x) \, dx = \int_0^1 u''(x)v(x) \, dx$
Conversely, if for a given $u$, (1) holds for every smooth function $v(x)$ then one may show that this $u$ will solve P1. (The proof is nontrivial and uses Sobolev spaces.)
By using integration by parts on the right-hand-side of (1), we obtain
(2)\begin{align} \int_0^1 f(x)v(x) \, dx & = &\int_0^1 u''(x)v(x) \, dx \\ & = &u'(x)v(x)|_0^1-\int_0^1 u'(x)v'(x) \, dx \\ & = &-\int_0^1 u'(x)v'(x) \, dx = -\phi (u,v). \end{align}
where we have used the assumption that $v(0) = v(1) = 0$.
### A proof outline of existence and uniqueness of the solution
We can define $H_0^1(0,1)$ to be the absolutely continuous functions of $(0,1)$ that are $0$ at $x = 0$ and $x = 1$. Such function are "once differentiable" and it turns out that the symmetric bilinear map $\!\,\phi$ then defines an inner product which turns $H_0^1(0,1)$ into a Hilbert space (a detailed proof is nontrivial.) On the other hand, the left-hand-side $\int_0^1 f(x)v(x)dx$ is also an inner product, this time on the Lp space $L2(0,1)$. An application of the Riesz representation theorem for Hilbert spaces shows that there is a unique $u$ solving (2) and therefore P1.
### The variational form of P2
If we integrate by parts using a form of Green's theorem, we see that if u solves P2, then for any v,
$\int_{\Omega} fv\,ds = -\int_{\Omega} \nabla u \cdot \nabla v \, ds = -\phi(u,v),$
where $\nabla$ denotes the gradient and $\cdot$ denotes the dot product in the two-dimensional plane. Once more $\,\!\phi$ can be turned into an inner product on a suitable space $H_0^1(\Omega)$ of "once differentiable" functions of $\Omega$ that are zero on $\partial \Omega$. We have also assumed that $v \in H_0^1(\Omega)$. The space $H_0^1(\Omega)$ can no longer be defined in terms of absolutely continuous functions, but see Sobolev spaces. Existence and uniqueness of the solution can also be shown.
## Discretization
A function in H10, with zero values at the endpoints (blue), and a piecewise linear approximation (red).
A piecewise linear function in two dimensions.
Basis functions vk (blue) and a linear combination of them, which is piecewise linear (red).
The basic idea is to replace the infinite dimensional linear problem:
Find $u \in H_0^1$ such that $\forall v \in H_0^1$ ; $-\phi(u,v)=\int fv$
with a finite dimensional version:
(3)Find $u \in V$ such that $\forall v \in V$ ; $-\phi(u,v)=\int fv$
where V is a finite dimensional subspace of $H_0^1$. There are many possible choices for V (one possibility leads to the spectral method). However, for the finite element method we take V to be a space of piecewise linear functions.
For problem P1, we take the interval (0,1), choose $x_n$ values 0 = $x0$ < $x_1$ < ... < $x_n$ < $x_{n+1}$ = 1 and we define V by
$\begin{matrix} V=\{v:[0,1] \rightarrow \Bbb R\;: v\mbox{ is continuous, }v|_{[x_k,x_{k+1}]} \mbox{ is linear for }\\ k=0,...,n \mbox{, and } v(0)=v(1)=0 \} \end{matrix}$
where we define $x_0 = 0$ and $x_{n + 1} = 1$. Observe that functions in V are not differentiable according to the elementary definition of calculus. Indeed, if $v \in V$ then the derivative is typically not defined at any $x = x_k, k = 1,...,n$. However, the derivative exists at every other value of x and one can use this derivative for the purpose of integration by parts.
For problem P2, we need V to be a set of functions of Ω. In the figure on the right, we have illustrated a triangulation of a 15 sided polygonal region Ω in the plane (below), and a piecewise linear function (above, in color) of this polygon which is linear on each triangle of the triangulation; the space V would consist of functions that are linear on each triangle of the chosen triangulation.
One often reads $V_h$ instead of V in the literature. The reason is that one hopes that as the underlying triangular grid becomes finer and finer, the solution of the discrete problem (3) will in some sense converge to the solution of the original boundary value problem P2. The triangulation is then indexed by a real valued parameter h > 0 which one takes to be very small. This parameter will be related to the size of the largest or average triangle in the triangulation. As we refine the triangulation, the space of piecewise linear functions V must also change with h, hence the notation $V_h$. Since we do not perform such an analysis, we will not use this notation.
### Choosing a basis
To complete the discretization, we must select a basis of V. In the one-dimensional case, for each control point $x_k$ we will choose the piecewise linear function $v_k \in V$ whose value is 1 at $x_k$ and zero at every $x_j,\;j \neq k$, i.e.,
$\Large v_{k}(x)=\begin{cases} {x-x_{k-1} \over x_k\,-x_{k-1}} & \mbox{ if } x \in [x_{k-1},x_k], \\ {x_{k+1}\,-x \over x_{k+1}\,-x_k} & \mbox{ if } x \in [x_k,x_{k+1}], \\ 0 & \mbox{ otherwise},\end{cases}$
for k = 1,...,n; this basis is a shifted and scaled tent function. For the two-dimensional case, we choose again one basis function $v_k$ per vertex $x_k$ of the triangulation of the planar region Ω. The function $v_k$ is the unique function of V whose value is 1 at $x_k$ and zero at every $x_j,\;j \neq k$.
Depending on the author, the word "element" in "finite element method" refers either to the triangles in the domain, the piecewise linear basis function, or both. So for instance, an author interested in curved domains might replace the triangles with curved primitives, in which case he might describe his elements as being curvilinear. On the other hand, some authors replace "piecewise linear" by "piecewise quadratic" or even "piecewise polynomial". The author might then say "higher order element" instead of "higher degree polynomial." Finite element method is not restricted to triangles (or tetrahedra in 3-D, or higher order simplexes in multidimensional spaces), but can be defined on quadrilateral sub-domains (hexahedra, prisms, or pyramids in 3-D, and so on). Higher order shapes (curvilinear elements) can be defined with polynomial and even non-polynomial shapes (e.g. ellipse or circle).
Methods that use higher degree piecewise polynomial basis functions are often called spectral element methods, especially if the degree of the polynomials increases as the triangulation size h goes to zero.
More advanced implementations (adaptive finite element methods) utilize a method to assess the quality of the results (based on error estimation theory) and modify the mesh during the solution aiming to achieve approximate solution within some bounds from the 'exact' solution of the continuum problem. Mesh adaptivity may utilize various techniques, the most popular are:
• refining (and un-refining) elements (h-adaptivity)
• changing order of base functions (p-adaptivity)
• combinations of the above (e.g. hp-adaptivity)
### Small support of the basis
Solving the two-dimensional problem u_{xx}+u_{yy}=-4 in the disk centered at the origin and radius 1, with zero boundary conditions.(a) The triangulation.
(b) The sparse matrix L of the discretized linear system.
(c) The computed solution, u(x, y)=1-x^2-y^2.
The primary advantage of this choice of basis is that the inner products
$\langle v_j,v_k \rangle=\int_0^1 v_j v_k\,dx$
and
$\phi(v_j,v_k)=\int_0^1 v_j' v_k'\,dx$
will be zero for almost all j,k. In the one dimensional case, the support of $v_k$ is the interval $[x_{k-1},x_{k+1}]$. Hence, the integrands of $\langle v_j,v_k \rangle$ and $Φ(vj,vk)$ are identically zero whenever $| j - k | > 1$.
Similarly, in the planar case, if $x_j$ and $x_k$ do not share an edge of the triangulation, then the integrals
$\int_{\Omega} v_j v_k\,ds$
and
$\int_{\Omega} \nabla v_j \cdot \nabla v_k\,ds$
are both zero.
### Matrix form of the problem
If we write $u(x)=\sum_{k=1}^n u_k v_k(x)$ and $f(x)=\sum_{k=1}^n f_k v_k(x)$ then problem (3) becomes
(4)$-\sum_{k=1}^n u_k \phi (v_k,v_j) = \sum_{k=1}^n f_k \int v_k v_j for j = 1,...,n.$
If we denote by $\mathbf{u}$ and $\mathbf{f}$ the column vectors $(u_1,...,u_n)t$ and $(f_1,...,f_n)t$, and if let $L = (L_{ij})$ and $M = (M_{ij})$ be matrices whose entries are $L_{ij} = φ(v_i,v_j)$ and $M_{ij}=\int v_i v_j$ then we may rephrase (4) as
(5)$-L \mathbf{u} = M \mathbf{f}.$
As we have discussed before, most of the entries of $L$ and $M$ are zero because the basis functions $v_k$ have small support. So we now have to solve a linear system in the unknown $\mathbf{u}$ where most of the entries of the matrix $L$, which we need to invert, are zero.
Such matrices are known as sparse matrices, and there are efficient solvers for such problems (much more efficient than actually inverting the matrix.) In addition, $L$ is symmetric and positive definite, so a technique such as the conjugate gradient method is favored. For problems that are not too large, sparse LU decompositions and Cholesky decompositions still work well. For instance, Matlab's backslash operator (which is based on sparse LU) can be sufficient for meshes with a hundred thousand vertices.
The matrix $L$ is usually referred to as the stiffness matrix, while the matrix $M$ is dubbed the mass matrix. Compare this to the simplistic case of a single spring governed by the equation is Kx = f, where K is the stiffness, x (or u) is the displacement and f is force.
### General form of the finite element method
In general, the finite element method is characterized by the following process.
• One chooses a grid for $\Omega$. In the preceding treatment, the grid consisted of triangles, but one can also use squares or curvilinear polygons.
• Then, one chooses basis functions. In our discussion, we used piecewise linear basis functions, but it is also common to use piecewise polynomial basis functions.
A separate consideration is the smoothness of the basis functions. For second order elliptic boundary value problems, piecewise polynomial basis function that are merely continuous suffice (i.e., the derivatives are discontinuous.) For higher order partial differential equations, one must use smoother basis functions. For instance, for a fourth order problem such as $u_{xxxx} + u_{yyyy} = f$, one may use piecewise quadratic basis functions that are C1.
Typically, one has an algorithm for taking a given mesh and subdividing it. If the main method for increasing precision is to subdivide the mesh, one has an h-method (h is customarily the diameter of the largest element in the mesh.) In this manner, if one shows that the error with a grid h is bounded above by Chp, for some $C<\infty$ and $p > 0$, then one has an order p method. Under certain hypotheses (for instance, if the domain is convex), a piecewise polynomial of order d method will have an error of order p = d + 1.
If instead of making h smaller, one increases the degree of the polynomials used in the basis function, one has a p-method. If one simultaneously makes h smaller while making p larger, one has an hp-method. High order method (with large p) are called spectral element methods, which are not to be confused with spectral methods.
For vector partial differential equations, the basis functions may take values in $\mathbb{R}^n$.
## Comparison to the finite difference method
The finite difference method (FDM) is an alternative way for solving PDEs. The differences between FEM and FDM are:
• The finite difference method is an approximation to the differential equation; the finite element method is an approximation to its solution.
• The most attractive feature of the FEM is its ability to handle complex geometries (and boundaries) with relative ease. While FDM in its basic form is restricted to handle rectangular shapes and simple alterations thereof, the handling of geometries in FEM is theoretically straightforward.
• The most attractive feature of finite differences is that it can be very easy to implement.
• There are several ways one could consider the FDM a special case of the FEM approach. One might choose basis functions as either piecewise constant functions or Dirac delta functions. In both approaches, the approximations are defined on the entire domain, but need not be continuous. Alternatively, one might define the function on a discrete domain, with the result that the continuous differential operator no longer makes sense, however this approach is not FEM.
• There are reasons to consider the mathematical foundation of the finite element approximation more sound, for instance, because the quality of the approximation between grid points is poor in FDM.
• The quality of a FEM approximation is often higher than in the corresponding FDM approach, but this is extremely problem dependent and several examples to the contrary can be provided.
Generally, FEM is the method of choice in all types of analysis in structural mechanics (i.e. solving for deformation and stresses in solid bodies or dynamics of structures) while computational fluid dynamics (CFD) tends to use FDM or other methods (e.g., finite volume method). CFD problems usually require discretization of the problem into a large number of cells/gridpoints (millions and more), therefore cost of the solution favors simpler, lower order approximation within each cell. This is especially true for 'external flow' problems, like air flow around the car or airplane, or weather simulation in a large area.
There are many finite element software packages, some free and some proprietary. | {"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": 56, "/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": 0.9245824813842773, "perplexity": 299.550226869185}, "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-2020-29/segments/1593655882634.5/warc/CC-MAIN-20200703153451-20200703183451-00134.warc.gz"} |
https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/An_Introduction_to_Number_Theory_(Veerman)/01%3A_A_Quick_Tour_of_Number_Theory/1.04%3A_Countable_and_Uncountable_Sets | # 1.4: Countable and Uncountable Sets
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Definition 1.18
A set $$S$$ is countable if there is a bijection $$f : \mathbb{N} \rightarrow S$$. An infinite set for which there is no such bijection is called uncountable.
Proposition 1.19
Every infinite set $$S$$ contains a countable subset.
Proposition 1.19
Every infinite set $$S$$ contains a countable subset.
Proof
Choose an element $$s_{1}$$ from $$S$$. Now $$S-\{s_{1}\}$$ is not empty because $$S$$ is not finite. So, choose $$s_{2}$$ from $$S-\{s_{1}\}$$. Then $$S-\{s_{1}, s_{2}\}$$ is not empty because $$S$$ is not finite. In this way, we can remove $$s_{n+1}$$ from $$S-\{s_{1}, s_{2}, \cdots, s_{n}\}$$ for all $$n$$. Then set $$\{s_{1}, s_{2}, \cdots\}$$ is countable and is contained in $$S$$.
So countable sets are the smallest infinite sets in the sense that there are no infinite sets that contain no countable set. But there certainly are larger sets, as we will see next.
Theorem 1.20
The set $$\mathbb{R}$$ is uncountable.
Proof
The proof is one of mathematics’ most famous arguments: Cantor’s diagonal argument [8]. The argument is developed in two steps .
Let $$T$$ be the set of semi-infinite sequences formed by the digits 0 and 2. An element $$t \in T$$ has the form $$t = t_{1}t_{2}t_{3} \dots$$ where $$t_{i} \in \{0, 2\}$$. The first step of the proof is to prove that $$T$$ is uncountable. So suppose it is countable. Then a bijection $$t$$ between $$\mathbb{N}$$ and $$T$$ allows us to uniquely define the sequence $$t(n)$$, the unique sequence associated to $$n$$. Furthermore, they form an exhaustive list of the elements of $$T$$. For example,
$t(1) = \textbf{0},0,0,0,0,0,0,0,0,0,0, \dots \nonumber$
$t(2) = 2, \textbf{0},2,0,2,0,2,0,2,2,2, \dots \nonumber$
$t(3) = 0,0, \textbf{0},2,2,2,2,2,2,2,2, \dots \nonumber$
$t(4) = 2,2,2, \textbf{2},2,2,0,0,0,0,0, \dots \nonumber$
$t(5) = 0,0,0,2, \textbf{2},0,2,0,0,2,0, \dots \nonumber$
$t(6) = 2,0,0,0,0, \textbf{2},0,0,0,2,2, \dots \nonumber$
$\vdots \nonumber$
Construct $$t^{*}$$ as follows: for every $$n$$, its nth digit differs from the nth digit of $$t(n)$$. In the above example, $$t^{*} = \textbf{2}, \textbf{2}, \textbf{2}, \textbf{0}, \textbf{2}, \textbf{0}, \dots$$. But now we have a contradiction, because the element $$t^{*}$$ cannot occur in the list. In other words, there is no surjection from $$\mathbb{N}$$ to $$T$$. Hence there is no bijection between $$\mathbb{N}$$ and $$T$$.
The second step is to show that there is a subset $$K$$ of $$\mathbb{R}$$ such that there is no surjection (and thus no bijection) from $$\mathbb{N}$$ to $$K$$. Let $$t$$ be a sequence with digits $$t_{i}$$. Define $$f : T \rightarrow \mathbb{R}$$ as follows
$f(t) = \sum_{i=1}^{\infty}t_{i}3^{-i} \nonumber$
If $$s$$ and $$t$$ are two distinct sequences in $$T$$, then for some $$k$$ they share the first $$k-1$$ digits but $$t_{k} = 2$$ and $$s_{k} = 0$$. So
$f(t)-f(s) = 2 \cdot 3^{-k}+\sum_{i=k+1}^{\infty} (t_{i}-s_{i}) 3^{-i} \ge 2 \cdot 3^{-k}-2 \sum_{i=k+1}^{\infty} 3^{-i} = 3^{-k} \nonumber$
Thus $$f$$ is injective. Therefore $$f$$ is a bijection between $$T$$ and the subset $$K = f(T)$$ of $$\mathbb{R}$$. If there is a surjection $$g$$ from $$\mathbb{N}$$ to $$K = f(T)$$, then,
$\mathbb{N} \xrightarrow{\text{g}} K \xleftarrow{\text{f}} T \nonumber$
And so $$f^{-1} g$$ is a surjection from $$\mathbb{N}$$ to $$T$$. By the first step, this is impossible. Therefore, there is no surjection $$g$$ from $$\mathbb{N}$$ to $$K$$, much less from $$\mathbb{N}$$ to $$\mathbb{R}$$.
Corollary 1.21
(i) The set of infinite sequences in $$\{1,2,\cdots, b-1\}^{\mathbb{N}}$$ is uncountable.
(ii) The set of finite sequences (but without bound) in $$\{1, 2, \cdots, b-1\}^{\mathbb{N}}$$ is countable.
Proof
The proof of (i) is the same as the proof that $$T$$ is uncountable in the proof of Theorem 1.20. The proof of (ii) consists of writing first all $$b$$ words of length 1, then all $$b^{2}$$ words of length 2, and so forth. Every finite string will occur in the list.
Theorem 1.22
(i) The set $$\mathbb{Z}^2$$ is countable.
(ii) $$\mathbb{Q}$$ is countable.
Proof
(i) The proof relies on Figure 2. In it, a directed path $$\gamma$$ is traced out that passes through all points of $$\mathbb{Z}^2$$. Imagine that you start at $$(0, 0)$$ and travel along $$\gamma$$ with unit speed. Keep a counter $$c \in \mathbb{N}$$ that marks the point $$(0, 0)$$ with a “1”. Up the value of the counter by 1 whenever you hit a point of $$\mathbb{Z}^2$$. This establishes a bijection between $$\mathbb{N}$$ and $$\mathbb{Z}^2$$.
Figure 2. A directed path $$\gamma$$ passing through all points of $$\mathbb{Z}^2$$.
(ii) Again travel along $$\gamma$$ with unit speed. Keep a counter $$c \in \mathbb{N}$$ that marks the point $$(0, 1)$$ with a “1”. Up the value of the counter by 1. Continue to travel along the path until you hit the next point $$(p, q)$$ that is not a multiple of any previous and such $$q$$ is not zero. Mark that point with the value of the counter. $$\mathbb{Q}$$ contains $$\mathbb{N}$$ and so is infinite. Identifying each marked point $$(p, q)$$ with the rational number $$\frac{p}{q}$$ establishes the countability of $$\mathbb{Q}$$.
Notice that this argument really tells us that the product of a countable set and another countable set is still countable. The same holds for any finite product of countable set. Since an uncountable set is strictly larger than a countable, intuitively this means that an uncountable set must be a lot largerthan a countable set. In fact, an extension of the above argument shows that the set of algebraic numbers numbers is countable. And thus, in a sense, it forms small subset of all reals. All the more remarkable, that almost all reals that we know anything about are algebraic numbers, a situation we referred to at the end of Section 1.4.
It is useful and important to have a more general definition of when two sets “have the same number of elements”.
Definition 1.23
Two sets A and B are said to have the same cardinality if there is a bijection $$f : A \rightarrow B$$. It is written as $$|A| = |B|$$. If there is an injection $$f : A \rightarrow B$$, then $$|A| \le |B|$$.
Definition 1.24
An equivalence relation on a set $$A$$ is a (sub)set $$\mathbb{R}$$ of ordered pairs in $$A \times A$$ that satisfy three requirements.
• $$(a, a) \in \mathbb{R}$$ (reflexivity).
• If $$(a, b) \in \mathbb{R}$$, then $$(b, a) \in \mathbb{R}$$ (symmetry).
• If $$(a, b) \in \mathbb{R}$$ and $$(b, c) \in \mathbb{R}$$, then If $$(a, c) \in \mathbb{R}$$ (transitivity).
Usually $$(a, b) \in \mathbb{R}$$ is abbreviated to $$a \sim b$$. The mathematical symbol “$$=$$” is an equivalence.
It is easy to show that having the same cardinality is an equivalence relation on sets (exercise 1.23). Note that the cardinality of a finite set is just the number of elements it contains. An excellent introduction to the cardinality of infinite sets in the context of naive set theory can be found in [15].
This page titled 1.4: Countable and Uncountable Sets is shared under a CC BY-NC license and was authored, remixed, and/or curated by J. J. P. Veerman (PDXOpen: Open Educational Resources) . | {"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": 0.9752072691917419, "perplexity": 99.61678054747503}, "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/1679296949093.14/warc/CC-MAIN-20230330004340-20230330034340-00766.warc.gz"} |
http://mathhelpforum.com/number-theory/133425-binomial-theorem.html | # Math Help - Binomial Theorem
1. ## Binomial Theorem
Hey, another problem ,
Let p be an odd prime and let n be a positive integer. Use the Binomial Theorem to show that
1. $(1 + p)^{p^{n-1}} \equiv 1 \mod p^n$
2. $(1 + p)^{p^{n-2}} \not\equiv 1 \mod p^n$ .
Edit: Found the solution, ask if you want to know it. | {"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": 2, "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": 0.9338933229446411, "perplexity": 523.702298008348}, "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/1430448957257.55/warc/CC-MAIN-20150501025557-00055-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.arxiv-vanity.com/papers/1012.3310/ | arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.
# The asymptotical error of broadcast gossip averaging algorithms
Fabio Fagnani Paolo Frasca Fabio Fagnani and Paolo Frasca are with the Dipartimento di Matematica, Politecnico di Torino, Corso Duca degli Abruzzi 24, 10129 Torino, Italy.
June 21, 2020
###### Abstract
In problems of estimation and control which involve a network, efficient distributed computation of averages is a key issue. This paper presents theoretical and simulation results about the accumulation of errors during the computation of averages by means of iterative “broadcast gossip” algorithms. Using martingale theory, we prove that the expectation of the accumulated error can be bounded from above by a quantity which only depends on the mixing parameter of the algorithm and on few properties of the network: its size, its maximum degree and its spectral gap. Both analytical results and computer simulations show that in several network topologies of applicative interest the accumulated error goes to zero as the size of the network grows large.
## 1 Introduction
Distributed computation of averages is an important building block to solve problems of estimation and control over networks. As a reliable time-independent communication topology may be unlikely in the applications, a growing interest has been devoted to randomized “gossip” algorithms to compute averages. In such algorithms, at each time step, a random subset of the nodes communicates and performs computations. Unfortunately, some of these iterative algorithms do not deterministically ensure that the average is preserved through iterations, and due to the accumulation of errors, in general there is no guarantee that the typical algorithm realization will converge to a value which is close to the desired average.
In the present paper we focus on one of these randomized algorithm, the Broadcast Gossip Algorithm (BGA). In this algorithm, a node is randomly selected at each time step to broadcast its current value to its neighbors, which in turn update their values by a local averaging rule. Since these updates are not symmetric, it is clear that the average is not preserved, but instead is changed at each time step by some amount. In this paper we study how these errors accumulate, and how much the convergence value of the algorithm differs from the original average to be computed.
### 1.1 Contribution
In this paper, we study the bias, or asymptotical error, committed by a Broadcasting Gossip averaging algorithm, and we show that large neighborhoods and a large mixing parameter induce a large asymptotical error. As a theoretical contribution, we study the average of states as a martingale, and by this interpretation we prove that on symmetric graphs the expectation of the accumulated error can not be larger than a constant times where is the “mixing” parameter of the algorithm, is the network size, is the maximum degree of the nodes, and is the network spectral gap. For some families of graphs (e.g, expander graphs), this is enough to prove that the bias goes to zero as goes to infinity. Further, by means of simulations we show that, on some example graph topologies, the mean bias is an increasing function of the mixing parameter and is proportional, on large networks, to the ratio between degree and size of the network. In particular, whenever the simulated bias goes to zero as goes to infinity.
### 1.2 Related works
The paper [9] provides a general theory for randomized linear averaging algorithms, and presents a few example algorithms, some of which do not preserve the average of the states. Among these algorithms, the Broadcast Gossip Algorithm, studied in the present paper, has been attracting a wide interest, for its natural application to wireless networks: main references include the paper [2] and the recent survey [6]. While it is simple to give conditions to ensure that the expectation of the convergence value is equal to the initial average, the problem of estimating the difference between expectation and realizations is harder, and has received partial answers in a few papers. In [1] the authors study a related communication model, in which the broadcasted values may not be received with a probability which depends on the transmitter and receiver nodes, and claim that “aggressive updating combined with large neighborhoods […] result in more variance [of the convergence value] within the short time to convergence”. This intuition extends to the Broadcast Gossip Algorithm which we are considering in this paper. Actually, in [2] the variance of the limit value has been estimated for general graphs, with an upper bound which is proportional to where is the -th smallest eigenvalue of the graph Laplacian. This bound, however, is not useful to prove that the bias goes to zero as grows, a fact which has been proved in [7] for sequences of Abelian Cayley graphs with bounded degree, using tools from algebra and Markov chain theory. Analogous problems can be studied for other randomized algorithms which do not preserve the average. For instance, in [8] two related algorithms are studied, in which node values are sent to one random neighbor only. If at each time step one random node sends its value, then the variance has an upper bound which is proportional to , while if at each time step every node sends its value, then the bound is proportional to
## 2 Problem statement
Let a graph with be given, together with real numbers . For every node we denote its out-neighborhood by , and its in-neighborhood by . The following Broadcasting Gossip Algorithm (BGA) is run in order to estimate the average .
