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https://www.jobilize.com/course/section/length-4-scaling-coefficient-vector-by-openstax?qcr=www.quizover.com | # 0.5 The scaling function and scaling coefficients, wavelet and (Page 10/13)
Page 10 / 13
## Parameterization of the scaling coefficients
The case where $\phi \left(t\right)$ and $h\left(n\right)$ have compact support is very important. It aids in the time localization properties of the DWT andoften reduces the computational requirements of calculating the DWT. If $h\left(n\right)$ has compact support, then the filters described in Chapter: Filter Banks and the Discrete Wavelet Transform are simple FIR filters. We have stated that $N$ , the length of the sequence $h\left(n\right)$ , must be even and $h\left(n\right)$ must satisfy the linear constraint of [link] and the $\frac{N}{2}$ bilinear constraints of [link] . This leaves $\frac{N}{2}-1$ degrees of freedom in choosing $h\left(n\right)$ that will still guarantee the existence of $\phi \left(t\right)$ and a set of essentially orthogonal basis functions generated from $\phi \left(t\right)$ .
## Length-2 scaling coefficient vector
For a length-2 $h\left(n\right)$ , there are no degrees of freedom left after satisfying the required conditions in [link] and [link] . These requirements are
$h\left(0\right)\phantom{\rule{0.166667em}{0ex}}+\phantom{\rule{0.166667em}{0ex}}h\left(1\right)\phantom{\rule{0.166667em}{0ex}}=\phantom{\rule{0.166667em}{0ex}}\sqrt{2}$
and
${h}^{2}\left(0\right)+{h}^{2}\left(1\right)\phantom{\rule{0.166667em}{0ex}}=\phantom{\rule{0.166667em}{0ex}}1$
which are uniquely satisfied by
${h}_{D2}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\left\{h,\left(,0,\right),,,h,\left(,1,\right)\right\}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\left\{\frac{1}{\sqrt{2}},\phantom{\rule{0.166667em}{0ex}},,,\phantom{\rule{0.166667em}{0ex}},\frac{1}{\sqrt{2}}\right\}.$
These are the Haar scaling functions coefficients which are also the length-2 Daubechies coefficients [link] used as an example in Chapter: A multiresolution formulation of Wavelet Systems and discussed later in this book.
## Length-4 scaling coefficient vector
For the length-4 coefficient sequence, there is one degree of freedom or one parameter that gives all the coefficients that satisfy the requiredconditions:
$h\left(0\right)\phantom{\rule{0.166667em}{0ex}}+h\left(1\right)\phantom{\rule{0.166667em}{0ex}}+h\left(2\right)\phantom{\rule{0.166667em}{0ex}}+h\left(3\right)\phantom{\rule{0.166667em}{0ex}}=\phantom{\rule{0.166667em}{0ex}}\sqrt{2},$
${h}^{2}\left(0\right)+{h}^{2}\left(1\right)+{h}^{2}\left(2\right)+{h}^{2}\left(3\right)=1$
and
$h\left(0\right)\phantom{\rule{0.166667em}{0ex}}h\left(2\right)\phantom{\rule{0.166667em}{0ex}}+\phantom{\rule{0.166667em}{0ex}}h\left(1\right)\phantom{\rule{0.166667em}{0ex}}h\left(3\right)\phantom{\rule{0.166667em}{0ex}}=\phantom{\rule{0.166667em}{0ex}}0$
Letting the parameter be the angle $\alpha$ , the coefficients become
$\begin{array}{cc}h\left(0\right)=\hfill & \left(1-cos\left(\alpha \right)+sin\left(\alpha \right)\right)/\left(2\sqrt{2}\right)\hfill \\ h\left(1\right)=\hfill & \left(1+cos\left(\alpha \right)+sin\left(\alpha \right)\right)/\left(2\sqrt{2}\right)\hfill \\ h\left(2\right)=\hfill & \left(1+cos\left(\alpha \right)-sin\left(\alpha \right)\right)/\left(2\sqrt{2}\right)\hfill \\ h\left(3\right)=\hfill & \left(1-cos\left(\alpha \right)-sin\left(\alpha \right)\right)/\left(2\sqrt{2}\right).\hfill \end{array}$
These equations also give the length-2 Haar coefficients [link] for $\alpha =0,\pi /2,3\pi /2$ and a degenerate condition for $\alpha =\pi$ . We get the Daubechies coefficients (discussed later in this book) for $\alpha =\pi /3$ . These Daubechies-4 coefficients have a particularly clean form,
${h}_{D4}=\left\{\frac{1+\sqrt{3}}{4\sqrt{2}},\phantom{\rule{0.166667em}{0ex}},,,\phantom{\rule{0.166667em}{0ex}},\frac{3+\sqrt{3}}{4\sqrt{2}},\phantom{\rule{0.166667em}{0ex}},,,\phantom{\rule{0.166667em}{0ex}},\frac{3-\sqrt{3}}{4\sqrt{2}},\phantom{\rule{0.166667em}{0ex}},,,\phantom{\rule{0.166667em}{0ex}},\frac{1-\sqrt{3}}{4\sqrt{2}}\right\}$
## Length-6 scaling coefficient vector
For a length-6 coefficient sequence $h\left(n\right)$ , the two parameters are defined as $\alpha$ and $\beta$ and the resulting coefficients are
$\begin{array}{cc}h\left(0\right)=\hfill & \left[\left(1+cos\left(\alpha \right)+sin\left(\alpha \right)\right)\left(1-cos\left(\beta \right)-sin\left(\beta \right)\right)+2sin\left(\beta \right)\phantom{\rule{0.166667em}{0ex}}cos\left(\alpha \right)\right]/\left(4\sqrt{2}\right)\hfill \\ h\left(1\right)=\hfill & \left[\left(1-cos\left(\alpha \right)+sin\left(\alpha \right)\right)\left(1+cos\left(\beta \right)-sin\left(\beta \right)\right)-2sin\left(\beta \right)\phantom{\rule{0.166667em}{0ex}}cos\left(\alpha \right)\right]/\left(4\sqrt{2}\right)\hfill \\ h\left(2\right)=\hfill & \left[1+cos\left(\alpha -\beta \right)+sin\left(\alpha -\beta \right)\right]/\left(2\sqrt{2}\right)\hfill \\ h\left(3\right)=\hfill & \left[1+cos\left(\alpha -\beta \right)-sin\left(\alpha -\beta \right)\right]/\left(2\sqrt{2}\right)\hfill \\ h\left(4\right)=\hfill & 1/\sqrt{2}-h\left(0\right)-h\left(2\right)\hfill \\ h\left(5\right)=\hfill & 1/\sqrt{2}-h\left(1\right)-h\left(3\right)\hfill \end{array}$
Here the Haar coefficients are generated for any $\alpha =\beta$ and the length-4 coefficients [link] result if $\beta =0$ with $\alpha$ being the free parameter. The length-4 Daubechies coefficients are calculatedfor $\alpha =\pi /3$ and $\beta =0$ . The length-6 Daubechies coefficients result from $\alpha =1.35980373244182\phantom{\rule{4pt}{0ex}}\text{and}\phantom{\rule{4pt}{0ex}}\beta =-0.78210638474440$ .
The inverse of these formulas which will give $\alpha$ and $\beta$ from the allowed $h\left(n\right)$ are
$\alpha =arctan\left(\frac{2\left(h{\left(0\right)}^{2}+h{\left(1\right)}^{2}\right)-1+\left(h\left(2\right)+h\left(3\right)\right)/\sqrt{2}}{2\phantom{\rule{0.166667em}{0ex}}\left(h\left(1\right)\phantom{\rule{0.166667em}{0ex}}h\left(2\right)-h\left(0\right)\phantom{\rule{0.166667em}{0ex}}h\left(3\right)\right)+\sqrt{2}\left(h\left(0\right)-h\left(1\right)\right)}\right)$
$\beta =\alpha -arctan\left(\frac{h\left(2\right)-h\left(3\right)}{h\left(2\right)+h\left(3\right)-1/\sqrt{2}}\right)$
As $\alpha$ and $\beta$ range over $-\pi$ to $\pi$ all possible $h\left(n\right)$ are generated. This allows informative experimentation to better see whatthese compactly supported wavelets look like. This parameterization is implemented in the Matlab programs in Appendix C and in the Aware, Inc. software, UltraWave [link] .
Since the scaling functions and wavelets are used with integer translations, the location of their support is not important, only thesize of the support. Some authors shift $h\left(n\right)$ , ${h}_{1}\left(n\right)$ , $\phi \left(t\right)$ , and $\psi \left(t\right)$ to be approximately centered around the origin. This is achieved by having the initial nonzero scaling coefficient start at $n=-\frac{N}{2}+1$ rather than zero. We prefer to have the origin at $n=t=0$ .
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Got questions? Join the online conversation and get instant answers! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 50, "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": 0.8087192177772522, "perplexity": 1817.4301253902834}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991943.36/warc/CC-MAIN-20210513173321-20210513203321-00192.warc.gz"} |
https://www.bartleby.com/solution-answer/chapter-158-problem-26e-calculus-early-transcendentals-8th-edition/9781285741550/use-spherical-coordinates-26-evaluate-e-x2y2z2dv-where-e-lies-above-the-cone-z-x2y2and/14f0e057-52f4-11e9-8385-02ee952b546e | Chapter 15.8, Problem 26E
### Calculus: Early Transcendentals
8th Edition
James Stewart
ISBN: 9781285741550
Chapter
Section
### Calculus: Early Transcendentals
8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem
# Use spherical coordinates.26. Evaluate ∫∫∫E x 2 + y 2 + z 2 dV, where E lies above the cone z = x 2 + y 2 and between the spheres x2 + y2 + z2 = 1 and x2 + y2 + z2 = 4.
To determine
To evaluate: The given triple integral by using spherical coordinates.
Explanation
Given:
The function is f(x,y,z)=x2+y2+z2 .
The region B lies above the cone z=x2+y2 and between the spheres x2+y2+z2=1 and x2+y2+z2=4 .
Formula used:
If f is a spherical region E given by aρb,αθβ,cϕd , then, Ef(x,y,z)dV=αβabcdf(ρsinϕcosθ,ρsinϕsinθ,ρcosϕ)ρ2sinϕdϕdρdθ (1)
If g(x) is the function of x and h(y) is the function of y and k(z) is the function of z then, abcdefg(x)h(y)k(z)dzdydx=abg(x)dxcdh(y)dyefk(z)dz (2)
The spherical coordinates (ρ,θ,ϕ) corresponding to the rectangular coordinates (x,y,z) is,
ρ=x2+y2+z2ϕ=cos1(zρ)θ=cos1(xρsinϕ)
Calculation:
By the given conditions, it is observed that ρ varies from 1 to 2, θ varies from 0 to 2π and ϕ varies from 0 to π4 . Use the formula mentioned above to change the given problem into spherical coordinates
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Single Variable Calculus: Early Transcendentals, Volume I
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Mathematical Applications for the Management, Life, and Social Sciences
#### The gradient vector field for is:
Study Guide for Stewart's Multivariable Calculus, 8th
#### In O,OY=5 and XZ=6. If XWWY, find WZ.
Elementary Geometry for College Students | {"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": 0.9256017208099365, "perplexity": 3496.5206653191462}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986677412.35/warc/CC-MAIN-20191018005539-20191018033039-00382.warc.gz"} |
http://makox.com/ | ## Real Numbers – Revisiting Rational and Their Decimal Expansions
REVISITING RATIONAL NUMBERS AND THEIR DECIMAL EXPANSIONS :
We have already studied in the previous class that rational numbers have either a terminating decimal repeating decimal expansion. We have to consider a rational number as $\displaystyle \frac{p}{q}$ (where $\displaystyle q\ne 0$) as terminating or non-terminating repeating (or recurring) decimal expansion.
e.g.
(A) (i) 0.0527 =
(ii) 26.12489 =
(B) (i) 0.0875 =
(ii) 23.3408 =
Now, we have converted a real number whose decimal expansion terminates into a rational number of the form. $\displaystyle \frac{p}{q}$, where p and q are coprime, and prime factorization of denominator
(i.e., q) has only power of 2 or power of 5 or both. We should also understand that the denominator is a power of 10 and can be only prime factors 2 and 5.
Now, we get that this real number is a rational number of the form $\displaystyle \frac{p}{q}$, where the prime factorization of q is of the form 2n, 5m, and n, m are some non-negative integers.
Theorem : Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form $\displaystyle \frac{p}{q}$, where p and q are coprime, and the prime factorization of q is of the form 2n5m, where n, m are non-negative integers.
Let $\displaystyle \frac{a}{b}$ be a rational number in the lowest form such that the prime factorization of b is of the form $\displaystyle {{2}^{n}}\times {{5}^{m}}$ . Where n and m are non-negative integers.
Example:
(i) $\displaystyle \frac{3}{8}=\frac{3}{{{2}^{3}}}=\frac{3\times {{5}^{3}}}{{{2}^{3}}\times {{5}^{3}}}=\frac{375}{{{10}^{3}}}=0.375$
(ii) $\displaystyle \frac{13}{125}=\frac{13}{{{5}^{3}}}=\frac{13\times {{2}^{3}}}{{{5}^{3}}\times {{2}^{3}}}=\frac{104}{{{10}^{3}}}=0.104$
(iii) $\displaystyle \frac{2139}{1250}=\frac{2139}{2\times {{5}^{4}}}=\frac{2139\times {{2}^{3}}}{{{2}^{4}}\times {{5}^{4}}}=\frac{2139\times 8}{{{\left( 2\times 5 \right)}^{4}}}=\frac{17112}{{{10}^{4}}}=1.7112$
Theorem : Let $\displaystyle x=\frac{p}{q}$ be a rational number, such that the prime factorization of q is of the form 2n5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.
Example:
$\displaystyle \frac{1}{7}$
In this case, denominator 7 is not of the form 2n5m. So, from above mentioned theorems, $\displaystyle \frac{1}{7}$ will not have a terminating decimal expansion. Hence, 0 (zero) will not be remainder and the remainders will start repeating after a certain stage. We have a block of digits, 142857, repeating in the quotient of $\displaystyle \frac{1}{7}$.
Hence, it is is true that any rational number is not converted by the above theorems.
Theorem :Let $\displaystyle x=\frac{p}{q}$ be a rational number, such that the prime factorization of q is not of the form 2n5m, where n, m are non-negative integers. Then, x has a decimal expansion which is not-terminating repeating (recurring).
Hence, we can say that the decimal expansion of every rational number is either terminating or non-terminating repeating (recurring).
## Real Number-Revisiting Irrational Numbers
REVISITING IRRATIONAL NUMBERS :
We have studied in the previous class about rational numbers as well as irrational numbers. Both are the members of the REAL NUMBERS family. We have also studied how to locate irrational numbers on the number line.
Any rational number is represented by $\displaystyle \frac{p}{q}$; where $\displaystyle q\ne 0$ whereas irrational number is represented by ‘s’
0.10110111011110….. etc. are the examples of irrational numbers.
Theorem : Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer.
Proof :
Let the prime factorization of a be as follows :
a = p1, p2, p3, p4, p5, …., …. pn
$\displaystyle \therefore$ a2 = (p1, p2, p3, p4, …. Pn) (p1, p2, p3, p4, …. Pn)
= p12, p22, p32, p42, …………… pn2.
It is given that p divides a2, so from the Fundamental Theorem of Arithmetic p is one of the prime factors of a2. Also a2 having prime factors p1, p2, p3, ….. pn.
Hence, p is one of p1, p2, p3, …… pn.
Example 6: To prove $\displaystyle \sqrt{2}$ is irrational.
Solution. Let us assume that $\displaystyle \sqrt{2}$ is rational.
Now, we can find integers r and $\displaystyle s\left( \ne 0 \right)$ such that $\displaystyle \sqrt{2}=\frac{r}{s}$.
Again let r and s have a common factor other than 1. Then we can divide by common factor to get, where a and b are coprime.
$\displaystyle \Rightarrow$ $\displaystyle b\sqrt{2}=a$ …….. (i)
$\displaystyle \Rightarrow$ $\displaystyle {{\left( b\sqrt{2} \right)}^{2}}={{\left( a \right)}^{2}}$ [squaring both sides]
$\displaystyle \Rightarrow$ 2b2 = a2
$\displaystyle \therefore$ 2 divides a2.
Now, by above theorem
2 divides a also.
So, we can write a = 2c for some integer c.
$\displaystyle \Rightarrow$ $\displaystyle b\sqrt{2}=2c$ [From (i)]
$\displaystyle \Rightarrow$ 2b2 = 4c2 [Squaring both sides]
$\displaystyle \therefore$ b2 = 2c2 ……. (ii)
i.e., 2 divides b2
So, 2 divides b [by using above theorem with p = 2]
Again, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that $\displaystyle \sqrt{2}$ is rational.
Hence, we conclude that $\displaystyle \sqrt{2}$ is irrational.
## Real Numbers-The Fundamental Theorem of Arithmetic
THE FUNDAMENTAL THEOREM OF ARITHMETIC
We have already studied in the previous classes that any natural can be written as a product of its prime factors. e.g. = 3 = 3, 6 = 2 $\displaystyle \times$ 3, 275 = 11 $\displaystyle \times$ 25 etc. i.e, any natural number can be obtained by multiplying prime numbers.
If we take prime numbers like 2,3,5,7,11,13, and 29 and if we multiply some or all of these numbers, allowing them to repeat as we can, then we can get a large collection of positive integers (infinitely).
e.g. 11 $\displaystyle \times$ 13 $\displaystyle \times$ 29 = 4147
7 $\displaystyle \times$ 11 $\displaystyle \times$ 13 $\displaystyle \times$ 29 = 29029
5 $\displaystyle \times$ 7 $\displaystyle \times$ 11 $\displaystyle \times$ 13 $\displaystyle \times$ 29 = 145145
3 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$ 11 $\displaystyle \times$ 13 $\displaystyle \times$ 29 = 435435
2 $\displaystyle \times$3 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$ 11 $\displaystyle \times$ 13 $\displaystyle \times$ 29 = 870870
22 $\displaystyle \times$3 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$ 11 $\displaystyle \times$ 13 $\displaystyle \times$ 29 = 1741740
23 $\displaystyle \times$3 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$ 11 $\displaystyle \times$ 13 $\displaystyle \times$ 29 = 3483480
Now, we see there are infinitely many primes and so we can get an infinite collection of numbers by all the primes and all possible products of primes.
In the other hand, if we factorise positive integers, we have to do the opposite of what we have done so far.
We can use factor tree for factorise of 65520 as shown below.
Hence, we have factorised 65520 as 2 $\displaystyle \times$ 2 $\displaystyle \times$ 2 $\displaystyle \times$2 $\displaystyle \times$ 3 $\displaystyle \times$ 3 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$ 13 as a product of primes, i.e., 65520 = 24 $\displaystyle \times$ 32 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$ 13 as a product of powers of primes.
If we take 123456789 then it is equal to 32 $\displaystyle \times$ 3803 $\displaystyle \times$ 3607.
In these factors 3803 and 3607 are primes. It means we have to understand that every composite number can be written as the product of primes. It is called Fundamental Theorem of Arithmetic because of its basic crucial importance to the study of integers.
Hence, we can say Fundamental Theorem of Arithmetic as follows :
Every composite number can be expressed (factorised) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur, i.e., for any composite number there will be one and only one way to express it as a product of primes, as long as we are not particular about the order in which the primes occur.
e.g 2 $\displaystyle \times$ 3 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$ 11 as the same as 5 $\displaystyle \times$ 7 $\displaystyle \times$ 11 $\displaystyle \times$ 2 $\displaystyle \times$ 3 or any other possible order in which these primes are written. It can be defined as follows :
The prime factorization of a natural number is unique, except for the order of its factors.
If any composite number is x, then the factor of x = p1 , p2 , p3 , …. Pn, where p1, p2, p3, …. Pn are primes and written in ascending order, i.e., $\displaystyle {{p}_{1}}\le {{p}_{2}}\le {{p}_{3}}......\le {{p}_{n}}$.
If we combine the same primes, we will get powers of primes.
E.g. 65520 = 2 $\displaystyle \times$ 2 $\displaystyle \times$ 2 $\displaystyle \times$ 2 $\displaystyle \times$ 3 $\displaystyle \times$ 3 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$13
24 $\displaystyle \times$ 32 $\displaystyle \times$ 5 $\displaystyle \times$ 7 $\displaystyle \times$13
When we decide that the order will be ascending, then the way the number is factorised, is unique.
There are many applications of the Fundamental Theorem of Arithmetic in mathematics as well as in other fields.
Example 4:Consider the number 16n, where n is a natural number. Check whether there is any value of n for which 16n ends with the digit zero.
Solution. If the number 16n, for any n, were to end with the digit zero, then it would be divisible by 5 and so its prime factorization must contain the prime 5.
16n = (24)n = 24n
The only prime in the factorization of 16n is 2.
There is no other primes in the factorization of 16n = 24n
[By uniqueness of the Fundamental Theorem of Arithmetic]
5 does not occur in the prime factorization of 16n for any n.
4n does not end with the digit zero for any natural number n.
We have already learnt how to find the HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic in previous classes, without realising it. This method is also called the Prime factorization method.
Example :
Find the LCM and HCF of 18 and 60 by the prime factorization method.
Solution.
18 = 2 $\displaystyle \times$ 3 $\displaystyle \times$ 3 = 21 $\displaystyle \times$ 32
60 = 2 $\displaystyle \times$ 2 $\displaystyle \times$ 3 $\displaystyle \times$ 5 = 22 $\displaystyle \times$ 31 $\displaystyle \times$ 51
We get HCF of 18 and 60 as 2
And LCM of 18 and 60 as 180
But now, we have to understand that
HCF of 18 and 60 is 21 = Product of the smallest power of each common prime factor in the numbers.
LCM of 18 and 60 is 22 $\displaystyle \times$ 32 $\displaystyle \times$5 = Product of the greatest power of each Prime factor, involved in the numbers.
For any two positive integers a and b , (HCF of a and b) $\displaystyle \times$ (LCM of a and b) = a x b
## Real Numbers: Euclid’s Division Lemma
EUCLID’S DIVISION LEMMA :
Euclid was the first Greek Mathematician who gave a new way of thinking the study of geometry. He also made important contributions to the number theory. Euclid’s Lemma is one of them. It is a proven statement which is used to prove other statements.
Let ‘a’ and ‘b’ be any two positive integers. Then, there exist unique integers q and r such that
;
Now, we say ‘a’ as dividend, ‘b’ as divisor, ‘q’ as quotient and ‘r’ as remainder.
Dividend = (divisor $\displaystyle \times$ quotient) + remainder
Example: Let 578 be divided by 16.
36 is as quotient and 2 as remainder.
In this case :
Dividend = 578
Divisor = 16
Quotient = 36
And remainder = 2
Dividend = (quotient $\displaystyle \times$ divisor) + remainder
Hence, 578 = (36 $\displaystyle \times$ 16) + 2
To get the HCF of two positive integers, let ‘c’ and ‘d’, with c > d, follow as :
(i) Applying Euclid’s Lemma,
$\displaystyle c=dq+r;\,0\le r
(ii) If r = 0, d is the HCF of c and d.
If $\displaystyle r\ne 0,$ then applying the division lemma to d and r.
(iii) We can continue the process till the remainder is zero. The divisor at this stage will be the required HCF.
This algorithm works because HCF (c,d) = HCF (d,r)
Where the symbol HCF (c,d) denotes the HCF of c and d etc.
Example: Use Euclid’s algorithm to find HCF of 425 and 40
Solution: Let a = 425 and b = 40
By Euclid’s division lemma; we have
$\displaystyle 425=40\times 10+25$ …. (i)
By using the above theorem, we observe that the common divisors of a = 425 and b = 40 are also the common divisors of b = 40 and r1 = 25 and vice-versa.
Applying Euclid’s division lemma on divisor b = 40 and remainder r1 = 25,
We get $\displaystyle 40=25\times 1+15$ …. (ii)
$\displaystyle b=q_{2}^{r},+{{r}_{2}}$; where q2 = 1 and r2 = 15
Again using the above theorem, we find that, the common divisors of r1 = 25 and r2 = 15 are the common divisors of b = 40 and r1 = 25 and vice-versa. But the common divisors of b = 40 and r1 = 25 are the common divisors of a = 425 and b = 40 and vice-versa.
Applying Euclid’s division lemma on r1 = 25 and r2 = 15, we get
$\displaystyle 25=15\times 1+10$ …. (iii)
$\displaystyle \Rightarrow$ r1 = r2q3 + r3, where q3 = 1 and r3 = 10
Again by using the above theorem, we find that common divisors of r2 = 15 and r3 = 10 are the common divisors of a = 425 and b = 40 and vice-versa.
Now, Using Euclid’s division lemma on r2 = 15 and r3 = 10, we get
$\displaystyle 15=10\times 1+5$ …. (iv)
$\displaystyle \Rightarrow$ $\displaystyle r{{ & }_{2}}={{r}_{3}}\times {{q}_{4}}+{{r}_{4}}$; where q4 = 1 and r4 = 5
Again by using the above theorem, we find that, the common divisors of r2 = 15 and r3 = 10 are the common divisors of a = 425 and b = 40 and vice-versa.
Using Euclid’s division lemma on r3 = 10 and r4 = 5, we get
$\displaystyle 15=5\times 3+0$ …. (v)
Hence, r4 = 5 is a divisor of r3 = 10 and r4 = 5. Also, it is the greatest common divisor (or HCF) of r3 and r4. So, r4 = 5 is the greatest common divisor (or HCF) of a = 425 and b = 40. We also observe that r4 = 5 is the last non-zero remainder in the above process of repeated application of Euclid’s division lemma on the divisor and the remainder in the next step.
The set of equation (i) to (v) is called Euclid’s division algorithm for 425 and 40. The last divisor, or the last but non-zero remainder 4 is the HCF (or GCD) of 425 and 40.
The above process of finding HCF can also be carried out by successive divisions as follows:
(i) Euclid’s division lemma and algorithm are so closely interlinked that it is often called division algorithm.
(ii) Euclid’s Division Algorithm is stated for only positive integers except zero. i.e., $\displaystyle b\ne 0$.
Example 3:
A sweet seller has 840 kaju barfis and 260 badam barfis. He wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the number of barfis that can be placed in each stack for this purpose?
Solution. The area of the tray that is used up will be the least. For this, we find HCF (840, 260). Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least.
Now, applying Euclid’s algorithm to find their HCF, we have
840 = 260 $\displaystyle \times$ 3 + 60
260 = 60 $\displaystyle \times$ 4 + 20
60 = 20 $\displaystyle \times$ 3 + 0
Hence, HCF of 840 and 260 is 20.
So, the sweet seller can make stacks of 10 for both kinds of barfi
## Real Numbers
• Introduction
• Euclid’s Division Lemma
• The Fundamental Theorem of Arithmetic
• Revisiting Irrational Numbers
• Revisiting Rational Numbers and Their Decimal Expansions
• Summary
INTRODUCTION :
We have already studied about irrational numbers in 9th class. Now, we will study the real numbers and also about the important properties of positive integers for Euclid’s division algorithm and the Fundamental Theorem of Arithmetic.
Euclid’s division algorithm says us about divisibility of integers. It states that any positive integer ‘a’ can be divided by any other positive integer ‘b’ in such a way that it has ‘r’ remainder which is smaller than ‘b’. In fact it is a long division method. We also use it to compute the HCF of two positive integers.
The Fundamental Theorem of Arithmetic says us about the expression of positive integers as the product of prime integers. It states that every positive integer is either prime or a product of powers of prime integers. We have already known how to find HCF and LCM of positive integers by using the Fundamental Theorem of Arithmetic in previous class. Now, we will learn about irrationality of many numbers like etc. by applying this theorem. We know that the decimal representation of a rational number is either terminating or if repeating if not terminating. We have to use the Fundamental Theorem of Arithmetic to determine the nature of the decimal expansion of rational numbers. | {"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": 264, "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": 0.9324132204055786, "perplexity": 312.41934024399075}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256980.46/warc/CC-MAIN-20190522223411-20190523005411-00333.warc.gz"} |
http://physicstasks.eu/2173/nitrogen-in-a-vessel | ## Nitrogen in a Vessel
One mole of nitrogen is enclosed in a vessel. Average translational kinetic energy of the nitrogen molecules is 3.5 kJ. Determine the most probable speed of nitrogen molecules after increasing temperature in the vessel by 200 K.
• #### Hint 1
Using the formula for average molecular kinetic energy of a gas $$\bar{E}_k$$, express the temperature T1 of nitrogen before heating.
• #### Hint 2: The Most Probable Speed
To calculate the most probable speed vp of the heated nitrogen molecules, use the following relation:
$v_p=\sqrt{\frac{2RT_2}{M_m}},$
where Mm is the molar mass of nitrogen, T2 is the temperature of the heated nitrogen and R is the molar gas constant.
• #### Notation
$$\bar{E}_k=3.5\,\mathrm{kJ}=3500\,\mathrm{J}$$ average translational kinetic energy of nitrogen molecules n = 1 mol amount of nitrogen ΔT = 200 K increase in temperature vp = ? the most probable speed
From the Tables:
Mm = 28 g mol−1 = 0.028 kg mol−1 molar mass of nitrogen R = 8.31 J mol−1K−1 molar gas constant
• #### Analysis
First we express the temperature of nitrogen before heating from the formula for the average kinetic energy of a gas, according to which the energy is directly proportional to the thermodynamic temperature. After that we add the temperature change to the calculated initial temperature and obtain the temperature of the heated nitrogen, which we subsequently substitute in the formula for the most probable molecular speed.
• #### Solution
The average kinetic energy $$\bar{E}_k$$ of any gas is directly proportional to its thermodynamic temperature T. The following relation applies
$\bar{E}_k=\frac{3}{2}nRT,$
where n is the amount of the gas and R is the molar gas constant.
From this formula we can easily determine temperature T1 of the nitrogen before heating:
$T_1=\frac{2\bar{E}_k}{3nR}.$
For temperature T2 of the heated nitrogen it holds true
$T_2=T_1+\mathrm{\Delta}T=\frac{2\bar{E}_k}{3nR}+\mathrm{\Delta}T.$
Finally, to calculate the most probable speed vp of the nitrogen molecules, we use the known formula
$v_p=\sqrt{\frac{2RT_2}{M_m}},$
where Mm is the molar mass of the nitrogen, and substitute for the obtained expression of temperature T2:
$v_p=\sqrt{\frac{2R}{M_m}\left(\frac{2\bar{E}_k}{3nR}+\mathrm{\Delta} T\right)}.$
• #### Numerical Substitution
$v_p=\sqrt{\frac{2R}{M_m}\left(\frac{2\bar{E}_k}{3nR}+\mathrm{\Delta} T\right)}=\sqrt{\frac{2\cdot{8.31}}{0.028}\cdot\left(\frac{2\cdot{3500}}{3\cdot{1}\cdot{8.31}}+200\right)}\,\mathrm{m\,s}^{-1}$ $v_p\dot{=}530\,\mathrm{m\,s}^{-1}$
Just out of interest let's also try to determine the most probable speed of the nitrogen molecules before heating:
$v_p=\sqrt{\frac{4\bar{E}_k}{3nM_m}}=\sqrt{\frac{4\cdot{3500}}{3\cdot{1}\cdot{0.028}}}\,\mathrm{m\,s}^{-1}$ $v_p\dot{=}410\,\mathrm{m\,s}^{-1}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9430513381958008, "perplexity": 641.9444535686981}, "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/1568514573735.34/warc/CC-MAIN-20190919204548-20190919230548-00002.warc.gz"} |
https://thecuriousastronomer.wordpress.com/tag/infrared/ | Feeds:
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## Two nearby planets potentially habitable?
This story caught my attention in the last few weeks. It is from the Universe Today website, and here is the link to the story. The two exoplanets in the story, TRAPPIST-1b and TRAPPIST-1c, are only 40 light-years away, which by cosmic standards is very close. They have been studied in near-infrared light using the Hubble Space Telescope’s Wide Field Camera 3 (WFC3), which was put onto Hubble during a last servicing mission in May 2009.
This story appeared recently on the Universe Today website.
By studying the planets in near-infrared light we can look at how sunlight is absorbed by different gases in a planet’s atmosphere. This method was pioneered by Gerard Kuiper working in the 1940s. This article mentions that the atmospheres of these two exoplanets have been shown to be “compact” like the Earth and Venus, rather than “puffy” like the gas giants Jupiter and Saturn.
How can we determine this from near-infrared studies done here? Rather than looking at reflected light from the parent star, instead these studies used the passing of the two planets in front of a background star (not the host star). By looking at the absorption lines produced by the two planets’ atmospheres, not only can the gases in them be determined, but by looking at the details of the absorption lines one can determine the temperature and pressure of the gas. This is an example of how powerful a technique spectroscopy is in determining the physical nature of gases.
To find two planets so nearby which could potentially harbour life is quite exciting. I am surprised this has not been a bigger story in the press.
## What is the scale height of water vapour in the Earth’s atmosphere?
Someone recently asked me what was the scale height of water vapour in the Earth’s atmosphere, so I decided to see if I could find out. The scale height of water vapour is particularly important for infrared, sub-millimetre, millimetre and microwave astronomy, as it is the water vapour in the Earth’s atmosphere which prevents large fractions of these parts of the electromagnetic spectrum from reaching the ground. This is why we can, for example, only study the Cosmic Microwave Background from space or from a few particularly dry places on Earth such as Antarctica and the Atacama desert in Chile.
## What does the term ‘scale height’ mean?
First of all, let me explain what the term “scale height” means. It is the altitude by which one needs to go up for the quantity of something (water, nitrogen, oxygen, carbon dioxide) to go down by a factor of $1/e$, where $e$ is the base of natural logarithms, and $e=2.71828.....$. The scale height, usually written as $H$, is dependent upon the temperature of the gas, the mass of the molecules, and the gravity of the planet. We can write that
$H = \frac{ kT }{ Mg } \text{ (1)}$
where $k$ is Boltzmann’s constant, $T$ is the temperature (in Kelvin), $M$ is the mass of the molecule and $g$ is the value of the acceleration due to gravity. If we were to plot the atmospheric pressure as a function of altitude we find that it follows an exponential, this is because of the differential equation which produces Equation (1) above (I will go into the mathematics of how Equation (1) is derived in a separate blog).
In the case of air, which is some 80% nitrogen molecules and 20% oxygen molecules, the scale height has been well determined and is $7.64 \text{ kilometres}$ (or, to put it another way, it drops by a half every $5.6 \text{ km}$). So, if one were at an altitude of $5.6 \text{ km}$, half of the atmosphere would be below you. Go up another $5.6 \text{ km}$ and it drops by a half again, so at $11.2 \text{ km}$ 75% of the atmosphere is below you.
## What is the scale height of water vapour in the Earth’s atmosphere?
Determining the scale height of water vapour in the Earth’s atmosphere is, I have discovered, essentially impossible. Or, to put it better, it is a meaningless figure. This is because it varies too much. It depends on temperature, so even in a given place it can vary quite a bit. So, instead, we talk of precipitable water vapour (PWV) at a particular place (both location and altitude). PWV is the equivalent height of a column of water if we were to take all the water vapour in the atmosphere above a particular location and it were to precipitate as rain.
The Mount Abu Infrared Observatory in India, for example, is at an altitude of 1,680 metres, and quotes a PWV of 1-2mm in winter. The PWV would be higher in summer, as water sinks in the atmosphere when it is cold. For Kitt Peak in Arizona, which is at an altitude of 2,090 metres, the PWV varies between about 15mm and 25mm. This is why very little infrared astronomy is done at Kitt Peak. For Mauna Kea in Hawaii, which is at an altitude of just over 4,000 metres, it varies between 0.5mm and 2mm. This is why there are a number of infrared, sub-mm and millimetre wave telescopes there.
At the South Pole, which is at an altitude of 2,835 metres, the PWV is measured to be between 0.25 and 0.4 in the middle of the Austral winter (June/July/August). Why is this so much lower than Mauna Kea, even though it it is at a lower altitude? It is because it is so much colder.
The average Precipitable Water Vapour at the South Pole averaged over a 50-year period from 1961 to 2010. Even in the Austral summer it is low, but in the Austral winter (June/July/August) it drops to as low as 0.25 to 0.35mm, one of the lowest values found anywhere on Earth.
High in the Atacama desert, on the Llano de Chajnantor (the Chajnantor plateau), which is at an altitude of 5,000 metres and where ALMA and other millimetre and microwave telescopes are being located, the PWV is typically about 1mm, and drops to as low as 0.25mm some 25% of the time (see e.g. this website). This is why Antarctica and the Atacama desert (in particular the Chajnantor plateau) have become places to study the Cosmic Microwave Background from the Earth’s surface; we need exceptionally dry air for the microwaves to reach the ground.
## Summary
To summarise, it is meaningless to talk about a scale height for water vapour in the Earth’s atmosphere, as the vertical distribution of water vapour not only varies from location to location, but varies at a given location. So, instead, we talk about Precipitable Water Vapour (PWV); the lower this number the drier the air is above our location. To be able to do infrared, sub-millimetre, millimetre and microwave astronomy we need the PWV to be as low as possible, the best sites (Antarctica and the Atacama) get as low as 0.25mm and are usually below 1mm. The exceptionally dry air above Antarctica and the Atacama desert enable us to study the Cosmic Microwave Background from the ground, something we usually have to do from space. | {"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": 14, "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": 0.8509668111801147, "perplexity": 632.3230152615447}, "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/1575540504338.31/warc/CC-MAIN-20191208021121-20191208045121-00005.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=45867&page=4 | by soul814
Tags: physics, vector
P: 35 how did u get the resultant vector?
Sci Advisor HW Helper PF Gold P: 12,016 $$\cos(30)=\frac{\sqrt{3}}{2},\sin(30)=\frac{1}{2}$$ (This is a rather well-known relation; you'll it later on) Hence A has coordinates $$(\frac{3\sqrt{3}}{2},\frac{3}{2})$$ Summing A with B (coord. (0,3)) yields the resultant vector).
P: 35 ahh indeed it does.. this makes sense now :D yes I learned cos(30) = 3/2 in precalculus the 30-60-90 triangle
P: 35 wohoo got B) correct how would you get the degree though?
Sci Advisor HW Helper PF Gold P: 12,016 How do you think?
P: 35 ummm I know all the measurements of the sides yet no angles except that 30 degrees, but its on the outside
Sci Advisor HW Helper PF Gold P: 12,016 But you know its horizontal and vertical coordinates, right? How can you calculate the tangent of the angle using that info?
P: 35 the figure isnt a right triangle though
Sci Advisor HW Helper PF Gold P: 12,016 You are to find the resultant vectur's angle to the x-axis; how can you construct a triangle in such a manner that the coordinates you've been given will help you find that angle?
P: 35 OOO I got it :D extend the Line downward so the verticles of the triangle are $$tan^-^1 = 4.5/2.6$$ (0,0) --> (2.6,0) --> (2.6,4.5)
Sci Advisor HW Helper PF Gold P: 12,016 You should get 60 degrees..
P: 35 Yep I got 60 degrees 59.9 i think. Thanks for your help :D nice i think i got section 3.2 and 3.3 of my text book down ahah... tommo I will probably ask about projectile motion, hopefully you'll help me again. The textbook leaves out alot of information. Like it leaves out little steps.
Sci Advisor HW Helper PF Gold P: 12,016 No problem..
P: 1 its an emergency...i really need to know this..... can two vectors representing two different physical quantities be equal???? can force and displacement be equal vectors if they are in same direction and have same magnitude?? does'nt representing different phy. quantities makes them different vectors???
P: 317 I hate vectors! I'm in physics A and I don't get it, even though I understand all of the force, energy, and projectile motion stuff.
Related Discussions General Math 3 Introductory Physics Homework 3 Calculus & Beyond Homework 9 Introductory Physics Homework 13 Calculus 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": 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": 0.9323037266731262, "perplexity": 2716.1148349808977}, "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/1409535920694.0/warc/CC-MAIN-20140909054037-00432-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/192334-finding-radius-convergence.html | # Thread: Finding the Radius of Convergence
1. ## Finding the Radius of Convergence
"Find the radius of convergence for g(x) = the series x^n divided by 3n + 1, where n = 0 and goes to infinity. Be sure to check the endpoints for convergence, and state which tests you are using."
I'm pretty lost on this guys. My notes don't really elaborate on what the radius of convergence is or how you figure it out. Any help is really appreciated. Thanks in advance everyone!
2. ## Re: Finding the Radius of Convergence
Originally Posted by bobsanchez
"Find the radius of convergence for g(x) = the series x^n divided by 3n + 1, where n = 0 and goes to infinity. Be sure to check the endpoints for convergence, and state which tests you are using."
Use the nth root test $\sqrt[n]{{\frac{{\left| x \right|^n }}{{3n + 1}}}} \to \left| x \right| < 1$.
Now solve that for x. BUT check the endpoints.
3. ## Re: Finding the Radius of Convergence
I'm sorry, but I still don't really understand. I don't know how to solve for x in that situation, or what that really means. I know it's dense of me, but this material hasn't been covered at all and I can't find a good source on it.
4. ## Re: Finding the Radius of Convergence
Originally Posted by bobsanchez
I'm sorry, but I still don't really understand. I don't know how to solve for x in that situation, or what that really means. I know it's dense of me, but this material hasn't been covered at all and I can't find a good source on it.
You are expected to be able to do first year algebra before working with series.
Can you solve $|x|<1~?$
If not, you need to work on your basic algebra skills before you try to proceed with series.
5. ## Re: Finding the Radius of Convergence
Originally Posted by Plato
You are expected to be able to do first year algebra before working with series.
Can you solve $|x|<1~?$
If not, you need to work on your basic algebra skills before you try to proceed with series.
What do I set it equal to to solve for the x in that formula you posted? That's what I don't understand.
6. ## Re: Finding the Radius of Convergence
Alright, sorry for wasting your time. I'll look elsewhere for help.
7. ## Re: Finding the Radius of Convergence
what do you know about the ratio test for convergence/divergence and using it to find the interval of convergence?
8. ## Re: Finding the Radius of Convergence
Originally Posted by skeeter
what do you know about the ratio test for convergence/divergence and using it to find the interval of convergence?
I know the ratio test works for determining whether a series has convergence/divergence, but I'm not familiar with using it to determine the interval of convergence.
10. ## Re: Finding the Radius of Convergence
Originally Posted by skeeter
Nice, excellent video. So basically do the ratio test, and under those circumstances you can pull the x out, turn it into an inequality, and solve, then plug in to check endpoints...okay, I think I understand it now.
Thanks so much! | {"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": 0.9330219626426697, "perplexity": 382.9381653565152}, "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-2017-39/segments/1505818687906.71/warc/CC-MAIN-20170921205832-20170921225832-00620.warc.gz"} |
https://www.physicsforums.com/threads/tension-and-wave-motion.267925/ | # Tension and wave motion
1. Oct 29, 2008
### beanieb
1. The problem statement, all variables and given/known data
A large ant is standing on the middle of a circus tightrope that is stretched with tension T. The rope has mass per unit length mu (no symbl). Wanting to shake the ant off the rope, a tightrope walker moves her foot up and down near the end of the tightrope, generating a sinusoidal transverse wave of wavelength lambda and amplitude A . Assume that the magnitude of the acceleration due to gravity is g. What is the minimum wave amplitude such that the ant will become momentarily weightless at some point as the wave passes underneath it? Assume that the mass of the ant is too small to have any effect on the wave propagation.
Express the minimum wave amplitude in terms of T, mu, lambda, g and .
2. Relevant equations
I am assuming the wave is moving with y(x,t)=Asin(wt-kx)
3. The attempt at a solution
I knowthe ant will become weightless when the normal force between the string and the ant becomes zero. This means that I have to find when the maximum accel = -g. So if i differentiate the wave twice to get accel and let this equal to -g i get (unless im wrong)
-Aw2(Sin wt)= -g. Dont know how im supposed to relate this back to tension and mass per unit length or even if im doing this right. Please help
2. Oct 29, 2008
### beanieb
also just thought v=sqrt(T/mu). is Amax = 4(pi)^2 *v^2*?
3. Oct 29, 2008
### beanieb
w^2= 4(pi)^2*(T/mu)
subbing into -g=-Aw^2 Sin wt and rearranging gives:
(4(pi)^2*T)/mu*A=g*Sin (wt)
How do i relater sin(wt) back into the variables I have?
4. Oct 29, 2008
### beanieb
relate sine wt back to lambda? sorry im posting so muchh im working this out as i go
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https://chemistry.stackexchange.com/questions/5098/tetrahedron-is-minimum-energy-conformation-for-atom-with-4-electron-domains-in-v | Background
A student asked me to prove that the regular tetrahedron is the minimum energy geometry available to describe the locations of the electron domains in three dimensions – the prediction of valence shell electron pair repulsion (VSEPR) theory for a central atom with 4 mutually repelling electron domains. With the tetrahedron demonstrated, it could then be shown using vector multiplication that the angles between all substituent atoms is $109.5^{\circ}$.
Some ideas
I can think of a few ways to start this proof:
• As mentioned above, we could define a regular tetrahedron and work backwards to see if the distance between all vertices is equal and maximized. Start with the vertices: A(0, 0, 0), B(1, 1, 0), C(1, 0, 1) and D(0, 1, 1), and a central point O($\frac{1}{2}$, $\frac{1}{2}$, $\frac{1}{2}$). With these, we can get vectors that point to each vertex. Call them $\vec{r_1}$ through $\vec{r_4}$. We remember that we can calculate the angle between any two vectors using the geometric and algebraic definitions of the dot product. I won't bore you with the details of this calculation but if anyone wishes to see it, I would be happy to share it. If you do it you will see that all the angles are equal and satisfy: $\cos^{-1}\left(-\frac{1}{3}\right) = \theta_i \approx 109.47122^{\circ}$. From here we would need to calculate the six distances bewteen the four vertices to show that the distances are all equal.
• Using multivariable calculus, we could treat this as an optimization problem subject to the constraint matrix, $g$; the sum square distance function, $f$; and use the optimization formalism described by Lagrange. This results in a system of 13 unknowns – the 4 points with three coordinates each plus $\lambda$ – and 13 equations – the four constraint equations plus the nine partial derivative equations from the Lagrange method:
$$g = 1 = \left\{\begin{array}{c}g_1(x_1,y_1,z_1) = {x_1}^2 + {y_1}^2 + {z_1}^2\\ g_2(x_2,y_2,z_2) = {x_2}^2 + {y_2}^2 + {z_2}^2\\ g_3(x_3,y_3,z_3) = {x_3}^2 + {y_3}^2 + {z_3}^2\\ g_2(x_4,y_4,z_4) = {x_4}^2 + {y_4}^2 + {z_4}^2\\ \end{array}\right.$$ $$f = \mathop{\sum_{i=1}^{3}\sum_{j=1}^{3}}_{i<j}\left[\left({x_i}-{x_j}\right)^2+\left({y_i}-{y_j}\right)^2+\left({z_i}-{x_j}\right)^2\right]$$ $$\nabla f=-\lambda \nabla g$$
• An approach similar to the one outlined above could be used with spherical coordinates instead. This version reduces the set of variables from 13 to 8 (the four sets of two angles for each point $\theta$'s and $\phi$'s; note the spherical radius could be set to 1 or any other bond distance, $a$, so is not a variable). These could be further reduced using symmetry arguments. For example, if we take the first vertex at the point $(0, 0, 1)$, we know based on symmetry that if we look down the z axis at the xy-plane, the angle between the three other vectors must have an angle of $120^{\circ}$ with respect to one another. Let one of the other vertices have the point $(x, 0, -\sqrt{1 - x^2})$ – satisfying the radius of 1 unit constraint we have imposed. Now based on symmetry, we know that the other two points must be $(x\cos(\frac{2\pi}{3}),x\sin(\frac{2\pi}{3}),-\sqrt{1-x^2})$ and $(x\cos(\frac{4\pi}{3}),x\sin(\frac{4\pi}{3}),-\sqrt{1-x^2})$ using spherical symmetry. From here, we can write a formula for the total sum square distance between all the points and set the derivative with respect to $x$ to zero to find the maximum sum square distance. Notice we have reduced the problem with these symmetry arguments from a multivariable calculus problem to a calculus BC problem. Alternatively, we could use some other guess and check method to solve for $x$. Note, we should find the same answer using this method as the others above (and below). | {"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": 0.891074001789093, "perplexity": 172.594859867509}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257767.49/warc/CC-MAIN-20190524204559-20190524230559-00205.warc.gz"} |
https://math.stackexchange.com/questions/3008454/determining-finitude-or-infinitude-from-a-simple-geometric-construction | # Determining finitude or infinitude from a simple geometric construction
Playing with a pencil on a checkered sheet I encountered this construction:
1) take a point $$A$$ on the grid and a point $$B$$ that is distant from $$A$$ $$n=2,3,4...$$ horizontal steps and $$1$$ vertical step, so that $$\overline{AB}=\sqrt{n^2+1}$$.
2) complete a square $$ABCD$$ on the grid starting from the segment $$AB$$. This is the square $$Q_0$$.
3) Starting from $$A$$ take a point $$A_1$$ at the first intersection of $$AB$$ with a line of the grid. And do the same starting from the other points $$B,C,D$$ finding the points $$B_1,C_1,D_1$$ .
4) connect the points $$A_1, B_1,C_1,D_1$$, so that we have a new square $$Q_1$$.
5) redo the same points $$3$$ and $$4$$ and find the square $$Q_2$$ and so again...
The result is a sequence $$S_n$$ of squares whose sides, orientation, and number of different elements depend from $$n$$.
The figures show the construction for $$n=2$$, and for $$n=3$$.
Note that it seems that the sequence $$S_2$$ has infinite different elements that converge to a square of the grid, but $$S_3$$ has a finite number of different elements.
So a first question is:
How we can prove that the number of different elements of $$S_n$$ is finite or not and, if finite, what is this number?
A second question is to find the length of the last square in the sequence, or the limit of the lengths if the sequence has infinite different elements.
More general, the side lengths of all the squares in a sequence $$S_n$$ can be determined in some way that does not require a terribly boring calculation?
Added: construction of the first six elements of $$S_4$$.
• I already love this question before understanding it. But to help the understanding: could you please mark the original $A$ and $B$ (and perhaps a few of the mentioned other points) in the drawing? – Vincent Nov 21 '18 at 22:17
• Given the symmetry of the construction you can mark as $A$ and $B$ any two extremes of a side of the greatest square, and $C, D$ are the other two vertices. $A_1$ is than the vertex of the second square that stay on $AB$ , $B_1$ the vertex on $BC$ and so one... – Emilio Novati Nov 21 '18 at 22:23
• I added the construction for $S_4$ with some marks. – Emilio Novati Nov 21 '18 at 22:59
• Yes, thank you! Now it is completely clear. I somehow thought you were working form inside to outside and got confused, but now that I see you work from outside in, it all makes sense – Vincent Nov 22 '18 at 9:48 | {"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": 28, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8493713140487671, "perplexity": 240.57450438828388}, "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/1563195525374.43/warc/CC-MAIN-20190717181736-20190717203736-00061.warc.gz"} |
http://sankhya.isical.ac.in/articles/377 | ## Article
#### Title: Testing Composite Hypothesis Based on the Density Power Divergence
##### Issue: Volume 80 Series B Part 2 Year 2018
###### Abstract
In any parametric inference problem, the robustness of the procedure is a real concern. A procedure which retains a high degree of efficiency under the model and simultaneously provides stable inference under data contamination is preferable in any practical situation over another procedure which achieves its efficiency at the cost of robustness or vice versa. The density power divergence family of Basu et al. (Biometrika 85, 549–559 1998) provides a flexible class of divergences where the adjustment between efficiency and robustness is controlled by a single parameter β. In this paper we consider general tests of parametric hypotheses based on the density power divergence. We establish the asymptotic null distribution of the test statistic and explore its asymptotic power function. Numerical results illustrate the performance of the theory developed. | {"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": 0.8276592493057251, "perplexity": 751.3992690098413}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257432.60/warc/CC-MAIN-20190523224154-20190524010154-00084.warc.gz"} |
https://link.springer.com/chapter/10.1007/978-1-4612-0289-9_2 | # Formal Languages and Formal Logic
• Howard Straubing
Part of the Progress in Theoretical Computer Science book series (PTCS)
## Abstract
Throughout this book we use sentences of formal logic to describe properties of words over a finite alphabet A. A sentence will thus define a language $$L \subseteq A^{*}$$; L is the set of all words that have the property described by the sentence.
## Keywords
Formal Language Free Variable Atomic Formula Predicate Symbol Numerical Relation
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves. | {"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": 0.8441544771194458, "perplexity": 821.7037234571129}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376829568.86/warc/CC-MAIN-20181218184418-20181218210418-00483.warc.gz"} |
http://aas.org/archives/BAAS/v27n4/aas187/S088002.html | Session 88 - Structure and Kinematics of Galaxies.
Oral session, Wednesday, January 17
La Condesa, Hilton
## [88.02] Chaos in Triaxial Elliptical Galaxies
M. Valluri, D. Merritt (Rutgers U.)
HST observations of elliptical galaxies have revealed the presence of strong cusps or black holes at their centers. We investigate the development of chaos in triaxial elliptical galaxies arising from the presence of central concentrations. The chief influence of a cusp or black hole is the disruption of the regular box orbits which are believed to form the ``backbone'' of integrable triaxial models. The Liapunov characteristic numbers for a large number of stochastic orbits tend to converge to a single value after a long time (\sim 10^4 orbital times). The Liapunov times (which measure the time scales for divergence of two nearby orbits) are typically 3-5 crossing times. This confirms the conventional wisdom that all stochastic regions of phase space are connected and that slow diffusion occurs between the different regions. Large ensembles of stochastic orbits chosen to have identical energies, mix on timescales of less than \sim 100 orbital times and reach a near-invarient density distribution. These results suggest that the stochastic regions of phase space in most ellipticals are likely to be fully mixed, which strongly constrains the existance of self-consistent triaxial equilibria. | {"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": 0.8566760420799255, "perplexity": 1721.1827872051358}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997894289.49/warc/CC-MAIN-20140722025814-00101-ip-10-33-131-23.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/260705/isomorphic-cyclic-quotient-group | # Isomorphic cyclic quotient group
Denote $R=\langle6\rangle$ and $S=\langle30\rangle$.
Consider the subgroups $R$ and $S$ 0f the set of integers $\mathbb{Z}$.
Show that $R/S \cong \mathbb{Z}_5$.
$(R/S$ is the set of left cosets of $S$ over $R$, $\mathbb{Z}_5$ denotes congruence class $\pmod 5.)$
Define a map $f:R/S \to \mathbb{Z}_5$ by $f(6m+30\mathbb{Z})=(m+5\mathbb{Z})$.
I have shown the following:
1. $f$ is well-defined.
2. $f$ is a homomorphism.
3. $f$ is surjective or onto.
Now, to show that $f$ is injective, I assume that $$f(6m+30\mathbb{Z})\equiv f(6n+30\mathbb{Z}) \pmod 5$$ My question is: will it imply that $m \equiv n \pmod 5$?
If yes, please show me how.
-
Unless I am misunderstanding, you are assuming that $$(*)\,\,\,\,\,f(6m+30\mathbb{Z}) = f(6n+30\mathbb{Z})$$ in $\mathbb{Z}/5\mathbb{Z}$. By definition, $f(6m + 30\mathbb{Z}) = m + 5\mathbb{Z}$ and $f(6n+30\mathbb{Z}) = n + 5\mathbb{Z}$, so assumption ($*$) says exactly that $m + 5\mathbb{Z} = n + 5\mathbb{Z}$, or that $m\equiv n\pmod{5}$.
But that doesn't show that $f$ is injective. To show that $f$ is injective, you must show that $6m + 30\mathbb{Z} = 6n + 30\mathbb{Z}$ in $R/S$. Indeed, since $m\equiv n\pmod{5}$, there is some integer $k$ such that $n = m + 5k$. Thus $6n = 6m + 30k$, so that $6m+30\mathbb{Z} = 6n + 30\mathbb{Z}$. | {"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": 0.9973586797714233, "perplexity": 102.7952538023257}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783396100.16/warc/CC-MAIN-20160624154956-00010-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/454102/divergence-transforms-as-scalar-under-rotation-in-2d-intuition | # Divergence transforms as scalar under rotation in 2D + intuition
Problem is as follows:
In two dimensions, show that the divergence transforms as a scalar under rotations.
Aim is to determine $\bar{v}_{y}$ and $\bar{v}_{z}$, and show that
$\frac{\partial\bar{v}_{y}}{\partial\bar{y}}$ + $\frac{\partial\bar{v}_{z}}{\partial\bar{z}} = \frac{\partial v_{y}}{\partial y} + \frac{\partial v_{y}}{\partial z}$
I'm not sure at all why the above shows that divergence transforms as a scalar under rotations. It was a part of the question (given as a hint) so I was just trying to solve it without really understanding what I was doing. Any help on clarifying why I do need to show that would be appreciated.
Since this is a rotation in two dimensions (in the $y$ and $z$ axis),
$\left( {\begin{array}{cc} \bar{v}_{y} \\ \bar{v}_{z} \end{array} } \right)$ = $\left( {\begin{array}{cc} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi \end{array} } \right)$ $\left( {\begin{array}{cc} {v}_{y} \\ v_{z} \end{array} } \right)$.
Expanding this out, I found that $\bar{v}_{y} = v_{y}.\cos\phi + v_{z}.\sin\phi$ and $\bar{v}_{z} = -v_{y}.\sin\phi + v_{z}.\cos\phi$.
Solving for $v_{y}$ and $v_{z}$, by multiplying $\bar{v}_{y}$ and $\bar{v}_{z}$ by $\sin\phi$ and $\cos\phi$, I was able to use the $\sin^{2}\phi + \cos^{2}\phi = 1$ identity to get
$v_{z} = \bar{v}_{y}.\sin\phi + \bar{v}_{z}.\cos\phi$ and
$v_{y} = \bar{v}_{y}.\cos\phi - \bar{v}_{z}.\sin\phi$.
Next, I found the components of the original equation:
The first partial derivative in the equation was determined as
$\frac{\partial \bar{v}_{y}}{\partial \bar{y}} = (\frac{\partial \bar{v}_{y}}{\partial y})(\frac{\partial y}{\partial \bar{y}}) + (\frac{\partial \bar{v}_{y}}{\partial z})(\frac{\partial z}{\partial \bar{y}})$
and the second as
$\frac{\partial \bar{v}_{z}}{\partial \bar{z}} = (\frac{\partial \bar{v}_{z}}{\partial y})(\frac{\partial y}{\partial \bar{z}}) + (\frac{\partial \bar{v}_{z}}{\partial z})(\frac{\partial z}{\partial \bar{z}})$.
At this point, I become stuck because I cannot seem to be able to find $\partial \bar{v}_{y}$ and $\partial \bar{v}_{z}$ with respect to $\partial y$ and $\partial z$.
How to I continue? Thanks in advance.
-
Divergence of a vector is a scalar; and a scalar is a constant and doesn't change under rotations, so if you transform all your variables under a rotation and then calculate the divergence of the vector in the new coordinates, the divergence must remain unchanged.
