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http://physics.stackexchange.com/questions/56358/faradays-law-and-magnetic-monopoles | # Faraday's Law and magnetic monopoles
The magnetic monopoles does not exist which can be shown by $\int {\vec{B} \cdot d\vec{A}} = 0$.
But in Faraday's Law of electromagnetic induction, we clearly show the EMF induced is the time rate of change of the magnetic flux, which is $E = -\frac{d\Phi_B}{dt} = -\frac{d\int{\vec{B}\cdot d\vec{A}}} {dt}$.
Now if $\int {\vec{B} \cdot d\vec{A}} = 0$ then shouldn't the induced emf be zero?
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1
In one case you're integrating over a closed surface, in the other over an open surface. – Michael Brown Mar 9 at 10:49
1
Yep, what Michael Brown said: In a statement of Faraday's law you typically have a wire loop and integrate over the surface it bounds. In the "no monopoles statement" you integrate over a closed surface surrounding the would-be monopole. – twistor59 Mar 9 at 10:52
## 1 Answer
This can be resolved by being clear about what surface you're integrating over. In the first equation, $$\oint {\overrightarrow{B} . \overrightarrow{dA}} = 0,$$ you're integrating over any closed surface, i.e. a surface without a hole in it, such as a sphere. The equation says that the magnetic flux coming in must equal the magnetic flux going out.
But in the second equation, $$E = \frac{d}{dt} \int_\Sigma {\overrightarrow{B} . \overrightarrow{dA}},$$ you're integrating over a surface with a hole in it, where the hole is a loop of wire, as shown in this diagram from Wikipedia:
This equation says that the rate of change of flux passing through the surface must equal the EMF in the wire loop. You can imagine it as coming in through the hole, and out through the surface. (Or the other way round or a bit of both.)
You can see a connection between the two equations if you imagine making the wire loop smaller and smaller until the hole closes completely. Then you're back at a closed surface again, where the first equation applies - and this infinitely small loop of wire can never experience an EMF.
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http://math.stackexchange.com/questions/232845/how-many-ways-are-there-of-coloring-the-vertices-of-a-regular-n-gon | # How many ways are there of coloring the vertices of a regular $n$-gon
How many ways are there of coloring the vertices of a regular $n$-gon with all $p$ colors ($n,p \ge 2$), such that each vertex is given one color, and every color isn't used for two adjacent vertices?
If in a way to color not necessarily use all $p$ colors, then the answer is $a_n=(-1)^n(p-1)+(p-1)^n\;.$
If in a way to color must use all $p$ colors, then, by using include & exclude, the answer is $\sum_{k=0}^p(-1)^{p-k}\binom pk(k-1)^n\;.$
But I do not know how to use the include/exclude to get such results. Can you explain this? Thank you!
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– joriki Nov 8 '12 at 11:58
## 1 Answer
Consider the sum
$$\sum_{k=0}^p(-1)^{p-k}\binom pka_n(k)\;,$$
where $a_n(k)=(-1)^n(k-1)+(k-1)^n$ is the number of colourings with at most $k$ colours and $\binom pk$ counts the number of ways of choosing the $k$ colours among all $p$ colours. A colouring with exactly $r$ colours is counted in the terms with $k=r$ through $k=p$, and it is counted $\binom kr$ times, once for each way of choosing $r$ colours from among the $k$ colours. Thus it contributes with coefficient
$$\sum_{k=r}^p(-1)^{p-k}\binom pk\binom kr=\delta_{pr}\;.$$
Thus we are counting each colouring with exactly $p$ colours with coefficient $1$ and each colouring with exactly $r\ne p$ colours with coefficient $0$, that is, we are counting the colourings with exactly $p$ colours.
Upon substituting the result for $a_n(k)$ given in the question, the sum is
$$\sum_{k=0}^p(-1)^{p-k}\binom pk\left((-1)^n(k-1)+(k-1)^n\right)\;,$$
and the sum over the first term vanishes, leaving
$$\sum_{k=0}^p(-1)^{p-k}\binom pk(k-1)^n\;.$$
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http://math.stackexchange.com/questions/249585/is-this-an-irreducible-polynomial?answertab=active | # Is this an irreducible polynomial?