At each time step , one node is sampled from a uniform distribution over . Then, node broadcasts its state to its neighbors which in turn update their states as
xu(t+1)=(1−q)xu(t)+qxv(t). (1)
The parameter is said to be the mixing parameter of the algorithm. If instead , there is no update:
It is known from [9, Corollary 3.2] that the BGA converges, in the sense that there exists a random variable such that almost surely Let now . Although one can find conditions to ensure that , in general is not equal to . Then, it is worth to ask how far the convergence value is from the initial average. To study this bias in the computation of the average, we define
β(t)=|xave(t)−xave(0)|2.
The goal of this work is to study this quantity, and in particular its limit , with a special attention to its dependence on the size of the network. In particular, we shall say that the algorithm is asymptotically unbiased if
## 3 Analysis
### 3.1 A simplistic bound
Using (1), it is immediate to compute that, given to be the broadcasting node at time ,
xave(t+1)−xave(t)=qN∑u∈N+v(xv(t)−xu(t)). (2)
Then, we can obtain the following deterministic bound on the error introduced at each time step,
|xave(t+1)−xave(t)|≤qNd+vL≤qd+maxNL, (3)
where is the out-degree of node , and This simple bound is worth some informal remarks. Indeed, Equation (3) suggests that choosing a low value of the mixing parameter , and a graph with low degree and large size , may ensure a small bias in the computation of the average. However, by choosing , or , one affects the speed of convergence of the algorithm. Assume one is interested in an accurate computation, and chooses low values for and , compared to . This choice would likely imply a slow convergence, and in turn a slow convergence may enforce to run the algorithms for a larger number of steps, in order to meet the same precision requirement. These extra steps, however, would introduce extra errors, thus possibly wasting the desired advantage in the accuracy. We argue from this discussion that there is a delicate trade-off between speed and accuracy for the BGA algorithm. The results presented in the next sections will shed light on this trade-off.
### 3.2 The average as a martingale
In this section, we shall derive a general bound on in terms of the topology of the graph. The derivation is based on applying the theory of martingales to the stochastic processes and . The reader can find the essentials of martingale theory in [11] or in [13].
###### Definition 3.1
Given a sequence (filtration) of -algebras a sequence of random variables is a -martingale if , for any .
Our first result states that is a martingale with respect to the filtration induced by . Before the statement, we need some definitions. Let and be the out-degree and in-degree of node . The graph is said to be balanced if for all . Given a set of random variables , we denote by the sigma-algebra generated by the random variables in .
###### Proposition 3.1
Let us consider the BGA algorithm and the filtration If the graph is balanced, then the sequence of random variables is a square-integrable -martingale.
Proof: First, note that is -measurable. Moreover, Equation (2) implies that for all ,
E[xave(t+1)−xave(t)|Ft]= 1N∑v∈V⎛⎜⎝qN∑u∈N+v(xv(t)−xu(t))⎞⎟⎠ = qN2⎡⎢⎣∑v∈Vd+vxv(t)−∑v∈V∑u∈N+vxu(t)⎤⎥⎦ = qN2[∑v∈Vd+vxv(t)−∑u∈Vd−uxu(t)] = qN2[∑v∈V(d+v−d−v)xv(t)]=0,
since we are assuming for every . Then, the sequence of random variables is an -martingale. Moreover, the fact that implies that the martingale is bounded in for every , and in particular square-integrable.
Let us define the distance from the agreement as
d(t):=1N∑v∈V(xv(t)−xave(t))2.
Using this definition, we can prove an inequality which is a refinement of (3). Let and
###### Lemma 3.2
Let be balanced. Then, the increments of the martingale have bounded variance, in particular, for all ,
E[(xave(t+1)−xave(t))2|Ft]≤4q2d2maxN2d(t). (4)
Proof: By (2), we have
E[(xave(t+1)−xave(t))2|Ft]= 1N∑v∈V⎛⎜⎝qN∑u∈N+v(xv(t)−xu(t))⎞⎟⎠2 ≤ 1N∑v∈Vq2N2d+v∑u∈N+v(xv(t)−xu(t))2 ≤ 1N∑v∈Vq2N22((d+v)2(xv(t)−xave(t))2+d+v∑u∈N+v(xave(t)−xu(t))2) ≤ 2(q2N2(d+max)2d(t)+d+maxd−maxd(t)) ≤ 4q2d2maxN2d(t).
This completes the proof.
We define the rate of convergence of the algorithm as
R:=supx(0)limsupt→+∞E[d(t)]1/t.
Then, there exists a positive constant , depending on , such that This fact, combined with Lemma 3.2, implies that
E[(xave(t+1)−xave(t))2]= E[E[(xave(t+1)−xave(t))2|Ft]] ≤ 4CRq2d2maxN2Rt (5)
We recall that the spectral gap of the graph is the smallest (in modulus) non-zero eigenvalue of its Laplacian matrix, and we denote this quantity by . It is well-known that relates to the mixing rate of Markov chains, and to the speed of convergence of gossip algorithms [3, 9]: the larger the spectral gap, the faster the convergence. In particular, let us assume that the graph be symmetric, that is, such that for all Then, we know from [7, Equation (18)] that
R≤1−2q(1−q)Nλ1. (6)
Using these facts, we are going to prove the next result about the asymptotic behavior of the bias as .
###### Proposition 3.3
If is symmetric, then there exists a constant such that
E[β(∞)]≤Cq1−qd2maxNλ1.
Proof: Using the orthogonality of the increments of square-integrable martingales, we observe that
limt→+∞E[(xave(t+1)−xave(0))2]= limT→+∞E⎡⎣(T−1∑t=0(xave(t+1)−xave(t)))2⎤⎦ = limT→+∞E[T−1∑t=0(xave(t+1)−xave(t))2 +2T−1∑t=1∑s
By applying Equation (5)
limt→+∞E[(xave(t+1)−xave(0))2]≤ 4CRq2d2maxN2limT→+∞T−1∑t=0Rt=4CRq2d2maxN211−R.
The inequality in (6) implies that . The thesis then follows, with , by applying the dominated convergence theorem.
Note that the proof of Proposition 3.3 also implies that
suptE[(xave(t+1)−xave(0))2]≤Cq1−qd2maxNλ1.
On the other hand, for a convergent square-integrable martingale , we know by Doob’s maximal inequality that Then, we can immediately obtain the following finite-time counterpart of Proposition 3.3.
###### Theorem 3.4
If is symmetric, then there exists such that
E[supt∈Nβ(t)]≤C′q1−qd2maxNλ1.
Let us now consider a sequence of graphs of increasing size . In such a sequence, both and are functions of . In this context, Proposition 3.3 implies the following corollary.
###### Corollary 3.5
Let and be a sequence of symmetric graphs such that and . If
d2maxλ1=o(N)as N→+∞,
then the BGA algorithm is asymptotically unbiased.
Note that, since converges a.s. and , it is trivial to find an upper bound on which does not depend on . The interest of the above corollary is in giving a sufficient condition for to be as
###### Remark 3.6
Applying Markov’s inequality, we see from Proposition 3.3 that for any ,
P[β(∞)>c]≤Cq1−qd2maxNλ11c
In the applications, one is often interested in computing an average because the average is the maximum likelihood estimator of the expectation of a random variable. In such context, the average enjoys the property that the mean square error, committed by approximating the expectation by the average of N samples, is equal to 1/N. For this reason, one would like to ensure that the bias introduced by the Broadcast Gossip algorithm is not larger than . If we take , we get
P[β(∞)>1N]≤Cq1−qd2maxλ1.
Then, provided the right-hand-side of this inequality does not diverge, we can choose the mixing parameter so that with a positive given probability the bias is below . In such case, if our purpose is distributed estimation of an expected value, averages which are approximated by a Broadcast Gossip Algorithm are as good as averages computed by a centralized method.
## 4 Examples
In this section, we show that the BGA is asymptotically unbiased on several example topologies which have been considered in the literature. Given a graph of size , we denote by and its spectral gap and maximum degree, respectively. We consider the following example sequences of graphs.
• Expander graphs.
A sequence of graphs is said to be an expander sequence if there exist and such that for every , and In this case, is bounded, and this fact implies by Proposition 3.3 that
E[β(∞)]=O(1N) as N→+∞,
and then the BGA is asymptotically unbiased and Remark 3.6 applies. An example of an expander sequence is given by a sequence of de Bruijn graphs on symbols of increasing dimensions . A de Bruijn graph on symbols of dimension has vertices and edges from any to (all modulo ). Their expander properties have already been applied to efficient averaging algorithms in [5].
• Hypercube graphs.
The -dimensional hypercube graph is the graph obtained drawing the edges of a -dimensional hypercube. It has nodes which can be identified with the binary words of length , and two nodes are neighbors if the corresponding binary words differ in only one component. For these graphs it is known, for instance from [10, Example 7], that and Then,
E[β(∞)]=O(log3NN) as N→+∞
and the BGA is asymptotically unbiased.
• -dimensional square lattices.
We consider square lattices obtained by tiling a -dimensional torus, with nodes. For these graphs we know [4, Theorem 6], that and Then,
E[β(∞)]=O(1N1−2/k) as N→+∞,
and we argue that lattices are asymptotically unbiased if However, we know that the BGA is asymptotically unbiased also if : this has been proved in [7] using different techniques.
• Random geometric graphs.
We can also consider random sequences of geometric graphs constructed as follows. We sample points from a uniform distribution over the unit square, and we let nodes and be connected if the two corresponding points in the square are less than far apart, with . For these graphs, we know from [12] that with high probability and Then, with high probability and we can not conclude asymptotical unbiasedness.
• Complete graphs.
For these graphs, and
Then, we can not conclude from Proposition 3.3 that the BGA is asymptotically unbiased on complete graphs. Actually, in [9] it is shown that the BGA is not asymptotically unbiased on complete graphs, and in particular
E[β(∞)]=Var(x(0))q2−qN−1N, (7)
where by we denote the (sample) variance of the initial condition.
## 5 Simulations
We have extensively simulated the evolution of BGA algorithm on sequences of graphs, and in particular on the example topologies presented in Section 4. In this section, we account for our results about the dependence of the bias on the size and on the parameter .
Our simulation setup is as follows. Let , and the graph topology be chosen. For every run of the algorithm we generate a vector of initial conditions , sampling from a uniform distribution over 111If the topology is random, namely a random geometric topology as described above, it is also sampled at this stage. Disconnected realizations are discarded: however, disconnected realizations are very few in our random geometric setting and their number decreases as grows, so that they are less than 2% when . Then, we run the algorithm until the disagreement is below a small threshold , which we set at . At this time , the algorithm is stopped, and is evaluated. In order to simulate the expectation of , we average the outcome of 1000 realizations of . Our results about the dependence on are summarized in Figure 1, which plots the average bias against in a log-log diagram. As expected, complete graphs are not asymptotically unbiased, while all other topologies, in which the degree is are asymptotically unbiased. In particular, for de Bruijn graphs on 2 symbols, ring graphs and torus graphs, the bias is , whereas for hypercubes and random geometric graphs the bias is Overall, our set of simulations suggests that
E[β(∞)]=Θ(dmaxN)as N→∞.
Results presented in Figure 2 confirm that this asymptotical law is independent of the choice of , provided . Note indeed that if we run the BGA with in the update (1), the convergence value is always one of the initial values, sampled according to a uniform distribution. This implies that, if , then If instead , simulations in Figure 3 show that the bias is an increasing function of the mixing parameter.
## 6 Conclusion
In this paper, we have proven that on any symmetric graph, and in particular the BGA is asymptotically unbiased on expander graphs. On the other hand, simulations suggest that, on sequences of (almost) regular graphs with degree , the bias is such that Our future research will be devoted to find a bound on which ensure asymptotic unbiasedness on a wider set of topologies, and more in general to study the trade-offs between speed and accuracy in gossip algorithms.
## Acknowledgements
P. Frasca wishes to thank M. Boccuzzi for his advice in implementing the simulations and S. Zampieri for finding an error in an earlier draft.
## References
Want to hear about new tools we're making? Sign up to our mailing list for occasional updates. | {"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": 0.973392128944397, "perplexity": 422.01308275013696}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737050.56/warc/CC-MAIN-20200807000315-20200807030315-00385.warc.gz"} |
https://www.physicsforums.com/threads/newtons-laws-of-motion.125134/ | # Homework Help: Newton's Laws of Motion
1. Jul 3, 2006
### Harmony
Two bodies of masses m1 and m2, are released from the position shown in the figure. If the mass of the smooth-topped table is m3, find the reaction of the floor on the table while the two bodies are in motion. Assume the table does not move.
I attempt to find out all the forces acting. There's tension of the String, w1 and w2, w3, normal reaction force and maybe friction. But how does movement of m2 has anything to do with the normal reaction force?
#### Attached Files:
• ###### figure7.bmp
File size:
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Views:
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Last edited: Jul 3, 2006
2. Jul 3, 2006
### vijay123
did they which of m1 or m2 is heavier?
3. Jul 3, 2006
### vijay123
is there any friction present at all?
4. Jul 3, 2006
### d_leet
That should be irrelevant.
Since the problem says that the table is "smooth-topped" we should assume that there is no friction.
5. Jul 3, 2006
### vijay123
it would be relevant if friction wus present.
anyway, i think that if the table stay at one place, then the reactive force of the floor would be g(m1+m2+m3). that wut i think but i am not very sure at alll!
6. Jul 3, 2006
### Hootenanny
Staff Emeritus
As d_leet states this is irrelavent.
Last edited: Jul 3, 2006
7. Jul 3, 2006
### vijay123
yea, you can the ans to my ques from the problem...sorry....my mistake
8. Jul 4, 2006
### Harmony
I am not sure whether friction is present or not. The answer given is ((m1m2/m1+m2)+m2+m3)g. And friction is include in the answer. The friction is m1m2g/m1+m2
Why is friction involved since the table is smooth? Is there something wrong with the question?
And yes, it is not stated whether m1 is greater than m2 or not.
9. Jul 4, 2006
### Kurdt
Staff Emeritus
The quantity:
$$\frac{m_1 m_2} {m_1+m_2}$$
is in fact the reduced mass of the masses m1 and m2.
Last edited: Jul 4, 2006
10. Jul 4, 2006
### arildno
Friction is needed between the table and the floor, in order to keep the table stationary with respect to the floor. | {"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": 0.9564530253410339, "perplexity": 2047.871283063546}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2018-51/segments/1544376823348.23/warc/CC-MAIN-20181210144632-20181210170132-00048.warc.gz"} |
http://www.ibex-lib.org/doc/system.html | # Systems¶
A system in IBEX is a set of constraints (equalities or inequalities) with, optionnaly, a goal function to minimize and an initial domain for variables. It corresponds to the usual concept of system in mathematical programming. Here is an example of system:
Minimize $$x+y,$$
$$x \in[-1,1], y\in[-1,1]$$
such that
$$x^2+y^2\le1$$
$$y\ge x^2$$.
One is usually interested in solving the system while minimizing the criterion, if any.
## Class and Fields¶
The class for representing a system is System.
### Systems fields¶
A system is not as simple as a collection of any constraints because each constraint must exactly relates the same set of arguments. And this set must also coincide with that of the goal function. Many algorithms of IBEX are based on this assumption. This is why they requires a system as argument (and not just an array of constraints). This makes systems a central concept in IBEX.
A system is an object of the System class. This object is made of several fields that are detailed below.
• const int nb_var: the total number of variables or, in other words, the size of the problem. This number is basically the sum of all arguments’ components. For instance, if one declares an argument x with 10 components and an argument y with 5, the value of this field will be 15.
• const int nb_ctr: the number of constraints.
• Function* goal: a pointer to the goal function. If there is no goal function, this pointer is NULL.
• Function f: the (usually vector-valued) function representing the constraints. For instance, if one defines three constraints: $$x+y\leq0,\ x-y=1$$, and $$x-y\geq0$$, the function f will be $$(x,y)\mapsto (x+y,x-y-1,x-y)$$. Note that the constraints are automatically transformed so that the right side is 0 but, however, without changing the comparison sign. It is however possible to normalize a system so that all inequalities are defined with the $$\le$$ sign (see ).
• IntervalVector box: when a system is loaded from a file, a initial box can be specified. It is contained in this field.
• Array<NumConstraint> ctrs: the array of constraints. The Array class of IBEX can be used as a regular C array.
### Auxiliary functions¶
(to be completed)
## Creating systems (in C++)¶
The first alternative for creating a system is to do it programmatically, that is, directly in your C++ program. Creating a system in C++ resorts to a temporary object called a system factory. The task is done in a few simple steps:
• declare a new system factory (an object of SystemFactory)
• create the system simply by passing the factory to the constructor of System
Here is an example:
Variable x,y;
SystemFactory fac;
System sys(fac);
If you compare the declaration of the constraint here with the examples given here, you notice that we do not list here the arguments before writing sqr(x)+sqr(y)<=1. The reason is simply that, as said above, the goal function and the constraints in a system share all the same list of arguments. This list is defined via add_var once for all.
## System Transformation¶
We present in this section the different transformations that can be applied to a system.
### Copy¶
The first transformation you can apply on a system is a simple copy. Of course, this is done via the copy constructor of the System class.
When calling the copy constructor, you can decide to copy everything, only the equations or only the inequalities. For this, set the second parameter of the constructor to either:
value def
COPY duplicate all the constraints
INEQ_ONLY duplicate only inequalities
EQ_ONLY duplicate only equalities
The first argument of the constructor is the system to copy of course.
output << "original system:" << endl;
output << "-------------------------"<< endl;
output << sys;
output << "-------------------------"<< endl << endl;
System sys2(sys,System::INEQ_ONLY);
output << "system with only inequalities" << endl;
output << "-------------------------"<< endl;
output << sys2;
output << "-------------------------"<< endl << endl;
The display is:
original system:
-------------------------
variables:
x y
goal:
(none)
constraints:
((x^2+y^2)-1)<=0
(y-x^2)>=0
((y+x)-1)=0
-------------------------
system with only inequalities
-------------------------
variables:
x y
goal:
(none)
constraints:
((x^2+y^2)-1)<=0
(y-x^2)>=0
-------------------------
### Normalization¶
It is confortable in some situations to assume that a system is made of inequalities only, and that each inequality is under the forme $$g(x)\le0$$, that is, it is a “less or equal” inequality. This is called a “normalized” system.
This need arises, e.g., in optimization methods where the calculation of Lagrange multipliers is simplified when the normalization assumption holds.
It is possible to automatically transform a system into a normalized one. The process is immediate. If a constraint is:
• $$g(x)\le0$$
it is already normalized so it is left unchanged.
• $$g(x)<0$$
it is replaced by $$g(x)\le0$$ (yes, there is a little loss of precision here)
• $$g(x)>0$$ or $$g(x)\ge0$$
it is replaced by $$-g(x)\le0$$
• $$g(x)=0$$
it is replaced by two constraints: $$g(x)\le0$$ and $$-g(x)\le0$$. It is also possible to introduce an inflation value or “thickness”, that is, to replace the equality by $$g(x)\le\varepsilon$$ and $$-g(x)\le \varepsilon$$ where $$\varepsilon$$ can be fixed to any value.
Note: There is a special treatment for “thick equalities”, that is, equations of the form $$g(x)=[l,u]$$. This kind of equations appear often in, e.g., robust parameter estimation problems. In this case, the equality is replaced by two inequalities, $$g(x)\le u$$ and $$-g(x)\le-l$$, and the $$\varepsilon$$-inflation is not applied unless $$|u-l|<\varepsilon$$.
Normalization is done by calling the constructor of NormalizedSystem, a sub-class of System. Here is an example where sys is a system built previously:
output << "original system:" << endl;
output << "-------------------------"<< endl;
output << sys;
output << "-------------------------"<< endl << endl;
// normalize the system with a "thickness"
// set to 0.1
NormalizedSystem norm_sys(sys,0.1);
output << "normalized system:" << endl;
output << "-------------------------"<< endl;
output << norm_sys;
output << "-------------------------"<< endl;
We get the following display:
original system:
-------------------------
variables:
x y
goal:
(none)
constraints:
((x^2+y^2)-1)<=0
(y-x^2)>=0
((y+x)-1)=0
-------------------------
normalized system:
-------------------------
variables:
x y
goal:
(none)
constraints:
((x^2+y^2)-1)<=0
(-(y-x^2))<=0
(((y+x)-1)-0.1)<=0
((-((y+x)-1))-0.1)<=0
-------------------------
### Extended System¶
An extended system is a system where the goal function is transformed into a constraint.
For instance, the extension of the system given above:
Minimize $$x+y,$$
$$x \in[-1,1], y\in[-1,1]$$
such that
$$x^2+y^2\le1$$
$$y\ge x^2$$.
is the following unconstrained system of constraints:
$$x \in[-1,1], y\in[-1,1], \mbox{__goal__}\in(-\infty,\infty)$$
such that
$$x+y=\mbox{__goal__}$$
$$x^2+y^2\le1$$
$$y\ge x^2$$
Once built, an extended system is a system like any other one, but it has also some extra information:
• the name of the goal variable which is automatically generated (it is “__goal__” in our previous example).
• the index of the goal variable (the last (2) in our previous example)
• the index of the “goal constraint” (the first (0) in our previous example)
For this reason, an extended system is represented by a subclass of System named ExtendedSystem.
To create an extended system just use the constructor of ExtendedSystem. We assume in the following example that the variable sys is a System previously built.
output << "original system:" << endl;
output << "-------------------------"<< endl;
output << sys;
output << "-------------------------"<< endl;
output << " number of variables:" << sys.nb_var << endl;
output << " number of constraints:" << sys.nb_ctr << endl << endl;
ExtendedSystem ext_sys(sys);
output << "extended system:" << endl;
output << "-------------------------"<< endl;
output << ext_sys;
output << "-------------------------"<< endl;
output << " number of variables:" << ext_sys.nb_var << endl;
output << " number of constraints:" << ext_sys.nb_ctr << endl;
output << " goal name:" << ext_sys.goal_name() << endl;
output << " goal variable:" << ext_sys.goal_var() << endl;
output << " goal constraint:" << ext_sys.goal_ctr() << endl;
We get the following display:
original system:
-------------------------
variables:
x y
goal:
(x+y)
constraints:
((x^2+y^2)-1)<=0
(y-x^2)<=0
-------------------------
number of variables:2
number of constraints:2
extended system:
-------------------------
variables:
x y __goal__
goal:
(none)
constraints:
((x+y)-__goal__)=0
((x^2+y^2)-1)<=0
(y-x^2)<=0
-------------------------
number of variables:3
number of constraints:3
goal name:__goal__
goal variable:2
goal constraint:0
### Fritz-John (Khun-Tucker) conditions¶
The generalized Khun-Tucker (aka Fritz-John) conditions can be obtained from a system. This produces a new system of n+M+R+K+1 variables where
• n is the number of basic variables (the ones of the original system)
• M is the number of Lagrange multipliers for inequalities (i.e., the number of inequalities in the original system)
• R is the number of Lagrange multipliers for equalities (i.e., the number of equalities in the original system)
• K is the number of Lagrange multipliers for bounding constraints. These bounding constraints correspond to the box field of the original system which is taken into account as 2n additional inequalities.
• The last variable is the “special coefficient” of the goal function that is equal to 0 in the case where constraint qualification (linear independency of constraints gradients) does not hold.
Generation of the Fritz-John conditions is based on Applying a function with numerous arguments.
Example:
output << "original system:" << endl;
output << "-------------------------"<< endl;
output << sys;
output << "-------------------------"<< endl;
FritzJohnCond fj(sys);
output << "Fritz-John system:" << endl;
output << "-------------------------"<< endl;
output << fj << endl;
output << "-------------------------"<< endl;
output << " number of variables:" << fj.nb_var << endl;
We get the following display. The variable _u is the coefficient of the goal function. The variable _l is the multiplier of the constraint.
original system:
-------------------------
variables:
x y
goal:
(x+y)
constraints:
((x^2+y^2)-1)<=0
-------------------------
Fritz-John system:
-------------------------
variables:
x y _u _l
goal:
(none)
constraints:
((_u+_l[0])-1)=0
((_u*1)+(_l[0]*(2*x)))=0
((_u*1)+(_l[0]*(2*y)))=0
(_l[0]*((x^2+y^2)-1))=0
-------------------------
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https://en.wikisource.org/wiki/Translation:Two_Papers_of_Henri_Poincar%C3%A9_on_Mathematical_Physics | # Translation:Two Papers of Henri Poincaré on Mathematical Physics
Two Papers of Henri Poincaré on Mathematical Physics (1921)
by Hendrik Lorentz, translated from French by Wikisource
In French: Deux Mémoires de Henri Poincaré sur la Physique Mathématique, Acta Mathematica. 38, pp 293–308, (written in 1914, published 1921); reprinted in Oeuvres de Henri Poincaré, Vol. XI, 247-261, 1956
Two Papers of Henri Poincaré on Mathematical Physics
by H. A. Lorentz
The following pages cannot at all give a complete idea of what theoretical physics owes to Poincaré. I would have been happy to pay homage to his memory by presenting to the reader such a general picture, but I moved back in front of this task, that cannot be done with dignity without long and serious studies for which there was no time for me. I limited thus myself to two papers, that on the Dynamics of the electron, written in 1905 and published the following year in Rendiconti del Circolo Matematico di Palermo, and the study on the quantum theory which appeared in the Journal de Physique at the beginning of 1912.
To fully appreciate the first of this works, I will have to enter in some details on the ideas whose development led to the principle of relativity. Thus let us speak a little about the part that I contributed to this development, I must say first that I have found a valuable encouragement in the benevolent interest that Poincaré constantly took with my studies. Moreover, we will see soon by which degree he surpassed me.
It is known that Fresnel had based the explanation of the astronomical aberration on the assumption of a motionless ether that the celestial bodies would cross without entraining it. We also know his famous theorem, a necessary complement of this fundamental assumption, of the partial entrainment of light waves by moving matter. An transparent body animated by translation will communicate to the rays only a fraction of its own speed, a fraction which is determined by the "coefficient of Fresnel" $1\tfrac{1}{n^ {2}}$, in which $N$ is the index of refraction of the medium.
When, thanks to the work of Clerk Maxwell, our views on the nature of light had been profoundly changed, it was natural to try a deduction of this coefficient based on the principles of the electromagnetic theory. That's the goal I set myself, which could be achieved without too much difficulty in the theory of electrons.
The majority of the phenomena which are connected to aberration, and in particular the absence of an influence of the earth's motion in all the experiments where the collective system of devices is at rest in respect to our planet, could now be explained in a satisfactory way. It was only necessary to make the restriction, that the considered effects were of first order of magnitude compared to the speed of the Earth divided by the speed of light, terms of the second order have been neglected in calculations.