First find $\bar v_y$ and $\bar v_z$ from the matrix transformation relation:
$\left( {\begin{array}{cc} \bar{v}_{y} \\ \bar{v}_{z} \end{array} } \right)$ = $\left( {\begin{array}{cc} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi \end{array} } \right)$ $\left( {\begin{array}{cc} {v}_{y} \\ v_{z} \end{array} } \right) \to$
$$\bar{v}_{y} = v_{y}.\cos\phi + v_{z}.\sin\phi$$ $$\bar{v}_{z} = -v_{y}.\sin\phi + v_{z}.\cos\phi$$
Then find their derivative w.r.t. $\bar y$ and $\bar z$, as is needed: $$\frac{\partial \bar v_y}{\partial \bar y}=\frac{\partial v_y}{\partial \bar y}\cos \phi+\frac{\partial v_z}{\partial \bar y}\sin \phi \tag{1}$$
expanding the right hand side derivatives $\frac{\partial {v}_{y}}{\partial \bar{y}}$ and $\frac{\partial {v}_{z}}{\partial \bar{y}}$ as:
$$\cases{\frac{\partial {v}_{y}}{\partial \bar{y}} = (\frac{\partial {v}_{y}}{\partial y})(\frac{\partial y}{\partial \bar{y}}) + (\frac{\partial{v}_{y}}{\partial z})(\frac{\partial z}{\partial \bar{y}}) \\ \frac{\partial {v}_{z}}{\partial \bar{y}} = (\frac{\partial {v}_{z}}{\partial y})(\frac{\partial y}{\partial \bar{y}}) + (\frac{\partial{v}_{z}}{\partial z})(\frac{\partial z}{\partial \bar{y}})} \tag{2}$$ we should now find the derivatives $\frac{\partial y}{\partial \bar y}$ and $\frac{\partial z}{\partial \bar y}$. Using the rotation formula in 2D to write $y$ and $z$ in terms of $\bar y$ and $\bar z$, we can find $\frac{\partial y}{\partial \bar y}$ and $\frac{\partial z}{\partial \bar y}$ : $$\cases{ y=\bar y \cos \phi-\bar z \sin \phi\\z=\bar y \sin \phi+ \bar z \cos \phi}\to\cases{\frac{\partial y}{\partial \bar y}=\cos \phi\\ \frac{\partial z}{\partial \bar y}=\sin \phi}\tag{3}$$ now substituting $(3)$ in $(2)$ and then $(2)$ in $(1)$, we will have $$\frac{\partial \bar v_y}{\partial \bar y}=\left(\frac{\partial v_y}{\partial y}\cos\phi+\frac{\partial v_y}{\partial z}\sin\phi \right )\cos\phi + \left( \frac{\partial v_z}{\partial y}\cos\phi+\frac{\partial v_z}{\partial z}\sin\phi \right)\sin \phi$$
doing the same for $\frac{\partial \bar v_z}{\partial \bar z}$, we will arrive at:
$$\frac{\partial \bar v_z}{\partial \bar z}=-\left(-\frac{\partial v_y}{\partial y}\sin\phi+\frac{\partial v_y}{\partial z}\cos\phi \right )\sin\phi + \left(- \frac{\partial v_z}{\partial y}\sin\phi+\frac{\partial v_z}{\partial z}\cos\phi \right)\cos \phi$$
Now just sum up the two terms$\frac{\partial\bar{v}_{y}}{\partial\bar{y}}$ and $\frac{\partial\bar{v}_{z}}{\partial\bar{z}}$ and apply $\sin^{2}\phi + \cos^{2}\phi = 1$ to arrive at the final result: $$\frac{\partial\bar{v}_{y}}{\partial\bar{y}} + \frac{\partial\bar{v}_{z}}{\partial\bar{z}} = \frac{\partial v_{y}}{\partial y} + \frac{\partial v_{y}}{\partial z}$$
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Thanks, I understand why I am meant to show the equivalence. However, would you mind explaining your working? I'm afraid I'm having difficulty following from when you expanded the r.h.s. derivatives. – achacttn Jul 28 '13 at 15:30
@dlckd I think it is clear now. – Mostafa Jul 28 '13 at 17:39
Here is an answer from a completely different point of view (=the "geometric view"). For a transformation $\Phi:U\mapsto V$, define the push forward $\Phi_*$ for different types of objects as follows:
• $(\Phi_* f)(y):=f\circ \Phi^{-1}(y)$ for a function $f:U\mapsto\mathbb R$
• $(\Phi_* v)(y):=(\Phi'v)\circ \Phi^{-1}(y)$ for a vector field $v:U\mapsto\mathbb R^n$
• $(\Phi_* p)(y):=(\frac{p}{|\det \Phi'|})\circ \Phi^{-1}(y)$ for a density $p:U\mapsto\mathbb R$
For a density $p$ and a vector field $v$, define the divergence of $v$ relative to $p$ as the function $$\operatorname{div}_p v:=\frac{1}{p}\sum_{i=1}^n\frac{\partial[pv]_i}{\partial x_i}$$ The usual divergence is then just given by $\operatorname{div}_1 v$ (or more generally by $\operatorname{div}_\alpha v$ for any $\alpha\in\mathbb R$). Now we can prove that $$\Phi_*(\operatorname{div}_p v)=\operatorname{div}_{(\Phi_*p)} (\Phi_*v)$$ where $\operatorname{div}_p v$ is transformed like a function, $v$ is transformed like a vector field and $p$ is transformed like a density. The most straightforward way to prove this is to first show that $$\int pdx\;L_v\varphi = -\int pdx\;\varphi \operatorname{div}_p v$$ for arbitrary smooth test-functions $\varphi$ with compact support, and then use the known trivial transformation behavior of the directional derivative $L_v$ (or call it Lie-derivative if you want) and the known transformation behavior of the integral to derive the transformation behavior of the divergence $\operatorname{div}_p v$. A long time ago, I wrote a German text where this is explained in more detail with some pictures, some applications and a bit more context.
The connection to the question at hand is that $\det \Phi'=1$ if $\Phi$ is a rotation, hence $\Phi_*1=1$ if $1$ is considered as a density, and hence $\Phi_*(\operatorname{div}_1 v)=\operatorname{div}_1 (\Phi_*v)$.
More generally, we don't even need $\det \Phi'=1$, it's enough that $\det \Phi'$ is constant, which is satisfied for any linear tranformation $\Phi$.
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At this point, I become stuck because I cannot seem to be able to find $\partial \bar{v}_{y}$ and $\partial \bar{v}_{z}$ with respect to $\partial y$ and $\partial z$.
You already determined earlier that $\bar{v}_{y} = v_{y}.\cos\phi + v_{z}.\sin\phi$ and $\bar{v}_{z} = -v_{y}.\sin\phi + v_{z}.\cos\phi$. Hence
$\frac{\partial \bar{v}_{y}}{\partial y}=\frac{\partial v_{y}}{\partial y}.\cos\phi + \frac{\partial v_{z}}{\partial y}.\sin\phi$
$\frac{\partial \bar{v}_{y}}{\partial z}=\frac{\partial v_{y}}{\partial z}.\cos\phi + \frac{\partial v_{z}}{\partial z}.\sin\phi$
and
$\frac{\partial \bar{v}_{z}}{\partial y} = -\frac{\partial v_{y}}{\partial y}.\sin\phi + \frac{\partial v_{z}}{\partial y}.\cos\phi$
$\frac{\partial \bar{v}_{z}}{\partial z} = -\frac{\partial v_{y}}{\partial z}.\sin\phi + \frac{\partial v_{z}}{\partial z}.\cos\phi$
However, this is boringly obvious, so you probably meant to ask something else instead.
At this point, I become stuck because I cannot seem to be able to find $\partial \bar{y}$ and $\partial \bar{z}$ with respect to $\partial y$ and $\partial z$.
We have $\left( {\begin{array}{cc} \bar{y} \\ \bar{z} \end{array} } \right)$ = $\left( {\begin{array}{cc} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi \end{array} } \right)$ $\left( {\begin{array}{cc} y \\ z \end{array} } \right)$, hence
$\frac{\partial \bar{y}}{\partial y}=\cos\phi$
$\frac{\partial \bar{y}}{\partial z}=\sin\phi$
and
$\frac{\partial \bar{z}}{\partial y}=-\sin\phi$
$\frac{\partial \bar{z}}{\partial z}=\cos\phi$
- | {"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": 0.9399409294128418, "perplexity": 117.06580268148049}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049276780.5/warc/CC-MAIN-20160524002116-00027-ip-10-185-217-139.ec2.internal.warc.gz"} |
http://www.math-only-math.com/definition-of-ellipse.html | # Definition of Ellipse
We will discuss the definition of ellipse and how to find the equation of the ellipse whose focus, directrix and eccentricity are given.
An ellipse is the locus of a point P moves on this plane in such a way that its distance from the fixed point S always bears a constant ratio to its perpendicular distance from the fixed line L and if this ratio is less than unity.
An ellipse is the locus of a point in a plane which moves in the plane in such a way that the ratio of its distance from a fixed point (called focus) in the same plane to its distance from a fixed straight line (called directrix) is always constant which is always less than unity.
The constant ratio usually denoted by e (0 < e < 1) and is known as the eccentricity of the ellipse.
If S is the focus, ZZ' is the directrix and P is any point on the ellipse, then by definition
$$\frac{SP}{PM}$$ = e
⇒ SP = e PM
The fixed point S is called a Focus and the fixed straight line L the corresponding Directrix and the constant ratio is called the Eccentricity of the ellipse.
Solved example to find the equation of the ellipse whose focus, directrix and eccentricity are given:
Determine the equation of the ellipse whose focus is at (-1, 0), directrix is 4x + 3y + 1 = 0 and eccentricity is equal to $$\frac{1}{√5}$$.
Solution:
Let S (-1, 0) be the focus and ZZ' be the directrix. Let P (x, y) be any point on the ellipse and PM be perpendicular from P on the directrix. Then by definition
SP = e.PM where e = $$\frac{1}{√5}$$.
⇒ SP$$^{2}$$ = e$$^{2}$$ PM$$^{2}$$
⇒ (x + 1)$$^{2}$$ + (y - 0)$$^{2}$$ = $$(\frac{1}{\sqrt{5}})^{2}[\frac{4x + 3y + 1}{\sqrt{4^{2} + 3^{2}}}]$$
⇒ (x + 1)$$^{2}$$ + y$$^{2}$$ = $$\frac{1}{25}$$$$\frac{4x + 3y + 1}{5}$$
⇒ x$$^{2}$$ + 2x + 1 + y$$^{2}$$ = $$\frac{4x + 3y + 1}{125}$$
⇒ 125x$$^{2}$$ + 125y$$^{2}$$ + 250x + 125 = 0, which is the required equation of the ellipse.
The Ellipse | {"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": 0.9682300686836243, "perplexity": 334.40780463858835}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812405.3/warc/CC-MAIN-20180219052241-20180219072241-00071.warc.gz"} |
https://www.transtutors.com/questions/a-where-in-the-plane-of-these-coils-could-the-magnetic-field-be-zero-1-only-inside--3923896.htm | # (a) Where, in the plane of these coils, could the magnetic field be zero: (1) only inside the...
(a) Where, in the plane of these coils, could the magnetic field be zero: (1) only inside the smaller one, (2) only between the inner and outer one, (3) only outside the larger one, or (4) inside the smaller one and outside the larger one?
(b) The larger one is a 200-turn coil of wire with a radius of 9.50 cm and carries a current of 11.5 A. The second one is a 100-turn coil with a radius of 2.50 cm. Determine the current in the inner coil so the magnetic field at their common center is zero. Neglect the Earth’s field.
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Free Plagiarism Checker | {"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": 0.8122033476829529, "perplexity": 911.8445257124339}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487617599.15/warc/CC-MAIN-20210615053457-20210615083457-00009.warc.gz"} |
http://mathoverflow.net/questions/158434/moduli-of-flag-varieties | # Moduli of flag varieties
I work over an algebraically closed field $k$ of characteristic zero.
Recall that a flag variety is a projective variety which is a homogeneous space for some semisimple algebraic group. Every flag variety is of the form $G/P$, where $G$ is a semisimple algebraic group and $P$ is a parabolic subgroup. It seems to me that this data is "discrete", so I expect flag varieties to have discrete moduli. Moreover flag varieties are Fano varieties, and it is known that there are only finitely many deformation types of Fano varieties of fixed dimension. This leads to my question.
For each $n \in \mathbb{N}$, are there are only finitely flag varieties of dimension $n$, up to isomorphism?
A proof/disproof or a reference for this would be much appreciated. By "up to isomorphism" I mean up to isomorphism as an algebraic variety.
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Yes. Use classification of semisimple algebraic groups in terms of Dynkin diagrams + classification of parabolic subgroups in terms of subsets of Dynkin diagrams. Lower bound dimension of flag variety in terms of dimension of group, giving finiteness. – Will Sawin Feb 23 '14 at 18:15
The thing that is worrying me is something like the possible existence of an infinite sequence $P_i \subset G_i$ ($i=1,\ldots,\infty$) where $\dim G_i \to \infty$, but $\dim G_i/P_i = n$ is fixed. In which case one needs to show that there are only finitely many isomorphism classes of varieties in the set $\{G_i/P_i: i=1,\ldots,\infty\}$. Does your approach achieve this? – Daniel Loughran Feb 23 '14 at 18:39
Ridiculous observation: in positive characteristic $p$, for every integer $n$, the following smooth, projective, $3$-dimensional variety, $\{([X_0,X_1,X_2],[Y_0,Y_1,Y_2])\in \mathbb{P}^2\times \mathbb{P}^2 : X_0Y_0^{p^n} + X_1Y_1^{p^n} + X_2Y_2^{p^n}\}$, is homogeneous for an action of $\textbf{SL}_3$. – Jason Starr Feb 23 '14 at 20:11
@Daniel, Ben Webster has already given a slick conceptual proof, but at least in this setting (alg closed, char 0) an easy direct proof comes by treating semisimple $G$'s as a list of examples. You can just use Will's/Jim's suggestion. (After reducing the problem from semisimple to simple there are only three (or four) infinite series of flag varieties: $SL_n/P$, $Sp_{2n}/P$, and $SO_n/P$. By direct inspection, each of these has dimension $\to \infty$ as $n\to\infty$.) – Dave Anderson Feb 24 '14 at 0:14
@pmath. Flag varieties are examples of Fano varieties. Thus they live in the moduli of Fano varieties, i.e., "anti-canonically polarized varieties". Although the moduli of flag varieties is zero-dimensional, the moduli of Fano's is positive dimensional in general. Consider, for example, Del Pezzo surfaces of degree 1,2,3, or 4. – Ariyan Javanpeykar Feb 27 '14 at 8:15
Yes, there are only finitely many. One only needs to observe that for a given group, the variety $G/P$ has dimension at least the rank of the group $G$ (the dimension of a maximal torus). You can see this by inspecting cases by hand, but there's also a conceptual reason: the maximal torus of the adjoint group $G$ acts on $G/P$ faithfully with isolated fixed points, so it acts faithfully on the tangent space of some fixed point. Since a faithful module over a torus must have dimension $\geq$ that of the torus, this establishes the desired result.
Thus, for a given dimension $n$, there can only be flag manifolds for groups with rank $\leq n$, there are finitely many of these, and only finitely many flag manifolds for each one. Q.E.D. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8610535264015198, "perplexity": 225.18232436626192}, "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-2015-32/segments/1438042989018.48/warc/CC-MAIN-20150728002309-00094-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/243465/does-the-ring-of-global-sections-functor-on-the-category-of-locally-ringed-space | # Does the ring of global sections functor on the category of locally ringed spaces have an adjoint functor?
Let $Rng$ be the category of commutative rings. Let $Loc$ be the category of locally ringed spaces. Let $(X, \mathcal{O}_X)$ be an locally ringed space. Then $\Gamma(X) = \Gamma(X, \mathcal{O}_X)$ is an commutative ring. Hence $\Gamma(X)$ induces an functor $\Gamma\colon Loc \rightarrow Rng^o$, where $Rng^o$ is the oposite category of $Rng$. Does $\Gamma$ have an adjoint functor?
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See my answer to this question: math.stackexchange.com/questions/56854/… – Keenan Kidwell Nov 23 '12 at 23:51
Dear @Rankeya, Thanks for the kind words. – Keenan Kidwell Nov 24 '12 at 0:13
I just want to mention that when I learned this result, particularly the fact that, when you have a morphism $X\rightarrow\mathrm{Spec}(A)$ with $X$ an LRS, the map on global sections determines the underlying map of topological spaces in the only possible way it could, it completely changed my understanding of schemes, and how to work with affine opens of general (non-affine) schemes. I use to be bothered by the definition of an affine open of an abstract scheme as "an open which is isomorphic as an LRS to the spectrum of some ring." Does the isomorphism matter? Are there lots of them? – Keenan Kidwell Nov 24 '12 at 0:30
With this result you can prove that a LRS (not just a scheme!) $X$ admits a unique morphism of LRS $can:X\rightarrow\mathrm{Spec}(\mathscr{O}_X(X))$, and it is an isomorphism if and only $X$ is an affine scheme in the sense that it admits some isomorphism to the spectrum of some ring. I actually greatly prefer to think of the definition of an affine scheme as "a locally ringed space for which the canonical morphism to the spectrum of its global sections is an isomorphism." I felt like I broke through a major barrier in understanding of schemes when I learned this result in the context of LRS. – Keenan Kidwell Nov 24 '12 at 0:32
The $Spec$ functor is the desired adjoint functor to the category of locally ringed spaces (it is right adjoint to $\Gamma$). I think Hartshorne has an exercise where he asks us to prove this when $Loc$ is replaced by the category of schemes.
It's Lemma 01I1 in Stacks. – Zhen Lin Nov 24 '12 at 8:39 | {"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": 0.9836363196372986, "perplexity": 223.04066588394545}, "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/1469257824395.52/warc/CC-MAIN-20160723071024-00217-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://homework.cpm.org/category/ACC/textbook/gb8i/chapter/5%20Unit%206/lesson/INT1:%205.2.1/problem/5-53 | ### Home > GB8I > Chapter 5 Unit 6 > Lesson INT1: 5.2.1 > Problem5-53
5-53.
Draw a slope triangle and use it to write the equation of the line shown in the graph below.
Use the same method you used in problem 5-48.
$y = −10x + 170$ | {"extraction_info": {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8257384300231934, "perplexity": 1679.9271076800476}, "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/1679296949958.54/warc/CC-MAIN-20230401094611-20230401124611-00021.warc.gz"} |
http://math.stackexchange.com/questions/283278/set-notation-for-infinite-subsets | # Set notation for infinite subsets.
In set notation, how can one express an infinite set of subsets where each subset has exactly two elements $\{an-1, an+1\}$ where $a$ is a constant and $n\ge1$ and the $n$ value for each subset is one more than that of the previous subset. Example: $\{ \{a1-1, a1+1\},~\{a2-1, a2+1\},~\{a3-1, a3+1\},~. . . \}$
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What about $\{\{an-1, an+1\}\ |\ n \in \mathbb{N}\setminus\{0\}\}$? Alternatively, for $n \in \mathbb{N}\setminus\{0\}$ you could define $A_n = \{an-1, an+1\}$ and the set you're interested in is $\{A_n\ |\ n \in \mathbb{N}\setminus\{0\}\}$.
Thanks, what about the part about $n$ for each subset being one more than the $n$ value for the previous subset? – Babiker Jan 21 '13 at 6:28
A set has no order, so there is no sense of the previous subset. I've edited my answer so that it is clear that $n$ is a positive integer. Does that answer your question? – Michael Albanese Jan 21 '13 at 6:40 | {"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": 0.9408447742462158, "perplexity": 141.67029542903188}, "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-2015-06/segments/1422115869264.47/warc/CC-MAIN-20150124161109-00027-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://www.ordinalnumbers.com/tag/ordinal-numbers-1-100-worksheet-pdf/ | ## Ordinal Numbers 1-100 Worksheet
Ordinal Numbers 1-100 Worksheet – It is possible to enumerate infinite sets by using ordinal numbers. It is also possible to use them to generalize ordinal numbers. 1st One of the basic concepts of mathematics is the ordinal number. It is a number indicating the location of an object in a list. The ordinal number … Read more
## Numbers Ordinal 1 100
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## Ordinals Numbers 1 100
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Ordinal Numbers 1 10 – With ordinal numbers, you can count infinite sets. They are also able to broaden ordinal numbers.But before you use them, you must comprehend what they are and how they work. 1st The ordinal numbers are one of the fundamental concepts in mathematics. It is a number that identifies the location … Read more
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Ordinal Numbers Worksheets Pdf – You can count unlimited sets with ordinal numbers. They can also be utilized as a method to generalize ordinal numbers. 1st Ordinal numbers are among the most fundamental ideas in math. It is a numerical value that represents the location of an object within the list. The ordinal number is … Read more
## Ordinal Numbers 1 To 100 Worksheet
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Ordinals Numbers 1-100 – With ordinal numbers, it is possible to count infinite sets. They can also serve to generalize ordinal numbers. 1st The ordinal number is one the most fundamental ideas in math. It is a number that identifies the location of an object in the list. Ordinarily, the ordinal numbers are between one … Read more
## Ordinal Numbers Worksheet 1-100
Ordinal Numbers Worksheet 1-100 – With ordinal numbers, it is possible to count any number of sets. These numbers can be utilized as a method to generalize ordinal numbers. 1st The ordinal number is among the fundamental concepts in mathematics. It is a number which signifies the location of an object within a list. Ordinal … Read more
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Numbers Ordinal 1-100 – You can enumerate the infinite amount of sets making use of ordinal numbers as an instrument. They can also serve to generalize ordinal quantities. 1st The ordinal number is one of the most fundamental concepts in mathematics. It is a numerical value that indicates where an object is in a list … Read more
## Ordinal Numbers 1 100
Ordinal Numbers 1 100 – With ordinal numbers, it is possible to count infinite sets. They also aid in generalize ordinal numbers. 1st The ordinal number is one of the mathematical concepts that are fundamental to math. It is a number that indicates the position of an object within a set of objects. Typically, ordinal … Read more | {"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": 0.8183309435844421, "perplexity": 493.21255190901155}, "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/1679296949598.87/warc/CC-MAIN-20230331082653-20230331112653-00228.warc.gz"} |
http://mathoverflow.net/questions/103895/can-omega-1-be-supercompact | # Can $\omega_1$ be supercompact?
Is "ZF + $\omega_1$ is supercompact" consistent relative to "ZFC + there is a supercompact cardinal"?
In particular, if $\delta$ is supercompact, does it remain so in $V(\mathbb{R} \cap V[G])$ where $G \subset Col(\omega,<\delta)$ is $V$-generic? This seems to be the case for measurability but I am having trouble proving it for supercompactness. It seems likely that someone else has tried this, so I though I'd ask here.
The appropriate definition of supercompactness in ZF is the one in terms of normal fine measures, where normality is defined using diagonal intersections.
I am aware that $\omega_1$ has some amount of supercompactness under AD. I am interested in a more direct proof using forcing, which I hope will give (full) supercompactness.
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– Asaf Karagila Aug 3 '12 at 20:31
Ah, I thought I remembered a question like that on here but I couldn't find it. Sorry for posting a similar question. – Trevor Wilson Aug 3 '12 at 20:50
Trevor, I think that my question may be a bit overly broad. It is just like the time I asked about the ability to destroy weak choice principles (or add some of them) via generic extensions (unlike full choice), and later Stefan Geschke asked about a concrete example. It's a reasonable question when trying to tackle a broad question, I think. – Asaf Karagila Aug 3 '12 at 20:54
The Jech construction preserves supercompactness. This is Lemma 1.3 in Apter-Henle, Large cardinal structures below $\aleph_{\omega}$. – Tanmay Inamdar Aug 13 '12 at 12:50
@TanmayInamdar: Great! Could you post that as an answer? (Maybe including the statement of the relevant theorem?) – Trevor Wilson Aug 13 '12 at 15:43
Thanks, that mentions a way to extend an ultrapower of Ord taken under ZF to an ultrapower of $V$ in a class generic extension. This would be a useful thing to do if $\omega_1$ were supercompact. It looks like the only example given of supercompactness without choice given in the article comes from AD though. – Trevor Wilson Aug 3 '12 at 20:45
Trevor, I had several chats over a cup of coffee with Magidor on the problem of an inner model for supercompactness. If you think about Solovay's model it has an $L(\mathbb R)$ sort of construction, and in a sense if we make $\aleph_1$ measurable we essentially take some inner model of measurability. However supercompactness does not yet have such canonical inner model, so I don't think that it's that easy to construct a model of $\aleph_1$ being supercompact for all $\lambda$. – Asaf Karagila Aug 3 '12 at 20:51
I was hoping that the inner model would just be $V(\mathbb{R} \cap V[G])$. Doesn't this work for measurability? Every subset of $\delta$ in this model is added by some proper initial segment of $G$. So we have a measure on $\delta$ given by the union of the small-forcing extensions of the original measure. This is countably complete, again because every countable sequence is added by a proper initial segment of the forcing. But if there is a countably complete measure on $\delta$ then there is a normal measure. – Trevor Wilson Aug 3 '12 at 21:00
...This argument fails to adapt to supercompactness in two ways: first, the set being measured is not added by any proper initial segment of the forcing, and second, normality does not come for free with $\mathcal{P}_\kappa(\lambda)$. Could you explain how the question relates to the inner model problem for supercompact cardinals? – Trevor Wilson Aug 3 '12 at 21:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9443326592445374, "perplexity": 500.6817371592747}, "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-2014-10/segments/1393999656144/warc/CC-MAIN-20140305060736-00017-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.vedantu.com/question-answer/find-the-value-of-x-for-the-given-inverse-class-12-maths-cbse-5ef9f09897157a190a4cb07b | QUESTION
# Find the value of x for the given inverse trigonometric function $4{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \pi$.
Hint: In order to solve the problem first try to convert these two trigonometric functions into one by the use of different inverse trigonometric identities and then move on with the simplification part and calculate the value of x.
Given equation is: $4{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \pi$
We need to find x from this equation,
As we know that
${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2} \\ \Rightarrow {\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}x \\$
Let us substitute the value from above formula into the given equation
$\Rightarrow 4{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \pi \\ \Rightarrow 4{\sin ^{ - 1}}x + \left( {\dfrac{\pi }{2} - {{\sin }^{ - 1}}x} \right) = \pi \\$
Now let us simplify the equation to find the value of x
$\Rightarrow 3{\sin ^{ - 1}}x + \dfrac{\pi }{2} = \pi \\ \Rightarrow 3{\sin ^{ - 1}}x = \pi - \dfrac{\pi }{2} \\ \Rightarrow 3{\sin ^{ - 1}}x = \dfrac{\pi }{2} \\ \Rightarrow {\sin ^{ - 1}}x = \dfrac{1}{3} \times \dfrac{\pi }{2} \\ \Rightarrow {\sin ^{ - 1}}x = \dfrac{\pi }{6} \\$
Now, let us bring the ${\sin ^{ - 1}}$ function from the LHS to the RHS. So it will convert to $\sin$ function.
$\Rightarrow x = \sin \left( {\dfrac{\pi }{6}} \right) \\ \Rightarrow x = \sin \left( {\dfrac{{{{180}^0}}}{6}} \right){\text{ }}\left[ {\because \pi = {{180}^0}} \right] \\ \Rightarrow x = \sin \left( {{{30}^0}} \right) \\$
As we know the value of $\sin {30^0}$ from the trigonometric table. So we will directly substitute the value.
$x = \sin {30^0} = \dfrac{1}{2} \\ \Rightarrow x = \dfrac{1}{2} \\$
Hence, the value of x is $\dfrac{1}{2}$ .
Note: In order to solve such problems students must remember the formulas for the inverse trigonometry. Also students must learn the values of the trigonometric terms for some common angles. This question would have been impossible to solve if we would not have converted the two functions into one so students must recognize which functions to remove. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9534760117530823, "perplexity": 211.272975078445}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655921988.66/warc/CC-MAIN-20200711032932-20200711062932-00178.warc.gz"} |
https://economics.stackexchange.com/questions/41259/competitive-equilibrium-with-production | Competitive equilibrium with production
Consider an economy with four goods, two individuals and two firms. Firm 1 produces good $$x$$, firm 2 produces good $$y$$. Consumers' utilities are $$u_1(x,y,z,w)=\min\{x,2y\}$$ and $$u_2(x,y,z,w)=\min\{2x,y\}$$. Production function of firm 1 is $$x=(1+a)z$$, production function of firm 2 is $$y=w$$, where $$a$$ is a small positive number. Consumer 1(resp. 2) owns firm 1(resp. 2). Initially, consumer 1(resp. 2) has $$1$$ unit of good $$z$$ (resp. $$w$$).
The question asks for an equilibrium. My approach is first stipulating a price, then try to deduce the equilibrium, but I fails to do so, since there are 4 goods here and they are not symmetric. Any help is appreciated.
• Shouldn't price be endogenous. Given that this is homework (at least seems like), you can perhaps show your attempt. For example, writing objective functions for the two individuals. Or (hint:) consider boundary cases – Dayne Dec 2 '20 at 15:33
First, note that both firms have constant returns to scale. This implies that firms must make zero profit in equilibrium. If they could make a positive profit, they could make even more profit by scaling up production and no profit-maximizing production plan would exist. So who owns which firm does not really matter. We know from the first welfare theorem that every equilibrium must be Pareto efficient. With the given utility functions, this means that both firms have to produce a positive amount. But this pins down the relative prices of $$x$$ and $$z$$, and of $$y$$ and $$w$$, respectively. So this reduces the problem to one in which you have to find two prices. You can set one of the remaining prices to be $$1$$, and then calculate the excess demand for one good. FInd the price that makes this excess demand zero. As a consequence of Walras' law, this price will clear all markets.
• "If they could make a positive profit, they could make even more profit by scaling up production and no profit-maximizing production plan would exist" - but OP has mentioned that each of the individual has limited input/endowment ($z,w=1$). So how can the production be scaled beyond that? We can say however that since $z,w$ do not appear in utility function, they will exhaust their endowment to produce $x,y$ and then exchange. – Dayne Dec 3 '20 at 10:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.923783004283905, "perplexity": 738.9302599289657}, "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-2021-21/segments/1620243988955.89/warc/CC-MAIN-20210509032519-20210509062519-00608.warc.gz"} |
https://stats.stackexchange.com/questions/373416/log-odds-to-probability-transformation-confidence-interval-and-interpretation | # log odds to probability transformation confidence interval and interpretation
I'm just doing a sanity check for myself and others as it relates to constructing and interpreting confidence intervals with logistic regression. Suppose you specify a logistic regression of $$y$$ on a binary variable $$x$$ which returns a coefficient estimate $$\hat{\beta_{1}}=0.05$$, and a standard error $$se_{\hat{\beta_{1}}}=0.001$$.
Is it appropriate to claim:
1) The upper and lower bound for the 95% confidence interval for $$\hat{\beta}$$ is given by $$\hat{\beta}{\pm}(1.96\times{se_{\hat{\beta}}})$$
2) Since we can convert log odds into a probability with $$\frac{e^x}{1+e^x}$$, does that mean this transformation also applies to the upper and lower bounds above?
My feeling is that it's not a problem that our confidence interval is not symmetric about $${\hat{\beta}}$$. (You can demonstrate this by calculating probabilities for extreme values of the log odds).
I'm also of the opinion that with these kinds of estimates even if you have an estimated coefficient that is statistically significant, it doesn't necessarily imply practical significance. For instance, if you had a coin which you knew had a long run probability of heads equal to .5000000005, a large enough number of trials would find statistical significance of unfairness, but in a practical sense you could consider the coin "fair".
My final question (and this is the hard part), is it appropriate to report these kinds of binomial probabilities in terms of the difference from a fair coin? Like, for the example in the beginning, the log odds being 0.05 instead of saying "the probability $$P(y=1|x=1)=0.5125$$", we instead subtract 0.5 from that value to say something like "x being equal to 1 and not 0 is associated with a change in the probability" $${\Delta{P(y=1)}=0.0125}$$, just to illustrate that that the size of the effect of x on y is small even though there might be statistical significance. I apologize if my notation is screwed up. Hopefully you understand what I'm getting at though.
• log odds or log odds ratio? There are different. Oct 23, 2018 at 21:03
• Oh you're right. There are different. silly me. R spits out the log odds for its coefficients. so That's what I'm referring to Oct 23, 2018 at 21:20
2. Yes. Because $$log\left(\frac {e^x}{1+e^x}\right)$$ is monotonic increase function.
3. It is question about how to express your idea, so reporting the difference is OK. But I prefer p=0.5125, because I do not know what p is if you tell me the $$ΔP(y=1)=0.0125$$. If you tell me p=0.5125 and I want to know the different from 0.5, I can calculate easily. | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8779953122138977, "perplexity": 263.38967295086815}, "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/1656103920118.49/warc/CC-MAIN-20220701034437-20220701064437-00650.warc.gz"} |
https://icml.cc/Conferences/2020/ScheduleMultitrack?event=6269 | Timezone: »
Poster
Near-Tight Margin-Based Generalization Bounds for Support Vector Machines
Allan Grønlund · Lior Kamma · Kasper Green Larsen
Tue Jul 14 01:00 PM -- 01:45 PM & Wed Jul 15 12:00 AM -- 12:45 AM (PDT) @ None #None
Support Vector Machines (SVMs) are among the most fundamental tools for binary classification.
In its simplest formulation, an SVM produces a hyperplane separating two classes of data using the largest possible margin to the data. The focus on maximizing the margin has been well motivated through numerous generalization bounds.
In this paper, we revisit and improve the classic generalization bounds in terms of margins. Furthermore, we complement our new generalization bound by a nearly matching lower bound, thus almost settling the generalization performance of SVMs in terms of margins. | {"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": 0.8242970705032349, "perplexity": 2516.858555006588}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662531762.30/warc/CC-MAIN-20220520061824-20220520091824-00679.warc.gz"} |
http://math.stackexchange.com/questions/34198/bound-for-number-of-points-on-surface-over-mathbbf-p | # Bound for number of points on surface over $\mathbb{F}_p$
I know of the bound for the number of points on an elliptic curve over a finite field: $$|\# E(\mathbb{F}_q) - q - 1| < 2\sqrt{q}$$ where this includes the point at infinity. I have been told that there are higher analogues of this formula which involve $q^{(k-1)/2}$ (instead of $q^{1/2}$ here), but I know very little about the theory so I'm not exactly sure what the $k$ is and all that. Can any one help me out with the formula for this? Thanks!
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The general form that this bound takes is known under the term Weil conjectures, which is a theorem of Deligne. Here is how it goes:
if $X$ is non-singular n-dimensional projective variety over $\mathbb{F}_q$ of dimension $n$, (an elliptic curve is the special case $n=1$), then you collect information about the number of points of $X$ over $\mathbb{F}_{q^m}$ for all $m$ in the generating function $$\zeta(X, s) = \exp\left(\sum_{m = 1}^\infty \frac{N_m}{m} (q^{-s})^m\right).$$ This is known as the zeta function of $X$.
The Weil conjectures then say that $\zeta(X, s)$ is a rational function of $T = q^{−s}$ and can be written as $$\prod_{i=0}^{2n} P_i(q^{-s})^{(-1)^{i+1}} = \frac{P_1(T)\dotsb P_{2n-1}(T)}{P_0(T)\dotsb P_{2n}(T)},$$ where each $P_i(T)$ is an integral polynomial that factors over $\mathbb{C}$ as $\prod_j (1 - \alpha_{i,j}T)$. If $X$ came from a projective variety defined over a number field with good reduction at $p=\text{char } \mathbb{F}_q$, then the degree of the $i$-th polynomial is the $i$-th Betty number of $X$. Moreover, $P_0(T) = 1 − T$ and $P_{2n}(T) = 1 − q^nT$. The Riemann hypothesis for these varieties, which is part of Weil's conjectures, says that $$|\alpha_{i,j}| = q^{i/2},$$ and this is the source of the Hasse bound you just cited. Here is what happens in your special case:
Let's assume for the moment that $X$ is a curve, so $n=1$. The degree of $P_1$ is twice the genus $g$ of the curve. If you take the logarithmic derivative of the zeta function and do some rearranging, you will find that $$N_m = 1 + q^m - (\alpha_{1,1}^m+\cdots+\alpha_{1,2g}^m),$$ which gives you the bound $|N_m - q^m - 1|\leq 2g\sqrt{q}^m$ that you quoted in the special case $g=1$.
If $X$ is a higher dimensional variety, then you can try to do the same manipulations, but I believe that the expression for $N_m$ will be less clean. Nevertheless, you should get something to the effect that $N_m = q^{nm} + O(q^{(2n-1)m/2})$. So your $k$ is my $2n$.
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Dear Alex, I think that the bound in the last line actually goes back to Lang--Weil (and in particular, predates Deligne). (You can get it by fibreing a higher dimensional variety by curves, or something similar, and using the Hasse--Weil bound for curves.) Best wishes, – Matt E Apr 21 '11 at 5:47
@Matt Dear Matthew, thank you, that's very interesting! If you have time to write your comment up as an answer at some point or to provide a reference, I would be very interested. – Alex B. Apr 21 '11 at 6:43
Dear Alex, See e.g. Theorem 2.1 of these notes by Mustata (where you can also find a reference to the original paper). Best wishes, – Matt E Apr 21 '11 at 7:22
@Matt Dear Matt, wonderful, thank you! Best wishes, – Alex B. Apr 21 '11 at 8:35
I think what's going on is if you have an equation in $k$ variables over a field of $q$ elements then you expect it to have $q^{k-1}$ solutions and the error term, under suitable hypotheses, is some constant multiple of $q^{(k-1)/2}$.
- | {"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": 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": 0.9609556794166565, "perplexity": 106.45620680516876}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783393518.22/warc/CC-MAIN-20160624154953-00081-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/79166/cardinality-of-the-set-of-elements-of-fixed-order | # Cardinality of the set of elements of fixed order.
Let us consider the group $G:=\mathbb{Z}_N^a$ (the product of the cyclic group with $N$ elements with itself $a$ times). Suppose we are given a number $m$ that divides $N$.
I would like to know how many elements $x$ in $G$ have the property that $(N/m)x$ has order precisely $m$ (and not any number dividing $m$).
For example, if $a$ equals $1$, and $m=p^{\alpha}$ for $p$ a prime, the number of such elements is $N-N/p$, if $m= p^{\alpha} q^{\beta}$, it is $N-N/p-N/q+N/pq$.
I expect that in general there might be a closed formula for this number, perhaps involving the Euler function $\phi$. Can anyone produce it or give me a suggestion on a good reference where I could look for it?
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Questions suitable for intro group theory homework assignments are not suitable for MO. Please see the faq. Voting to close. – Gerry Myerson Oct 26 '11 at 21:26
Let $Ord(m)$ be the number of elements of order $m$ in $\mathbb{Z}_N^a$, where $m|N$. A nice way to compute $Ord(m)$ is to start from the observation
$$\sum_{m|N} Ord(m) = N^a$$
and to apply Möbius inversion:
$$Ord(m) = \sum_{d|m} \mu(m/d)d^a.$$
$Ord(m)$ is multiplicative (if $m$ and $n$ are relatively prime, then $Ord(mn) = Ord(m)Ord(n)$; cf. Chinese remainder theorem). Thus, it is not hard to work out $Ord(m)$ in terms of how $\mu$ behaves on the prime-power factors of $m$. After a little algebra,
$$Ord(m) = m^a \prod_{\text{prime } p|m} (1-\frac1{p^a}).$$
Notice that $N$ has receded from the picture. Indeed, for $m|N$, let
$$\mathbb{Z}_m^a \hookrightarrow \mathbb{Z}_N^a$$
be the obvious inclusion of abelian groups, mapping onto the subgroup of elements whose order divides $m$. Then the set of elements of order exactly $m$ in $\mathbb{Z}_m^a$ maps onto the set of elements of order exactly $m$ in $\mathbb{Z}_N^a$, so we are just counting the former set.
Edit: Oh wait, I didn't quite answer your exact question, did I? But no problem: the number of $x$ such that $(N/m)x$ has order exactly $m$ is just the inverse image of the set of order $m$ elements w.r.t. the map
$$(N/m) \cdot -: \mathbb{Z}_N^a \to \mathbb{Z}_N^a.$$
This map has kernel of size $(N/m)^a$, and the inverse image we want consists of $Ord(m)$ many distinct cosets of this kernel. Thus, the answer to the actual question is
$$N^a \prod_{\text{prime } p|m} (1 - \frac1{p^a}).$$
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Thank you very much! Your explanation is very clear and neat. – Math Oct 27 '11 at 12:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8624595999717712, "perplexity": 95.6817830419376}, "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/1448398447043.45/warc/CC-MAIN-20151124205407-00030-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/90748/explanation-of-generic-point-with-examples | # Explanation of “generic point” with examples?
Could someone please explain to me why $X$ and $Y$ are generic points of $\mathbb{R}[X, Y]/(XY)$?
And why is the ideal generated by irreducible polynomial is a generic point in $\mathbb{R}[X, Y]$?
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All this follows immediately from the definitions of irreducible element, prime ideal, spectrum, and generic point. Which part of the definitions do you have trouble with? – Zhen Lin Dec 12 '11 at 11:02
X is is generic point of ℝ[X, Y]/(XY) it means V(X)= Spec(ℝ[X, Y]/(XY))? – Name Dec 12 '11 at 11:17
help me please, I don't see the solution – Name Dec 12 '11 at 11:59
A scheme $S$ has a generic point if and only if its underlying topological space $|S|$ is irreducible, in which case there is a unique point $\eta\in S$ such that $\overline { \lbrace \eta \rbrace}=|S|$.
If $S=Spec(A)$ is an affine scheme, irreducibility amounts to the condition that $Nil(A)$, the nilradical, be prime or equivalently that the reduction $A_{red}=A/Nil(A)$ be a domain.
In your case $A=\mathbb R[X, Y]/(XY)=\mathbb R[x,y] \;$ is already reduced but is not a domain , so that $Spec(A)$ has no generic point.
End of story? No!
If a scheme $S$ is not irreducible, $|S|$ has a decomposition into irreducible components $S=\bigcup S_i$ , each $S_i$ having a dense point $\eta_i$. Those $\eta_i$ are called maximal points or even (by "abuse of language") generic points of $S$.
In the affine case, they correspond to the minimal ideals ${\mathfrak p_i}\subset A$.
In your case you have two maximal points $\eta_x, \eta_y$ corresponding to the only two minimal ideals $(x),(y)$ of $k[x,y]$.
They are the generic (=dense) points of the lines $V(x)$ and $V(y)$, which are the irreducible components of your scheme $S=Spec(k[x,y])$
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So (X) is generic point for V(x) and by abuse of language, we say (X) is generic point for ℝ[X, Y]/(XY)?. – Name Dec 12 '11 at 12:59
Dear @ name: yes, this is correct. To be absolutely precise it is better to write "$(x)$ is the generic point for $V(x)$ " , with a small letter $x$, so as to carefully distinguish between the polynomial $X\in k[X,Y]$ and its class $x$ in the quotient$k[x,y]=k[X,Y]/(X.Y)$ – Georges Elencwajg Dec 12 '11 at 13:17
Thank so much for your help Georges Elencwajg – Name Dec 12 '11 at 13:21 | {"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": 0.8091443777084351, "perplexity": 248.0476873219493}, "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-2014-35/segments/1408500815991.16/warc/CC-MAIN-20140820021335-00180-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/61928-proof-question.html | ## a proof question..
i am given a real and positive number a1 a2 a3 ...
which goes by a(n)<=c*a(n-1) for every n=>2 for a certain given number c>0 .
prove that
a(n)<=a(1)*c^(n-1)
??
a(n) is the n'th number of the series | {"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": 0.8512706756591797, "perplexity": 3051.5700888502315}, "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-2018-22/segments/1526794865702.43/warc/CC-MAIN-20180523180641-20180523200641-00389.warc.gz"} |
http://mathoverflow.net/questions/84910/what-is-the-bruhat-decomposition-of-the-affine-grassmannian | # What is the Bruhat decomposition of the affine Grassmannian?
We define the affine Grassmannian to be the quotient $Gr = GL_n(\mathbb{C}((t)))/GL_n(\mathbb{C}[[t]])$ where $\mathbb{C}((t))$ is the field of formal Laurent series and $\mathbb{C}[[t]]$ is the ring of formal power series. (The affine Grassmannian can be defined more generally, but here we restrict to a special case.) If we let $B$ be the Borel subgroup of upper triangular matrices in $GL_n(\mathbb{C})$, $T$ a maximal torus, then the Weyl group $W=N(T)/T$ is just $S_n$. Let $\widetilde{W} = \mathbb{Z}^{n-1} \rtimes W$ denote the affine Weyl group. Then for $i = 1, 2,...,n-1$ the affine permutations in $\widetilde{W}$ correspond to the usual permutation matrices in $GL_n(\mathbb{C})$, namely the identity matrix with columns $i$ and $i+1$ interchanged. The matrix for the affine permutation $s_0$ has ones along the diagonal in rows $2, 3, ..., n-1$, has $t$ in the right hand corner, and $t^{-1}$ in the bottom left corner. Let $I$ denote the Iwahori subgroup, that is, the inverse image of $B$ under the reduction map $GL_n(\mathbb{C}((t))) \rightarrow GL_n(\mathbb{C})$. Then $I$ is the set of upper triangular matrices mod $t$. I read somewhere that $GL_n(\mathbb{C}((t)))$ has a decomposition
$GL_n(\mathbb{C}((t))) = \cup IwGL_n(\mathbb{C}[[t]])$
where $w$ varies across the affine permutation matrices. This decomposition is supposed to induce the Bruhat decomposition of the affine Grassmannian into Schubert cells. Now something here is wrong. Since $I \subset GL_n(\mathbb{C}[[t]])$, and the determinant of any affine permutation matrix is 1 or -1, we have that for any $w \in \widetilde{W}$ the determinant of any matrix in $IwGL_n(\mathbb{C}[[t]])$ has power series determinant, but the matrix
t^{-1} 0
0 t^{-1}
is in $GL_n(\mathbb{C}((t)))$ with determinant $t^{-2}$ and inverse
t 0
0 t
So, my question is, what is wrong here? What is the correct decomposition and indexing set?
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There are two discrepancies: first, the "strict" affine Bruhat decomposition applies to SL(n), not GL(n); second, the decomposition you're wanting, in the strict case, would be $G=\bigcup_w IwI$, that is, with Iwahori on both sides. (This would make it a disjoint union.)
This doesn't depend so much on $\mathbb C((t))$ and $\mathbb C[[t]]$, as it applies to the field of fractions of a (maybe complete) discrete valuation ring.
I didn't see where you defined affine permutation matrix, but anyway, it should be a monomial matrix with powers of t. If you want the $IwK$ decomposition (which I would call $IwP$, but never mind) to be disjoint, let $w$ only vary over $diag(t^{d_1}, t^{d_2}, \ldots)$ with $(d_1 \geq d_2 \geq \ldots d_n) \in \integers^n$. – Allen Knutson Jan 16 '12 at 23:25 | {"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": 0.9963470697402954, "perplexity": 131.47764192855038}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375095346.56/warc/CC-MAIN-20150627031815-00023-ip-10-179-60-89.ec2.internal.warc.gz"} |
http://harvard.voxcharta.org/tag/polarization-measurements/ | # Posts Tagged polarization measurements
## Recent Postings from polarization measurements
### Evidence of a Mira-like tail and bow shock about the semi-regular variable V CVn from four decades of polarization measurements
Polarization is a powerful tool for understanding stellar atmospheres and circumstellar environments. Mira and semi-regular variable stars have been observed for decades and some are known to be polarimetrically variable, however, the semi-regular variable V Canes Venatici displays an unusually large, unexplained amount of polarization. We present ten years of optical polarization observations obtained with the HPOL instrument, supplemented by published observations spanning a total interval of about forty years for V CVn. We find that V CVn shows large polarization variations ranging from 1 – 6%. We also find that for the past forty years the position angle measured for V CVn has been virtually constant suggesting a long-term, stable, asymmetric structure about the star. We suggest that this asymmetry is caused by the presence of a stellar wind bow shock and tail, consistent with the star’s large space velocity.
### Polarization measurements analysis II. Best estimators of polarization fraction and angle
With the forthcoming release of high precision polarization measurements, such as from the Planck satellite, it becomes critical to evaluate the performance of estimators for the polarization fraction and angle. These two physical quantities suffer from a well-known bias in the presence of measurement noise, as has been described in part I of this series. In this paper, part II of the series, we explore the extent to which various estimators may correct the bias. Traditional frequentist estimators of the polarization fraction are compared with two recent estimators: one inspired by a Bayesian analysis and a second following an asymptotic method. We investigate the sensitivity of these estimators to the asymmetry of the covariance matrix which may vary over large datasets. We present for the first time a comparison among polarization angle estimators, and evaluate the statistical bias on the angle that appears when the covariance matrix exhibits effective ellipticity. We also address the question of the accuracy of the polarization fraction and angle uncertainty estimators. The methods linked to the credible intervals and to the variance estimates are tested against the robust confidence interval method. From this pool of estimators, we build recipes adapted to different use-cases: build a mask, compute large maps, and deal with low S/N data. More generally, we show that the traditional estimators suffer from discontinuous distributions at low S/N, while the asymptotic and Bayesian methods do not. Attention is given to the shape of the output distribution of the estimators, and is compared with a Gaussian. In this regard, the new asymptotic method presents the best performance, while the Bayesian output distribution is shown to be strongly asymmetric with a sharp cut at low S/N.Finally, we present an optimization of the estimator derived from the Bayesian analysis using adapted priors.
### Polarization measurements analysis I. Impact of the full covariance matrix on polarization fraction and angle measurements
With the forthcoming release of high precision polarization measurements, such as from the Planck satellite, the metrology of polarization needs to improve. In particular, it is crucial to take into account full knowledge of the noise properties when estimating polarization fraction and angle, which suffer from well-known biases. While strong simplifying assumptions have usually been made in polarization analysis, we present a method for including the full covariance matrix of the Stokes parameters in estimates for the distributions of the polarization fraction and angle. We thereby quantify the impact of the noise properties on the biases in the observational quantities. We derive analytical expressions for the pdf of these quantities, taking into account the full complexity of the covariance matrix, including the Stokes I intensity components. We perform simulations to explore the impact of the noise properties on the statistical variance and bias of the polarization fraction and angle. We show that for low variations of the effective ellipticity between the Q and U components around the symmetrical case the covariance matrix may be simplified as is usually done, with negligible impact on the bias. For S/N on intensity lower than 10 the uncertainty on the total intensity is shown to drastically increase the uncertainty of the polarization fraction but not the relative bias, while a 10\% correlation between the intensity and the polarized components does not significantly affect the bias of the polarization fraction. We compare estimates of the uncertainties affecting polarization measurements, addressing limitations of estimates of the S/N, and we show how to build conservative confidence intervals for polarization fraction and angle simultaneously. This study is the first of a set of papers dedicated to the analysis of polarization measurements.
### Compatibility of Planck and BICEP2 in the Light of Inflation
We investigate the implications for inflation of the detection of B-modes polarization in the Cosmic Microwave Background (CMB) by BICEP2. We show that the hypothesis of primordial origin of the measurement is only favored by the first four bandpowers, while the others would prefer unreasonably large values of the tensor-to-scalar ratio. Using only those four bandpowers, we carry out a complete analysis in the cosmological and inflationary slow-roll parameter space using the BICEP2 polarization measurements alone and extract the Bayesian evidences and complexities for all the Encyclopaedia Inflationaris models. This allows us to determine the most probable and simplest BICEP2 inflationary scenarios. Although this list contains the simplest monomial potentials, it also includes many other scenarios, suggesting that focusing model building efforts on large field models only is unjustified at this stage. We demonstrate that the sets of inflationary models preferred by Planck alone and BICEP2 alone are almost disjoint, indicating a clear tension between the two data sets. We address this tension with a Bayesian measure of compatibility between BICEP2 and Planck. We find that for models favored by Planck the two data sets tend to be incompatible, whereas there is a moderate evidence of compatibility for the BICEP2 preferred models. As a result, it would be premature to draw any conclusion on the best Planck models, such as Starobinsky and/or Kahler moduli inflation. For the subset of scenarios not exhibiting data sets incompatibility, we update the evidences and complexities using both data sets together.
### Compatibility of Planck and BICEP2 in the Light of Inflation [Cross-Listing]
We investigate the implications for inflation of the detection of B-modes polarization in the Cosmic Microwave Background (CMB) by BICEP2. We show that the hypothesis of primordial origin of the measurement is only favored by the first four bandpowers, while the others would prefer unreasonably large values of the tensor-to-scalar ratio. Using only those four bandpowers, we carry out a complete analysis in the cosmological and inflationary slow-roll parameter space using the BICEP2 polarization measurements alone and extract the Bayesian evidences and complexities for all the Encyclopaedia Inflationaris models. This allows us to determine the most probable and simplest BICEP2 inflationary scenarios. Although this list contains the simplest monomial potentials, it also includes many other scenarios, suggesting that focusing model building efforts on large field models only is unjustified at this stage. We demonstrate that the sets of inflationary models preferred by Planck alone and BICEP2 alone are almost disjoint, indicating a clear tension between the two data sets. We address this tension with a Bayesian measure of compatibility between BICEP2 and Planck. We find that for models favored by Planck the two data sets tend to be incompatible, whereas there is a moderate evidence of compatibility for the BICEP2 preferred models. As a result, it would be premature to draw any conclusion on the best Planck models, such as Starobinsky and/or Kahler moduli inflation. For the subset of scenarios not exhibiting data sets incompatibility, we update the evidences and complexities using both data sets together.
### Compatibility of Planck and BICEP2 in the Light of Inflation [Cross-Listing]
We investigate the implications for inflation of the detection of B-modes polarization in the Cosmic Microwave Background (CMB) by BICEP2. We show that the hypothesis of primordial origin of the measurement is only favored by the first four bandpowers, while the others would prefer unreasonably large values of the tensor-to-scalar ratio. Using only those four bandpowers, we carry out a complete analysis in the cosmological and inflationary slow-roll parameter space using the BICEP2 polarization measurements alone and extract the Bayesian evidences and complexities for all the Encyclopaedia Inflationaris models. This allows us to determine the most probable and simplest BICEP2 inflationary scenarios. Although this list contains the simplest monomial potentials, it also includes many other scenarios, suggesting that focusing model building efforts on large field models only is unjustified at this stage. We demonstrate that the sets of inflationary models preferred by Planck alone and BICEP2 alone are almost disjoint, indicating a clear tension between the two data sets. We address this tension with a Bayesian measure of compatibility between BICEP2 and Planck. We find that for models favored by Planck the two data sets tend to be incompatible, whereas there is a moderate evidence of compatibility for the BICEP2 preferred models. As a result, it would be premature to draw any conclusion on the best Planck models, such as Starobinsky and/or Kahler moduli inflation. For the subset of scenarios not exhibiting data sets incompatibility, we update the evidences and complexities using both data sets together.
### Testing CPT Symmetry with Current and Future CMB Measurements
In this paper we use the current and future cosmic microwave background (CMB) experiments to test the Charge-Parity-Time Reversal (CPT) symmetry. We consider a CPT-violating interaction in the photon sector $\mathcal{L}_{\rm cs}\sim p_\mu A_\nu \tilde{F}^{\mu\nu}$ which gives rise to a rotation of the polarization vectors of the propagating CMB photons. By combining current CMB polarization measurements, the nine-year WMAP, BOOMERanG 2003 and BICEP observations, we obtain a tight constraint on the isotropic rotation angle $\bar{\alpha} = -2.12 \pm 1.14$ ($1\sigma$), indicating an about $2\sigma$ detection of the CPT violation. Here, we particularly take the systematic errors of CMB measurements into account. Then, we study the effects of the anisotropies of the rotation angle [$\Delta{\alpha}({\bf \hat{n}})$] on the CMB polarization power spectra in detail. Due to the small effects, the current CMB polarization data can not constrain the related parameters very well. We obtain the 95\% C.L. upper limit of the variance of the anisotropies of the rotation angle $C^\alpha(0) < 0.035$ from all the CMB datasets. More interestingly, including the anisotropies of rotation angle could lower the best fit value of $r$ and relax the tension on the constraints of $r$ between BICEP2 and Planck. Finally, we investigate the capabilities of future Planck polarization measurements on $\bar{\alpha}$ and $\Delta{\alpha}({\bf \hat{n}})$. Benefited from the high precision of Planck data, the constraints of the rotation angle can be significantly improved.
### Testing CPT Symmetry with Current and Future CMB Measurements [Cross-Listing]
In this paper we use the current and future cosmic microwave background (CMB) experiments to test the Charge-Parity-Time Reversal (CPT) symmetry. We consider a CPT-violating interaction in the photon sector $\mathcal{L}_{\rm cs}\sim p_\mu A_\nu \tilde{F}^{\mu\nu}$ which gives rise to a rotation of the polarization vectors of the propagating CMB photons. By combining current CMB polarization measurements, the nine-year WMAP, BOOMERanG 2003 and BICEP observations, we obtain a tight constraint on the isotropic rotation angle $\bar{\alpha} = -2.12 \pm 1.14$ ($1\sigma$), indicating an about $2\sigma$ detection of the CPT violation. Here, we particularly take the systematic errors of CMB measurements into account. Then, we study the effects of the anisotropies of the rotation angle [$\Delta{\alpha}({\bf \hat{n}})$] on the CMB polarization power spectra in detail. Due to the small effects, the current CMB polarization data can not constrain the related parameters very well. We obtain the 95\% C.L. upper limit of the variance of the anisotropies of the rotation angle $C^\alpha(0) < 0.035$ from all the CMB datasets. More interestingly, including the anisotropies of rotation angle could lower the best fit value of $r$ and relax the tension on the constraints of $r$ between BICEP2 and Planck. Finally, we investigate the capabilities of future Planck polarization measurements on $\bar{\alpha}$ and $\Delta{\alpha}({\bf \hat{n}})$. Benefited from the high precision of Planck data, the constraints of the rotation angle can be significantly improved.
### Testing CPT Symmetry with Current and Future CMB Measurements [Cross-Listing]
In this paper we use the current and future cosmic microwave background (CMB) experiments to test the Charge-Parity-Time Reversal (CPT) symmetry. We consider a CPT-violating interaction in the photon sector $\mathcal{L}_{\rm cs}\sim p_\mu A_\nu \tilde{F}^{\mu\nu}$ which gives rise to a rotation of the polarization vectors of the propagating CMB photons. By combining current CMB polarization measurements, the nine-year WMAP, BOOMERanG 2003 and BICEP observations, we obtain a tight constraint on the isotropic rotation angle $\bar{\alpha} = -2.12 \pm 1.14$ ($1\sigma$), indicating an about $2\sigma$ detection of the CPT violation. Here, we particularly take the systematic errors of CMB measurements into account. Then, we study the effects of the anisotropies of the rotation angle [$\Delta{\alpha}({\bf \hat{n}})$] on the CMB polarization power spectra in detail. Due to the small effects, the current CMB polarization data can not constrain the related parameters very well. We obtain the 95\% C.L. upper limit of the variance of the anisotropies of the rotation angle $C^\alpha(0) < 0.035$ from all the CMB datasets. More interestingly, including the anisotropies of rotation angle could lower the best fit value of $r$ and relax the tension on the constraints of $r$ between BICEP2 and Planck. Finally, we investigate the capabilities of future Planck polarization measurements on $\bar{\alpha}$ and $\Delta{\alpha}({\bf \hat{n}})$. Benefited from the high precision of Planck data, the constraints of the rotation angle can be significantly improved.