I have this polynomial:
$x^8+x^4+x^3+x+1$
and I would like to know if it is irreducible over $\mathbb{F}_q$ with $q=2^8$. My book gives me it is irreducible but matlab says it is not irreducible.
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This is a polynomial over which ring? It is (certainly!) reducible as a polynomial over $\mathbb C$, for example. – Henning Makholm Dec 2 '12 at 23:34
I edited the post... I mean over Galois Field $GF(2^8)$ – Mazzy Dec 2 '12 at 23:35
Reducibility depends on the field. Every polynomial with real coefficients can be written as a product of terms of the type $(x^2 + b x + c)$ and other terms of the type $(x + c)$ with real numbers $b,c$ and perhaps an overall multiplier $a.$ Thi, of course, is just the statement that roots occur in complex conjugate pairs. – Will Jagy Dec 2 '12 at 23:36
@WillJagy I edited the post. I mean galois finite field – Mazzy Dec 2 '12 at 23:38
## 2 Answers
Theorem. Let $P$ be any polynomial of degree $k\ge 2$ with coefficients in $\mathbb F_p$. Then $P$ is reducible over $\mathrm{GF}(p^k)$.
Proof. If $P$ is reducible over $\mathbb F_p$, then the same factorization works over $\mathrm{GF}(p^k)$. So assume $P$ is irreducible over $\mathbb F_p$. Then $\mathbb F_p[X]/(P)$ is isomorphic to $\mathrm{GF}(p^k)$, so the image of $X$ under this isomorphism is a root of $P$. Therefore $P$ has a linear factor in $\mathrm{GF}(p^k)[X]$.
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So you want say me that my polynomial is reducible. right? – Mazzy Dec 2 '12 at 23:49
according what you say instead, $1+x^2+x^3+x^4+x^8$ is reducible over GF(2^8), right? – Mazzy Dec 2 '12 at 23:52
@Mazzy: Um, yes. – Henning Makholm Dec 2 '12 at 23:53
So why matlab says me that the last one is a primitive polynomial? – Mazzy Dec 2 '12 at 23:54
1
@Mazzy: Beats me. My immediate guess would be that it's answering a different question than you think it is (such as whether the polynomial is irreducible over $\mathbb F_2$). – Henning Makholm Dec 2 '12 at 23:59
If a polynomial of degree 8 over $\mathbb{F}_q$ is irreducible over the finite field $\mathbb{F}_q$, then its splitting field is $\mathbb{F}_{q^8}$, and is thus reducible over $\mathbb{F}_{q^8}$. If a polynomial of degree 8 over $\mathbb{F}_q$ is reducible over the finite field $\mathbb{F}_q$, then it remains reducible over $\mathbb{F}_{q^8}$.
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http://math.stackexchange.com/questions/254067/machine-numbers-in-ieee-single-precision?answertab=active | # machine numbers in IEEE single precision
Is the following numbers machines numbers on the IEEE single precision system?
$10^{304}$
$2^4+2^{27}.$
What do I have to do to know whether they are machine numbers on IEEE single precision?
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I'd say that the correct statement of the question is rather: Are the following numbers representable in the IEEE single-precision format? – leonbloy Dec 9 '12 at 0:38
## 1 Answer
A number $x$ is representable in IEEE single-precision format if it can be written in the form $$x = S\cdot 2^{e}$$ for an integer $S$ between $-(2^{24}-1)$ and $+(2^{24}-1)$ and an integer $e$ between -126 and +127.
In particular, $10^{304}$ can't, because it is much too big, but $2^4+2^{27}$ can: we can factor out $2^4$ from both terms to get:
$$2^4+2^{27} = (2^{23}+1)\cdot2^4$$
Where here $S = 2^{23}+1$ and $e=+4$.