However, in 1881 Mr. Michelson succeeded to interfere two light rays, that were departed from a single point and came back after following rectilinear and mutually perpendicular paths of equal length. He found that the observed phenomena are again insensitive to the earth's motion; the interference fringes preserved the same positions, whatever were the directions of the arms of the device.
This time it was indeed an effect of the second order and it was easy to see that the hypothesis of the stationary aether alone is not sufficient for the explanation of the negative result. I was obliged to make a new assumption which consists in admitting, that the translation of a body through the aether produces a slight contraction of the body in the direction of motion. This assumption was indeed the only possibility; it had also been imagined by Fitzgerald and it found the approval of Poincaré, who however did not conceal the lack of satisfaction that the theories gave him in which one multiplies special assumptions invented for particular phenomena. This criticism was for me an additional reason to seek a general theory, in which the same principles would lead to the explanation of the experiment of Mr. Michelson and all those that were undertaken after him to discover effects of the second order. In the theory that I proposed, the absence of phenomena due to the collective motion of a system should be demonstrated for any value of speed less than that of light.
The method to be followed was indeed indicated. It was obviously necessary to demonstrate that the phenomena which take place in a material system can be represented by equations of the same form, the system may be at rest or being animated by a uniform translatory motion, and this equality of form has to be obtained using a suitable substitution of new variables. It was a question of finding transformation formulas, suitable for the independent variables, the coordinates x, y, z and time t, as well as for the various physical magnitudes, speeds, forces, etc, and by showing the invariance of the equations for these transformations.
The formulas that I established for coordinates and time can be put as[1]
(1) $x'=kl(x+\epsilon t),\ y'=ly,\ z'=lz,\ t'=kl(t+\epsilon x),$
where ε, k, l are constants which are, however, reduced to one. We see immediately that the origin of the new coordinates ($x'=0$) is
$x=-\epsilon t\,$
so the point moves in the system x, y, z, t with speed in the direction of the x-axis. The coefficient k is defined by
$k=\left(1-\epsilon^{2}\right)^{-\frac{1}{2}}$
and ε is a function of l that has the value 1 for ε = 0. I initially left it undetermined, but I found in the course of my calculations, that to obtain the invariance (that I had in mind) we must put l = 1.
These were the considerations published by me in 1904 which gave place to Poincaré to write his paper on the dynamics of the electron, in which he attached my name to the transformation to which I will come to speak. I must notice on this subject that the same transformation was already present in an article of Mr. Voigt published in 1887, and that I did not draw from this artifice all the possible parts. Indeed, for some of the physical quantities which enter the formulas, I did not indicate the transformation which suits best. That was done by Poincaré and then by Mr. Einstein and Minkowski.
To find the "transformations of relativity", as I will call them now, it is sufficient in some cases to describe the phenomena in the system $x',y',z',t'$ exactly in the same way as we do it in system x, y, z, t. Let us consider, for example, the motion of a point. If, in time dt the coordinates x, y, z undergo the changes dx, dy, dz, then we have for the velocity components
$\xi=\frac{dx}{dt},\ \eta=\frac{dy}{dt},\ \zeta=\frac{dz}{dt}$
However, by these relations the variations dx, dy, dz, dt contain the changes
(2) $dx'=kl(dx+\epsilon\ dt),\ dy'=l\ dy,\ dz'=l\ dz,\ dt'=kl(dt+\epsilon\ dx)$
of the new variables. It is natural to define the velocity components in the new system by the formulas
(3) $\xi'=\frac{dx'}{dt'},\ \eta'=\frac{dy'}{dt'},\ \zeta'=\frac{dz'}{dt'}$
which gives us
(4) $\xi'=\frac{\xi+\epsilon}{1+\epsilon\xi},\ \eta'=\frac{\eta}{k(1+\epsilon\xi)},\ \zeta'=\frac{\zeta}{k(1+\epsilon\xi)}$
To have another example, we can imagine a great number of mobile points whose velocities are continuous functions of the coordinates and time. Let an element of volume located at point x, y, z and let us fix the attention to the points of the system which are in this element at one given moment t. Let $t'_0$ be the special value of $t'$ which corresponds to x, y, z, t by the equations (1), and consider for the various points the values of $x', y', z'$ which correspond to the given value $t'=t'_0$; in other words, let us consider the positions of the points in the new system, all taken for the same value of "time" $t'$. One might ask after the extension of the element $d\tau'$ of space $x', y', z'$, in which are at this moment $t'_0$ the selected points which are in at the time t. A simple calculation, which I omit here, led to the relation
(5) $d\tau'=\frac{l^{3}}{k}\frac{1}{1+\epsilon\xi}d\tau$
Finally, let us suppose that the points in question carry equal electric charges, and admit that in two systems x, y, z, t and $x', y', z', t'$ we attribute the same numerical values to these charges. If the points are sufficiently close to each other, we obtain a continuous distribution of electricity and it is clear that the charge contained in the element at the moment t is equal to that which is in $d\tau'$ at the moment $t'$. Consequently, if ρ and $\rho'$ are the densities of these charges,
(6) $\rho\ d\tau=\rho'd\tau'$
and, by virtue of (5)
(7) $\rho'=\frac{k}{l^{3}}(1+\epsilon\xi)\rho$
By this formula, combined with (4), we deduce again
$\rho'\xi'=\frac{k}{l^{3}}\rho(\xi+\epsilon),\ \rho'\eta'=\frac{1}{l^{3}}\rho\eta,\ \rho'\zeta'=\frac{1}{l^{3}}\rho\zeta$
These are the transformation formulas for the convection current.
For other physical quantities such as electric and magnetic forces, it is necessary to follow a less direct method; we will seek, perhaps with a little groping, the formulas of transformation suitable to ensure the invariance of the electromagnetic equations.
The formulas (4) and (7) are not in my memoir of 1904. Because I had not thought of the direct way which led there, and because I had the idea that there is an essential difference between systems x, y, z, t and $x', y', z', t'$. In one we use - such was my thought - coordinate axes which have a fixed position in the aether and which we can call "true" time; in the other system, on the contrary, we would deal with simple auxiliary quantities whose introduction is only a mathematical artifice. In particular, the variable $t'$ could not be called "time" in the same way as the variable t.
In this order of ideas I did not think of describing the phenomena in the system $x', y', z', t',$ exactly in the same way as in system x, y, z, t, and I did not define by the equations (3) and (7) the quantities ξ', η', ζ', ρ' which will correspond to ξ, η, ζ, ρ. It is rather by groping that I arrived at my formulas of transformation which, with our current notation, take the form
$\xi'=k^{2}(\xi+\epsilon),\ \eta'+k\eta,\ \zeta'=k\zeta,\ \rho'=\frac{1}{kl^{3}}\rho$
and that I wanted to choose, so as to obtain in the new system the simplest equations. Later, I could see in the paper of Poincaré that when proceeding more systematically I could have reached an even greater simplification. Not having noticed it, I did not succeed in obtaining the exact invariance of the equations; my formulas remained encumbered with certain terms which should have disappeared. These terms were too small to have an appreciable effect on the phenomena and I could thus explain the independence of the earth's motion that was revealed by observations, but I did not establish the principle of relativity as rigorously and universally true.
Poincaré, on the contrary, obtained a perfect invariance of the equations of electrodynamics, and he formulated the "postulate of relativity", terms which he was the first to employ. Indeed, stating from the point of view that I had missed, he found the formulas (4) and (7). Let us add that by correcting the imperfections of my work he never reproached me for them.
I can not explain here all the beautiful results obtained by Poincaré. Let us insist however on some points. Initially, he was not satisfied to show that the transformations of relativity leave intact the form of the electromagnetic equations. He explains the success of substitutions by noticing that these equations can be put in the form of the principle of least action and that the fundamental equation which expresses this principle, as well as the operations by which we deduce the field equations, are the same in systems x, y, z, t and $x', y', z', t'$.
In the second place, in accordance with the title of his paper, Poincaré particularly considers the way in which the deformation of a moving electron occurs, comparable with that of the arms of the device of Mr. Michelson, which is required by the postulate of relativity. Two different assumptions had been proposed on this subject. According to both an electron, presumably spherical in the state of rest, would change by a translation into an oblated ellipsoid of revolution, the axis of symmetry coincide with the direction of motion and the ratio of this axis to the diameter of the equator being given by $\sqrt{1-v^{2}},$, if v is the velocity. But the assumptions differed between them with regard to the length of the axes and consequently the volume of the electron. While I had been led to admit that the radius of the equator remains equal to that of the original sphere, Mr. Bucherer and Mr. Langevin rather wanted to assign a constant size with volume. The first assumption corresponds to l = 1, the second with $kl^3 =1$. Let us immediately add that the first value is the only one which is compatible with the postulate of relativity.
If one wants to realize the persistence and the balance of an electron while making use of the ordinary notions of mechanics, it is obviously not sufficient to consider the electrodynamic actions. The particle - that we consider here as a sphere carrying a surface charge - would immediately explode because of the mutual repulsions or, which is to same, of the stresses of Maxwell exerted on its surface. Therefore another concept should also be introduced, and Poincaré distinguishes at this place between the "bindings" and the "supplementary forces". He initially supposed that there is only the connection represented by the equation
$r = b\theta^m$
r is the semi-axis of the electron, its equatorial radius, b and m variables that remain constant when r and θ (or one of these quantities) vary with the translation speed v. This granted, we know for any value of v the dimensions of the electron - because we know that $\theta=\left(1-v^{2}\right)^{-\frac{1}{2}}$ - and by the ordinary formulas of electromagnetic field, the energy, momentum and the Lagrange function can be calculated. Between these values, considered as functions of v, there must be well known relations. Poincaré shows that they are verified only for $m=-\tfrac{2}{3}$, which brings us back to the constancy of volume, that is to say, the hypothesis of Mr. Bucherer and Langevin. But we know already that it is not this hypothesis, but only that of a constant equatorial radius, which is in agreement with the postulate of relativity. It is thus necessary to have recourse to additional forces.
By supposing that they depend on a potential of the form $Ar^\alpha \theta^\beta$, where A, α and β are constants, Poincaré finds that the constancy of the equatorial radius requires α = 3, β = 2, i.e. the potential in question must be proportional to volume. It results from it that the sought supplementary forces are equivalent to a pressure or a normal tension exerted on the surface and whose magnitude per unit of area remains constant, whatever the speed of translation. It is immediately seen that only a tension directed towards the interior is appropriate; we will determine the magnitude by the condition of an electron at rest and which has consequently the shape of a sphere, and it must be in equilibrium with the electrostatic repulsions. So when the particle is set into motion, the stress of Poincaré is united with the electrodynamic actions, and will inevitably produce the oblateness which is required by the principle of relativity.
After having found his supplementary force, Poincaré showed that the transformations of relativity do not change the form of the terms which it represents; thus he showed that arbitrary motions of a system of electrons can take place in the completely same manner in system x, y, z, t and in the system $x', y', z', t'$.
I already spoke about the necessity for posing l = 1 (constancy of the equatorial radius of the electron). I will not repeat here the demonstration given by Poincaré and I will only say that he showed the mathematical origin of this condition. One can consider all the transformations which are represented by formulas (1), with different values for speed , and the corresponding values of k and l, this last coefficient has to be regarded as a function of ε; we can add to it other similar transformations which we deduce from (1) by changing the directions of the axes, and finally by arbitrary rotations. The postulate of relativity requires that all these transformations form a group and that is only possible if l has the constant value 1.
The "group of relativity" obtained, consists of linear substitutions which do not affect the quadratic form
$x^2 + y^2 + z^2 - t^2$
The paper ends with the application of the postulate of relativity on the phenomena of gravitation. Here, it is the question of finding the rule which determines the propagation of it, and the formulas which express the components of the force according to the coordinates and the speed, as well as of the attracted body as of the attracting body. By considering these questions, Poincaré starts by seeking the invariants of the group of relativity; indeed, it is clear that it must be possible to represent the phenomena by equations which contain only these invariants. However, the problem is undetermined. It is natural to admit that the propagation velocity is equal to that of light and that the variations of the law of Newton must be of second-order magnitude in respect to the velocities. But even with these restrictions, there is the choice between several assumptions, among which there are two that were especially indicated by Poincaré.
In this last part of the article one finds some new concepts which I must especially announce. Poincaré notices, for example, if x, y, z and $t\sqrt{-1}$ are considered as the coordinates of a point in four-dimensional space, the transformations of relativity are reduced to rotations in this space. He also had the idea of adding to the three force-components X, Y, Z the magnitude
$T = X\xi + Y\eta + Z\zeta\,$
which is nothing but the work of the force per unit time and which we can (to some extent) regard as a fourth component. When we ask after the force that a body experiences per unit volume, the magnitudes X, Y, Z, T$\sqrt{-1}$ are affected by a transformation of relativity in the same way as the magnitudes x, y, z, t$\sqrt{-1}$.
I recall these ideas of Poincaré because they are similar to methods that Minkowski and other scientists have later used to facilitate mathematical operations that arise in the theory of relativity.
***
Let us pass now to the paper on the quantum theory. Towards the end of 1911 Poincaré had attended the meeting of the Council of Physics convened in Brussels by Mr. Solvay, in which we had especially dealt with the phenomena of the calorific radiation and the hypothesis of the elements or quanta of energy imagined by Mr. Planck to explain them. In the discussions, Poincaré had shown all the promptness and the penetration of his spirit and we had admired the facility with which he could enter the most difficult questions of Physics, even in those which were new for him. At the return to Paris, he did not cease dealing with the problem of which he felt the high importance. If the hypothesis of Mr. Planck were true, "the physical phenomena would cease obeying laws expressed by differential equations, and it would be, undoubtedly, the greatest and most profound revolution that natural philosophy suffered since Newton".
But are these new conceptions really inevitable and is there no way to arrive at the radiation law without introducing these discontinuities which are in direct opposition with the notions of traditional mechanics? Here is the question that Poincaré poses in his paper and to which he gives an answer that I will allow myself to briefly summarize.
Let us consider a system made up of n resonators of Planck and p molecules, n and p being very great numbers; let us suppose that all the resonators are equal between them and that it is the same for the molecules. Let us indicate by $\xi_{1},\dots,\xi_{p}$ the energies of the molecules and by $\eta_{1},\dots,\eta_{n}$ those of the resonators; each one of these variables will be able to take all the positive values.
Poincaré showed first that the probability so that the quantities of energy are between the limits $\xi_{1}$ and $\xi_{1}+d\xi_{1},\dots,\xi_{p}$, and $\xi_{p}+d\xi_{p}$, $\eta_{1}$ and $\eta_{1}+d\eta_{1},\dots,\eta_{n}$, $\eta_{n}$ and $\eta_{n}+d\eta_{n}$, can be represented by
$\omega\left(\eta_{1}\right)\dots\omega\left(\eta_{n}\right)d\eta_{1}\dots d\eta_{n}d\xi_{1}\dots d\xi_{p}$
where ω is a function for which we can make different hypotheses.
Once we know this function we can tell how much energy h will be distributed over the molecules and resonators. For this purpose, we can imagine a space of p + n dimensions, $\xi_{1},\dots,\xi_{\rho},\eta_{1},\dots,\eta_{n}$, the infinitely thin layer S, in which the total energy
$\xi_{1}+\dots+\xi_{p}+\eta_{1}+\dots+\eta_{n}$
lies between h and an infinitely close value h + dh. The three integrals will be calculated
$\begin{array}{l} I=\int\omega\left(\eta_{1}\right)\dots\omega\left(\eta_{n}\right)d\eta_{1}\dots d\eta_{n}d\xi_{1}\dots d\xi_{p}\\ \\I'=\int x\omega\left(\eta_{1}\right)\dots\omega\left(\eta_{n}\right)d\eta_{1}\dots d\eta_{n}d\xi_{1}\dots d\xi_{p}\\ \\I''=\int(h-x)\omega\left(\eta_{1}\right)\dots\omega\left(\eta_{n}\right)d\eta_{1}\dots d\eta_{n}d\xi_{1}\dots d\xi_{p}\end{array}$
$\left(x=\eta_{1}+\dots+\eta_{n}\right)$
extended to the layer S, and we have $\tfrac{I'}{I}$ for the energy that the resonators take, and $\tfrac{I''}{I}$ for that of all the molecules. Therefore, if Y is the mean energy of a resonator, and X is that of a molecule,
$nYI=I',\ pXI=I''$
To calculate the integral I, we may first give fixed values to variables $\eta_{1},\dots,\eta_{n}$ and consequently to their sum x, and extend the integration over ξ for all positive values of these variables, for which the sum $\xi_{1}+\dots+\xi_{p}$ is between h - x and h - x + dh. This gives us
$\int d\xi_{1}\dots d\xi_{p}=\frac{1}{(p-1)!}(h-x)^{p-1}dh$
Then we can calculate the integral
$\int\omega\left(\eta_{1}\right)\dots\omega\left(\eta_{n}\right)d\eta_{1}\dots d\eta_{n}$
extended to positive values of η such that $\eta_{1}+\dots+\eta_{p}$ lies between x and x + dx. Let
(8) $\int\omega\left(\eta_{1}\right)\dots\omega\left(\eta_{n}\right)d\eta_{1}\dots d\eta_{n}=\varphi(x)dx$
$\varphi$ is a function that depends on the function ω and we have
$I=\frac{dh}{(p-1)!}\int_{0}^{h}(h-x)^{p-1}\varphi(x)dx$
$I'$ and $I''$ are calculated in the same manner, we only need to introduce under the sign of integration the factor x or the factor h - x. Ultimately, we can write
(9) $nY=C\int_{0}^{h}x(h-x)^{p-1}\varphi(x)dx$
(10) $pX=C\int_{0}^{h}(h-x)^{p}\varphi(x)dx$
where the factor C is the same in both cases. We do not have to deal with it because it is sufficient to determine the ratio of X to Y.
Now we obtain the Planck formula - which can be regarded as an expression of reality - if we make on the function ω the following hypothesis, which is consistent with quantum theory.
Let ε be the magnitude of the quantum of energy which is specific to the resonators considered, and denote by δ an infinitely small quantity[2]. The function ω is zero, except in the intervals
$k\epsilon < \eta < k\epsilon + \delta\,$
and for each of these intervals the integral $\int_{k\epsilon}^{k\epsilon+\delta}\omega\ d\eta$ has the value 1.
These data are sufficient for determining the function $\varphi$ and the ratio $\tfrac{Y}{X}$ for which we find, as I said before, the value given by Planck's theory. I did not stop at these calculations and I pass immediately to the principal question, whether the discontinuities that I just mentioned must necessarily be admitted.
I will reproduce the reasoning of Poincaré, but I will at first say that in the formulas that we will encounter, α indicates a complex variable of which the real part $\alpha_r$ is always positive. In the representation we will limit ourselves to the half of plane α characterized by $\alpha_r>0$, and in integrations in respect to α we will follow a straight line l perpendicular to the axis of real α, and prolonged indefinitely on the two sides. The values of the integrals will be independent of the length of the distance $\alpha_r>0$ of this line at the origin of α.
Poincaré introduced an auxiliary function that defines the equation
(11) $\Phi(\alpha)=\int_{0}^{\infty}\omega(\eta)e^{-\alpha\eta}d\eta$
and demonstrated that the function ω and the derived function $\varphi$ can be be expressed by using Φ.
We obtain at first, by inverting (11)
(12) $\omega(\eta)=\frac{1}{2i\pi}\int_{(L)}\Phi(\alpha)e^{\alpha\eta}d\alpha$
For a similar formula for $\varphi(x)$ we notice that in equation (11) we can replace η by any of the variables $\eta_{1},\dots,\eta_{n}$. Multiplying the n equations which we obtained, we find
$\left[\Phi(\alpha)\right]^{n}=\int_{0}^{\infty}\dots\int_{0}^{\infty}\omega\left(\eta_{1}\right)\dots\omega\left(\eta_{n}\right)e^{-\alpha x}d\eta_{1}\dots d\eta_{n}$
or, by virtue of the formula (8)
$\left[\Phi(\alpha)\right]^{n}=\int_{0}^{\infty}\varphi(x)^{-\alpha x}dx$
and by inversion
$\varphi(x)=\frac{1}{2\pi i}\int_{(L)}\left[\Phi(\alpha)\right]^{n}e^{\alpha x}d\alpha$
The formulas (9) and (10) now become
$\begin{array}{l} nY=\frac{C}{2i\pi}\int_{0}^{h}\int_{(L)}x(h-x)^{p-1}[\Phi(\alpha)]^{n}e^{\alpha x}dx\ d\alpha\\ \\pX=\frac{C}{2i\pi}\int_{0}^{h}\int_{(L)}(h-x)^{p}[\Phi(\alpha)]^{n}e^{\alpha x}dx\ d\alpha\end{array}$
and Poincaré again transforms them by substitutions
$x=n\omega,\ h=n\beta,\ p=nk$
which give
$\begin{array}{l} nY=\frac{Cn^{p+1}}{2i\pi}\int_{0}^{\beta}\int_{(L)}\frac{\omega}{\beta-\omega}\Theta^{n}d\omega\ d\alpha\\ \\pX=\frac{Cn^{p+1}}{2i\pi}\int_{0}^{\beta}\int_{(L)}\Theta^{n}d\omega\ d\alpha\end{array}$
he posed
$\Theta=\Phi(\alpha)e^{\alpha\omega}(\beta-\omega)^{k}$
Note that ω is nothing else than the average energy of a single resonator for the case
$\eta_{1}+\dots+\eta_{n}=x$
that β is the value that would be ω if all available energy h was in the resonator, and that k is the ratio between the number of molecules and the resonators.
When, in the applications of the probability theory to the molecular theories, we seek the state of a system that presents the maximum of probability, we always find that, thanks to the immense number of the molecules, this maximum is so pronounced that one can neglect the probability of all the states which deviate appreciably from the most probable state. In the case which occupies us, there is something similar.
Let us admit with Poincaré that, for values given of h and β, the function Θ has a maximum for α = α0, ω = ω0 and passes through the point α0, the place of the maximum, the line l whose distance α0 in the beginning could be selected at will. As the exponent n is very high, the maximum of Θn is extremely pronounced and the only elements of the integrals which we have to take into account, are those who are in the immediate vicinity of α0 and of ω0. That immediately gives us for the sought ratio
$\frac{nY}{pX}=\frac{\omega_{0}}{\beta-\omega_{0}}$
and, by virtue of the equation
$nY = pX = h = n\beta$
(13) $Y = \omega 0$
(14) $X=\frac{\beta-\omega_{0}}{k}$
To determine the values of α0 and ω0, we can use equations
$\frac{\partial\log\Theta}{\partial\alpha}=0,\ \frac{\partial\log\Theta}{\partial\omega}=0$
from which we derive
(15) $\frac{\Phi'\left(\alpha_{0}\right)}{\Phi\left(\alpha_{0}\right)}+\omega_{0}=0$
and
(16) $\alpha_{0}-\frac{k}{\beta-\omega_{0}}=0$
We see from these formulas that α0 and ω0 depend on the quantity β, that is to say the total amount of energy h which was communicated to the system; this is a result which was to be expected. Equation (16) tells us further that α0 will always be real. This quantity determines immediately the average energy of a molecule as it follows from (14) and (16)
$X=\frac{I}{\alpha_{0}}$
Now we see that the average energy of a molecule is proportional to absolute temperature T. We can write
$\alpha_{0}=\frac{c}{T}$
where c is a known constant, and equation
(17) $Y=-\frac{\Phi'\left(\alpha_{0}\right)}{\Phi\left(\alpha_{0}\right)}$
which we draw from (13) and (15), gives us the average energy as a function of temperature. We see that this result is independent of the ratio between the numbers n and p.
Suppose now that we know for all temperatures the average energy of a resonator. By (17) we will thus know for all positive values of α the derivative $\tfrac{d\log\Phi(\alpha)}{d\alpha}$; we will deduce from them Φ(α) except for a constant factor. Of course, these findings will at first be limited to real values of α, but the function Φ(α) is assumed to be as determined throughout the semi-plane α about which we spoke, when it is given at all points of the real and positive semi-axis.
Finally, the formula (12) will provide us the function of probability ω for an unspecified positive value of η. It is true that the unspecified factor of function Φ(α) will be found in ω, but such a factor does not have any importance.
We can thus say that the probability ω is entirely given as soon as we know the distribution of energy for all temperatures. There is only one function ω for a distributions which is given as a function of the temperature. Consequently, the assumptions that we made on ω and which lead to the law of Planck are the only ones that we can admit.
That is the reasoning by which Poincaré established the necessity of the quantum hypothesis.
We see that the conclusion depends on the assumption that Planck's formula is an accurate image of reality. This could be drawn into question, and the formula could only be approximate. It is for this reason that Poincaré takes up the problem by abandoning Planck's law and using only the relationship that this physicist has found between the energy of a resonator and that of black body radiation. The reassessment led to the conclusion that the total energy of the radiation will be infinite unless the integral $\int_{0}^{\eta_{0}}\omega\ d\eta$ does not tend to zero with η0. The function ω must have at least one discontinuity (for η = 0), similar to those given by quantum theory[3].
1. I follow here the notations of Poincaré and I choose the units of length and time so that the speed of light is equal to 1.
2. This is the first theory of Planck, in which it is assumed that the energy of a resonator can only have values 0, ε, 2ε, 3ε, etc..
3. This result was found by P. Ehrenfest, see Ann. Physik, t. 36, 1911, p. 91. | {"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": 105, "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": 0.9480326771736145, "perplexity": 318.48598172826917}, "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/1443736676092.10/warc/CC-MAIN-20151001215756-00150-ip-10-137-6-227.ec2.internal.warc.gz"} |
http://www.physicspages.com/2017/02/21/uncertainties-in-the-harmonic-oscillator-and-hydrogen-atom/ | # Uncertainties in the harmonic oscillator and hydrogen atom
Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 9, Exercises 9.4.1 – 9.4.2.
Here we’ll look at a couple of calculations relevant to the application of the uncertainty principle to the hydrogen atom. When calculating uncertainties, we need to find the average values of various quantities. First, we’ll look at an average in the case of the harmonic oscillator.