### Tracing the ISM magnetic field morphology: The potential of multi-wavelength polarization measurements
$\textit{Aims.}$ We present a case study to demonstrate the potential of multi-wavelength polarization measurements. The aim is to investigate the effects that dichroic polarization and thermal re-emission have on tracing the magnetic field in the interstellar medium (ISM). Furthermore, we analyze the crucial influence of imperfectly aligned compact dust grains on the resulting synthetic continuum polarization maps.$\\ \textit{Methods.}$ We developed an extended version of the well-known 3D Monte-Carlo radiation transport code MC3D for multi-wavelength polarization simulations running on an adaptive grid.We investigated the interplay between radiation, magnetic fields and dust grains. Our results were produced by post-processing both ideal density distributions and sophisticated magnetohydrodynamic (MHD) collapse simulations with radiative transfer simulations. We derived spatially resolved maps of intensity, optical depth, and linear and circular polarization at various inclination angles and scales in a wavelength range from 7 $\mu m$ to 1 $mm$.$\\ \textit{Results.}$ We predict unique patterns in linear and circular polarization maps for different types of density distributions and magnetic field morphologies for test setups and sophisticated MHD collapse simulations. We show that alignment processes of interstellar dust grains can significantly influence the resulting synthetic polarization maps. Multi-wavelength polarization measurements allow one to predict the morphology of the magnetic field inside the ISM. The interpretation of polarization measurements of complex structures still remains ambiguous because of the large variety of the predominant parameters in the ISM.
### A compendium of AGN inclinations with corresponding UV/optical continuum polarization measurements
The anisotropic nature of active galactic nuclei (AGN) is thought to be responsible for the observational differences between type-1 (pole-on) and type-2 (edge-on) nearby Seyfert-like galaxies. In this picture, the detection of emission and/or absorption features is directly correlated to the inclination of the system. The AGN structure can be further probed by using the geometry-sensitive technique of polarimetry, yet the pairing between observed polarization and Seyfert type remains poorly examined. Based on archival data, I report here the first compilation of 53 estimated AGN inclinations matched with ultraviolet/optical continuum polarization measurements. Corrections, based on the polarization of broad emission lines, are applied to the sample of Seyfert-2 AGN to remove dilution by starburst light and derive information about the scattered continuum alone. The resulting compendium agrees with past empirical results, i.e. type-1 AGN show low polarization degrees (P < 1%) predominantly associated with a polarization position angle parallel to the projected radio axis of the system, while type-2 objects show stronger polarization percentages (P > 7%) with perpendicular polarization angles. The transition between type-1 and type-2 inclination occurs between 45 and 60 degrees without noticeable impact on P. The compendium is further used as a test to investigate the relevance of four AGN models. While an AGN model with fragmented regions matches observations better than uniform models, a structure with a failed dusty wind along the equator and disc-born, ionized, polar outflows is by far closer to observations. However, although the models correctly reproduce the observed dichotomy between parallel and perpendicular polarization, as well as correct polarization percentages at type-2 inclinations, further work is needed to account for some highly polarized type-1 AGN
### The CMB flexes its BICEPs while walking the Planck
Recent microwave polarization measurements from the BICEP2 experiment may reveal a long-sought signature of inflation. However, these new results appear inconsistent with the best-fit model from the Planck satellite. We suggest a particularly simple idea for reconciling these data-sets, and for explaining a wide range of phenomena on the cosmic microwave sky.
### Quarkonium production in the LHC era: a polarized perspective
Polarization measurements are usually considered as the most difficult challenge for the QCD description of quarkonium production. In fact, global data fits for the determination of the non-perturbative parameters of bound-state formation traditionally exclude polarization observables and use them as a posteriori verifications of the predictions, with perplexing results. With a change of perspective, we move polarization data to the centre of the study, advocating that they actually provide the strongest fundamental indications about the production mechanisms, even before we explicitly consider perturbative calculations. Considering psi(2S) and Y(3S) measurements from LHC experiments and state-of-the-art NLO short-distance calculations in the framework of non-relativistic QCD factorization (NRQCD), we perform a search for a kinematic domain where the polarizations can be correctly reproduced together with the cross sections, by systematically scanning the phase space and accurately treating the experimental uncertainties. This strategy provides a straightforward solution to the "quarkonium polarization puzzle" and reassuring signs that the theoretical framework is reliable. At the same time, the results expose unexpected hierarchies in the non-perturbative NRQCD parameters, that open new paths towards the understanding of bound-state formation in QCD.
### Probing the radio emission from air showers with polarization measurements [Replacement]
The emission of radio waves from air showers has been attributed to the so-called geomagnetic emission process. At frequencies around 50 MHz this process leads to coherent radiation which can be observed with rather simple setups. The direction of the electric field induced by this emission process depends only on the local magnetic field vector and on the incoming direction of the air shower. We report on measurements of the electric field vector where, in addition to this geomagnetic component, another component has been observed which cannot be described by the geomagnetic emission process. The data provide strong evidence that the other electric field component is polarized radially with respect to the shower axis, in agreement with predictions made by Askaryan who described radio emission from particle showers due to a negative charge-excess in the front of the shower. Our results are compared to calculations which include the radiation mechanism induced by this charge-excess process.
### Probing the radio emission from air showers with polarization measurements
The emission of radio waves from air showers has been attributed to the so-called geomagnetic emission process. At frequencies around 50 MHz this process leads to coherent radiation which can be observed with rather simple setups. The direction of the electric field induced by this emission process depends only on the local magnetic field vector and on the incoming direction of the air shower. We report on measurements of the electric field vector where, in addition to this geomagnetic component, another component has been observed which cannot be described by the geomagnetic emission process. The data provide strong evidence that the other electric field component is polarized radially with respect to the shower axis, in agreement with predictions made by Askaryan who described radio emission from particle showers due to a negative charge-excess in the front of the shower. Our results are compared to calculations which include the radiation mechanism induced by this charge-excess process.
### Parkes full polarization spectra of OH masers - II. Galactic longitudes 240 to 350
Full polarization measurements of 1665 and 1667-MHz OH masers at 261 sites of massive star formation have been made with the Parkes radio telescope. Here we present the resulting spectra for 157 southern sources, complementing our previously published 104 northerly sources. For most sites, these are the first measurements of linear polarization, with good spectral resolution and complete velocity coverage. Our spectra exhibit the well-known predominance of highly circularly polarized features, interpreted as $\sigma$ components of Zeeman patterns. Focusing on the generally weaker and rarer linear polarization, we found three examples of likely full Zeeman triplets (a linearly polarized $\pi$ component, straddled in velocity by $\sigma$ components), adding to the solitary example previously reported. We also identify 40 examples of likely isolated $\pi$ components, contradicting past beliefs that $\pi$ components might be extremely rare. These were recognised at 20 sites where a feature with high linear polarization on one transition is accompanied on the other transition by a matching feature, at the same velocity and also with significant linear polarization. Large velocity ranges are rare, but we find eight exceeding 25 km/s, some of them indicating high velocity blue-shifted outflows. Variability was investigated on timescales of one year and over several decades. More than 20 sites (of 200) show high variability (intensity changes by factors of four or more) in some prominent features. Highly stable sites are extremely rare.
### Parkes full polarization spectra of OH masers - I. Galactic longitudes 350 through the Galactic Centre to 41
Full polarization measurements of 1665 and 1667-MHz OH masers at sites of massive star formation have been made with the Parkes 64-m radio telescope. Here we present the resulting spectra for 104 northerly sources. For more than 20 masers we made new measurements with the ATCA (which also revealed several hitherto unreported masers), in most cases yielding arcsecond precision to match the majority of sites. Position improvements assist in distinguishing OH masers with accompanying methanol masers from those without (thought to be at a later stage of evolution). There was no existing linear polarization information at many sites, and spectral resolution was sometimes poor, or velocity coverage incomplete. These inadequacies are addressed by the present Parkes spectra. The whole OH maser sample exhibit the well-known predominance of highly circularly polarized features. We find that linear polarization is also common, but usually much weaker, and we highlight the rare cases of very pronounced linear polarization that can extend to 100 per cent. Unusually large velocity ranges of at least 25 km/s are present at seven sites. Our spectra measurements for most sources are at two epochs spaced by nearly one year, and reveal high stability at most sites, and marked variability (more than factors of two in the strongest feature) at only five sites. The spectra also provide a valuable reference for longer term variability, with high stability evident over the past decades at 10 sites and marked variability for four of the sample. Future systematic monitoring of these variables may uncover further examples of periodicity, a phenomenon so far recognised in only one source.
### Exoplanetary searches with gravitational microlensing: polarization issues
There are different methods for finding exoplanets such as radial spectral shifts, astrometrical measurements, transits, timing etc. Gravitational microlensing (including pixel-lensing) is among the most promising techniques with the potentiality of detecting Earth-like planets at distances about a few astronomical units from their host star or near the so-called snow line with a temperature in the range $0-100^0$ C on a solid surface of an exoplanet. We emphasize the importance of polarization measurements which can help to resolve degeneracies in theoretical models. In particular, the polarization angle could give additional information about the relative position of the lens with respect to the source.
### Exoplanetary searches with gravitational microlensing: polarization issues [Cross-Listing]
There are different methods for finding exoplanets such as radial spectral shifts, astrometrical measurements, transits, timing etc. Gravitational microlensing (including pixel-lensing) is among the most promising techniques with the potentiality of detecting Earth-like planets at distances about a few astronomical units from their host star or near the so-called snow line with a temperature in the range $0-100^0$ C on a solid surface of an exoplanet. We emphasize the importance of polarization measurements which can help to resolve degeneracies in theoretical models. In particular, the polarization angle could give additional information about the relative position of the lens with respect to the source.
### The multi-wavelength polarization of Cygnus X-1
Polarization measurements of the microquasar Cygnus X-1 exist at gamma-ray, X-ray, UV, optical and radio frequencies. The gamma-ray emission has been shown to be highly linearly polarized. Here, we present new infrared polarimetric data of Cygnus X-1 taken with the 10.4-m Gran Telescopio Canarias and the 4.2-m William Herschel Telescope. We show that the broadband, radio to gamma-ray flux spectrum and polarization spectrum in the hard state are largely consistent with a simple phenomenological model of a strongly polarized synchrotron jet, an unpolarized Comptonized corona and a moderately polarized interstellar dust component. In this model, the origin of the gamma-ray, X-ray and some of the infrared polarization is the optically thin synchrotron power law from the inner regions of the jet. The model requires the magnetic field in this region to be highly ordered and perpendicular to the axis of the resolved radio jet. This differs to studies of some other X-ray binaries, in which the magnetic field is turbulent, variable and aligned with the jet axis. The model is able to explain the approximate polarization strength and position angle at all wavelengths including the detected X-ray (3 – 5 keV) polarization, except the observed position angle of the gamma-ray polarization, which differs to the model by ~ 60 degrees. Past numerical modelling has shown that a curved synchrotron spectrum can produce a shift in position angle by ~ 60 degrees, which may account for this.
### Probing magnetars magnetosphere through X-ray polarization measurements
The study of magnetars is of particular relevance since these objects are the only laboratories where the physics in ultra-strong magnetic fields can be directly tested. Until now, spectroscopic and timing measurements at X-ray energies in soft gamma-repeaters (SGRs) and anomalous X-ray pulsar (AXPs) have been the main source of information about the physical properties of a magnetar and of its magnetosphere. Spectral fitting in the ~ 0.5-10 keV range allowed to validate the "twisted magnetosphere" model, probing the structure of the external field and estimating the density and velocity of the magnetospheric currents. Spectroscopy alone, however, may fail in disambiguating the two key parameters governing magnetospheric scattering (the charge velocity and the twist angle) and is quite insensitive to the source geometry. X-ray polarimetry, on the other hand, can provide a quantum leap in the field by adding two extra observables, the linear polarization degree and the polarization angle. Using the bright AXP 1RXS J170849.0-400910 as a template, we show that phase-resolved polarimetric measurements can unambiguously determine the model parameters, even with a small X-ray polarimetry mission carrying modern photoelectric detectors and existing X-ray optics. We also show that polarimetric measurements can pinpoint vacuum polarization effects and thus provide an indirect evidence for ultra-strong magnetic fields.
### Long-term polarization observations of Mira variable stars suggest asymmetric structures
Mira and semi-regular variable stars have been studied for centuries but continue to be enigmatic. One unsolved mystery is the presence of polarization from these stars. In particular, we present 40 years of polarization measurements for the prototype o Ceti and V CVn and find very different phenomena for each star. The polarization fraction and position angle for Mira is found to be small and highly variable. On the other hand, the polarization fraction for V CVn is large and variable, from 2 – 7 %, and its position angle is approximately constant, suggesting a long-term asymmetric structure. We suggest a number of potential scenarios to explain these observations.
### Polarized synchrotron radiation from the Andromeda Galaxy M31 and background sources at 350 MHz [Replacement]
Low-frequency radio continuum observations are ideally suited to search for radio halos of inclined galaxies. Polarization measurements at low frequencies allow detection of small Faraday rotation measures caused by regular magnetic fields in galaxies and in the foreground of the Milky Way. M31 was observed in two overlapping pointings with the Westerbork Synthesis Radio Telescope (WSRT) resulting in about 4′ resolution in total intensity and linearly polarized emission. The frequency range 310-376 MHz was covered by 1024 channels which allowed the application of RM Synthesis on the polarization data. For the first time, diffuse polarized emission from a nearby galaxy is detected below 1 GHz. The degree of polarization is only 0.23 +/- 0.04 %, consistent with extrapolation of internal depolarization from data at higher radio frequency. A catalogue of 33 polarized sources and their Faraday rotation in the M31 field is presented. Their average depolarization is DP(90,20) = 0.14 +/- 0.02, 7 times stronger depolarized than at 1.4 GHz. We argue that this strong depolarization originates within the sources, e.g. in their radio lobes, or in intervening galaxies on the line of sight. On the other hand the Faraday rotation of the sources is mostly produced in the foreground of the Milky Way and varies significantly across the ~9 square degree M31 field. As expected, polarized emission from nearby galaxies and extragalactic background sources is much weaker at low frequencies compared to the GHz range. Future observations with LOFAR, with high sensitivity and high angular resolution to reduce depolarization, may reveal diffuse polarization from the outer disks and halos of galaxies.
### Polarized synchrotron radiation from the Andromeda Galaxy M31 and background sources at 350 MHz [Replacement]
Polarization measurements at low radio frequencies allow detection of small Faraday rotation measures caused by regular magnetic fields in galaxies and in the foreground of the Milky Way. M31 was observed in two overlapping pointings with the Westerbork Synthesis Radio Telescope (WSRT) resulting in ~4′ resolution in total intensity and linearly polarized emission. The frequency range 310-376 MHz was covered by 1024 channels which allowed the application of RM synthesis. We derived a data cube in Faraday depth and compared two symmetric ranges of negative and positive Faraday depths. This new method avoids the range of high instrumental polarization and allows the detection of very low degrees of polarization. For the first time, diffuse polarized emission from a nearby galaxy is detected below 1 GHz. The degree of polarization is only 0.23 +/- 0.04 %, consistent with extrapolation of internal depolarization from data at higher radio frequency. A catalogue of 33 polarized sources and their Faraday rotation in the M31 field is presented. Their average depolarization is DP(90,20) = 0.14 +/- 0.02, 7 times stronger depolarized than at 1.4 GHz. We argue that this strong depolarization originates within the sources, e.g. in their radio lobes, or in intervening galaxies on the line of sight. On the other hand, the Faraday rotation of the sources is mostly produced in the foreground of the Milky Way and varies significantly across the ~9 square degree M31 field. As expected, polarized emission from nearby galaxies and extragalactic background sources is much weaker at low frequencies compared to the GHz range. Future observations with LOFAR, with high sensitivity and high angular resolution to reduce depolarization, may reveal diffuse polarization from the outer disks and halos of galaxies.
### Polarized synchrotron radiation from the Andromeda Galaxy M31 and background sources at 350 MHz [Replacement]
Polarization measurements at low radio frequencies allow detection of small Faraday rotation measures caused by regular magnetic fields in galaxies and in the foreground of the Milky Way. The galaxy M31 was observed in two overlapping pointings with the Westerbork Synthesis Radio Telescope (WSRT) resulting in ~4′ resolution in total intensity and linearly polarized emission. The frequency range 310-376 MHz was covered by 1024 channels which allowed the application of RM synthesis. We derived a data cube in Faraday depth and compared two symmetric ranges of negative and positive Faraday depths. This new method avoids the range of high instrumental polarization and allows the detection of very low degrees of polarization. For the first time, diffuse polarized emission from a nearby galaxy is detected below 1 GHz. The degree of polarization is only 0.21 +/- 0.05 %, consistent with extrapolation of internal depolarization from data at higher radio frequency. A catalogue of 33 polarized sources and their Faraday rotation in the M31 field is presented. Their average depolarization is DP(90,20) = 0.14 +/- 0.02, 7 times stronger depolarized than at 1.4 GHz. We argue that this strong depolarization originates within the sources, e.g. in their radio lobes, or in intervening galaxies on the line of sight. On the other hand, the Faraday rotation of the sources is mostly produced in the foreground of the Milky Way and varies significantly across the ~9 square degree M31 field. As expected, polarized emission from M31 and extragalactic background sources is much weaker at low frequencies compared to the GHz range. Future observations with LOFAR, with high sensitivity and high angular resolution to reduce depolarization, may reveal diffuse polarization from the outer disks and halos of galaxies.
### Polarized synchrotron radiation from the Andromeda Galaxy M31 and background sources at 350 MHz
Low-frequency radio continuum observations are ideally suited to search for radio halos of inclined galaxies. Polarization measurements at low frequencies allow detection of small Faraday rotation measures caused by regular magnetic fields in galaxies and in the foreground of the Milky Way. M31 was observed in two overlapping pointings with the Westerbork Synthesis Radio Telescope (WSRT) resulting in about 4′ resolution in total intensity and linearly polarized emission. The frequency range 310-376 MHz was covered by 1024 channels which allowed the application of RM Synthesis on the polarization data. For the first time, diffuse polarized emission from a nearby galaxy is detected below 1 GHz. The degree of polarization is only 0.23 +/- 0.04 %, consistent with extrapolation of internal depolarization from data at higher radio frequency. A catalogue of 33 polarized sources and their Faraday rotation in the M31 field is presented. Their average depolarization is DP(90,20) = 0.14 +/- 0.02, 7 times stronger depolarized than at 1.4 GHz. We argue that this strong depolarization originates within the sources, e.g. in their radio lobes, or in intervening galaxies on the line of sight. On the other hand the Faraday rotation of the sources is mostly produced in the foreground of the Milky Way and varies significantly across the ~9 square degree M31 field. As expected, polarized emission from nearby galaxies and extragalactic background sources is much weaker at low frequencies compared to the GHz range. Future observations with LOFAR, with high sensitivity and high angular resolution to reduce depolarization, may reveal diffuse polarization from the outer disks and halos of galaxies.
### Optical linear polarization measurements of WR massive binary and single stars
We present optical (UBVRI) linear polarimetric observations of 8 Wolf-Rayet (WR) massive binaries and single stars. We have corrected the observed values for the interstellar extinction and polarization by the interstellar medium to obtain the intrinsic polarization and position angle. We find three highly polarization stars between 5% and 10% (WR1, WR5 and WR146), three between 3% and 4% (WR2, WR3 and WR4), and two between 1% and 2% (WR137 and WR140). Moreover, 5 stars show increasing degree of polarization to shorter wavelengths (e.g WR 146) indicative with asymmetric circumstellar envelope and 3 have nearly constant polarization within the errors (e.g WR 140).
### Polarization of GRB Prompt Emission
We review the recent observational results of the gamma-ray linear polarization of Gamma-Ray Bursts (GRBs), and discuss some theoretical implications for the prompt emission mechanism and the magnetic composition of GRB jets. We also report a strict observational verification of CPT invariance in the photon sector as a result of the GRB polarization measurements.
### Candidate Type II Quasars at 2 < z < 4.3 in the Sloan Digital Sky Survey III
At low redshifts, dust-obscured quasars often have strong yet narrow permitted lines in the rest-frame optical and ultraviolet, excited by the central active nucleus, earning the designation Type II quasars. We present a sample of 145 candidate Type II quasars at redshifts between 2 and 4.3, encompassing the epoch at which quasar activity peaked in the universe. These objects, selected from the quasar sample of the Baryon Oscillation Spectroscopic Survey of the Sloan Digital Sky Survey III, are characterized by weak continuum in the rest-frame ultraviolet (typical continuum magnitude of i \approx 22) and strong lines of CIV and Ly \alpha, with Full Width at Half Maximum less than 2000 kms-1. The continuum magnitudes correspond to an absolute magnitude of -23 or brighter at redshift 3, too bright to be due exclusively to the host galaxies of these objects. Roughly one third of the objects are detected in the shorter-wavelength bands of the WISE survey; the spectral energy distributions (SEDs) of these objects appear to be intermediate between classic Type I and Type II quasars seen at lower redshift. Five objects are detected at rest frame 6\mu m by Spitzer, implying bolometric luminosities of several times 10^46 erg s-1. We have obtained polarization measurements for two objects; they are roughly 3% polarized. We suggest that these objects are luminous quasars, with modest dust extinction (A_V ~ 0.5 mag), whose ultraviolet continuum also includes a substantial scattering contribution. Alternatively, the line of sight to the central engines of these objects may be partially obscured by optically thick material.
### Candidate Type II Quasars at 2 < z < 4.3 in the Sloan Digital Sky Survey III [Replacement]
At low redshifts, dust-obscured quasars often have strong yet narrow permitted lines in the rest-frame optical and ultraviolet, excited by the central active nucleus, earning the designation Type II quasars. We present a sample of 145 candidate Type II quasars at redshifts between 2 and 4.3, encompassing the epoch at which quasar activity peaked in the universe. These objects, selected from the quasar sample of the Baryon Oscillation Spectroscopic Survey of the Sloan Digital Sky Survey III, are characterized by weak continuum in the rest-frame ultraviolet (typical continuum magnitude of i \approx 22) and strong lines of CIV and Ly \alpha, with Full Width at Half Maximum less than 2000 kms-1. The continuum magnitudes correspond to an absolute magnitude of -23 or brighter at redshift 3, too bright to be due exclusively to the host galaxies of these objects. Roughly one third of the objects are detected in the shorter-wavelength bands of the WISE survey; the spectral energy distributions (SEDs) of these objects appear to be intermediate between classic Type I and Type II quasars seen at lower redshift. Five objects are detected at rest frame 6\mu m by Spitzer, implying bolometric luminosities of several times 10^46 erg s-1. We have obtained polarization measurements for two objects; they are roughly 3% polarized. We suggest that these objects are luminous quasars, with modest dust extinction (A_V ~ 0.5 mag), whose ultraviolet continuum also includes a substantial scattering contribution. Alternatively, the line of sight to the central engines of these objects may be partially obscured by optically thick material.
### Cosmic-ray leptons, magnetic fields and interstellar synchrotron emission
Interstellar synchrotron emission depends on Galactic magnetic fields and on cosmic-ray leptons. Observations of radio emission are an important tool for studying cosmic-ray propagation models and interstellar electron spectrum and distribution in the Galaxy. We present the latest developments in our modeling of Galactic synchrotron emission with the GALPROP code, including polarization, absorption, and free-free emission. Using surveys over a wide range of radio frequencies and polarization measurements, we derive constraints on the low-energy interstellar cosmic-ray electron spectrum, magnetic fields and cosmic-ray propagation models. This work is of interest for studies of interstellar gamma-ray emission with Fermi-LAT, and synchrotron for the Planck mission.
### Cosmic-ray leptons, magnetic fields and interstellar synchrotron emission [Replacement]
Interstellar synchrotron emission depends on Galactic magnetic fields and on cosmic-ray leptons. Observations of radio emission are an important tool for studying cosmic-ray propagation models and interstellar electron spectrum and distribution in the Galaxy. We present the latest developments in our modeling of Galactic synchrotron emission with the GALPROP code, including polarization, absorption, and free-free emission. Using surveys over a wide range of radio frequencies and polarization measurements, we derive constraints on the low-energy interstellar cosmic-ray electron spectrum, magnetic fields and cosmic-ray propagation models. This work is of interest for studies of interstellar gamma-ray emission with Fermi-LAT, and synchrotron for the Planck mission.
### Constraints on neutrino masses from Planck and Galaxy Clustering data
We present here bounds on neutrino masses from the combination of recent Planck Cosmic Microwave Background measurements and galaxy clustering information from the Baryon Oscillation Spectroscopic Survey (BOSS), part of the Sloan Digital Sky Survey-III. We use the full shape of either the photometric angular clustering (Data Release 8) or the 3D spectroscopic clustering (Data Release 9) power spectrum in different cosmological scenarios. In the Lambda\$CDM scenario, spectroscopic galaxy clustering measurements improve significantly the existing neutrino mass bounds from Planck data. We find sum m_nu< 0.39 eV at 95% confidence level for the combination of the 3D power spectrum with Planck CMB data (with lensing included) and Wilkinson Microwave Anisoptropy Probe 9-year polarization measurements. Therefore, robust neutrino mass constraints can be obtained without the addition of the prior on the Hubble constant from HST. In extended cosmological scenarios with a dark energy fluid or with non flat geometries, galaxy clustering measurements are essential to pin down the neutrino mass bounds, providing in the majority of cases better results than those obtained from the associated measurement of the Baryon Acoustic Oscillation scale only. In the presence of a freely varying (constant) dark energy equation of state, we find sum m_nu<0.49 eV at 95% confidence level for the combination of the 3D power spectrum with Planck CMB data (with lensing included) and Wilkinson Microwave Anisoptropy Probe 9-year polarization measurements. This same data combination in non flat geometries provides the neutrino mass bound sum m_nu<0.35 eV at 95% confidence level.
### On the Statistical Analysis of X-ray Polarization Measurements
In many polarimetry applications, including observations in the X-ray band, the measurement of a polarization signal can be reduced to the detection and quantification of a deviation from uniformity of a distribution of measured angles. We explore the statistics of such polarization measurements using Monte Carlo simulations and chi-squared fitting methods. We compare our results to those derived using the traditional probability density used to characterize polarization measurements and quantify how they deviate as the intrinsic modulation amplitude grows. We derive relations for the number of counts required to reach a given detection level (parameterized by beta, the "number of sigma’s" of the measurement) appropriate for measuring the modulation amplitude by itself (single interesting parameter case) or jointly with the position angle (two interesting parameters case). We show that for the former case when the intrinsic amplitude is equal to the well known minimum detectable polarization (MDP) it is, on average, detected at the 3-sigma level. For the latter case, when one requires a joint measurement at the same confidence level, then more counts are needed than that required to achieve the MDP level. This additional factor is amplitude-dependent, but is approximately 2.2 for intrinsic amplitudes less than about 20%. It decreases slowly with amplitude and is 1.8 when the amplitude is 50%. We find that the position angle uncertainty at 1-sigma confidence is well described by the relation 28.5 (deg) / beta.
### The Submillimeter Polarization Spectrum of M17
We present 450 {\mu}m polarimetric observations of the M17 molecular cloud obtained with the SHARP polarimeter at the Caltech Submillimeter Observatory. Across the observed region, the magnetic field orientation is consistent with previous submillimeter and far-infrared polarization measurements. Our observations are centered on a region of the molecular cloud that has been compressed by stellar winds from a cluster of OB stars. We have compared these new data with previous 350 {\mu}m polarimetry and find an anti-correlation between the 450 to 350 {\mu}m polarization magnitude ratio and the ratio of 21 cm to 450 {\mu}m intensity. The polarization ratio is lower near the east end of the studied region where the cloud is exposed to stellar winds and radiation. At the west end of the region, the polarization ratio is higher. We interpret the varying polarization spectrum as evidence supporting the radiative alignment torque (RAT) model for grain alignment, implying higher alignment efficiency in the region that is exposed to a higher anisotropic radiation field.
### Three-dimensional magnetic and abundance mapping of the cool Ap star HD 24712 I. Spectropolarimetric observations in all four Stokes parameters
High-resolution spectropolarimetric observations provide simultaneous information about stellar magnetic field topologies and three-dimensional distributions of chemical elements. Here we present analysis of a unique full Stokes vector spectropolarimetric data set, acquired for the cool magnetic Ap star HD 24712. The goal of our work is to examine circular and linear polarization signatures inside spectral lines and to study variation of the stellar spectrum and magnetic observables as a function of rotational phase. HD 24712 was observed with the HARPSpol instrument at the 3.6-m ESO telescope over a period of 2010-2011. The resulting spectra have S/N ratio of 300-600 and resolving power exceeding 100000. The multiline technique of least-squares deconvolution (LSD) was applied to combine information from the spectral lines of Fe-peak and rare-earth elements. We used the HARPSPol spectra of HD 24712 to study the morphology of the Stokes profile shapes in individual spectral lines and in LSD Stokes profiles corresponding to different line masks. From the LSD Stokes V profiles we measured the longitudinal component of the magnetic field, <Bz>, with an accuracy of 5-10 G. We also determined the net linear polarization from the LSD Stokes Q and U profiles. We determined an improved rotational period of the star, P_rot = 12.45812 +/- 0.00019d. We measured <Bz> from the cores of Halpha and Hbeta lines. The analysis of <Bz> measurements showed no evidence for a significant radial magnetic field gradient in the atmosphere of HD 24712. We used our <Bz> and net linear polarization measurements to determine parameters of the dipolar magnetic field topology. We found that magnetic observables can be reasonably well reproduced by the dipolar model. We discovered rotational modulation of the Halpha core and related it a non-uniform surface distribution of rare-earth elements.
### CMB Faraday rotation as seen through the Milky Way
Faraday Rotation (FR) of CMB polarization, as measured through mode-coupling correlations of E and B modes, can be a promising probe of a stochastic primordial magnetic field (PMF). While the existence of a PMF is still hypothetical, there will certainly be a contribution to CMB FR from the magnetic field of the Milky Way. We use existing estimates of the Milky Way rotation measure (RM) to forecast its detectability with upcoming and future CMB experiments. We find that the galactic RM will not be seen in polarization measurements by Planck, but that it will need to be accounted for by CMB experiments capable of detecting the weak lensing contribution to the B-mode. We then discuss prospects for constraining the PMF in the presence of FR due to the galaxy under various assumptions that include partial de-lensing and partial subtraction of the galactic FR. We find that a realistic future sub-orbital experiment, covering a patch of the sky near the galactic poles, can detect a scale-invariant PMF of 0.1 nano-Gauss at better than 95% confidence level, while a dedicated space-based experiment can detect even smaller fields.
### CMB Faraday rotation as seen through the Milky Way [Replacement]
Faraday Rotation (FR) of CMB polarization, as measured through mode-coupling correlations of E and B modes, can be a promising probe of a stochastic primordial magnetic field (PMF). While the existence of a PMF is still hypothetical, there will certainly be a contribution to CMB FR from the magnetic field of the Milky Way. We use existing estimates of the Milky Way rotation measure (RM) to forecast its detectability with upcoming and future CMB experiments. We find that the galactic RM will not be seen in polarization measurements by Planck, but that it will need to be accounted for by CMB experiments capable of detecting the weak lensing contribution to the B-mode. We then discuss prospects for constraining the PMF in the presence of FR due to the galaxy under various assumptions that include partial de-lensing and partial subtraction of the galactic FR. We find that a realistic future sub-orbital experiment, covering a patch of the sky near the galactic poles, can detect a scale-invariant PMF of 0.1 nano-Gauss at better than 95% confidence level, while a dedicated space-based experiment can detect even smaller fields.
### Polarimetric Properties of Flux-Ropes and Sheared Arcades in Coronal Prominence Cavities
The coronal magnetic field is the primary driver of solar dynamic events. Linear and circular polarization signals of certain infrared coronal emission lines contain information about the magnetic field, and to access this information, either a forward or an inversion method must be used. We study three coronal magnetic configurations that are applicable to polar-crown filament cavities by doing forward calculations to produce synthetic polarization data. We analyze these forward data to determine the distinguishing characteristics of each model. We conclude that it is possible to distinguish between cylindrical flux ropes, spheromak flux ropes, and sheared arcades using coronal polarization measurements. If one of these models is found to be consistent with observational measurements, it will mean positive identification of the magnetic morphology that surrounds certain quiescent filaments, which will lead to a greater understanding of how they form and why they erupt.
### Comparison of force-free coronal magnetic field modeling using vector fields from Hinode and Solar Dynamics Observatory
Photospheric magnetic vector maps from two different instruments are used to model the nonlinear force-free coronal magnetic field above an active region. We use vector maps inferred from polarization measurements of the Solar Dynamics Observatory/Helioseismic and Magnetic Imager (HMI) and the Solar Optical Telescope Spectropolarimeter (SP) aboard Hinode. Besides basing our model calculations on HMI data, we use both, SP data of original resolution and scaled down to the resolution of HMI. This allows us to compare the model results based on data from different instruments and to investigate how a binning of high-resolution data effects the model outcome. The resulting 3D magnetic fields are compared in terms of magnetic energy content and magnetic topology. We find stronger magnetic fields in the SP data, translating into a higher total magnetic energy of the SP models. The net Lorentz forces of the HMI and SP lower boundaries verify their force-free compatibility. We find substantial differences in the absolute estimates of the magnetic field energy but similar relative estimates, e.g., the fraction of excess energy and of the flux shared by distinct areas. The location and extension of neighboring connectivity domains differs and the SP model fields tend to be higher and more vertical. Hence, conclusions about the magnetic connectivity based on force-free field models are to be drawn with caution. We find that the deviations of the model solution when based on the lower-resolution SP data are small compared to the differences of the solutions based on data from different instruments.
### Interferometric Upper Limits on Millimeter Polarization of the Disks around DG Tau, GM Aur, and MWC 480
Millimeter-wavelength polarization measurements offer a promising method for probing the geometry of magnetic fields in circumstellar disks. Single dish observations and theoretical work have hinted that magnetic field geometries might be predominantly toroidal, and that disks should exhibit millimeter polarization fractions of 2-3%. While subsequent work has not confirmed these high polarization fractions, either the wavelength of observation or the target sources differed from the original observations. Here we present new polarimetric observations of three nearby circumstellar disks at 2" resolution with the Submillimeter Array (SMA) and the Combined Array for Research in Millimeter Astronomy (CARMA). We reobserve GM Aur and DG Tau, the systems in which millimeter polarization detections have been claimed. Despite higher resolution and sensitivity at wavelengths similar to the previous observations, the new observations do not show significant polarization. We also add observations of a new HAeBe system, MWC 480. These observations demonstrate that a very low (<0.5%) polarization fraction is probably common at large (>100 AU) scales in bright circumstellar disks. We suggest that high-resolution observations may be worthwhile to probe magnetic field structure on linear distances smaller than the disk scale height, as well as in regions closer to the star that may have larger MRI-induced magnetic field strengths.
### The AGN phenomenon: open issues
The aim of this short paper is to motivate and encourage research in the field of Active Galactic Nuclei (AGN). Here we summarize the main open questions concerning the central engine. Is the central black hole rapidly spinning and can we prove this? What is the dominant accretion mechanism in AGN? Why do some AGN form jets while others don’t and how do the jets originate? What keeps jets collimated out to distances of 100 kpc? Is the emission of blazars dominated rather by synchrotron self-Compton or by external Compton processes? Which parameters are important in the unified model? We outline the status of related research, formulate the questions and try to hint at research projects able to tackle these fundamental topics. Deep surveys, polarization measurements, improved models, faster and more accurate simulations as well as bridging the gap in the MeV range can be part of the tools to bring us closer to an understanding of AGN.
### Unveiling a network of parallel filaments in the Infrared Dark Cloud G14.225-0.506
We present the results of combined NH3(1,1) and (2,2) line emission observed with the Very Large Array and the Effelsberg 100m telescope of the Infrared Dark Cloud G14.225-0.506. The NH3 emission reveals a network of filaments constituting two hub-filament systems. Hubs are associated with gas of rotational temperature Trot \sim 25 K, non-thermal velocity dispersion ~1.1 km/s, and exhibit signs of star formation, while filaments appear to be more quiescent (Trot \sim 11 K, non-thermal velocity dispersion ~0.6 km/s). Filaments are parallel in projection and distributed mainly along two directions, at PA \sim 10 deg and 60 deg, and appear to be coherent in velocity. The averaged projected separation between adjacent filaments is between 0.5 pc and 1pc, and the mean width of filaments is 0.12 pc. Cores within filaments are separated by ~0.33 pc, which is consistent with the predicted fragmentation of an isothermal gas cylinder due to the ‘sausage’-type instability. The network of parallel filaments observed in G14.225-0.506 is consistent with the gravitational instability of a thin gas layer threaded by magnetic fields. Overall, our data suggest that magnetic fields might play an important role in the alignment of filaments, and polarization measurements in the entire cloud would lend further support to this scenario.
### Non-Zeeman Circular Polarization of Molecular Rotational Spectral Lines
We present measurements of circular polarization from rotational spectral lines of molecular species in Orion KL, most notably 12CO (J=2 – 1), obtained at the Caltech Submillimeter Observatory with the Four-Stokes-Parameter Spectra Line Polarimeter. We find levels of polarization of up to 1 to 2% in general, for 12CO (J=2 – 1) this level is comparable to that of linear polarization also measured for that line. We present a physical model based on resonant scattering in an attempt to explain our observations. We discuss how slight differences in scattering amplitudes for radiation polarized parallel and perpendicular to the ambient magnetic field, responsible for the alignment of the scattering molecules, can lead to the observed circular polarization. We also show that the effect is proportional to the square of the magnitude of the plane of the sky component of the magnetic field, and therefore opens up the possibility of measuring this parameter from circular polarization measurements of Zeeman insensitive molecules.
### Summer Mesosphere Temperature Distribution from Wide-Angle Polarization Measurements of the Twilight Sky [Cross-Listing]
The paper contains the results of wide-angle polarization camera (WAPC) measurements of the twilight sky background conducted in summer 2011 and 2012 at 55.2 degs.N, 37.5 degs.E, southwards from Moscow. The method of single scattering separation based on polarization data is suggested. The obtained components of scattering matrixes show the domination of Rayleigh scattering in the mesosphere for all observation days. It made possible to retrieve the altitude distribution of temperature in the mesosphere. The results are compared with the temperature data by TIMED/SABER and EOS Aura/MLS instruments for nearby dates and locations.
### Reionization on Large Scales III: Predictions for Low-ell Cosmic Microwave Background Polarization and High-ell Kinetic Sunyaev-Zel'dovich Observables
We present new predictions for temperature (on small angular scales) and polarization (on large angular scales) CMB anisotropies induced during the epoch of reionization (EoR). Using a novel method calibrated from Radiation-Hydrodynamic simulations we model the EoR in large volumes (L >~ 2 Gpc/h) in the context of galactic reionization. We find that the EoR contribution to the kinetic Sunyaev- Zel’dovich power spectrum (patchy kSZ) ranges between ~0.6 – 2.8 muK^2 at ell = 3000, for the parameter space we explored. These patchy kSZ power spectra are calculated from large 15 Deg x 15 Deg maps that are found to be necessary. Decreasing the size of these maps biases the overall patchy kSZ power to higher values. We find that the amplitude of the patchy kSZ power spectrum at ell = 3000 follows simple scalings of D_ell=3000^kSZ propto <z> and D_ell=3000^kSZ propto Delz^0.47 for the mean redshift (<z>) of reionization and the duration (dz). Using the constraints on <z> from WMAP 7-year results and the lower limit on dz from EDGES we find a lower limit of ~ 0.4 muK^2 on the kSZ at ell = 3000. Planck will constrain the mean redshift and the Thomson optical depth from the low-ell polarization power spectrum. Future measurements of the high-ell CMB power spectrum from the South Pole Telescope (SPT) and the Atacama Cosmology Telescope (ACT) should detect the patchy kSZ signal if the cross correlation between the cosmic infrared background and the thermal Sunyaev Zel’dovich effect is constrained. We show that the combination of temperature and polarization measurements constrains both <z> and dz. The patchy kSZ maps, power spectra templates and the polarization power spectra will be publicly available.
### SPTpol: an instrument for CMB polarization measurements with the South Pole Telescope
SPTpol is a dual-frequency polarization-sensitive camera that was deployed on the 10-meter South Pole Telescope in January 2012. SPTpol will measure the polarization anisotropy of the cosmic microwave background (CMB) on angular scales spanning an arcminute to several degrees. The polarization sensitivity of SPTpol will enable a detection of the CMB "B-mode" polarization from the detection of the gravitational lensing of the CMB by large scale structure, and a detection or improved upper limit on a primordial signal due to inflationary gravity waves. The two measurements can be used to constrain the sum of the neutrino masses and the energy scale of inflation. These science goals can be achieved through the polarization sensitivity of the SPTpol camera and careful control of systematics. The SPTpol camera consists of 768 pixels, each containing two transition-edge sensor (TES) bolometers coupled to orthogonal polarizations, and a total of 1536 bolometers. The pixels are sensitive to light in one of two frequency bands centered at 90 and 150 GHz, with 180 pixels at 90 GHz and 588 pixels at 150 GHz. The SPTpol design has several features designed to control polarization systematics, including: single-moded feedhorns with low cross-polarization, bolometer pairs well-matched to difference atmospheric signals, an improved ground shield design based on far-sidelobe measurements of the SPT, and a small beam to reduce temperature to polarization leakage. We present an overview of the SPTpol instrument design, project status, and science projections.
### Feedhorn-coupled TES polarimeter camera modules at 150 GHz for CMB polarization measurements with SPTpol
The SPTpol camera is a dichroic polarimetric receiver at 90 and 150 GHz. Deployed in January 2012 on the South Pole Telescope (SPT), SPTpol is looking for faint polarization signals in the Cosmic Microwave Background (CMB). The camera consists of 180 individual Transition Edge Sensor (TES) polarimeters at 90 GHz and seven 84-polarimeter camera modules (a total of 588 polarimeters) at 150 GHz. We present the design, dark characterization, and in-lab optical properties of the 150 GHz camera modules. The modules consist of photolithographed arrays of TES polarimeters coupled to silicon platelet arrays of corrugated feedhorns, both of which are fabricated at NIST-Boulder. In addition to mounting hardware and RF shielding, each module also contains a set of passive readout electronics for digital frequency-domain multiplexing. A single module, therefore, is fully functional as a miniature focal plane and can be tested independently. Across the modules tested before deployment, the detectors average a critical temperature of 478 mK, normal resistance R_N of 1.2 Ohm, unloaded saturation power of 22.5 pW, (detector-only) optical efficiency of ~ 90%, and have electrothermal time constants < 1 ms in transition.
### Separation of two contributions to the high energy emission of Cygnus X-1: Polarization measurements with INTEGRAL SPI
Operational since 2002 on-board the INTEGRAL observatory, the SPI spectrometer can be used to perform polarization measurements in the hard X-ray/soft gamma-ray domain (~ 130 keV – 8 MeV). However, this phenomenon is complex to measure at high energy and requires high fluxes. Cyg X-1 appears as the best candidate amongst the X-ray binaries since it is one of the brightest persistent sources in this energy domain. Furthermore, a polarized component has recently been reported above 400 keV from IBIS data. We have therefore dedicated our efforts to develop the required tools to study the polarization in the INTEGRAL SPI data and have first applied them to 2.6 Ms of Cyg X-1 observations, covering 6.5 years of the INTEGRAL mission. We have found that the high energy emission of Cyg X-1 is indeed polarized, with a mean polarization fraction of 76 % +/- 15 % at a position angle estimated to 42 +/- 3 degrees, for energies above 230 keV. The polarization fraction clearly increases with energy. In the 130-230 keV band, the polarization fraction is lower than 20 %, but exceeds 75 % between 370 and 850 keV, with the (total) emission vanishing above this energy. This result strongly suggests that the emission originates from the jet structure known to emit in the radio domain. The same synchrotron process could be responsible for the emission from radio to MeV, implying the presence of high energy electrons. This illustrates why the polarization of the high energy emission in compact objects is an increasingly important observational objective.
### Discerning the location of the gamma-ray emission region in blazars from multi-messenger observations
Relativistic jets in AGN in general, and in blazars in particular, are the most energetic and among the most powerful astrophysical objects known so far. Their relativistic nature provides them with the ability to emit profusely at all spectral ranges from radio wavelengths to gamma-rays, as well as to vary extremely at time scales from hours to years. Since the birth of gamma-ray astronomy, locating the origin of gamma-ray emission has been a fundamental problem for the knowledge of the emission processes involved. Deep and densely time sampled monitoring programs with the Fermi Gamma-ray Space Telescope and other facilities at most of the available spectral ranges (including millimeter interferometric imaging and polarization measurements wherever possible) are starting to shed light for the case of blazars. After a short review of the status of the problem, we summarize two of our latest results -obtained from the comprehensive monitoring data compiled by the Boston University Blazar monitoring program – that locate the GeV flaring emission of the BL Lac objects AO 0235+164 and OJ287 within the jets of these blazars, at >12 parsecs from the central AGN engine. | {"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": 0.8873763084411621, "perplexity": 1528.9041104189805}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510273350.41/warc/CC-MAIN-20140728011753-00453-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://mathhelpforum.com/math-topics/29670-wavelength-radio-print.html | • Mar 1st 2008, 08:30 PM
lovinhockey26
Find the wavelength for a typical AM radio with a frequency of 900 kHz?
Find the wavelength for a typical FM radio with a frequency of 99.6 MHz
• Mar 2nd 2008, 01:25 AM
earboth
Quote:
Originally Posted by lovinhockey26
Find the wavelength for a typical AM radio with a frequency of 900 kHz?
Find the wavelength for a typical FM radio with a frequency of 99.6 MHz
I'll use the value of the light speed in vacuum which I have at hand (this value is from 1972 so it may differ a little bit today): $c = 299,792,456\ \frac ms$
Let f denote the frequency and l the wavelength then you know that
$f \cdot l = c$
Plug in the values you already know:
$900 \cdot 10^3\ \frac1s \cdot l = 299,792,456\ \frac ms$
$99.6 \cdot 10^6\ \frac1s \cdot l = 299,792,456\ \frac ms$ . Solve for l.
(For confrmation only: 333.103 m; 3.01 m) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.989747941493988, "perplexity": 1608.6995100968813}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891812247.50/warc/CC-MAIN-20180218173208-20180218193208-00076.warc.gz"} |
https://link.springer.com/chapter/10.1007%2F978-3-642-31692-0_1 | # Introduction
Chapter
Part of the Foundations in Signal Processing, Communications and Networking book series (SIGNAL, volume 8)
## Keywords
Channel State Information Broadcast Channel Perfect Channel State Information Multiple Access Channel Dirty Paper Code
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves. | {"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": 0.8043026924133301, "perplexity": 3598.2791373477885}, "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/1524125946256.50/warc/CC-MAIN-20180423223408-20180424003408-00636.warc.gz"} |
http://www.ibiblio.org/kuphaldt/electricCircuits/Devel/tutorial.html | # The SubML markup language
## Introduction
SubML stands for Substitutionary Markup Language. Similar in structure to an SGML-based language, SubML is intended for simple text formatting with very few frills, but providing the capability of standard font emphasis modes, itemized lists, and image inclusion.
SubML is designed so that it may be translated into practically any markup language with nothing more than some search-and-replace commands (hence the term substitutionary), executed in the sed stream editor. Rather than rely on complex translational algorithms (i.e. a Perl or Python script), the philosophy here is to design ease of conversion into the structure of the original markup so that any fool can write a sed script to convert to any new markup. So far, the following conversions are provided in a set of sed scripts supplied with this tutorial:
• SubML to TEX
• SubML to LATEX
• SubML to HTML
• SubML to plain text (ASCII)
More conversion routines are planned. As far as I can see, none of them should present any unordinary difficulties in conversion. I simply haven't got around to writing and testing all the scripts yet. These include:
• SubML to nroff/troff/groff
• SubML to Texinfo
• SubML to DocBook (SGML and/or XML)
• SubML to Lout
• SubML to Qwertz
Also, it should be fairly easy to write an XML DTD for SubML, making it directly readable by XML-compatible browsers and other software.
Platform compatibility is limited only to the availability of a sed binary to perform the conversion. And since sed is such a widely used and robust utility (free, too, thanks to the Free Software Foundation!), this should not be a problem. I've successfully “compiled” SubML documents on both Linux and Microsoft Windows 95 with equal ease.
Characters usually interpreted as escape characters in other markup languages like \, &, $, %, |, ~, ^, and _ are handled without special tagging as well (100% of the time, too -- this makes SubML worth$1,000,000 & that's not all!). The only characters SubML requires you to specially code (not type verbatim in your source document) are the < and > symbols, simply because SubML itself uses them as escape characters to mark the beginning and end of tags.
## Levels of sections under each chapter
This is text contained in the first true section of this tutorial.
### This is the first subsection (titlebar)
This is text contained in the first subsection of this tutorial.
### This is the second subsection (titlebar)
This is text contained in the second subsection of this tutorial.
#### This is the first subsubsection (titlebar)
This is text contained in the first subsubsection of this tutorial, which is within the second subsection.
## Gallery of inline text formatting tricks
In this section, we will explore the various inline (embedded within a sentence) formatting commands provided by SubML.
Note that this may not be the fanciest array of formatting commands, but it should suffice for most common formatting requirements.
If the standard SubML philosophy is followed, additional formatting capabilities may be included at a later date. The only real restriction is that whatever formatting capability is added must be translatable to the desired output type (TEX, HTML, DocBook, etc.) using nothing more than simple search-and-replace algorithms.
### Sub- and super-scripting
This is a test of the subscripting and superscripting capabilities of SubML. This is useful to create simple mathematical (-2-3 = -0.125) and chemical (H2O, 92U235) expressions.
While the following displays in html, it does not display properly in ps or pdf due to tex/latex errors when using the normal <subscript>, <superscript>, as above. Instead, we use <subscript2>, <superscript2>.
10log10(VI)
10log10(VI/VO)
Un-comment line here to create error.
Note the <math> </math> around the whole subscript and superscript line in the tutorial.sml source above.(You need to be looking at tutorial.sml) Only use this if you have tex/Latex errors, no ps or pdf. Complex mixtures of both superscripts and subscripts are a reason.
### Boolean overline negation
Boolean negation (not) is supported in LaTeX by the \overline{ } command, available in the math environment. HTML provides no such support for overline. However, it is customary in some texts to indicate negation with a single quote (') post-fixed to the negated variable. Thus, we support Boolean negation in SML with the <ob> and </ob> tags (overbar) enclosing the negated variable.The sed processed Latex output will show (dvi, ps, pdf) overline negated variables, the html has the post-fixed single quote form of negation. Equations with any negated variables must be surrounded by <math> and </math> tags to activate the "math" environment for latex.
Any extensive use of Boolean equations should be xcircuit images so that real overlines will be available in html as well as LaTeX. The methods here are meant to support simple in-sentence Boolean expressions, not free-standing equations.
<math>Y = (<ob>A</ob> + <ob>B</ob>)</math> This markup gives the result below:
Y = (A' + B' ) This result.
<math>Y = <ob>(<ob>A</ob> + <ob>B</ob>)</ob></math> This markup gives:
Y =(A' + B' )' This result with long overline is due to outer tags. The span of the overline is analogous to the span of a pair of bold tags. While the parenthesis are not necessary in the LaTeX rendition, they are mandatory in the "single quoted" html version to indicate the extent of the negation.
Some other examples follow:
Y = (A' + B' )' =((A B)' )'
Y = (A' B' C' ED' )' Incorrect in LaTeX, we wanted broken bar BC like AB.
Y = ( A ' B ' C ' E D ' )' This is incorrect in LaTeX, OK on html. We wanted broken bar between ABC.
Y = (A' B' C' ED' )' Like this by putting spaces between ABC. See tutorial.sml
Y = ((A (B C' )' )' )' This is better as an xcircuit image; html is difficult to follow.
### Emphasis fonts
Italicized, boldface, and underlined type are also available in SubML.
### Special dashes
The regular dash, such as that used for hyphenation, looks-like-this. A dash specifically used for subtraction is typeset using a special SubML tag, so that 5-3 (math dash) looks distinct from 5-3 (ordinary dash). Some people don't care too much about this, so use this tag at your discretion.
Sometimes it is useful to show a pair of dashes -- not the “em-dash” used in setting off a section of text like this -- but a real pair of dashes. In this case, another special SubML tag has been created to do this -- and you just read over it! I use it to denote series-connected electronic components in symbolic form. For example, a pair of resistors (R1 and R2) are connected in parallel with each other, but together they're in series with R3. Symbolically, I represent such a configuration like this: (R1//R2)--R3.
## Block formatting
An important feature I've found in document processing is the ability to typeset a literal segment of text. That is, a section of print in a monospaced font with all normal paragraph formatting features of the target markup language turned off.
One common usage of this feature is for the typesetting of computer programming code. An example follows:
File listing: hello.c
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
. #include <stdio.h>
.
. int main(void)
. {
. printf("\nHello, world!\n");
. return (0);
. }
.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The dots are inserted manually within the SubML document to “set off” the literal block of text from the rest of the document. Also, the leading dots (at very left of each line) help overcome a problem I'm having with TEX formatting where leading spaces get discarded and everything ends up smashed against the left margin.
Without the dots, it looks like this:
#include <stdio.h>
int main(void)
{
printf("\nHello, world!\n");
return (0);
}
The "set off" leading dot may be replaced by the <sp> tag to avoid the dot in your literal block.
#include <stdio.h>
int main(void)
{
printf("\nHello, world!\n");
return (0);
}
Another kind of block formatting is the inclusion of offset quotations. Note the following example:
"Vague and insignificant forms of speech, and abuse of language, have so long passed for mysteries of science; and hard or misapplied words with little or no meaning have, by prescription, such a right to be mistaken for deep learning or height of speculation, that it will not be easy to persuade either those who speak or those who hear them, that they are but the covers of ignorance and hindrance of true knowledge." - John Locke
Italics may also be added to “set off” a quotation from the rest of the text, especially in HTML. Combining the italic and bold tag sets inside of the quotation tag set accomplishes this goal nicely:
"Vague and insignificant forms of speech, and abuse of language, have so long passed for mysteries of science; and hard or misapplied words with little or no meaning have, by prescription, such a right to be mistaken for deep learning or height of speculation, that it will not be easy to persuade either those who speak or those who hear them, that they are but the covers of ignorance and hindrance of true knowledge." - John Locke
While perhaps not a true block-formatting feature, itemized lists can be created using SubML. Take the following example:
• This is the first item
• This is the second item
• This is the third item
• Isn't this fun?
In the spirit of simplicity, I haven't created the option of enumerated lists, indented lists, or anything fancy like that within the language of SubML.
## Including graphic images in a document
Graphic image inclusion is perhaps the best feature of SubML. Note the following example:
You must be sure to specify an HTML-compatible image in the markup code. This means an image file specified with a filename ending in .png, .jpg, .bmp, or .gif (three-character extensions only: .jpg, not .jpeg!). For TEX or LATEX output, there must be an Encapsulated Postscript image file .eps in the same directory, but not specified in the markup code.
For example, the markup code necessary to place the "happy face" image shown above is as follows:
<image>test.png</image>
Two versions of the image exist: test.png for inclusion into the HTML output, and test.eps for inclusion into the TEX or LATEX output, but only the HTML-compatible file need be specified in the SubML source code.
This is a fine caption.
Below is the markup code necessary to place the "happy face" image with a caption shown in figure above. A "Figure x.x" string precedes the caption in LATEX. It also generates LATEX code for a //lable test.eps, which is used to reference the figure. The caption is included in the html without the "Figure x.x" designation.
<image>test.png<caption>This is a fine caption.</caption></image>
Note that in the previous paragraph, we reference "figure 1.1" or "figure above" in tutorial.ps and tutorial.html, respectively. The markup below, between the ref tags, is for referencing the above image as a figure. The image name, test.png, is a symbolic reference, replaced by 1.1, 1.2, etc., during "latex tutorial.latex" processing. Put the image name between the tags.
See figure<ref>test.png</ref> for a "happy face".
If you read about Latex figures, labels, and references, you will find that the label is completely arbitrary. The only requirement is that the //ref command must call out the label associated with the figure. In our case the sml2latx.sed file contains substitutions which fill in the image number, eg: test.png, 02041.png, for the label. Thus, we do not have to manually fill that in for each of our images, which we may or may not reference. If we do wish to reference a figure we reference the image number. It may be necessary to run "latex tutorial.latex" twice to resolve the references.
As an option for html, a word may follow the image name as below. Eg., "test.png above" will put "above" into the tutorial.html. We have no way to generate numbered figures in the html. So, figure above, figure below, may be usefull. View tutorial.html vs tutorial.ps for "figure 1.1" vs "figure above", respectively. Here we reference figure again, but only in tex/latex, no html as in the above markup. The markup below shows the optional html word.
See figure<ref>test.png above</ref> for a "happy face".
In the case of html, we do not have the referencing facilities provided by LATEX. The best we can do is refer to the figure above or below as shown in the above markup.
Unrelated, take a look at tutorial.html to see how we have indented the above markup code without a leading dot. Compare to previous unindented markups.
See caution in next section: only one reference per line (pair of <ref> tags). Else, split line with (return).
### Labeling a figure
Do not confuse the "Labeling" with the caption on a figure. In most all cases you can skip this section and let the sed processing automatically generate the label which the "figure" requires so that it may be referenced. The automatic label is the same as the image file name (eg 02221.png). The previous section covers this. The only reason to read this section is in the rare event that a second instance of a figure is being used. In which case, it needs a new, unique, not automatically generated label, not the (automatic) label for the first instance of the image. You may also skip this, if there is no caption for the figure. We will give the second instance of the image a unique label so that it will not be confused with the first instance when we reference it. See Figure below
Caption for the second instance of our image.
Note that our new figure is captioned as Figure above. The caption is different than the caption for the previous Figure 2nd-above. We are able to assign a label to it:
<image>test.png<caption>This is a fine caption.</caption>
<label>newtest.png</label></image>
Note that the above markup must be on one line. It is too wide for our page. So, we wrapped it. It may wrap in the text editor. But there cannot be a (return) except at the end of the line. The sed script processes a line at a time for each command. We process all the tags in the line with one command for image, caption, label, and ref tags.
Once it has a label, we can distinguish it from the other figure by referencing it the same way we reference other figures (just a different label):
If we compare the above image caption for newtest.png to the previous caption for for test.png, we find that both specify the same image, test.png. The latter has a different label "newtest.png" This is just a label. There is no image by that name.
See figure<ref>newtest.png above</ref> for a 2nd "happy face".