The tricky thing about the second number is that the significand ($S$ value) of an IEEE single-precision number must fit into 23 bits, and here $S = 2^{23}+1$ = `1000,00000,00000,00000,00001`, which would seem to require 24 bits. But in IEEE format, the initial `1` is not stored, and is always implicit. So we can squeeze in that extra bit.
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http://mathhelpforum.com/trigonometry/202133-xyz-coordinate-sphere-known-info-radius-latitude-longitude.html | # Thread:
1. ## XYZ Coordinate on Sphere. Known Info: Radius, latitude, longitude
Title says it all. I need to find the x,y,z coordinate on the surface of a sphere where the origin is 0,0,0 at the center of the sphere. The known information is the radius, latitude, and longitude.
2. ## Re: XYZ Coordinate on Sphere. Known Info: Radius, latitude, longitude
If $\theta$ is the longitude and $\phi$ is the latitude, the coordinate transformations are similar to if you were using spherical polar coordinate convention.
$x = R \cos{\theta} \cos{\phi}$
$y = R \cos{\theta} \sin{\phi}$
$z = R \sin{\theta}$
The difference between geographic convention (G) and spherical polar convention (SP) is in the definition of $\theta$. In G, this angle takes on the value 0 at the equator, 90 at the north pole and -90 at the south pole. In SP theta has 0 at the north pole and increases to 180 at the south pole. The difference in the transformations is that $\sin{\theta}$ and $\cos{\theta}$ are switched between conventions. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8905166387557983, "perplexity_flag": "middle"} |
http://mathoverflow.net/revisions/71691/list | ## Return to Question
5 We'll see if this succeeds in making "align" work on this site. (It's unproblematic on stackexchange.)
It follows from the law of cosines that if $a,b,c$ are the lengths of the sides of a triangle with respective opposite angles $\alpha,\beta,\gamma$, then $$a^2+b^2+c^2 = 2ab\cos\gamma + 2ac\cos\beta + 2bc\cos\alpha.$$
For a cyclic (i.e. inscribed in a circle) polygon, consider the angle "opposite" a side to be the angle between adjacent diagonals whose endpoints are those of that side (it doesn't matter which vertex of the polygon serves as the vertex of that angle). Then for a cyclic quadrilateral with sides $a,b,c,d$ and opposite angles $\alpha,\beta,\gamma,\delta$, one can show that
\begin{align} a^2+b^2+c^2+d^2 & = 2ab\cos\gamma\cos\delta + 2ac\cos\beta\cos\delta + 2ad\cos\beta\cos\gamma{} 2ad\cos\beta\cos\gamma \\ & {} +2bc\cos\alpha\cos\delta+2bd\cos\alpha\cos\gamma+2cd\cos\alpha\cos\beta$$2bc\cos\alpha\cos\delta+2bd\cos\alpha\cos\gamma+2cd\cos\alpha\cos\beta \\ & {}-4\frac{abcd}{(\text{diameter})^2} \end{align} $ ${}-4\frac{abcd}{(\text{diameter})^2}$$
And for a cyclic pentagon, with sides $a,b,c,d,e$ and respective opposite angles $\alpha,\beta,\gamma,\delta,\varepsilon$,
a^2 \begin{align} a^2 + \cdots + e^2 & = 2ab\cos\gamma\cos\delta\cos\varepsilon+\text{9 more terms} terms} \\ & {} - 4\frac{abcd}{(\text{diameter})^2}\cos\varepsilon+ \text{4 more terms}. terms}. \end{align} \$
And for a cyclic $n$-gon with sides $a_i$ and opposite angles $\alpha_i$,
\sum_{i=1}^n \begin{align} \sum_{i=1}^n a_i^2 & = \text{a sum of }\binom{n}{2}\text{ terms each with coefficient 2}{} 2} \\ & {} - \text{a sum of }\binom{n}{4}\text{ terms each with coefficient 4}{} 4} \\ & {} + \text{a sum of }\binom{n}{6}\text{ terms each with coefficient 6}{} 6} \\ & {} - \cdots \text{ and so on} $\end{align}$
The number of terms depends on $n$ and the power of the diameter on the bottom is in each case what is needed to make the term homogeneous of degree 2 in the side lengths ("dimensional correctness" if you like physicists' language), and the alternation of signs continues.