$\displaystyle \psi_{n}(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (1)$
where ${H_{n}}$ is the ${n}$th Hermite polynomial. For ${n=1,}$ we have
$\displaystyle H_{1}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)=2\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (2)$
so
$\displaystyle \psi_{1}(x)=\frac{\sqrt{2}}{\pi^{1/4}}\left(\frac{m\omega}{\hbar}\right)^{3/4}x\;e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (3)$
For this state, we can calculate the average
$\displaystyle \left\langle \frac{1}{X^{2}}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\psi_{1}^{2}(x)\frac{1}{x^{2}}dx\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\int_{-\infty}^{\infty}e^{-m\omega x^{2}/\hbar}dx\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\sqrt{\frac{\pi\hbar}{m\omega}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2m\omega}{\hbar} \ \ \ \ \ (7)$
where we evaluated the Gaussian integral in the second line.
We can compare this to ${1/\left\langle X^{2}\right\rangle }$ as follows:
$\displaystyle \left\langle X^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\psi_{1}^{2}(x)x^{2}dx\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\int_{-\infty}^{\infty}e^{-m\omega x^{2}/\hbar}x^{4}dx\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\frac{3\sqrt{\pi}}{4}\left(\frac{\hbar}{m\omega}\right)^{5/2}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{2}\frac{\hbar}{m\omega}\ \ \ \ \ (11)$ $\displaystyle \frac{1}{\left\langle X^{2}\right\rangle }$ $\displaystyle =$ $\displaystyle \frac{2}{3}\frac{m\omega}{\hbar} \ \ \ \ \ (12)$
Thus ${\left\langle \frac{1}{X^{2}}\right\rangle }$ and ${\frac{1}{\left\langle X^{2}\right\rangle }}$ have the same order of magnitude, although they are not equal.
In three dimensions, we consider the ground state of hydrogen
$\displaystyle \psi_{100}\left(r\right)=\frac{1}{\sqrt{\pi}a_{0}^{3/2}}e^{-r/a_{0}} \ \ \ \ \ (13)$
where ${a_{0}}$ is the Bohr radius
$\displaystyle a_{0}\equiv\frac{\hbar^{2}}{me^{2}} \ \ \ \ \ (14)$
with ${m}$ and ${e}$ being the mass and charge of the electron. The wave function is normalized as we can see by doing the integral (in 3 dimensions):
$\displaystyle \int\psi_{100}^{2}(r)d^{3}\mathbf{r}$ $\displaystyle =$ $\displaystyle \frac{4\pi}{\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r^{2}dr \ \ \ \ \ (15)$
We can use the formula (given in Shankar’s Appendix 2)
$\displaystyle \int_{0}^{\infty}e^{-r/\alpha}r^{n}dr=\frac{n!}{\alpha^{n+1}} \ \ \ \ \ (16)$
We get
$\displaystyle \int\psi_{100}^{2}(r)d^{3}\mathbf{r}=\frac{4\pi}{\pi a_{0}^{3}}\frac{2!}{2^{3}}a_{0}^{3}=1 \ \ \ \ \ (17)$
as required.
For a spherically symmetric wave function centred at ${r=0}$,
$\displaystyle \left(\Delta X\right)^{2}=\left\langle X^{2}\right\rangle -\left\langle X\right\rangle ^{2}=\left\langle X^{2}\right\rangle \ \ \ \ \ (18)$
with identical relations for ${Y}$ and ${Z}$. Since
$\displaystyle r^{2}$ $\displaystyle =$ $\displaystyle x^{2}+y^{2}+z^{2}\ \ \ \ \ (19)$ $\displaystyle \left\langle r^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle x^{2}\right\rangle +\left\langle y^{2}\right\rangle +\left\langle z^{2}\right\rangle =3\left\langle X^{2}\right\rangle \ \ \ \ \ (20)$ $\displaystyle \left\langle X^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{3}\left\langle r^{2}\right\rangle \ \ \ \ \ (21)$
Thus
$\displaystyle \left\langle X^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{3}\int\psi_{100}^{2}(r)r^{2}d^{3}\mathbf{r}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi}{3\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r^{4}dr\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4}{3a_{0}^{3}}\frac{4!}{2^{5}}a_{0}^{5}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a_{0}^{2}\ \ \ \ \ (25)$ $\displaystyle \Delta X$ $\displaystyle =$ $\displaystyle a_{0}=\frac{\hbar^{2}}{me^{2}} \ \ \ \ \ (26)$
We can also find
$\displaystyle \left\langle \frac{1}{r}\right\rangle$ $\displaystyle =$ $\displaystyle \int\psi_{100}^{2}(r)\frac{1}{r}d^{3}\mathbf{r}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi}{\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r\;dr\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4}{a_{0}^{3}}\frac{a_{0}^{2}}{4}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{a_{0}}\ \ \ \ \ (30)$ $\displaystyle \left\langle r\right\rangle$ $\displaystyle =$ $\displaystyle \int\psi_{100}^{2}(r)r\;d^{3}\mathbf{r}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi}{\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r^{3}dr\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4}{a_{0}^{3}}\frac{6a_{0}^{4}}{16}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{2}a_{0} \ \ \ \ \ (34)$
Thus both ${\left\langle \frac{1}{r}\right\rangle }$ and ${\frac{1}{\left\langle r\right\rangle }}$ are of the same order of magnitude as ${1/a_{0}=me^{2}/\hbar^{2}}$. | {"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": 0, "wp_latex": 101, "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": 0.9971311688423157, "perplexity": 66.66577917409971}, "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-2017-13/segments/1490218189686.56/warc/CC-MAIN-20170322212949-00612-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.likera.com/blog/wp/archives/5288 | What: Royal blue latex “zentai” catsuit with a hood and shoulder zipper
Where: the item, the shop
Why: first of all I like latex, catsuits and royal blue colour. But that’s not it. Have a closer look at the zipper.
I see the following pro’s:
• more flexibility and freedom
• the zipper is not under stress in comparison with a zipper located at the back
• the suit beneath the zipper can be made absolutely hermetical (you need to ask to attach feet and remove the crotch zip)
• the zippers can easily be made lockable (just ask for it)
On the other hand, there are some contra’s:
• it will be not that easy to get in
• arms and shoulders are very mobile and that will put lots of pressure on the shoulder part of the zipper
• Anyway, I haven’t seen such a construction before and it look very interesting. | {"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": 0.8318949341773987, "perplexity": 2191.1364923492283}, "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-2017-22/segments/1495463607848.9/warc/CC-MAIN-20170524152539-20170524172539-00540.warc.gz"} |
https://www.math.ias.edu/delellis/research | Broadly speaking my research focuses on geometric analysis and partial differential equations. As a student I started working in the calculus of variations (more specifically in geometric measure theory) under the supervision of Luigi Ambrosio. With time my interests have branched towards more classic partial differential equations (systems of conservation laws, transport equations, Hamilton-Jacobi equations, the incompressible Euler equations) and minimal surfaces in Riemannian manifolds. In all these fields I am typically interested in the formation and behavior of singularities (or, in some lucky cases, in the absence of singularities).
Let me list three very broad subjects in which the presence (or absence) of singularities seem to be the most compelling issue.
Area-minimizing surfaces and Geometric measure theory
Consider the famous Plateau's problem: given a surface N of dimension k in a certain ambient space, we look for the surface(s) of dimension k+1 which span N and have the least area possible. It is well known that, in general, these least area (or area-minimizing) surfaces are singular. The very formulation of Plateau's problem poses fundamental question like: what is a surface, what is its content, what does it mean for a surface to span a contour? Geometric measure theory is a branch of analysis which addresses the above questions and many others.
Transport equations and Hyperbolic systems of conservation laws
Several classical models in continuum physics end up in formulating partial differential equations describing the evolution of a given physical systems. Often these systems are the mathematical description of conservation laws for some relevant physical quantities. Often these conservation laws appear in the form of quantities which are transported along the flows. Several of these systems of partial differential equations (especially in compressible fluid dynamics) fall in a class called "hyperbolic systems of conservation laws". For these systems a typical phenomenon is the creation of singularities which travel in time, called shock waves.
Incompressible fluid dynamics
Consider the Euler equations for incompressible fluids (derived by Euler more than 250 years ago) or the Navier-Stokes equations (roughly speaking a modification of the Euler equations proposed in the nineteenth century to take into account the effect of viscosity). It was proved in the last century that in 2 space dimensions there is no formation of singularities for these equations*. It is presently not known whether this is true also in 3 dimensions. For the Navier-Stokes equation this is one of the famous "millennium problems" and hence regarded as one of the major challenges in mathematics.
*I am obviously speaking of the Cauchy problem in the whole space: it is well known that for more complicated boundary conditions the description of boundary layers for the Euler equations is one of the formidable problems of fluid-dynamics. | {"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": 0.8407386541366577, "perplexity": 342.89850980366396}, "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/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00507.warc.gz"} |
http://math.stackexchange.com/questions/1844/how-to-calculate-reflected-light-angle?answertab=active | # How to calculate reflected light angle?
On a two dimensional plane, line $X$ is at an angle of $x$ radians and an incoming light travels at an angle of $y$ radians. How can I calculate the angle of the outgoing light reflected off of the line $X$? How can I cover all possible cases?
Edit: I was trying to figure out Project Euler problem 144.
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Can you please make a drawing? It's hard to see which angle you're referring to. – kennytm Aug 8 '10 at 14:58
There are no angles x and y in the link. – kennytm Aug 8 '10 at 19:15
@KennyTM, you are right. That was how I wanted to solve the problem. Now think of it, it might be better to just try to figure out the outgoing light's tangent. – grokus Aug 8 '10 at 20:10
Since you've stated all three angles in similar terms, and want a formula that works in all cases, lets use the angle with the x axis, in 360 degree terms, which accomplishes both purposes and is good for computation. So here's the picture, with z the angle you are seeking, and a the angle of reflection...
Then, using the two triangles with the x axis as base and the fact that an exterior angle of a triangle is the sum of the other interior angles, you get
z = x + a
y = $\pi$ - 2a + z
And solving those for z in terms of x and y gives
z = $\pi$ + 2x - y
OK, this is actually far from a general analysis -- angles x, y and z could occur in any order on the x axis, but we may assume that y is to the right of x without losing generality, so there are only two other cases. Also angle x could > $\frac{\pi}{2}$, but since the lines are undirected, we may assume that x < $\pi$. Finally, angle x could be 0.
Hmmm... thinking about this some more. When z falls to the right of y, the same formula follows because the roles of z and y are interchanged. But when z falls to the left of x, it's because the line of the reflected light intersects the x axis "backwards". And then the geometry yields z = 2x - y, or the prior formula if you take the angle of the reflected light as the supplement of z.
So we really need vectors for the light rays, not undirected lines, and/or the original problem is not altogether well-formed, that is the notion "angle" of a light ray needs to be defined as the direction of its vector. If you do that, then the angle labeled z in my diagram is not the correct vector directional angle. It should be $\pi$ + z, so the true formula, in vector direction is z = 2x - y, and that works for all cases. (Haven't checked the degenerate cases, but no reason to expect them to fail).
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Seeing your answer, I tried to figure out how to guarantee that it was general, and couldn't easily do so. That's what led me to my answer, where I started by moving the x-axis to the point where the light hits the mirror (or, alternately, moving that point to the "origin") and used directed angles to guarantee the generality – Isaac Aug 8 '10 at 20:21
See my hmmm comment added to the post. – David Lewis Aug 8 '10 at 21:47
I'm curious about what software you used to draw a picture like this. Thanks. – grokus Aug 9 '10 at 3:34
Ok, here's another way to look at it, using vector directions and kinda intuitive, but I think it is pretty close to a real proof.
Start with the mirror line X horizontal, that is, its angle x = 0. The it's clear that the vector angle, z, of the reflected light is the negative of the vector angle, y, of the light itself: z = -y.
Now rotate line X by d degrees around the point of reflection, either way, leaving the original light ray (angle y) fixed and letting the reflected ray (angle z) rotate around the point of reflection to maintain the angle of reflection equal to the angle of incidence. Assuming counterclockwise rotation, d > 0, this "pushes" the reflected line by 2d, one d for a direct push to maintain the angle fo reflection, and another d because the angle of incidence also increases by d, so the reflected light must rotate that much more to keep the two angles of the light equal. Likewise when d < 0 for clockwise rotation.
So we are increasing (or decreasing, for counterclockwise rotation) angle x by d, but angle z (vector angle) by 2d. Hence...
z = 2d - y = 2x - y
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Note that regardless of what line the angles are measured from, there is a parallel line through the point of reflection on the mirror for which the angle measures will be the same. My diagram is not fully general as drawn (in terms of the angles), but the argument below is fully general, provided that all angle measures are directed—that is, counterclockwise angles are positive and clockwise angles are negative. x is the angle from a reference line to (one half of) the mirror, y is the angle from that same reference line to the incoming light ray.
The angle from the measured half of the mirror to the incoming light ray is $y-x$ (note that if the diagram is as I'd drawn, except that x were to the other half of the mirror, $x>y$ so $y-x<0$, which makes sense since $y-x$ would be going counterclockwise from the other half of the mirror to the incoming light ray). Because the magnitude of the angles of incidence and reflection are the same, but the angle of reflection is the opposite direction from the other half of the mirror, the angle from the other half of the mirror to the outgoing light ray is $-(y-x)=x-y$.
This makes the directed angle from the measured half of the mirror to the outgoing light ray $\pi+(x-y)$ (start at the measured half of the mirror, rotate $\pi$ counterclockwise, then rotate $x-y$, which is directed back clockwise). From the angle reference line to the outgoing light ray is then $x+\pi+x-y=\pi+2x-y$ (as in David Lewis's answer).
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Let the line has equation $ax + by + c = 0$, point where light comes on the line has coordinates $(x_0, y_0)$ and direction of the light is $(v_x, v_y)$. Suppose also that line and light direction are normed: $a^2 + b^2 = 1$ and $v_x^2 + v_y^2 = 1$.
Vector $(-b, a)$ is a vector parallel to the line. Let $d = -bv_x + av_y$ --- projection of light's direction to the line.
Then new direction of the light is $((-b)\cdot(2d) - v_x, (a)\cdot(2d) - v_y)$.
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sir I didn't forget you. I will take sometime to understand what you say here. The Wikipedia article on Specular reflection seems to talk about the same approach. – grokus Aug 9 '10 at 21:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8070672750473022, "perplexity": 290.0043295108076}, "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-2015-27/segments/1435375095671.53/warc/CC-MAIN-20150627031815-00128-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/206202/transient-vs-steadystate?answertab=votes | Let
$$s(t) = \cos(wt)\cdot u(t)$$
with $u(t)$ being the unit step.
Suppose we can represent such a signal as the sum of a transient and of a "steady state". A transient is a short-time wide-band spectral component of part of the signal. In this case the transient occurs at $t = 0$.
My question is, given $s(t)$ above, what are $s_t(t)$ and $s_s(t)$ in the decomposition
$$s(t) = s_t(t) + s_s(t)$$
where $s_t(t)$ is localized around t = 0 with minimum support in time and $s_s(t)$ has minimum support in frequency. Essentially we are trying to maximize the time-locality for $s_t(t)$ and maximize the frequency-locality for $s_s(t)$.
Note that solutions such as $s_t(t) = -\cos(wt)\cdot u(-t)$ and $s_s(t) = \cos(wt)$ are not allowed as $s_t(t)$ is not localized around t. In fact I think we should require the transient to be causal in nature. Another possibility is $s_t(t) = \cos(wt)\cdot u(t) \cdot u(1 + \alpha - t)$ and $s_s(t) = \cos(wt)\cdot u(t - 1 - \alpha)$ but $s_s(t)$ does not have minimal bandwidth.
I would expect the above problem to have a unique solution. I imagine $s_s(t)$ have a Gaussian like attack. Any ideas?
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How about $$s_t(t)= u(t) \cos(\omega t) (1-e^{t_0/t})$$ and $$s_s(t)= u(t) \cos(\omega t) e^{-t_0/t}?$$ The constant $t_0$ you can still adjust to fit your needs.
@JonSlaughter: there is no mathematical definition there. I would assume that you have to define the support as the the size of the interval where the function is nonzero? Then if $\sigma_t$ is the (time)support of the transient and $\sigma_\omega$ the (frequency)support of the steady state. Would you want to minimize $\sigma_t \sigma_\omega$ or $\sigma_t + \sigma_\omega$ or $\text{max}(\sigma_t, \sigma_\omega)$ or ...? Furthermore, why you care about the support and not of a weaker (but maybe more suitable measure) like standard deviation? – Fabian Oct 9 '12 at 17:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9321022033691406, "perplexity": 180.17070096518003}, "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-2013-48/segments/1387345758904/warc/CC-MAIN-20131218054918-00009-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/575597/is-there-any-simple-way-to-predict-beta-decay-half-lives?noredirect=1 | Is there any simple way to predict beta decay half lives?
Question
For nuclides that decay by alpha emission, the Geiger-Nuttall law gives a simple and reasonably accurate estimate of the half-life. Essentially, one can model the alpha particle as a particle in a "box" – the nucleus – and calculate the probability of tunneling out of the box using basic quantum mechanics. The result is that the half-life depends exponentially on the decay energy:
$$\ln \lambda \approx a_0 - a_1 \frac{Z}{\sqrt{E}}$$
where $$\lambda$$ is the half-life, $$Z$$ is the atomic number, and $$E$$ is the decay energy.
Is there any similar simple relationship for beta decay?
Beta decay half-lives appear to be very poorly correlated with:
• Decay energy. Rhenium-187 and lutetium-176 have similar half-lives (about 40 billion years), but lutetium-176 has a decay energy of 1.2 MeV while rhenium-187 has a decay energy of just 2.6 keV.
• Atomic number or number of neutrons. Cadmium-113 has a half-life of more than 8 quadrillion years, while cadmium-115 has a half-life of just 2 days – even though isotopes 112, 114, and 116 are all stable or have extremely long half-lives.
• Nuclear shell closure. Even though potassium-40 decays to calcium-40, which has magic numbers of both protons and neutrons, it has a half-life of over a billion years. Meanwhile, potassium-42 has a half-life of only 12 hours.
In short, beta decay half-lives vary widely, and I don't see any obvious rules to help understand why. Is there any way of explaining these results without a very complex calculation?
Review of related Physics.SE questions
That leaves the phase space available to the products as the thing that almost solely determined the lifetime.
In short the phase space available to the interaction depends to a large degree on the total energy and to a much smaller degree on the mass of the recoiling remnant nucleus.
However, while the theoretical argument that the phase space should depend strongly on the total energy seems plausible, it doesn't seem to match experimental results very well. As I noted above, nuclei with similar half-lives often have completely different decay energies, and vice versa.
The most upvoted answer to Can we predict the half-lives of radioactive isotopes from theory? mentions a
quantity "$$ft$$" which convolves the half-life of the decay with the electrical interaction between the emitted electron and the positively-charged daughter nucleus.
It is supposed to vary within a narrow range. However, I don't understand this description; also, I looked up some tables of $$ft$$ values and found that $$ft$$ actually varies by many orders of magnitude between different nuclei. For example, this textbook chapter (chapter 8 from Modern Nuclear Chemistry by Loveland, Morrissey, and Seaborg) cites $$\log(ft)$$ values ranging from $$-0.27$$ to $$+7.36$$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.960260808467865, "perplexity": 585.7495375313965}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487612154.24/warc/CC-MAIN-20210614105241-20210614135241-00552.warc.gz"} |
https://www.aimsciences.org/journal/2156-8472/2012/2/2 | # American Institute of Mathematical Sciences
ISSN:
2156-8472
eISSN:
2156-8499
All Issues
## Mathematical Control & Related Fields
June 2012 , Volume 2 , Issue 2
Special Issue dedicated to Professor Charles Pearce on the occasion of his 70th birthday
Special Issue Papers: 223-375; Regular Papers: 377-435
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Export/Reference:
2012, 2(2): 101-120 doi: 10.3934/mcrf.2012.2.101 +[Abstract](2182) +[PDF](827.1KB)
Abstract:
Leakage represents a large part of the supplied water in Water Distribution Systems (WDS). Consequently, it is important to develop some efficient strategies to manage such a phenomenon. In this paper an improved formulation of the hydraulic network equations that incorporate pressure-dependent leakage, is presented and validated. The formulation is derived from the Navier-Stokes equations and solved using an adequate splitting method. Then, this formulation is used to study a constrained optimization problem with the objective to minimize the distributed water volume reducing the leakage. The problem is described and validated for academic case studies and real networks.
2012, 2(2): 121-140 doi: 10.3934/mcrf.2012.2.121 +[Abstract](2120) +[PDF](436.5KB)
Abstract:
This paper is devoted to prove the local exact controllability to the trajectories for a coupled system, of the Boussinesq kind, with a reduced number of controls. In the state system, the unknowns are the velocity field and pressure of the fluid $(\mathbf{y},p)$, the temperature $\theta$ and an additional variable $c$ that can be viewed as the concentration of a contaminant solute. We prove several results, that essentially show that it is sufficient to act locally in space on the equations satisfied by $\theta$ and $c$.
2012, 2(2): 141-170 doi: 10.3934/mcrf.2012.2.141 +[Abstract](2297) +[PDF](536.1KB)
Abstract:
The notion of semilinear parabolic equation of normal type is introduced. The structure of dynamical flow corresponding to equation of this type with periodic boundary condition is investigated. Stabilization of mentioned equation with arbitrary initial condition by start control supported in prescribed subset is constructed.
2012, 2(2): 171-182 doi: 10.3934/mcrf.2012.2.171 +[Abstract](2837) +[PDF](367.6KB)
Abstract:
In this paper we prove the approximate controllability of the a broad class of semilinear reaction diffusion equation in a Hilbert space, with application to the semilinear $n$D heat equation, the Ornstein-Uhlenbeck equation, amount others.
2012, 2(2): 183-194 doi: 10.3934/mcrf.2012.2.183 +[Abstract](3359) +[PDF](343.0KB)
Abstract:
In this paper, we study the finite element method for constrained optimal control problems governed by nonlinear elliptic PDEs. Instead of the standard error estimates under $L^2$- or $H^1$- norm, we apply the goal-oriented error estimates in order to avoid the difficulties which are generated by the nonsmoothness of the problem. We derive the a priori error estimates of the goal function, and the error bound is $O(h^2)$, which is the same as one for some well known quadratic optimal control problems governed by linear elliptic PDEs. Moreover, two kinds of practical algorithms are introduced to solve the underlying problem. Numerical experiments are provided to confirm our theoretical results.
2012, 2(2): 195-215 doi: 10.3934/mcrf.2012.2.195 +[Abstract](3645) +[PDF](469.1KB)
Abstract:
Traditionally, the time domains that are widely used in mathematical descriptions are limited to real numbers for the case of continuous-time optimal control problems or to integers for the case of discrete-time optimal control problems. In this paper, based on a family of "needle variations", we derive maximum principle for optimal control problem on time scales. The results not only unify the theory of continuous and discrete optimal control problems but also conclude problems involving time domains in partly continuous and partly discrete ingredients. A simple optimal control problem on time scales is discussed in detail. Meanwhile, the results also unify the theory of some hybrid systems, for example, impulsive systems.
2020 Impact Factor: 1.284
5 Year Impact Factor: 1.345
2020 CiteScore: 1.9 | {"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": 0.8764364719390869, "perplexity": 406.45074298985827}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154126.73/warc/CC-MAIN-20210731203400-20210731233400-00468.warc.gz"} |
http://cms.math.ca/cjm/msc/42C10?fromjnl=cjm&jnl=CJM | Search results
Search: MSC category 42C10 ( Fourier series in special orthogonal functions (Legendre polynomials, Walsh functions, etc.) )
Expand all Collapse all Results 1 - 4 of 4
1. CJM 2010 (vol 62 pp. 1182)
Yue, Hong
A Fractal Function Related to the John-Nirenberg Inequality for $Q_{\alpha}({\mathbb R^n})$ A borderline case function $f$ for $Q_{\alpha}({\mathbb R^n})$ spaces is defined as a Haar wavelet decomposition, with the coefficients depending on a fixed parameter $\beta>0$. On its support $I_0=[0, 1]^n$, $f(x)$ can be expressed by the binary expansions of the coordinates of $x$. In particular, $f=f_{\beta}\in Q_{\alpha}({\mathbb R^n})$ if and only if $\alpha<\beta<\frac{n}{2}$, while for $\beta=\alpha$, it was shown by Yue and Dafni that $f$ satisfies a John--Nirenberg inequality for $Q_{\alpha}({\mathbb R^n})$. When $\beta\neq 1$, $f$ is a self-affine function. It is continuous almost everywhere and discontinuous at all dyadic points inside $I_0$. In addition, it is not monotone along any coordinate direction in any small cube. When the parameter $\beta\in (0, 1)$, $f$ is onto from $I_0$ to $[-\frac{1}{1-2^{-\beta}}, \frac{1}{1-2^{-\beta}}]$, and the graph of $f$ has a non-integer fractal dimension $n+1-\beta$. Keywords:Haar wavelets, Q spaces, John-Nirenberg inequality, Greedy expansion, self-affine, fractal, Box dimensionCategories:42B35, 42C10, 30D50, 28A80
2. CJM 2004 (vol 56 pp. 431)
Rosenblatt, Joseph; Taylor, Michael
Group Actions and Singular Martingales II, The Recognition Problem We continue our investigation in [RST] of a martingale formed by picking a measurable set $A$ in a compact group $G$, taking random rotates of $A$, and considering measures of the resulting intersections, suitably normalized. Here we concentrate on the inverse problem of recognizing $A$ from a small amount of data from this martingale. This leads to problems in harmonic analysis on $G$, including an analysis of integrals of products of Gegenbauer polynomials. Categories:43A77, 60B15, 60G42, 42C10
3. CJM 1998 (vol 50 pp. 1236)
Kalton, N. J.; Tzafriri, L.