Caution, a limitation of the sed script for caption processing is that only one figure reference ( eg.: <ref>newtest.png</ref>) may be processed properly per line. Typically, there is only one line, all the words up to the end-of-line between <para> tags. If we need more than one <ref></ref> in a paragraph, the paragraph may be split into two or more lines between the two paragraph tags. See tutorial.sml for an example of this in the paragraph "Note that our new figure is captioned. . . "
### Scaling an image
Once in a long while, an image which is of satisfactory size in the html version of a document is too small in the LaTeX produced pdf document. The solution is to make the image the "right size" for the html document, then scale it to a suitable size in the LaTeX file. This is done by a sed (string substitution program) command in sml2latx.sed. When the sml source is processed, a scale factor is added to the .latex file, but not the .html file.
The scale factor must be added to the .sml as a modification between the <image> tag and the file name of the image. This markup produces Figure below).
<image>[scale=0.5]test.png<caption>This is scaled down in LaTeX.</
caption><label>smalltest.png</label></image>
The image command must be on a single line, a CR only at the end, none in the middle. Though, we wrapped it above for appearance. And, don't put two on one line– split into two lines. This scale parameter, [scale=0.5], only works if the <caption> tags are used, due to sed script limitations. The same is true of the <label> tags. The caption tags generate a figure number, even if there is nothing between the tag. There must be a unique label between the <label> tags, else LaTeX give an error. There must be no space between [scale=0.5] and test.png. LaTeX doesn't want a space in front of the image file name. It must be like this [scale=0.5]test.png.
This is scaled down in LaTeX.
## Special characters
In addition to special logos like TEX, SubML provides for certain often-used characters of the Greek alphabet.
The ratio of a circle's circumference to its diameter is symbolized by the Greek letter “pi,” which SubML represents like this: π. The area of a circle is given as A=πr2. Not many people realize that the standard symbol π is actually the lower-case version of the Greek letter. The capital version looks like this: Π, and it does not represent the same thing in mathematics.
But there are other useful Greek characters for us to use in SubML as well. When SubML is converted to plain ASCII text, some of the Greek characters like µ and ρ will be represented by the closest-resembling Roman (English alphabet) character available. If there is no Roman character close enough, the Greek character's name will be spelled in parentheses. TEX, on the other hand, is very Greek-literate and requires no “fudging” to obtain perfect representation. HTML output from SubML conversion renders these characters using Unicode. In order for a web browser to properly display them, it must be set up with Unicode character support. For your viewing pleasure, we have:
• Alpha (lower-case): α
• Beta (lower case): β
• Gamma (lower case): γ . . . . . . Gamma (capital): Γ
• Delta (lower case): δ . . . . . . Delta (capital): Δ
• Epsilon (lower case): ε
• Varepsilon (lower case): ɛ
• Zeta (lower case): ζ
• Eta (lower case): η
• Theta (lower case): θ . . . . . . Theta (capital): Θ
• Vartheta (lower case): ϑ
• Iota (lower case): ι
• Kappa (lower case): κ
• Lambda (lower case): λ . . . . . . Lambda (capital): Λ
• Mu (lower case): µ
• Nu (lower case): ν
• Xi (lower case): ξ . . . . . . Xi (capital): Ξ
• Pi (lower case): π . . . . . . Pi (capital): Π
• Rho (lower case): ρ
• Varrho (lower case): ϱ
• Sigma (lower case): σ . . . . . . Sigma (capital): Σ
• Varsigma (lower case): ς
• Tau (lower case): τ
• Upsilon (lower case): υ . . . . . . Upsilon (capital) ϒ
• Phi (lower case): φ . . . . . . Phi (capital): Φ
• Varphi (lower case): ϕ
• Chi (lower case): χ
• Psi (lower case): ψ . . . . . . Psi (capital): Ψ
• Omega (lower case): ω . . . . . . Omega (capital): Ω
• non-breaking space 1 1 1 1 2 2 2 3 3 3 4 4 4 4
• Tau (lower case): τ
• bigtriangledown: ∇
• oplus, exclusive or sign: ⊕
• almostequal: ≈
• approxequal, approximately equal: ≅
• neq, not equal: ≠
• plusminus, plus or minus: ±
• cdot, centered dot, times dot: ·
• leq, less than or equal: ≤
• geq, greater than or equal: ≥
• times, times sign: ×
• registered, registration sign: ®
Another special symbol available in SubML is the ∠ symbol (<angle>), used in mathematical statements to designate an angle. This is useful for expressing complex numbers in polar form. Take for example this impedance: 500 Ω ∠ -34.61o. By the way, the way I typeset the "degree" symbol is with a superscript letter "o".
Other mathematical symbols included in SubML's vocabulary are the integration symbol (∫), partial derivative symbol (∂), and the infinity symbol (∞). Here are some examples of these symbols in use:
V = ∫Q dt + C
∂x/∂t
∞ is bigger than BIG!
Note that you cannot show upper and lower integration limits for a definite integral using the "∫" markup tag. It is useful for crude in-line formatting of an integral equation only. If you want to show lower and upper integration limits in a SubML document, you must use a graphic image -- sorry!
For special characters used in other languages (Spanish, French, German, etc.), the following are available in the SubML vocabulary:
• "a" with grave (back prime): à . . . . . . À
• "a" with acute (forward prime): á . . . . . . Á
• "a" with circumflex (caret): â . . . . . . Â
• "a" with umlaut/dieresis/tremat: ä . . . . . . Ä
• "a" with tilde: ã . . . . . . Ã
• "a" with "ring" above: å . . . . . . Å
• "c" with cedilla: ç . . . . . . Ç
• "e" with grave (back prime): è . . . . . . È
• "e" with acute (forward prime): é . . . . . . É
• "e" with circumflex (caret): ê . . . . . . Ê
• "e" with umlaut/dieresis/tremat: ë . . . . . . Ë
• "i" with grave (back prime): ì . . . . . . Ì
• "i" with acute (forward prime): í . . . . . . Í
• "i" with circumflex (caret): î . . . . . . Î
• "i" with umlaut/dieresis/tremat: ï . . . . . . Ï
• "n" with tilde: ñ . . . . . . Ñ
• "o" with grave (back prime): ò . . . . . . Ò
• "o" with acute (forward prime): ó . . . . . . Ó
• "o" with circumflex (caret): ô . . . . . . Ô
• "o" with umlaut/dieresis/tremat: ö . . . . . . Ö
• "o" with tilde: õ . . . . . . Õ
• "u" with grave (back prime): ù . . . . . . Ù
• "u" with acute (forward prime): ú . . . . . . Ú
• "u" with circumflex (caret): û . . . . . . Û
• "u" with umlaut/dieresis/tremat: ü . . . . . . Ü
• Inverted question mark ¿
• Inverted exclamation mark ¡
So, now you may impress all your Español-speaking amigos with the following phrases in your documents:
"¿Dónde está el cuarto de baño?"
"¡Más cerveza, por favor!"
"¿Puede indicarme dónde está en el mapa?"
"Por favor, dígale tu amigo que voy a llegar cinco minutos tarde."
"Aquí tiene mi casa."
And when your friend asks you this . . .
"¿Qué procesador de textos usted utiliza?"
. . . you may respond with pride:
"No utilizo un procesador de textos.¡En lugar, utilizo un lenguaje de marcas!"
## Tex/Latex only, HTML only
Tags <tex>, </tex>, <htmlo>, </htmlo> are provided to include text from .sml selectively only in .latex, .tex or only in .html. The <tex> </tex> tags mark text that is only included in the .latex and .tex outputs of "sed -f sml2latx.sed" and "sed -f sml2tex.sed". Text that is only to be included in the .html is marked of by the <htmlo>, </htmlo> tags.
This following markup is to only show text in tutorial.latex and tutorial.tex. Following the markup, see text in tutorial.latex, tutorial.tex, but not in tutorial.html
<tex>This only shows in tutorial.latex and tutorial.tex</tex>
This following markup is to only show text in tutorial.html. Following the markup we see the text in tutorial.html but not tutorial.latex, tutorial.tex.
<htmlo>This only shows in tutorial.html</htmlo>
This only shows in tutorial.html
Given both a portrait and landscape version on a same-size image, a practical application of the <tex>, <htmlo> tags is to selectively direct those images to tutorial.latex or tutorial.html. We do not actually do this in tutorial.sml, but show the markup. For example, we wish to send the landscape version of a big image to the html version of our book so that readers do no have to rotate their monitors. This landscape is too big for our .latex, .tex, .ps, .pdf 6-inch wide book pages. We cannot reduce the size of the landscape, which would be unreadable. So, we rotate our big landscape to a portrait. It started out 4-inches tall and is now 4-inches wide. It fits side ways nicely on a book page. We have not reduced the size, just rotated it. A book reader can easily rotate the book to view the large image.
<htmlo><image>landscape.png</image></htmlo>
<tex><image>portrait.png</image></tex>
## Hyperlinks and targets
link at end of this section.
sample target located here, jump here from a link (Click) near the bottom of this section
The <url>, </url> tags provide clickable links to URLs in both the html and pdf versions of a document. The pdf is derived from LaTeX. Internal links are provided by <hyperlink>, </hyperlink> tags, which link to targets defined by the <hypertarget>, </hypertarget> tags. The syntax for these tags takes the following form:
<url>url_link[text]</url>
The "link" for <hyperlink> must match the "link" at the <hypertarget> to actually jump there on clicking. The links for <hypertarget> in the case of multiple targets needs to be unique– no two targets the same. The "link" for<hyperlink> and <hypertarget> may not contain any underscores, eg., invisible_link. Though, it works in html, the pdf links will be dead. And, no errors are generated. The <url> and <hyperlink> text will appear colored in both html and pdf when viewed. The <hypertarget> text is not colored, and is optional.
The following markup provides an external link to a URL in both html and pdf documents:
Go to
<url>http:www.ibiblio.org/obp/electricCircuits/index.htm[Lessons in Electric Circuits]</url>
to learn about electricity.
Go to Lessons in Electric Circuits to learn about electricity.
Why are there no quotes around the URL above? While the quotes are needed in html code, they are not used in LATEX. Therefore, we do not include them here. They are added by the sml2html.sed script to the html document.
Click this link to jump to invisible target at end of section. At the top of this section click on "link at" to also jump to the end of the section.
The following markup provides the link below it to the top of this section:
<hyperlink>LINK[Click]</hyperlink> to go to target at top of section.
Click to go to target at top of section.
Here is the markup for an "invisible" target at the end of this section:
<hyperlink>invisibleTarget[]</hyperlink>
## Bibliography and citations
The <thebibliography>,</thebibliography> tags mark a section of text to be treated as a list of bibliographic references. Contained therein are individual bibliographic entries delimited by <bibitem></bibitem> tags. Theses entries may be referenced from the body of the main text by <cite></cite> tags. The syntax of these tags is as follows:
<thebibliography>
<bibitem[ref]text</bibitem>
<bibitem[ref2]text2</bibitem>
</thebibliography>
The purpose of this paragraph is to reference the bibliography below. This paragraph is broken into several lines terminated by a return.[footnotes] You should skip to the bibliography and look at the first entry, here.[latex] The second entry in the bibliography is here.[1d] Note that the fourth bibitem contains a url to link to home of this project.[4]
The bracketed reference, [ref], in the bibitem needs to be matched by the corresponding citation reference <cite>ref</cite> in the body of the text. See above and below. In LaTeX, this is usually an easy to remember mnemonic. This is replaced by bracketed a number, eg. [2], in the processed LaTeX version of the document. However, the html version of the document will not have numbers unless the reference is a number, eg. <cite>4</cite>. The bibliography in html is a numbered list. However, these numbers do not necessarily correspond to the sml bibitem reference. Use numbers instead of mnemonics in the bibitem reference for numbers in the html.[4].
A sample bibliography with four items follows:
## Bibliography
Note that the last entry above contains a url. The whole bibterm must be on one line, only one return, at the end.
## What SubML won't do
SubML is designed to be a simple markup language, and as such it lacks certain advanced features found in other, more capable languages like TEX or DocBook. One of these missing features is tables. However, I have found that it often works well to create a table using a graphics editor and then insert it into the document as an image. One advantage to doing tables this way is consistency in appearance between different outputs (TEX, HTML, etc.).
Another thing SubML makes no provision for is easy, verbatim display of its own markup code. In order to show verbatim SubML code, you must mark all < and > symbols with the appropriate <lt> and <gt> tags. The following paragraph shows the markup required for this paragraph. For a really wild experience, view the source code of this file to see how I mark up that paragraph:
<para>
Another thing SubML makes no provision for is easy, verbatim display
of its own markup code. In order to show verbatim SubML code, you
must mark all <lt> and <gt> symbols with the appropriate
<lt>lt<gt> and <lt>gt<gt> tags. The
following paragraph shows the markup required for this paragraph.
For a really wild experience, view the source code of this file to
see how I mark up <italic>that</italic> paragraph:
</para>
I could carry the recursion one step further, but that would be cruel and unusual punishment for both of us.
## How to do the conversion
First, you need to have sed installed and operational on your computer. Next, be sure that all conversion scripts (sml2tex.sed, sml2html.sed, etc.) have been installed in the same directory as the SubML document that you wish to convert. If you wish to convert your SubML document to TEX, groff, or some other markup language requiring further processing, you must of course have the necessary software installed on your computer to process the markup format(s) of choice.
For instance, if you converted your SubML document into a TEX document using the sml2tex.sed script provided with this tutorial, but didn't have Donald Knuth's TEX processing system installed on your computer, all the sed script will do is produce a TEX source file: a new document marked up with TEX commands and tags in place of SubML tags. In other words, these scripts simply convert SubML source code into source code for other markup languages. With the exceptions of HTML and plain ASCII text, none of the output formats generated by these sed scripts will be ready-to-use.
If you wish to convert your source document (entitled foo.sml) to HTML, here is what you would have to type at the command prompt:
sed -f sml2html.sed foo.sml > foo.htm
The -f option tells sed to look to file sml2html.sed for instructions rather than take direct search-and-replace commands from the command prompt when processing the input file foo.sml. The output file is named foo.htm.
The redirection command ( > ) is necessary, otherwise sed will simply send the converted text to standard output (the computer's command-line screen) and all of it will flash before your very eyes instead of being saved in a file. Of course, you can name the target file anything you wish, so long as the extension is appropriate to the type of converted document that it is (i.e. .htm or .html for HTML output, so that a browser will recognize the filename).
The use of standard input and standard output in a sed script allows for great flexibility in the use of SubML. For instance, I have a book I'm writing (Lessons In Electric Circuits), in which I'm using Makefiles to direct compilation from SubML to LATEX and HTML. By using stdin/stdout redirection within the Makefile commands, I'm able to prepend and append files containing special LATEX and HTML code to the basic text (written in SubML format) to achieve markup capabilities beyond the basic scope of SubML. For instance, I may want to generate a coverpage for my book using a series of special LATEX commands. SubML doesn't specify detailed layout tags, and so I write the necessary LATEX code in a file that gets prepended to the sed-converted output of the main text body. Same for the generation of an index: a special file containing the necessary LATEX commands gets appended to the very end, after sed has converted the main body of the text. Same for navigation buttons at the beginning and end of each HTML file generated from SubML.
## How mini TOC works
A mini Table of Contents (TOC) is automatically inserted after the chapter title for (1) html, (2) LATEX which provices dvi, ps, and pdf. There is no mini TOC support for other formats: txt, tex, or groff. This requires different packages for (1) html, (2) LATEX. Thus, the method of generation of the mini TOC is different for the two case. In both cases the automatic generation is initiated by the sed command file substitution for the </chaptertitle> tag. Other features in headers or makefiles cause the required software to generate and insert the mini TOC after the chapter title.
In the case of html, the sml2htm.sed file contains the </chpatertitle> tag substitution: <!--minitoc–> which flags the html for inclusion of the mini TOC. We use a a perl script, htmltoc, modified for our requirements to htmltoc2 for placing a mini TOC at the <!--minitoc–> tag. The original script placed the mini TOC before the chapter title. So, we modified it to place the mini TOC at our tag, which is after the title. The Makefile has a line calling minitoc with appropriate parameters:
./htmltoc2 -inline -noorg -toclabel " " -tocmap toc.map \
-minitoc "<\!\-\-\minitoc\-\->" AC_1.html
See the minitoc documentation for details. We added the -minitoc parameter to the htmltoc perl script for our htmltoc2 so that it looks for the quoted tag which follows it. In our case we want the mini TOC at the <!--minitoc–> tag, so that tag with escaping backslashes follows.
The makefile for each book has a make target for each of the book chapters. The chapters for which we want a mini TOC require the above htmltoc2 command in the make targets. We include it in chapter targets, _1, _2, etc., but not the appendix targets, _A1, _A2, _A3. Thus, all chapters but the appendices have a mini TOC after the chapter title. Eg., see AC/Makefile targets: AC_14.html, AC_A1.latex for chapter vs appendix.
In the case of the LATEX translation, .latex, the </chaptertitle> in .sml is replaced by /minitoc. See sml2latx.sed. This /minitoc tells LATEX where to place the mini TOC.
Also, the header, hi.latex, contains \usepackage{minitoc} and \dominitoc to load the minitoc package and "do" the minitable of contents respectively. The table will be placed where the /minitoc command is encountered in the chapter text.
Nothing unusual is required of the makefile to generate the mini TOC. However, if we do not want the mini TOC in the appendices, a sed script in each of the latex appendix targets, removes the /minitoc command from the .latex. Normal target processing, puts a chapter mini TOC in for all chapters but appendices. Eg., see AC/Makefile targets: lines.latex, about.latex for chapter vs appendix. | {"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": 0.8848941922187805, "perplexity": 3200.0205937468736}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982965886.67/warc/CC-MAIN-20160823200925-00267-ip-10-153-172-175.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/191047/integration-with-using-gauss-law | # Integration with using Gauss law
Task: find the vector $\mathbf E$ in the center of the sphere with radius $R$, which has charge volume distribution $\rho , \rho = (\mathbf a \cdot \mathbf r ), \mathbf a = \operatorname{const}, \mathbf r$ - radius-vector from the center of the sphere.
My solution: I used Gauss theorem, $$\oint \mathbf E \cdot d\mathbf S = \int \nabla \cdot \mathbf E dV = 4\pi \int\rho\ dV = 4\pi a \int r\cos(\mathbf a , \mathbf r)\ dV .$$ Unfortunately, I don't know how to find $\mathbf E$ using written above: $|\mathbf E| \neq const$ into the surface of the sphere, so I can't write something like $$\oint \mathbf E \cdot d \mathbf S = E \int dS = ...$$ and equate it to (1).
Can you help me?
I created a routinely solution. In the beginning
1. I can consider only surface of the sphere with surface denstity $\rho (S) = (\mathbf a \cdot \mathbf r)$.
2. I "moved" $\mathbf a$ to the center of the sphere by parallel transfer (look to the picture 1 below).
So, I can divide the sphere into thin rings with charge $$dQ = \rho (S)dS = \rho (S) 2\pi r Rd\varphi$$ (R - length from the ring to the center of the sphere, a - length from the ring plane to the center of the sphere, $\varphi$ - an angle between $\mathbf a , \mathbf R$, and $\mathbf r = \sqrt{R^{2} - a^{2}})$. Using that, $d|\mathbf E|$ creating by this ring in the center of the sphere (using the formula for the ring E = \frac{kaQ}{R}), can be write as $$dE = \frac{kadQ}{R^{3}} = \frac{ak\rho (S) 2\pi r R d\varphi}{R^{3}}.$$ According to the spherical symmetry, $$d\mathbf E_{center} = -\frac{\mathbf a}{a}dE,$$ so, according, that $$r = Rsin(\varphi ), \quad a = Rcos(\varphi ), \quad \rho (S) = (\mathbf a \cdot \mathbf r) = aRcos(\varphi ),$$ I get $$d\mathbf E = -\frac{k R cos(\varphi) R cos(\varphi ) 2\pi R sin(\varphi) R d\varphi}{R^{3}}\mathbf a = -2\pi k R cos^{2}(\varphi )sin(\varphi)\mathbf a \Rightarrow$$
$$\Rightarrow \mathbf E = 2\pi k \mathbf a \int \limits_{-1}^{1}cos^{2}(\varphi )d(cos(\varphi )) = \frac{4 \pi k R \mathbf a}{3}.$$ By the next step I accept, that R is a variable (takes into account the volume distribution of charge), so, with the new vector $\mathbf a$ (formally, it's changed only by a dimension, $[|\mathbf a_{new}|] = \frac{[|\mathbf a_{old}|]}{meter}$), $$d\mathbf E = \frac{4 \pi k RdR}{3}\mathbf a \Rightarrow \mathbf E = \frac{2\pi k R^{2}}{3}\mathbf a.$$
-
Two questions: 1) Why do you assume E is constant on the surface of the sphere? 2) Why are you integrating to the surface of the sphere when it is E at the center that you seek? – Robert Miller Sep 4 '12 at 18:38
Oh yes, I made a mistake. I forgot about a scalar product. – John Taylor Sep 4 '12 at 18:45
That charge density is interesting. What's the total charge? – James S. Cook Sep 4 '12 at 18:47
So, this task cannot be solved by using Gauss law? – John Taylor Sep 4 '12 at 18:47
"...I was reading that as a dot-product :)..." It was a dot product with $\mathbf a = const$ as a vector. Maybe I didn't understand your words. [:(]. – John Taylor Sep 4 '12 at 18:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.8618625998497009, "perplexity": 459.0093322291765}, "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/1448398466260.18/warc/CC-MAIN-20151124205426-00342-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://getrevising.co.uk/resources/aqa_physics_unit_two | # Aqa Physics Unit Two
All of Unit Two notes condensed into a powerpoint. Enjoy :)
HideShow resource information
• Created by: Cora H
• Created on: 29-05-12 14:20
## Other slides in this set
### Slide 2
Here's a taster:
Velocity and Acceleration
Speed and Velocity are both how fast you are going
· Speed and velocity are both measured in m/s. They both simply say
how fast you are going however, speed is just how fast you are going,
with no regards to the direction. Velocity however must also have
the direction specified.
Speed, Distance and Time
· Speed=Distance/time
Acceleration is how quickly you are speeding up
· This is not the same as velocity or speed.
· Acceleration is how quickly the velocity is changing
· Acceleration =Change in velocity / Time taken…read more
### Slide 3
Here's a taster:
D-T and V-T Graphs
Distance-time graphs
· Speed = Gradient (vertical/horizontal)
· Flat sections are where it stopped.
· The steeper the graph, the faster its going
· Downhill sections means its going back towards the starting point
· Curves represent acceleration or deceleration
· A steepening curve means its speeding up
· A levelling off curve means its slowing down.
Velocity-time graphs
· Acceleration = gradient
· Flat sections represent steady speed
· The steeper the graph, the greater the acceleration or deceleration
· Uphill sections are acceleration
· Downhill sections are deceleration
· The area under any section of the graph is equal to the distance travelled
in that time interval.
· A curve means changing acceleration.…read more
### Slide 4
Here's a taster:
Mass, Weight and Gravity
Gravity is the force of attraction between all masses
· Gravity attracts all masses, but you only notice it when one of the masses is really big. E.g. a
planet; anything near a planet is attracted to it.
· This has 3 important effects:
· On the surface of a planet, it makes all things accelerate towards the ground.
· It gives everything a weight
· It keeps planets, moons, and satellites in their orbit. The orbit is a balance between the
forward motion if the object and the force of gravity pulling it inwards.
Weight and mass are not the same
· Mass is just the amount of stuff in an object. For any given object the mass will be the same
anywhere in the universe.
· Weight is caused by the pull of gravity. In most questions, the weight of an object is just the
force of gravity pulling it towards the centre of the earth.
· An object has the same mass whether it is on earth or on the moon-but its weight will be
different.
· Weight is a force measured in newton's. Its measured using a spring balance or newton
metre. Mass is not a force. Its measured in kilograms with a mass balance.
The very important formula Mass, weight and gravity.
· Weight = mass x gravitational field strength (W = m x g)…read more
### Slide 5
Here's a taster:
The 3 Laws of Motion
First law balanced forces mean no change in velocity
· So long as the forces on an object are all balanced, then it will stay still, or else if it is already
moving it will just carry on at the same velocity so long as the forces are all balanced.
· When a train or car or bus or anything else is moving at a constant velocity then the forces on it
must all be balanced.
· Never let yourself entertain the ridiculous idea that things need a constant overall force to
keep them moving.
· To keep going at a steady speed, there must be zero resultant force.
Second Law A resultant force means Acceleration
· If there is an unbalanced force, then the object will accelerate in that direction
· An unbalanced force will always produce acceleration
· This acceleration can take five different forms: starting, stopping, speeding up, slowing down,
and changing direction
· On a force diagram the arrows will be unequal.
Three points which should be obvious
· The bigger the force, the greater the acceleration or deceleration
· The bigger the mass the smaller the acceleration
· The get a big mass to accelerate as fast as a small mass it needs a bigger force. Just to think
about pushing heavy trolleys and it should seem fairy obvious.
The overall unbalanced force is often called the resultant force
· Any resultant force will produce acceleration and the formula is:
· F= ma or a=F/m…read more
### Slide 6
Here's a taster:
The 3 Laws of Motion Cont.
Resultant force is really important especially for F=ma
· The notion of resultant force is a really important one for you to get your head
around.
· in most real situations there are at least two forces acting on an object along with
any direction. The overall effect of these forces will decide the motion of an object
accelerate, decelerate or stay at a steady speed.
· If the forces all point along the same direction the overall effect is found by just
adding or subtracting them.
· The overall force you get is called the resultant force. F is always the resultant
force.
The Third Law Reaction Forces
· If object A exerts a force on object B then object B exerts the exact opposite force
on object A.
· This means if you push something, the object will push back against you just as
hard.
· As soon as you stop pushing, so does the object
· One object normally has a smaller mass, so will accelerate more.
· E.g. in swimming you push against the water with your arms and
legs, and the water pushes you forward with an equal- sized force in
the opposite direction.…read more | {"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": 0.9039410948753357, "perplexity": 2048.4782711076036}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698544140.93/warc/CC-MAIN-20161202170904-00107-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://mymathforum.com/math/46421-what-lies-beyond-infinity-3.html | My Math Forum > Math what lies beyond infinity?
Math General Math Forum - For general math related discussion and news
September 18th, 2014, 05:54 AM #21
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Posts: 16,046
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Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms
Quote:
Originally Posted by v8archie I'd actuallly be interested in finding out about respectable theories of infinitesimals.
Maybe you should start a new thread, lest it be lost in the detritus here.
September 18th, 2014, 07:23 AM #22 Member Joined: Aug 2014 From: Somewhere between order and chaos Posts: 44 Thanks: 5 Math Focus: Set theory, abstract algebra, analysis, computer science I agree with v8archie. You are exhibiting a fundamental misunderstanding of the concept of infinity. You are trying to treat infinite and infinitesimal numbers like real numbers. There is no difference between 0, +0, and -0 for instance. Surreal numbers can not be expressed in the traditional way (don't ask me what the proper way of expressing them is, because I know very little about surreal numbers other than that they are a theoretical possibility). You also show a fundamental misunderstanding of limits. Your statement about the limit of zero being something other than zero is meaningless. If you want to truly understand limits, I suggest you look up the Cauchy convergence test. You might want to look up the Cauchy test for convergent sequences first, since it uses fewer variables and is somewhat easier to understand; it's a good intermediate step towards understanding limits of functions. Thanks from CRGreathouse and topsquark
September 18th, 2014, 09:17 AM #23 Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Let me take this in a different direction. There are lots of systems other than the real numbers out there. For example, there is the free monoid on, say, ASCII characters. In that system "0" is different from "+0" since the latter is two letters and the former is only one, and "-0" is different from both since the first letters differ. Or take sequences of real numbers relative to some fixed nonprincipal ultrafilter U. You could define $\alpha=(1,1/2,1/3,1/4,\ldots)$ and (as usual) identify the real number $x$ with the sequence $(x,x,x,\ldots)$. Then you have $-\alpha<0<\alpha,$ just as you like (though you can't use the symbol +0 for what I have called $\alpha$ since +0 has its own meaning in that system), regardless of your choice of U. Note that for any real number $\varepsilon>0$ you have $0<\alpha<\varepsilon$ so this $\alpha$ is an actual infinitesimal. But it doesn't make sense to ask for its decimal expansion -- it's a hyperreal, not a real. (You could ask for the decimal expansion of its shadow, which is the real number 0 = 0.000000000....) Thanks from topsquark
September 20th, 2014, 12:40 AM #24
Senior Member
Joined: Nov 2013
Posts: 160
Thanks: 7
Quote:
Originally Posted by CRGreathouse I suspect that you could not write a formal proof, though, so the process is not as easy as I make it sound.
I think I have never written a formal proof. I have proofs, what makes them formal I don't know, I could call them just proofs, normal proofs. I have also what I could call "prophetic proofs".
My method has not been mainstream. I have my own way. I need to believe
in what I do, because I don't have always proofs at my disposal.
I have noticed that it is possible to arrive at right answers in two way:
even a wrong method can lead to a correct answer, as well as a right method
of course.
All my "prophetic proofs" have been systematically denied in all forums
where I have written them. When I drop them, and when I choose to write
instead my "normal" proofs which I consider as facts, these are also the same way systematically denied. Might it be better not to write any proofs at all?
September 20th, 2014, 03:49 AM #25 Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra If you want to have a proper discussion about your mathematical ideas, you need proofs to show why your ideas aren't just wishful thinking and hand-waving that are based on a misunderstanding of theories that are well-known, rigorouslesslly tested, formally proved and completely understood by the mathematical community. I don't recall reading any proof of yours in this thread. Thanks from CRGreathouse
September 20th, 2014, 04:48 AM #26 Banned Camp Joined: Feb 2013 Posts: 224 Thanks: 6 Another Dimension What lies beyond infinity is another dimension. 1/3 is done by infinite binary divisions in 1 dimensional space (a number line). But can easily be accomplished in 2 dimensions. Angle trisection can be done by infinite angle bisection. But can easily be done in 3 dimensional space.
September 20th, 2014, 08:26 AM #27
Senior Member
Joined: Nov 2013
Posts: 160
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Quote:
Originally Posted by long_quach What lies beyond infinity is another dimension. 1/3 is done by infinite binary divisions in 1 dimensional space (a number line). But can easily be accomplished in 2 dimensions. Angle trisection can be done by infinite angle bisection. But can easily be done in 3 dimensional space.
I think you are demonstrating that 1/3=0.3333............first by infinite steps of binary divisions, something like a never ending long division. Then you show some plots which show how to divide a line and an angle into three parts. The last one of your pictures is low quality and it is hard to see exactly what is going on there.
I can also tell about my approach to how to divide by three, and what happens if the division is continued for ever. It is an example which is posted on sciforums, I avoid writing all again so I will write the link here
It is also an example of how I try to prove what I say so that my ideas aren't just wishful thinking and hand-waving. I am writing that dividing
by three leaves a nonzero infinitesimal, but usually it is "swept under the rug" because it is so small that no-one knows or sees it, but dividing for example by 2 does not leave an infinitesimal. The decimal number
system allows an infinitely small decimal number, which is nonzero.
The infinitesimal x has no exact value, therefore it is not equal to 0,
it has a limit which is 0. Limit of x is 0. The mathematical community
has not accepted my ideas because they think that the infinitesimals
are always equal to zero, therefore they don't exist.
My ideas are not accepted, but I don't see what is wrong with my
ideas. I might as well stop writing on these forums if I realized what is wrong. This is after all just my hobby. Life would be much easier for
me if I did not need face the flamewars into which my ideas lead. I am not trying to teach anyone to accept my ideas, I am not a teacher, if my ideas are not accepted, it does not matter.
Last edited by TwoTwo; September 20th, 2014 at 08:33 AM.
September 20th, 2014, 09:37 AM #28 Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra The most important thing I can say about that post is that 0.333333... (and the whole of the decimal system) is no more than a representation of numbers. For your argument to hold (that division by three leaves some some small remainder) you will have to explain $\frac{1}{3}$. Where is the remainder now? How is this different to $\frac{1}{2}$. I rather suspect that the only difference is that 3 is coprime to 10, which means that your infinitesimals depend entirely on the arbitrary choice of base. If we chose to use base 3, we'd have $\frac12 = 0.111\cdots$. Does that mean that division by 2 both leaves an infinitesimal and that it doesn't? Your idea seems to be either inconsistent or to introduce a quantity that can be made to disappear whenever we want by changing our notation. And since changing the notation doesn't change the object, I'm forced to conclude that it never existed anyway. As a mathematician, I have always disliked using decimals, because of the loss of accuracy that comes from using finite representations. I always prefer to use fractions and symbols that represent irrationals such as $\pi$ and $\mathbb e$.
September 20th, 2014, 10:44 AM #29
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Joined: Nov 2013
Posts: 160
Thanks: 7
Quote:
Originally Posted by v8archie The most important thing I can say about that post is that 0.333333... (and the whole of the decimal system) is no more than a representation of numbers.
True.
Quote:
Originally Posted by v8archie For your argument to hold (that division by three leaves some some small remainder) you will have to explain $\frac{1}{3}$. Where is the remainder now?
There would be no remainder if $\frac{1}{3}$ had no decimal representation. Maybe it is wrong to ask what is the decimal representation of
$\frac{1}{3}$, because the question assumes a priori that $\frac{1}{3}$
has a decimal representation. The right question is "Does $\frac{1}{3}$ have a decimal representation?"
Quote:
Originally Posted by v8archie How is this different to $\frac{1}{2}$.
$\frac{1}{2}$ is not different than $\frac{1}{3}$ if they both had no decimal representation.
Quote:
Originally Posted by v8archie I rather suspect that the only difference is that 3 is coprime to 10, which means that your infinitesimals depend entirely on the arbitrary choice of base. If we chose to use base 3, we'd have $\frac12 = 0.111\cdots$. Does that mean that division by 2 both leaves an infinitesimal and that it doesn't?
I did not think what happens to infinitesimals if we chose to use a different base. It is possible that they disappear, their existence depends on the base.
The base-10 system that we use, seems to allow the existence of infinitely
small decimal numbers. They seem to exist only because of the decimal system that we have chosen to use, the base-10 with its decimals.
Quote:
Originally Posted by v8archie If we chose to use base 3, we'd have $\frac12 = 0.111\cdots$. Does that mean that division by 2 both leaves an infinitesimal and that it doesn't?
Again, the problem can be avoided by asking only right questions.
Don't ask what is the decimal represention of $\frac{1}{2}$, don't
ask what is the decimal represention of $\frac{1}{2}$ in base 3.
Ask only "does $\frac{1}{2}$ have a decimal representation?"
Quote:
Originally Posted by v8archie Your idea seems to be either inconsistent or to introduce a quantity that can be made to disappear whenever we want by changing our notation. And since changing the notation doesn't change the object, I'm forced to conclude that it never existed anyway.
Yes, it is possible that they don't exist at all as real objects, it would be very hard to think that there would be an elementary particle whose size
was infinitesimally small. Unlikely is also that matter is infinitely divisible. It is unthinkable that matter disappears if it is divided into infinitely small pieces. Therefore there must be a smallest size of everything, even though the smallest size was not infinitesimally small.The number system that we have chosen to use just seems to allow infinitesimals. They
are probably just products of our choice of number system.
If we used only fractions like $\frac{1}{3}$ we would not need to deal
with infinitesimals.
Quote:
Originally Posted by v8archie As a mathematician, I have always disliked using decimals, because of the loss of accuracy that comes from using finite representations. I always prefer to use fractions and symbols that represent irrationals such as $\pi$ and $\mathbb e$.
I understand. Decimal numbers belong more to physics than mathematics.
Physicists need usually deal with approximations and decimal numbers
suit for writing approximations, for example value of $\pi$ . There
are fractional representations for $\pi$ but maybe no-one uses them anymore. I have read that they were used in antiquity.
Maybe we should abandon the decimals.
September 20th, 2014, 10:58 AM #30 Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms To add to v8archie's point: one of the most important meta-skills for a mathematician to learn is the ability to switch between different representations of a problem. For example, when working on my paper on odd perfect numbers, I often had to switch between thinking of numbers in an 'Archimedian' sense (real numbers which can be compared with < and >) and what I now call a 'supernatural' sense (integers are a particular case of possibly infinite products of primes raised to possibly infinite exponents). In the first case the natural order is total while in the second case the natural order (divisibility) is partial. For real numbers there are many forms. Binary (or decimal) expansion, continued fraction, the Engel or Pierce expansion, etc. If the number is rational you can also write it as a fraction of integers; if it's an integer the Zekendorff expansion; and so forth. There's nothing special about the decimal expansion. In fact, it's poorly suited to most purposes. Use it when it works, but please don't mistake it for anything more than it is.
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Contact - Home - Forums - Cryptocurrency Forum - Top | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9029523730278015, "perplexity": 749.5850565339861}, "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-2019-35/segments/1566027330913.72/warc/CC-MAIN-20190826000512-20190826022512-00393.warc.gz"} |
https://www.physicsforums.com/threads/gravitational-force-among-4-spheres.223145/ | Gravitational force among 4 spheres
1. Mar 20, 2008
gillyr2
1. The problem statement, all variables and given/known data
Four 7.5kg spheres are located at the corners of a square of side 0.70 m.
Calculate the magnitude of the gravitational force exerted on one sphere which is in left and down corner of the square by the other three.
2. Relevant equations
F_tot = F_2 + sqrt(2) * F1
3. The attempt at a solution
came up with that equation, but it says i am wrong.
can anyone help/
2. Mar 20, 2008
Sourabh N
3. Mar 20, 2008
gillyr2
well since they are in a square and have equal masses F_1 = F_3 in magnitude. and for our triangle we have 1/sqrt2 are the sides and 1 is the hypotenuse. i got that. but i don't think its right
4. Mar 20, 2008
gillyr2
am i way offf? can anybody help
5. Mar 20, 2008
Staff: Mentor
If I understand what you're saying here, this is correct. But it's not the final answer.
What are F_2 and F_1?
6. Mar 21, 2008
gillyr2
F1 and F3 are the 90 degree forces and F2 is the 45 degree force. what do i still need to do?
7. Mar 21, 2008
Sourabh N
Still your solution solution looks bit wrong. Show the substitution process (the values for F1 ..) and and calculation, so that I can help you better.
8. Mar 21, 2008
Staff: Mentor
That's just the schematic of the answer. Now you have to figure out F1 & F2 and plug in the numbers. I assume they want an actual answer for the force in Newtons.
9. Mar 22, 2008
gillyr2
F1 = ((6.67 * 10 ^-11)* 7.5 *7.5 )/ .70^2 = 7.66 * 10^-9
length of hypotnuse
sqrt(.70^2 +.70^2); = .99
F2 = ((6.67 * 10 ^-11)* 7.5 *7.5 )/ .99^2 = 3.83*10^-9
F_tot = 3.83*10^-9 + sqrt(2) * 7.66*10^-9 =
1.46 * 10^-8
PROBLEM SOLVED. I MUST HAVE JUST PLUGGED IN SOME WRONG NUMBERS. THANKS ANYWAY | {"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": 0.9065144658088684, "perplexity": 2129.588306696404}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560282140.72/warc/CC-MAIN-20170116095122-00246-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://hjm.math.uzh.ch/Vol44-1.html | ## HOUSTON JOURNAL OF MATHEMATICS
### Electronic Edition Vol. 44, No. 1, 2018
Editors: D. Bao (San Francisco, SFSU), D. Blecher (Houston), B. G. Bodmann (Houston), H. Brezis (Paris and Rutgers), B. Dacorogna (Lausanne), K. Davidson (Waterloo), M. Dugas (Baylor), M. Gehrke (LIAFA, Paris7), C. Hagopian (Sacramento), R. M. Hardt (Rice), Y. Hattori (Matsue, Shimane), W. B. Johnson (College Station), M. Rojas (College Station), Min Ru (Houston), S.W. Semmes (Rice).
Managing Editors: B. G. Bodmann and K. Kaiser (Houston)
Contents
May, Coy L., Towson University, Towson, MD., USA, ([email protected]) and Zimmerman, Jay, Towson University, Towson, MD., USA. ([email protected]).
The symmetric genus of large odd order groups, pp. 1-19.
ABSTRACT. Let L be the set of integers n for which there is a solvable group G of order n generated by elements of prime order p. Then the set L has density 0 in the set of positive integers. Finite quotients of the Fuchsian triangle groups of type (3, 3, n), for n = 5, 7, 9 and 11, are generated by two elements of order 3. We call odd order groups of these four types LO-1 groups through LO-4 groups. The symmetric genus s(G) is the minimum genus of any Riemann surface on which G acts faithfully. If G acts on a Riemann surface X of genus g > 1, we say that G is a large odd order group since |G| > 8(g - 1). We obtain restrictions on the powers of the primes dividing the orders of such groups. In addition, we study the metabelian LO-3 groups, and classify the integers that are the orders of metabelian LO-3 groups.
Calugareanu, Grigore, Babes-Bolyai University, Cluj-Napoca, Romania ([email protected]).
A new class of semiprime rings, pp. 21-30.
ABSTRACT. A ring R is called unit-semiprime if for any element a in R, a=0 whenever aua=0 for all units u in U(R). This turns out to be a proper class of semiprime rings which, among others, includes the reduced rings and the unit-regular rings, is closed to matrix extensions but is not Morita invariant. The usual Ring theoretic constructions are investigated and connections with some other known classes of rings are established.
Hesam Safa
, Department of Mathematics, Faculty of Basic Sciences, University of Bojnord, Bojnord, Iran ([email protected]), M. Farrokhi D. G., Institute for Advanced Studies in Basic Sciences (IASBS), and the Center for Research in Basic Sciences and Contemporary Technologies, IASBS, P.O.Box 45195-1159, Zanjan 66731-45137, Iran ([email protected]), ([email protected]), and Mohammad Reza R. Moghaddam, Department of Mathematics, Khayyam University, International Campus and Centre of Excellence in Analysis on Algebraic Structures, Ferdowsi University of Mashhad, Mashhad, Iran ([email protected]) .
Some properties of 2-auto-Engel groups, pp. 31-48.
ABSTRACT. For a given group G, with an element x in G and automorphism α in Aut(G), the nth autocommutator [x,nα] is defined recursively by [x,α]=x-1xα and [x,nα]=[[x,n- 1α],α] for all n>1. The group G is said to be n-auto-Engel if [x,nα]=[α,nx]=1, for all x∈G and all α∈Aut(G), where [α,x]=[x,α]-1. We study the structure of 2-auto-Engel groups and show that 2-auto-Engel groups indeed satisfy the equation α(x)α-1(x)=x2, for all x∈G and α∈Aut (G). Also, we give a precise description of all abelian 2-auto-Engel groups of finite 2-rank as well as 2-auto-Engel 2-groups, which are not purely non-abelian, and construct an infinite family of purely non-abelian 2-auto-Engel 2-groups.
Luck Darniére, Département de mathématiques, Facult\é des sciences, Université d'Angers, France ([email protected]) and Markus Junker, Mathematisches Institut, Abteilung für mathematische Logik, Albert-Ludwigs-Universität Freiburg, Deutschland ([email protected])
Model completion of varieties of co-Heyting algebras, pp. 49-82.
ABSTRACT. It is known that exactly eight varieties of Heyting algebras have a model-completion. However no concrete axiomatization of these model-completions were known by now except for the trivial variety (reduced to the one-point algebra) and the variety of Boolean algebras. For each of the six remaining varieties we introduce two axioms and show that 1) these axioms are satisfied by all the algebras in the model-completion, and 2) all the algebras in this variety satisfying these two axioms satisfy a certain remarkable embedding theorem. For four of these six varieties (those which are locally finite) these two results provide a new proof of the existence of a model-completion with, in addition, an explicit and finite axiomatization.
Janusz Sokół, Faculty of Mathematics and Natural Sciences, University of Rzeszow, ul. Prof. Pigonia 1, 35-310 Rzeszow, Poland ([email protected]) and Derek K. Thomas, Department of Mathematics, Swansea University, Singleton Park, Swansea, SA2 8PP, UK ([email protected]).
Further results on a class of starlike functions related to the Bernoulli lemniscate, pp. 83-95.
ABSTRACT. The purpose of this paper is to provide further results for the class of starlike functions S L consisting of analytic functions f normalized by f(0) = f’(0) - 1 = 0 in the open unit disk |z|<1 satisfying the subordination condition zf’(z)/f(z) is subordinate to square root of 1+z. Various new results for the coe cients of f(z) are obtained, together with sharp inequalities for the Fekete-Szegö functional, the Hankel determinant, the coefficients of the inverse function, and the coefficients of log f(z)/z.
Mishev, Ilia, Department of Mathematics, University of Colorado at Boulder, Campus Box 395, Boulder, CO 80309-0395, U.S.A. ([email protected]).
On a class of transformations of sequences of complex numbers, pp. 97-119.
ABSTRACT. In this paper we consider a transformation La of sequences of complex numbers. We find the inverse transformation of La as well as the inverse of a related transformation. We explore a connection to the binomial transform and significantly generalize a previously known result. We also obtain new relations among many classical hypergeometric orthogonal polynomials as well as new formulas for sums involving terminating hypergeometric series.
Memarian Yashar,
Department of Mathematics, University of Notre-Dame, IN 46556, U.S.A. ([email protected])
A note on the geometry of positively-curved Riemannian manifolds, pp. 121-145.
ABSTRACT. In this paper I present a comparison theorem for the waist of Riemannian manifolds with positive sectional curvature. The main theorem of this paper gives a partial positive answer to a conjecture formulated by M.Gromov in 2003. The content of this paper combines two aspects: classical volume comparison theorems of Riemannian geometry, and geometric measure theoretic ideas stemming from Almgren-Pitts Min-Max theory. I present the background which is needed in Riemannian Geometry as well as in geometric measure theory in different sections. One of my wishes was to give the reader a broad idea on the subject: which is estimation of metric invariants on concrete geometric spaces.
Deaconu, Valentin, University of Nevada, Reno, NV 89557-0084 ([email protected]).
Group actions on graphs and C*-correspondences, pp. 147-168.
ABSTRACT. If G acts on a C*-correspondence H over the C*-algebra A (see Definition 3), then by the universal property G acts on the Cuntz-Pimsner algebra O(H) and we study the crossed product O(H)xG and the fixed point algebra. Using intertwiners, we define the Doplicher-Roberts algebra of a representation of a compact group G on H and prove that under certain conditions the fixed point algebra is isomorphic to the Doplicher-Roberts algebra. The action of G commutes with the gauge action on O(H), therefore G acts also on the core algebras. We give applications for the action of a group G on the C*-correspondence associated to a topological graph E. If G is finite and E is discrete and locally finite, we prove that the crossed product C*(E)xG is isomorphic to the C*-algebra of a graph of C*-correspondences and stably isomorphic to a locally finite graph algebra. If C*(E) is simple and purely infinite and the action of G is outer, then the fixed point algebra and the crossed product are also simple and purely infinite with the same K-theory groups. We illustrate with several examples.
Linghui Kong, School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, China ([email protected]) and Yufeng Lu, School of Mathematical Sciences, Dalian University of Technology, Dalian 116024, China ([email protected]).
Some algebraic properties of dual Toeplitz operators, pp. 169-185.
ABSTRACT. In this paper, we study some algebraic properties of dual Toeplitz operators, and prove that the dual Toeplitz operator commuting with an analytic dual Toeplitz operator must be also an analytic dual Toeplitz operator on the orthogonal complement of the Bergman space on the unit ball. Furthermore, we characterize when the sum of products of two dual Toeplitz operators is equal to a dual Toeplitz operator on the orthogonal complement of the Bergman space on the unit ball or the Hardy space on the sphere.
Turdebek N. Bekjan,, L. N. Gumilyov Eurasian National University, Astana 010008, Kazakhstan} ([email protected]).
On dominated convergence in noncommutative integration, pp. 187-200.
ABSTRACT. Let M be a von Neumann algebra and let T: L0(M) → L0(M) be a linear bounded positive map. If (xn)n≥1 is a sequence in L0(M) converging to x almost uniformly in Lance's sense and (xn)n≥1 satisfies some kind of domination condition, then (T(xn))n≥1 converges to T(x) almost uniformly.
Kevin Beanland, Department of Mathematics, Washington and Lee University, Lexington, VA 24450 ([email protected] and Ryan M. Causey, Department of Mathematics, University of South Carolina, Columbia, SC 29208 ([email protected]).
On a generalization of Bourgain's tree index, pp. 201-208.
ABSTRACT. For a Banach space X, a sequence of Banach spaces (Yn) and a Banach space Z with an unconditional basis, D. Alspach and B. Sari introduced a generalization of a Bourgain tree called a (⊕nYn)Z-tree in X. These authors also prove that any separable Banach space admitting a (⊕nYn)Z-tree with order ω1 admits a subspace isomorphic to (⊕nYn)Z. In this paper we give two new proofs of this result.
Boedihardjo, March, Texas A&M University, College Station, TX 77843-3368 ([email protected]) and Dykema, Ken, Texas A&M University, College Station, TX 77843-3368 ([email protected]).
On algebra-valued R-diagonal elements, pp. 209-252.
ABSTRACT. For an element in an algebra-valued ∗-noncommutative probability space, equivalent conditions for algebra-valued R-diagonality (a notion introduced by Sniady and Speicher) are proved. Formal power series relations involving the moments and cumulants of such R-diagonal elements are proved. Decompositions of algebra-valued R-diagonal elements into products of the form unitary times self-adjoint are investigated; sufficient conditions, in terms of cumulants, for ∗-freeness of the unitary and the self-adjoint part are proved, and a tracial example is given where ∗-freeness fails. The particular case of algebra-valued circular elements is considered; as an application, the polar decompostion of the quasinilpotent DT-operator is described.
Benoît Collins,
Department of Mathematics, Kyoto University ([email protected]).
Haagerup's inequality and additivity violation of the Minimum Output Entropy, pp. 253-261.
ABSTRACT.We give a simple and conceptual proof of the fact that random unitary channels yield violation of the Minimum Output Entropy additivity. The proof relies on strong convergence of random unitary matrices and Haagerup's inequality.
Gehér, György Pál, MTA-SZTE Analysis and Stochastics Research Group, Bolyai Institute, University of Szeged, H-6720 Szeged, Aradi vértanúk tere 1., Hungary; and MTA-DE "Lendület" Functional Analysis Research Group, Institute of Mathematics, University of Debrecen, H-4010 Debrecen, P.O. Box 12, Hungary ([email protected]) or ([email protected]).
Surjective Kuiper isometries, pp. 263-281.
ABSTRACT. Characterisations of surjective isometries with respect to the Kuiper distance on three classes of Borel probability measures on the real line (or equivalently, probability distribution functions) are presented here. These classes are the set of continuous, absolute continuous and general measures.
Thomas R. Hoffman and James P. Solazzo, Wall Building 124 F, Mathematics Department, Coastal Carolina University, Conway, SC 29528, ([email protected]), ([email protected]).
Complex two-graphs, pp. 283-300.
ABSTRACT. In A survey of two-graphs, J.J. Seidel lays out the connections between simple graphs, two-graphs, equiangular lines in Rk and strongly regular graphs. It is well known that there is a one-to-one correspondence between two-graphs and sets of equiangular lines in Rk. This article gives a generalization of two-graphs by allowing the entries of the Seidel matrix to be roots of unity beyond ± 1. Many of the results regarding \emph{real} two-graphs have a natural generalization in the complex setting including equiangular lines in Ck.
Sneh Lata, Department of Mathematics, Shiv Nadar University, School of Natural Sciences, Gautam Budh Nagar - 203207, Uttar Pradesh, India ([email protected]) and Dinesh Singh, Department of Mathematics, University of Delhi, Delhi 110007, India ([email protected])
A class of sub-Hardy Hilbert spaces associated with weighted shifts, pp. 301-308.
ABSTRACT. In this note we study sub-Hardy Hilbert spaces on which the the action of the operator of multiplication by the coordinate function $z$ is assumed to be weaker than that of an isometry. We identify such operators with a class of weighted shifts. The well known results of de Branges and Beurling are deduced as corollaries.
Kilbane, James, Department of Pure Mathematics and Mathematical Statistics, University of Cambridge, Cambridge, United Kingdom ([email protected]) and Ostrovskii, M. I., Department of Mathematics and Computer Science, St John's University, NY 11439 ([email protected]).
There is no finitely isometric Krivine's theorem, pp. 309-317.
ABSTRACT. We prove that for every real number p>1 except p=2 there exist a Banach space X isomorphic to lp and a finite subset U in lp, such that U is not isometric to a subset of X. This result shows that the finite isometric version of the Krivine theorem (which would be a strengthening of the Krivine theorem (1976)) does not hold.
Gokieli, Maria ([email protected]), Kenmochi, Nobuyuki ([email protected]), and Niezgódka, Marek ([email protected]), Interdisciplinary Centre for Mathematical and Computational Modelling, University of Warsaw, Tyniecka 15/17, 02-630 Warsaw, Poland.
A new compactness theorem for variational inequalities of parabolic type, pp. 319-350.
ABSTRACT. This paper deals with the weak solvability of fully nonlinear parabolic variational inequalities with time dependent convex constraints. As possible approaches to such problems, one has for instance the time-discretization method and the fixed point method of Schauder type with appropriate compactness theorems. In this paper, our attention is paid to the latter approach. However, no appropriate compactness theorem, that would enable a direct application of the fixed point method to variational inequalities of parabolic type, has been established up to now. We thus start with setting up a new compactness theorem, and then apply it to prove existence of solutions for a wide class of parabolic variational inequalities.
Zhangyong Cai, Department of Mathematics, Guangxi Teachers Education University, Nanning 530023, P.R. China, School of Mathematics, Sichuan University, Chengdu 610065, P.R. China ([email protected]), Shou Lin, Institute of Mathematics, Ningde Normal University, Ningde 352100, P. R. China, Department of Mathematics, Minnan Normal University, Zhangzhou 363000, P.R. China ([email protected]), and Chuan Liu, Department of Mathematics, Ohio University Zanesville Campus, Zanesville, OH 43701, USA ([email protected]).
Copies of special spaces in free (abelian) paratopological groups, pp. 351-362.
ABSTRACT. Let FP(X) (AP(X)) denote the free paratopological group (free Abelian paratopological group) over a topological space X. In this paper, a homeomorphism theorem for the free Abelian paratopological group over a topological space X is established, which extends a result of A. Arhangel'skii. As an application, it is shown that if X is a Tychonoff space and P is a densely self-embeddable prime space with a q-point, then AP(X) contains a copy of P if and only if FP(X) contains a copy of P if and only if X contains a copy of P, which generalizes a theorem of K. Eda, H. Ohta and K. Yamada. At last, it is shown that if the free paratopological group FP(X) (the free Abelian paratopological group AP(X)) over a Tychonoff space X contains a non-trivial convergent sequence, then FP(X) (AP(X)) contains a closed copy of Arens' space, further, which gives an affirmative answer to a question in literature.
F. Azarpanah, Department of mathematics, Shahid Chamran University of Ahvaz, Ahvaz, Iran([email protected] ) and M. Ghirati., Department of Mathematics, Yasouj University, Yasouj, Iran([email protected] ) and A. Taherifar, Department of Mathematics, Yasouj University, Yasouj, Iran([email protected] and [email protected]).
, pp. 363-383.
ABSTRACT. In [M. Ghirati and A. Taherifar, Intersections of essential (resp., free) maximal ideals of C(X), Topol. Appl. 167(2014) 62-68], the authors have given a new representation for some closed ideals of C(X). For instance, they have shown that the intersection of all free maximal ideals of C(X), can be represented by the set of all continuous functions f such that Z(1-fg) is compact for each continuous function g. In this paper we generalize this representation for all closed ideals in C(X) and the intersections of all maximal ideals in C*(X) . For a completely regular Hausdorff space X, we construct a space λX in βX containing νX and show that X is Lindelöf if and only if X coincides with λX.
Lin, Fucai, Minnan Normal University, Zhangzhou 363000, PR China ([email protected]), Lin, Shou, Ningde Normal University, Ningde 352100, PR China([email protected]), and Sakai, Masami, Kanagawa University, Hiratsuka 259-1293, Japan ([email protected])
Point-countable covers and sequence-covering maps, pp. 385-397.
ABSTRACT. We answer some questions on the theory of generalized metric spaces posed in the book: S. Lin, Point-Countable Covers and Sequence-Covering Mappings (second edition), Beijing, China Science Press, 2015. | {"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": 0.8488139510154724, "perplexity": 1493.9716406839493}, "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/1664030335362.18/warc/CC-MAIN-20220929163117-20220929193117-00431.warc.gz"} |
https://en.wikipedia.org/wiki/Quiver_%28mathematics%29 | Quiver (mathematics)
In mathematics, a quiver is a directed graph where loops and multiple arrows between two vertices are allowed, i.e. a multidigraph. They are commonly used in representation theory: a representation V of a quiver assigns a vector space V(x) to each vertex x of the quiver and a linear map V(a) to each arrow a.
In category theory, a quiver can be understood to be an underlying structure of a category, but without identity morphisms and composition. That is, there is a forgetful functor from Cat to Quiv. Its left adjoint is a free functor which, from a quiver, makes the corresponding free category.
Definition
A quiver Γ consists of:
• The set V of vertices of Γ
• The set E of edges of Γ
• Two functions: s: EV giving the start or source of the edge, and another function, t: EV giving the target of the edge.
This definition is identical to that of a multidigraph.
A morphism of quivers is defined as follows. If ${\displaystyle \Gamma =(V,E,s,t)}$ and ${\displaystyle \Gamma '=(V',E',s',t')}$ are two quivers, then a morphism ${\displaystyle m=(m_{v},m_{e})}$ of quivers consist of two functions ${\displaystyle m_{v}:V\to V'}$ and ${\displaystyle m_{e}:E\to E'}$ such that following diagrams commute:
${\displaystyle m_{v}\circ s=s'\circ m_{e}}$
and
${\displaystyle m_{v}\circ t=t'\circ m_{e}}$
Category-theoretic definition
The above definition is based in set theory; the category-theoretic definition generalizes this into a functor from the free quiver to the category of sets.
The free quiver (also called the walking quiver, Kronecker quiver, 2-Kronecker quiver or Kronecker category) Q is a category with two objects, and four morphisms: The objects are V and E. The four morphisms are s: EV, t: EV, and the identity morphisms idV: VV and idE: EE. That is, the free quiver is
${\displaystyle E\;{\begin{matrix}s\\[-6pt]\rightrightarrows \\[-4pt]t\end{matrix}}\;V}$
A quiver is then a functor Γ: QSet.
More generally, a quiver in a category C is a functor Γ: QC. The category Quiv(C) of quivers in C is the functor category where:
Note that Quiv is the category of presheaves on the opposite category Qop.
Path algebra
If Γ is a quiver, then a path in Γ is a sequence of arrows an an−1 ... a3 a2 a1 such that the head of ai+1 = tail of ai, using the convention of concatenating paths from right to left.
If K is a field then the quiver algebra or path algebra KΓ is defined as a vector space having all the paths (of length ≥ 0) in the quiver as basis (including, for each vertex i of the quiver Γ, a trivial path ${\displaystyle e_{i}}$ of length 0; these paths are not assumed to be equal for different i), and multiplication given by concatenation of paths. If two paths cannot be concatenated because the end vertex of the first is not equal to the starting vertex of the second, their product is defined to be zero. This defines an associative algebra over K. This algebra has a unit element if and only if the quiver has only finitely many vertices. In this case, the modules over KΓ are naturally identified with the representations of Γ. If the quiver has infinitely many vertices, then KΓ has an approximate identity given by ${\displaystyle e_{E}:=\sum _{v\in E}1_{v}}$ where E ranges over finite subsets of the vertex set of Γ.