I showed this by induction. It should work for infinitely many sides, too, by taking limits. Each term would then have a product of infinitely many cosines.
My question is: Is there some reasonable geometric interpretation of the sum of squares of sides of a polygon inscribed in a circle?
Later note:The question was in part inspired by a similar situation in which there is a simple geometric interpretation of the right side of the identity. That looks like this:
\begin{align} & 2\left(\frac{abc}{\text{diameter}}\cos\delta\cos\gamma\cos\zeta\cos\eta\cdots\right) + \text{all other terms with 3 sides} \\ & {} - 4 \left( \frac{abcde}{(\text{diameter})^3} \cos\zeta\cos\eta\cdots \right) + \text{other terms with 5 sides} \\ & {} + 6 \left(\text{similarly with 7}\right) - 8(\text{similarly with 9}) + \cdots $\end{align}$
That's one side of the identity. The other is $$\text{diameter}\cdot (a\cos\alpha + b\cos\beta + c\cos\gamma + \cdots).$$ The expression on both sides is 4 times the area of the polygon. That's the simple geometric interpretation. The terms on the "other" side of the identity are signed areas of triangles with a vertex at the center of the circle. Notice that at most one of the cosines can be negative, and that happens precisely if the center of the circle is not in the interior of the polygon.
4 added 99 characters in body; deleted 99 characters in body
3 note on how the question arose; added 165 characters in body
Later note:The question was in part inspired by a similar situation in which there is a simple geometric interpretation of the right side of the identity. That looks like this:2\left(\frac{abc}{\text{diameter}}\cos\delta\cos\gamma\cos\zeta\cos\eta\cdots\right) + \text{all other terms with 3 sides}{} - 4 \left( \frac{abcde}{(\text{diameter})^3} \cos\zeta\cos\eta\cdots \right) + \text{other terms with 5 sides}{} + 6 \left(\text{similarly with 7}\right) - 8(\text{similarly with 9}) + \cdotsThat's one side of the identity. The other is\text{diameter}\cdot (a\cos\alpha + b\cos\beta + c\cos\gamma + \cdots).The expression on both sides is 4 times the area of the polygon. That's the simple geometric interpretation. The terms on the "other" side of the identity are signed areas of triangles with a vertex at the center of the circle. Notice that at most one of the cosines can be negative, and that happens precisely if the center of the circle is not in the interior of the polygon.
2 fixed a typo
It follows from the law of cosines that if $a,b,c$ are the lengths of the sides of a triangle with respective opposite angles $\alpha,\beta,\gamma$, then $$a^b+b^2+c^2 a^2+b^2+c^2 = 2ab\cos\gamma + 2ac\cos\beta + 2bc\cos\alpha.$$
For a cyclic (i.e. inscribed in a circle) polygon, consider the angle "opposite" a side to be the angle between adjacent diagonals whose endpoints are those of that side (it doesn't matter which vertex of the polygon serves as the vertex of that angle). Then for a cyclic quadrilateral with sides $a,b,c,d$ and opposite angles $\alpha,\beta,\gamma,\delta$, one can show that $$a^2+b^2+c^2+d^2 = 2ab\cos\gamma\cos\delta + 2ac\cos\beta\cos\delta + 2ad\cos\beta\cos\gamma$$ $${} +2bc\cos\alpha\cos\delta+2bd\cos\alpha\cos\gamma+2cd\cos\alpha\cos\beta$$ $${}-4\frac{abcd}{(\text{diameter})^2}$$ And for a cyclic pentagon, with sides $a,b,c,d,e$ and respective opposite angles $\alpha,\beta,\gamma,\delta,\varepsilon$, $$a^2 + \cdots + e^2 = 2ab\cos\gamma\cos\delta\cos\varepsilon+\text{9 more terms}$$ $${} - 4\frac{abcd}{(\text{diameter})^2}\cos\varepsilon+ \text{4 more terms}.$$ And for a cyclic $n$-gon with sides $a_i$ and opposite angles $\alpha_i$, $$\sum_{i=1}^n a_i^2 = \text{a sum of }\binom{n}{2}\text{ terms each with coefficient 2}$$ $${} - \text{a sum of }\binom{n}{4}\text{ terms each with coefficient 4}$$ $${} + \text{a sum of }\binom{n}{6}\text{ terms each with coefficient 6}$$ $${} - \cdots \text{ and so on}$$ The number of terms depends on $n$ and the power of the diameter on the bottom is in each case what is needed to make the term homogeneous of degree 2 in the side lengths ("dimensional correctness" if you like physicists' language), and the alternation of signs continues.