The behaviour of Legendre and ultraspherical polynomials in $L_p$-spaces We consider the analogue of the $\Lambda(p)-$problem for subsets of the Legendre polynomials or more general ultraspherical polynomials. We obtain the best possible'' result that if $2 Categories:42C10, 33C45, 46B07 4. CJM 1997 (vol 49 pp. 175) Xu, Yuan Orthogonal Polynomials for a Family of Product Weight Functions on the Spheres Based on the theory of spherical harmonics for measures invariant under a finite reflection group developed by Dunkl recently, we study orthogonal polynomials with respect to the weight functions$|x_1|^{\alpha_1}\cdots |x_d|^{\alpha_d}$on the unit sphere$S^{d-1}$in$\RR^d\$. The results include explicit formulae for orthonormal polynomials, reproducing and Poisson kernel, as well as intertwining operator. Keywords:Orthogonal polynomials in several variables, sphere, h-harmonicsCategories:33C50, 33C45, 42C10 | {"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": 0.9383797645568848, "perplexity": 827.1821383084095}, "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-2013-48/segments/1386164032288/warc/CC-MAIN-20131204133352-00007-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://theinformaticists.com/2019/03/24/sensory-processing-in-the-retina/ | # Sensory Processing in the Retina
EE376A (Winter 2019)
By Saarthak Sarup, Raman Vilkhu, and Louis Blankemeier
## Modeling the Vertebrate Retina
“To suppose that the eye with all its inimitable contrivances for adjusting the focus to different distances, for admitting different amounts of light, and for the correction of spherical and chromatic aberration, could have been formed by natural selection, seems, I confess, absurd in the highest degree…The difficulty of believing that a perfect and complex eye could be formed by natural selection , though insuperable by our imagination, should not be considered subversive of the theory.” – Charles Darwin
The sentiment behind this quote truly captures the miracle that is the human eye… even Darwin had to specifically address its unique case in his argument for natural selection. In this project, we aim to capture some of this complexity in the structure and function of the eye and study it under the lens of information theory.
#### A short biology lesson…
The human eye is composed of various complex structures, however the one studied and modeled throughout this blog is the retina. The retina is responsible for converting light responses captured by the cornea into electrical signals sent to the visual cortex in the brain via the optic nerve. As seen in Figure 1, the retina lies in the back of the eye and can be further decomposed into layers of neurons and photoreceptors. As a brief introduction to the biology, the light passes through the layers from the leftmost to the rightmost: nerve fiber layer (RNFL), retinal ganglion cells (RGC), amacrine cells (AC), bipolar cells (BC), horizontal cells (HC), photoreceptors (PR), and the retinal pigment epithelium (RPE). For the scope of this blog, it is sufficient having a high level understanding that the retina translates varying light responses to electrical spikes (known as action potentials) which can be transmitted along the neurons in the optic nerve to the visual cortex, where perception takes place. For those interested in these specific retina layers and their unique contributions to this encoding process, please refer to [4].
#### Communicating through the eye
The primary objective of the project was to model the eye under the lens of information theory and see what analytical results can be derived to answer questions such as: what is the maximum rate achievable by the eye? What is the best model for the encoding/decoding done in the human visual system? In order to study these overarching questions, we had to start by modeling the system as an information processing channel.
First, the eye takes in a light response captured at the cornea and encodes that into electrical spikes defined by a spiking rate. In an information theory sense, this is equivalent to encoding a source using some encoder that takes intensity and maps it to discrete spikes. Here, as seen in Figure 2, we assume that the eye does rate-encoding, where the majority of the useful “information” is captured by the spiking rate rather than precise spatio-temporal information about each individual spike. This particular assumption is one heavily debated in literature and the true encoding scheme used by the eye is still a mystery, however this assumption is nonetheless a good starting point. Additionally, for the scope of our project, we generated the visual inputs with known statistics matching that of natural scenes, as discussed in later sections. Therefore, mapping the biology to the information theory, the encoder in the system models the retina — mapping light responses to spikes.
In the visual system, the communication channel is represented by the optic nerve, which runs from the retina to the visual cortex in the brain. In our model, this communication channel is modeled as a parallel, inhomogeneous Poisson channel with some additive Poisson noise. Specifically, the number of parallel channels corresponds to the number of retinal ganglion cells connected to the optic nerve bundle, each transmitting some spike rate on the channel.
Finally, in our model, the communication channel feeds into a decoder which implements a strategy to capture the spike rates and reconstruct the perceived visual inputs from these estimated rates. In biology, this corresponds to the visual cortex of the brain, which perceives the spikes as some visual image.
Overall, this general framework of modelling the visual system as an information theory problem is particularly interesting because we can apply well-known information theory techniques to gain insights into otherwise mysterious processes.
## Joint Source-Channel Coding: A Review
Our model of the early visual system depicts both compression via neural downsampling and communication via Poisson spiking. All together, our work fits into the joint source-channel coding framework. The optimal strategy in this framework according to Shannon’s separation theorem is to independently optimize both our source coding compression scheme and our channel coding communication protocol.
#### Source coding for optimal compression
In source coding, we deal with a sequence of inputs $latex X_1,X_2,\dots,X_N ~ p(x), x\in \mathcal{X}$, which is encoded by an index $latex f_n(X^n)\in {1,2,\dots,2^{nR}}$ and then decoded by an estimate $latex \hat{X}^n$ from the output alphabet $latex \hat{\mathcal{X}}$. This procedure is outlined in Figure 4 below. An important quantity in this framework is the compression-rate $latex R$, defined as the average number of bits per input symbol. According to Shannon’s source coding theorem, it is impossible to perfectly compress input data such that the compression-rate is less than the entropy of the source. In the case where lossy compression is tolerated, the tradeoff between minimizing the compression-rate and minimizing the loss is given by the rate-distortion curve. Points along this curve $latex R(D)$ define the smallest achievable compression-rates (largest compression factors) for a given distortion $latex D$.
#### Channel coding for reliable communication
Channel coding consists of finding a set of channel input symbol sequences such that each sequence is sufficiently “far” from the other sequences and each input sequence maps to a disjoint set of output sequences. In this case, the channel input sequence can be decoded without error. Shannon showed that for every channel, there exists some channel code such that $latex max_{p(x)} I(Y;X)$ bits can be sent per channel use. $latex max_{p(x)} I(Y;X)$ is known as the channel capacity. This idea can be generalized to a channel which takes a continuous waveform as input. In this case, channel coding consists of finding waveforms, which last for duration $latex T$, and are spaced sufficiently far apart such that they can be decoded without error. This is shown in Figure 4 for the case of four independent Poisson channels. Here, the codebook consists of waveforms which take the form of pulse modulated signals. As discussed in sections “Channel Capacity Under Poisson Noise” and a “Mathematical Aside”, a code generated with such waveforms can reach capacity. Thus, each continuous input is mapped to a pulse modulated signal which is then transmitted across the Poisson channel.
## Compressing a Gaussian Source with Finite Bandwidth
How can we statistically characterize our visual inputs? In the space of continuous and unbounded alphabets, Gaussian distributions maximize the available entropy under a bounded power (technically, second moment) constraint. As a result, we take a Gaussian source to be a general description of our 2D image data. To model a general bandwidth assumption on our images, we employ shift-invariant autocorrelation function given by the squared-exponential kernel $latex K(\textbf{x}-\textbf{x’}) = \exp\Big(\frac{-|\textbf{x}-\textbf{x’}|^2}{2l^2}\Big)$. This Gaussian process allows us to draw random $latex S\times S$ images as $latex S^2$-length vectors from $latex \mathcal{N} \big(\textbf{0}, \Sigma_l \big)$, where the covariance matrix $latex \Sigma_l$ is parameterized by the length constant $latex l$. This process gives a non-uniform power spectrum with analytic form given below:
$latex S(f) = \sqrt{2\pi l^2}\exp(-2\pi^2 l^2 f^2)$
To further simplify our analysis, we consider images in gray-scale, as color representation in the retinal code can be handled somewhat orthogonally to intensity by color-tuned photoreceptors and retinal ganglion cells. Example frames from this procedure can be seen in Fig. 5 below.
#### How good can we theoretically do?
When compressing a memoryless Gaussian source with a uniform power spectrum $latex \sigma^2$, the maximum achievable compression rate for a minimum mean-squared distortion $latex D$ is given by the well-known expression, $latex R(D) = \frac{1}{2}\log\frac{\sigma^2}{D}$.
For our non-memoryless case, the optimal compression strategy can be clarified in the spectral domain. Here, the single spatially-correlated source can be decomposed into infinitely many parallel independent sources, each with variance given by the power $latex S(f)$. This transformation inspires a reverse water-filling solution [7], where we use a given distortion budget $latex D$ to determine a threshold $latex \theta$, such that compression is done by only encoding the components where $latex S(f)>\theta$. In this parametric form, the new rate-distortion curve is given by the following equations:
$latex D(\theta) = \int_{-\frac{1}{2}}^{\frac{1}{2}}\min(\theta,S(f)) \ df$
$latex R(\theta) = \int_{-\frac{1}{2}}^{\frac{1}{2}}\max\Big(0,\frac{1}{2}\log\frac{S(f)}{\theta}\Big) \ df$
To reiterate an important point, these $latex R(D)$ curves depend on the length scale of the source $latex l$. In the limit where $latex l \rightarrow \infty$, our image becomes uniform and its power spectrum is concentrated at $latex f=0$. This source distribution corresponds to the largest rate-distortion curve, since the all the input pixels can be encoded with the single uniform intensity and all the distortion is concentrated in the quantization of that value. In the limit where $latex l\rightarrow 0$, our bandwidth becomes unbounded and we recover the white Gaussian source, with a constant power spectrum $latex S(f) = \sigma^2$. At this extreme, each input pixel is uncorrelated with the other pixels, and for compression under the same fixed distortion $latex D$, each symbol must share this budget with a more aggressive quantization. This intuition is validated in Fig. 6a, where the rate-distortion curves are computed for a variety of length-scales.
#### How good do we practically do?
Given these fundamental limits, we now describe our neural compression scheme and compare our results. In the pioneering work by Dr. Haldan Hartline in 1938 [3], researchers discovered evidence of a receptive field in retinal ganglion cells, corresponding to a particular region of sensory space in which a stimulus would modify the firing properties of the neuron. In more mathematical terms, this property is modeled through a convolutional spatial filter which the neuron applies to its input to determine its response.
We thus interpret each pixel value as the encoded firing intensity of the primary photoreceptors, which we compress by applying spatial filters $latex k(x-x_c,y-y_c)$ at each neuron location $latex (x_c,y_c)$. Our model uses an isotropic Gaussian for each neuron’s spatial filter, producing an output $latex r(x_c,y_c) = k(x-x_c,y-y_c) * I(x,y)$ which is used as the neuron’s firing rate. If the neurons are spaced $latex \Delta x$ pixels apart, a $latex S\times S$ image is compressed into a $latex \frac{S}{\Delta x}\times\frac{S}{\Delta x}$ output, corresponding to a compression rate of $latex \frac{1}{\Delta x^2}$.
To measure the distortion of this compression scheme, we propose decoding schemes for reconstructing the image. These schemes are not biologically motivated, as its unclear what neural process, if any, attempts to reconstruct the visual inputs to the photoreceptors. It is nevertheless instructive to consider how our biologically-inspired encoding scheme could be reversed. A simple procedure is one which deconvolves a neuron’s firing rate $latex r$ with its receptive field, producing an estimate $latex \hat{I}(x,y) = \sum_{(x_c,y_c)} r(x_c,y_c)k(x-x_c,y-y_c)$. This decoding procedure is nearly optimal under the fewest assumptions on the input source distribution, as is shown in our appendix below. A more complex procedure which is allowed perfect knowledge of the distribution’s covariance and global firing rate information can decode with much lower distortion, though its estimate $latex \hat{I}(x,y)$ is difficult to express analytically and is instead the solution to a convex optimization problem. It’s derivation is tied closely to the minimization of the Mahalanobis distance $latex d_M(\textbf{x};\mu, \Sigma)^2 = (\textbf{x}-\mu)^TW(\textbf{x}-\mu)$, and is also shown more explicitly in the appendix below. An example compression and decompression is shown in Fig. 7.
#### Are we even measuring this right?
Commonly used image fidelity metrics, such as mean-squared error (MSE), are simple to calculate and have clear physical meaning, however these metrics often do not reflect perceived visual quality. Therefore, modern research has been done on fidelity metrics that incorporate high-level properties of the human visual system. One such metric is known as the structural similarity (SSIM) metric. For the scope of this blog post it is enough knowing that SSIM incorporates image properties such as averages, variances, co-variances, and dynamic range in order to better gauge perceived similarity between images.
## Channel Capacity under Poisson Noise
After compression, how well can this information be transmitted?
Here we outline some key results from Wyner [1][2]. The Poisson channel takes as input a continuous waveform input, $latex \lambda(t)$, with domain $latex 0\leq t < \infty$. We assume the following constraints: $latex 0\leq \lambda(t) \leq A$ and $latex (1/T)\int_0^T \lambda(t)dt \leq \sigma A$ where $latex 0<\sigma \leq 1$. Now, the channel output is a Poisson process with intensity $latex \lambda(t)+\lambda_0$ where $latex \lambda_0$ represents additive Poisson noise. $latex \nu (t)$ defines a Poisson counting function which gives the number of spikes received before time $latex t$. More specifically, $latex \nu (t)$ is a staircase function where each step represents detection of a spike at the output. We then have that $latex Pr\{\nu (t+\tau)-\nu (t)=j\}=\frac{\exp^{-\Lambda}\Lambda^j}{j!}$ where $latex \Lambda=\int_t^{t+\tau}(\lambda(t’)+\lambda_0)dt’$.
We now describe a general Poisson channel code. The input codebook consists of a set of M waveforms $latex \lambda_m(t)$ where $latex 0\leq t \leq T$ which satisfy $latex 0\leq \lambda_m(t) \leq A$ and $latex (1/T)\int_0^T \lambda_m(t)dt \leq \sigma A$. The output alphabet consists of the set $latex S(T)$ where $latex S(T)$ is the set of all $latex \nu (t)$ for $latex 0\leq t \leq T$. The channel decoder is the mapping: $latex S(T) \rightarrow \{1,2,…,M\}$. The output of the channel takes on a value $latex \nu_0^T$ from the output alphabet $latex S(T)$. $latex \nu_0^T$ is a staircase function with domain $latex [0,T)$. We then have an error probability given by: $latex P_e = \frac{1}{M}\sum _{m=1}^M Pr\{D(\nu_0^T)\neq m\}$. The rate, $latex R$ of the code in bits per second is therefore $latex \frac{\log M}{T}$.
We say the $latex R$ is achievable if for all $latex \epsilon>0$ there exists a code such that $latex M\geq 2^{RT}$ and $latex P_e \leq \epsilon$. The channel capacity is then the supremum of all such achievable rates. For any set of codes satisfying these constraints, we have the following capacity:
$latex C=A[q^*(1+s)\log (1+s)+(1-q^*)s \log s- (q^* + s) \log (q^* + s)]$
Here, $latex s=\lambda_0/A$, $latex q^*=\min{\sigma,q_0(s)}$, and $latex q_0(s)=\frac{(1+s)^{1+s}}{s^s}-s$.
#### But what if we had more noise?
Interestingly, work by Frey [5] has shown that in the presence of random, nondeterministic background noise $latex \lambda_0$, feedback can improve channel capacity (under mild assumptions on the distribution of the noise). In particular the This is surprising because the Poisson channel is still memoryless, and in general, memoryless channels cannot increase their capacity with the addition of feedback. Nevertheless, biologically speaking, this is a more realistic setting, given that seemingly stochastic vesicle release appears to inject random noise into the propagation of spikes within the nervous system. Thus this is a particularly encouraging result from information theory as recurrent connections and backpropagating dendritic action potentials are a common feature within biological neural networks, and may jointly be facilitating this feedback mechanism to counter the intrinsic background noise.
## A Mathematical Aside
For the more mathematically inclined:
#### Optimal decoders
We first set up the problem. Each $latex S\times S$ image frame $latex I$ is vectorized into $latex S^2$-length vectors $latex f$. The receptive field of neuron $latex i$ as a $latex w\times w$ patch, when centered at its location $latex (x_c,y_c)$ within the $latex S\times S$ frame, is also vectorized into an $latex S^2$-length vector $latex k_i$. In this framework, we can compute the rates of all $latex N$ neurons by arranging $latex k_1 \dots k_N$ receptive field vectors into a matrix $latex C \in \mathbb{R}^{N\times S^2}$ such that the rates $latex r$ are computed as $latex r = Cf$.
This operation necessarily loses information as the solution to $latex r=C\hat{f}$ is underdetermined. A common technique in probabilistic inference is determining the maximum likelihood estimate (MLE). Since our source is a Gaussian source with $latex \mu =0$ and covariance $latex \Sigma_l$ (where the subscript $latex l$ designates its parametrization by the length constant $latex l$), we write out our log-likelihood function $latex L(\mu,\Sigma)$:
$latex L(0,\Sigma_l) = -\frac{1}{2}\log(|\Sigma_l|) – \frac{1}{2}x^T\Sigma_l^{-1}x – \frac{S^2}{2}\log(2\pi)$
Within this expression we recognize the square of the Mahalanobis distance $latex d$, which can be considered an expression of the distance between a point $latex x$ to the distribution $latex \mathcal{N}(0, \Sigma_l)$. Rewriting this expression along with a constant term $latex c = \frac{1}{2}\log(|\Sigma_l|) + \frac{S^2}{2}\log(2\pi)$, we get:
$latex L(\mu,\Sigma_l) = -\frac{1}{2}d^2 – c$
With this, we see we can maximize the log-likelihood by minimizing the Mahalanobis distance $latex d(\hat{f};0,\Sigma_l) = (\hat{f}-\mu)^T\Sigma_l^{-1}(\hat{f}-\mu)$ subject to the constraint $latex C\hat{f} = r$. Moreover, this is a simple convex optimization problem, since the matrix $latex \Sigma_l^{-1}$ is a positive definite precision matrix.
$latex min_f \quad f^T\Sigma_l^{-1}f$
$latex s.t. \quad Cf = r$
This decoder is optimal under the assumption that the source is a stationary Gaussian distribution with $latex \mu=0$, covariance $latex \Sigma_l$, and the decoder has perfect knowledge of the distribution. We also consider the case where the decoder makes no assumptions on a finite bandwidth of the source, i.e., $latex \Sigma_l \rightarrow \mathbb{I}$. The resulting objective corresponds to a least-norm problem on $latex f$:
$latex min_f \quad f^T f$
$latex s.t. \quad Cf = r$
This optimization problem has an explicit solution $latex f = C^T(CC^T)^{-1}r$, which we now compare to our naive deconvolution solution $latex \hat{I}(x,y) = \sum_{(x_c,y_c)} r(x_c,y_c) k(x-x_c,y-y_c)$. When vectorized into the format of this framework, this is equivalent to $latex f = C^Tr$. Now the comparison becomes more clear, since the least-norm solution and the deconvolution solution are equivalent if $latex CC^T = \mathbb{I}$. This is the case when each neuron’s receptive field $latex k_i$ is unit length and pairwise orthogonal ($latex k_i^Tk_j = 0$), or in other words, have no overlap over the image.
Important to note is that this decoder is optimal in the noiseless channel case. In the case where the receiver’s estimate $latex \hat{r} \neq r$, the decompression process minimizes the objective in the nullspace of $latex Cf = \hat{r}$ and fails to find the appropriately reconstruction. We thus modify our constraints to respect our noisy channel by assuming the true rate $latex r$ falls in the set $latex [\hat{r}-\sigma_{\hat{r}}, \hat{r}+\sigma_{\hat{r}} ]$. The new stochastic optimization problem is reformulated below, where the channel noise is known to take a Poisson distribution.
$latex min_f \quad f^T\Sigma_l^{-1}f$
$latex s.t. \quad Cf \leq \hat{r}+\sqrt{\hat{r}}, Cf \geq \hat{r}-\sqrt{\hat{r}}$
#### Channel capacity lower bound
We now derive an expression for the lower bound of the Poisson channel capacity [1][2]. The main idea of this proof is that we can analyze $latex max_{p(x)} I(Y;X)$ when $latex X$ is constrained to a subset all possible inputs to the Poisson channel. Clearly, if this constraint is lifted, the same distribution over the inputs can be achieved by setting $latex p(x)=0$ for any $latex x$ which is not a member of the constrained subset. Therefore, $latex max_{p(x)} I(Y;X)$ when $latex X$ is constrained to a subset of inputs to the channel is a lower bound on the capacity of the Poisson channel.
Specifically, we impose the constraint that $latex \lambda(t)$ is piecewise constant so that it takes one of two values, 0 or A, over each interval $latex ((n-1)\Delta,n\Delta]$. We then define the discrete signal $latex x_n$ such that $latex x_n = 1$ if $latex \lambda(t)=A$ in the interval $latex ((n-1)\Delta,n\Delta]$. Otherwise, $latex x_n=0$. Additionally, we assume that the detector sees only the values $latex \nu(n\Delta)$. Thus, it sees the number of spikes received in the interval $latex ((n-1)\Delta,n\Delta]$ which we define as $latex y_n=\nu(n\Delta)-\nu((n-1)\Delta)$. Although there is a finite probability that $latex \nu(n\Delta)-\nu((n-1)\Delta)\geq 2$, we take $latex y_n = 0$ for all cases where $latex \nu(n\Delta)-\nu((n-1)\Delta)\geq\neq 1$. This is a reasonable approximation for small enough $latex \Delta$. Thus, $latex y_n$ is binary valued. We have now reduced the channel to a binary input, binary output channel characterized by $latex P(y_n|x_n)$. Any capacity achieved by this channel will be a lower bound on the capacity achieved by the general Poisson channel as we are dealing with a restriction of the input and output spaces. However, it will turn out that this bound is tight.
From the Poisson distribution, we have that $latex P(y_n=1|x_n=0)=\lambda_0\Delta\exp^{-\lambda_0\Delta}$. We have that $latex \lambda_0=s A$. Therefore, $latex P(y_n=1|x_n=0)=s A\Delta \exp^{-s A\Delta}$. Likewise, $latex P(y_n=1|x_n=1)=(1+s)A\Delta \exp^{-(1+s)A\Delta}$. Now, the average power constraint takes the form $latex \frac{1}{N}\sum_{n=1}^N x_{mn} \leq \sigma$. Where $latex m$ indexes the codeword. Therefore, the lower bound on capacity is given by $latex \max I(x_n;y_n)/\Delta$ where the maximum is taken over all distributions of $latex x_n$. Clearly, $latex p(x_n)$ must satisfy $latex E(x_n)\leq \sigma$. Now we define $latex q=Pr\{x_n=1\}$, $latex a=sA\Delta\exp^{-sA\Delta}$, and $latex b=(1+s)A\Delta\exp^{-(1+s)A\Delta}$. We then have that $latex I(x_n;y_n)=h(qb+(1-q)a)-qh(b)-(1-q)h(a)$ where $latex h(.)$ is the binary entropy. We then have $latex \Delta C\geq \max_{0\leq q \leq \sigma} I(x_n;y_n)$. After making a few simplifying approximations, we have an expression which is convex in $latex q$. We then maximize with respect to $latex q$ subject to the constraint $latex q\leq \sigma$. The mutual information is maximized when $latex q=\frac{(1+s)^{s+1}}{s^s}-s$. We define $latex q_0(s)=\frac{(1+s)^{s+1}}{s^s}-s$. We then have that the capacity is lower bounded by $latex C=A[q^*(1+s)\log (1+s)+(1-q^*)s \log s- (q^* + s) \log (q^* + s)]$ where $latex q^*=\min{\sigma,q_0(s)}$.
Taking our channel to consist of $latex m$ neurons, each modeled as an independent Poisson channel, we have the total capacity $latex m A[q^*(1+s)\log (1+s)+(1-q^*)s \log s- (q^* + s) \log (q^* + s)]$.
## Outreach Activity
For the outreach activity at Nixon Elementary school, our group decided to create a board game called A Polar (codes) Expedition. The purpose of this outreach activity was to help the kids walk away with an ability to XOR, and, furthermore, use that ability to decode a simple 4-bit polar code.
The overarching theme of the board game is that the player is trapped in a polar storm and needs to use polar codes to figure out where the rescue team is coming. The outreach activity was a huge success (see some pictures from the event below). It was especially amazing to see kids even smaller than we anticipated being the target age for our game (3rd and 4th graders) pick up XOR-ing really fast and walk through the game. Additionally, it was inspiring to see the kids so excited after solving the polar code and then realizing this decoding scheme is an actual one used in modern day. Overall, our group believes the outreach component of the course was truly a wholesome experience and we all came out of it hoping to continue taking part in STEM outreach in some capacity throughout the rest of our PhDs. It was special to see the genuine curiosity that the kids had while learning about polar codes through our game and end the end of the day it really reminded us why we are here to do a PhD in the first place — it boils down to the simple fact that we find these topics to be cool and interesting and we derive happiness from learning more and more about EE.
The transcript for our board game can be found at, please take a look if you would like to read the story arcs that the kids got to take on their polar expedition: https://docs.google.com/document/d/1gQk1USQOqbW8xdxMENY7QQ4kr4LFIazgFS149aJIhuQ/edit?usp=sharing
## References
[1] A.D. Wyner, “Capacity and error exponent for the direct detection photon channel–Part 1,” in IEEE Trans. Inform. Theory, vol. 34, pp. 1449-1461, 1988.
[2] A.D. Wyner, “Capacity and error exponent for the direct detection photon channel–Part 2,” in IEEE Trans. Inform. Theory, vol. 34, pp. 1462-1471, 1988.
[3] H.K. Hartline, “The response of single optic nerve fibers to illumination of the retina,” in American Journal of Physiology, vol. 121, no. 2, pp. 400-415, 1938.
[4] Kolb H, Fernandez E, Nelson R, editors. Webvision: The Organization of the Retina and Visual System [Internet]. Salt Lake City (UT): University of Utah Health Sciences Center; 1995-.
[5] M. Frey, “Information capacity of the Poisson channel,” in IEEE Trans. Inform. Theory, vol. 37, no. 2, pp. 244-256, 1991.
[6] Palanker D – Own work, CC BY-SA 4.0 https://commons.wikimedia.org/w/index.php?curid=49156032.
[7] R. Zamir, Y. Kochman, U. Erez, “Achieving the Gaussian rate-distortion function by prediction,” in IEEE Trans. Inform. Theory, vol. 54, no. 7, pp. 3354-3364, 2008.
[8] Z. Wang et al., “Image Quality Assessment: From Error Visibility to Structural Similarity,” in IEEE Transactions on Image Processing, vol. 13, no. 4, pp. 600-612, 2004.