If the quiver has finitely many vertices and arrows, and the end vertex and starting vertex of any path are always distinct (i.e. Q has no oriented cycles), then KΓ is a finite-dimensional hereditary algebra over K. Conversely, if K is algebraically closed, then any finite-dimensional, hereditary, associative algebra over K is Morita equivalent to the path algebra of its Ext quiver (i.e., they have equivalent module categories).
Representations of quivers
A representation of a quiver Q is an association of an R-module to each vertex of Q, and a morphism between each module for each arrow.
A representation V of a quiver Q is said to be trivial if V(x) = 0 for all vertices x in Q.
A morphism, fV → V′, between representations of the quiver Q, is a collection of linear maps ${\displaystyle f(x):V(x)\rightarrow V'(x)}$ such that for every arrow a in Q from x to y ${\displaystyle V'(a)f(x)=f(y)V(a)}$, i.e. the squares that f forms with the arrows of V and V′ all commute. A morphism, f, is an isomorphism, if f(x) is invertible for all vertices x in the quiver. With these definitions the representations of a quiver form a category.
If V and W are representations of a quiver Q, then the direct sum of these representations, ${\displaystyle V\oplus W}$, is defined by ${\displaystyle (V\oplus W)(x)=V(x)\oplus W(x)}$ for all vertices x in Q and ${\displaystyle (V\oplus W)(a)}$ is the direct sum of the linear mappings V(a) and W(a).
A representation is said to be decomposable if it is isomorphic to the direct sum of non-zero representations.
A categorical definition of a quiver representation can also be given. The quiver itself can be considered a category, where the vertices are objects and paths are morphisms. Then a representation of Q is just a covariant functor from this category to the category of finite dimensional vector spaces. Morphisms of representations of Q are precisely natural transformations between the corresponding functors.
For a finite quiver Γ (a quiver with finitely many vertices and edges), let KΓ be its path algebra. Let ei denote the trivial path at vertex i. Then we can associate to the vertex i the projective KΓ-module KΓei consisting of linear combinations of paths which have starting vertex i. This corresponds to the representation of Γ obtained by putting a copy of K at each vertex which lies on a path starting at i and 0 on each other vertex. To each edge joining two copies of K we associate the identity map.
Quiver with relations
To enforce commutativity of some squares inside a quiver a generalization is the notion of quivers with relations (also named bound quivers). A relation on a quiver ${\displaystyle Q}$ is a ${\displaystyle K}$ linear combination of paths from ${\displaystyle Q}$. A quiver with relation is a pair ${\displaystyle (Q,I)}$ with ${\displaystyle Q}$ a quiver and ${\displaystyle I\subseteq K\Gamma }$ an ideal of the path algebra. The quotient ${\displaystyle K\Gamma /I}$ is the path algebra of ${\displaystyle (Q,I)}$.
Quiver Variety
Given the dimensions of the vector spaces assigned to every vertex, one can form a variety which characterizes all representations of that quiver with those specified dimensions, and consider stability conditions. These give quiver varieties, as constructed by King (1994).
Gabriel's theorem
A quiver is of finite type if it has only finitely many isomorphism classes of indecomposable representations. Gabriel (1972) classified all quivers of finite type, and also their indecomposable representations. More precisely, Gabriel's theorem states that:
1. A (connected) quiver is of finite type if and only if its underlying graph (when the directions of the arrows are ignored) is one of the ADE Dynkin diagrams: ${\displaystyle A_{n}}$, ${\displaystyle D_{n}}$, ${\displaystyle E_{6}}$, ${\displaystyle E_{7}}$, ${\displaystyle E_{8}}$.
2. The indecomposable representations are in a one-to-one correspondence with the positive roots of the root system of the Dynkin diagram.
Dlab & Ringel (1973) found a generalization of Gabriel's theorem in which all Dynkin diagrams of finite dimensional semisimple Lie algebras occur. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 28, "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": 0.959619402885437, "perplexity": 369.3799884039351}, "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-2018-34/segments/1534221218189.86/warc/CC-MAIN-20180821132121-20180821152121-00056.warc.gz"} |
http://math.stackexchange.com/questions/120613/need-help-with-this-primitive-roots-question | # Need help with this primitive roots question
Question:
If p and q are odd primes and $({a^p+1})/q$, show that either $(a+1)/q$ or $q= 2kp + 1$ for some integer $k$
I read the theorem that says If p and q are odd primes and $({a^p-1})/q$, show that either $(a-1)/q$ or $q= 2kp + 1$ for some integer $k$. I tried to follow a similar method as the proof of that theorem, but I can't seem to be able to come up with a solution.
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Sorry: what do you mean by "and $(a^p+1)/q$"? And $\frac{a^p+1}{q}$ what? Or is that supposed to mean "$q$ divides $a^p+1$"? (I guessed it might mean "$a^p+1$ divides $q$, but that would be silly: it would mean $a=0$ or $a^p+1=q$). – Arturo Magidin Mar 15 '12 at 19:31
Apparently Google thinks you mean $q \mid (a^p + 1).$ This exact problem is verbatim If p and q are odd primes and q | a^(p) + 1, Show that either q | a+1 or q=2kp +1 for some integer k.? here on: answers.yahoo.com/question/index?qid=20090714022447AAXZQrK – user2468 Mar 15 '12 at 19:36
Isn't this just a matter of replacing $a$ with $-a$? – anon Mar 15 '12 at 19:36
I think the notation may be supposed to mean that the fractions are integers - but it needs clarifying to make sense of the problem. – Mark Bennet Mar 15 '12 at 19:37
Suppose that $q$ divides $a^p+1$. Then $a^p \equiv -1 \pmod q$. Note that $(a^p)^2=a^{2p}$. So $a^{2p}\equiv 1 \pmod q$.
Let $e$ be the smallest positive integer such that $a^e\equiv 1 \pmod q$. Then $e$ divides $2p$. In principle, there are $4$ possibilities to examine: (i) $e=1$; (ii) $e=2$; (iii) $e=p$; (iv) $e=2p$.
Possibility (i): If $e=1$, then $a \equiv 1\pmod{q}$. This cannot happen, because we were told that $a^p \equiv -1\pmod{q}$. But $1\not\equiv -1\pmod{q}$, since $q$ is odd.
Possibility (ii): If $e=2$, then $a\equiv -1\pmod{q}$, so $q$ divides $a+1$.
Possibility (iii) If $e=p$, we reach a contradiction, since $a^p\equiv -1 \pmod{q}$, and $1\not\equiv -1\pmod{q}$.
Possibility (iv): Recall that by the minimality of $e$, the number $e$ must divide any $k$ such that $a^k\equiv 1\pmod{q}$. Since $a$ cannot be divisible by $p$, we have, by Fermat's Theorem, $a^{q-1}\equiv 1\pmod{q}$. Thus if $e=2p$, the number $2p$ must divide $q-1$. Let $q-1=(2p)k$. Then $q=2kp+1$.
So the only live possibilities are (ii) and (iv). One yields that $q$ divides $a+1$, and the other yields that $q$ is of the shape $2kp+1$. Both can happen.
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Suppose by hypothesis that $a^p+1\equiv0 \;(q)$. Then $(-a)^p-1\equiv0 \;(q)$, so by the original theorem we may conclude that either $(-a)-1\equiv 0 \;(q)$, which is $a+1\equiv0\;(q)$, or $q=2kp+1$, as desired.
- | {"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": 0.9934444427490234, "perplexity": 160.25246273496643}, "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/1408500813241.26/warc/CC-MAIN-20140820021333-00393-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/512272/is-there-a-transitive-set-which-is-not-inductive | # Is there a transitive set which is not inductive?
Is there a transitive set which is not inductive? Appreciate any advice , thank you.
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## 2 Answers
$\{\varnothing\}$ is not inductive. Can you see why?
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$\varnothing$ is not inductive. Can you see why?
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(Sorry, Asaf. I couldn't resist.) – Trevor Wilson Oct 2 '13 at 15:46
Assuming the pattern continues, this implies that $-1$ is not inductive... Can you see why? ;) (Sorry Trevor, I couldn't resist) – goblin Oct 2 '13 at 15:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.846541166305542, "perplexity": 1336.6196843532614}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2016-07/segments/1454701153736.68/warc/CC-MAIN-20160205193913-00034-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/129224-differentiation-problem.html | Math Help - Differentiation Problem
1. Differentiation Problem
Hi
I have a question on a paper that I have to solve though I have become unsure as to what to do.
The problem is as follows
Given that y=16x+x^-1, find the two values of x for which dy/dx = 0.
When I differentiate I get 16 + x^-2 or 16 + 1/x^2
to remove the fraction I multiply both sides by x^2 giving
16x^2 + 1 = 0
so dy/dx = 0 -> 16x^2 + 1 = 0.
I thought to obtain the two values of x you have to factorise the quadratic, and I am not sure how to do this given the form of the derived equation.
Any help with this would be gratefully appreciated.
Thank You
David
2. Originally Posted by DavidRUK
Hi
I have a question on a paper that I have to solve though I have become unsure as to what to do.
The problem is as follows
Given that y=16x+x^-1, find the two values of x for which dy/dx = 0.
When I differentiate I get 16 + x^-2 or 16 + 1/x^2 Mr F says: This is wrong. It should be 16 - x^-2 or 16 - 1/x^2.
to remove the fraction I multiply both sides by x^2 giving
16x^2 + 1 = 0
so dy/dx = 0 -> 16x^2 + 1 = 0.
I thought to obtain the two values of x you have to factorise the quadratic, and I am not sure how to do this given the form of the derived equation.
Any help with this would be gratefully appreciated.
Thank You
David
..
3. Originally Posted by DavidRUK
Given that y=16x+x^-1, find the two values of x for which dy/dx = 0.
When I differentiate I get 16 + x^-2 or 16 + 1/x^2
This is not completly correct
$
y=16x+x^{-1} \implies \frac{dy}{dx}= 16-x^{-2}
$
Now finish it off as you were.
4. Other than that error in differentiating it looks like your main doubt was:
I thought to obtain the two values of x you have to factorise the quadratic, and I am not sure how to do this given the form of the derived equation.
Since there is no x^1 term in the quadratic, you don't need to factorise anything.
You have $\frac{dy}{dx}= 16- \frac{1}{x^2} = 0$
I wont tell you the full solution unless you want me too; I'm sure you can see what to do. You need to do some simple algebra and then remember to think about whether you need +.
5. Sorry about that. I got so caught up in the problem i forgot to bring the -1 down to change the sign.
I am still not sure how to find two values for x though.
6. Originally Posted by DavidRUK
Sorry about that. I got so caught up in the problem i forgot to bring the -1 down to change the sign.
I am still not sure how to find two values for x though.
I don't see how you can be studying calculus and not know how to solve an equation like $16- \frac{1}{x^2} = 0$. That is very worrying and suggests life is going to be very tough for you unless you get remedial help in pre-calculus ....
The most obvious approach is
$\frac{1}{x^2} = 16$
$\Rightarrow x^2 = \frac{1}{16}$
etc.
7. You have no idea...
Try doing anything with a screaming baby and a sick wife..
Also I have been up since 4:30AM.
Not good.
Thanks though.. I will go and practice.
8. Originally Posted by DavidRUK
You have no idea...
Try doing anything with a screaming baby and a sick wife..
Also I have been up since 4:30AM.
Not good.
Thanks though.. I will go and practice.
I'm sorry to hear that.
(Note that 99 times out of 100 there is no "screaming baby and a sick wife" ....) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8490053415298462, "perplexity": 519.7604975868471}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375095806.49/warc/CC-MAIN-20150627031815-00037-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/circular-orbit-of-spacecraft.69725/ | # Circular orbit of spacecraft
1. Apr 2, 2005
### bmoran08
An unmanned spacecraft is in a circular orbit around the moon, observing the lunar surface from an altitude of 56.0 km. To the dismay of scientists on earth, an electrical fault causes an on-board thruster to fire, decreasing the speed of the spacecraft by 16.0 m/s.
If nothing is done to correct its orbit, with what speed (in km/h) will the spacecraft crash into the lunar surface?
I tried finding the velocity to keep it in orbit, subtracted the decrease in speed, found the new radius and then found the new velocity it crashes with, but it is wrong. can anyone help?
2. Apr 2, 2005
### SpaceTiger
Staff Emeritus
Try using conservation of energy. What is its total energy after the slowing? What would its kinetic energy be upon impact? | {"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": 0.8697957992553711, "perplexity": 1746.1451591489747}, "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-2017-04/segments/1484560282631.80/warc/CC-MAIN-20170116095122-00541-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/141611-prove-n-e-all-g.html | # Thread: Prove that a^(n) = e for all a in G.
1. ## Prove that a^(n) = e for all a in G.
Let G be a group of finite order n. Prove that a^(n) = e for all a in G.
2. Originally Posted by rainyice
Let G be a group of finite order n. Prove that a^(n) = e for all a in G.
By Lagrange's theorem, and $a^k=e\Longrightarrow a^n=(a^k)^x=\ldots$
Tonio | {"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": 0.8180680274963379, "perplexity": 1105.9229158010169}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121153.91/warc/CC-MAIN-20170423031201-00145-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://blog.computationalcomplexity.org/2003/01/foundations-of-complexity-lesson-13.html | ## Sunday, January 12, 2003
### Foundations of Complexity Lesson 13: Satisfiability
Previous Lesson | Next Lesson
Boolean-Formula Satisfiability (SAT) is the single most important language in computational complexity. Here is an example of a Boolean formula.
(u OR v) AND (u OR v)
u and v are variables that take on values from {TRUE, FALSE}. u means the negation of u. A literal is either a variable or its negation.
An assignment is a setting of the variables to true and false, for example (u→TRUE, v→FALSE). Once all of the variables are assigned a truth value, the formula itself has a truth value. The assignment (u→TRUE, v→FALSE) makes the formula above false. A satisfying assignment is an assignment that makes the formula true. For the formula above, the assignment (u→TRUE, v→TRUE) is satisfying. If a formula has a satisfying assignment we say the formula is satisfiable.
SAT is the set of satisfiable formula. The formula above is in SAT. This formula is not.
u AND (u OR v) AND (u or v)
A formula is in conjunctive normal form (CNF) if it is the AND of several clauses, each consisting of an OR of literals, like the formulas above. A disjunctive normal form (DNF) formula is the same with AND and OR reversed. A formula is k-CNF if every clause has exactly k literals. The first formula above is in 2-CNF.
CNF-SAT is the set of satisfiable CNF formulas. k-CNF-SAT or k-SAT is the set of satisfiable formulas in k-CNF.
Cook and Levin independently showed that SAT is NP-complete. The problem remains NP-complete if we restrict to CNF-SAT or even 3-SAT.
Next lesson we will show that CNF-SAT is NP-complete. | {"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": 0.8385486602783203, "perplexity": 810.1187582618827}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999656144/warc/CC-MAIN-20140305060736-00024-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/a-point-charge-in-a-sphere-with-v0.386520/ | # A point charge in a sphere with V0
• Start date
• #1
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If there is a point charge at the center of a hollow sphere that has voltage of V0. How would I go by solving the voltage and electric field distribution?
I know I can use Poisson (Laplace) to find distribution had there been no charge inside the sphere, and could use Gauss had there been no V on the sphere. But how about the combination?
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Hmmm. So there is a spherical shell with a voltage, and now we want to find the voltage inside the shell?
So it seems that what you'll have to do, unless I'm overlooking something, is
$$\nabla^2 V = \frac{q}{\epsilon_0}$$
You'll be solving a non-homogeneous PDE. It may be a tricky solution to work out. Of course, once you solve V, you can E from the gradient (which also may be not easy task).
• #3
671
2
The spherical symmetry makes this into a simple problem.
Since the point charge is placed at the very center of the sphere, that means that whatever charge redistribution it causes on the hollow sphere, retains the spherical symmetry. That means that the sphere provides no inwards E-field and acts as a point charge outwards (Remember that no charge has flowed off the sphere or onto it).
Use Gauss' Law and the capacitance of the sphere to solve for the charges, potentials and fields.
• #4
378
2
Hmmm. So there is a spherical shell with a voltage, and now we want to find the voltage inside the shell?
So it seems that what you'll have to do, unless I'm overlooking something, is
$$\nabla^2 V = \frac{q}{\epsilon_0}$$
You'll be solving a non-homogeneous PDE. It may be a tricky solution to work out. Of course, once you solve V, you can E from the gradient (which also may be not easy task).
There is a spherical shell with a point charge Q at its center and I want to know how to find E/V distribution over all the space inside/outside.
$$\nabla^2 V = \frac{q}{\epsilon_0}$$
So the "q" will be just the charge that is located inside the sphere?
The spherical symmetry makes this into a simple problem.
Since the point charge is placed at the very center of the sphere, that means that whatever charge redistribution it causes on the hollow sphere, retains the spherical symmetry. That means that the sphere provides no inwards E-field and acts as a point charge outwards (Remember that no charge has flowed off the sphere or onto it).
Use Gauss' Law and the capacitance of the sphere to solve for the charges, potentials and fields.
Because, there is a boundary condition that V0 is placed on the shell doesn't that affect the way Q induces charge over the shell or it can be just
Q/A
• #5
1,860
0
There is a spherical shell with a point charge Q at its center and I want to know how to find E/V distribution over all the space inside/outside.
$$\nabla^2 V = \frac{q}{\epsilon_0}$$
So the "q" will be just the charge that is located inside the sphere?
Right, this is just verbatim of Maxwell's Equation. It's been a while since I've solved a non-homogenous PDE, but it's just like for differential equations: the solution is the particular solution plus the homogeneous solution (the general solution is a well-known solution for this type of problem). Now generally, you'd use eigenfunction expansion to solve this, but I'm not sure how well that will work with Legendre polynomials. However, you should be able to just guess the particular solution. After all, you only need to figure out something that will give you a constant after two derivatives.
Again, I haven't yet worked it out myself, but maybe it won't be so bad now that I think about it. Let me know how it goes (or went if this is already past your due date).
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3K | {"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": 0.9179049730300903, "perplexity": 356.6530152183586}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991650.73/warc/CC-MAIN-20210518002309-20210518032309-00468.warc.gz"} |
https://www.futurelearn.com/courses/logic-the-philosophical-science-of-truth/0/steps/157490 | ## Want to keep learning?
This content is taken from the University of York's online course, Logic: The Language of Truth. Join the course to learn more.
4.6
# Arrow and ‘if … then … ’: an argument for equivalence
Here is an argument for the view that arrow and ‘if … then … ’ have the same truth-table. This argument uses some of the key resources we’ve developed in this course.
First, note that where ‘If α then β’ is true, that’s enough to guarantee that ‘(α $$\rightarrow$$ β)’ is true: we’ve already noted that where ‘If α then β’ is true, it can’t be the case that ‘α’ is true and ‘β’ is false, and that’s the only kind of situation in which‘(α $$\rightarrow$$ β)’ can be false. So, ‘If α then β’ entails ‘(α $$\rightarrow$$ β)’.
Now, look at this simple argument:
• Boris did it or Dominic did it
• Therefore, if it’s not the case that Boris did it, then Dominic did it
Note that this argument is expressed in English and uses ‘If … then … ’. It seems that it is valid. More than that, it seems to be formally valid. Any argument with this shape
• α or β
• Therefore, if ~α, then β
will be valid.
But now notice that ‘or’ and vel have the same truth-table. So, we can substitute vel for ‘or’ in the premise, like this:
• (α $$\vee$$ β)
• Therefore, if ~α, then β
(If you are worried about exclusive-‘or’ here, note that there do seem to be cases of ‘or’ in English which are inclusive, and someone could plausibly say the premise of our example argument intending inclusive-‘or’ and the argument would be valid.)
But now look at the truth-tables for ‘(α $$\vee$$ β)’ and ‘(~α $$\rightarrow$$ β)’.
#### Figure 1. Truth-table for ‘(α $$\vee$$ β)’ and ‘(~α $$\rightarrow$$ β)’
What does this show about ‘(α $$\vee$$ β)’ and ‘(~α $$\rightarrow$$ β)’?
It shows they are logically equivalent. So, as far as valid arguments go, we can swap one for the other and, if our original argument was valid, the argument resulting from the substitution will be valid too.
Making the substitution gives us this:
• (~α $$\rightarrow$$ β)
• Therefore, if ~α, then β
But now look what we’ve got. We started with an argument in English that we are confident is formally valid. We’ve substituted logically equivalent sentences for the original premise, but the result is that we have an argument (a formally valid argument) from an arrow-sentence to the corresponding ‘If … then … ’-sentence. Putting our two results together, we have:
• If α then β $$\models$$ (α $$\rightarrow$$ β)
and
• (α $$\rightarrow$$ β) $$\models$$ If α then β
This means ‘If α then β’ and ‘(α $$\rightarrow$$ β)’ are logically equivalent. Arrow and ‘If … then … ’ do have the same truth-table. | {"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": 0.9193545579910278, "perplexity": 3068.9251032821776}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2020-45/segments/1603107889651.52/warc/CC-MAIN-20201025183844-20201025213844-00076.warc.gz"} |
http://mathoverflow.net/questions/36848/what-category-without-initial-object-do-you-care-about?sort=newest | # What category without initial object do you care about?
Recently I have been listening to some constructions that have been designed to accommodate categories without an initial object. The speaker has given some idea of a category or two that he cares about, and thus why he was thinking in this direction, but I am now wondering;
As working mathematicians, what category are you concerned with that does not have an initial object?
I am sorry if this question is slightly strange, I have made it CW because it seems appropriate.
Thanks!
-
Can you, please, indicate examples of constructions "designed to accommodate categories without an initial object" you had in mind? Otherwise, your question risks turning into a not particularly illuminating fishing expedition. – Victor Protsak Aug 27 '10 at 8:37
I am also curious. If a category has no initial object, you can always just formally adjoin an initial object such that nothing maps into it (except itself). This is basically the role the empty set plays in the concrete categories listed. – Kevin Ventullo Aug 27 '10 at 9:50
@Andrew I don't understand your comment, perhaps it relates to a comment that has been deleted. – B. Bischof Aug 27 '10 at 18:38
@B.Bischof: it did. There was a comment which had three suggestions of categories without initial objects. I thought that at least two of these did have initial objects and so said so. It appears that the original comment has been deleted. To avoid further confusion, I have now deleted my comment! Once you've read this, I recommend that we delete these two comments to remove all sources of confusion! – Andrew Stacey Aug 27 '10 at 19:32
But if you do that, then this comment will look really out of place... – Cam McLeman Aug 27 '10 at 22:35
show 1 more comment
Isn't it so that the category of fields doesn't have an initial object? I seem to recall seeing a lot of F_uss about this on the web from time to time.
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I guess the question is meant to ask abut "classical concrete categories" only, those whose objects are sets with structure. For otherwise, the class of examples that we certainly care about is vast and uninteresting (as a class).
For instance a groupoid has an initial object precisely if it is equivalent to the trivial groupoid.
Any disjoint union of two categories has no initial object.
Any poset without smallest elements has no initial object.
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Actually, I don't mean only concrete categories. The question was not in the hope of studying these categories as a class, it was instead to understand where Working mathematicians run into categories where no initial object exists, and thus they lose the power of things like homological algebra. Thank you for the examples you did provide, in particular the first one of groupoids. – B. Bischof Aug 27 '10 at 18:40
The bordism categories which arise in the study of topological quantum field theories are extremely interesting but don't have initial or terminal objects (except in degenerate cases). These categories have some flavor of d-dimensional manifolds for objects and (d+1)-dimensional cobordisms between them for morphisms. See the Wikipedia and n-lab articles for further details and references.
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The category of varieties (or just integral schemes) has no initial object. (The empty set is not irreducible; this does not prove my claim, but indicates that the naive choice for an initial object does not work).
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Why the empty set is not irreducible? It has no proper subsets, therefore it cannot be union of two ones. Or am I wrong? – Lennart Meier Aug 27 '10 at 8:57
I think this is the same as the question 'Why is one not prime?' It cannot be written as the product of two proper factors. – Simon Wadsley Aug 27 '10 at 9:30
The category of fields has no initial object: It has one connected component for each characteristic, each of which has an initial object (this is a special case of Urs Schreiber's general remark about non-connected categories, but one I care about...)
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Let R be the hyperfinite type III1 factor (if you don't know what that is, "let R be a ring" is a good enough approximation). The following category is equivalent to the category of R-R-bimodules on infinite dimensional separable Hilbert spaces.
Objects: Unital ring homomorphisms RR.
Morphisms: Hom(φ,ψ) = {xR : ∀yR, x φ(y) = ψ(y) x }
Composition of morphisms is given by multiplication in R. This category is actually a strict monoidal, with monoidal structure given by composition of ring homomorphisms.
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The category of CDG-rings (curved DG-rings) does not have an initial object. (A CDG-ring B = (B,d,h) is a graded ring B endowed with an odd derivation d of degree 1 and an element h of degree 2 such that d2(b) = [h,b] for any b from B and d(h)=0. A morphism of CDG-rings (B,d_B,h_B) → (A,d_A,h_A) is a pair (f,a), where f is a morphism of graded rings A → B and a is an element of A of degree 1 such that f(d_B(b)) = d_A(f(b)) + [a,f(b)] for any b from B and f(h_B) = h_A + d_A(a) + a2, where [,] denotes the supercommutator (with signs). I leave the definition of the composition of morphisms to the readers of this answer.)
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Any programming languages based on lambda calculus (Haskell, (OCa)ML, Clean, Coq, Agda) forms a category -- in many different ways, in fact. One way is to have an object for each type of the language and a morphism for each well-typed expression with exactly one free variable. If the programming language has linear types then this category will not have a terminal object, since such an object would imply the ability to "discard" a value of any type.
Linear types are extremely useful in giving pure functional programming languages the ability to express side-effecting operations. Loosely, you have a linear type called "world"; every impure function both takes and returns a value of that type (and perhaps other values too). Since you can't duplicate or discard a "world", its handling imposes a deterministic evaluation order on all of the impure functions, and program transformations will not alter this order. The aggregate result of the side effects can then be reasoned about (formally, even).
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With a non-strict language like Haskell, there is no initial object either (without the need to bring linearity into the picture). The naive choice would be the empty type, but a lazy function defined on the empty type can still return a value (any value it likes) if it never tries to inspect its argument. – Mike Shulman Sep 3 '10 at 7:00
Egad. Somehow I managed to answer the question "... without a terminal object" when the question asked clearly said "initial object". – Adam Sep 3 '10 at 18:19
In universal algebra and model theory, one usually requires that the underlying set be non-empty. If the signature has constants in it, then this is no restriction, but otherwise, there is no initial object. For example, under this definition there is no initial object in the category of semigroups. The reason for this restriction is that bad things happen to first-order logic when the underying carrier set is empty. Many standard theorems break down when you allow the empty domain. For example, the following theorem of first-order logic $$\forall x P(x) \rightarrow \exists x P(x)$$ becomess false. | {"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": 0.8661138415336609, "perplexity": 711.1230078376758}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1393999654315/warc/CC-MAIN-20140305060734-00098-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.studypug.com/sg/sg-primary6/multiplying-and-dividing-fractions/divide-fractions-and-mixed-numbers | # Dividing fractions and mixed numbers
### Dividing fractions and mixed numbers
In this section, we will use diagrams to divide fractions and mixed numbers (a.k.a. compound fractions) for the purpose of helping you getting a better understanding about the concepts of fraction division. We will also teach you how to divide fractions by using multiplications.
#### Lessons
• Introduction
a)
Simplify fractions: Method A - By using greatest common factors
b)
Simplify fractions: Method B - By using common factors
c)
How to divide fractions with fractions?
• Why do we "flip and multiply" when we divide fractions?
d)
How to do cross-cancelling?
e)
How to convert between mixed numbers and improper fractions?
• 1.
Find each of the following quotient by using diagrams.
a)
$\frac{5}{6} \div \frac{3}{4}$
b)
$2\frac{1}{3} \div \frac{1}{8}$
c)
$\frac{7}{5} \div \frac{1}{2}$
• 2.
Find each quotient by using multiplication.
a)
$\frac{5}{8} \div \frac{1}{4}$
b)
$2 \div \frac{3}{5}$
c)
$3\frac{2}{7} \div 1\frac{9}{{10}}$
• 3.
Three cups of water can fill up $\frac{2}{3}$ of a kettle. How many cups of water are required to fill up 5 kettles? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 9, "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": 0.8188428282737732, "perplexity": 2455.6996907976436}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794864657.58/warc/CC-MAIN-20180522092655-20180522112655-00321.warc.gz"} |
http://math.stackexchange.com/questions/94778/what-do-these-terms-mean-commutative-associative-distributive?answertab=oldest | # What do these terms mean: commutative, associative, distributive
I am reading a book, and I am trying to understand what the writer really mean by the following terms. I would like to understand what these words mean in relation to the examples.
In regular algebra, addition and multiplication are commutative: $$A + B = B + A$$ $$A \times B = B \times A$$ they are also associative: $$A + (B + C) = (A + B) + C$$ $$A \times (B \times C) = (A\times B) \times C$$ And multiplication is said to be distributive: $$A \times (B + C) = (A \times B) + (A \times C)$$
In Boolean algebra, the $+$ operator is distributive over the $\times$ operator: $$W + (B \times F) = (W + B)\times (W + F)$$ $$W = \text{white}\qquad B = \text{black}\qquad F = \text{female}$$
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What part of this don't you understand? – Thomas Andrews Dec 28 '11 at 20:46
There is a typo, when you say multiplication is said to be distributive, in symbols that should be $A\times (B+C)=A\times B+A\times C$. – André Nicolas Dec 28 '11 at 20:48
The meaning of the words are the formulas that come right after the colon in each case -- no more, no less. – Henning Makholm Dec 28 '11 at 21:28
@HananN. Are you asking how the mathematical concepts relate to the meaning of the words in the English language? – Bill Dubuque Dec 28 '11 at 21:46
Commutative: the result is the same if operands are commutated. $x + y$ is the same as $y + x.$ Assosicative: $x + y + z$ is the same as $(x + y) + z$ as $x + (y + z)$. It doesn't matter if you assigned an order (also: associate) to $(x+y)$ before $(y + z)$ and vice versa. Distributive: results are the same when you distribute. – user2468 Dec 29 '11 at 0:04
You've quoted the definitions right there; there's not much more to understanding beyond just getting a lot of experience with manipulating expressions using those rules to gain some intuition of their utility.
For instance, you may figure out that if you have an operation * which is commutative but not associative, then the following manipulations are valid:
$a * (b * c) = a*(c*b) = (b*c) * a = (c*b)*a$
but the following is not:
$a*(b*c) = b*(a*c)$
The effect of associativity is to allow you to drop the parentheses altogether without worrying about ambiguity of the expression. So there are 5 different interpretations of the expression $a*b*c*d$, namely, $(a*b)*(c*d), ((a*b)*c)*d, (a*(b*c))*d, a*((b*c)*d)$ and $a*(b*(c*d))$. Associativity guarantees that these are all the same.
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Let us examine the associative property first.
1. Associative property of addition is how numbers are associated or
grouped together. We need a convention on how to compute 1+(2+3). The
associative property shows that 1+(2+3) yields same value as (1+2)+3.
For instance,
( * **) *** * (** ***)
(1 + 2 ) + 3 = 1 + (2 + 3 )
( A + B ) + C = A + (B + C )
2. Consider the following diagram where we have three 4s or four 3s which depends on if we group by rows or by columns.
* * * *
* * * *
* * * *
Thus commutative principle for multiplication shows that the order does not matter and 3 X 4 = 4 x 3 or the general form:
(A)(B)$\equiv$(B)(A)
3. For distributive principle of multiplication consider the following array of counters to represent (3)(12):
* * * * * * * * * * | * *
* * * * * * * * * * | * *
* * * * * * * * * * | * *
The rectangular array is divided into two rectangles such as the
first one is (3)(10) and the next one is (3)(2). The multiplication has been broken into two parts ie distributed. The product (3)(10+2) can be replaced as (3)(10) + (3)(2).
The full statement is:
Multiplication is distributive with respect to addition.
Symbolically, it can written as: (A) (B+C) $\equiv$ (A) (B) + (A) (C)
Multiplication is also distributive with respect to subtraction.
Thus, (A) (B-C) $\equiv$ (A) (B) - (A) (C)
Reference:
Lay, L.Clark. The Study of Arithmetic. The Macmillan Company. 1966.
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Commutatitive:Is a condition where by the number's and letter's are adding or multiply. Example addition; x+y=y+x or 2+3=3+2 and multiply; xy=yx or 2*3=3*2
Associative:Is a condition that a group of quantities connecte by operators gives the same result whatever their grouping. Example addition, a+(b+c)=(a+b)+c or multiplication, a*(bc)=(ab)*c.
Distributive:Is a matter of separate or break it into parts,and it makes the number to a group or letter to a group to multiply and add easly. Example a+(bc)=(a+b)(a+c) or 2+(3*4)=(2+3)*(2+4).
- | {"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": 0.8516358137130737, "perplexity": 448.83160572896844}, "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/1432207924799.9/warc/CC-MAIN-20150521113204-00250-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_15&curid=17410&oldid=148408 | # 2019 AIME II Problems/Problem 15
## Problem
In acute triangle points and are the feet of the perpendiculars from to and from to , respectively. Line intersects the circumcircle of in two distinct points, and . Suppose , , and . The value of can be written in the form where and are positive integers, and is not divisible by the square of any prime. Find .
## Solution
First we have , and Similarly, and dividing these each by gives .
It is known that the sides of the orthic triangle are , and its angles are ,, and . We thus have the three sides of the orthic triangle now. Letting be the foot of the altitude from , we have, in , similarly, we get Our final answer is then
The requested sum is .
༺\\crazyeyemoody9❂7//༻
## Solution 1
Let
Therefore
By power of point, we have Which are simplified to
Or
(1)
Or
Let Then,
In triangle , by law of cosine
Pluging (1)
Or
Substitute everything by
The quadratic term is cancelled out after simplified
Which gives
Plug back in,
Then
So the final answer is
By SpecialBeing2017
## Solution 2
Let and
By power of point, we have and
Therefore, substituting in the values:
Notice that quadrilateral is cyclic.
From this fact, we can deduce that and
Therefore is similar to .
Therefore:
Now using Law of Cosines on we get:
Notice
Substituting and Simplifying:
Now we solve for using regular algebra which actually turns out to be very easy.
We get and from the above relations between the variables we quickly determine , and
Therefore
So the answer is
By asr41
## See Also
2019 AIME II (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byLast Question 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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Login to AoPS | {"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": 0.9470025897026062, "perplexity": 1454.5021202114299}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "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-2021-17/segments/1618038060603.10/warc/CC-MAIN-20210411000036-20210411030036-00011.warc.gz"} |
https://math.stackexchange.com/questions/903508/bogus-proof-that-the-liouville-form-on-the-cotangent-bundle-is-nondegenerate | # Bogus proof that the Liouville Form on the cotangent bundle is nondegenerate.
Suppose we have a manifold $M$ of dimension $n$ and its cotangent bundle $T^*M$. The Liouville form $\lambda$ on $T^*M$ is defined as $\lambda_{\omega_p} = \pi^*(\omega_p)$ where $\pi$ is the standard projection map from $T^*M$ to $M$. The problem is to prove that $d\lambda$ is a nondegenerate 2-form on $T^*M$. I was able to solve this problem by simply expressing the form with the aid of local coordinates and deriving the result by straightforward computation. However, I also have a bogus proof where I can't identify the incorrect step. Exponents are repeated wedge products.
$d\lambda_{\omega_p} = d\pi^*(\omega_p)$
$(d\lambda_{\omega_p})^n = (d\pi^*(\omega_p))^n = \pi^*(d\omega_p)^n$
(Because pullbacks commute with exterior differentiation and wedge products).
However, $\pi^*(d\omega_p)^n = 0$ because it is the image of a $2n$ form on a $n$ dimensional manifold.
Where am I going wrong? I'm guessing I'm missing some hypothesis in order to use the commutative fact but I am not sure. Thank you.
You can't differentiate a pointwise expression. In particular, $\lambda$ is not the pullback of a form on $M$.
• I think I understand what you're saying but just to make it clear I want to see if what I'm saying in my own words is correct. You can't differentiate a specific covector i.e. $d\omega_p$ makes no sense. However, if $\omega$ is a form on $M$ we do have $(d\omega)_p$. Aug 20, 2014 at 0:48 | {"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": 0.9785712957382202, "perplexity": 95.67930048706461}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662594414.79/warc/CC-MAIN-20220525213545-20220526003545-00569.warc.gz"} |
https://physicsoverflow.org/26082/underlying-graphical-calculus-interactions-round-lattice?show=26083 | # What is the underlying graphical calculus of the Interactions-Round-a-Face lattice model?
+ 3 like - 0 dislike
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## Background
Let $\mathcal{L}$ be an $m \times n$ square lattice on a torus, and let $\Sigma$ be a finite set. We think of $\Sigma$ as the possible spin values that can be assigned to the points of the lattice; a state will therefore be a map $\mathcal{L} \to \Sigma$. In the Interactions-Round-a-Face (IRF) model, there is a weight function $W: \Sigma^4 \to \mathbb{R}$, and the total Boltzmann weight of a state $s: \mathcal{L} \to \Sigma$ will be $\overline{W}(s) = \prod_{\text{faces } (i, j, k, l)} W(s_i, s_j, s_k, s_l)$. (The vertices of a face are labeled starting from the upper left and going clockwise, say.) The partition function is defined to be $\sum_{\text{states } s} \overline{W}(s)$. This model has many exactly solvable 2-dimensional lattice models as special cases.
The standard analysis of this partition function begins by considering the row transfer matrix, defined as follows. Given a pair of adjacent rows and spin vectors $\phi = (s_1, \ldots, s_n)$, $\phi' = (s_1', \ldots, s_n')$, we define $T_{\phi, \phi'} = \prod_{i = 1}^n W(s_i, s_{i + 1}, s_{i + 1}', s_i')$. We think of $T$ as defining a $\lvert \Sigma \rvert^n \times \lvert \Sigma \rvert^n$ matrix. One then observes that the partition function is simply the trace of $T^m$.
More generally, one could take each face to have a different weight function $W$; we would then get a different transfer matrix for each row, and the partition function would be the trace of the product of all these transfer matrices.
## Question
Computations in the IRF model can be viewed as a sort of graphical calculus, where we think of the row transfer matrix as a map from $V = \mathbb{C}^{\Sigma^n}$ to itself, where the input is given by the assignment of spins to the top row, and the output is given by the assignment of spins to the bottom row. More explicitly, $T$ should be thought of as a multilinear form on $V \otimes V$, which we then convert into a map $V \to V$ using the given basis to identify $V$ with $V^\ast$. Stacking rows corresponds to composing $T$ with itself, and identifying the $(m + 1)$-st row with the first row corresponds to taking the trace. This interpretation also works when $T$ varies from row to row.
What's not clear to me is how to extend this graphical calculus to horizontal composition. One would like to start not from the row transfer matrix $T$, whose definition seems somewhat artificial, but the face weight $W$, which can be thought of as a 4-linear form on $V' = \mathbb{C}^{\Sigma}$, or as a map $W: V'^{\otimes 2} \to V'^{\otimes 2}$ by identifying the "bottom" two copies of $V'$ with $V'^{\ast}$ as above. Vertical composition is again given by powering $W$, but the horizontal composition is strange. It sends the $(2, 2)$-tensors $W_{s_1's_2'}^{s_1s_2}$, $W_{s_2's_3'}^{s_2s_3}$ to the "$(3, 3)$-tensor" $X_{s_1's_2's_3'}^{s_1s_2s_3} = W_{s_1's_2'}^{s_1s_2} W_{s_2's_3'}^{s_2s_3}$, which doesn't seem to be a "categorical" construction. The problem is that we need to use the middle input (and output) twice, so the resulting map will be quadratic, rather than linear, in these variables.
What is the correct categorical framework for the IRF model?
This post imported from StackExchange MathOverflow at 2015-01-16 22:09 (UTC), posted by SE-user Evan Jenkins
retagged Jan 16, 2015
You're looking for an answer beyond "the calculus of 4-tensors", I assume. Note that in the Penrose notation, you should use the Poincare dual, so that your faces become 4-valent vertices. In any case, probably what you're looking for is some word like "planar algebra" or "spider" or "pivotal monoidal category" or...?
This post imported from StackExchange MathOverflow at 2015-01-16 22:09 (UTC), posted by SE-user Theo Johnson-Freyd
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https://www.geneconvenevi.org/articles/analysis-of-the-trojan-y-chromosome-eradication-strategy-for-an-invasive-species/ | # Analysis of the Trojan Y-Chromosome eradication strategy for an invasive species
X. Y. Wang, J. R. Walton, R. D. Parshad, K. Storey and M. Boggess, Journal of Mathematical Biology, 68:1731-1756. 2013. The Trojan Y-Chromosome (TYC) strategy, an autocidal genetic biocontrol method, has been proposed to eliminate invasive alien species. In this work, we analyze the dynamical system model of the TYC strategy, with the aim of studying the viability of the TYC eradication and control strategy of an invasive species. In particular, because the constant introduction of sex-reversed trojan females for all time is not possible in practice, there arises the question: What happens if this injection is stopped after some time? Can the invasive species recover? To answer that question, we perform a rigorous bifurcation analysis and study the basin of attraction of the recovery state and the extinction state in both the full model and a certain reduced model. In particular, we find a theoretical condition for the eradication strategy to work. Additionally, the consideration of an Allee effect and the possibility of a Turing instability are also studied in this work. Our results show that: (1) with the inclusion of an Allee effect, the number of the invasive females is not required to be very low when the introduction of the sex-reversed trojan females is stopped, and the remaining Trojan Y-Chromosome population is sufficient to induce extinction of the invasive females; (2) incorporating diffusive spatial spread does not produce a Turing instability, which would have suggested that the TYC eradication strategy might be only partially effective, leaving a patchy distribution of the invasive species. | {"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": 0.8000472187995911, "perplexity": 1104.697014888958}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500158.5/warc/CC-MAIN-20230205000727-20230205030727-00756.warc.gz"} |
https://www.physicsforums.com/threads/re-eigenvalue-inequality-problem.91310/ | # Homework Help: Re(eigenvalue) inequality problem
1. Sep 28, 2005
### N2
Hello,
if a diff.eqn. has the characteristic equation $\lambda^2 + (3-K) \lambda + 1 = 0$
the eigenvalues solves to $\lambda=-3/2 + K/2 \pm 1/2 \sqrt{5-6K+K^2}$. No problem there. But when is the diff.eqn. asymp. stable, meaning $\Re(\lambda)<0$ ?
I can only get this far
$\Re(-3/2 + K/2 \pm 1/2*\sqrt{5-6K+K^2})<0$
$-3/2+1/2 \Re(K \pm \sqrt{5-6K+K^2})<0$
How can i find the values for K, where this inequality is true?
Thanks
2. Sep 29, 2005
### CarlB
My inclination would be to first consider the case when $$5-6K+K^2$$ is positive, and then consider the case when it is negative. If you separate them out, it shouldn't be too hard.
Carl
3. Sep 30, 2005
### HallsofIvy
Solving K2- 6K+ 5> 0 tells us that there will be complex roots for K between 1 and 5 and real roots for K<= 1, >= 5.
I there are complex roots, the real part is just -3/2+ K/2. That is 0 for K= 3, negative for K< 3, positive for K> 3. The solution will be stable for 1< K< 3, unstable for 3< K< 5.
For K<= 1 or K>= 5, we need to look at all of $-3/2 + K/2 \pm 1/2 \sqrt{5-6K+K^2}$
The best way to determine where that is positive or negative is to set it equal to 0 and solve the resulting quadratic equation. Those will separate "< 0" from "> 0". Choose one value of K in each resulting interval to see whether this is positive or negative. | {"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": 0.8168007731437683, "perplexity": 963.8798858930323}, "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/1537267156724.6/warc/CC-MAIN-20180921013907-20180921034307-00548.warc.gz"} |
http://davegiles.blogspot.com/2011/09/micronumerosity.html | ## Thursday, September 15, 2011
### Micronumerosity
In a much earlier post I took a jab at the excessive attention paid to the concept of "multicollinearity", historically, in econometrics text books.
Art Goldberger (1930-2009) made numerous important contributions to econometrics, and modelling in the social sciences in general. He wrote several great texts, the earliest of which (Goldberger, 1964) was one of the very first to use the matrix notation that we now take as standard for the linear regression model.
In one of his text books, Art also poked fun at the attention given to multicollinearity, and I'm going to share his parody with you here in full. In a couple of places I've had to replace formulae with words. What follows is from Chapter 23.3. of Goldberger (1991):
"Econometrics texts devote many pages to the problem of multicollinearity in multiple regression, but they say little about the closely analogous problem of small sample size in estimation a univariate mean. Perhaps that imbalance is attributable to the lack of an exotic polysyllabic name for 'small sample size'. If so, we can remove that impediment by introducing the term micronumerosity.
Suppose an econometrician set out to write a chapter about small sample size in sampling from a univariate population. Judging from what is now written about multicollinearity, the chapter might look like this:
1. Micronumerosity
The extreme case, 'exact micronumerosity', arises when n = 0; in which case the sample estimate of μ is not unique. (Technically, there is a violation of the rank condition n > 0: the matrix 0 is singular.) The extreme case is easy enough to recognize. 'Near micronumerosity' is more subtle, and yet very serious. It arises when the rank condition n > 0 is barely satisfied. Near micronumerosity is very prevalent in empirical economics.
2. Consequences of micronumerosity
The consequences of micronumerosity are serious. Precision of estimation is reduced. There are two aspects of this reduction: estimates of μ may have large errors, and not only that, but [the variance of the sample mean; DG] will be large.
Investigators will sometimes be led to accept the hypothesis μ = 0 because [the ratio of the sample mean to its standard error; DG] is small, even though the true situation may be not that μ = 0 but simply that the sample data have not enabled us to pick μ up.
The estimate of μ will be very sensitive to sample data, and the addition of a few more observations can sometimes produce drastic shifts in the sample mean.
The true μ may be sufficiently large for the null hypothesis μ= 0 to be rejected, even though [the variance of the sample mean; DG] = σ2/n is large because of micronumerosity. But if the true μ is small (although nonzero) the hypothesis μ = 0 may mistakenly be accepted.
3. Testing for micronumerosity
Tests for the presence of micronumerosity require the judicious use of various fingers. Some researchers prefer a single finger, others use their toes, still others let their thumbs rule.
A generally reliable guide may be obtained by counting the number of observations. Most of the time in econometric analysis, when n is close to zero, it is also far from infinity.
Several test procedures develop critical values n*; such that micronumerosity is a problem only if n is smaller than n*: But those procedures are questionable.
4. Remedies for micronumerosity
If micronumerosity proves serious in the sense that the estimate of μ has an unsatisfactorily low degree of precision, we are in the statistical position of not being able to make bricks without straw. The remedy lies essentially in the acquisition, if possible, of larger samples from the same population.
But more data are no remedy for micronumerosity if the additional data are simply 'more of the same'. So obtaining lots of small samples from the same population will not help."
If you check the data that go with that earlier post of mine, you'll see that in this text book, Goldberger devoted 8 pages (2.12% of the book) - including what you've just read - to the topic of multicollinearity. The average for the introductory texts in my sample was 2.15%. This was probably the only time that Art was (slightly) below average!
References
Goldberger, A. S. (1964). Econometric Theory. Wiley, New York.
Goldberger, A. S. (1991). A Course in Econometrics. Harvard University Press, Cambridge MA.
© 2011, David E. Giles
1. I think the reason multicollinearity is emphasised so much is because its easy to understand the problem - so people talk about it lot. (Makes people look knowledgeable when really they're not).
2. Really enjoyed this post, thank you. There can sometimes be a bit of a disconnect between the contents of stats/econometrics courses and the nitty gritty of empirical research...
3. Sinclair, Frances: Thanks for the comments!
4. in multiple regession, if two or more explanatory variables are highly correlated with each other, multicollinearity will occur. Is this the standard assumption ?
1. Assumption????? No!
2. Why is this not the case ?
3. Sorry - you've lost me! Why would we assume that the variables are collinear??
4. The question that was aked to me was: is the following statement correct "in multiple regession, if two or more explanatory variables are highly correlated with each other, multicollinearity will occur" explain your reasoning, I assumed that this statement would be correct? But now that you have said no I'm confused and I don't have great sources of information to guide me
Thanks for your time!
5. I didn't say "no"; I said why would you ASSUME that they are collinear. Of course, if the variable are highly correlated, multicollinearity will occur. That's just the definition of multicollinearity! What more can I say?
5. Thanks for sharing | {"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": 0.8019813299179077, "perplexity": 1303.537617881601}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698540932.10/warc/CC-MAIN-20161202170900-00248-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://stem.mrozarka.com/calculus-1/lessons/unit-2/day-20 | ### Day 20 - Product/Quotient Rule - 02.01.16
Update Unit 2 TestFriday, 02.12.16Calculus Equations/TheoremsDerivative Operator$\dpi{100} \frac{d}{dx}$ Find the derivative of the following: none of the aboveFind the second derivative (derivative of the derivative) of the following: none of the aboveFind the point(s) at which the following function has a horizontal tangent line: none of the aboveFind the derivative of the following function using the Product Rule: none of the aboveFind the derivative of the following function: none of the aboveReviewPre-calculusSlopeEquation of a LineSecant Line vs. Tangent Line (video)DerivativeHow can the slope of a tangent line be found? Equation of a Tangent Line (video)Tangent Line GraphDerivative (video)CheckpointsBasic Differentiation RulesConstant Rule (video) (example)Power Rule (video) (example)Constant Multiple Rule (video) (example)Sum and Difference Rule (video)Derivative of Sine and Cosine FunctionsCheckpointsLessonProduct Rule (video) and Quotient Rule (video)How can derivatives of the product/quotient of functions be calculated?Checkpoints Exit Ticket Posted on the board at the end of the block.HomeworkWNQ the following:Chain Rule (video)Study! Help Request List Standard(s) APC.2Define and apply the properties of limits of functions.Limits will be evaluated graphically and algebraically.Includes:limits of a constantlimits of a sum, product, and quotientone-sided limitslimits at infinity, infinite limits, and non-existent limitsAPC.3Use limits to define continuity and determine where a function is continuous or discontinuous.Includes:continuity in terms of limitscontinuity at a point and over a closed intervalapplication of the Intermediate Value Theorem and the Extreme Value Theoremgeometric understanding and interpretation of continuity and discontinuity | {"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": 1, "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": 0.9425227046012878, "perplexity": 2022.1017958052767}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600402130531.89/warc/CC-MAIN-20200930235415-20201001025415-00081.warc.gz"} |
http://mathhelpforum.com/advanced-applied-math/56957-3d-equilibrium-rigid-body.html | # Thread: 3D Equilibrium of a rigid body
1. ## 3D Equilibrium of a rigid body
I am stuck on this question
I generally know how to solve the problem but the force is throwing me off. Since it is parallel to the XY plane there will no be k component of the force. But how should I about breaking the 970N force into i,j components? Thanks.
2. The distance from $O$ to the point where $\vec{F}$ takes effect is $\vec{r} = (250,\ 200,\ -150)\ mm$. $\vec{F}$ is in the x-y plane, so $\vec{F}=(F_x,\ F_y,\ 0)$. The momentum will then be
$\vec{M}=\vec{r}\times\vec{F}$
Now, we probably want to charge the support at $O$ as much as possible, so we set $|\vec{F}| = 970\ N$. Besides, we want $\vec{r}$ and $\vec{F}$ to be orthogonal to each other, so
$\vec{r}\cdot\vec{F} = 0 \Leftrightarrow 250 F_x + 200 F_y = 0 \Leftrightarrow F_y = -\frac{5}{4}F_x$
$F_z = 0$, so $\vec{F}$ is already in the x-y plane.
So we could parametrisize $\vec{F}$ and write $\vec{F} = t\cdot(4,\ -5,\ 0)$, implying that $|\vec{F}| = \sqrt{41}\ |t|$, which means that $\vec{F}$ could have basically any non-negative length.
Now we know that there exist $\vec{F}$ with the in the x-y plane with the length 970 N, orthogonal to $\vec{r}$. Let's use that in our momentum formula:
$\vec{M} = \vec{r}\times\vec{F} \Rightarrow$
$|\vec{M}| = |\vec{r}\times\vec{F}| = |\vec{r}|\cdot|\vec{F}|\cdot\sin(\alpha) = \sqrt{250^2+200^2+(-150)^2}\ mm\ \cdot\ 970\ N\cdot\ \sin(\alpha) =$ $\frac{485}{\sqrt{2}}\ Nm\cdot\sin(\alpha)$, where $\alpha$ is the angle between $\vec{r}$ and $\vec{F}$
Now, since $\vec{r}$ and $\vec{F}$ are orthogonal, $\sin(\alpha) = 1$, so $|\vec{M}| = \frac{485}{\sqrt{2}}\ Nm$, which should be the maximum magnitude of the momentum.
Alternativelly, you could turn $\vec{F}$ into poolar coordinates:
$F_x=970\ N\cdot\cos(\varphi)$
$F_y=970\ N\cdot\sin(\varphi)$
$F_z=0$
perform the cross product between $\vec{r}$ and $\vec{F} = 970\ N\cdot(\cos(\varphi),\ \sin(\varphi),\ 0)$ to get the momentum, extract the magnitude of the momentum (the vector lenth), and analyse the expression to se which $\varphi$ that gives the greatest magnitude of the momentum or to directly extract the greatest value of the expression.