I showed this by induction. It should work for infinitely many sides, too, by taking limits. Each term would then have a product of infinitely many cosines.
My question is: Is there some reasonable geometric interpretation of the sum of squares of sides of a polygon inscribed in a circle?
1
# Geometric meaning of a trigonometric identity
It follows from the law of cosines that if $a,b,c$ are the lengths of the sides of a triangle with respective opposite angles $\alpha,\beta,\gamma$, then $$a^b+b^2+c^2 = 2ab\cos\gamma + 2ac\cos\beta + 2bc\cos\alpha.$$
For a cyclic (i.e. inscribed in a circle) polygon, consider the angle "opposite" a side to be the angle between adjacent diagonals whose endpoints are those of that side (it doesn't matter which vertex of the polygon serves as the vertex of that angle). Then for a cyclic quadrilateral with sides $a,b,c,d$ and opposite angles $\alpha,\beta,\gamma,\delta$, one can show that $$a^2+b^2+c^2+d^2 = 2ab\cos\gamma\cos\delta + 2ac\cos\beta\cos\delta + 2ad\cos\beta\cos\gamma$$ $${} +2bc\cos\alpha\cos\delta+2bd\cos\alpha\cos\gamma+2cd\cos\alpha\cos\beta$$ $${}-4\frac{abcd}{(\text{diameter})^2}$$ And for a cyclic pentagon, with sides $a,b,c,d,e$ and respective opposite angles $\alpha,\beta,\gamma,\delta,\varepsilon$, $$a^2 + \cdots + e^2 = 2ab\cos\gamma\cos\delta\cos\varepsilon+\text{9 more terms}$$ $${} - 4\frac{abcd}{(\text{diameter})^2}\cos\varepsilon+ \text{4 more terms}.$$ And for a cyclic $n$-gon with sides $a_i$ and opposite angles $\alpha_i$, $$\sum_{i=1}^n a_i^2 = \text{a sum of }\binom{n}{2}\text{ terms each with coefficient 2}$$ $${} - \text{a sum of }\binom{n}{4}\text{ terms each with coefficient 4}$$ $${} + \text{a sum of }\binom{n}{6}\text{ terms each with coefficient 6}$$ $${} - \cdots \text{ and so on}$$ The number of terms depends on $n$ and the power of the diameter on the bottom is in each case what is needed to make the term homogeneous of degree 2 in the side lengths ("dimensional correctness" if you like physicists' language), and the alternation of signs continues.
I showed this by induction. It should work for infinitely many sides, too, by taking limits. Each term would then have a product of infinitely many cosines.
My question is: Is there some reasonable geometric interpretation of the sum of squares of sides of a polygon inscribed in a circle? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "mathjax_display_tex": 30, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8860685229301453, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/20773/is-there-a-notion-of-galois-extension-for-z-p2 | ## Is there a notion of Galois extension for Z / p^2?
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The above title is in fact a special case of what I want to ask.
Certainly we have a well defined notion of Galois extension for $\mathbb{Q}_p$. The intersections of these extensions to the ring of integer of the absolute algebraic closure of $\mathbb{Q}_p$ give us a notion of Galois extensions for $\mathbb{Z}_p$. ( I know that there is a notion of Galois extension for commutative rings, and I believe that it should give us this. Am I correct?)