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https://asmedigitalcollection.asme.org/energyresources/article-abstract/107/1/18/406039/In-Line-Forces-on-Vertical-Cylinders-in-Deepwater?redirectedFrom=fulltext | Laboratory measurements of the total in-line forces on a fixed vertical 2-in-dia cylinder in deep-water regular and random waves are given and compared with predictions from the Morison equation. Results show, for regular waves with heights ranging from 2 to 22 in. and frequencies ranging from 0.4 to 0.9 Hz that the Morison equation, with Stokes wave theory and constant drag and inertia coefficients of 1.2 and 1.8, respectively, provides good agreement with the measured maximum wave forces. The force variation over the entire wave cycle is also well represented. The linearized Morison equation, with linear wave theory and the same coefficients likewise provides close agreement with the measured rms wave forces for irregular random waves having approximate Bretschneider spectra and significant wave heights from 5 to 14 in. The success of the constant-coefficient approximation is attributed to a decreased dependence of the coefficients on dimensionless flow parameters as a result of the circular particle motions and large kinematic gradients of the deep-water waves.
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http://math.stackexchange.com/users/30518/viet-hoang-quoc?tab=activity&sort=all | Viet Hoang Quoc
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Sep 6 answered What should my $u$ be for substitution? Jun 5 comment Inequality with exponents The differentiation step does not look right to me, we have either $(a^x)' = \ln (a) \times a^x$ or $(x^a)'=a \times x^{a-1}$. Jun 4 answered For what values of $a$ the equation $(ax)^2-x^4=e^{|x|}$ have no solution Jun 4 comment Inequality with exponents This happens for all positive real number and induction clearly does not work nicely here. Jun 4 comment Inequality with exponents Can you elaborate a bit more because I have used AM-GM but the difference in the power could be a major issue? Jun 4 awarded Commentator Jun 4 comment For what values of $a$ the equation $(ax)^2-x^4=e^{|x|}$ have no solution What do you mean by $|a| < \sqrt{2}$?. Do we have to prove that when $|a|<\sqrt{2}$, the equation has no solution? Jun 4 comment For what values of $a$ the equation $(ax)^2-x^4=e^{|x|}$ have no solution The $\ln$ operation is not correct because it is applied for division or multiplication. Jun 4 asked Inequality with exponents Dec 6 awarded Teacher Jul 21 answered Percentages - Find Maximum value. May 20 comment Differentiablity and Taylor polynomials @Thomas So, what do I need to prove then. Cheers May 20 asked Differentiablity and Taylor polynomials May 5 comment Continuity leads to constant function (Assignment question) Here is another similar type of question proving the function is constant using the continuity definition, Let $f: [-1,1] \to \mathbb{R}$ be a function with $f(x)=f(x^2)$ for all $x \in (-1,1)$. Suppose that $f$ is continuous at 0. Prove that $f$ is a constant. May 5 comment Continuity leads to constant function (Assignment question) Thanks all the the great effort made, let's me just recap what we have had so far: $f$ is continuous iff for all $\epsilon >0$, $x \in B_{\delta}(x_0)$ then $f(x) \in B_{\epsilon}(f(x_0))$ May 5 comment Continuity leads to constant function (Assignment question) To my understanding, if we can pick a rational point $x_1$ satisfying $|x_1-x_0|< \delta$ such that $|f(x_1)-f(x_0)| \ge \epsilon$ then this will give a contradiction. May 5 comment Continuity leads to constant function (Assignment question) @Benjamin Lim : This is the first course in Analysis so we have not learnt anything such as Topological spaces and other things, just simply the definition and way to go. Thanks. May 5 awarded Student May 5 comment Continuity leads to constant function (Assignment question) The formal one as follows For all $\epsilon >0$, there exists $\delta >0$ such that for all $x \in \mathbb{R}$ satisfying $|x-x_0| < \delta$, one has $|f(x)-f(x_0)| < \epsilon$. May 5 comment Continuity leads to constant function (Assignment question) Yes, it was my initial idea of using Density Theorem. However, I do not see the connection between that with the function. | {"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": 0.9602994322776794, "perplexity": 444.6116684933879}, "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-2016-07/segments/1454702039825.90/warc/CC-MAIN-20160205195359-00256-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/43581/are-schematic-fixed-points-of-a-cohen-macaulay-scheme-cohen-macaulay/43588 | # Are schematic fixed-points of a Cohen-Macaulay scheme Cohen-Macaulay?
I'm not sure how long this iterative questions can go on, but let me try again. Let's say $X$ is a Cohen-Macaulay scheme with an action of $\mathbb{G}_m$ (i.e. if $X$ is affine, a grading on the coordinate ring). Are the schematic fixed points $X^{\mathbb{G}_m}$ of $X$ Cohen-Macaulay?
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I am not sure what the usual business is, but it is not true that if $\mathbb G_{\rm m}$ acts on a variety $X$ with a fixed point $p$, then this induces an action of $\mathbb G_{\rm m}$ on $\mathop{\rm Spec}\mathcal O_{X,p}$. – Angelo Oct 25 '10 at 21:39
Is this an issue of not being able to find a $\mathbb{G}_m$-invariant affine open containing $p$? – Ben Webster Oct 25 '10 at 22:23
Those invariant affines are not cofinal in all neighborhoods. Algebraically, the localization of C[x] at the ideal (x) does not admit a grading...does it? – David Treumann Oct 25 '10 at 22:34
Ah, right. That was complete nonsense. Removed. I think you probably you can reduce to the graded local case, but let me not worry about that. – Ben Webster Oct 25 '10 at 23:03
Let $T^1$ act on ${\mathbb A}^3$ with weights $0,1,1$. Then on the ${\mathbb P}^2$, the fixed-point set is not equidimensional. So you can only hope to have a local statement. – Allen Knutson Oct 26 '10 at 0:39
Here is a counterexample. Consider the action of $\mathbb G_{\rm m}$ on $\mathbb A^4$ defined by $t \cdot(x,y,z,w) = (x, y, tz, t^{-1}w)$, and let $X$ be the invariant closed subscheme with ideal $(xy, y^2 + zw)$; this is a complete intersection, hence it is Cohen-Macaulay. The fixed point subscheme is obtained by intersecting with the fixed point subscheme in $\mathbb A^4$, which is given by $z = w = 0$; hence it is the subscheme of $\mathbb A^2$ given by $xy = y^2 = 0$, which is of course the canonical example of a non Cohen-Macaulay scheme.
Developing this idea a little, one can show that any kind of horrible singularity can appear in the fixed point subscheme of a $\mathbb G_{\rm m}$-action on a complete intersection variety.
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Murphy's law strikes again! – Ben Webster Oct 27 '10 at 6:50
Edit: the following does not answer Ben's question. It gives an example of the subring fixed by $G_m$ being not CM, while the question asked about the subscheme of fixed points, see the comments for more details.
Let $R$ be the (homogenous) cone of a curve $C$ of genus $g>0$, for example $R=\mathbb C[x,y,z]/(x^3+y^3+z^3)$. Let $S=R[u,v]$, $X=\text{Spec}(S)$ and $G_m$ acts by
$a.(x,y,z,u,v) = (ax,ay,az,a^{-1}u, a^{-1}v)$.
Then $A= S^{G_m}$ would be a homogenous coordinate ring for $Y= C\times \mathbb P^1$, so it is not Cohen-Macaulay (if $A$ is CM, it would mean that $H^1(Y,\mathcal O_Y)=0$, impossible, see here for an explanation).
(I learned this idea from Hochster, let me try to find a reference)
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Hailong- this is the categorical quotient, not schematic fixed points. The schematic fixed points here are a single point with reduced structure and thus Cohen-Macaulay. – Ben Webster Oct 25 '10 at 23:12
Hmm, sorry, I always thought of this notation as the invariants. What do you mean by schematic fixed pts? Are they just literally the pts of $X$ fixed by the group action? – Hailong Dao Oct 26 '10 at 0:10
@Hailong: you applied the fixed-point functor to the algebra rather than to the space. That is, the notation itself has a consistent meaning. (I'm not surely how to reasonable define schematic fixed points; in characteristic zero, with a connected group, one could look where the generating vector fields vanish.) – Allen Knutson Oct 26 '10 at 0:43
Dear Hailong: No ad hoc constructions/definitions are required. For any scheme $X$ over a ring $k$ and action on $X$ by a $k$-gp scheme $G$, define functor $X^G$ on $k$-algebras as follows: for $k$-algebra $R$, $X^G(R)$ is set of $x \in X(R)$ fixed by the $G_R$-action on $X_R$ (i.e., for any $R$-algebra $R'$, $x$ viewed in $X(R')$ is $G(R')$-invariant). Is this represented by a closed subscheme of $X$? Yes, provided $X$ is locally of finite type and separated over $k$ and $G$ is affine and fppf over $k$ with connected fibers. (See Prop. A.8.10(1)ff. in "Pseudo-reductive groups" for details.) – BCnrd Oct 26 '10 at 1:28
@ Allen and BCnrd: thank you! I will leave my answer, so people might benefit from your explanations and avoid my mistakes! – Hailong Dao Oct 26 '10 at 2:07 | {"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": 0.9647451043128967, "perplexity": 342.6973121349545}, "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-2015-06/segments/1422122086930.99/warc/CC-MAIN-20150124175446-00190-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://planetmath.org/KroneckersJugendtraum | # Kronecker’s Jugendtraum
Kronecker’s Jugendtraum (Jugendtraum is German for “youthful dream”) describes a central problem in class field theory, to explicitly describe the abelian extensions of an arbitrary number field $K$ in of values of transcendental functions.
Class field theory gives a solution to this problem in the case where $K=\mathbb{Q}$, the field of rational numbers. Specifically, the Kronecker-Weber theorem gives that any number field sits inside one of the cyclotomic fields $\mathbb{Q}(\zeta_{n})$ for some $n$. Refining this only slightly gives that we can explicitly generate all abelian extensions of $\mathbb{Q}$ by adjoining values of the transcendental function $e^{2\pi iz}$ for certain points $z\in\mathbb{Q}/\mathbb{Z}$.
A slightly more complicated example is when $K$ is a quadratic imaginary extension of $\mathbb{Q}$, in which case Kronecker’s Jugendtraum has been solved by the theory of “complex multiplication” (see CM-field). The specific transcendental functions which generate all these abelian extensions are the $j$-function (as in elliptic curves) and Weber’s $w$-function.
Though there are partial results in the cases of CM-fields or real quadratic fields, the problem is largely still open (http://planetmath.org/OpenQuestion), and earned great prestige by being included as Hilbert’s twelfth problem.
Title Kronecker’s Jugendtraum KroneckersJugendtraum 2013-03-22 15:01:08 2013-03-22 15:01:08 mathcam (2727) mathcam (2727) 7 mathcam (2727) Definition msc 11R37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 11, "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": 0.9019852876663208, "perplexity": 673.293524684907}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039743968.63/warc/CC-MAIN-20181118052443-20181118074443-00164.warc.gz"} |
https://www.physicsforums.com/threads/sample-topologies.272191/ | # Sample topologies
1. ### cothranaimeel
2
Think of a very simple topological space in the following examples:
a) (X,t) is a topological space, G is open, and G Does not equal Int(Clo(G))?
b) (X,t) is a topological space, F is closed, and F Does not equal Clo(Int (F))?
thanks for helping
2,020 | {"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": 0.8231528997421265, "perplexity": 2373.6261560329444}, "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/1448398454160.51/warc/CC-MAIN-20151124205414-00221-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-12-data-analysis-and-probability-12-6-permutations-and-combinations-standardized-test-prep-page-756/67 | ## Algebra 1
You have 7.3 $\times$ $10^{-2}$ and you have to find the standard form. Since the exponent is negative, we move the decimal point to the left. The exponent is -2 so we move the decimal point two places to the left to find the standard form.So 7.3 $\times$ $10^{-2}$=0.073 | {"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": 0.9062718749046326, "perplexity": 230.3141889133888}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540486979.4/warc/CC-MAIN-20191206073120-20191206101120-00551.warc.gz"} |
https://community.ivanti.com/thread/5436 | 5 Replies Latest reply on Apr 13, 2009 5:36 PM by danpixley
# Is it possible to see batch files run?
Good Day,
I was wondering if anybody knew how to make a batch file run visually?
I'm Provisioning using the Provisioning Templates:
My "setup" based drivers are installed via batch file when Windows XP first Launches using Run Once.
This works well but other then the provisioning window this leave little indication that anything is happening.
As this tasks takes about 10-15 minutes to run I have had people turn off machines thinking the process got hungup.
I added Echo command in the file at each stage but they only show in the provisioning window when the batch finishes it's run.
BONUS question: Rather then seeing the batch file run would it be possible to have the batch file send updates to the provisioning agent (which would show up in the provisioning window)?
• ###### 1. Re: Is it possible to see batch files run?
have you tried to make the batch file called by the provisioning agent spawn a separate batch file? if provisioning spawns a 'startrun.bat' which then runs cmd /c yourbatchfile.bat I think it'll defeat the UI supression.
• ###### 2. Re: Is it possible to see batch files run?
have you tried to make the batch file called by the provisioning agent spawn a separate batch file? if provisioning spawns a 'startrun.bat' which then runs cmd /c yourbatchfile.bat I think it'll defeat the UI supression.
Sorry about the delay in my response,
I had tried to call it from another batch file but not in it's own shell, I tried this as it seemed promising but no dice
=o(
• ###### 3. Re: Is it possible to see batch files run?
You could script a very easy Autoitscript/vbscript that pops up a progress and kill it at the end.
1 of 1 people found this helpful
• ###### 4. Re: Is it possible to see batch files run?
zman wrote:
You could script a very easy Autoitscript/vbscript that pops up a progress and kill it at the end.
Good Call,
While this does not give an exact statues of what's going on it certainly let's the end user know something is happening.
Which solves the root of my problem =o)
• ###### 5. Re: Is it possible to see batch files run?
I found a way to run Echo commands from a batch file in OSD. I am not sure if this will work in Provisioning or not. For example, in my hardware independent imaging OSD script, I have the following line:
REMEXEC414=cmd /c s:\tools\copydrivers.bat
Echo is supressed in OSD the same way it is in Provisioning, so I got around it by using the following commands for the copydrivers.bat file:
@ECHO OFF
cmd /C "ECHO Please wait for the drivers to copy... && s:\tools\copydrivers.exe /c"
Calling another command window from the OSD command is another way to escape the supression of Echo. | {"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": 0.8844032883644104, "perplexity": 4428.944333064618}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267161638.66/warc/CC-MAIN-20180925123211-20180925143611-00335.warc.gz"} |
https://stacks.math.columbia.edu/tag/02S8 | Lemma 42.23.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be a scheme locally of finite type over $S$. The categories $\textit{Coh}_{\leq k}(X)$ are Serre subcategories of the abelian category $\textit{Coh}(X)$.
Proof. The definition of a Serre subcategory is Homology, Definition 12.10.1. The proof of the lemma is straightforward and omitted. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | {"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": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9952202439308167, "perplexity": 336.04914234258297}, "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-2022-33/segments/1659882571993.68/warc/CC-MAIN-20220814022847-20220814052847-00302.warc.gz"} |
http://math.andrej.com/2009/01/23/on-the-failure-of-fixed-point-theorems-for-chain-complete-lattices-in-the-effective-topos/ | # On the Failure of Fixed-point Theorems for Chain-complete Lattices in the Effective Topos
Abstract: In the effective topos there exists a chain-complete distributive lattice with a monotone and progressive endomap which does not have a fixed point. Consequently, the Bourbaki-Witt theorem and Tarski’s fixed-point theorem for chain-complete lattices do not have constructive (topos-valid) proofs.
p.l.lumsdaine
Very nice paper!
I think there may be a typo in the first paragraph of Sect. 2, by the way: you define that X is discrete if the diagonal map X --> X^{\nabla 2} is constant. But if internally this is the condition that every map fron \nabla 2 to X is constant, i.e. in the image of the diagonal map, should the original condition be that the diagonal map is epimorphic?
Thank you for spotting that one. It should say that the diagonal map is an isomorphism, or equivalently epimorphism, as you suggest.
Write your comment using Markdown. Use $⋯$ for inline and $$⋯$$ for display LaTeX formulas, and <pre>⋯</pre> for display code. Your E-mail address is only used to compute your Gravatar and is not stored anywhere. Comments are moderated through pull requests. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9624143838882446, "perplexity": 989.6420032578351}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496671245.92/warc/CC-MAIN-20191122065327-20191122093327-00462.warc.gz"} |
http://nag.com/numeric/fl/nagdoc_fl24/html/D03/d03pjf.html | D03 Chapter Contents
D03 Chapter Introduction
NAG Library Manual
# NAG Library Routine DocumentD03PJF/D03PJA
Note: before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.
## 1 Purpose
D03PJF/D03PJA integrates a system of linear or nonlinear parabolic partial differential equations (PDEs), in one space variable with scope for coupled ordinary differential equations (ODEs). The spatial discretization is performed using a Chebyshev ${C}^{0}$ collocation method, and the method of lines is employed to reduce the PDEs to a system of ODEs. The resulting system is solved using a backward differentiation formula (BDF) method or a Theta method (switching between Newton's method and functional iteration).
D03PJA is a version of D03PJF that has additional parameters in order to make it safe for use in multithreaded applications (see Section 5).
## 2 Specification
### 2.1 Specification for D03PJF
SUBROUTINE D03PJF ( NPDE, M, TS, TOUT, PDEDEF, BNDARY, U, NBKPTS, XBKPTS, NPOLY, NPTS, X, NCODE, ODEDEF, NXI, XI, NEQN, UVINIT, RTOL, ATOL, ITOL, NORM, LAOPT, ALGOPT, RSAVE, LRSAVE, ISAVE, LISAVE, ITASK, ITRACE, IND, IFAIL)
INTEGER NPDE, M, NBKPTS, NPOLY, NPTS, NCODE, NXI, NEQN, ITOL, LRSAVE, ISAVE(LISAVE), LISAVE, ITASK, ITRACE, IND, IFAIL REAL (KIND=nag_wp) TS, TOUT, U(NEQN), XBKPTS(NBKPTS), X(NPTS), XI(*), RTOL(*), ATOL(*), ALGOPT(30), RSAVE(LRSAVE) CHARACTER(1) NORM, LAOPT EXTERNAL PDEDEF, BNDARY, ODEDEF, UVINIT
### 2.2 Specification for D03PJA
SUBROUTINE D03PJA ( NPDE, M, TS, TOUT, PDEDEF, BNDARY, U, NBKPTS, XBKPTS, NPOLY, NPTS, X, NCODE, ODEDEF, NXI, XI, NEQN, UVINIT, RTOL, ATOL, ITOL, NORM, LAOPT, ALGOPT, RSAVE, LRSAVE, ISAVE, LISAVE, ITASK, ITRACE, IND, IUSER, RUSER, CWSAV, LWSAV, IWSAV, RWSAV, IFAIL)
INTEGER NPDE, M, NBKPTS, NPOLY, NPTS, NCODE, NXI, NEQN, ITOL, LRSAVE, ISAVE(LISAVE), LISAVE, ITASK, ITRACE, IND, IUSER(*), IWSAV(505), IFAIL REAL (KIND=nag_wp) TS, TOUT, U(NEQN), XBKPTS(NBKPTS), X(NPTS), XI(*), RTOL(*), ATOL(*), ALGOPT(30), RSAVE(LRSAVE), RUSER(*), RWSAV(1100) LOGICAL LWSAV(100) CHARACTER(1) NORM, LAOPT CHARACTER(80) CWSAV(10) EXTERNAL PDEDEF, BNDARY, ODEDEF, UVINIT
## 3 Description
D03PJF/D03PJA integrates the system of parabolic-elliptic equations and coupled ODEs
$∑j=1NPDEPi,j ∂Uj ∂t +Qi=x-m ∂∂x xmRi, i=1,2,…,NPDE, a≤x≤b,t≥t0,$ (1)
$Fit,V,V.,ξ,U*,Ux*,R*,Ut*,Uxt*=0, i=1,2,…,NCODE,$ (2)
where (1) defines the PDE part and (2) generalizes the coupled ODE part of the problem.
In (1), ${P}_{i,j}$ and ${R}_{i}$ depend on $x$, $t$, $U$, ${U}_{x}$, and $V$; ${Q}_{i}$ depends on $x$, $t$, $U$, ${U}_{x}$, $V$ and linearly on $\stackrel{.}{V}$. The vector $U$ is the set of PDE solution values
$U x,t = U 1 x,t ,…, U NPDE x,t T ,$
and the vector ${U}_{x}$ is the partial derivative with respect to $x$. Note that ${P}_{i,j}$, ${Q}_{i}$ and ${R}_{i}$ must not depend on $\frac{\partial U}{\partial t}$. The vector $V$ is the set of ODE solution values
$Vt=V1t,…,VNCODEtT,$
and $\stackrel{.}{V}$ denotes its derivative with respect to time.
In (2), $\xi$ represents a vector of ${n}_{\xi }$ spatial coupling points at which the ODEs are coupled to the PDEs. These points may or may not be equal to some of the PDE spatial mesh points. ${U}^{*}$, ${U}_{x}^{*}$, ${R}^{*}$, ${U}_{t}^{*}$ and ${U}_{xt}^{*}$ are the functions $U$, ${U}_{x}$, $R$, ${U}_{t}$ and ${U}_{xt}$ evaluated at these coupling points. Each ${F}_{i}$ may only depend linearly on time derivatives. Hence the equation (2) may be written more precisely as
$F=G-AV.-B Ut* Uxt* ,$ (3)
where $F={\left[{F}_{1},\dots ,{F}_{{\mathbf{NCODE}}}\right]}^{\mathrm{T}}$, $G$ is a vector of length NCODE, $A$ is an NCODE by NCODE matrix, $B$ is an NCODE by $\left({n}_{\xi }×{\mathbf{NPDE}}\right)$ matrix and the entries in $G$, $A$ and $B$ may depend on $t$, $\xi$, ${U}^{*}$, ${U}_{x}^{*}$ and $V$. In practice you need only supply a vector of information to define the ODEs and not the matrices $A$ and $B$. (See Section 5 for the specification of ODEDEF.)
The integration in time is from ${t}_{0}$ to ${t}_{\mathrm{out}}$, over the space interval $a\le x\le b$, where $a={x}_{1}$ and $b={x}_{{\mathbf{NBKPTS}}}$ are the leftmost and rightmost of a user-defined set of break points ${x}_{1},{x}_{2},\dots ,{x}_{{\mathbf{NBKPTS}}}$. The coordinate system in space is defined by the value of $m$; $m=0$ for Cartesian coordinates, $m=1$ for cylindrical polar coordinates and $m=2$ for spherical polar coordinates.
The PDE system which is defined by the functions ${P}_{i,j}$, ${Q}_{i}$ and ${R}_{i}$ must be specified in PDEDEF.
The initial values of the functions $U\left(x,t\right)$ and $V\left(t\right)$ must be given at $t={t}_{0}$. These values are calculated in UVINIT.
The functions ${R}_{i}$ which may be thought of as fluxes, are also used in the definition of the boundary conditions. The boundary conditions must have the form
$βix,tRix,t,U,Ux,V=γix,t,U,Ux,V,V., i=1,2,…,NPDE,$ (4)
where $x=a$ or $x=b$. The functions ${\gamma }_{i}$ may only depend linearly on $\stackrel{.}{V}$.
The boundary conditions must be specified in BNDARY.
The algebraic-differential equation system which is defined by the functions ${F}_{i}$ must be specified in ODEDEF. You must also specify the coupling points $\xi$ in the array XI. Thus, the problem is subject to the following restrictions:
(i) in (1), ${\stackrel{.}{V}}_{\mathit{j}}\left(t\right)$, for $\mathit{j}=1,2,\dots ,{\mathbf{NCODE}}$, may only appear linearly in the functions ${Q}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$, with a similar restriction for $\gamma$; (ii) ${P}_{\mathit{i},j}$ and the flux ${R}_{\mathit{i}}$ must not depend on any time derivatives; (iii) ${t}_{0}<{t}_{\mathrm{out}}$, so that integration is in the forward direction; (iv) the evaluation of the functions ${P}_{i,j}$, ${Q}_{i}$ and ${R}_{i}$ is done at both the break points and internally selected points for each element in turn, that is ${P}_{i,j}$, ${Q}_{i}$ and ${R}_{i}$ are evaluated twice at each break point. Any discontinuities in these functions must therefore be at one or more of the mesh points; (v) at least one of the functions ${P}_{i,j}$ must be nonzero so that there is a time derivative present in the PDE problem; (vi) if $m>0$ and ${x}_{1}=0.0$, which is the left boundary point, then it must be ensured that the PDE solution is bounded at this point. This can be done either by specifying the solution at $x=0.0$ or by specifying a zero flux there, that is ${\beta }_{i}=1.0$ and ${\gamma }_{i}=0.0$.
The parabolic equations are approximated by a system of ODEs in time for the values of ${U}_{i}$ at the mesh points. This ODE system is obtained by approximating the PDE solution between each pair of break points by a Chebyshev polynomial of degree NPOLY. The interval between each pair of break points is treated by D03PJF/D03PJA as an element, and on this element, a polynomial and its space and time derivatives are made to satisfy the system of PDEs at ${\mathbf{NPOLY}}-1$ spatial points, which are chosen internally by the code and the break points. The user-defined break points and the internally selected points together define the mesh. The smallest value that NPOLY can take is one, in which case, the solution is approximated by piecewise linear polynomials between consecutive break points and the method is similar to an ordinary finite element method.
In total there are $\left({\mathbf{NBKPTS}}-1\right)×{\mathbf{NPOLY}}+1$ mesh points in the spatial direction, and ${\mathbf{NPDE}}×\left(\left({\mathbf{NBKPTS}}-1\right)×{\mathbf{NPOLY}}+1\right)+{\mathbf{NCODE}}$ ODEs in the time direction; one ODE at each break point for each PDE component, ${\mathbf{NPOLY}}-1$ ODEs for each PDE component between each pair of break points, and NCODE coupled ODEs. The system is then integrated forwards in time using a Backward Differentiation Formula (BDF) method or a Theta method.
## 4 References
Berzins M (1990) Developments in the NAG Library software for parabolic equations Scientific Software Systems (eds J C Mason and M G Cox) 59–72 Chapman and Hall
Berzins M and Dew P M (1991) Algorithm 690: Chebyshev polynomial software for elliptic-parabolic systems of PDEs ACM Trans. Math. Software 17 178–206
Berzins M, Dew P M and Furzeland R M (1988) Software tools for time-dependent equations in simulation and optimization of large systems Proc. IMA Conf. Simulation and Optimization (ed A J Osiadcz) 35–50 Clarendon Press, Oxford
Berzins M and Furzeland R M (1992) An adaptive theta method for the solution of stiff and nonstiff differential equations Appl. Numer. Math. 9 1–19
Zaturska N B, Drazin P G and Banks W H H (1988) On the flow of a viscous fluid driven along a channel by a suction at porous walls Fluid Dynamics Research 4
## 5 Parameters
1: NPDE – INTEGERInput
On entry: the number of PDEs to be solved.