3. I got the answer, thanks man
4. I'm studuing mechanics myself right 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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 36, "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": 0.971402645111084, "perplexity": 242.16103212544834}, "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/1502886110792.29/warc/CC-MAIN-20170822143101-20170822163101-00084.warc.gz"} |
https://clevelandlodge.org/wxdye/83cf62-john-3%3A1-7-meaning | The general methodology for this involves assuming a solution of the form $$y = \\sum_{n=0}^\\infty a_nx^{n+r}.$$ One normally keeps the index $0$ for the first and second derivatives. 500 1000 500 500 500 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 In the case the point is ordinary, we can find solution around that point by power series.The solution around singular points has been left to explain. endobj 12 0 obj /LastChar 196 Step 3: Use the system of equations , {\displaystyle r_{2}} /Subtype/Type1 A. 2 694.5 295.1] is a rational function, the power series can be written as a generalized hypergeometric series. View Chapter 4.3 The Method of Frobenius from MATHEMATIC 408s at University of Texas. 33 0 obj 0 /LastChar 196 endobj /FirstChar 33 875 531.3 531.3 875 849.5 799.8 812.5 862.3 738.4 707.2 884.3 879.6 419 581 880.8 This is the extensive document regarding the Frobenius Method. 460.7 580.4 896 722.6 1020.4 843.3 806.2 673.6 835.7 800.2 646.2 618.6 718.8 618.8 All the three cases (Values of 'r' ) are covered in it. 36 0 obj Hence adjoining a root Ï of it to the field of 3-adic numbers Q 3 gives an unramified extension Q 3 (Ï) of Q 3. z Can't Go There 6. 777.8 777.8 1000 500 500 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 /FontDescriptor 23 0 R (2.13) 2.1 Possible problems Let me give you a couple of examples to compare. which will not be solvable with regular power series methods if either p(z)/z or q(z)/z2 are not analytic at z = 0. SINGULAR POINTS AND THE METHOD OF FROBENIUS 287 7.3.2 ThemethodofFrobenius Beforegivingthegeneralmethod,letusclarifywhenthemethodapplies.Let In particular there are three questions in my text book that I have attempted. My question show (§4.3) that one obtains in this way a Frobenius structure on M. (0.6) We illustrate this method with two examples: (1) the universal deformation of a connection on a bundle F o on the aï¬ne line A 1 , ⦠1002.4 873.9 615.8 720 413.2 413.2 413.2 1062.5 1062.5 434 564.4 454.5 460.2 546.7 /BaseFont/BPIREE+CMR6 Room With a View Some of this music was created 20 years ago and it was time to curate a collection and make them public. r+ ~c( ) ~a( ) = 0; (18) which is called the indicial equation for (14). the recurrence relation places no restriction on the coefficient for the term k /Widths[1000 500 500 1000 1000 1000 777.8 1000 1000 611.1 611.1 1000 1000 1000 777.8 As before, if $$p(x_0) = 0$$, then $$x_0$$ is a singular point. PDF | On Jan 1, 2020, Asadullah Torabi published Frobenius Method for Solving Second-Order Ordinary Differential Equations | Find, read and cite all the research you need on ResearchGate The Frobenius method on a second-order... 1147 3 The Solution of a Second-Order Homoge-neous Linear ODE using Method of Frobe-nius Lemma 3.1. k Browse other questions tagged complex-analysis singularity frobenius-method or ask your own question. The LibreTexts libraries are Powered by MindTouch ® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. /Widths[300 500 800 755.2 800 750 300 400 400 500 750 300 350 300 500 500 500 500 1277.8 811.1 811.1 875 875 666.7 666.7 666.7 666.7 666.7 666.7 888.9 888.9 888.9 >> Method of Frobenius: Equal Roots to the Indicial Equation We solve the equation x2 y''+3 xy'+H1-xL y=0 using a power series centered at the regular singular point x=0. Method of Frobenius General Considerations L. Nielsen, Ph.D. Department of Mathematics, Creighton University Di erential Equations, Fall 2008 L. Nielsen, Ph.D. /FirstChar 33 295.1 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 295.1 295.1 << /Name/F8 The Method Of Frobenius 2. 300 325 500 500 500 500 500 814.8 450 525 700 700 500 863.4 963.4 750 250 500] Since the ratio of coefficients / In the following we solve the second-order differential equation called the hypergeometric differential equation using Frobenius method, named after Ferdinand Georg Frobenius. 2 Whatever Happened 3. Featured on Meta New Feature: Table Support Ascolta senza pubblicità oppure acquista CD e MP3 adesso su Amazon.it. /FirstChar 33 >> Let $p(x) y'' + q(x) y' + r(x) y = 0$ be an ODE. /FontDescriptor 26 0 R /Subtype/Type1 I find the Frobenius Method quite beautiful, and I would like to be able to apply it. 3.2 The Frobenius method for second-order equations In this section, we will consider second-order linear equations u00+ p(z)u0+ q(z)u= 0: Clearly, everything we know from the real case (superposition principle, etc.) 0 0 0 0 0 0 0 0 0 0 777.8 277.8 777.8 500 777.8 500 777.8 777.8 777.8 777.8 0 0 777.8 − im having a hard time problem in the indicial equations. Cul-De-Sac 7. 783.4 872.8 823.4 619.8 708.3 654.8 0 0 816.7 682.4 596.2 547.3 470.1 429.5 467 533.2 Regular and Irregular Singularities As seen in the preceding example, there are situations in which it is not possible to use Frobeniusâ method to obtain a series solution. The other solution will be of a form indicated by the indicial equation. Ascolta senza pubblicità oppure acquista CD e MP3 adesso su Amazon.it. Suppose the roots of the indicial equation are r 1 and r 2. These equations will allow us to compute r and the c n. 6. 0 t = is a singular point of the ordinary differential âEquation (4) ... Case 3: kk. stream /FirstChar 33 x��ZYo�6~�_�G5�fx�������d���yh{d[�ni"�q�_�U$����c�N���E�Y������(�4�����ٗ����i�Yvq�qbTV.���ɿ[�w��:��ȿo��{�XJ��7��}��jj?�o���UW��k�Mp��/���� The Method of Frobenius III. In a power series starting with r /Widths[609.7 458.2 577.1 808.9 505 354.2 641.4 979.2 979.2 979.2 979.2 272 272 489.6 << 380.8 380.8 380.8 979.2 979.2 410.9 514 416.3 421.4 508.8 453.8 482.6 468.9 563.7 If the difference between the roots is not an integer, we get another, linearly independent solution in the other root. /FontDescriptor 29 0 R ACM95b/100b Lecture Notes Caltech 2004 The Method of Frobenius Consider the equation x2 y 00 + xp(x)y 0 + q(x)y = 0, (1) where x = 0 is a regular singular point. /BaseFont/XKICMY+CMSY10 Introduction The âna¨Ä±veâ Frobenius method The general Frobenius method Remarks Under the hypotheses of the theorem, we say that a = 0 is a regular singular point of the ODE. 666.7 666.7 666.7 666.7 611.1 611.1 444.4 444.4 444.4 444.4 500 500 388.9 388.9 277.8 List the three cases of the Frobenius method. 756.4 705.8 763.6 708.3 708.3 708.3 708.3 708.3 649.3 649.3 472.2 472.2 472.2 472.2 In Trench 7.5 and 7.6 we discussed methods for finding Frobenius solutions of a homogeneous linear second order equation near a regular singular point in the case where the indicial equation has a repeated root or distinct real roots that donât differ by an integer. /Widths[342.6 581 937.5 562.5 937.5 875 312.5 437.5 437.5 562.5 875 312.5 375 312.5 Contents 1. / 295.1 826.4 531.3 826.4 531.3 559.7 795.8 801.4 757.3 871.7 778.7 672.4 827.9 872.8 /LastChar 196 Note: 1 or 1.5 lectures, §8.4 and §8.5 in , §5.4â§5.7 in . 545.5 825.4 663.6 972.9 795.8 826.4 722.6 826.4 781.6 590.3 767.4 795.8 795.8 1091 b(sub 3) = -3/128. In ⦠820.5 796.1 695.6 816.7 847.5 605.6 544.6 625.8 612.8 987.8 713.3 668.3 724.7 666.7 carries over to the complex case and we know that the solutions are analytic whenever the coe cients p(z) and q(z) are. 761.6 679.6 652.8 734 707.2 761.6 707.2 761.6 0 0 707.2 571.2 544 544 816 816 272 From (r â 1)2 = 0 we get a double root of 1. logo1 Method of Frobenius Example First Solution Second Solution (Fails) What is the Method of Frobenius? For each value of r (typically there are two), we can 656.3 625 625 937.5 937.5 312.5 343.8 562.5 562.5 562.5 562.5 562.5 849.5 500 574.1 1 Method of Frobenius. Section 7.3 Singular points and the method of Frobenius. /Type/Font 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 606.7 816 748.3 679.6 728.7 811.3 765.8 571.2 0 0 0 0 0 0 0 0 0 0 0 0 675.9 937.5 875 787 750 879.6 812.5 875 812.5 875 0 0 812.5 5. Get 1:1 help now from expert Calculus tutors Solve it with our calculus problem solver and calculator The general methodology for this involves assuming a solution of the form $$y = \\sum_{n=0}^\\infty a_nx^{n+r}.$$ One normally keeps the index$0\$ for the first and second derivatives. The Method of Frobenius. ~b( ) ~a( ) 1 ! 734 761.6 666.2 761.6 720.6 544 707.2 734 734 1006 734 734 598.4 272 489.6 272 489.6 767.4 767.4 826.4 826.4 649.3 849.5 694.7 562.6 821.7 560.8 758.3 631 904.2 585.5 5. {\displaystyle (e^{z})/z,} 708.3 708.3 826.4 826.4 472.2 472.2 472.2 649.3 826.4 826.4 826.4 826.4 0 0 0 0 0 Introduction The âna¨Ä±veâ Frobenius method The general Frobenius method Remarks Under the hypotheses of the theorem, we say that a = 0 is a regular singular point of the ODE. {\displaystyle B_{k}.} we get linear combination of some elementary functions like x^2, lnx, e^ax, sin(ax), cos(ax) etc as general & particular solution. This function ~y(x) will not in general be a solution to (14), but we expect that ~y(x) will be close to being a solution. << No headers. endobj Evaluation of Real Definite Integrals, Case II: Singular Points of Linear Second-Order ODEs (4.3) The Method of Frobenius (4.4) Handout 2 on An Overview of the Fobenius Method : 16-17: Evaluation of Real Definite Integrals, Case III Evaluation of Real Definite Integrals, Case IV: The Method of Frobenius - Exceptional Cases (4.4, 4.5, 4.6) 18-19 888.9 888.9 888.9 888.9 666.7 875 875 875 875 611.1 611.1 833.3 1111.1 472.2 555.6 /Name/F5 − 3. /Type/Font The Method of Frobenius If either p(x) or q(x) in y00+ p(x)y0+ q(x)y = 0 isnot analyticnear x 0, power series solutions valid near x 0 may or may not exist. 0 These equations will allow us to compute r and the c n. 6. The Set-Up The Calculations and Examples The Main Theorems Inserting the Series into the DE Getting the Coe cients Observations Roots Di ering by a Positive Integer Here we have r 1 =r 2 +N for some positive integer N . /BaseFont/TBNXTN+CMTI12 462.4 761.6 734 693.4 707.2 747.8 666.2 639 768.3 734 353.2 503 761.2 611.8 897.2 Substituting the above differentiation into our original ODE: is known as the indicial polynomial, which is quadratic in r. The general definition of the indicial polynomial is the coefficient of the lowest power of z in the infinite series. ) A similar method of solution can be used for matrix equations of the first order, too. 450 500 300 300 450 250 800 550 500 500 450 412.5 400 325 525 450 650 450 475 400 {\displaystyle y_{1}(x)} /FontDescriptor 32 0 R /BaseFont/KNRCDC+CMMI12 >> Before giving the general method, let us clarify when the method applies. − 726.9 726.9 976.9 726.9 726.9 600 300 500 300 500 300 300 500 450 450 500 450 300 Method of Frobenius â A Problematic Case. endobj Singular points y" + p(x)y' + p(x)y = 272 272 489.6 544 435.2 544 435.2 299.2 489.6 544 272 299.2 516.8 272 816 544 489.6 >> 489.6 489.6 489.6 489.6 489.6 489.6 489.6 489.6 489.6 489.6 272 272 272 761.6 462.4 {\displaystyle z=0} case : sensitive by Method of Frobenius, released 14 September 2019 1. 11 .3 Frobenius Series Solutions 659 The Method of Frobenius We now approach the task of actually finding solutions of a second-order linear dif ferential equation near the regular singular point x = 0. /LastChar 196 I'm not sure if I'm doing this right. 444.4 611.1 777.8 777.8 777.8 777.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Regular singular points1 2. 2 597.2 736.1 736.1 527.8 527.8 583.3 583.3 583.3 583.3 750 750 750 750 1044.4 1044.4 /Type/Font /BaseFont/FQHLHM+CMBX12 /BaseFont/XZJHLW+CMR12 491.3 383.7 615.2 517.4 762.5 598.1 525.2 494.2 349.5 400.2 673.4 531.3 295.1 0 0 For the Love of Jayne 10. /Subtype/Type1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 826.4 295.1 826.4 531.3 826.4 k /Type/Font 334 405.1 509.3 291.7 856.5 584.5 470.7 491.4 434.1 441.3 461.2 353.6 557.3 473.4 324.7 531.3 531.3 531.3 531.3 531.3 795.8 472.2 531.3 767.4 826.4 531.3 958.7 1076.8 << /Widths[660.7 490.6 632.1 882.1 544.1 388.9 692.4 1062.5 1062.5 1062.5 1062.5 295.1 761.6 272 489.6] 708.3 795.8 767.4 826.4 767.4 826.4 0 0 767.4 619.8 590.3 590.3 885.4 885.4 295.1 so we see that the logarithm does not appear in any solution. My question /FirstChar 33 /Subtype/Type1 The simplest such equation is the constantâcoefficient equidimensional equation 2 ⦠also Fuchsian equation). The Frobenius method has been used very successfully to develop a theory of analytic differential equations, especially for the equations of Fuchsian type, where all singular points assumed to be regular (cf. /Subtype/Type1 / In this Big Guitar 4. Case (d) Complex conjugate roots If c 1 = λ+iμ and c 2 = λâiμ with μ = 0, then in the intervals âd < x < 0 and 0 < x < d the two linearly independent solutions of the differential equation are /Name/F2 A Solution at singular point. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Two independent solutions are /Type/Font /Type/Font Case 3. While behavior of ODEs at singular points is more complicated, certain singular points are not especially difficult to solve. In particular, this can happen if the coe cients P(x) and Q(x) in the ODE y00+ P(x)y0+ Q(x)y = 0 fail to be de ned at a point x 0. 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 312.5 312.5 342.6 624.1 928.7 753.7 1090.7 896.3 935.2 818.5 935.2 883.3 675.9 870.4 896.3 896.3 1220.4 Method of Frobenius: Equal Roots to the Indicial Equation We solve the equation x2 y''+3 xy'+H1-xL y=0 using a power series centered at the regular singular point x=0. Frobeniusâ method for curved cracks 63 At the same time the unknowns B i must satisfy the compatibility equations (2.8), which, after linearization, become 1 0 B i dξ=0. We introduce the Frobenius series method to solve second order linear equations, and illustrate it by concrete examples. 343.8 593.8 312.5 937.5 625 562.5 625 593.8 459.5 443.8 437.5 625 593.8 812.5 593.8 384.3 611.1 675.9 351.8 384.3 643.5 351.8 1000 675.9 611.1 675.9 643.5 481.5 488 Case (d) Complex conjugate roots If c 1 = λ+iμ and c 2 = λâiμ with μ = 0, then in the intervals âd < x < 0 and 0 < x < d the two linearly independent solutions of the differential equation are This problem has been solved! 805.5 896.3 870.4 935.2 870.4 935.2 0 0 870.4 736.1 703.7 703.7 1055.5 1055.5 351.8 The method of frobenius 1. − >> endobj For each value of r (typically there are two), we can /LastChar 196 /LastChar 196 826.4 295.1 531.3] z << 295.1 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 295.1 carries over to the complex case and we know that the solutions are analytic whenever the coe cients p(z) and q(z) are. Subject:- Mathematics Paper:-Ordinary Differential Equations and Special Functions Principal Investigator:- Prof. M.Majumdar /Name/F6 See the answer. >> The LibreTexts libraries are Powered by MindTouch ® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Math 338 Notes: Illustration to Case 3 of the Frobenius Theorem. In some cases the constant C must be zero. This is the extensive document regarding the Frobenius Method. Example 1 Take ï¬rst the case of dy dx = αy x. /Widths[295.1 531.3 885.4 531.3 885.4 826.4 295.1 413.2 413.2 531.3 826.4 295.1 354.2 << The method of Frobenius is to seek a power series solution of the form. << 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 663.6 885.4 826.4 736.8 /Type/Font which has the requisite singularity at z = 0. In this case it happens to be that this is the rth coefficient but, it is possible for the lowest possible exponent to be r â 2, r â 1 or, something else depending on the given differential equation. << ) Kim [3] used the the method of Frobenius to. Let y=Ún=0 ¥a xn+r. 384.3 611.1 611.1 611.1 611.1 611.1 896.3 546.3 611.1 870.4 935.2 611.1 1077.8 1207.4 endobj If, furthermore, the limits /FirstChar 33 L. Nielsen, Ph.D. Consider a 2nd order linear homogeneous ODE y00(x)+ b(x) x y0(x)+ b(x) x y(x) = 0: (1) To ï¬nd basis of solutions y1(x);y2(x) of (1), one seeks them in the form of generalized power series y(x) = xr X1 n=0 anx n; (2) where without loss of generality, a0 6= 0. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 458.3 458.3 416.7 416.7 324.7 531.3 590.3 295.1 324.7 560.8 295.1 885.4 590.3 531.3 590.3 560.8 414.1 419.1 Formulation of the method2 3. If the root is repeated or the roots differ by an integer, then the second solution can be found using: where /BaseFont/IMGAIM+CMR8 Frobenius Method ( All three Cases ) - Free download as PDF File (.pdf), Text File (.txt) or read online for free. and so is unramified at the prime 3; it is also irreducible mod 3. 9 0 obj /Widths[791.7 583.3 583.3 638.9 638.9 638.9 638.9 805.6 805.6 805.6 805.6 1277.8 Application of Frobeniusâ method In order to solve (3.5), (3.6) we start from a plausible representation of B x,B y that is 24 0 obj has a power series starting with the power zero. The one solution of the second-order homogeneous linear di er- ... this paper, we consider the case for which is a prime number and because. One of the two solutions will always be of the form (2), where r is a root of (4). /FontDescriptor 8 0 R 544 516.8 380.8 386.2 380.8 544 516.8 707.2 516.8 516.8 435.2 489.6 979.2 489.6 489.6 7.3. /FontDescriptor 35 0 R {\displaystyle 1/z} In the process of synchronizing all the series of the differential equation to start at the same index value (which in the above expression is k = 1), one can end up with complicated expressions. /FirstChar 33 )()()()( ''' xfyxqyxpyxr =++ â )( )( )( )( )( )( ''' xr xf y xr xq y xr xp y =++ The points where r(x)=0 are called as singular points. The Frobenius method yields a basis of solutions. If r 1 âr 2 â Z, then both r = r 1 and r = r 2 yield (linearly /BaseFont/NPKUUX+CMMI8 = 18 0 obj 500 500 500 500 500 500 500 300 300 300 750 500 500 750 726.9 688.4 700 738.4 663.4 1 638.4 756.7 726.9 376.9 513.4 751.9 613.4 876.9 726.9 750 663.4 750 713.4 550 700 z The Frobenius function is a placeholder for representing the Frobenius form (or Rational Canonical form) of a square matrix. << 27 0 obj r Best Answer 100% (1 rating) Previous question Next question Get more help from Chegg. {\displaystyle B_{0}} 687.5 312.5 581 312.5 562.5 312.5 312.5 546.9 625 500 625 513.3 343.8 562.5 625 312.5 im very confused. In this section we discuss a method for finding two linearly independent Frobenius solutions of a homogeneous linear second order equation near a regular singular point in the case where the indicial equation has a repeated real root. /Subtype/Type1 The simplest such equation is the constantâcoefficient equidimensional equation 2 ⦠1111.1 1511.1 1111.1 1511.1 1111.1 1511.1 1055.6 944.4 472.2 833.3 833.3 833.3 833.3 Suppose the roots of the indicial equation are r 1 and r 2. 413.2 590.3 560.8 767.4 560.8 560.8 472.2 531.3 1062.5 531.3 531.3 531.3 0 0 0 0 >> It was explained in the last chapter that we have to analyse first whether the point is ordinary or singular. We may find the image of Ï under the Frobenius map by locating the root nearest to Ï 3, which we may do by Newton's method. {\displaystyle z^{0},} Using this, the general expression of the coefficient of zk + r is, These coefficients must be zero, since they should be solutions of the differential equation, so. If it is set to zero then with this differential equation all the other coefficients will be zero and we obtain the solution 1/z. 833.3 1444.4 1277.8 555.6 1111.1 1111.1 1111.1 1111.1 1111.1 944.4 1277.8 555.6 1000 Method of Frobenius. Methods of Frobenius ⢠If x is not analytic, it is a singular point. The right hand side blows up at x = 0 but not too badly. {\displaystyle B_{k}} It is used in conjunction with either mod or evala. /FontDescriptor 17 0 R 675.9 1067.1 879.6 844.9 768.5 844.9 839.1 625 782.4 864.6 849.5 1162 849.5 849.5 to obtain a differential equation of the form. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 753.7 1000 935.2 831.5 FROBENIUS SERIES SOLUTIONS 3. where ris a root of r2+. B Using this root, we set the coefficient of zk + r â 2 to be zero (for it to be a solution), which gives us: Given some initial conditions, we can either solve the recurrence entirely or obtain a solution in power series form. Let y=Ún=0 ¥a xn+r. Frobenius Method ( All three Cases ) - Free download as PDF File (.pdf), Text File (.txt) or read online for free. Question: List The Three Cases Of The Frobenius Method. FROBENIUS SERIES SOLUTIONS TSOGTGEREL GANTUMUR Abstract. 791.7 777.8] SU/KSK MA-102 (2018) Substituting this series in (1), we obtain the recursion formula a n+1 = n2 n 1 n+1 a n: ... Case I:When (3) has two distinct roots r 1, r 2. This then determines the rest of the y /FontDescriptor 11 0 R 1000 1000 1055.6 1055.6 1055.6 777.8 666.7 666.7 450 450 450 450 777.8 777.8 0 0 >> /FontDescriptor 20 0 R {\displaystyle (e^{z}-1)/z} , which can be set arbitrarily. 495.7 376.2 612.3 619.8 639.2 522.3 467 610.1 544.1 607.2 471.5 576.4 631.6 659.7 1 ( 295.1 826.4 501.7 501.7 826.4 795.8 752.1 767.4 811.1 722.6 693.1 833.5 795.8 382.6 The necessary conditions for solving equations of the form of (2) However, the method of Frobenius provides us with a method ⦠ACM95b/100b Lecture Notes Caltech 2004 ( (You should check that zero is really a regular singular point.) For example DE $$(x-1)^2x^4y'' + 2(x-1)xy' - y = 0$$ 21 0 obj If r 1 âr 2 â Z, then both r = r 1 and r = r 2 yield (linearly independent) solutions. {\displaystyle z^{2}} 15 0 obj endobj /FontDescriptor 14 0 R 299.2 489.6 489.6 489.6 489.6 489.6 734 435.2 489.6 707.2 761.6 489.6 883.8 992.6 812.5 875 562.5 1018.5 1143.5 875 312.5 562.5] << / 699.9 556.4 477.4 454.9 312.5 377.9 623.4 489.6 272 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 The Set-Up The Calculations and Examples The Main Theorems Outline 1 The Set ⦠489.6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 611.8 816 Section 1.1 Frobenius Method. Everything Is Platinum 8. Section 8.4 The Frobenius Method 467 where the coefï¬cients a n are determined as in Case (a), and the coefï¬cients α n are found by substituting y(x) = y 2(x) into the differential equation. 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Independent solution in the other coefficients will be of a form indicated by the indicial equation r! Now from expert Calculus tutors solve it with our Calculus problem solver and calculator the Method of solution can used! Nielsen, Ph.D 2.1 Possible problems let me give You a couple of examples to compare for ( 14.... The two solutions will always be of a form indicated by the indicial equation for ( 14.! Is called the indicial equation â 1 ) 2 = 0 we get a double root of ( )! Haarsa and S. Pothat nd a solution of the Euler-Cauchy equation expressed by erential! Ferdinand Georg Frobenius is ordinary or singular 5 - Frobenius Step 2 Set... Acquista CD e MP3 adesso su Amazon.it: Use the system of equations the Method of Frobenius MATHEMATIC. R ' ) are covered in it which is called the indicial equation are r 1 and r 2 You! 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With our Calculus problem solver and calculator the Method of Frobenius Step by Step from ESE 319 Washington! The B k this looks wrong, can anyone explain where I might be going wrong first order too... Off it has three cases of the indicial equation 0 but not too badly get another, linearly solution... Find the Frobenius Method a regular singular point of the two solutions will always of... The two solutions will always be of the Euler-Cauchy equation expressed by di operator... P ( x_0 ) = 0 but not too badly is not an,. Of a form indicated by the indicial equation ) previous question Next question get more help from Chegg a to! Before, if \ ( p ( x_0 ) = 0 but too. Pothat nd a solution of the lowest power of z ï¬rst the Case of dy =. The roots are integer an infinite series solution of the ordinary differential âEquation 4... Apply it under grant numbers 1246120, 1525057, and 1413739 also acknowledge previous National Science support. Introduce the Frobenius Theorem previous question Next question get more help from Chegg other root last Case if. Possible problems let me give You a couple of examples to compare find the Frobenius Method beautiful! Of ' r ' ) are covered in it but not too badly 4 ) - Frobenius Step:. 2.1 Possible problems let me give You a couple of examples to compare in conjunction with either or.
Beaglebone Black I2c C Example, Annoying Office Phrases, Beautyrest 16" Plush Pillow Top Mattress King, Openssl Signature Verification, Small Computer Desks For Small Spaces, When Do Figs Ripen In Israel, Bringing Up Bébé Epub, Eugene Field School, Deer Sunset Wallpaper, Strain Gauge Types, Marriage License In Florida For Non Resident, | {"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": 2, "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": 0.8792106509208679, "perplexity": 564.2669332245554}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00524.warc.gz"} |
http://cpr-hepex.blogspot.com/2013/07/13073518-christian-weinheimer-et-al.html | Neutrino Masses [PDF]
Christian Weinheimer, Kai Zuber
The various experiments on neutrino oscillation evidenced that neutrinos have indeed non-zero masses but cannot tell us the absolute neutrino mass scale. This scale of neutrino masses is very important for understanding the evolution and the structure formation of the universe as well as for nuclear and particle physics beyond the present Standard Model. Complementary to deducing constraints on the sum of all neutrino masses from cosmological observations two different methods to determine the neutrino mass scale in the laboratory are pursued: the search for neutrinoless double $\beta$-decay and the direct neutrino mass search by investigating single $\beta$-decays or electron captures. The former method is not only sensitive to neutrino masses but also probes the Majorana character of neutrinos and thus lepton number violation with high sensitivity. Currently quite a few experiments with different techniques are being constructed, commissioned or are even running, which aim for a sensitivity on the neutrino mass of {\cal O}(100) meV. The principle methods and these experiments will be discussed in this short review.
View original: http://arxiv.org/abs/1307.3518 | {"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": 0.9798257350921631, "perplexity": 1030.0894757060664}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948513512.31/warc/CC-MAIN-20171211125234-20171211145234-00573.warc.gz"} |
http://www.zora.uzh.ch/99539/ | # A polynomial optimization approach to principal-agent problems - Zurich Open Repository and Archive
Schmedders, Karl; Renner, Philipp Johannes (2015). A polynomial optimization approach to principal-agent problems. Econometrica, 83(2):729-769.
## Abstract
This paper presents a new method for the analysis of moral hazard principal-agent problems. The new approach avoids the stringent assumptions on the distribution of outcomes made by the classical first-order approach and instead only requires the agent's expected utility to be a rational function of the action. This assumption allows for a reformulation of the agent's utility maximization problem as an equivalent system of equations and inequalities. This reformulation in turn transforms the principal's utility maximization problem into a nonlinear program. Under the additional assumptions that the principal's expected utility is a polynomial and the agent's expected utility is rational in the wage, the final nonlinear program can be solved to global optimality. The paper also shows how to first approximate expected utility functions that are not rational by polynomials, so that the polynomial optimization approach can be applied to compute an approximate solution to non-polynomial problems. Finally, the paper demonstrates that the polynomial optimization approach extends to principal-agent models with multi-dimensional action sets.
## Abstract
This paper presents a new method for the analysis of moral hazard principal-agent problems. The new approach avoids the stringent assumptions on the distribution of outcomes made by the classical first-order approach and instead only requires the agent's expected utility to be a rational function of the action. This assumption allows for a reformulation of the agent's utility maximization problem as an equivalent system of equations and inequalities. This reformulation in turn transforms the principal's utility maximization problem into a nonlinear program. Under the additional assumptions that the principal's expected utility is a polynomial and the agent's expected utility is rational in the wage, the final nonlinear program can be solved to global optimality. The paper also shows how to first approximate expected utility functions that are not rational by polynomials, so that the polynomial optimization approach can be applied to compute an approximate solution to non-polynomial problems. Finally, the paper demonstrates that the polynomial optimization approach extends to principal-agent models with multi-dimensional action sets.
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Item Type: Journal Article, refereed, original work 03 Faculty of Economics > Department of Business Administration 330 Economics English March 2015 17 Oct 2014 10:59 05 Apr 2016 18:25 Wiley-Blackwell Publishing, Inc. 0012-9682 https://doi.org/10.3982/ECTA11351 http://onlinelibrary.wiley.com/doi/10.3982/ECTA11351/abstract merlin-id:10309
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You can navigate and zoom the map. Mouse-hovering a term displays its timeline, clicking it yields the associated documents. | {"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": 0.8749908804893494, "perplexity": 832.9397855887576}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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-2017-30/segments/1500549425254.88/warc/CC-MAIN-20170725142515-20170725162515-00689.warc.gz"} |
https://www.lessonplanet.com/teachers/the-large-and-small-of-it | # The Large and Small of It
##### This The Large and Small of It lesson plan also includes:
Students explore the extreme between the distance to the moon in comparison to the size of a particle of moon dust and solve problems related to the Apollo space missions.
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https://homework.cpm.org/category/CCI_CT/textbook/apcalc/chapter/6/lesson/6.5.1/problem/6-172 | ### Home > APCALC > Chapter 6 > Lesson 6.5.1 > Problem6-172
6-172.
Evaluate each expression.
1. $\int _ { 0 } ^ { x ^ { 2 } } \operatorname { cos } ( e ^ { t } ) d t$
Fundamental Theorem of Calculus states that the derivative of an INDEFINITE integral gives you the original function.
But, this is a definite integral.
Be sure to multiply by the derivative of the bounds.
$2x\cos\left(e^{x^2}\right)−\left(0\right)\cos\left(e^0\right)=2x\cos\left(e^{x^2}\right)$
1. $\frac { d } { d x } \int _ { 0 } ^ { \pi } \sqrt { \operatorname { sin } ( t ) } d t$
Refer to the hints in part (a). Or, notice that the definite integral equals a constant. What is the derivative of a constant?
1. $\int _ { 1 } ^ { 10 } f ^ { \prime } ( x ) d x$, where $f ( x ) = \operatorname { cos } ( \pi e ^ { x ^ { 2 } - 100 } ) ( x ) \operatorname { log } ( x )$
$\text{Translation: }\int_{1}^{10}\frac{d}{dx}\left(x\log(x)\left [ \text{cos}\left ( \pi e^{x^{2}-100} \right ) \right ]\right)dx$ | {"extraction_info": {"found_math": true, "script_math_tex": 6, "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": 0.9991753101348877, "perplexity": 891.8446866754691}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141686635.62/warc/CC-MAIN-20201202021743-20201202051743-00106.warc.gz"} |
https://www.physicsforums.com/threads/conical-representation-of-sphere.925134/ | B Conical Representation of Sphere
1. Sep 11, 2017
Leo Authersh
Is Sphere a more generalized form of Cone i.e. formed by 2 dimensional rotation to 360° of a cone?
Or is Cone a more generalized form of Sphere since sphere can be formed by rotating about Z axis a zero eccentric planar intersection of a cone?
@fresh_42 @FactChecker @WWGD
2. Sep 11, 2017
phinds
A sphere is a degenerate case of an ellipsoid just as a circle is a degenerate case of an ellipse.
3. Sep 11, 2017
Staff: Mentor
4. Sep 11, 2017
Leo Authersh
This representation is excellent. I remember the term Hyperboloid.
But what confuses me is that, if the hyperboloid is rotated to 90°, we get a cube. How is a cube which is a linear geometric form that has one variable be formed by a Hyperboloid that has three variables?
5. Sep 11, 2017
Leo Authersh
@Infrared
6. Sep 12, 2017
Staff: Mentor
Why do you think this?
This isn't right, either. Let's look at two dimensions first. The unit square in the first quadrant does not have a single equation. Instead, it has four equations, one for each side, along with inequalities that indicate the minimum and maximum values of the variable on each side. For example, the upper horizontal side would be represented by the equation y = 1, and the inequality $0 \le x \le 1$. There would be an equation/inequality pair for each side.
For a cube you would need equation/inequality pairs for each of the six faces.
Back to the hyperboloid. There are actually two kinds of hyperboloids -- hyperboloid of one sheet (or surface) and hyperboloid of two sheets (two distinct surfaces). If you take the hyperbola $x^2 - y^2 = 1$ and rotate it about the x-axis, you get a hyperboloid of two sheets (in three dimensions). If you rotate the same hyperbola about the y-axis, you get a hyperboloid of one sheet.
Do a web search on, say, wikipedia for hyperboloid to see the formulas and graphs of these quadric surfaces.
Mod note: Thread moved to General Mathematics -- the question isn't really about topology or analysis.
Draft saved Draft deleted
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https://math.libretexts.org/Bookshelves/Differential_Equations/Book%3A_Partial_Differential_Equations_(Miersemann)/2%3A_Equations_of_First_Order/2.2.2%3A_Initial_Value_Problem_of_Cauchy | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 2.2.2: Initial Value Problem of Cauchy
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Consider again the quasilinear equation
($$\star$$) $$a_1(x,y,u)u_x+a_2(x,y,u)u_y=a_3(x,y,u)$$.
Let
$$\Gamma:\ \ x=x_0(s),\ y=y_0(s),\ z=z_0(s), \ s_1\le s\le s_2,\ -\infty<s_1<s_2<+\infty$$
be a regular curve in $$\mathbb{R}^3$$ and denote by $$\mathcal{C}$$ the orthogonal projection of $$\Gamma$$ onto the $$(x,y)$$-plane, i. e.,
$$\mathcal{C}:\ \ x=x_0(s),\ \ y=y_0(s).$$
Initial value problem of Cauchy: Find a $$C^1$$-solution $$u=u(x,y)$$ of $$(\star)$$ such that $$u(x_0(s),y_0(s))=z_0(s)$$, i. e., we seek a surface $$\mathcal{S}$$ defined by $$z=u(x,y)$$ which contains the curve $$\Gamma$$.
Figure 2.2.2.1: Cauchy initial value problem
Definition. The curve $$\Gamma$$ is said to be non-characteristic if
$$x_0'(s)a_2(x_0(s),y_0(s))-y_0'(s)a_1(x_0(s),y_0(s))\not=0.$$
Theorem 2.1. Assume $$a_1,\ a_2,\ a_2\in C^1$$ in their arguments, the initial data $$x_0,\ y_0,\ z_0\in C^1[s_1,s_2]$$ and $$\Gamma$$ is non-characteristic.
Then there is a neighborhood of $$\cal{C}$$ such that there exists exactly one solution $$u$$ of the Cauchy initial value problem.
Proof. (i) Existence. Consider the following initial value problem for the system of characteristic equations to ($$\star$$):
\begin{eqnarray*}
x'(t)&=&a_1(x,y,z)\\
y'(t)&=&a_2(x,y,z)\\
z'(t)&=&a_3(x,y,z)
\end{eqnarray*}
with the initial conditions
\begin{eqnarray*}
x(s,0)&=&x_0(s)\\
y(s,0)&=&y_0(s)\\
z(s,0)&=&z_0(s).
\end{eqnarray*}
Let $$x=x(s,t)$$, $$y=y(s,t)$$, $$z=z(s,t)$$ be the solution, $$s_1\le s\le s_2$$, $$|t|<\eta$$ for an $$\eta>0$$. We will show that this set of curves, see Figure 2.2.2.1, defines a surface. To show this, we consider the inverse functions $$s=s(x,y)$$, $$t=t(x,y)$$ of $$x=x(s,t)$$, $$y=y(s,t)$$ and show that $$z(s(x,y),t(x,y))$$ is a solution of the initial problem of Cauchy. The inverse functions $$s$$ and $$t$$ exist in a neighborhood of $$t=0$$ since
$$\det \frac{\partial(x,y)}{\partial(s,t)}\Big|_{t=0}= \left|\begin{array}{cc}x_s&x_t\\y_s&y_t\end{array}\right|_{t=0} =x_0'(s)a_2-y_0'(s)a_1\not=0,$$
and the initial curve $$\Gamma$$ is non-characteristic by assumption.
Set
$$u(x,y):=z(s(x,y),t(x,y)),$$
then $$u$$ satisfies the initial condition since
$$u(x,y)|_{t=0}=z(s,0)=z_0(s).$$
The following calculation shows that $$u$$ is also a solution of the differential equation ($$\star$$).
\begin{eqnarray*}
a_1u_x+a_2u_y&=&a_1(z_ss_x+z_tt_x)+a_2(z_ss_y+z_tt_y)\\
&=&z_s(a_1s_x+a_2s_y)+z_t(a_1t_x+a_2t_y)\\
&=&z_s(s_xx_t+s_yy_t)+z_t(t_xx_t+t_yy_t)\\
&=&a_3
\end{eqnarray*}
since $$0=s_t=s_xx_t+s_yy_t$$ and $$1=t_t=t_xx_t+t_yy_t$$.
(ii) Uniqueness. Suppose that $$v(x,y)$$ is a second solution. Consider a point $$(x',y')$$ in a neighborhood of the curve $$(x_0(s),y(s))$$, $$s_1-\epsilon\le s\le s_2+\epsilon$$, $$\epsilon>0$$ small. The inverse parameters are $$s'=s(x',y')$$, $$t'=t(x',y')$$, see Figure 2.2.2.2.
Figure 2.2.2.2: Uniqueness proof
Let
$${\mathcal{A}}:\ \ x(t):=x(s',t),\ y(t):=y(s',t),\ z(t):=z(s',t)$$
be the solution of the above initial value problem for the characteristic differential equations with the initial data
$$x(s',0)=x_0(s'),\ y(s',0)=y_0(s'),\ z(s',0)=z_0(s').$$
According to its construction this curve is on the surface $$\mathcal{S}$$ defined by $$u=u(x,y)$$ and $$u(x',y')=z(s',t')$$. Set
$$\psi(t):=v(x(t),y(t))-z(t),$$
then
\begin{eqnarray*}
\psi'(t)&=&v_xx'+v_yy'-z'\\
&=&x_xa_1+v_ya_2-a_3=0
\end{eqnarray*}
and
$$\psi(0)=v(x(s',0),y(s',0))-z(s',0) =0$$
since $$v$$ is a solution of the differential equation and satisfies the initial condition by assumption. Thus, $$\psi(t)\equiv0$$, i. e.,
$$v(x(s',t),y(s',t))-z(s',t)=0.$$
Set $$t=t'$$, then
$$v(x',y')-z(s',t')=0,$$
which shows that $$v(x',y')=u(x',y')$$ because of $$z(s',t')=u(x',y')$$.
$$\Box$$
Remark. In general, there is no uniqueness if the initial curve $$\Gamma$$ is a characteristic curve, see an exercise and Figure 2.2.2.3, which illustrates this case.
Figure 2.2.2.3: Multiple solutions
## Examples
Example 2.2.2.1:
Consider the Cauchy initial value problem
$$u_x+u_y=0$$
with the initial data
$$x_0(s)=s,\ y_0(s)=1,\ z_0(s)\ \mbox{is a given}\ C^1\mbox{-function}.$$
These initial data are non-characteristic since $$y_0'a_1-x_0'a_2=-1$$. The solution of the associated system of characteristic equations
$$x'(t)=1,\ y'(t)=1,\ u'(t)=0$$
with the initial conditions
$$x(s,0)=x_0(s),\ y(s,0)=y_0(s),\ z(s,0)=z_0(s)$$
is given by
$$x=t+x_0(s),\ y=t+y_0(s),\ z=z_0(s) ,$$
i. e.,
$$x=t+s,\ y=t+1,\ z=z_0(s).$$
It follows $$s=x-y+1,\ t=y-1$$ and that $$u=z_0(x-y+1)$$ is the solution of the Cauchy initial value problem.
Example 2.2.2.2:
A problem from kinetics in chemistry. Consider for $$x\ge0$$, $$y\ge0$$ the problem
$$u_x+u_y=\left(k_0e^{-k_1x}+k_2\right)(1-u)$$
with initial data
$$u(x,0)=0,\ x>0,\ \mbox{and}\ u(0,y)=u_0(y),\ y>0.$$
Here the constants $$k_j$$ are positive, these constants define the velocity of the reactions in consideration, and the function $$u_0(y)$$ is given. The variable $$x$$ is the time and $$y$$ is the height of a tube, for example, in which the chemical reaction takes place, and $$u$$ is the concentration of the chemical substance.
In contrast to our previous assumptions, the initial data are not in $$C^1$$. The projection $${\mathcal C}_1\cup {\mathcal C}_2$$ of the initial curve onto the $$(x,y)$$-plane has a corner at the origin, see Figure 2.2.2.4.
Figure 2.2.2.4: Domains to the chemical kinetics example
The associated system of characteristic equations is
$$x'(t)=1,\ y'(t)=1,\ z'(t)=\left(k_0e^{-k_1x}+k_2\right)(1-z).$$
It follows $$x=t+c_1$$, $$y=t+c_2$$ with constants $$c_j$$. Thus the projection of the characteristic curves on the $$(x,y)$$-plane are straight lines parallel to $$y=x$$. We will solve the initial value problems in the domains $$\Omega_1$$ and $$\Omega_2$$, see Figure 2.2.2.4, separately.
(i) The initial value problem in $$\Omega_1$$. The initial data are
$$x_0(s)=s,\ y_0(s)=0, \ z_0(0)=0,\ s\ge 0.$$
It follows
$$x=x(s,t)=t+s,\ y=y(s,t)=t.$$
Thus
$$z'(t)=(k_0e^{-k_1(t+s)}+k_2)(1-z),\ z(0)=0.$$
The solution of this initial value problem is given by
$$z(s,t)=1-\exp\left(\frac{k_0}{k_1}e^{-k_1(s+t)}-k_2t-\frac{k_0}{k_1}e^{-k_1s}\right).$$
Consequently
$$u_1(x,y)=1-\exp\left(\frac{k_0}{k_1}e^{-k_1x}-k_2y-{k_0}{k_1}e^{-k_1(x-y)}\right)$$
is the solution of the Cauchy initial value problem in $$\Omega_1$$. If time $$x$$ tends to $$\infty$$, we get the limit
$$\lim_{x\to\infty} u_1(x,y)=1-e^{-k_2y}.$$
(ii) The initial value problem in $$\Omega_2$$. The initial data are here
$$x_0(s)=0,\ y_0(s)=s, \ z_0(0)=u_0(s),\ s\ge 0.$$
It follows
$$x=x(s,t)=t,\ y=y(s,t)=t+s.$$
Thus
$$z'(t)=(k_0e^{-k_1t}+k_2)(1-z),\ z(0)=0.$$
The solution of this initial value problem is given by
$$z(s,t)=1-(1-u_0(s))\exp\left(\frac{k_0}{k_1}e^{-k_1t}-k_2t-\frac{k_0}{k_1}\right).$$
Consequently
$$u_2(x,y)=1-(1-u_0(y-x))\exp\left(\frac{k_0}{k_1}e^{-k_1x}-k_2x-\frac{k_0}{k_1}\right)$$
is the solution in $$\Omega_2$$.
If $$x=y$$, then
\begin{eqnarray*}
u_1(x,y)&=&1-\exp\left(\frac{k_0}{k_1}e^{-k_1x}-k_2x-\frac{k_0}{k_1}\right)\\
u_2(x,y)&=&1-(1-u_0(0))\exp\left(\frac{k_0}{k_1}e^{-k_1x}-k_2x-\frac{k_0}{k_1}\right).
\end{eqnarray*}
If $$u_0(0)>0$$, then $$u_1<u_2$$ if $$x=y$$, i. e., there is a jump of the concentration of the substrate along its burning front defined by $$x=y$$.
Remark. Such a problem with discontinuous initial data is called Riemann problem. See an exercise for another Riemann problem.
The case that a solution of the equation is known
Here we will see that we get immediately a solution of the Cauchy initial value problem if a solution of the homogeneous linear equation
$$a_1(x,y)u_x+a_2(x,y)u_y=0$$
is known.
Let
$$x_0(s),\ y_0(s),\ z_0(s),\ s_1<s<s_2$$
be the initial data and let $$u=\phi(x,y)$$ be a solution of the differential equation. We assume that
$$\phi_x(x_0(s),y_0(s))x_0'(s)+\phi_y(x_0(s),y_0(s))y_0'(s)\not=0$$
is satisfied. Set
$$g(s)=\phi(x_0(s),y_0(s))$$
and let $$s=h(g)$$ be the inverse function.
The solution of the Cauchy initial problem is given by $$u_0\left(h(\phi(x,y))\right)$$.
This follows since in the problem considered a composition of a solution is a solution again, see an exercise, and since
$$u_0\left(h(\phi(x_0(s),y_0(s))\right)=u_0(h(g))=u_0(s).$$
Example 2.2.2.3:
Consider equation
$$u_x+u_y=0$$
with initial data
$$x_0(s)=s,\ y_0(s)=1,\ u_0(s)\ \mbox{is a given function}.$$
A solution of the differential equation is $$\phi(x,y)=x-y$$. Thus
$$\phi((x_0(s),y_0(s))=s-1$$
and
$$u_0(\phi+1)=u_0(x-y+1)$$
is the solution of the problem.
## Contributors
• Integrated by Justin Marshall. | {"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": 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": 0.9954074025154114, "perplexity": 220.74055144955454}, "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/1575540579703.26/warc/CC-MAIN-20191214014220-20191214042220-00531.warc.gz"} |
http://mathhelpforum.com/algebra/193163-averages-print.html | # Averages.
• Dec 1st 2011, 05:27 AM
cmpprice
Averages.
For the life of me I'm unable to locate a way to calculate this issue.. I'm trying to take the average of 3 numbers, however each number has more or less weight on the total average than the other 2 numbers.. For example.
(Number 1) 50% total weight
(Number 2) 37.5% total weight
(Number 3) 12.5% total weight
All equaling 1 total average. Any help would be appreciated. Thank you!
• Dec 1st 2011, 07:37 AM
Soroban
Re: An easy question? Averages.
Hello, cmpprice!
Quote:
I'm trying to take the average of 3 numbers. .However each number
has more or less weight on the total average than the other 2 numbers.
For example:
Number 1: 50% total weight
Number 2: 37.5% total weight
Number 3: 12.5% total weight
All equaling one total average.
Look at it this way:
. . $\begin{array}{cccccc}\text{Number} & \text{Weight} \\ X & 50\% &=& \frac{4}{8} \\ \\[-3mm] Y & 37.5\% &=& \frac{3}{8} \\ \\[-3mm] Z & 12.5\% &=& \frac{1}{8} \end{array}$
Their weights are in the ratio . $X:Y:Z \:=\:4:3:1$
Their average is: . $\dfrac{4X + 3Y + Z}{8}$
• Jan 1st 2012, 04:28 PM
olivianewton
Re: An easy question? Averages.
I am fix , cannot get this.
• Jan 3rd 2012, 08:15 AM
okokjae
Re: Averages.
Quote:
Originally Posted by cmpprice
For the life of me I'm unable to locate a way to calculate this issue.. I'm trying to take the average of 3 numbers, however each number has more or less weight on the total average than the other 2 numbers.. For example.
(Number 1) 50% total weight
(Number 2) 37.5% total weight
(Number 3) 12.5% total weight
All equaling 1 total average. Any help would be appreciated. Thank you!
Hello.
There are 3 different people, one with 50% weight, 37.5% weight, and 12.5% weight. For this example, the 50% weight person is X, 37.5 weight person is W, and 12.5% person is U. So we have 50%, X. 37.5% ,W. 12.5% U.
Now we know that 50% = 50/100 = 1/2 = 4/8
37.5% = 3/8
12.5% = 1/8, You can divide 3/8 by 3.
Now lets refer back to X, W, and U. For simple purposes, lets give them values. X = 100, W = 150, and U = 200.
Now to find the average... [ (4/8)100 + (3/8)150 + (1/8)200 ]/3 = y
You divide everything by 3, because you're trying to find the average of all three people.
Or you can use, [ (4/8)X + (3/8)W + (1/2)U ]/3 = Y
• Jan 4th 2012, 04:23 AM
HallsofIvy
Re: Averages.
Or, simply 0.50X+ 0.375Y+ 0.125Z. | {"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": 0.8617027401924133, "perplexity": 3792.1383808151445}, "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/1500549423901.32/warc/CC-MAIN-20170722042522-20170722062522-00117.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=483544 | # Electric Field of Non-Conducting Sheets.
by dvsumosize
Tags: electric, field, nonconducting, sheets
P: 12 1. The problem statement, all variables and given/known data 3 Infinitely large non conductiong sheets are uniformly charged with surface charge densities Sigma1 = +2x10^-6c/m^2, Sigma2 = +4x10^-6c/m^2, Sigma3 = -5.0x10^-6c/m^2. Distance L = 1.6cm. What is the magnitude and direction of the net electric field at point P? * P | L/2 ________________ Sigma3 | | | 2 *L | ________________ Sigma2 | L | ________________ Sigma 1 2. Relevant equations The only 1 i know of is E = Sigma/(2*εnot) 3. The attempt at a solution Es1 + Es2 + Es3 = Etot at P. So my teacher never really taught us about non-conducting sheets over any distance, and instead told us to google it and ask on forums instead. Please me through how you do this, i can't find a formula that deals with distance between non-conducting sheets.
PF Gold P: 280 electric fields are added as vectors. just add them with sign.
Emeritus Sci Advisor HW Helper PF Gold P: 7,404 Use Gauss's Law.
P: 12
## Electric Field of Non-Conducting Sheets.
Which formula from that law will i use, the only one i know is the E = Sigma/(2 * e-not)
Emeritus
HW Helper
PF Gold
P: 7,404
Quote by dvsumosize Which formula from that law will i use, the only one i know is the E = Sigma/(2 * e-not)
That's a result which can be obtained from Gauss's Law.
Gauss's Law: $$\oint_S\vec{E}\cdot d\vec{A}=\frac{q_{inside}}{\epsilon_0}$$
P: 12 Cramster tells me what i did at the beginning was right, distance doesn't matter in this case. Why did none of you tell me so?
Emeritus Sci Advisor HW Helper PF Gold P: 7,404 It's rather hard to read your mind well enough to know what you do & don't understand.
PF Gold P: 280 distance doesn't matter here, as electric field is independent of distance for uniformly charged infinite sheet.
P: 12 I see.
PF Gold P: 280 so did you get the answer?
P: 12 yeah
P: 121 the same formula sigma/(2epsilon) can be obtained as follows: A Gaussian surface in the form of cylindrical surface can be taken.The field is perpendicular to sheet so only end caps contribute to flux. =>(epsilon)*(sufaceintegral(E.dA))=q =>(epsilon)*(EA+EA)=sigma*A =>(epsilon)*(2EA)=sigma*A canceling A from both sides & rearranging: =>E=sigma/(2*epsilon) Thus for a very large non-conducting sheet distance doesn't matter at all!what matters is sigma.so just as per vectors the electric fields depending upon nature of charge(positive or negative).If negative sigma then E is generally taken as negative(however you can take anything!!!).So the data regarding distance is irrelevant here!!!
Related Discussions Classical Physics 5 Introductory Physics Homework 1 Introductory Physics Homework 1 Introductory Physics Homework 3 Introductory Physics Homework 1 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8870977163314819, "perplexity": 2434.2035362728434}, "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/1397609539705.42/warc/CC-MAIN-20140416005219-00378-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://byjus.com/slant-asymptote-formula/ | # Slant Asymptote Formula
When a polynomial has a numerator higher than the denominator, then a slant asymptote occurs. Slant asymptote can also be referred to an oblique. To find the oblique, we need to divide the numerator to the denominator using synthetic division method or long division. The numerator being stronger, “pulls” the graph far from the x-axis or other fixed y value. The distance of the curve is so close that they approach if extended until infinity.
### Formula For Slant Asymptote
If for example we have a function : f(x), then the slant asymptote will be in the form:
$\large f(x)=\frac{ax^{n}+bx^{n-1}+…}{cx^{n-1}+….}$
### Solved Example
Question: Find the Slant Asymptote for the function: y = $\frac{x^2 + 3x + 5}{x + 5}$
Solution:
Given function is y = $\frac{x^2 + 3x + 5}{x + 5}$. Here the degree of numerator is more than that of denominator by 1. Hence the given function is a slant asymptote.
x – 2
= x + 5x2 + 3x + 5
= x2 2 + 5x
= -2x + 5
= – 2x – 10
5.
The slant asymptote for the function y = $\frac{x^2 + 3x + 5}{x + 5}$ is (x – 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": 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": 0.9773339629173279, "perplexity": 696.2003739283862}, "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/1573496670512.94/warc/CC-MAIN-20191120060344-20191120084344-00170.warc.gz"} |
https://www.physicsforums.com/threads/fourier-series-how-do-i-simpligy-this-integral.344125/ | Fourier Series: How do i simpligy this integral?
1. Oct 9, 2009
exidez
Fourier Series: How do i simplify this integral?
1. The problem statement, all variables and given/known data
http://img18.imageshack.us/img18/4586/matlabfouriercomplex.jpg [Broken]
2. Relevant equations
$$g(t)=\sum_{n=-\infty}^{\infty}c_{n}e^{jn \omega t}$$
$$c_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}g(t)e^{-jn \omega T}$$
3. The attempt at a solution
I need to find the complex fourier series of the above function extended as an odd function
I Just want to know how to simplify the final solution i have. So far i have
$$c_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}g(t)e^{-jn \omega T}$$
$$c_{n}=\frac{1}{T}(\int_{-T/10}^{0}-Ae^{-jn \omega T}dt + \int_{0}^{T/10}Ae^{-jn \omega T}dt)$$
$$c_{n}=\frac{A}{T}( \frac{1}{jn \omega } - \frac{e^{(jn \omega T/10)}}{jn \omega } - \frac{e^{(-jn \omega T/10})}{jn \omega } + \frac{1}{jn \omega})$$
$$\omega = \frac{2 \pi }{T}$$
$$c_{n}=\frac{A}{jn2 \pi }(2 - e^{(jn2 \pi /10)} - e^{(-jn2 \pi /10)})$$
Now if i put this in matlab i will put
for n = -N:???????:N, % loop over series index n
cn = A/(j*n*2*pi)*(2-exp(j*n*(pi/5))-exp(-j*n*(pi/5))); % Fourier Series Coefficient
yce = yce + cn*exp(j*n*wo*t); % Fourier Series computation
end
I dont know what to incriment N by.. All the other fourier examples i have done have been in the form:
$$c_{n}=\frac{1}{jn \pi }(1 - e^{(jn \pi )})$$
Here i understand $$(1 - e^{(jn \pi )})$$ will be 0 when n is even 2 when n is odd
Hence the series becomes:
$$\sum_{n=odd}^{\infty}\frac{2}{ \pi jn}e^{jn \omega t}$$
and the loop will go:
for n = -N:2:N, % loop over series index n (odd)
cn = 1/(j*n*pi)*(1-exp(j*n*pi)); % Fourier Series Coefficient
yce = yce + cn*exp(j*n*wo*t); % Fourier Series computation
end
But what is it for my example??? It is not so straight forward. Can it be simplified like above? How do i progress from here?
Last edited by a moderator: May 4, 2017
Can you offer guidance or do you also need help?
Draft saved Draft deleted
Similar Discussions: Fourier Series: How do i simpligy this integral? | {"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": 0.9534235596656799, "perplexity": 1788.9966836685373}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948542031.37/warc/CC-MAIN-20171214074533-20171214094533-00534.warc.gz"} |
http://math.stackexchange.com/questions/466198/algorithm-to-get-the-maximum-size-of-n-squares-that-fit-into-a-rectangle-with-a | # Algorithm to get the maximum size of n squares that fit into a rectangle with a given width and height
I am looking for an algorithm that can return the number of size of n squares that fit into a a rectangle of a given width and height, maximizing the use of space (thus, leaving the least amount of leftover space for squares that do not fit). Neither the rectangle nor the squares can be rotated.
For example, let's say I have a rectangle that is 5 inches by 7 inches, and I need it to fit 35 squares. The algorithm needs to tell me that the 35 squares fit if they are 1 inch wide/tall (as they could be laid out inside the rectangle in a 5 x 7 grid). Another example is if I need to divide a rectangle 35 inches by 1 inch into 35 squares. It should still tell me that the squares will fit if they are 1-inch wide/tall (as they could be laid out in the rectangle in a 35 x 1 grid).
The tricky part is that sometimes there may be leftover space, as the squares cannot be divided into partial squares. Let's say for either of the two examples above I need to lay out 34 squares and not 35 (in which case the answers would still be 1 inch), or maybe 33, or 7 squares. Or, perhaps the rectangle width and height aren't whole numbers. With the number of squares being a variable I need an algorithm that can tell me the size of the squares for a given rectangle width and height.
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Starting point: If you take all the squares and move them to a corner, the leftover space is much easier to calculate.(I am assuming all the squares are of equal size) – chubakueno Aug 12 '13 at 22:00
Also, as you note in your tags, I see a formula kind of unlikely: this problems are usually solved by algorithms. The trigonometry tag is not relevant, I think. – chubakueno Aug 12 '13 at 22:06
See squares in squares and squares in dominoes from Erich's Packing Center for some specific cases. Finding these automatically seems like a pretty hard problem. – MvG Aug 12 '13 at 22:18
@MvG thanks for the links. I just realized that I need to clarify that squares cannot be rotated. – Anton Aug 12 '13 at 22:25
Let's analyze a single case:
If we want to cover a $6x5$ rectangle with $7$ squares, we first note that each square at most has an area of $30/7$. We also know that for our filling to be optimal, we need to fill at least an axis.
So now we know that we will fill either the $6$ or the $5$ completely. We will first try with the 5. Since the maximal side of our squares is $\sqrt{30/7}$, we now need the smallest integer more than $5/\sqrt{30/7}$, that is equal to $\lceil5/\sqrt{30/7}\rceil=3$
So we would divide the $5$ in $3$ parts $p_x$, so each side will have $5/3$ . If so, I would cover $(5/3)^2*7=19\frac49$ of the thirty squares. If they don't fit vertically, we try with more parts $p_x$, until the squares fit in the other axis. $$\lfloor6/(5/p_x)\rfloor*\lfloor5/(5/p_x)\rfloor=\lfloor6/(5/p_x)\rfloor*p_x\ge7 \,\,\,\,\,\,\,\,\,\, (5/p_x=\text{the actual size of our squares)}$$ Also, since $p_x$ is integer $\lfloor p_x \rfloor =p_x$
If we do the same with the $6$, we get $\lceil6/\sqrt{30/7}\rceil$ again covering $19\frac49$ of the surface. We then do the same as with the $5$ and save it as $p_y$
Let's now recall the formulas we had.
We had an $x∗y$ rectangle and filled it with $N$ squares. We started with $p_x=⌈x/\sqrt{xy/N}⌉$=$⌈\sqrt{Nx/y}⌉$ or $p_y=y/⌈\sqrt{Nx/y}⌉$ and then we made them fit by shrinking them until they fit in the other axis. But we know we want the area maximised, so $Side=max(x/p_x,y/p_y)$.
A better method:
Once we have the start value $p_x=⌈x/\sqrt{xy/N}⌉=⌈\sqrt{Nx/y}⌉$,if it does not fit in the $y$ axis, we need to make it fit in the $y$ axis. For that, we use that $$a=x/p_x$$ where $a$ is the value of the actual side of our squares. Then we know that for our side to fit we need to reduce it: $$S_x=y/(\lceil y/a \rceil)$$ We do the same with $S_y$ and calculate $max(S_x,S_y)$ The advantage of this is that it needs a constant number of operations, the most expensive one being the square root.
Plain, unoptimized C Code
Some multiplications may be reused but for code readability and practical uses this is enough. Input: values of $x$,$y$, and $n$. Handcoded.
Output: The optimal side of our squares
#include <math.h>
#include <stdio.h>
#define MAX(x,y) (x)>(y)?(x):(y)
int main(){
double x=5, y=6, n=7;//values here
double px=ceil(sqrt(n*x/y));
double sx,sy;
if(floor(px*y/x)*px<n) //does not fit, y/(x/px)=px*y/x
sx=y/ceil(px*y/x);
else
sx= x/px;
double py=ceil(sqrt(n*y/x));
if(floor(py*x/y)*py<n) //does not fit
sy=x/ceil(x*py/y);
else
sy=y/py;
printf("%f",MAX(sx,sy));
return 0;
}
-
Thank you for the detailed answer. I'm going to give it a go tonight/tomorrow morning. Right now I'm having a little trouble wrapping my head around what to do after calculating P(x), when it comes to reducing. – Anton Aug 13 '13 at 0:20
@Anton I posted now a method to reduce it without any iterations, so it is $O(\text{the time your sqrt function takes})$ – chubakueno Aug 13 '13 at 0:32
Sorry, I don't understand what you mean. What is O and what time are you referring to for the sqrt function? – Anton Aug 13 '13 at 1:20
@Anton The notation $O(n)$ is the notation of asymptotic complexity. One of it's basic properties is that you just care about the highest order term and you drop all the constants, for example $O(5x^2+x+1)$=$O(x^2)$. It is a useful way of noting the time complexity of your algorithm. In this case, the slowest operation is the square root. Since there are varying implementations, I wrote $O(\text{the time your sqrt function takes})$ because I don't know your code implementation. Assuming you will code this, this makes sense. If you are doing this by hand, this is unimportant :). – chubakueno Aug 13 '13 at 1:36
show 1 more comment | {"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": 0.8114230036735535, "perplexity": 406.0157801424417}, "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-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00464-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/306042/countable-field-extension-of-a-countable-field | # Countable Field Extension of a Countable Field
Okay, first question on this site, apologies in advance for any mistakes I may make.
Question: So I need to show that an algebraic field extension $E:F$, with $F$ being countable, is countable.
My Thoughts: As far as I am aware, $E$ being algebraic means that every element in $E$ is algebraic (root of some polynomial in $F[x]$ ) and being countable means that there is a natural mapping of the integers onto $E$.
At the moment I am thinking that as every element in $E$ is algebraic then it has some corresponding polynomial in $F[x]$, that it is a root of, and then because $F$ is countable I need a bijection (or possibly just a surjection) of $F$ onto $E$ and that that may be achievable through the connection between an element of $E$ and it's polynomial in $F[x]$. I just can't see how to make this rigorous.
Note: I did try to find something resembling this question but mostly found analysis based questions. This may be down to my unfamiliarity with the website though.
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Let $L/K$ be an algebraic extension. We have a natural map $\phi: L \rightarrow K[t]$ given by $\phi(\alpha)=\mathrm{min}_K(\alpha,t)$. Sadly this map need not be injective, for instance if $K=\mathbb Q$ and $L=\mathbb Q(\sqrt{2})$ then $\phi(\sqrt{2})=\phi(-\sqrt{2})$. So we have to work a bit harder. If $p(t) \in K[t]$ then hat can we say about the cardinality of $\phi^{-1}(p(t))$.
Now work in your original setting and recall that a countable union of finite sets is countable.
Edit: The argument concludes by noting that for each $p(t) \in K[t]$ we have that $|\phi^{-1}(p(t)) | \leq \deg p(t)$ so the fibers of $\phi$ are finite. In particular
$$L=\bigcup_{p(t) \in K[t]} \phi^{-1}(p(t))$$ so $L$ is a countable union of finite sets and thereby countable. This proof shows in general that any algebraic extension of an infinite field has the same cardinality as the base field.