Let's go further. Let $A_K$ be the ring of integer in a finite Galois extension $K$ of $\mathbb{Q}_p$. Let $e$ be the ramification degree of $K$ over $\mathbb{Q}_p$. The injection of $\mathbb{Z}_p$ into $A_K$ will induce an injection of $\mathbb{Z} / p^n$ into $A_K / \mathfrak{p}^{en}$. In this picture, there seems to be some desire to say that $A_K / \mathfrak{p}^{en}$ is the correct notion Galois of extension of $\mathbb{Z} / p^n$. But there are problems; taking this notion of Galois extension, if $K$ is has ramification degree $e >1$, the corresponding extension $A_K /p^e$ is not a field (it is not even an integral domain).
Question 1: Is there any notion of Galois extensions corresponding to what I desire?
Question 2: Can a class field theory (i.e a nice description of absolute abelian Galois extension) of $\mathbb{Z}/p^n$ be developed in this context? Is there any relationship between this and the local class field theory of $\mathbb{Q}_p$ ( which is the same as that of $\mathbb{Z}_p$)?
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Re Q2. Class field theory only describes abelian extensions. – Felipe Voloch Apr 8 2010 at 20:22
Ah, yes thank you. – Tran Chieu Minh Apr 9 2010 at 1:18
I removed the LaTeX from the title as it does not display properly on the front page. – Harry Gindi Apr 9 2010 at 2:21
I have accepted the answer by dke, still I hope for an answer to question 2. – Tran Chieu Minh Apr 9 2010 at 14:00
## 2 Answers
Perhaps not directly answering your questions but something along those lines is Deligne's theory of truncated valuation rings, given in Les corps locaux de caractéristique $p$, limites de corps locaux de caractéristique 0.
A truncated valuation ring is an Artin local ring with principal maximal ideal and finite residue field - by Cohen's structure theorem, these are precisely the quotients of rings of integers in local fields by a power of the maximal ideal. Deligne sets up a category using these truncated valuation rings and provides definitions of extensions aswell as a ramification theory for them.
He goes on to show an equivalence between the category of "at most $e$-ramified" separable extensions of a local field $K$ and the category of "at most $e$-ramified" extensions of the length $e$ truncation of the ring of integers of $K$.
The main point of all this is that the behaviour of objects defined over discrete valuation rings is often determined by their reduction modulo a power of the maximal ideal i.e. on truncated data. It also ties in with Krasner's idea (hence the title of Deligne's paper) that local fields of characteristic $p$ are limits of local fields of characteristic 0 as the absolute ramification index tends to infinity.
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Thank you for the answer. This seems very promising. – Tran Chieu Minh Apr 9 2010 at 13:59
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There's the notion of Galois ring. Let $K$ be the degree $m$ unramified extension of $\mathbb{Q}_p$ and let $\mathcal{O}_K$ be its ring of integers. Then the quotient $R=\mathcal{O}_K/p^n\mathcal{O}_K$ is called the Galois ring of characteristic $p^m$ and residue field $\mathbf{F}_{p^m}$.
The Frobenius map of $\mathbb{F}_{p^m}$ lifts to an automorphism of $R$ whose fixed ring is $\mathbb{Z}/p^n\mathbb{Z}$. There is a whole book on the topic: http://www.worldscibooks.com/mathematics/5350.html .
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This is fine for the unramified extensions, i.e. those that p generates the maximal ideal. – Felipe Voloch Apr 8 2010 at 20:18
If I recall correctly, the Galois rings are precisely the finite local rings with maximal ideal generated by a prime $p$. – Robin Chapman Apr 8 2010 at 20:28
Thank you for the answer and the link. – Tran Chieu Minh Apr 9 2010 at 14:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 39, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8988168239593506, "perplexity_flag": "head"} |
http://mathhelpforum.com/math-software/176383-loglog-plot-matlab.html | # Thread:
1. ## loglog plot in MATLAB
Hi,
I was reading some notes on-line on linear multi-step methods and saw this graph:
I thought it would be good MATLAB-practice to replicate the graph and gave it a go.