Constraint: ${\mathbf{NPDE}}\ge 1$.
2: M – INTEGERInput
On entry: the coordinate system used:
${\mathbf{M}}=0$
Indicates Cartesian coordinates.
${\mathbf{M}}=1$
Indicates cylindrical polar coordinates.
${\mathbf{M}}=2$
Indicates spherical polar coordinates.
Constraint: ${\mathbf{M}}=0$, $1$ or $2$.
3: TS – REAL (KIND=nag_wp)Input/Output
On entry: the initial value of the independent variable $t$.
On exit: the value of $t$ corresponding to the solution values in U. Normally ${\mathbf{TS}}={\mathbf{TOUT}}$.
Constraint: ${\mathbf{TS}}<{\mathbf{TOUT}}$.
4: TOUT – REAL (KIND=nag_wp)Input
On entry: the final value of $t$ to which the integration is to be carried out.
5: PDEDEF – SUBROUTINE, supplied by the user.External Procedure
PDEDEF must compute the functions ${P}_{i,j}$, ${Q}_{i}$ and ${R}_{i}$ which define the system of PDEs. The functions may depend on $x$, $t$, $U$, ${U}_{x}$ and $V$; ${Q}_{i}$ may depend linearly on $\stackrel{.}{V}$. The functions must be evaluated at a set of points.
The specification of PDEDEF for D03PJF is:
SUBROUTINE PDEDEF ( NPDE, T, X, NPTL, U, UX, NCODE, V, VDOT, P, Q, R, IRES)
INTEGER NPDE, NPTL, NCODE, IRES REAL (KIND=nag_wp) T, X(NPTL), U(NPDE,NPTL), UX(NPDE,NPTL), V(NCODE), VDOT(NCODE), P(NPDE,NPDE,NPTL), Q(NPDE,NPTL), R(NPDE,NPTL)
The specification of PDEDEF for D03PJA is:
SUBROUTINE PDEDEF ( NPDE, T, X, NPTL, U, UX, NCODE, V, VDOT, P, Q, R, IRES, IUSER, RUSER)
INTEGER NPDE, NPTL, NCODE, IRES, IUSER(*) REAL (KIND=nag_wp) T, X(NPTL), U(NPDE,NPTL), UX(NPDE,NPTL), V(NCODE), VDOT(NCODE), P(NPDE,NPDE,NPTL), Q(NPDE,NPTL), R(NPDE,NPTL), RUSER(*)
1: NPDE – INTEGERInput
On entry: the number of PDEs in the system.
2: T – REAL (KIND=nag_wp)Input
On entry: the current value of the independent variable $t$.
3: X(NPTL) – REAL (KIND=nag_wp) arrayInput
On entry: contains a set of mesh points at which ${P}_{i,j}$, ${Q}_{i}$ and ${R}_{i}$ are to be evaluated. ${\mathbf{X}}\left(1\right)$ and ${\mathbf{X}}\left({\mathbf{NPTL}}\right)$ contain successive user-supplied break points and the elements of the array will satisfy ${\mathbf{X}}\left(1\right)<{\mathbf{X}}\left(2\right)<\cdots <{\mathbf{X}}\left({\mathbf{NPTL}}\right)$.
4: NPTL – INTEGERInput
On entry: the number of points at which evaluations are required (the value of ${\mathbf{NPOLY}}+1$).
5: U(NPDE,NPTL) – REAL (KIND=nag_wp) arrayInput
On entry: ${\mathbf{U}}\left(\mathit{i},\mathit{j}\right)$ contains the value of the component ${U}_{\mathit{i}}\left(x,t\right)$ where $x={\mathbf{X}}\left(\mathit{j}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{NPTL}}$.
6: UX(NPDE,NPTL) – REAL (KIND=nag_wp) arrayInput
On entry: ${\mathbf{UX}}\left(\mathit{i},\mathit{j}\right)$ contains the value of the component $\frac{\partial {U}_{\mathit{i}}\left(x,t\right)}{\partial x}$ where $x={\mathbf{X}}\left(\mathit{j}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{NPTL}}$.
7: NCODE – INTEGERInput
On entry: the number of coupled ODEs in the system.
8: V(NCODE) – REAL (KIND=nag_wp) arrayInput
On entry: if ${\mathbf{NCODE}}>0$, ${\mathbf{V}}\left(\mathit{i}\right)$ contains the value of the component ${V}_{\mathit{i}}\left(t\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NCODE}}$.
9: VDOT(NCODE) – REAL (KIND=nag_wp) arrayInput
On entry: if ${\mathbf{NCODE}}>0$, ${\mathbf{VDOT}}\left(\mathit{i}\right)$ contains the value of component ${\stackrel{.}{V}}_{\mathit{i}}\left(t\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NCODE}}$.
Note: ${\stackrel{.}{V}}_{\mathit{i}}\left(t\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NCODE}}$, may only appear linearly in ${Q}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,{\mathbf{NPDE}}$.
10: P(NPDE,NPDE,NPTL) – REAL (KIND=nag_wp) arrayOutput
On exit: ${\mathbf{P}}\left(\mathit{i},\mathit{j},\mathit{k}\right)$ must be set to the value of ${P}_{\mathit{i},\mathit{j}}\left(x,t,U,{U}_{x},V\right)$ where $x={\mathbf{X}}\left(\mathit{k}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$, $\mathit{j}=1,2,\dots ,{\mathbf{NPDE}}$ and $\mathit{k}=1,2,\dots ,{\mathbf{NPTL}}$ .
11: Q(NPDE,NPTL) – REAL (KIND=nag_wp) arrayOutput
On exit: ${\mathbf{Q}}\left(\mathit{i},\mathit{j}\right)$ must be set to the value of ${Q}_{\mathit{i}}\left(x,t,U,{U}_{x},V,\stackrel{.}{V}\right)$ where $x={\mathbf{X}}\left(\mathit{j}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{NPTL}}$.
12: R(NPDE,NPTL) – REAL (KIND=nag_wp) arrayOutput
On exit: ${\mathbf{R}}\left(\mathit{i},\mathit{j}\right)$ must be set to the value of ${R}_{\mathit{i}}\left(x,t,U,{U}_{x},V\right)$ where $x={\mathbf{X}}\left(\mathit{i}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{NPTL}}$.
13: IRES – INTEGERInput/Output
On entry: set to $-1\text{ or }1$.
On exit: should usually remain unchanged. However, you may set IRES to force the integration routine to take certain actions as described below:
${\mathbf{IRES}}=2$
Indicates to the integrator that control should be passed back immediately to the calling (sub)routine with the error indicator set to ${\mathbf{IFAIL}}={\mathbf{6}}$.
${\mathbf{IRES}}=3$
Indicates to the integrator that the current time step should be abandoned and a smaller time step used instead. You may wish to set ${\mathbf{IRES}}=3$ when a physically meaningless input or output value has been generated. If you consecutively set ${\mathbf{IRES}}=3$, then D03PJF/D03PJA returns to the calling subroutine with the error indicator set to ${\mathbf{IFAIL}}={\mathbf{4}}$.
Note: the following are additional parameters for specific use with D03PJA. Users of D03PJF therefore need not read the remainder of this description.
14: IUSER($*$) – INTEGER arrayUser Workspace
15: RUSER($*$) – REAL (KIND=nag_wp) arrayUser Workspace
PDEDEF is called with the parameters IUSER and RUSER as supplied to D03PJF/D03PJA. You are free to use the arrays IUSER and RUSER to supply information to PDEDEF as an alternative to using COMMON global variables.
PDEDEF must either be a module subprogram USEd by, or declared as EXTERNAL in, the (sub)program from which D03PJF/D03PJA is called. Parameters denoted as Input must not be changed by this procedure.
6: BNDARY – SUBROUTINE, supplied by the user.External Procedure
BNDARY must compute the functions ${\beta }_{i}$ and ${\gamma }_{i}$ which define the boundary conditions as in equation (4).
The specification of BNDARY for D03PJF is:
SUBROUTINE BNDARY ( NPDE, T, U, UX, NCODE, V, VDOT, IBND, BETA, GAMMA, IRES)
INTEGER NPDE, NCODE, IBND, IRES REAL (KIND=nag_wp) T, U(NPDE), UX(NPDE), V(NCODE), VDOT(NCODE), BETA(NPDE), GAMMA(NPDE)
The specification of BNDARY for D03PJA is:
SUBROUTINE BNDARY ( NPDE, T, U, UX, NCODE, V, VDOT, IBND, BETA, GAMMA, IRES, IUSER, RUSER)
INTEGER NPDE, NCODE, IBND, IRES, IUSER(*) REAL (KIND=nag_wp) T, U(NPDE), UX(NPDE), V(NCODE), VDOT(NCODE), BETA(NPDE), GAMMA(NPDE), RUSER(*)
1: NPDE – INTEGERInput
On entry: the number of PDEs in the system.
2: T – REAL (KIND=nag_wp)Input
On entry: the current value of the independent variable $t$.
3: U(NPDE) – REAL (KIND=nag_wp) arrayInput
On entry: ${\mathbf{U}}\left(\mathit{i}\right)$ contains the value of the component ${U}_{\mathit{i}}\left(x,t\right)$ at the boundary specified by IBND, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$.
4: UX(NPDE) – REAL (KIND=nag_wp) arrayInput
On entry: ${\mathbf{UX}}\left(\mathit{i}\right)$ contains the value of the component $\frac{\partial {U}_{\mathit{i}}\left(x,t\right)}{\partial x}$ at the boundary specified by IBND, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$.
5: NCODE – INTEGERInput
On entry: the number of coupled ODEs in the system.
6: V(NCODE) – REAL (KIND=nag_wp) arrayInput
On entry: if ${\mathbf{NCODE}}>0$, ${\mathbf{V}}\left(\mathit{i}\right)$ contains the value of the component ${V}_{\mathit{i}}\left(t\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NCODE}}$.
7: VDOT(NCODE) – REAL (KIND=nag_wp) arrayInput
On entry: if ${\mathbf{NCODE}}>0$, ${\mathbf{VDOT}}\left(\mathit{i}\right)$ contains the value of component ${\stackrel{.}{V}}_{\mathit{i}}\left(t\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NCODE}}$.
Note: ${\stackrel{.}{V}}_{\mathit{i}}\left(t\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NCODE}}$, may only appear linearly in ${Q}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,{\mathbf{NPDE}}$.
8: IBND – INTEGERInput
On entry: specifies which boundary conditions are to be evaluated.
${\mathbf{IBND}}=0$
BNDARY must set up the coefficients of the left-hand boundary, $x=a$.
${\mathbf{IBND}}\ne 0$
BNDARY must set up the coefficients of the right-hand boundary, $x=b$.
9: BETA(NPDE) – REAL (KIND=nag_wp) arrayOutput
On exit: ${\mathbf{BETA}}\left(\mathit{i}\right)$ must be set to the value of ${\beta }_{\mathit{i}}\left(x,t\right)$ at the boundary specified by IBND, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$.
10: GAMMA(NPDE) – REAL (KIND=nag_wp) arrayOutput
On exit: ${\mathbf{GAMMA}}\left(\mathit{i}\right)$ must be set to the value of ${\gamma }_{\mathit{i}}\left(x,t,U,{U}_{x},V,\stackrel{.}{V}\right)$ at the boundary specified by IBND, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$.
11: IRES – INTEGERInput/Output
On entry: set to $-1\text{ or }1$.
On exit: should usually remain unchanged. However, you may set IRES to force the integration routine to take certain actions as described below:
${\mathbf{IRES}}=2$
Indicates to the integrator that control should be passed back immediately to the calling (sub)routine with the error indicator set to ${\mathbf{IFAIL}}={\mathbf{6}}$.
${\mathbf{IRES}}=3$
Indicates to the integrator that the current time step should be abandoned and a smaller time step used instead. You may wish to set ${\mathbf{IRES}}=3$ when a physically meaningless input or output value has been generated. If you consecutively set ${\mathbf{IRES}}=3$, then D03PJF/D03PJA returns to the calling subroutine with the error indicator set to ${\mathbf{IFAIL}}={\mathbf{4}}$.
Note: the following are additional parameters for specific use with D03PJA. Users of D03PJF therefore need not read the remainder of this description.
12: IUSER($*$) – INTEGER arrayUser Workspace
13: RUSER($*$) – REAL (KIND=nag_wp) arrayUser Workspace
BNDARY is called with the parameters IUSER and RUSER as supplied to D03PJF/D03PJA. You are free to use the arrays IUSER and RUSER to supply information to BNDARY as an alternative to using COMMON global variables.
BNDARY must either be a module subprogram USEd by, or declared as EXTERNAL in, the (sub)program from which D03PJF/D03PJA is called. Parameters denoted as Input must not be changed by this procedure.
7: U(NEQN) – REAL (KIND=nag_wp) arrayInput/Output
On entry: if ${\mathbf{IND}}=1$ the value of U must be unchanged from the previous call.
On exit: the computed solution ${U}_{\mathit{i}}\left({x}_{\mathit{j}},t\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{NPTS}}$, and ${V}_{\mathit{k}}\left(t\right)$, for $\mathit{k}=1,2,\dots ,{\mathbf{NCODE}}$, evaluated at $t={\mathbf{TS}}$, as follows:
• ${\mathbf{U}}\left({\mathbf{NPDE}}×\left(\mathit{j}-1\right)+\mathit{i}\right)$ contain ${U}_{\mathit{i}}\left({x}_{\mathit{j}},t\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{NPTS}}$, and
• ${\mathbf{U}}\left({\mathbf{NPTS}}×{\mathbf{NPDE}}+\mathit{i}\right)$ contain ${V}_{\mathit{i}}\left(t\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NCODE}}$.
8: NBKPTS – INTEGERInput
On entry: the number of break points in the interval $\left[a,b\right]$.
Constraint: ${\mathbf{NBKPTS}}\ge 2$.
9: XBKPTS(NBKPTS) – REAL (KIND=nag_wp) arrayInput
On entry: the values of the break points in the space direction. ${\mathbf{XBKPTS}}\left(1\right)$ must specify the left-hand boundary, $a$, and ${\mathbf{XBKPTS}}\left({\mathbf{NBKPTS}}\right)$ must specify the right-hand boundary, $b$.
Constraint: ${\mathbf{XBKPTS}}\left(1\right)<{\mathbf{XBKPTS}}\left(2\right)<\cdots <{\mathbf{XBKPTS}}\left({\mathbf{NBKPTS}}\right)$.
10: NPOLY – INTEGERInput
On entry: the degree of the Chebyshev polynomial to be used in approximating the PDE solution between each pair of break points.
Constraint: $1\le {\mathbf{NPOLY}}\le 49$.
11: NPTS – INTEGERInput
On entry: the number of mesh points in the interval $\left[a,b\right]$.
Constraint: ${\mathbf{NPTS}}=\left({\mathbf{NBKPTS}}-1\right)×{\mathbf{NPOLY}}+1$.
12: X(NPTS) – REAL (KIND=nag_wp) arrayOutput
On exit: the mesh points chosen by D03PJF/D03PJA in the spatial direction. The values of X will satisfy ${\mathbf{X}}\left(1\right)<{\mathbf{X}}\left(2\right)<\cdots <{\mathbf{X}}\left({\mathbf{NPTS}}\right)$.
13: NCODE – INTEGERInput
On entry: the number of coupled ODE components.
Constraint: ${\mathbf{NCODE}}\ge 0$.
14: ODEDEF – SUBROUTINE, supplied by the NAG Library or the user.External Procedure
ODEDEF must evaluate the functions $F$, which define the system of ODEs, as given in (3).
If you wish to compute the solution of a system of PDEs only (${\mathbf{NCODE}}=0$), ODEDEF must be the dummy routine D03PCK/D53PCK for D03PJF (or D53PCK for D03PJA). D03PCK/D53PCK and D53PCK are included in the NAG Library.
The specification of ODEDEF for D03PJF is:
SUBROUTINE ODEDEF ( NPDE, T, NCODE, V, VDOT, NXI, XI, UCP, UCPX, RCP, UCPT, UCPTX, F, IRES)
INTEGER NPDE, NCODE, NXI, IRES REAL (KIND=nag_wp) T, V(NCODE), VDOT(NCODE), XI(NXI), UCP(NPDE,*), UCPX(NPDE,*), RCP(NPDE,*), UCPT(NPDE,*), UCPTX(NPDE,*), F(NCODE)
The specification of ODEDEF for D03PJA is:
SUBROUTINE ODEDEF ( NPDE, T, NCODE, V, VDOT, NXI, XI, UCP, UCPX, RCP, UCPT, UCPTX, F, IRES, IUSER, RUSER)
INTEGER NPDE, NCODE, NXI, IRES, IUSER(*) REAL (KIND=nag_wp) T, V(NCODE), VDOT(NCODE), XI(NXI), UCP(NPDE,*), UCPX(NPDE,*), RCP(NPDE,*), UCPT(NPDE,*), UCPTX(NPDE,*), F(NCODE), RUSER(*)
1: NPDE – INTEGERInput
On entry: the number of PDEs in the system.
2: T – REAL (KIND=nag_wp)Input
On entry: the current value of the independent variable $t$.
3: NCODE – INTEGERInput
On entry: the number of coupled ODEs in the system.
4: V(NCODE) – REAL (KIND=nag_wp) arrayInput
On entry: if ${\mathbf{NCODE}}>0$, ${\mathbf{V}}\left(\mathit{i}\right)$ contains the value of the component ${V}_{\mathit{i}}\left(t\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NCODE}}$.
5: VDOT(NCODE) – REAL (KIND=nag_wp) arrayInput
On entry: if ${\mathbf{NCODE}}>0$, ${\mathbf{VDOT}}\left(\mathit{i}\right)$ contains the value of component ${\stackrel{.}{V}}_{\mathit{i}}\left(t\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NCODE}}$.
6: NXI – INTEGERInput
On entry: the number of ODE/PDE coupling points.
7: XI(NXI) – REAL (KIND=nag_wp) arrayInput
On entry: if ${\mathbf{NXI}}>0$, ${\mathbf{XI}}\left(\mathit{i}\right)$ contains the ODE/PDE coupling points, ${\xi }_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,{\mathbf{NXI}}$.
8: UCP(NPDE,$*$) – REAL (KIND=nag_wp) arrayInput
On entry: if ${\mathbf{NXI}}>0$, ${\mathbf{UCP}}\left(\mathit{i},\mathit{j}\right)$ contains the value of ${U}_{\mathit{i}}\left(x,t\right)$ at the coupling point $x={\xi }_{\mathit{j}}$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{NXI}}$.
9: UCPX(NPDE,$*$) – REAL (KIND=nag_wp) arrayInput
On entry: if ${\mathbf{NXI}}>0$, ${\mathbf{UCPX}}\left(\mathit{i},\mathit{j}\right)$ contains the value of $\frac{\partial {U}_{\mathit{i}}\left(x,t\right)}{\partial x}$ at the coupling point $x={\xi }_{\mathit{j}}$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{NXI}}$.
10: RCP(NPDE,$*$) – REAL (KIND=nag_wp) arrayInput
On entry: ${\mathbf{RCP}}\left(\mathit{i},\mathit{j}\right)$ contains the value of the flux ${R}_{\mathit{i}}$ at the coupling point $x={\xi }_{\mathit{j}}$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{NXI}}$.
11: UCPT(NPDE,$*$) – REAL (KIND=nag_wp) arrayInput
On entry: if ${\mathbf{NXI}}>0$, ${\mathbf{UCPT}}\left(\mathit{i},\mathit{j}\right)$ contains the value of $\frac{\partial {U}_{\mathit{i}}}{\partial t}$ at the coupling point $x={\xi }_{\mathit{j}}$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{NXI}}$.
12: UCPTX(NPDE,$*$) – REAL (KIND=nag_wp) arrayInput
On entry: ${\mathbf{UCPTX}}\left(\mathit{i},\mathit{j}\right)$ contains the value of $\frac{{\partial }^{2}{U}_{\mathit{i}}}{\partial x\partial t}$ at the coupling point $x={\xi }_{\mathit{j}}$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{NXI}}$.
13: F(NCODE) – REAL (KIND=nag_wp) arrayOutput
On exit: ${\mathbf{F}}\left(\mathit{i}\right)$ must contain the $\mathit{i}$th component of $F$, for $\mathit{i}=1,2,\dots ,{\mathbf{NCODE}}$, where $F$ is defined as
$F=G-AV.-B Ut* Uxt* ,$ (5)
or
$F=-AV.-B Ut* Uxt* .$ (6)
The definition of $F$ is determined by the input value of IRES.
14: IRES – INTEGERInput/Output
On entry: the form of $F$ that must be returned in the array F.
${\mathbf{IRES}}=1$
Equation (5) must be used.
${\mathbf{IRES}}=-1$
Equation (6) must be used.
On exit: should usually remain unchanged. However, you may reset IRES to force the integration routine to take certain actions as described below:
${\mathbf{IRES}}=2$
Indicates to the integrator that control should be passed back immediately to the calling (sub)routine with the error indicator set to ${\mathbf{IFAIL}}={\mathbf{6}}$.
${\mathbf{IRES}}=3$
Indicates to the integrator that the current time step should be abandoned and a smaller time step used instead. You may wish to set ${\mathbf{IRES}}=3$ when a physically meaningless input or output value has been generated. If you consecutively set ${\mathbf{IRES}}=3$, then D03PJF/D03PJA returns to the calling subroutine with the error indicator set to ${\mathbf{IFAIL}}={\mathbf{4}}$.
Note: the following are additional parameters for specific use with D03PJA. Users of D03PJF therefore need not read the remainder of this description.
15: IUSER($*$) – INTEGER arrayUser Workspace
16: RUSER($*$) – REAL (KIND=nag_wp) arrayUser Workspace
ODEDEF is called with the parameters IUSER and RUSER as supplied to D03PJF/D03PJA. You are free to use the arrays IUSER and RUSER to supply information to ODEDEF as an alternative to using COMMON global variables.
ODEDEF must either be a module subprogram USEd by, or declared as EXTERNAL in, the (sub)program from which D03PJF/D03PJA is called. Parameters denoted as Input must not be changed by this procedure.
15: NXI – INTEGERInput
On entry: the number of ODE/PDE coupling points.
Constraints:
• if ${\mathbf{NCODE}}=0$, ${\mathbf{NXI}}=0$;
• if ${\mathbf{NCODE}}>0$, ${\mathbf{NXI}}\ge 0$.
16: XI($*$) – REAL (KIND=nag_wp) arrayInput
Note: the dimension of the array XI must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{NXI}}\right)$.
On entry: ${\mathbf{XI}}\left(\mathit{i}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NXI}}$, must be set to the ODE/PDE coupling points.
Constraint: ${\mathbf{XBKPTS}}\left(1\right)\le {\mathbf{XI}}\left(1\right)<{\mathbf{XI}}\left(2\right)<\cdots <{\mathbf{XI}}\left({\mathbf{NXI}}\right)\le {\mathbf{XBKPTS}}\left({\mathbf{NBKPTS}}\right)$.
17: NEQN – INTEGERInput
On entry: the number of ODEs in the time direction.
Constraint: ${\mathbf{NEQN}}={\mathbf{NPDE}}×{\mathbf{NPTS}}+{\mathbf{NCODE}}$.
18: UVINIT – SUBROUTINE, supplied by the user.External Procedure
UVINIT must compute the initial values of the PDE and the ODE components ${U}_{\mathit{i}}\left({x}_{\mathit{j}},{t}_{0}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{NPTS}}$, and ${V}_{\mathit{k}}\left({t}_{0}\right)$, for $\mathit{k}=1,2,\dots ,{\mathbf{NCODE}}$.
The specification of UVINIT for D03PJF is:
SUBROUTINE UVINIT ( NPDE, NPTS, X, U, NCODE, V)
INTEGER NPDE, NPTS, NCODE REAL (KIND=nag_wp) X(NPTS), U(NPDE,NPTS), V(NCODE)
The specification of UVINIT for D03PJA is:
SUBROUTINE UVINIT ( NPDE, NPTS, X, U, NCODE, V, IUSER, RUSER)
INTEGER NPDE, NPTS, NCODE, IUSER(*) REAL (KIND=nag_wp) X(NPTS), U(NPDE,NPTS), V(NCODE), RUSER(*)
1: NPDE – INTEGERInput
On entry: the number of PDEs in the system.
2: NPTS – INTEGERInput
On entry: the number of mesh points in the interval $\left[a,b\right]$.
3: X(NPTS) – REAL (KIND=nag_wp) arrayInput
On entry: ${\mathbf{X}}\left(\mathit{i}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPTS}}$, contains the current values of the space variable ${x}_{\mathit{i}}$.
4: U(NPDE,NPTS) – REAL (KIND=nag_wp) arrayOutput
On exit: if ${\mathbf{NXI}}>0$, ${\mathbf{U}}\left(\mathit{i},\mathit{j}\right)$ contains the value of the component ${U}_{\mathit{i}}\left({x}_{\mathit{j}},{t}_{0}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NPDE}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{NPTS}}$.
5: NCODE – INTEGERInput
On entry: the number of coupled ODEs in the system.
6: V(NCODE) – REAL (KIND=nag_wp) arrayOutput
On exit: ${\mathbf{V}}\left(\mathit{i}\right)$ contains the value of component ${V}_{\mathit{i}}\left({t}_{0}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NCODE}}$.
Note: the following are additional parameters for specific use with D03PJA. Users of D03PJF therefore need not read the remainder of this description.
7: IUSER($*$) – INTEGER arrayUser Workspace
8: RUSER($*$) – REAL (KIND=nag_wp) arrayUser Workspace
UVINIT is called with the parameters IUSER and RUSER as supplied to D03PJF/D03PJA. You are free to use the arrays IUSER and RUSER to supply information to UVINIT as an alternative to using COMMON global variables.
UVINIT must either be a module subprogram USEd by, or declared as EXTERNAL in, the (sub)program from which D03PJF/D03PJA is called. Parameters denoted as Input must not be changed by this procedure.
19: RTOL($*$) – REAL (KIND=nag_wp) arrayInput
Note: the dimension of the array RTOL must be at least $1$ if ${\mathbf{ITOL}}=1$ or $2$ and at least ${\mathbf{NEQN}}$ if ${\mathbf{ITOL}}=3$ or $4$.