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Is your notation L/K the same as L:K? Also, I assume by the way you say it need not be injective that we do in fact need a bijective map for the countability? – Nytram12 Feb 17 '13 at 8:31
@Nytram12 Yes $L/K$ means that $K \subset L$ is an extension of fields. Finding an explicition bijection to a countable set is one way to show countability but it's not the only way. – JSchlather Feb 17 '13 at 8:36
Oh? What other ways are there? – Nytram12 Feb 17 '13 at 8:38
@Nytram12 Well countable sets are closed under countable unions and finite products for instance. So if you can write your field $E$ as a countable union of countable sets then you've shown its countable. – JSchlather Feb 17 '13 at 8:52
Fair enough. I suspect I am meant to go with the bijection route here. Thanks for your help – Nytram12 Feb 17 '13 at 18:25 | {"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": 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": 0.8864313960075378, "perplexity": 154.48486747652478}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860111365.36/warc/CC-MAIN-20160428161511-00093-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://teachingcalculus.com/2014/07/28/limacons/ | # Limaçons
When I first started getting interested in roulettes I began in polar form graphing limaçons. Without going into as much detail as with the roulettes, I offer just one today.
I found this Winplot illustration instructive as to how polar graphs are formed and just how the graphs work and relate to rectangular form graphs of the same functions. So this could be used as a first example, or an investigation of its own.
For my example we will consider the limaçon, or a cardioid with an inner loop, given below (using t for the usual theta, since the LaTex translator doesn’t seem to be able to handle thetas).
$r\left( t \right)=1.4-2\sin \left( t \right)$
In polar form ${{r}_{1}}\left( t \right)=1.4$ is a circle with center at the pole and radius of 1.4.This is shown in light blue in the figures.
The polar curve ${{r}_{2}}\left( t \right)=2\sin \left( t \right)$ is a circle with center at $\left( 1,\tfrac{\pi }{2} \right)$ and radius of 1. This is shown in orange in the figures below.
The graph in the example is$r\left( t \right)={{r}_{2}}\left( t \right)-{{r}_{1}}\left( t \right)$ the directed distance (length of the vector) from ${{r}_{1}}\left( t \right)$ to ${{r}_{2}}\left( t \right)$ and is shown by the green arrow in figures 1, 3, and 4.
The blue arrow is congruent to the green with its tail at the pole; its point traces the limaçon shown in black. (Click to enlarge.)
• In figure 1, the distance is positive and both arrows point in the same direction along the rotating ray.
• In figure 2, the two circles intersect and the distance between them is zero. The limaçon goes through the origin.
• In figure 3, the curves have changed position and the directed distance is negative. The blue arrow points in the negative direction opposite to the rotating ray drawing the inner loop.
• In figure 4, the arrows return to pointing in the same direction, but are longer due to the fact that the distance runs from the orange circle to the far side of the light blue circle forming the bottom outside loop
In the clip below the limaçon is drawn as the black ray rotates from 0 to $2\pi$ radians. Watch how the green and blue arrows (always the same length, but not the same direction), work to draw the limaçon.
On the right side of the clip are the two functions graphed in rectangular form. The blue arrow on the right is the same length as the blue arrow on the left and gives the directed distance from ${{r}_{1}}\left( t \right)$ to ${{r}_{2}}\left( t \right)$. This form is probably more familiar to students and may help them see the relationships.
A limaçon being graphed.
This form is probably more familiar to students and may help them see the relationships. The Winplot file may be downloaded here. If you or your students what to investigate further, click on the Winplot graph and then CTRL+SHIFT+N to see the notes; they will also tell you how to change the A, B, and R sliders to change the 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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 10, "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": 0.9055587649345398, "perplexity": 536.4366673892831}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320491.13/warc/CC-MAIN-20170625115717-20170625135717-00099.warc.gz"} |
https://proceedings.mlr.press/v31/meng13a.html | # Distributed Learning of Gaussian Graphical Models via Marginal Likelihoods
Zhaoshi Meng, Dennis Wei, Ami Wiesel, Alfred Hero III
Proceedings of the Sixteenth International Conference on Artificial Intelligence and Statistics, PMLR 31:39-47, 2013.
#### Abstract
We consider distributed estimation of the inverse covariance matrix, also called the concentration matrix, in Gaussian graphical models. Traditional centralized estimation often requires iterative and expensive global inference and is therefore difficult in large distributed networks. In this paper, we propose a general framework for distributed estimation based on a maximum marginal likelihood (MML) approach. Each node independently computes a local estimate by maximizing a marginal likelihood defined with respect to data collected from its local neighborhood. Due to the non-convexity of the MML problem, we derive and consider solving a convex relaxation. The local estimates are then combined into a global estimate without the need for iterative message-passing between neighborhoods. We prove that this relaxed MML estimator is asymptotically consistent. Through numerical experiments on several synthetic and real-world data sets, we demonstrate that the two-hop version of the proposed estimator is significantly better than the one-hop version, and nearly closes the gap to the centralized maximum likelihood estimator in many situations. | {"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": 0.9714856743812561, "perplexity": 498.0789618848636}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991378.48/warc/CC-MAIN-20210515070344-20210515100344-00108.warc.gz"} |
http://mathhelpforum.com/calculus/177460-derivative-logarithmic-function.html | # Math Help - Derivative of Logarithmic Function
1. ## Derivative of Logarithmic Function
Differentiate $h(x) = \ln{(x + \sqrt{x^2-6})}$
My work:
let $u = x + \sqrt{x^2-6}$ then $y = \ln{u}$
$\displaystyle\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}=\frac{1}{u}\frac{du}{dx } = \frac{1}{x + \sqrt{x^2-6}}(1+\frac{1}{2}(x^2-6)^{-1/2}(2x))$
$\displaystyle=\frac{2x}{(x+\sqrt{x^2-6})(2)\sqrt{x^2-6}}$
$\displaystyle=\frac{x}{(x+\sqrt{x^2-6})\sqrt{x^2-6}}$
$\displaystyle=\frac{x}{(x\sqrt{x^2-6} + x^2-6}$
Where did I go wrong??? Thanks.
2. Originally Posted by joatmon
Differentiate $h(x) = \ln{(x + \sqrt{x^2-6})}$
My work:
let $u = x + \sqrt{x^2-6}$ then $y = \ln{u}$
$\displaystyle\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}=\frac{1}{u}\frac{du}{dx } = \frac{1}{x + \sqrt{x^2-6}}(\boxed{1}+\frac{1}{2}(x^2-6)^{-1/2}(2x))$
$\displaystyle=\frac{2x}{(x+\sqrt{x^2-6})(2)\sqrt{x^2-6}}$
$\displaystyle=\frac{x}{(x+\sqrt{x^2-6})\sqrt{x^2-6}}$
$\displaystyle=\frac{x}{(x\sqrt{x^2-6} + x^2-6}$
Where did I go wrong??? Thanks.
You forgot about the term I boxed in your work.
The final answer should be $\dfrac{1}{x + \sqrt{x^2-6}}+\dfrac{x}{x\sqrt{x^2-6} + x^2-6}=\dfrac{x+\sqrt{x^2-6}}{x\sqrt{x^2-6}+x^2-6}=\dfrac{1}{\sqrt{x^2-6}}$
I hope this makes sense. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9922683835029602, "perplexity": 2734.1371148307467}, "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-2015-32/segments/1438042989507.42/warc/CC-MAIN-20150728002309-00298-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/437236/effects-of-different-species-in-shock-waves | # Effect(s) of different species in shock waves
In air, transitioning from a local mach number ($$M$$) of $$M > 1$$ to $$M < 1$$ produces a shock. But $$M$$ is defined simply based on the velocity of an object relative to the local speed of sound. If a positron-electron gas is introduced into air, moving at supersonic speeds, will a shock be produced? The original air and injected gas species no long have similar masses, and I'm simply curious if a shock would be produced in such a scenario where the individual positrons and electrons may have significantly lower momentum than the gas particles, but are nevertheless moving faster than the local air's speed of sound.
I think you are thinking about partial pressures (if not, you should be) because the speed of sound is defined as: $$C_{s}^{2} = \frac{ \partial P }{ \partial \rho }$$ where $$P = \sum_{s} P_{s}$$ is the total gas pressure summed over all species $$s$$ and $$\rho$$ is the average mass density of the gas. | {"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": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8455686569213867, "perplexity": 311.0716707806397}, "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/1575540481281.1/warc/CC-MAIN-20191205164243-20191205192243-00495.warc.gz"} |
https://voer.edu.vn/c/relativistic-momentum/0e60bfc6/d51d7251 | Giáo trình
# College Physics
Science and Technology
## Relativistic Momentum
Tác giả: OpenStaxCollege
In classical physics, momentum is a simple product of mass and velocity. However, we saw in the last section that when special relativity is taken into account, massive objects have a speed limit. What effect do you think mass and velocity have on the momentum of objects moving at relativistic speeds?
Momentum is one of the most important concepts in physics. The broadest form of Newton’s second law is stated in terms of momentum. Momentum is conserved whenever the net external force on a system is zero. This makes momentum conservation a fundamental tool for analyzing collisions. All of Work, Energy, and Energy Resources is devoted to momentum, and momentum has been important for many other topics as well, particularly where collisions were involved. We will see that momentum has the same importance in modern physics. Relativistic momentum is conserved, and much of what we know about subatomic structure comes from the analysis of collisions of accelerator-produced relativistic particles.
The first postulate of relativity states that the laws of physics are the same in all inertial frames. Does the law of conservation of momentum survive this requirement at high velocities? The answer is yes, provided that the momentum is defined as follows.
Note that we use $u$ for velocity here to distinguish it from relative velocity $v$ between observers. Only one observer is being considered here. With $p$ defined in this way, total momentum ${p}_{\text{tot}}$ is conserved whenever the net external force is zero, just as in classical physics. Again we see that the relativistic quantity becomes virtually the same as the classical at low velocities. That is, relativistic momentum $\text{γmu}$ becomes the classical $\text{mu}$ at low velocities, because $\gamma$ is very nearly equal to 1 at low velocities.
Relativistic momentum has the same intuitive feel as classical momentum. It is greatest for large masses moving at high velocities, but, because of the factor $\gamma$, relativistic momentum approaches infinity as $u$ approaches $c$. (See [link].) This is another indication that an object with mass cannot reach the speed of light. If it did, its momentum would become infinite, an unreasonable value.
Relativistic momentum is defined in such a way that the conservation of momentum will hold in all inertial frames. Whenever the net external force on a system is zero, relativistic momentum is conserved, just as is the case for classical momentum. This has been verified in numerous experiments.
In Relativistic Energy, the relationship of relativistic momentum to energy is explored. That subject will produce our first inkling that objects without mass may also have momentum.
# Section Summary
• The law of conservation of momentum is valid whenever the net external force is zero and for relativistic momentum. Relativistic momentum $p$ is classical momentum multiplied by the relativistic factor $\gamma$.
• $p=\text{γmu}$, where $m$ is the rest mass of the object, $u$ is its velocity relative to an observer, and the relativistic factor $\gamma =\frac{1}{\sqrt{1-\frac{{u}^{2}}{{c}^{2}}}}$.
• At low velocities, relativistic momentum is equivalent to classical momentum.
• Relativistic momentum approaches infinity as $u$ approaches $c$. This implies that an object with mass cannot reach the speed of light.
• Relativistic momentum is conserved, just as classical momentum is conserved.
# Conceptual Questions
How does modern relativity modify the law of conservation of momentum?
Is it possible for an external force to be acting on a system and relativistic momentum to be conserved? Explain.
# Problem Exercises
Find the momentum of a helium nucleus having a mass of $6\text{.}\text{68}×{\text{10}}^{\text{–27}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ that is moving at $0\text{.}\text{200}c$.
$4\text{.}\text{09}×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$
What is the momentum of an electron traveling at $0\text{.}\text{980}c$?
(a) Find the momentum of a $1\text{.}\text{00}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{kg}$ asteroid heading towards the Earth at $\text{30.0 km/s}$. (b) Find the ratio of this momentum to the classical momentum. (Hint: Use the approximation that $\gamma =1+\left(1/2\right){v}^{2}/{c}^{2}$ at low velocities.)
(a) $3\text{.}\text{000000015}×{\text{10}}^{\text{13}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$.
(b) Ratio of relativistic to classical momenta equals 1.000000005 (extra digits to show small effects)
(a) What is the momentum of a 2000 kg satellite orbiting at 4.00 km/s? (b) Find the ratio of this momentum to the classical momentum. (Hint: Use the approximation that $\gamma =1+\left(1/2\right){v}^{2}/{c}^{2}$ at low velocities.)
What is the velocity of an electron that has a momentum of $3.04×{\text{10}}^{\text{–21}}\phantom{\rule{0.25em}{0ex}}\text{kg⋅m/s}$? Note that you must calculate the velocity to at least four digits to see the difference from $c$.
$\text{2.9957}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}$
Find the velocity of a proton that has a momentum of $4\text{.}\text{48}×{\text{–10}}^{-\text{19}}\phantom{\rule{0.25em}{0ex}}\text{kg⋅m/s}\text{.}$
(a) Calculate the speed of a $1.00-\mu \text{g}$ particle of dust that has the same momentum as a proton moving at $0\text{.}\text{999}c$. (b) What does the small speed tell us about the mass of a proton compared to even a tiny amount of macroscopic matter?
(a) $1\text{.}\text{121}×{\text{10}}^{\text{–8}}\phantom{\rule{0.25em}{0ex}}\text{m/s}$
(b) The small speed tells us that the mass of a proton is substantially smaller than that of even a tiny amount of macroscopic matter!
(a) Calculate $\gamma$ for a proton that has a momentum of $\text{1.00 kg⋅m/s}\text{.}$ (b) What is its speed? Such protons form a rare component of cosmic radiation with uncertain origins. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 49, "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": 0.9915509819984436, "perplexity": 267.45936779884516}, "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/1573496671260.30/warc/CC-MAIN-20191122115908-20191122143908-00508.warc.gz"} |
http://dave.thehorners.com/tech-talk/unix-linux-bsd-osx-etc/90-control-panels | Dave Horner's Website - Yet another perspective on things...
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Don't worry about what anybody else is going to do. The best way to predict the future is to invent it. -- Alan Kay
\begin{bmatrix} 1 & 0 & \ldots & 0 \\ 0 & 1 & 0 & \vdots \\ \vdots & 0 & \ddots & 0\\ 0 & \ldots & 0 & 1_{n} \end{bmatrix}
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Last Updated on Friday, 31 October 2008 08:44 | {"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": 0.9311713576316833, "perplexity": 2407.349690990763}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501172000.93/warc/CC-MAIN-20170219104612-00282-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://www.mathematicalfoodforthought.com/2006/10/divvy-it-up-topic-number-theory-level_12.html | ## Thursday, October 12, 2006
### Divvy It Up. Topic: Number Theory. Level: AMC.
Problem: Find the number of divisors of a postive integers.
Solution: For all you people at math club who didn't understand it... here it is. Suppose our number is $n = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k}$, where the $p$'s are the prime divisors of $n$ and the $e$'s are the number of times (power) each $p$ divides $n$. For example, $36 = 2^2 \cdot 3^2$.
We will show that the number of divisors is $(e_1+1)(e_2+1) \cdots (e_k+1)$. Let $d$ be a divisor of $n$. Write $d$ as
$d = p_1^{a_1} \cdot p_2^{a_2} \cdots p_k^{a_k}$.
We know that $d$ can only have the prime divisors of $n$ or it obviously won't divide $n$. So now it remains to choose the $a$'s. Well, what can $a_1$ be? It can be $0, 1, 2, \ldots, e_1$ but if $a_1 > e_1$ then $d$ will have more $p_1$'s than $n$ and won't divide it. So there are $e_1+1$ choices there. Similarly, $a_2$ can be $0, 1, 2, \ldots, e_2$ and $a_j$ can be $0, 1, 2, \ldots, e_j$ for $j = 3, 4, \ldots, k$.
So if we just multiply the number of ways we can pick the power on each prime factor, we get the total number of divisors, i.e. $(e_1+1)(e_2+1) \cdots (e_k+1)$. QED.
--------------------
Comment: This is very important for general number theory; it is one of the fundamental ideas. Plus, it's logical and easy to remember so it should always be accessible during competitions.
--------------------
Practice Problem: (2002-2003 MIC Championships Team Test - #2) How many natural numbers less than $50,000$ have exactly $35$ positive integer factors?
1. WOW THAT"S SOO COOL~
so 36 has (2+1)(2+1)=9 factors!
35... mm.. so 5*7=35... and (2^5)(2^7) is.. uh.. 4096
(3^5)(2^7)=30014
(2^7)(3^5)=69984
So 2 numbers?
but how am i supposed to do this with out a calc??
2. the powers are one less than the factors; those should be 4 and 6, not 5 and 7.
calculating powers without a calculator isn't so hard, xuan :)
3. also, the primes have to be different. (2^5) (2^7) counts as 2^12, which has 13 factors.
4. .. oh yea. 4 and 6.. right....
cool cool thanks tOrajirOu | {"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": 0.9394247531890869, "perplexity": 543.5369848999261}, "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-2018-34/segments/1534221208676.20/warc/CC-MAIN-20180814062251-20180814082251-00223.warc.gz"} |
http://mathhelpforum.com/differential-equations/127915-diff-eq-help-model-disease-print.html | # Diff. EQ. Help - model for disease
• Feb 8th 2010, 09:08 PM
tactical
Diff. EQ. Help - model for disease
The following problem is an SIS disease problem:
Calling: I(t) = number of infectives at time t
N = the total population (assumed constant)
b = infection rate (here, a positive constant)
v = recovery rate (also, a positive constant)
a model for this disease is given bu the following:
dI/dt = bI(N-I) - vI
And since the population is assumed constant, we can just take S(t) to be N -I(t). Derive a condition for when the number of infectives goes to zero.
Is there anyone out there than can help me, even if it's just a little bit?
• Feb 9th 2010, 03:47 AM
shawsend
Hi. That's a Bernoulli equation which can be solved exactly but wouldn't the recovery rate have to be greater than the infection rate to drop the number of infected down to zero? Not sure. May need to just solve the equation with test values of the constants to see what's happening.
• Feb 9th 2010, 11:24 AM
tactical
Quote:
Originally Posted by shawsend
Hi. That's a Bernoulli equation which can be solved exactly but wouldn't the recovery rate have to be greater than the infection rate to drop the number of infected down to zero? Not sure. May need to just solve the equation with test values of the constants to see what's happening.
My teacher gave us a hint that said solving this equation is definitely not the best way, and for us to think of a way to model long-term equations.
By this I think he wants us to use the phase diagram.
So i found that I is zero when I*= 0 and I*= (bN-v)/B, but this is where I get stuck. I choose values less than 0, but that doesn't tell me if the function is increasing of decreasing.
Is there anything that I am missing?
• Feb 9th 2010, 01:25 PM
shawsend
1 Attachment(s)
"Differential Equations" by Blanchard, Devaney, and Hall is what I use. It goes over this really well. So you figured out the equilibrium points of zero and $N-v/b$. That's when the derivative is zero and since this is an autonumous equation, the value of I completely determines the value of the derivative so once I reaches one of the equilibrium points, then the infections stay at that value (although below it's negative which is not applicable in real life). So we have:
$\frac{dI}{dt}=bI(N-I)-vI$
and I want to study just one particular case:
$\frac{dI}{dt}=0.005 I(100-I)-0.55I$
That means of course the recovery rate is much higher than the infection rate. Suppose I then plot $y=0.005x(100-x)-0.55x$. That's the first plot and it represents the derivative as a function of the number of infections (I).. Note the derivative is positive between -10 and zero. That means if we start I anywhere in that range (not realistic in actual practice), the derivative is positive and I will increase until it reaches the nearest equilibrium point of zero. If I<-10, then the derivative is negative and the value of I will continue to decrease. And if I is greater than zero, then again the derivative is negative and so I will decrease again towards it's nearest equilibrium point. I've shown all these cases in the second plot which is I(t) with the red lines as the equilibrium points. Note that if under these conditions, the initial value of I>0, since it's derivative is negative, the value of I will continue to drop until it reaches the nearest equilibrium point which is zero (the number of infections is then zero and stays there right?). So what happens if I then use different values of b and v (just leave N=100)? How will this change the first plot and by the shape of that plot, since it's the value of the derivative, how then will the solutions behave for various starting values of I based on the value of the derivative there? Just try a few different values and see what happens
• Feb 9th 2010, 05:49 PM
tactical
Quote:
Originally Posted by shawsend
"Differential Equations" by Blanchard, Devaney, and Hall is what I use. It goes over this really well. So you figured out the equilibrium points of zero and $N-v/b$. That's when the derivative is zero and since this is an autonumous equation, the value of I completely determines the value of the derivative so once I reaches one of the equilibrium points, then the infections stay at that value (although below it's negative which is not applicable in real life). So we have:
$\frac{dI}{dt}=bI(N-I)-vI$
and I want to study just one particular case:
$\frac{dI}{dt}=0.005 I(100-I)-0.55I$
That means of course the recovery rate is much higher than the infection rate. Suppose I then plot $y=0.005x(100-x)-0.55x$. That's the first plot and it represents the derivative as a function of the number of infections (I).. Note the derivative is positive between -10 and zero. That means if we start I anywhere in that range (not realistic in actual practice), the derivative is positive and I will increase until it reaches the nearest equilibrium point of zero. If I<-10, then the derivative is negative and the value of I will continue to decrease. And if I is greater than zero, then again the derivative is negative and so I will decrease again towards it's nearest equilibrium point. I've shown all these cases in the second plot which is I(t) with the red lines as the equilibrium points. Note that if under these conditions, the initial value of I>0, since it's derivative is negative, the value of I will continue to drop until it reaches the nearest equilibrium point which is zero (the number of infections is then zero and stays there right?). So what happens if I then use different values of b and v (just leave N=100)? How will this change the first plot and by the shape of that plot, since it's the value of the derivative, how then will the solutions behave for various starting values of I based on the value of the derivative there? Just try a few different values and see what happens
can you please explain where you got .005, 100 and .55 from?
• Feb 10th 2010, 05:00 AM
shawsend
That's just any old numbers. Just a set of numbers to illustrate a particular case of the differential equation. Your mission, should you choose to accept it, is to try other values of the infection and recovery rate and substitute it into the equation $y=bI(100-I)-vI$, generate the parabola, and since that parabola represents the value of the derivative, use that information like I did above to explain the long-term behavior of the infection function I(t). Alright then, here's some to try:
$\begin{array}{cc} \text{b} & \text{v} \\
.1 & .2 \\
.2 & .1 \\
.01 & 0.01 \\
.2 & .2 \\
.79 & .8
\end{array}
$
So I plot $y=.1 I(100-I)-0.2 I$ then I say, ok, it's negative here. That means if I start the initial conditions with this value, the derivative is negative so the infections will continue to drop. Ok, here, it's positive. So if I start the initial conditions with this value, since the derivative is positive, the infections will increase. And just do that like I did above and from studying those examples, make some conclusion about the relationship between infection and recovery rate that is necessary for the infection to drop to zero based on the values of b and v. For example, if the recovery rate is ANY value above the infection rate like b=0.1 and v=0.10001 would that still result in the infections eventually dropping to zero? I don't know. It depends entirely on what the parabola $0.1 I(100-I)-0.10001 I$ tells you. So just try doing that for the first set of values for b and v and risk doing it wrong cus' look, I'll tell you a secret: sometimes the wrong ones are on the road to the right ones. :)
• Feb 10th 2010, 05:53 AM
HallsofIvy
Quote:
Originally Posted by tactical
The following problem is an SIS disease problem:
Calling: I(t) = number of infectives at time t
N = the total population (assumed constant)
b = infection rate (here, a positive constant)
v = recovery rate (also, a positive constant)
a model for this disease is given bu the following:
dI/dt = bI(N-I) - vIAnd since the population is assumed constant, we can just take S(t) to be N -I(t). Derive a condition for when the number of infectives goes to zero.
Is there anyone out there than can help me, even if it's just a little bit?
In order that I go to 0, it must decrease. And for that to be true, the derivative must be negative. Under what conditions is
$\frac{dI}{dt}= bI(N-I)- vI< 0$? That's the same as $I(bN- v- I)< 0$. The only way a product can be negative is if one factor is positive and the other is negative. Since I, the number of infected persons cannot be negative, we must have bN- v-I< 0. That is the same as I> bN- v. Of course, if I< bN- v it will start going up again. So a condition that I go to 0 is that bN- v be negative; that is, that v> bN. | {"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": 14, "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": 0.8853229880332947, "perplexity": 401.94098290233103}, "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-2017-26/segments/1498128329372.0/warc/CC-MAIN-20170629154125-20170629174125-00291.warc.gz"} |
https://www.physicsforums.com/threads/some-basic-algebra-using-isomorphism-theorems.178190/ | Some basic algebra (using Isomorphism Theorems)
1. Jul 25, 2007
bham10246
1. The problem statement, all variables and given/known data
Let $G$ be a group with a normal subgroup $N$ and subgroups $K \triangleleft H \leq G.$
If $H/K$ is nontrivial, prove that at least one of $HN/KN$ and $(H\cap N)/(K\cap N)$ must be nontrivial.
2. Relevant equations
The Three (or Four) Isomorphism Theorems.
3. The attempt at a solution
By the first isomorphism theorem, we saw that $HN/KN \cong H/K$. So if $H/K$ is nontrivial, then $HN/KN$ is also nontrivial.
Now to show that $(H\cap N)/(K\cap N)$ is also nontrivial, what normal subgroup of $H/K$ is this quotient group $(H\cap N)/(K\cap N)$ isomorphic to?
Because of the "and" in the statement of the problem, should both be nontrivial?
2. Jul 25, 2007
CompuChip
The problem asks you to prove that at least one of those two is non-trivial. If I read your post correctly, you've proven that the first one is always non trivial (given the err.. given data), so automatically at least one of them is. In which case, though, the question seems rather pointless (you've kind of reduced it to "show that if A always holds, then at least one of A or B holds"). But I don't know if your first steps are correct, since I don't know the isomorphism theorems
3. Jul 25, 2007
Kummer
2nd Isomorphism Theorem:
$$H/(H\cap K)\simeq HK/K$$ and $$H/(H\cap N)\simeq HN/N$$
Form factor groups,
$$(H/(H\cap K))/(H/(H\cap N))\simeq (HK/K)/(HN/N)$$
3rd Isomorphism Theorem:
$$(H\cap N)/(H\cap K)\simeq (HK/K)/(HN/N)$$
Are you sure you are looking for,
$$(H\cap N)/(K\cap N)$$
And not for,
$$(H\cap N)/(H\cap K)$$
Because we do not know if $$K\leq N$$ and so cannot use theorem.
4. Jul 25, 2007
bham10246
Hi CompuChip, yes, your reasoning is correct and this part of the problem doesn't really make sense. But this is how I proved the first part of this problem which states: prove that $HN/KN$ is isomorphic with a quotient group of $H/K$.
So let $fN\rightarrow H/K$ be a homomorphism where $f: hn\longmapsto hK$ where $h \in H, n \in N$. So $f(hn)=K \Rightarrow h \in K \Rightarrow hn\in KN$. This shows that $HN/KN \cong H/K$. So am I right?
And Kummer, I'm not sure... but isn't $H\cap K = K$?
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https://www.physicsforums.com/threads/increasing-force.300503/ | # Increasing Force.
1. Mar 17, 2009
1. The problem statement, all variables and given/known data
(First of all sorry for my English, I don't know any scientific expressions in English...)
A Force given by the type F=5x + 5 (x at metres, F at Newtons) is applied to a body of mass m while it creates (the Force) an angle φ with the horizontal level. The coefficient of friction is μ. (the body will move by doing the numbers) First it asks the velocity of the body, when it takes off the ground. Then, it asks the hight, h of the object, if the force stops after 2 meters (after the takeoff).
Givens: m=1kg, μ=0,2 , φ=30, g=10m/s^2
2. Relevant equations
F=5x + 5 , T=μ*N (T is friction), N=m*g - F*sinφ
I guess that's it.
3. The attempt at a solution
First of all I don't know how much advanced this problem is therefore I couldn't classify it.
The first question is rather easy.
We know that the body will take off when N=0 (N is the reaction from the ground)
So: F*sinφ + N = mg (the ΣF for the Vertical axis)
By zeroing N we get x=3. The by doing the diagramm of (F*cosφ)(x) and T(x)=mg - F*sinφ
and by calculating the acreage we find both Force's work. By subtracting them we get the Kinetical Energy and thus the velocity.
The second question I think it's impossible, or there is something wrong with the exercise.
I use Principal of Energy Conservation to get: WF - mgh = 1/2mv22 - 1/2mv12.
I know v(1) from the previous question and I can find WF with the same way, with the diagramm. But I can't find the velocity at the final state, or the hight to find the velocity afterwards.
Last edited: Mar 17, 2009
Can you offer guidance or do you also need help?
Draft saved Draft deleted
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https://repository.uantwerpen.be/link/irua/112260 | Title Natural orbitals in relation to quantum information theory : from model light atoms through to emergent metallic properties Author March, N.H. Angilella, G.G.N. Pucci, R. Faculty/Department Faculty of Sciences. Physics Publication type article Publication 2013 Singapore , 2013 Subject Physics Source (journal) International journal of modern physics: B: condensed matter physics, statistical physics, applied physics. - Singapore Volume/pages 27(2013) :30 , p. 1-26 ISSN 0217-9792 Article Reference 1330021 ISI 000326678000001 Carrier E-only publicatie Target language English (eng) Full text (Publishers DOI) Affiliation University of Antwerp Abstract The review begins with a consideration of three forms of quantum information entropy associated with Shannon and Jaynes. For model two-electron spin compensated systems, some analytic progress is first reported. The Jaynes entropy is clearly related to correlation kinetic energy. A way of testing the usefulness of a known uncertainty principle inequality is proposed for a whole class of model two-electron atoms with harmonic confinement but variable electron-electron interaction. Emerging properties are then studied by reference to bcc Na at ambient pressure and its modeling by "jellium". Jellium itself has collective behavior with changes of the density, especially noteworthy being the discontinuity of the momentum distribution at the Fermi surface. This has almost reduced to zero at r(s) = 100 a.u., the neighborhood in which the quantal Wigner electron solid transition is known to occur. However, various workers have studied crystalline Na under pressure and their results are compared and contrasted. Work by DFT on K, Rb and Cs is discussed, but now with reduced density from the ambient pressure value. The crystalline results for the cohesive energy of these metals as a function of lattice parameters and local coordination number are shown to be closely reproduced by means of ground and excited states for dimer potential energy curves. Then, pair potentials for liquid Na and Be are reviewed, and compared with the results of computer simulations from the experimental structure factor for Na. Finally, magnetic field effects are discussed. First a phenomenological model of the metal-to-insulator transition is presented with an order parameter which is the discontinuity in the Fermi momentum distribution. Lastly, experiments on a two-dimensional electron assembly in a GaAs/AlGaAs heterojunction in a perpendicular magnetic field are briefly reviewed and then interpreted. E-info http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:000326678000001&DestLinkType=RelatedRecords&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:000326678000001&DestLinkType=FullRecord&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:000326678000001&DestLinkType=CitingArticles&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 Handle | {"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": 0.8159258961677551, "perplexity": 2429.633615425207}, "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-2017-17/segments/1492917120881.99/warc/CC-MAIN-20170423031200-00642-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://grocid.net/category/uncategorized/ | # Breaking affine ciphers – a matrix approach
An affine cipher is a one the remnants of classical cryptography, easily broken with todays computational power. The cipher defines a symbol mapping from $f :\{A,B,\ldots,\} \mapsto \mathbb{Z}_n$. Each cipher symbol is then computed as $a \cdot x + b \rightarrow y$, where $a \in \mathbb{Z}^*_n$ and $b \in \mathbb{Z}_n$. Decryption is then done by computing $x= (y - b) \cdot a^{-1}$.
In this blog post, I will show how to break this cipher in time faster than trying all keys. Let us first sketch the general idea. Consider an expected distribution $\hat{P}$ of the symbols and a given distribution $P$, the integral $\int (\hat{P}(x) - P(x))^2 dx$ defines a statistical distance between the distributions (this would correspond to the Euclidian distance), which we would like to minimize. Now, clearly
$(\hat{P}(x) - P(x))^2 = \hat{P}(x)^2 - \hat{P}(x)P(x) + P(x)^2$.
Trivially, $\hat{P}(x)^2$ and $P(x)^2$ remains constant over any keypair $(a,b)$, so instead of minimizing the above, we can maximize $\hat{P}(x)P(x)$. Therefore, the minimization problem can be turned into a maximization problem $\max_{a,b} \int \hat{P}(x)P_{a,b}(x) dx$. Cool.
In terms of our cipher, which is discrete, the minimization problem is a sum $\max_{a,b} \sum \hat{P}(x)P_{a,b}(x)$. The observant reader may notice that this looks like a term in a matrix multiplication. There is just one caveat; the indices corresponding appear only in one term. There is an easy way to get around this. Instead of applying transformations on only $P$, we may split them among the two. So by instead computing
$\max_{a,b} \sum \hat{P}_a(x) P_{b}(x)$,
we have achieved what we desired. This means that we shuffle $\hat{P}$ with $a$ and ${P}$ with $b$. Let us interpret this as Python. The expected distribution of an alphabet ABCDEFGHIJKLMNOPQRSTUVWXYZ ,. may be as follows (depending on the observation):
P_hat = {' ': 0.05985783763561542, ',': 0.0037411148522259637, '.': 0.0028058361391694723, 'A': 0.0764122708567153, 'C': 0.02600074822297044, 'B': 0.012065095398428732, 'E': 0.11878039655817432, 'D': 0.03974934530490086, 'G': 0.018892630003741116, 'F': 0.020856715301159744, 'I': 0.0651889263000374, 'H': 0.05695847362514029, 'K': 0.00720164609053498, 'J': 0.0014029180695847362, 'M': 0.02254021698466143, 'L': 0.03769173213617658, 'O': 0.07023943135054246, 'N': 0.06313131313131314, 'Q': 0.0009352787130564909, 'P': 0.01805087916199027, 'S': 0.05920314253647587, 'R': 0.0560231949120838, 'U': 0.025813692480359144, 'T': 0.08473625140291807, 'W': 0.022072577628133184, 'V': 0.00916573138795361, 'Y': 0.01842499064721287, 'X': 0.0014029180695847362, 'Z': 0.0006546950991395436}
The transformations are done by computing the matrices
# compute first matrix for transformed P_hat
for i in range(1, N):
for element in priori_dist:
X[i, (look_up.index(element) * i) % N] = priori_dist[element]
# compute second matrix for transformed P
for j in range(N):
for element in dist:
Y[(look_up.index(element) - j) % N, j] = dist[element]
Here, the $i$th row in $X$ corresponds to $\hat{P}$ transformed by $a = i$. Moreover, the $j$th row in $Y$ corresponds ${P}$ transformed by $b = j$.
For some distribution, they may look like
As we can see, $X$ is only shifted (by the addition), while in $Y$ the indices are reordered by multiplication with row index $i$. Taking advantage of the matrix multiplication property, we may now compute $Z=XY$.
Any entry in $Z$ is $Z_{a,b} = \sum_x X_{a,x} Y_{x,b}$ so finding a maximum element in $Z$ is equivalent to saying
$\max_{a,b} \sum_x X_{a,x} Y_{x,b}$.
Looks familiar? It should. This is our maximization problem, which we stated earlier.
Therefore, we may solve the problem using
Z = numpy.dot(X, Y)
a, b = numpy.unravel_index(Z.argmax(), Z.shape)
This breaks the affine cipher.
Some notes on complexity
So, what is the complexity of the matrix approach? Computing the matrices takes $O(N^2)$ modular operations. The matrix multiplication takes naively $O(N^3)$ operations, but for large $N$ this can be achieved faster. For instance Strassen takes $O(N^{2.807})$ but faster algorithms exist. Also, taking advantage of symmetry and structure could probably decrease the complexity further. This is the total complexity of this approach.
Compare this with brute-force guessing of the key (taking $O(N^2)$ guesses) and for each guess, compute the distance taking $O(N)$ operations, which in total yields $O(N^3)$. It should be noted that complexity of this approach may be reduced by picking $a,b$ in an order which minimizes the number of tries.
Example implementation for the id0-rsa.pub: github
# Custom terminal for Vagrant/SSH
Short story: I wanted to distinguish my terminal windows between local sessions, ssh sessions and vagrant sessions.
SSH_THEME="SSH"
VAGRANT_THEME="Vagrant"
set_th () {
osascript -e "tell app \"Terminal\" to set current settings of first window to settings set \"$1\"" } set_id () { osascript -e "tell app \"Terminal\" to set current settings of first window to$1 $2$3 $4" #$@ does not work!
}
get_id () {
cur_id=$(osascript -e "tell app \"Terminal\" to get current settings of first window") } ssh(){ #!/bin/sh get_id set_th$SSH_THEME
/usr/bin/ssh "$@" set_id$cur_id
}
vagrant(){
#!/bin/sh
if [ $1 = "ssh" ]; then get_id set_th$VAGRANT_THEME
/opt/vagrant/bin/vagrant "$@" set_id$cur_id
else
/opt/vagrant/bin/vagrant "$@" fi } The code creates a temporary variable of the current theme before switching. So, when ending the session, the original theme changes back instead of a fixed one. Putting the above code in your .bash_profile: gives the following nice behavior: Color coding your sessions is a great way to visualize things and make sure you do not take any unwanted action by mistake 🙂 Of course, the code can be used to wrap any application. For instance, one could use it to make the interactive interpreter of Python/Sage or terminal sessions using torsocks appear in different colors or even fonts. # Re-mapping KBT Pure Pro in OS X For my everyday-use computer, I use a modded KBT Pure Pro; this is a small mechanical keyboard with aluminium base and background lightning, perfect for programming and typing. The size of the keyboard is 60 % of a normal one, making it suitable for spatially constrained workspaces. To my experience, it is also more ergonomic. Below is a comparison of the Pure Pro and a wireless Apple keyboard. For those being the in the process of buying a keyboard, I recommend this one 🙂 For quite a while, I have used Linux on this computer. But after installing OS X, the keyboard map went wack, so to speak. Many keys were mapped incorrectly. Using Ukulele, I created a customized layout with correct mapping (don’t mind the duplicate keys): The layout covers all keys and can be found here. NOTE: this is a layout for KBT Pure Pro with British ISO layout and not ANSI. # BackdoorCTF16 – Collision Course With 350 points and a description as follows: In today’s world, hash collisions are becoming more and more popular. That is why, one must rely on standardized hashing techniques, such as bcrypt. However, n00bster shall never learn, and he has implemented his own hash function which he proudly calls foobar. Attached is an implementation of the hash function and a file with which you are supposed to find a collision. He believes that you will not be able to find a collision for the file, especially since he hasn’t even given you the hashing algorithm, but has packaged it as a black box application. Prove to him that he is wrong. Note: Multiple collisions are possible, but only one of them is a valid flag. You will realize you’ve gotten it once you do. The hash is given as follows: So, we start off by looking at the binary. Using Hopper, we obtain the following pseudo code by decompilation: int hash(int input) { eax = _rotr(input ^ 0x24f50094, (input ^ 0x24f50094) & 0xf); eax = _rotl(eax + 0x2219ab34, eax + 0x2219ab34 & 0xf); eax = eax * 0x69a2c4fe; return eax; } int main() { esp = (esp & 0xfffffff0) - 0x20; puts(0x80486d0); gets(0x804a060); stack[2039] = "\nBar:"; puts(stack[2039]); while (stack[2039] < *(esp + 0x18)) { stack[2039] = *(stack[2039] + stack[2039] * 0x4); *(esp + 0x14) = *(esp + 0x14) ^ hash(stack[2039]); eax = _rotr(stack[2039], 0x7); printf("%08lx", stack[2039]); *(esp + 0x10) = *(esp + 0x10) + 0x1; } eax = putchar(0xa); return eax; } We sketch the above code as block scheme below: The first thing to note is that we can find an infinite number of collisions just by appending arbitrary data after 10 blocks. However, this is not interesting to us, but completely defeats the conditions for a safe cryptographic hash function. This Merkle-Damgård-like structure allows us to solve blocks iteratively, starting from the first. Here is how. Starting from the first block, we can find an input to the function $H$ such that when rotated 7 steps is equal to block 0 (here, denoted $B_0$). Hence, the problem we solve is to find an $x$ such that $H(x) \ll 7 = B_0$. This is a simple thing for Z3. Then, we take the next block and solve for $(H(x) \oplus B_0) \ll 7 = B_1$ and so forth. Implemented in Python/Z3, it may look like the following: from z3 import * import binascii, string, itertools bits = 32 mask = 2**bits - 1 allowed_chars = string.printable def convert_to_hex(s): return ''.join([hex(ord(x))[2:].zfill(2) for x in s[::-1]]) def convert_to_string(h): return ''.join([chr(int(x, 16)) for x in list(map(''.join, zip(*[iter(hex(h)[2:])]*2)))[::-1]]) def rot(val, steps): return (val << (bits-steps)) | LShR(val, steps) def hash_foobar(input): eax = rot(input ^ 0x24f50094, (input ^ 0x24f50094) & 0xf) eax = rot(eax + 0x2219ab34, bits - (eax + 0x2219ab34 & 0xf)) eax = eax * 0x69a2c4fe return eax & mask def break_iteratively(hashdata, i): if i == 0: prev_block = 0 else: prev_block = hashdata[i-1] s = Solver() j = BitVec('current_block', bits) eax = rot(prev_block ^ hash_foobar(j), 7) s.add(eax == hashdata[i]) block_preimages = [] while s.check() == sat: sol = s.model() s.add(j != sol[j].as_long()) block_string = convert_to_string(sol[j].as_long()) if all(c in allowed_chars for c in block_string): block_preimages.append(block_string) return block_preimages known = '9513aaa552e32e2cad6233c4f13a728a5c5b8fc879febfa9cb39d71cf48815e10ef77664050388a3' # this the hash of the file data = list(map(''.join, zip(*[iter(known)]*8))) hashdata = [int(x, 16) for x in data] print '[+] Hash:', ''.join(data) print '[+] Found potential hashes:\n' for x in itertools.product(*[break_iteratively(hashdata, i) for i in range(10)]): print ' * ' + ''.join(x) This code is surprisingly fast, thanks to Z3, and runs in 0.3 seconds. Taking all possible collisions into consideration… [+] Hash: 9513aaa552e32e2cad6233c4f13a728a5c5b8fc879febfa9cb39d71cf48815e10ef77664050388a3 [+] Found potential hashes: * CTFEC0nstra1nts_m4keth_fl4g} * CTFEC0nstra1nts_m4keth_nl4g} * CTFEC0nstra1nws_m4keth_fl4g} * CTFEC0nstra1nws_m4keth_nl4g} * CTFEC0nstra9nts_m4keth_fl4g} * CTFEC0nstra9nts_m4keth_nl4g} * CTFEC0nstra9nws_m4keth_fl4g} * CTFEC0nstra9nws_m4keth_nl4g} * CTF{C0nstra1nts_m4keth_fl4g} * CTF{C0nstra1nts_m4keth_nl4g} * CTF{C0nstra1nws_m4keth_fl4g} * CTF{C0nstra1nws_m4keth_nl4g} * CTF{C0nstra9nts_m4keth_fl4g} * CTF{C0nstra9nts_m4keth_nl4g} * CTF{C0nstra9nws_m4keth_fl4g} * CTF{C0nstra9nws_m4keth_nl4g} …we finally conclude that the flag is the SHA-256 of C0nstra1nts_m4keth_fl4g. # BackdoorCTF16 – Baby Worth 200 points, this challenge was presented with the following: z3r0c00l has a safe repository of files. The filename is signed using z3r0c00l’s private key (using the PKCS-1 standard). Anyone willing to read a file, has to ask for a signature from z3r0c00l. But z3r0c00l is currently unavailable. Can you still access a file named “flag” on z3rc00l’s repository? nc hack.bckdr.in 9001 Let us take a look at the public key… 3072 bits and public exponent $e = 3$. Hmm… having a small exponent is usually not a good practice. First, I tried computing the roots to $x^3 - s \bmod n$, where $s$ is the signature and $n$ is the modulus, but then I realized that this was not the way to go. What if we use non-modular squareroot, plain old Babylonian style? After looking around, I also realized that this is Bleicherbacher’s $e = 3$ attack, which I probably should have known about. There is a lot of information about this attack (therefore, I will not describe it here) and, of course, lots of people have already written code for this. Being lazy/efficient, I rewrote a functional code into the the following: from libnum import * from gmpy2 import mpz, iroot, powmod, mul, t_mod import hashlib, binascii, rsa, os def get_bit(n, b): """ Returns the b-th rightmost bit of n """ return ((1 << b) & n) >> b def set_bit(n, b, x): """ Returns n with the b-th rightmost bit set to x """ if x == 0: return ~(1 << b) & n if x == 1: return (1 << b) | n def cube_root(n): return int(iroot(mpz(n), 3)[0]) snelhest = hashlib.sha256('flag') ASN1_blob = rsa.pkcs1.HASH_ASN1['SHA-256'] suffix = b'\x00' + ASN1_blob + snelhest.digest() sig_suffix = 1 for b in range(len(suffix)*8): if get_bit(sig_suffix ** 3, b) != get_bit(s2n(suffix), b): sig_suffix = set_bit(sig_suffix, b, 1) while True: prefix = b'\x00\x01' + os.urandom(3072//8 - 2) sig_prefix = n2s(cube_root(s2n(prefix)))[:-len(suffix)] + b'\x00' * len(suffix) sig = sig_prefix[:-len(suffix)] + n2s(sig_suffix) if b'\x00' not in n2s(s2n(sig) ** 3)[:-len(suffix)]: break print hex(s2n(sig))[2:-1] Ok, so lets try it: Great! # Defcon CTF – b3s23 (partial?) The server runs a program (game of life) which has a $110 \times 110$ board with cells (bits). After a fixed number $n$ of iterations, the simulation stops and the program jumps to the first bit of the memory containing the board. We want to create an input which contains shellcode in this area after $n$ iterations. Obviously, we could choose any shellcode, and run game of life backwards. Cool, let us do that then! Uh-oh, inverting game of life is in fact a very hard problem… so it is not really feasible 😦 What to do, then? Game of life Game of life a cellular automata, found by Jon Conway, and is based on the following rules: 1. A cell is born if it has exactly 3 neighbours. Neighbors are defined as adjacent cells in vertical, horistontal and diagonal. 2. A cell dies if it has less than two or more than three neighbors. Stable code (still life) Still life consists of cell structures with repeating cycles having period 1. Here are the building blocks I used to craft the shellcode. Of course, the still life is invariant of rotation and mirroring. Shellcode So, I tried to find the shortest shellcode that would fit one line (110 bits). This one is 8 bytes. Great. 08048334 <main>: 8048334: 99 cltd 8048335: 6a 0b push$0xb
8048337: 58 pop %eax
8048338: 60 pusha
8048339: 59 pop %ecx
804833a: cd 80 int \$0x80
In binary, this translates to:
000001101001100101101010000010110101100001100000010110011100110110000000
Ok, so we note that
110101 ... 01110
cannot be constructed by our building blocks (there most certainly exist such blocks, but I didn’t consider them). So, I use a padding trick. By inserting an operation which does nothing specific
10110011 00000000 mov bl,0x0
we are able to use the blocks given in previous section. This Python code gives the binary (still-life-solvable) sequence:
from pwn import *
binary_data = ''.join([bin(ord(opcode))[2:].zfill(8) for opcode in shellcode])
context(arch = 'i386', os = 'linux')
print disasm(shellcode)
print binary_data[0:110]
which is
0000011010011001101100110000000001101010000010110101100001100000010110011100110110000000
The following cellular automata is stable, and the first line contains our exploit:
As can be seen in the animation below, we have found a still-life shellcode.
When feeding it to the program, we find that it remains in memory after any number of iterations:
Nice! Unfortunately, the code did not give me a shell, but at least the intended code was executed. I had a lot of fun progressing this far 🙂
# TU CTF – Secure Auth
This was a 150 point challenge with the description:
We have set up this fancy automatic signing server!
We also uses RSA authentication, so it’s super secure!
nc 104.196.116.248 54321
Connecting to the service, we get the following
Obviously, we cannot feed the message get_your_hands_off_my_RSA! to the oracle. So, we will only receive signatures, but no way to verify them; this means we don’t know either the public modulus, nor the public exponent. But, of course, we could guess the public exponent… there are a few standard ones: $3, 17, 65537...$
First, I obtained the signatures for $3$ and $4$ from the provided service. Denote these $s_3, s_4$, respectively. We note that given a correct public exponent $e$, we may compute $s_3^e = 3 + k \cdot N$ and $s_4^e = 4 + l \cdot N$. Inevitably, $\textnormal{gcd}(s_3^e-3,s_4^e-4) = \textnormal{gcd}(k,l)\cdot N$. Hoping for $\textnormal{gcd}(k,l)$ to be small, we can use serveral pairs until we find one that works.
Trying all the listed (guessed) public exponents, we find that $e = 65537$ (this was performed surprisingly fast in Sage with my Intel hexacore). Hence, we have now determined the modulus
$\begin{array}{rl} N = & 24690625680063774371747714092931245796723840632401231916590850908498671935961736 \\ &33219586206053668802164006738610883420227518982289859959446363584099676102569045 \\ &62633701460161141560106197695689059405723178428951147009495321340395974754631827 \\ &95837468991755433866386124620786221838783092089725622611582198259472856998222335 \\ &23640841676931602657793593386155635808207524548748082853989358074360679350816769 \\ &05321318936256004057148201070503597448648411260389296384266138763684110173009876\\ &82339192115588614533886473808385041303878518137898225847735216970008990188644891 \\ &634667174415391598670430735870182014445537116749235017327.\end{array}$
Now, note that
libnum.strings.s2n('get_your_hands_off_my_RSA!') % 3 == 0
OK, so we may split this message $m$ into a product of two message factors: $m_1 = 3$ and $m_2 = 166151459290300546021127823915547539196280244544484032717734177$ and sign them. Then, we compute the final signature $s = m^d = (m_1 \cdot m_2)^d = m_1^d \cdot m_2^d = s_1 \cdot s_2 \bmod N$. Mhm, so what now?
Phew 🙂
# TU CTF – Hash’n’bake
This challenge, worth 200 points, exhibits a trivial (and, obviously, non-secure) hash function with the objective to find a keyed hash. The description:
A random goat from Boston hashed our password!
Can you find the full output?
The hash function is defined as:
def to_bits(length, N):
return [int(i) for i in bin(N)[2:].zfill(length)]
def from_bits(N):
return int("".join(str(i) for i in N), 2)
CONST2 = to_bits(65, (2**64) + 0x1fe67c76d13735f9)
def hash_n_bake(mesg):
mesg += CONST
shift = 0
while shift < len(mesg) - 64:
if mesg[shift]:
for i in range(65):
mesg[shift + i] ^= CONST2[i]
shift += 1
return mesg[-64:]
def xor(x, y):
return [g ^ h for (g, h) in zip(x, y)]
The following computations will give the hash
PLAIN_1 = "goatscrt"
PLAIN_2 = "tu_ctf??"
def str_to_bits(s):
return [b for i in s for b in to_bits(8, ord(i))]
def bits_to_hex(b):
return hex(from_bits(b)).rstrip("L")
if __name__ == "__main__":
with open("key.txt") as f:
print PLAIN_1, "=>", bits_to_hex(hash_n_bake(xor(KEY, str_to_bits(PLAIN_1))))
print "TUCTF{" + bits_to_hex(hash_n_bake(xor(KEY, str_to_bits(PLAIN_2)))) + "}"
# Output
# goatscrt => 0xfaae6f053234c939
# TUCTF{****REDACTED****}
So, the problem is: we need to compute the hash without knowing the key (or brute forcing it). The first observation we make is that the hash function is a truncated affine function, i.e., $h(m) = f((m \cdot 2^{64} \oplus \texttt{CONST})\cdot \texttt{CONST}_2)$, with $f(a \oplus b) = f(a) \oplus f(b)$ . There is a quite simple relation emerging: $h(k \oplus m) = h(k) \oplus h(m) \oplus h(0)$ (note: $h(0)$ denotes empty input here). Using this relation, we can do the following. We know $h(k \oplus m_1)$ and $h(m_2)$ and want to determine $h(k \oplus m_2)$. Consider the following relation:
$\begin{array}{rl} h(k \oplus m_2) = & h(k) \oplus h(m_2) \oplus h(0) \\ = & h(k) \oplus h(m_1) \oplus h(0) \oplus h(m_1) \oplus h(m_2) \phantom{\bigg(} \\ = & h(k \oplus m_1) \oplus h(m_1) \oplus h(m_2). \end{array}$
All terms on the last line of the above equation are known. So, we can easily compute the hash, even without knowing the key. Ha-ha! Computing the above relation using Python can be done in the following manner:
xor(xor(to_bits(64, 0xfaae6f053234c939), hash_n_bake(str_to_bits(PLAIN_1))),hash_n_bake(str_to_bits(PLAIN_2)))
This gives the flag TUCTF{0xf38d506b748fc67}. Sweet 🙂
# TU CTF – Pet Padding Inc.
A web challenge worth 150 points, with description
We believe a rouge whale stole some data from us and hid it on this website. Can you tell us what it stole?
http://104.196.60.112/
Visiting the site, we see that there is a cookie youCantDecryptThis. Alright… lets try to fiddle with it.
We run the following command
curl -v --cookie "youCantDecryptThis=aaaa" http://104.196.60.112/
and we observe that there is an error which is not present compared to
when running it with the correct cookie is set, i.e.,
curl -v --cookie "youCantDecryptThis=0KL1bnXgmJR0tGZ/E++cSDMV1ChIlhHyVGm36/k8UV/3rmgcXq/rLA==" http://104.196.60.112/
Clearly, this is a padding error (actually, there is an explicit padding error warning but it is not shown by curl). OK, so decryption can be done by a simple padding oracle attack. This attack is rather simple to implement (basically, use the relation $P_i = D_K(C_i) \oplus C_{i-1}$ and the definition of PCKS padding, see the wikipedia page for a better explanation), but I decided to use PadBuster. The following (modified example) code finds the decryption:
class PadBuster(PaddingOracle):
def __init__(self, **kwargs):
self.session = requests.Session()
self.wait = kwargs.get('wait', 2.0)
def oracle(self, data, **kwargs):
while 1:
try:
response = self.session.get('http://104.196.60.112',
stream=False, timeout=5, verify=False)
break
except (socket.error, requests.exceptions.RequestException):
logging.exception('Retrying request in %.2f seconds...', self.wait)
time.sleep(self.wait)
continue
self.history.append(response)
return
if __name__ == '__main__':
import logging
import sys
logging.basicConfig(level=logging.DEBUG)
The decrypted flag we get is TUCTF{p4dding_bec4use_5ize_m4tt3rs}!
# Byte-wise decryption of AES-CBC
The journey continues. This time, I am looking at AES-CBC. Consider the scenario where we have an encryption oracle such that outputs ciphertexts $\Pi(s) = \textsf{Enc}_\text{AES}(s \| d, k) \rightarrow c$, where $\|$ denotes concatenation. The oracle uses PKCS7 padding and encrypts with fixed but unknown key $k$ and IV. The CBC-mode of operation is defined as follows:
Embodied in Python code, the oracle is
from Crypto.Cipher import AES
import base64
import random
plaintext = b'''This is an encrypted message.'''
key = None
cprefix = None
iv = None
def randbytes(k):
missing = abs(len(s) - (len(s)/blksize+1) * blksize)
return s+(chr(missing)*missing)
def encryption_oracle(s):
global key
global iv
if key is None:
key = randbytes(16)
if iv is None:
iv = randbytes(16)
cipher = AES.new(key, AES.MODE_CBC, IV=iv)
s = pkcs7pad(bytes(s) + plaintext, 16)
return cipher.encrypt(s)
So, for instance, when querying the oracle with $\Pi(\texttt{We can break this!})$, we obtain the encryption of $\texttt{We can break this!This is an encrypted message.}$ We note the following:
1. If the input is a fixed string, say, $q = \texttt{aaaaaaaaaaaaaaa}$, with length 15 bytes (blocklength-1), we will include exactly one byte from the (unknown) plaintext $d$. This block will be independent of the subsequent unknown plaintext bytes. For the sake of simplicity, assume that $IV= \texttt{0x0}$.
2. So, if we query the oracle for all 256 strings $\texttt{aaaaaaaaaaaaaaa}\|x$, where $x$ runs over all possible bytes, we can compute a look-up table.
3. We may then look at our first encryption and, using the look-up table from before, see which $x$ that satisfies the first block of $\Pi(q)$. This determines the first byte of $d$.
4. This may repeated in several steps, until we find all bytes in the first block. Let us return to the IV issue implicitly stated earlier. We do not know the IV, but does that really matter? No! The IV will be constant in the first block and, thus, the input string fed into it. Now, as long as we can control the IV in the coming block (keep it constant), it has no effect on the result.
5. Using the same padding methodology as before, we may now compute the bytes of the coming block, but looking at the next block.
Again, with Python as our weapon of choice, we impale the decryption problem as follows
def attack(blocksize, known):
index = len(known) / blocksize
prefix = 'a'*(blocksize-len(known)%blocksize-1)
lookup = {}
for char in range(0, 256):
lookup[encryption_oracle(prefix+known+chr(char))[index*16:index*16+16]] = chr(char)
substring = encryption_oracle(prefix)[16*index:16*index+16]
if lookup.has_key(substring):
return lookup[substring]
else:
return None
Running this until we obtain a None-flag, i.e.,
plain = ''
while attack(16, plain) != None:
plain += attack(16, plain)
print 'Found plaintext:\n', plain
we obtain the sought plaintext $\texttt{This is an encrypted message.}$! | {"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": 96, "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": 0.8306084275245667, "perplexity": 3171.5078260785367}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056548.77/warc/CC-MAIN-20210918154248-20210918184248-00152.warc.gz"} |
https://forum.allaboutcircuits.com/threads/volumes-of-revolution-help.117918/ | # volumes of revolution ...help!!
#### oussama123443
Joined Nov 2, 2014
27
hey guys...!! what is the volume of the solid when rotating the areas between y=(x^3)-x and the x-axis around the line x=1 ???
#### djsfantasi
Joined Apr 11, 2010
7,693
Hey Oussama,
Does the solid extend into infinity? At what points along $$y=x^3-x$$ and $$x=0$$ (if at all) does the solid end?
#### oussama123443
Joined Nov 2, 2014
27
no boundaries were given for the integral , the function has x=1 and x=-1 as roots so i'm trying to figure if the shape generated will be from " -1 to 1"
#### WBahn
Joined Mar 31, 2012
26,398
hey guys...!! what is the volume of the solid when rotating the areas between y=(x^3)-x and the x-axis around the line x=1 ???
The bounds are definitely important and they are not at all obvious.
no boundaries were given for the integral , the function has x=1 and x=-1 as roots so i'm trying to figure if the shape generated will be from " -1 to 1"
What about the root at x=0?
I would recommend giving your answer in three parts. First, figure out the solid of revolution between each adjacent pairs of roots and then also give the sum of them. In addition, provide an explanation why you are using those limits as opposed to something even broader.
#### WBahn
Joined Mar 31, 2012
26,398
MOD NOTE: Moved to Homework Help from Math. | {"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": 0.8336693048477173, "perplexity": 1064.6917752373708}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487643703.56/warc/CC-MAIN-20210619051239-20210619081239-00202.warc.gz"} |
https://mothur.org/wiki/read.otu/ | NOTE: Starting with mothur version 1.18.0 the read.otu command no longer exists. You can enter your file names directly into the commands that use them. For example:
mothur > read.otu(list=98_sq_phylip_amazon.fn.list)
mothur > collect.single(list=98_sq_phylip_amazon.fn.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": 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": 0.8489233255386353, "perplexity": 1461.589584309087}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991378.48/warc/CC-MAIN-20210515070344-20210515100344-00270.warc.gz"} |
https://en.wikipedia.org/wiki/Leapfrog_integration | # Leapfrog integration
In mathematics leapfrog integration is a method for numerically integrating differential equations of the form
${\displaystyle {\ddot {x}}=d^{2}x/dt^{2}=F(x)}$,
or equivalently of the form
${\displaystyle {\dot {v}}=dv/dt=F(x),\;{\dot {x}}=dx/dt=v}$,
particularly in the case of a dynamical system of classical mechanics.
The method is known by different names in different disciplines. In particular, it is similar to the velocity Verlet method, which is a variant of Verlet integration. Leapfrog integration is equivalent to updating positions ${\displaystyle x(t)}$ and velocities ${\displaystyle v(t)={\dot {x}}(t)}$ at interleaved time points, staggered in such a way that they "leapfrog" over each other.
Leapfrog integration is a second-order method, in contrast to Euler integration, which is only first-order, yet requires the same number of function evaluations per step. Unlike Euler integration, it is stable for oscillatory motion, as long as the time-step ${\displaystyle \Delta t}$ is constant, and ${\displaystyle \Delta t\leq 2/\omega }$.[1]
In leapfrog integration, the equations for updating position and velocity are
{\displaystyle {\begin{aligned}x_{i}&=x_{i-1}+v_{i-1/2}\,\Delta t,\\a_{i}&=F(x_{i}),\\v_{i+1/2}&=v_{i-1/2}+a_{i}\,\Delta t,\end{aligned}}}
where ${\displaystyle x_{i}}$ is position at step ${\displaystyle i}$, ${\displaystyle v_{i+1/2\,}}$ is the velocity, or first derivative of ${\displaystyle x}$, at step ${\displaystyle i+1/2\,}$, ${\displaystyle a_{i}=F(x_{i})}$ is the acceleration, or second derivative of ${\displaystyle x}$, at step ${\displaystyle i}$, and ${\displaystyle \Delta t}$ is the size of each time step. These equations can be expressed in a form that gives velocity at integer steps as well:[2]
{\displaystyle {\begin{aligned}x_{i+1}&=x_{i}+v_{i}\,\Delta t+{\tfrac {1}{2}}\,a_{i}\,\Delta t^{\,2},\\v_{i+1}&=v_{i}+{\tfrac {1}{2}}(a_{i}+a_{i+1})\,\Delta t.\end{aligned}}}
However, even in this synchronized form, the time-step ${\displaystyle \Delta t}$ must be constant to maintain stability.[3]
One use of this equation is in gravity simulations, since in that case the acceleration depends only on the positions of the gravitating masses (and not on their velocities), although higher-order integrators (such as Runge–Kutta methods) are more frequently used.
There are two primary strengths to leapfrog integration when applied to mechanics problems. The first is the time-reversibility of the Leapfrog method. One can integrate forward n steps, and then reverse the direction of integration and integrate backwards n steps to arrive at the same starting position. The second strength is its symplectic nature, which implies that it conserves the (slightly modified) energy of dynamical systems. This is especially useful when computing orbital dynamics, as many other integration schemes, such as the (order-4) Runge-Kutta method, do not conserve energy and allow the system to drift substantially over time.
Because of its time-reversibility, and because it is a symplectic integrator, leapfrog integration is also used in Hamiltonian Monte Carlo, a method for drawing random samples from a probability distribution whose overall normalization is unknown.[4] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 18, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9558408260345459, "perplexity": 543.7316621255588}, "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/1469257823670.44/warc/CC-MAIN-20160723071023-00321-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/a-problem-on-linear-transformation-and-standard-basis.543563/ | # A problem on linear transformation and standard basis
1. Oct 24, 2011
Problem
Given a transformation T : P(t) -> (2t + 1)P(t) where P(t) ϵ P3
(a) Show that transformation is linear.
(b) Find the image of P(t) = 2 t^2 - 3 t^3
(c) Find the matrix of T relative to the standard basis ε = {1, t, t^2, t^3}
(d) Find the matrix of T relative to the basis β1 = {1, (1-t), (1-t)^2, (1-t)^3}
(e) Suppose [x]β1 = a column vector of 5 rows in which all entries = 2
find [x]ε
Attempt
For part a, should I show that the transformation of P(t) + transformation of Q(t) = transformation of P(t) + Q(t) ?
For part b, do I simply multiply P(t) by (2t+1) ?
I have no idea about parts c, d, e
Note: This is not a homework problem. In fact, I finished schooling years back. I need answers to these, with explanations please, to help somebody else. This thing was never in my syllabus. I had matrix but nothing related to basis.
So, could you explain to me or refer me to some web page from where I can learn this stuff? Thank you.
2. Oct 24, 2011
### Deveno
for part (a) that's PART of what you need to do. you also need to show that if you multiply P(t) by a scalar c, that T(cP(t)) = c(T(P(t)).
for part (b), yes, you apply T to the given polynomial.
(c) actually, a matrix depends on TWO bases, that of the domain, and that of the co-domain (often called "the range" although that is not 100% accurate).
T:P3 → P4, i would assume you will use the basis {1,t,t2,t3,t4} for P4.
the columns of the matrix for T will be the coefficients of T(1), T(t), T(t2), and T(t3) (constant first, in increasing order of powers of t, since that is how the basis is listed).
(d) will take a long time to answer, especially if you are shaky on what a basis is. but the general idea is this:
we define a "change of basis" matrix P that converts β1-coordinates to "standard basis coordinates". then you apply the matrix of part (c), let's call it M, then another matrix that changes ε-coordinates back to β1-coordinates (i use these terms loosely, because your not really making it clear what the 2nd basis is for the image space of T, you're not specifying ANY basis for P4, and you really need to).
QMP, which will be the matrix for T in the new basis (β1-coordinates to β1-coordinates).
for part (e) use P to change from β1-coordinates to ε-coordinates (standard coordinates).
3. Oct 24, 2011
@Deveno,
Thank you. So, are the following correct?
a) T(P(t)) = (2t+1)P(t)
T(Q(t)) = (2t+1)Q(t)
T(P(t) + Q(t)) = (2t+1)(P(t) + Q(t))
T(P(t) + Q(t)) = (2t+1)P(t) + (2t+1)Q(t)
T(P(t) + Q(t)) = T(P(t)) + T(Q(t))----------(1)
Also, T(cP(t)) = (2t+1) (c P(t)) = c (2t+1) P(t)
T(cP(t)) = c T(P(t)) ------------(2)
From (1) and (2), the transformation T is linear.
(b) T(P(t)) = (2t+1)P(t)
= (2t+1) (2 t^2 - 3 t^3)
= 4t^3 - 6t^4 + 2t^2 - 3t^3
= -6t^4 + t^3 + 2t^2
(c)
T(1) = 2t+1 = 1 + 2t + 0t^2 + 0t^3 + 0t^4
T(t) = (2t+1)(t) = 2t^2 + t = 0 + t + 2t^2 + 0t^3 + 0t^4
T(t^2) = (2t+1) (t^2) = 2t^3 + t^2 = 0 + 0t + t^2 + 2t^3 + 0t^4
T(t^3) = (2t+1)(t^3) = 2t^4 + t^3 = 0 + 0t + 0t^2 + t^3 + 2t^4
The matrix =
1 ... 0 ... 0 ... 0
2 ... 1 ... 0 ... 0
0 ... 2 ... 1 ... 0
0 ... 0 ... 2 ... 1
0 ... 0 ... 0 ... 2
Could you clarify answers of parts d and e more? I have posted the question in part d exactly as given.
Part (e) says
Suppose [x]β1 =
2
2
2
2
find [x]ε
Earlier I had said 5 rows. I just now realized there are 4 rows.
4. Oct 25, 2011
### Deveno
the thing is, it makes no sense to just specify one basis for a matrix.
a matrix is just numbers (field entries), we need the bases to tell us what those numbers stand for.
β1 is a basis for P3, what basis are we supposed to express the column entries of the matrix for T in? is it the standard basis, or {1,1-t,(1-t)2,(1-t)3,(1-t)4} do you see the problem? we'll get "different coordinates" depending on which basis we use (and a different matrix for T).
if your book doesn't say, they've left a very important piece of information out.
5. Oct 25, 2011
@Deveno, the book does not say. I typed the complete question above. So, could you answer based on what could we most likely ?
Thank you
6. Oct 25, 2011
What if the basis for P4 is given? What if it is {1, t, t^2, t^3, t^4} ?
7. Oct 25, 2011
What if the basis for P4 is given? What if it is {1, t, t^2, t^3, t^4} ?
8. Oct 25, 2011
### Deveno
if the basis for P4 is just the standard one, then you only need to calculate
the change of basis matrix P, which is:
$$\begin{bmatrix}1&1&1&1\\0&-1&-2&-3\\0&0&1&3\\0&0&0&-1\end{bmatrix}$$
and then calculate MP (where M is the matrix for T you already found).
P changes β1-coordinate vectors to their ε-coordinate form:
P(1) = P([1,0,0,0]β1) = [1,0,0,0]ε = 1+0t+0t2+t3
P((1-t)) = P([0,1,0,0]β1) = [1,-1,0,0]ε = 1-t+0t2+0t3
P((1-t)2) = P([0,0,1,0]β1) = [1,-2,-1,0]ε = 1-2t+t2+0t3
P((1-t)3) = P([0,0,0,1]β1) = [1,-3,3,-1]ε = 1-3t+3t2-t3 | {"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": 0.8723223209381104, "perplexity": 1168.348783268464}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647649.70/warc/CC-MAIN-20180321121805-20180321141805-00615.warc.gz"} |
https://www.physicsforums.com/threads/magnetic-flux-through-solenoid.591206/ | # Homework Help: Magnetic Flux through Solenoid
1. Mar 28, 2012
### tnbstudent
1. The problem statement, all variables and given/known data
A wire circle of radius 0.050 m is embedded in a solenoid of length 0.20 m with 1000 turns that carries a current of 0.50 A. If a vector that is normal to the plane of the circle makes a 40° angle with the axis of the solenoid, what is the magnetic flux through the loop?
2. Relevant equations
B (solenoid) = μ*I*n
Flux = B*A*cosθ
3. The attempt at a solution
n=1000/.2
I'm using the loops per unit length
l=.20m
r=0.05m
I=0.5A
I plugged in the values to solve for the magnetic field:
B=(4*∏*10^-7)*(0.5A)*(1000/.2)
B=0.031T
A=(∏r^2)*(l)
A=1.5e-4 m^3
I'm having some trouble visualizing the angle, but I think the angle should be 50?
Can someone point me in the right direction?
2. Mar 28, 2012
### dikmikkel
Plugged in? You have to do a flux integral.
the flux If the solenoid points alon an axis, z, the angle between the magnetic field and the circle is still 40deg. So
$\Phi = \int \vec{B}_{solenoid}\cdot d\vec{a} = \int |B_{solenoid}|\cos(\theta) da =|B_{solenoid}|\int \cos(\theta)da$
B goes out in front because it is constant over the circle. So do the surface integral yourself.
I would always suggest to do the solution algebraically first, then insert numbers, and then evaluate your result. Is it realistic and does it diverge at any point.
And always draw your situation, saves a lot of trouble.
3. Mar 28, 2012
### tnbstudent
Thanks - I'm sure you are correct, but we will not get to integrals until a later section. I would like to get the answer using the tools from this section which are the formulas I listed above.
Thanks again - I should have been more specific with my question. | {"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": 0.9541193246841431, "perplexity": 544.7191408656307}, "config": {"markdown_headings": true, "markdown_code": false, "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-2018-34/segments/1534221209755.32/warc/CC-MAIN-20180815004637-20180815024637-00090.warc.gz"} |
https://arxiver.wordpress.com/2017/03/09/alma-constraints-on-star-forming-gas-in-a-prototypical-z1-5-clumpy-galaxy-the-dearth-of-co5-4-emission-from-uv-bright-clumps-ga/ | # ALMA constraints on star-forming gas in a prototypical z=1.5 clumpy galaxy: the dearth of CO(5-4) emission from UV-bright clumps [GA]
We present deep ALMA CO(5-4) observations of a main sequence, clumpy galaxy at z=1.5 in the HUDF. Thanks to the ~0.5″ resolution of the ALMA data, we can link stellar population properties to the CO(5-4) emission on scales of a few kpc. We detect strong CO(5-4) emission from the nuclear region of the galaxy, consistent with the observed $L_{\rm IR}$-$L^{\prime}_{\rm CO(5-4)}$ correlation and indicating on-going nuclear star formation. The CO(5-4) gas component appears more concentrated than other star formation tracers or the dust distribution in this galaxy. We discuss possible implications of this difference in terms of star formation efficiency and mass build-up at the galaxy centre. Conversely, we do not detect any CO(5-4) emission from the UV-bright clumps. This might imply that clumps have a high star formation efficiency (although they do not display unusually high specific star formation rates) and are not entirely gas dominated, with gas fractions no larger than that of their host galaxy (~50%). Stellar feedback and disk instability torques funnelling gas towards the galaxy centre could contribute to the relatively low gas content. Alternatively, clumps could fall in a more standard star formation efficiency regime if their actual star-formation rates are lower than generally assumed. We find that clump star-formation rates derived with several different, plausible methods can vary by up to an order of magnitude. The lowest estimates would be compatible with a CO(5-4) non-detection even for main-sequence like values of star formation efficiency and gas content.
A. Cibinel, E. Daddi, F. Bournaud, et. al.
Thu, 9 Mar 17
12/54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8964266777038574, "perplexity": 4475.450109842522}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917118552.28/warc/CC-MAIN-20170423031158-00334-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/64526/sums-of-uncountably-many-real-numbers | # Sums of uncountably many real numbers [closed]
Suppose $S$ is an uncountable set, and $f$ is a function from $S$ to the positive real numbers. Define the sum of $f$ over $S$ to be the supremum of $\sum_{x \in N} f(x)$ as $N$ ranges over all countable subsets of $S$. Is it possible to choose $S$ and $f$ so that the sum is finite? If so, please exhibit such $S$ and $f$.
-
## closed as too localized by Andres Caicedo, Andreas Thom, Felipe Voloch, S. Carnahan♦May 10 '11 at 20:15
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.
There has to be something eluding me, why did you tag your question with the large cardinals and set theory tags? Does the following argument not work? $S=\Cup_{n\in\mathbb{N}\setminus \lbrace 0\rbrace} S_n$ where $S_n=\lbrace s\in S\mathrm{~s.t.~} f(s)>\frac{1}{n}\rbrace$, thus one of them is non denumerable and taking a denumerable subset of said $S_{n_0}$ will yield an infinite sum. – Olivier Bégassat May 10 '11 at 18:45
I changed the tags. The question is certainly not about large cardinals, nor even about set theory as that is understood on this site. – Pete L. Clark May 10 '11 at 19:04
Even though you already have your answer, I'm going to close since this particular question isn't quite what we want here. Also, it is phrased in a style suspiciously resembling a homework problem. – S. Carnahan May 10 '11 at 20:20
It looks like something from Concrete Mathematics (Knuth, et.al.). Gerhard "Ask Me About System Design" Paseman, 2011.05.10 – Gerhard Paseman May 10 '11 at 20:51
@David Roberts: the identically $0$ function does not fit the conditions imposed by the OP. (Maybe that's part of your joke; if so, okay, but I didn't get it...) – Pete L. Clark May 11 '11 at 2:33
No. $S$ is the union of the countably many sets $A_n=\{s\in S:f(s)>1/n\}$, so some $A_n$ must be infinite (in fact uncountable). Thus, your sum contains infinitely many terms all of which are at least $1/n$.
-
Actually, I just realized how to solve the problem. The answer is that it is not possible.
Suppose the sum is finite. Let $S_n$, for positive integer $n$, be the set of $x \in S$ such that $f(x) \ge \frac{1}{n}$. Then for each $n$, $S_n$ must be finite, if the sum is finite. But $S = \bigcup_n S_n$, meaning that $S$ is at most countable.
In other words, the sum of uncountably-many non-negative real numbers is finite only if all but countably many of those real numbers are $0$.
-
This is a standard result in undergraduate analysis, although it is admittedly somewhat hard to find in the standard references. The following is a very non-standard reference: see the last exercise in II.9.4 of these notes on sequences and series (see p. 69...for now; page numbers are subject to change). They occur in the context of a larger discussion on unordered summation, which is what you are looking into above. The general definition of unordered summability is a bit more complicated (it is a nice special case of convergence with respect to a net, although one needn't use the term), but in the case where the values of the "$S$-indexed sequence" are non-negative, it coincides with what you have given: see Proposition 82.
Note that this fact comes up sometimes in practice. In this math.SE question I set as a challenge to give a proof of the following fact -- there is no function $f: \mathbb{R} \rightarrow \mathbb{R}$ with a removable discontinuity at every point -- which does not use the kind of uncountable pigeonhole principle argument that you need to answer the current question. And I got a very nice answer!
- | {"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": 0.9014650583267212, "perplexity": 188.3606174853776}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644068098.37/warc/CC-MAIN-20150827025428-00049-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://www.tennisforum.com/156-aig-japan-open-tennis-championships-2002/39983-go-spanish-players.html?t-39983.html= | Go Spanish players! - TennisForum.com
Old Oct 1st, 2002, 07:54 PM Thread Starter
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Go Spanish players!
Come on Marta, Arantxa and Vivi!
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Go Girls!!
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Old Oct 1st, 2002, 08:40 PM
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Why isn't Ruano Pascual playing doubles here?
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Old Oct 1st, 2002, 10:58 PM
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cause she is so stupid. it seems she does not want to become #1.
¡¡Vamos chicas!!
CoNcHiTa ArAnChA mArTa MaRiAjO
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Old Oct 2nd, 2002, 05:44 AM
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Or is eveyone else so stupid not to convince her to play with them?
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Old Oct 2nd, 2002, 09:59 AM
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They all won in singles
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WOW ! Congrats Marta
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In order to verify that you are a human and not a spam bot, please enter the answer into the following box below based on the instructions contained in the graphic. | {"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": 0.9963878989219666, "perplexity": 1958.7221516456311}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443737911339.44/warc/CC-MAIN-20151001221831-00219-ip-10-137-6-227.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/6472/parbox-vs-minipage-differences-in-applicability/6475 | # \parbox vs. minipage: Differences in applicability
Lamport, LaTeX: A document preparation system, states on p. 104:
There are two ways to make a parbox at a given point in the text: with the \parbox command and the minipage environment. They can be used to put one or more paragraphs of text inside a picture on in a table item.
\parbox and minipage share one mandatory argument (width of the parbox) and the optional argument (vertical alignment). (The second mandatory argument of \parbox "is the text to be put in the parbox" [p. 105].) Lamport recommends the use of minipages instead of parboxes in some cases (e.g. a parbox containing a tabbing or a list-making environment), but doesn't substantiate his advice (or at least I skipped that part). Finally, from Hendrik Vogt's comment to this answer, I gather that one reason to prefer minipages is that "[y]ou don't have to wait that long for the matching closing brace".
I'm aware that the \footnote command doesn't work with \parbox; by contrast, it "puts a footnote at the bottom of the parbox produced by the [minipage] environment" (Lamport, p. 105). Are there other differences in applicability between \parbox and the minipage environment?
P.S.: Kopka and Daily, A guide to LaTeX, state on p. 89:
The text in a \parbox may not contain any of the centering, list, or other environments described in Sections 4.2 through 4.5. These may, on the other hand, appear within a minipage environment.
However, I did some tests using center, itemize and tabbing environments within a \parbox, and LaTeX did not throw error messages. Are Kopka and Daly wrong, or did I miss something?
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Perhaps I meant syntactic? – Hendrik Vogt Dec 3 '10 at 17:16
@Hendrik: Perhaps I should only quote statements I get the hang of? (But I do grasp the concept of matching braces, I swear!) – lockstep Dec 3 '10 at 17:24
I'm rather sure you fully got what I meant. I'm just not so sure which word would have been adequate where I wrote "semantically" ... – Hendrik Vogt Dec 3 '10 at 18:17
The main reason I see to use minipage over \parbox is to allow verbatim (\verb, verbatim, etc.) text inside the box (unless, of course, you also put the minipage inside a macro argument).
EDIT Here are other differences between minipage and \parbox (from the comments to Yiannis' answer and from looking at the source code of both these macros in source2e).
A first difference, as already mentioned by lockstep in his question, is in the footnote treatment: minipage handles them by putting them at the bottom of the box while footnotes are lost in a \parbox (to avoid this, you must resort to the \footnotemark/footnotetext trick):
\documentclass{article}
\begin{document}
\parbox[t]{3cm}{text\footnote{parbox footnote}}
\begin{minipage}[t]{3cm}text\footnote{minipage footnote}\end{minipage}
\end{document}
A second difference is in that minipage resets the \@listdepth counter, meaning that, inside a minipage, you don't have to worry about the list nesting level when using them. Here's an example which illustrates the point:
\documentclass{article}
\begin{document}
\begin{list}{}{}\item\begin{list}{}{}\item\begin{list}{}{}\item\begin{list}{}{}\item
\begin{list}{}{}\item\begin{list}{}{}
\item %\parbox{5cm}{\begin{list}{}{}\item \end{list}}% error
\item %\begin{minipage}{5cm}\begin{list}{}{}\item \end{list}\end{minipage}% no error
\end{list}\end{list}\end{list}\end{list}\end{list}\end{list}
\end{document}
A third difference is that minipage sets the boolean \@minipagefalse which in turn deactivates \addvspace if it's the first thing to occur inside a minipage. This means that minipage will have better spacing and allow better alignment compared to \parbox in some cases like the following (left is minipage, right is \parbox):
\documentclass{article}
\begin{document}
Pros: \begin{minipage}[t]{3cm}\begin{itemize}\item first \item second%
\end{itemize}\end{minipage}
Cons: \parbox[t]{3cm}{\begin{itemize}\item first \item second\end{itemize}}
\end{document}
-
One can (apparently, I've never tried it) use the footnote package to get sane behavior with footnotes in both \parbox and minipage. – TH. Dec 4 '10 at 22:23
@TH: the footnote package does the right thing (or did 15 years ago when i tried it), but it's pretty old now and one might expect it to interact badly with other sophisticated packages. – wasteofspace Jan 30 '12 at 19:51
@lockstep: I know it's not a common practice, but I did put those braces around environments on purpose (clearer distinction between commands and environments). Oh, and there's still one left (the line beginning with EDIT). – Philippe Goutet Sep 22 '12 at 14:55
@PhilippeGoutet I corrected the one left. If you feel strongly about the braces, please use the "rollback" feature to revert to the former version of the post, and I will refrain from editing it again. – lockstep Sep 22 '12 at 14:57
@Christian: not really, no. An additional difference, though: the values of \textwidth and \columnwidth are untouched by \parbox but changed to the {minipage} width. – Philippe Goutet Feb 27 '13 at 18:42
When your text you wish to enclose contains only a small piece of text then use \parbox. It should have nothing fancy inside such as any of the paragraph making environments. For larger pieces of text, such as those that do contain paragraph-making environments you should use a minipage environment.
There are also more subtle differences between the two such as the amount of space left on top etc. In general a minipage acts like a full page, whereas a parbox acts more like a fancy paragraph. Both of them offer very little semantic meaning - in my opinion. You can add semantic meaning by creating new environments with macro names that reflect their function in the document structure. For example you could define one of those boxes that one finds in manuals with a framed warning, a warning box using parbox and call it warning.
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@Yannis: You're right that neither \parbox nor minipage offer semantic meaning. As for "\parbox[...] should have nothing fancy inside such as any of the paragraph making environments", I'd like to hear why this is so. – lockstep Dec 3 '10 at 17:05
@lockstep It's not as if the fate of the earth depends on it, but if you do you will get all sorts of errors. If there is an answer to the why it can be found in ltboxes.dtx in the source2e (it beat me I don't really make heads or tails out of it!) but from what I can gather the parbox is just a box while the minipage is an environment based on parbox. To simplify my life, I mostly use minipage. – Yiannis Lazarides Dec 3 '10 at 18:29
@Yannis: I'd like to see an example where \parbox results in an error but minipage doesn't. (That is, an example without using verbatim stuff.) – lockstep Dec 3 '10 at 18:33
@Yannis: Thanks for pointing me to source2e. – lockstep Dec 3 '10 at 18:44
@lockstep: as you can see by looking at source2e, {minipage} resets \@listdepth whereas \parbox does not. So if you nest six \begin{list}{}{}\item...\end{list} inside each other and then use another one in a \parbox, there will be a too deeply nested error which would not appear if you had used {minipage}. – Philippe Goutet Dec 3 '10 at 19:26
If you make a new environment, you must enclose the body of the environment using lrbox first, then use the saved box in \parbox. But if you use minipage instead of \parbox, lrbox is no longer needed.
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depending on what you're doing, that's only partially true. For example, if you want to put the content of an environment inside a macro (let's say \fbox), you need the {lrbox} whether you use \parbox or {minipage}. – Philippe Goutet Dec 3 '10 at 16:49
@Philippe: You can't use \parbox macro to collect the content of an environment in any case. It is true, that you need to box the content first to use it in a macro later, but that is not the fault of minipage. – Martin Scharrer Jun 28 '11 at 13:37 | {"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": 0.8972612023353577, "perplexity": 2406.1785764699544}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802768397.103/warc/CC-MAIN-20141217075248-00072-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://tex.stackexchange.com/questions/31645/latex-color-package-wrong-spacing | # LaTeX color package: wrong spacing
I'm making a simple drawing, on which high school students have to complete the drawing.It's a drawing in R², so I also made a raster. If I use black lines for my raster, it is too dark, so I loaded the color package and changed the color to cyan. But the lines that are colored got shifted by about 1mm.
The weird thing is, if I remove the color for the horizontal lines, the vertical lines are blue but correctly placed. If I make the Horizontal lines blue, the vertical ones or shifted. The only change in code I make is commenting out the color.
\begin{picture}(82,100)
\linethickness{0.02mm}
\textcolor{cyan}
{\multiput(0,0)(5,0){17}%
{\line(0,1){100}}}
\textcolor{cyan}
{\multiput(0,0)(0,5){21}%
{\line(1,0){80}}}
\linethickness{0.3mm}
\multiput(10,9)(10,0){7}%
{\line(0,1){2}}
\multiput(19,10)(0,10){9}%
{\line(1,0){2}}
\linethickness{0.3mm}
\put(20,0){\vector(0,1){98}}
\put(0,10){\vector(1,0){78}}
\put(0,10){\line(2,1){80}}
\put(40,80){\circle*{0.6}}
\put(28,21){$r$}
\put(41,81){$P$}
\put(6,5){$-1$}
\put(16,5){$0$}
\put(29,5){$1$}
\put(39,5){$2$}
\put(49,5){$3$}
\put(59,5){$4$}
\put(69,5){$5$}
\put(74,5){$x$}
\put(16,18){$1$}
\put(16,28){$2$}
\put(16,38){$3$}
\put(16,48){$4$}
\put(16,58){$5$}
\put(16,68){$6$}
\put(16,78){$7$}
\put(16,88){$8$}
\put(16,94){$y$}
\end{picture}
\begin{picture}(82,100)
\linethickness{0.02mm}
\textcolor{cyan}
{\multiput(0,0)(5,0){17}%
{\line(0,1){100}}}
%\textcolor{cyan}
{\multiput(0,0)(0,5){21}%
{\line(1,0){80}}}
\linethickness{0.3mm}
\multiput(10,9)(10,0){7}%
{\line(0,1){2}}
\multiput(19,10)(0,10){9}%
{\line(1,0){2}}
\linethickness{0.3mm}
\put(20,0){\vector(0,1){98}}
\put(0,10){\vector(1,0){78}}
\put(0,10){\line(2,1){80}}
\put(40,80){\circle*{0.6}}
\put(28,21){$r$}
\put(41,81){$P$}
\put(6,5){$-1$}
\put(16,5){$0$}
\put(29,5){$1$}
\put(39,5){$2$}
\put(49,5){$3$}
\put(59,5){$4$}
\put(69,5){$5$}
\put(74,5){$x$}
\put(16,18){$1$}
\put(16,28){$2$}
\put(16,38){$3$}
\put(16,48){$4$}
\put(16,58){$5$}
\put(16,68){$6$}
\put(16,78){$7$}
\put(16,88){$8$}
\put(16,94){$y$}
\end{picture}
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Welcome to TeX.sx! :) – Paulo Cereda Oct 15 '11 at 10:55
Thanks, I wasn't able to add the screenshots (a missing privilege). – sanderd17 Oct 15 '11 at 11:35
You have two spurious spaces
\begin{picture}(82,100)
\linethickness{0.02mm}
\textcolor{cyan}
{\multiput(0,0)(5,0){17}%
{\line(0,1){100}}}% <<<<<<< Otherwise a space will appear
\textcolor{cyan}
{\multiput(0,0)(0,5){21}%
{\line(1,0){80}}}% <<<<<<< Otherwise a space will appear
\linethickness{0.3mm}
...
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+1. Picture environments were my first exposure to the insidious effects of spurious spaces in my LaTeX source! – Niel de Beaudrap Oct 15 '11 at 11:20
Thanks for the fast reply. It worked – sanderd17 Oct 15 '11 at 11:31 | {"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": 0.8914911150932312, "perplexity": 2481.23707144817}, "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/1448398446535.72/warc/CC-MAIN-20151124205406-00005-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://www.allaboutcircuits.com/video-tutorials/op-amp-applications/ | # Common Operational Amplifier (Op-Amp) Applications
September 13, 2020 by Robert Keim
In this video, we’ll look at some circuits that represent common op-amp applications. We’ll start with the voltage follower, then we’ll move on to an inverting amplifier, an active filter, and a current-to-voltage converter.
The term “operational amplifier” refers to a diverse category of integrated circuits that are used very frequently in analog and mixed-signal applications. The operational amplifier is undoubtedly one of the most useful and versatile components available to the electrical engineer.
These devices are relatively easy to understand and implement, and they can be incorporated into circuits ranging from the most basic analog buffer to high-order filters and complex signal generators.
### Voltage Follower
As the name implies, the voltage follower is a circuit in which the output voltage follows the input voltage. In other words, $$V_{OUT} = V_{IN}$$. As shown in the diagram below, an operational amplifier is the only required component.
The voltage follower is a good reminder that the value of operational amplifiers goes far beyond amplification. In fact, we frequently design op-amp circuits that are not intended to increase the amplitude of an input signal. In the case of the voltage follower, the desired functionality is buffering.
An op-amp makes an excellent buffer because it provides very high input impedance and very low output impedance. This is exactly what we want when the objective is to efficiently transfer a voltage signal: the high input impedance makes the voltage follower compatible with a wide variety of source circuits, and the low output impedance makes it compatible with a wide variety of load circuits.
The following list summarizes the characteristics of the op-amp voltage follower:
• Unity gain (i.e., $$V_{OUT}/V_{IN} = 1$$)
• High input impedance
• Low output impedance
• No phase inversion
### Inverting Amplifier
An operational amplifier, when considered as a standalone component, is a differential amplifier with an extremely high gain. However, we typically do not use op-amps as high-gain amplifiers. Instead, we use a negative-feedback configuration to convert the operational amplifier into a low-gain amplifier circuit in which the input-to-output relationship depends on external passive components.
The configuration shown below, called an inverting amplifier, is one of the most fundamental op-amp-based amplification circuits.
The negative-feedback action causes the gain of this circuit to be almost completely independent of the gain of the op-amp itself. Furthermore, we can precisely control the gain simply by choosing the values of the two resistors. The inverting amplifier also inverts the input signal—that is, it creates 180° of phase shift between the input and the output.
The behavior of the inverting amplifier is summarized as follows:
• Inverts and amplifies the input signal
• Gain = $$–R_F/R_1$$
• Low output impedance
• Input impedance is equal to $$R_1$$ and therefore is not necessarily high
### Active Filter
In many applications, we can adequately attenuate the high-frequency components of a signal with nothing more than a resistor and a capacitor. One way to improve upon the basic RC low-pass filter is to add a buffer, as shown in the diagram below.
This circuit is really just an RC filter that has been combined with a voltage follower in order to improve the output impedance, but it does bring us one step closer to an active filter—that is, a circuit in which the filtering action relies upon both an amplifying component and passive components.
Active filters are important because they provide an effective and convenient means of achieving the improved frequency response of a second-order filter. Engineers often work with signals in which the important frequencies are close to the frequencies that need to be suppressed, and second-order (or higher-order) filters are used to achieve a more rapid transition between the portion of the frequency response that has low attenuation and the portion of the frequency response that has high attenuation.
The diagram below shows an example of an active low-pass filter based on the widely used Sallen–Key topology.
### Current-to-Voltage Converter
Operational amplifiers are used as a simple and effective means of converting a current signal into a voltage signal. The most basic implementation, shown in the diagram below, requires only one resistor in addition to the op-amp.
The input current is applied to the inverting input terminal, and the op-amp generates an output voltage whose magnitude is equal to the current multiplied by the feedback resistance ($$R_F$$).
Perhaps the most common application of the current-to-voltage converter, also known as a transimpedance amplifier, is photodiode circuits such as the one shown below.
The photodiode generates a current that is proportional to light intensity, and consequently, the overall circuit generates a voltage signal that is proportional to light intensity.
### Summary
• Op-amps are extremely versatile and are used in a wide variety of electronic circuits.
• The voltage follower is a simple circuit that requires only an operational amplifier; it functions as an effective buffer because it has high input impedance and low output impedance.
• An inverting amplifier consists of an op-amp and two resistors. The op-amp provides the amplification, but the values of the resistors determine the gain.
• An op-amp can work in conjunction with resistors and capacitors to generate a second-order frequency response; these circuits are called active filters.
• An op-amp combined with one feedback resistor creates a circuit that accepts an input signal from a current source and produces a corresponding output voltage.
1 Comment
• K
keepitsimplestupid September 14, 2020
The current to voltage converter is stupidly drawn. Vout should be the normal -I*Rf
The inverting amplifier is always drawn the way you did, but it’s harder to understand. When it’s drawn as a divider, it’s much easier to understand where gain comes from.
Like. | {"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": 0.8208206295967102, "perplexity": 929.6064202366546}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711200.6/warc/CC-MAIN-20221207153419-20221207183419-00530.warc.gz"} |
http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples?answertab=oldest | What are your favorite instructional counterexamples?
Related: question #879, Most interesting mathematics mistake. But the intent of this question is more pedagogical.
In many branches of mathematics, it seems to me that a good counterexample can be worth just as much as a good theorem or lemma. The only branch where I think this is explicitly recognized in the literature is topology, where for example Munkres is careful to point out and discuss his favorite counterexamples in his book, and Counterexamples in Topology is quite famous. The art of coming up with counterexamples, especially minimal counterexamples, is in my mind an important one to cultivate, and perhaps it is not emphasized enough these days.
So: what are your favorite examples of counterexamples that really illuminate something about some aspect of a subject?
Bonus points if the counterexample is minimal in some sense, bonus points if you can make this sense rigorous, and extra bonus points if the counterexample was important enough to impact yours or someone else's research, especially if it was simple enough to present in an undergraduate textbook.
As usual, please limit yourself to one counterexample per answer.
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@Regenbogen - I am familiar with the proof that Selmer's curve has points everywhere locally but not globally. But that counterexample led many people to study the manner in which the Hasse Prinicple could fail. For example, there is the Brauer-Manin Obstruction. However Skorobogatov has found examples of curves with trivial Manin obstruction and everywhere local points but no global points, so the story is not finished...In my comment I was suggesting that someone more familiar with the current work might use this example. – Ben Linowitz Mar 2 '10 at 17:23
In question #14739, I asked whether the product of two ideals of a commutative ring $R$ could be defined lattice-theoretically the same way the sum and intersection can. Bjorn Poonen gave a great counterexample that shows the answer is no! This supports a point fpqc had been trying to make to me earlier that the relationship between $R$ and the Zariski topology on $\text{Spec } R$ was more subtle than I had thought: in particular, it has more structure than just the Galois connection.
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A polynomial $p(x) \in \mathbb{Z}[x]$ is irreducible if it is irreducible $\bmod l$ for some prime $l$. This is an important and useful enough sufficient criterion for irreducibility that one might wonder whether it is necessary: in other words, if $p(x)$ is irreducible, is it necessarily irreducible $\bmod l$ for some prime $l$?
The answer is no. For example, the polynomial $p(x) = x^4 + 16$ is irreducible in $\mathbb{Z}[x]$, but reducible $\bmod l$ for every prime $l$. This is because for every odd prime $l$, one of $2, -2, -1$ is a quadratic residue. In the first case, $p(x) = (x^2 + 2 \sqrt{2} x + 4)(x^2 - 2 \sqrt{2} x + 4)$. In the second case, $p(x) = (x^2 + 2 \sqrt{-2} x - 4)(x^2 - 2 \sqrt{-2} x - 4)$. In the third case, $p(x) = (x^2 + 4i)(x^2 - 4i)$. This result can be thought of as a failure of a local-global principle, and the counterexample is minimal in the sense that the answer is yes for quadratic and cubic polynomials.
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Similarly, if the units modulo n are not cyclic, then the nth cyclotomic polynomial \Phi_n(x) will be reducible mod p for all p. – Ben Linowitz Mar 2 '10 at 6:15
Even better, the polynomial $x^4-72x^2+4$ is irreducible in $\mathbb{Z}[x]$, but reducible modulo every <I>integer</I>. (Dummit and Foote, 3rd edition, page 309) – Alfonso Gracia-Saz Mar 2 '10 at 16:41
The polynomial $(x^2+31)(x^3+x+1)$ has a root modulo every prime but no roots in Q. No polynomial of degree < 5 has this property. – AVS Apr 4 '10 at 20:21
Counterexamples are very important when a student learns how to think in intuitionistic logic (and he has already been "spoiled" by classical logic). The counterexamples destroy the classical intuition, and when properly explained they help the student understand how to think intuitionistically. Some that seem to work praticularly well in my experience involve finite sets. Intuitionistically the following are not provable:
1. A subset of a finite set is finite.
2. The powerset of a finite set is finite.
3. If a subset of $\mathbb{N}$ is not finite then it is infinite.
4. The elements of a finite set may be listed without repetition.
All of these can be rescued with the additional assumption that the sets involved have decidable equality and that the subsets involved have decidable membership.
However, it does not really help the student to just know that certain "obvious" facts are not provable. He really needs to see how the "facts" can be false. The ones listed above are all false in the effective topos, but that's a complicated gadget for a beginner. It turns out informal explanations work well enough because most students know a little bit of programming. They just needs to know that the Halting Oracle does not exist.
My favorite counterexample in intuitionistic logic is that it is consistent to assume the so-called Axiom of Enumerability, which says that there are countably many countable subsets of $\mathbb{N}$. (Explanation: in the effective topos this just means that there is an effective enumeration of computably enumerable subsets of $\mathbb{N}$.) Many basic theorems of computability theory can be proved, phrased in a suitable form, from the axiom of enumerability using just constructive logic and no mention of machines of any kind.
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Which definitions of "finite set" and "infinite set" are you using? – aorq Mar 2 '10 at 14:33
A set $A$ is finite if there exists a natural number $n$ and an onto map $e : \lbrace{1, ..., n\rbrace} \to A$. A set $B$ is infinite if there exists 1-1 map $m : \mathbb{N} \to B$, i.e., $B$ contains an infinite sequence without repetitions. – Andrej Bauer Mar 2 '10 at 21:34
A counter-example in graph theory - the Petersen graph.
In many ways it is the most simple graph with many strange properties. See the article on Wiki.
Quote from our professor who teaches graph theory:
If you think you've proved any lemma about graphs, try Petersen first!
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The matrix $\left(\begin{smallmatrix}0 & 1\\ 0 & 0\end{smallmatrix}\right)$ has the following wonderful properties. (Feel free to add or edit; I can't remember all the reason I loathed it when I was learning linear algebra. It's funny how unexciting they all now seem, but it's a counterexample for almost every wrong linear algebra proof I tried to give.)
• Only zeroes as eigenvalues, but non-zero minimal polynomial (in particular, the minimal polynomial has bigger degree than the number of eigenvalues). Probably my favorite way to state this fact: the minimal polynomial is not irreducible or square-free. The same thing in a fancier language: the Jordan canonical form is not diagonal.
• Not diagonalizable, even over an algebraically closed field.
• Not divisible over $\mathbb C$. There are no matrices $M$ and integers $n\ge2$ so that $M^n = \left(\begin{smallmatrix}0 & 1\\\ 0 & 0\end{smallmatrix}\right).$ All diagonalizable and most non-diagonalizable complex matrices have $n$th roots.
(This is because, if there was a square root, it'd have minimal polynomial x4, but since it's a two-by-two matrix, Cayley-Hamilton implies that the characteristic polynomial has degree 2).
• The matrix is nilpotent but not zero.
• It's one of the best examples when you need to remember why matrix multiplication is not commutative.
• Thinking of k2 as a k[x]-module where x acts as this matrix should give wonderful (counter)-examples of modules for all the same reasons.
Also, $\left(\begin{smallmatrix}1 & 1\\ 0 & 1\end{smallmatrix}\right)$ is an example of an invertible matrix with the first three properties above. Its action on k2 is in some sense the simplest example of a representation of a group ($\mathbb{Z}$) which is indecomposable but not irreducible.
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"the minimal polynomial is not irreducible" I am not sure I get your point. Minimal polynomials are usually not irreducible. Did you mean "does not have to be multiplicity-free"? – Vladimir Dotsenko Mar 3 '10 at 14:38
The Moulton plane is a projective plane that is a counterexample to the Desargues theorem, the little Desargues theorem, and just about every "nice" property of projective planes.
Its discoverer, F.R. Moulton, is best known as an astronomer. He apparently came up with the Moulton plane after sitting in on a projective geometry course as a graduate student.
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Part of why Desargues "Theorem" is intriguing is that it holds in some projective planes and not in others. There are finite planes where it holds and finite planes where it does not hold. If there is a way to introduce coordinates for the plane with numbers from a division ring then then Desargues Theorem will hold. It also holds for projective planes sitting in higher dimensional projective spaces. In the real projective plane the theorem holds. The Moulton plane is a fascinating example. – Joseph Malkevitch Mar 3 '10 at 14:05
The blowup of $\mathbb{P}^2$ in the 9 points of intersection of two generic cubics admits infinitely many $(-1)$ curves. This example is very important in getting rid of the naif picture of algebraic surfaces.
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Even though the idea of "blowing down" is to get rid "a" (-1) curve. You can always blow down finitely many times to get rid of all of them, even if you start infinitely many. This is a nice example of that case. – Matt Apr 4 '10 at 17:38
There exists a $3$-dimensional smooth projective variety $X$ which cannot be birational to a smooth variety with nef canonical bundle. This is because $K_X$ is big; if it was also nef it would have no cohomology and we could compute its self-intersection with Riemann-Roch by looking at the number of sections of its powers. It turns out that the self-intersection would be $3/2$.
This example (by Reid, I think) shows that if you want to have minimal models you have to allow singular varieties, so that $K_X$ can still be defined, but is not a Cartier divisor. This has led to the whole branch of birational geometry studying the type of singularities which are allowed in the minimal model program, like terminal, canonical, log-terminal, KLT and so on.
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My favourite counterexample is purely academic: it does not have any applications, but I think it is pretty.
Let $X = \mathbb{N} \times \mathbb{N}$. Define a non-empty set $U \subseteq X$ to be open if for cofinitely many $x \in \mathbb{N}$ the set $\{ y \in \mathbb{N} \vert (x,y) \in U\}$ is cofinite.
Construct a sequence in $X$ that hits every point in $X$ exactly once. In other words, take a bijection $\mathbb{N} \rightarrow X$. Then:
• $X$ is countable;
• every point in $X$ is an accumulation point of this sequence, but
• the sequence has no convergent subsequences.
In particular, this is an example in a countable set that accumulation point of a sequence does not have to be a limit of a subsequence. I call this the Herreshoff topology for the (high-school) student of mine who came up with it. (I could not find it anywhere else, although I do not discard that I did not look hard enough.)
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This is sort of Arens-Fort space: en.wikipedia.org/wiki/Arens-Fort_space. Here you have the neighbourhoods of (0,0) in that space (considering (0,0) to be not in N x N), where there N x N is discrete. Your sequence then has the same properties. – Henno Brandsma Mar 6 '10 at 8:40
The Baumslag--Solitar groups have presentations
$BS(p,q)=\langle a,b\mid a^p=b^{-1}a^q b\rangle$.
They have the following nice properties:
1. they're two generator, one relator groups;
2. they can be written as an HNN extension of $\mathbb{Z}$ over $\mathbb{Z}$. (This means that they're constructed by 'gluing' $\mathbb{Z}$ to itself in some way.)
So from the point of view of combinatorial group theory, they could hardly be simpler. And yet, for suitable values of $p$ and $q$ (typically $p,q$ relatively prime integers greater than 1 will do), we find that:
1. they're non-Hopfian, meaning that they admit a self-epimorphism with non-trivial kernel;
2. hence they're not even residually finite;
3. they have exponential Dehn function (meaning that the word problem can be solved, but only very slowly);
4. their virtual first Betti number is one (meaning that every finite-index subgroup has abelianisation of rank one)...
I could go on.
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And for p != q they can't be the fundamental group of a 3-manifold -- see mathoverflow.net/questions/6132/… for some references. – Steven Sivek Mar 2 '10 at 20:32
Finite topological spaces often provide nice and simple counterexamples in topology, including algebraic topology (check J. Barmak's thesis). After getting familiar with those spaces one easily comes up with examples of phenomena such as weakly homotopy equivalent spaces which are not homotopy equivalent (spaces consisting of 4 points and 6 points suffice) or homomorphisms between homology/homotopy groups that are not induced by continuous maps.
Of course, other counterexamples are available, but finite ones are certainly minimal in a sense.
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The Poincaré homology sphere, a spherical 3-manifold with fundamental group the binary isosahedral group, was Poincaré's counterexample to the original formulation (in terms of homology) of his conjecture. Due to its countless descriptions -- as a spherical 3-manifold, via Dehn surgery, as the configuration space of an isosahedron, etc -- it's still a motivational example in geometry and topology.
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The Cantor set is a nice source of counterexamples:
The first measure zero sets you meet are usually countable. However, the Cantor set is uncountable and measure zero.
It is totally disconnected, yet it is not a discrete space. In particular, this shows that connected components of a topological space need not be open sets.
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This is a very good answer. I would go so far as to say that if you're studying general topology but haven't encountered the Cantor set, your ideas of what a topological space can be are fundamentally incomplete. – Pete L. Clark Dec 29 '10 at 6:49
“It is totally disconnected, yet it is not a discrete space.” As a professional matter, I like $\mathbb Q_p$ even better as an example of this behaviour. (Of course, topologically it's nearly the same!) If I may piggy-back on Pete's comment, if you haven't encountered $\mathbb Q_p$, then your ideas of what a complete metric space can be are fundamentally incomplete. :-) – L Spice Mar 28 '11 at 16:03
also positive measure cantor set is a very nice example to difference betwean meagre and null sets – Ostap Chervak Apr 10 '11 at 9:57
The following are, I think, the "worst possible" counterexamples in measure theory. They would benefit from a nice list of properties -- I have a feeling that I'm forgetting a lot. Feel free to improve!
The Cantor set and its friend the Cantor function are standard counterexamples. Keeps increasing regardless of the zero derivative almost everywhere... Also, the corresponding measure $\mu$, defined so that the measure of the interval [a,b] is f(a)-f(b) where f is the Cantor function is supported on a Lebesgue-zero set.
Another good source of examples is the measurable set $A \subset [0,1]$ such that for any interval I, $\lambda(I\cap A) > 0$ and $\lambda(I\cap A^c) > 0$. ($\lambda$ is the Lebesgue measure, c denotes complement).
Here's a construction of A that I heard from Ulrik Buchholtz. Instead of just constructing A, we'll make two disjoint sets A and B which have intersection of positive measure with any interval. Consider the set of all subintervals of [0, 1] with rational endpoints. It is countable, so let In be the n-th interval in the list. Put two fat (positive-measure) disjoint Cantor sets (one for A and one for B) inside I1. (We can just put the second inside some gap of the first). By the main property of Cantor sets, every interval In minus the Cantor sets is a non-empty union of intervals. So, we can put two fat disjoint Cantor sets (also disjoint from the previous ones) inside I2, and keep going forever. Every time, we add one Cantor set to A and one to B.
Now, each subinterval of [0,1] will contain one of the In-s, and therefore its intersection with both A and B has positive measure. Both A and B are countable unions of measurable sets, and therefore measurable. We are done.
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@llya The fat Cantor set is one of the great teaching examples of both analysis and topology.Most professors just go over the plain vanilla Cantor set. This is really doing the class a disservice because they don't really get the depth of the sheer diversity of pathology that can occur the real line simply by varying the details of the method of construction of the set, – Andrew L Jul 29 '10 at 18:36
The statement S "every injective endomap is also surjective" can be formalized in terms of second-order logic (and, of course, precisely states that the strcture in question is finite). This is a counterexample to any kind of compactness result for second-order logic, because if such a result existed, one would be able to get infinite sets satisfying S.
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I'd say the Tutte Graph, which is a counterexample to
Tait's conjecture: Every 3-connected cubic planar graph has a Hamiltonian cycle.
Initially, I thought this counterexample was extremely non-instructive, since I assumed that Tutte discovered it via some ingenious trial and error. But, after seeing a talk by Bill Cunningham, I discovered how Tutte came up with his counterexample and why it is a counterexample (it's unclear from looking at Tutte's graph that it is not Hamiltonian). The idea is quite simple but useful. Tutte assumed that Tait's conjecture was true and proceeded to prove a sequence of stronger (yet equivalent) conjectures. He then found a very small counterexample to the strongest conjecture, and then deconstructed the sequence of proofs to obtain a counterexample to Tait's conjecture.
I really like this method because it shows that there is hope for a mathematical caveman like me as long as I use my brain. That is, the counterexample was actually not pulled out of thin air like I initially thought.
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Very few results are pulled out of thin air. The problem is, mathematical culture has not favored showing where they do come from. – Mariano Suárez-Alvarez Mar 3 '10 at 16:56
I'd add that one of the reason for this is that journals have a strong preference for short articles, although today, with internet distribution, I do not see any reason for this. As a result, mathematicians cut out the why from the proofs, leaving only the necessary steps. But this is of course discussed in other threads... – Andrea Ferretti Mar 3 '10 at 17:09
The Weierstrass function - which I guess is a counterexample to the conjecture that a function which is continuous everywhere must be differentiable somewhere. I remember being pretty amazed when I first encountered it. It made me realize that continuity and differentiability are really different notions.
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The basic fact that there are smooth non-analytic functions on $\mathbb R$, and that there are compactly supported smooth functions, is important in real analysis and functional analysis.
$f(x) =\begin{cases} \exp(-1/(1-x^2)),& x \in (-1,1) \\\ 0& \text {otherwise} \end{cases}$
The usual examples of these functions often seem contrived. Here are examples of smooth nowhere analytic functions.
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Volterra's function has a derivative everywhere which is bounded, discontinuous, and cannot be Riemann-integrated. It depends on the Cantor sets, of course, already mentioned.
Possible reference: Bernard R. Gelbaum, John M. H. Olmsted: Counterexamples in Analysis.
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The 8-element quaternion group. It can't be reconstructed from its character table (D_4 has the same one), and every subgroup is normal but it's not abelian.
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And it has quaternionic representations - something that serves as counterexample to many beginner's conjectures in representation theory (a la "any representation can be constructed in the smallest field where its character lives"). – darij grinberg Apr 4 '10 at 15:59
I like the Sorgenfrey line. It's finer than the metric topology on R, and hereditarily Lindelöf, hereditarily separable, first countable, but not second countable. It's non-orderable, but generalised orderable, etc. It's a popular example for metrisation theorems, e.g. All its compact subsets are at most countable.
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"Every finitely-branching tree with infinitely many nodes has an infinite branch" is constructively false, as witnessed by the following counterexample:
Andrej Bauer's exposition (above) is especially nice; most textbooks take a far less direct route to the result, which makes it harder to see what's really going on past the level of "yeah, the proof is correct step-by-step."
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The Fabius function, everywhere $C^\infty$, nowhere analytic.
see... sci.math post
references:
J. Fabius, "A probabilistic example of a nowhere analytic $C^\infty$-function". Z. Wahrsch. Verw. Geb. 5 (1966) 173--174.
K. Stromberg, PROBABILITY FOR ANALYSTS (Chapman & Hall, 1994), pp. 117--120.
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Very cool! I'm going to print out the definition and bring it to every physics class I'm in from now on. :) – Vectornaut Apr 4 '10 at 22:02
The link has gone bad (takes you to "google groups"). – Victor Protsak Sep 30 '10 at 4:01
Another link (math forum): mathforum.org/kb/message.jspa?messageID=508877&tstart=0 – Gerald Edgar Nov 4 '12 at 13:56
The elliptic curve 960d1 in Cremona's tables is the smallest conductor example of an optimal elliptic curve with nontrivial Shafarevich-Tate group which is isogenous to an elliptic curve with trivial Shafarevich-Tate group.
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Any classical counter-example to inversion of a limit and an integral, $f_n:[0,1[\to\mathbb{R} ; x\mapsto n^2 x^n$ say. Basic, but important to motivate the dominated convergence theorem.
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The function $x\mapsto x^3\sin(1/x)$ has a second-order Taylor series but is not twice differentiable at $0$. The circumstances where I came across this example are too embarrassing to tell here...
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I flunked an exam in ordinary differential equations getting that one wrong-trust me,the circumstances can't be worse then that. – Andrew L Jul 29 '10 at 18:28
Here is some simple counterexample in commutative algebra, which I found really cute when I first meet it:
Let $k$ be a field, $A = k[X_{1},X_{2},X_{3}\ldots],$ $I = (X_{1}, X_{2}^{2}, X_{3}^{3},\ldots)$ and $R = A/I.$ Then $\text{Spec}(R)$ consists of one point (because $\text{rad}(I)$ is maximal ideal of $A$); in particular $\text{Spec}(R)$ is a noetherian space, and $\dim R = 0$; although $R$ is not noetherian ring (since $\text{nil}(R)^{n}\neq 0$ for every $n$).
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Nevertheless, I think that it's not obvious that there exist commutative rings with only one prime (not only with one maximal) ideal that are not noetherian. – ifk Apr 4 '10 at 21:35
@ifk: There are simpler examples of that: Consider the direct sum $R=k\oplus V$ of a field $k$ and an infinite dimensional vector space $V$, made into a ring so that $V$ is an ideal which squares to zero, $k$ and $V$ multiply as you expect, and $k$ is a subring (this is called a trivial extension, in some contexts) Then $R$ is commutative, has only one prime, and it is not noetherian. – Mariano Suárez-Alvarez Apr 5 '10 at 6:05
I've always been fond of the popcorn function (aka Thomae's Function), which is given by $f\colon \mathbb{R} \to \mathbb{R}$ via
$f(x) = \begin{cases} \frac{1}{n} & \mbox{if } x = \frac{m}{n} \in \mathbb{Q} \\ 0 & \mbox{if } x \notin \mathbb{Q}. \end{cases}$
This function has a couple of amusing properties.
(1) It is upper semicontinuous on $\mathbb{R}$, yet has a dense set of discontinuities (every one of which is removable) (namely $\mathbb{Q})$.
(2) Since it is bounded and has a set of measure zero as its set of discontinuities, it is Riemann integrable. So if we consider $g(x) = \int_0^x f(t)\ dt$, we see that $g \equiv 0$, so that $g'(x) \not \hskip 2pt = f(x)$ on a dense set.
References: http://en.wikipedia.org/wiki/Thomae%27s_function and of course "Counterexamples in Analysis" (Sec 2.15-2.17)
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The alternating group on 4 letters is nice because it provides a counterexample to the converse of Lagrange's theorem: It has order 12, but it does not have a subgroup of order 6.
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Similarly, the symmetric group on 5 letters has no subgroup of order 15, since (up to isomorphism) the only group of order 15 is the cyclic group, and $S_5$ has no element of order 15. – Gerry Myerson Mar 27 '11 at 23:35
Rotations $\rho_\alpha$ of the unit circle by an angle $2\pi\alpha$ are nice examples in the theory of discrete dynamical systems.
If $\alpha=m/n$ is rational, then every point on the circle is periodic of prime period $n$ for $\rho_\alpha$, but has no fixed points. This shows that Sharkowskii's theorem does not hold in general for functions continuous $f\colon X\to X$ if $X$ is not the real line or an interval of the real line.
If $\alpha$ is irrational, then the orbit under $\rho_\alpha$ of every point of the circle is dense, but $\rho_\alpha$ has nor sensitive dependence on initial conditions, and in particular is not caotic.
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http://web.cs.elte.hu/egres/www/tr-02-09.html | ## The generalized Kaneko theorem
### Abstract
The following problem was first studied by Kaneko: which simple, connected graphs have vertex disjoint paths of length at least 2, that altogether cover all the vertices of the graph. He gave a good characterization to the problem. Hell and Kirkpatrick observed that a possible generalization of the problem of Kaneko is to characterize those graphs which have a spanning forest such that each tree component of this forest has highest degree k for a fixed integer k \geq 1. In this paper we prove a generalization of the theorem of Kaneko in this sense, show an algorithm solving the generalized problem and prove a Berge-Tutte-type theorem on the minimum number of nodes which are missed by a tree packing.
We can solve the following problem too: given two bounds l,u: V(G) \rightarrow N, does G have a spanning subgraph having degrees at most u such that each component of this subgraph covers a vertex w the degree of which in this component is at least l(w). Many known and new results follow from this formulation.
Bibtex entry:
@techreport{egres-02-09,
AUTHOR = {Szab{\'o}, J{\'a}cint},
TITLE = {The generalized Kaneko theorem},
NOTE= {{\tt www.cs.elte.hu/egres}},
INSTITUTION = {Egerv{\'a}ry Research Group, Budapest},
YEAR = {2002},
NUMBER = {TR-2002-09}
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# Which of the following statements is not correct ?
$\begin {array} {1 1} (A)\;\text{Copper can displace silver and gold from their respective salt solutions} \\(B)\;\text{Silver can displace copper and gold from their respective salt solutions} \\ (C)\;\text{Silver can displace gold from its salt solution and not copper from its salt solution} \\ (D)\;\text{Gold cannot displace copper and silver from their respective salt solutions} \end {array}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8125695586204529, "perplexity": 347.5024191850868}, "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/1568514574765.55/warc/CC-MAIN-20190922012344-20190922034344-00469.warc.gz"} |
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