On entry: the relative local error tolerance.
Constraint: ${\mathbf{RTOL}}\left(i\right)\ge 0.0$ for all relevant $i$.
20: ATOL($*$) – REAL (KIND=nag_wp) arrayInput
Note: the dimension of the array ATOL must be at least $1$ if ${\mathbf{ITOL}}=1$ or $3$ and at least ${\mathbf{NEQN}}$ if ${\mathbf{ITOL}}=2$ or $4$.
On entry: the absolute local error tolerance.
Constraint: ${\mathbf{ATOL}}\left(i\right)\ge 0.0$ for all relevant $i$.
Note: corresponding elements of RTOL and ATOL cannot both be $0.0$.
21: ITOL – INTEGERInput
On entry: a value to indicate the form of the local error test. ITOL indicates to D03PJF/D03PJA whether to interpret either or both of RTOL or ATOL as a vector or scalar. The error test to be satisfied is $‖{e}_{i}/{w}_{i}‖<1.0$, where ${w}_{i}$ is defined as follows:
ITOL RTOL ATOL ${w}_{i}$ 1 scalar scalar ${\mathbf{RTOL}}\left(1\right)×\left|{U}_{i}\right|+{\mathbf{ATOL}}\left(1\right)$ 2 scalar vector ${\mathbf{RTOL}}\left(1\right)×\left|{U}_{i}\right|+{\mathbf{ATOL}}\left(i\right)$ 3 vector scalar ${\mathbf{RTOL}}\left(i\right)×\left|{U}_{i}\right|+{\mathbf{ATOL}}\left(1\right)$ 4 vector vector ${\mathbf{RTOL}}\left(i\right)×\left|{U}_{i}\right|+{\mathbf{ATOL}}\left(i\right)$
In the above, ${e}_{\mathit{i}}$ denotes the estimated local error for the $\mathit{i}$th component of the coupled PDE/ODE system in time, ${\mathbf{U}}\left(\mathit{i}\right)$, for $\mathit{i}=1,2,\dots ,{\mathbf{NEQN}}$.
The choice of norm used is defined by the parameter NORM.
Constraint: $1\le {\mathbf{ITOL}}\le 4$.
22: NORM – CHARACTER(1)Input
On entry: the type of norm to be used.
${\mathbf{NORM}}=\text{'M'}$
Maximum norm.
${\mathbf{NORM}}=\text{'A'}$
Averaged ${L}_{2}$ norm.
If ${{\mathbf{U}}}_{\mathrm{norm}}$ denotes the norm of the vector U of length NEQN, then for the averaged ${L}_{2}$ norm
$Unorm=1NEQN∑i=1NEQNUi/wi2,$
while for the maximum norm
$U norm = maxi Ui / wi .$
See the description of ITOL for the formulation of the weight vector $w$.
Constraint: ${\mathbf{NORM}}=\text{'M'}$ or $\text{'A'}$.
23: LAOPT – CHARACTER(1)Input
On entry: the type of matrix algebra required.
${\mathbf{LAOPT}}=\text{'F'}$
Full matrix methods to be used.
${\mathbf{LAOPT}}=\text{'B'}$
Banded matrix methods to be used.
${\mathbf{LAOPT}}=\text{'S'}$
Sparse matrix methods to be used.
Constraint: ${\mathbf{LAOPT}}=\text{'F'}$, $\text{'B'}$ or $\text{'S'}$.
Note: you are recommended to use the banded option when no coupled ODEs are present (i.e., ${\mathbf{NCODE}}=0$).
24: ALGOPT($30$) – REAL (KIND=nag_wp) arrayInput
On entry: may be set to control various options available in the integrator. If you wish to employ all the default options, then ${\mathbf{ALGOPT}}\left(1\right)$ should be set to $0.0$. Default values will also be used for any other elements of ALGOPT set to zero. The permissible values, default values, and meanings are as follows:
${\mathbf{ALGOPT}}\left(1\right)$
Selects the ODE integration method to be used. If ${\mathbf{ALGOPT}}\left(1\right)=1.0$, a BDF method is used and if ${\mathbf{ALGOPT}}\left(1\right)=2.0$, a Theta method is used. The default value is ${\mathbf{ALGOPT}}\left(1\right)=1.0$.
If ${\mathbf{ALGOPT}}\left(1\right)=2.0$, then ${\mathbf{ALGOPT}}\left(\mathit{i}\right)$, for $\mathit{i}=2,3,4$ are not used.
${\mathbf{ALGOPT}}\left(2\right)$
Specifies the maximum order of the BDF integration formula to be used. ${\mathbf{ALGOPT}}\left(2\right)$ may be $1.0$, $2.0$, $3.0$, $4.0$ or $5.0$. The default value is ${\mathbf{ALGOPT}}\left(2\right)=5.0$.
${\mathbf{ALGOPT}}\left(3\right)$
Specifies what method is to be used to solve the system of nonlinear equations arising on each step of the BDF method. If ${\mathbf{ALGOPT}}\left(3\right)=1.0$ a modified Newton iteration is used and if ${\mathbf{ALGOPT}}\left(3\right)=2.0$ a functional iteration method is used. If functional iteration is selected and the integrator encounters difficulty, then there is an automatic switch to the modified Newton iteration. The default value is ${\mathbf{ALGOPT}}\left(3\right)=1.0$.
${\mathbf{ALGOPT}}\left(4\right)$
Specifies whether or not the Petzold error test is to be employed. The Petzold error test results in extra overhead but is more suitable when algebraic equations are present, such as ${P}_{i,\mathit{j}}=0.0$, for $\mathit{j}=1,2,\dots ,{\mathbf{NPDE}}$, for some $i$ or when there is no ${\stackrel{.}{V}}_{i}\left(t\right)$ dependence in the coupled ODE system. If ${\mathbf{ALGOPT}}\left(4\right)=1.0$, then the Petzold test is used. If ${\mathbf{ALGOPT}}\left(4\right)=2.0$, then the Petzold test is not used. The default value is ${\mathbf{ALGOPT}}\left(4\right)=1.0$.
If ${\mathbf{ALGOPT}}\left(1\right)=1.0$, then ${\mathbf{ALGOPT}}\left(\mathit{i}\right)$, for $\mathit{i}=5,6,7$, are not used.
${\mathbf{ALGOPT}}\left(5\right)$
Specifies the value of Theta to be used in the Theta integration method. $0.51\le {\mathbf{ALGOPT}}\left(5\right)\le 0.99$. The default value is ${\mathbf{ALGOPT}}\left(5\right)=0.55$.
${\mathbf{ALGOPT}}\left(6\right)$
Specifies what method is to be used to solve the system of nonlinear equations arising on each step of the Theta method. If ${\mathbf{ALGOPT}}\left(6\right)=1.0$, a modified Newton iteration is used and if ${\mathbf{ALGOPT}}\left(6\right)=2.0$, a functional iteration method is used. The default value is ${\mathbf{ALGOPT}}\left(6\right)=1.0$.
${\mathbf{ALGOPT}}\left(7\right)$
Specifies whether or not the integrator is allowed to switch automatically between modified Newton and functional iteration methods in order to be more efficient. If ${\mathbf{ALGOPT}}\left(7\right)=1.0$, then switching is allowed and if ${\mathbf{ALGOPT}}\left(7\right)=2.0$, then switching is not allowed. The default value is ${\mathbf{ALGOPT}}\left(7\right)=1.0$.
${\mathbf{ALGOPT}}\left(11\right)$
Specifies a point in the time direction, ${t}_{\mathrm{crit}}$, beyond which integration must not be attempted. The use of ${t}_{\mathrm{crit}}$ is described under the parameter ITASK. If ${\mathbf{ALGOPT}}\left(1\right)\ne 0.0$, a value of $0.0$ for ${\mathbf{ALGOPT}}\left(11\right)$, say, should be specified even if ITASK subsequently specifies that ${t}_{\mathrm{crit}}$ will not be used.
${\mathbf{ALGOPT}}\left(12\right)$
Specifies the minimum absolute step size to be allowed in the time integration. If this option is not required, ${\mathbf{ALGOPT}}\left(12\right)$ should be set to $0.0$.
${\mathbf{ALGOPT}}\left(13\right)$
Specifies the maximum absolute step size to be allowed in the time integration. If this option is not required, ${\mathbf{ALGOPT}}\left(13\right)$ should be set to $0.0$.
${\mathbf{ALGOPT}}\left(14\right)$
Specifies the initial step size to be attempted by the integrator. If ${\mathbf{ALGOPT}}\left(14\right)=0.0$, then the initial step size is calculated internally.
${\mathbf{ALGOPT}}\left(15\right)$
Specifies the maximum number of steps to be attempted by the integrator in any one call. If ${\mathbf{ALGOPT}}\left(15\right)=0.0$, then no limit is imposed.
${\mathbf{ALGOPT}}\left(23\right)$
Specifies what method is to be used to solve the nonlinear equations at the initial point to initialize the values of $U$, ${U}_{t}$, $V$ and $\stackrel{.}{V}$. If ${\mathbf{ALGOPT}}\left(23\right)=1.0$, a modified Newton iteration is used and if ${\mathbf{ALGOPT}}\left(23\right)=2.0$, functional iteration is used. The default value is ${\mathbf{ALGOPT}}\left(23\right)=1.0$.
${\mathbf{ALGOPT}}\left(29\right)$ and ${\mathbf{ALGOPT}}\left(30\right)$ are used only for the sparse matrix algebra option, ${\mathbf{LAOPT}}=\text{'S'}$.
${\mathbf{ALGOPT}}\left(29\right)$
Governs the choice of pivots during the decomposition of the first Jacobian matrix. It should lie in the range $0.0<{\mathbf{ALGOPT}}\left(29\right)<1.0$, with smaller values biasing the algorithm towards maintaining sparsity at the expense of numerical stability. If ${\mathbf{ALGOPT}}\left(29\right)$ lies outside this range then the default value is used. If the routines regard the Jacobian matrix as numerically singular then increasing ${\mathbf{ALGOPT}}\left(29\right)$ towards $1.0$ may help, but at the cost of increased fill-in. The default value is ${\mathbf{ALGOPT}}\left(29\right)=0.1$.
${\mathbf{ALGOPT}}\left(30\right)$
Is used as a relative pivot threshold during subsequent Jacobian decompositions (see ${\mathbf{ALGOPT}}\left(29\right)$) below which an internal error is invoked. If ${\mathbf{ALGOPT}}\left(30\right)$ is greater than $1.0$ no check is made on the pivot size, and this may be a necessary option if the Jacobian is found to be numerically singular (see ${\mathbf{ALGOPT}}\left(29\right)$). The default value is ${\mathbf{ALGOPT}}\left(30\right)=0.0001$.
25: RSAVE(LRSAVE) – REAL (KIND=nag_wp) arrayCommunication Array
If ${\mathbf{IND}}=0$, RSAVE need not be set on entry.
If ${\mathbf{IND}}=1$, RSAVE must be unchanged from the previous call to the routine because it contains required information about the iteration.
26: LRSAVE – INTEGERInput
On entry: the dimension of the array RSAVE as declared in the (sub)program from which D03PJF/D03PJA is called. Its size depends on the type of matrix algebra selected.
If ${\mathbf{LAOPT}}=\text{'F'}$, ${\mathbf{LRSAVE}}\ge {\mathbf{NEQN}}×{\mathbf{NEQN}}+{\mathbf{NEQN}}+\mathit{nwkres}+\mathit{lenode}$.
If ${\mathbf{LAOPT}}=\text{'B'}$, ${\mathbf{LRSAVE}}\ge \left(3×\mathit{mlu}+1\right)×{\mathbf{NEQN}}+\mathit{nwkres}+\mathit{lenode}$.
If ${\mathbf{LAOPT}}=\text{'S'}$, ${\mathbf{LRSAVE}}\ge 4×{\mathbf{NEQN}}+11×{\mathbf{NEQN}}/2+1+\mathit{nwkres}+\mathit{lenode}$.
Where
$\mathit{mlu}$ is the lower or upper half bandwidths such that $\mathit{mlu}=3×{\mathbf{NPDE}}-1$, for PDE problems only (no coupled ODEs); or $\mathit{mlu}={\mathbf{NEQN}}-1$, for coupled PDE/ODE problems. $\mathit{nwkres}=\left\{\begin{array}{ll}3×{\left({\mathbf{NPOLY}}+1\right)}^{2}+\left({\mathbf{NPOLY}}+1\right)×\left[{{\mathbf{NPDE}}}^{2}+6×{\mathbf{NPDE}}+{\mathbf{NBKPTS}}+1\right]+8×{\mathbf{NPDE}}+{\mathbf{NXI}}×\left(5×{\mathbf{NPDE}}+1\right)+{\mathbf{NCODE}}+3\text{,}& \text{when }{\mathbf{NCODE}}>0\text{ and }{\mathbf{NXI}}>0\text{; or}\\ 3×{\left({\mathbf{NPOLY}}+1\right)}^{2}+\left({\mathbf{NPOLY}}+1\right)×\left[{{\mathbf{NPDE}}}^{2}+6×{\mathbf{NPDE}}+{\mathbf{NBKPTS}}+1\right]+13×{\mathbf{NPDE}}+{\mathbf{NCODE}}+4\text{,}& \text{when }{\mathbf{NCODE}}>0\text{ and }{\mathbf{NXI}}=0\text{; or}\\ 3×{\left({\mathbf{NPOLY}}+1\right)}^{2}+\left({\mathbf{NPOLY}}+1\right)×\left[{{\mathbf{NPDE}}}^{2}+6×{\mathbf{NPDE}}+{\mathbf{NBKPTS}}+1\right]+13×{\mathbf{NPDE}}+5\text{,}& \text{when }{\mathbf{NCODE}}=0\text{.}\end{array}\right\$ $\mathit{lenode}=\left\{\begin{array}{ll}\left(6+\mathrm{int}\left({\mathbf{ALGOPT}}\left(2\right)\right)\right)×{\mathbf{NEQN}}+50\text{,}& \text{when the BDF method is used; or}\\ 9×{\mathbf{NEQN}}+50\text{,}& \text{when the Theta method is used.}\end{array}\right\$
Note: when ${\mathbf{LAOPT}}=\text{'S'}$, the value of LRSAVE may be too small when supplied to the integrator. An estimate of the minimum size of LRSAVE is printed on the current error message unit if ${\mathbf{ITRACE}}>0$ and the routine returns with ${\mathbf{IFAIL}}={\mathbf{15}}$.
27: ISAVE(LISAVE) – INTEGER arrayCommunication Array
If ${\mathbf{IND}}=0$, ISAVE need not be set on entry.
If ${\mathbf{IND}}=1$, ISAVE must be unchanged from the previous call to the routine because it contains required information about the iteration required for subsequent calls. In particular:
${\mathbf{ISAVE}}\left(1\right)$
Contains the number of steps taken in time.
${\mathbf{ISAVE}}\left(2\right)$
Contains the number of residual evaluations of the resulting ODE system used. One such evaluation involves computing the PDE functions at all the mesh points, as well as one evaluation of the functions in the boundary conditions.
${\mathbf{ISAVE}}\left(3\right)$
Contains the number of Jacobian evaluations performed by the time integrator.
${\mathbf{ISAVE}}\left(4\right)$
Contains the order of the ODE method last used in the time integration.
${\mathbf{ISAVE}}\left(5\right)$
Contains the number of Newton iterations performed by the time integrator. Each iteration involves residual evaluation of the resulting ODE system followed by a back-substitution using the $LU$ decomposition of the Jacobian matrix.
28: LISAVE – INTEGERInput
On entry: the dimension of the array ISAVE as declared in the (sub)program from which D03PJF/D03PJA is called. Its size depends on the type of matrix algebra selected:
• if ${\mathbf{LAOPT}}=\text{'F'}$, ${\mathbf{LISAVE}}\ge 24$;
• if ${\mathbf{LAOPT}}=\text{'B'}$, ${\mathbf{LISAVE}}\ge {\mathbf{NEQN}}+24$;
• if ${\mathbf{LAOPT}}=\text{'S'}$, ${\mathbf{LISAVE}}\ge 25×{\mathbf{NEQN}}+24$.
Note: when using the sparse option, the value of LISAVE may be too small when supplied to the integrator. An estimate of the minimum size of LISAVE is printed on the current error message unit if ${\mathbf{ITRACE}}>0$ and the routine returns with ${\mathbf{IFAIL}}={\mathbf{15}}$.
On entry: specifies the task to be performed by the ODE integrator.
${\mathbf{ITASK}}=1$
Normal computation of output values U at $t={\mathbf{TOUT}}$.
${\mathbf{ITASK}}=2$
One step and return.
${\mathbf{ITASK}}=3$
Stop at first internal integration point at or beyond $t={\mathbf{TOUT}}$.
${\mathbf{ITASK}}=4$
Normal computation of output values U at $t={\mathbf{TOUT}}$ but without overshooting $t={t}_{\mathrm{crit}}$ where ${t}_{\mathrm{crit}}$ is described under the parameter ALGOPT.
${\mathbf{ITASK}}=5$
Take one step in the time direction and return, without passing ${t}_{\mathrm{crit}}$, where ${t}_{\mathrm{crit}}$ is described under the parameter ALGOPT.
Constraint: ${\mathbf{ITASK}}=1$, $2$, $3$, $4$ or $5$.
30: ITRACE – INTEGERInput
On entry: the level of trace information required from D03PJF/D03PJA and the underlying ODE solver. ITRACE may take the value $-1$, $0$, $1$, $2$ or $3$.
${\mathbf{ITRACE}}=-1$
No output is generated.
${\mathbf{ITRACE}}=0$
Only warning messages from the PDE solver are printed on the current error message unit (see X04AAF).
${\mathbf{ITRACE}}>0$
Output from the underlying ODE solver is printed on the current advisory message unit (see X04ABF). This output contains details of Jacobian entries, the nonlinear iteration and the time integration during the computation of the ODE system.
If ${\mathbf{ITRACE}}<-1$, then $-1$ is assumed and similarly if ${\mathbf{ITRACE}}>3$, then $3$ is assumed.
The advisory messages are given in greater detail as ITRACE increases. You are advised to set ${\mathbf{ITRACE}}=0$, unless you are experienced with sub-chapter D02M–N.
31: IND – INTEGERInput/Output
On entry: indicates whether this is a continuation call or a new integration.
${\mathbf{IND}}=0$
Starts or restarts the integration in time.
${\mathbf{IND}}=1$
Continues the integration after an earlier exit from the routine. In this case, only the parameters TOUT and IFAIL should be reset between calls to D03PJF/D03PJA.
Constraint: ${\mathbf{IND}}=0$ or $1$.
On exit: ${\mathbf{IND}}=1$.
32: IFAIL – INTEGERInput/Output
Note: for D03PJA, IFAIL does not occur in this position in the parameter list. See the additional parameters described below.
On entry: IFAIL must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\mathbf{1}\text{ or }\mathbf{1}$ is used it is essential to test the value of IFAIL on exit.
On exit: ${\mathbf{IFAIL}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
Note: the following are additional parameters for specific use with D03PJA. Users of D03PJF therefore need not read the remainder of this description.
32: IUSER($*$) – INTEGER arrayUser Workspace
33: RUSER($*$) – REAL (KIND=nag_wp) arrayUser Workspace
IUSER and RUSER are not used by D03PJF/D03PJA, but are passed directly to PDEDEF, BNDARY, ODEDEF and UVINIT and may be used to pass information to these routines as an alternative to using COMMON global variables.
34: CWSAV($10$) – CHARACTER(80) arrayCommunication Array
35: LWSAV($100$) – LOGICAL arrayCommunication Array
36: IWSAV($505$) – INTEGER arrayCommunication Array
37: RWSAV($1100$) – REAL (KIND=nag_wp) arrayCommunication Array
38: IFAIL – INTEGERInput/Output
Note: see the parameter description for IFAIL above.
## 6 Error Indicators and Warnings
If on entry ${\mathbf{IFAIL}}={\mathbf{0}}$ or $-{\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF).
Errors or warnings detected by the routine:
${\mathbf{IFAIL}}=1$
On entry, ${\mathbf{TOUT}}-{\mathbf{TS}}$ is too small, or ${\mathbf{ITASK}}\ne 1$, $2$, $3$, $4$ or $5$, or ${\mathbf{M}}\ne 0$, $1$ or $2$, or at least one of the coupling point in array XI is outside the interval [${\mathbf{XBKPTS}}\left(1\right),{\mathbf{XBKPTS}}\left({\mathbf{NBKPTS}}\right)$], or ${\mathbf{NPTS}}\ne \left({\mathbf{NBKPTS}}-1\right)×{\mathbf{NPOLY}}+1$, or ${\mathbf{NBKPTS}}<2$, or ${\mathbf{NPDE}}\le 0$, or ${\mathbf{NORM}}\ne \text{'A'}$ or $\text{'M'}$, or ${\mathbf{ITOL}}\ne 1$, $2$, $3$ or $4$, or ${\mathbf{NPOLY}}<1$ or ${\mathbf{NPOLY}}>49$, or NCODE and NXI are incorrectly defined, or ${\mathbf{NEQN}}\ne {\mathbf{NPDE}}×{\mathbf{NPTS}}+{\mathbf{NCODE}}$, or ${\mathbf{LAOPT}}\ne \text{'F'}$, $\text{'B'}$ or $\text{'S'}$, or ${\mathbf{IND}}\ne 0$ or $1$, or break points ${\mathbf{XBKPTS}}\left(i\right)$ are badly ordered, or LRSAVE is too small, or LISAVE is too small, or the ODE integrator has not been correctly defined; check ALGOPT parameter, or either an element of RTOL or ${\mathbf{ATOL}}<0.0$, or all the elements of RTOL and ATOL are zero.
${\mathbf{IFAIL}}=2$
The underlying ODE solver cannot make any further progress, with the values of ATOL and RTOL, across the integration range from the current point $t={\mathbf{TS}}$. The components of U contain the computed values at the current point $t={\mathbf{TS}}$.
${\mathbf{IFAIL}}=3$
In the underlying ODE solver, there were repeated error test failures on an attempted step, before completing the requested task, but the integration was successful as far as $t={\mathbf{TS}}$. The problem may have a singularity, or the error requirement may be inappropriate.
${\mathbf{IFAIL}}=4$
In setting up the ODE system, the internal initialization routine was unable to initialize the derivative of the ODE system. This could be due to the fact that IRES was repeatedly set to $3$ in at least PDEDEF, BNDARY or ODEDEF, when the residual in the underlying ODE solver was being evaluated.
${\mathbf{IFAIL}}=5$
In solving the ODE system, a singular Jacobian has been encountered. You should check your problem formulation.
${\mathbf{IFAIL}}=6$
When evaluating the residual in solving the ODE system, IRES was set to $2$ in at least PDEDEF, BNDARY or ODEDEF. Integration was successful as far as $t={\mathbf{TS}}$.
${\mathbf{IFAIL}}=7$
The values of ATOL and RTOL are so small that the routine is unable to start the integration in time.
${\mathbf{IFAIL}}=8$
In one of PDEDEF, BNDARY or ODEDEF, IRES was set to an invalid value.
${\mathbf{IFAIL}}=9$ (D02NNF)
A serious error has occurred in an internal call to the specified routine. Check the problem specification and all parameters and array dimensions. Setting ${\mathbf{ITRACE}}=1$ may provide more information. If the problem persists, contact NAG.
${\mathbf{IFAIL}}=10$
The required task has been completed, but it is estimated that a small change in ATOL and RTOL is unlikely to produce any change in the computed solution. (Only applies when you are not operating in one step mode, that is when ${\mathbf{ITASK}}\ne 2$ or $5$.)
${\mathbf{IFAIL}}=11$
An error occurred during Jacobian formulation of the ODE system (a more detailed error description may be directed to the current error message unit).
${\mathbf{IFAIL}}=12$
In solving the ODE system, the maximum number of steps specified in ${\mathbf{ALGOPT}}\left(15\right)$ have been taken.
${\mathbf{IFAIL}}=13$
Some error weights ${w}_{i}$ became zero during the time integration (see the description of ITOL). Pure relative error control (${\mathbf{ATOL}}\left(i\right)=0.0$) was requested on a variable (the $i$th) which has become zero. The integration was successful as far as $t={\mathbf{TS}}$.
${\mathbf{IFAIL}}=14$
The flux function ${R}_{i}$ was detected as depending on time derivatives, which is not permissible.
${\mathbf{IFAIL}}=15$
When using the sparse option, the value of LISAVE or LRSAVE was not sufficient (more detailed information may be directed to the current error message unit).
## 7 Accuracy
D03PJF/D03PJA controls the accuracy of the integration in the time direction but not the accuracy of the approximation in space. The spatial accuracy depends on both the number of mesh points and on their distribution in space. In the time integration only the local error over a single step is controlled and so the accuracy over a number of steps cannot be guaranteed. You should therefore test the effect of varying the accuracy parameter ATOL and RTOL.
The parameter specification allows you to include equations with only first-order derivatives in the space direction but there is no guarantee that the method of integration will be satisfactory for such systems. The position and nature of the boundary conditions in particular are critical in defining a stable problem.
The time taken depends on the complexity of the parabolic system and on the accuracy requested.
## 9 Example
This example provides a simple coupled system of one PDE and one ODE.
$V 1 2 ∂ U 1 ∂ t -x V 1 V . 1 ∂ U 1 ∂ x = ∂ 2 U 1 ∂ x 2 V . 1 = V 1 U 1 + ∂ U 1 ∂ x +1 +t ,$
for $t\in \left[{10}^{-4},0.1×{2}^{i}\right]\text{, }i=1,2,\dots ,5,x\in \left[0,1\right]$.
The left boundary condition at $x=0$ is
$∂U1 ∂x =-V1expt.$
The right boundary condition at $x=1$ is
$U1=-V1V.1.$
The initial conditions at $t={10}^{-4}$ are defined by the exact solution:
$V1=t, and U1x,t=expt1-x-1.0, x∈0,1,$
and the coupling point is at ${\xi }_{1}=1.0$.
### 9.1 Program Text
Note: the following programs illustrate the use of D03PJF and D03PJA.
Program Text (d03pjfe.f90)
Program Text (d03pjae.f90)
### 9.2 Program Data
Program Data (d03pjfe.d)
Program Data (d03pjae.d)
### 9.3 Program Results
Program Results (d03pjfe.r)
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