To start off with, I did not include Adams-Bashfort, and just used Euler's to solve the equation $y'=y$. What I end up with is quite far from what i wanted, namely:
Not very linear... The code that includes the loglog plot is;
Code:
```yprime = @(t,y) y;
tspan = [0 1];
y0 = 1;
h = logspace(-3,0);
n = length(h);
absErrorEuler = zeros(n,1);
for i = 1:n
[teuler,yeuler] = euler(yprime,tspan,y0,h(i));
absErrorEuler(i) = abs(yeuler(end)-exp(1));
end
figure(2), clf,hold on
set(gca,'XDir','reverse')
loglog(h,absErrorEuler,'kd')
xlabel('h'), ylabel('error at t=1')```
The implementation of Euler's is;
Code:
```function [t,y] = euler(yprime, tspan, y0, h)
t0 = tspan(1); tfinal = tspan(end);
% set up the t values at which we will approximate the solution
t=[t0:h:tfinal];
% include tfinal even if h does not evenly divide tfinal-t0
if t(end)~=tfinal, t = [t tfinal]; end
% execute Euler's method
y = [y0 zeros(length(y0),length(t)-1)];
for j=1:length(t)-1
y(:,j+1) = y(:,j) + h*feval(yprime,t(j),y(:,j));
end```
Before dealing with the details of how to put $10^0,10^1...$ on the x and y axes, I need to deal with the fact that the plot should show a linear relationship.
Any suggestions are welcome.
Thanks.
2. Originally Posted by Mollier
Hi,
I was reading some notes on-line on linear multi-step methods and saw this graph:
I thought it would be good MATLAB-practice to replicate the graph and gave it a go.
To start off with, I did not include Adams-Bashfort, and just used Euler's to solve the equation $y'=y$. What I end up with is quite far from what i wanted, namely:
Not very linear... The code that includes the loglog plot is;
Code:
```yprime = @(t,y) y;
tspan = [0 1];
y0 = 1;
h = logspace(-3,0);
n = length(h);
absErrorEuler = zeros(n,1);
for i = 1:n
[teuler,yeuler] = euler(yprime,tspan,y0,h(i));
absErrorEuler(i) = abs(yeuler(end)-exp(1));
end
figure(2), clf,hold on
set(gca,'XDir','reverse')
loglog(h,absErrorEuler,'kd')
xlabel('h'), ylabel('error at t=1')```
The implementation of Euler's is;
Code:
```function [t,y] = euler(yprime, tspan, y0, h)
t0 = tspan(1); tfinal = tspan(end);
% set up the t values at which we will approximate the solution
t=[t0:h:tfinal];
% include tfinal even if h does not evenly divide tfinal-t0
if t(end)~=tfinal, t = [t tfinal]; end
% execute Euler's method
y = [y0 zeros(length(y0),length(t)-1)];
for j=1:length(t)-1
y(:,j+1) = y(:,j) + h*feval(yprime,t(j),y(:,j));
end```
Before dealing with the details of how to put $10^0,10^1...$ on the x and y axes, I need to deal with the fact that the plot should show a linear relationship.
Any suggestions are welcome.
Thanks.
I would suggest you check that you are running the right version of the script, then try using the GUI version (use the plot selector) of loglog.
CB
3. Originally Posted by CaptainBlack
I would suggest you check that you are running the right version of the script, then try using the GUI version (use the plot selector) of loglog.
CB
When I use the plot selector I get:
Running the plot selector gives loglog(h,absErrorEuler,'DisplayName','h,absErrorEu ler');figure(gcf)
in the command window, which I guess is not enough the give the scale I want. That is, when I use that command in my script, I get the "wrong" scale.
Any suggestions? | {"extraction_info": {"found_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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.947350800037384, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/42882/sub-hopf-algebras-of-group-algebras | ## Sub-Hopf algebras of group algebras
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $k$ be a field and $G$ a finite group. Is every sub-Hopf algebra over $k$ of the group algebra $k[G]$ of the form $k[U]$ for a subgroup $U$ of $G$ ?
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## 2 Answers
Yes: The dual `$H^*$` is a quotient algebra of `$k[G]$`, and the latter is the commutative function algebra `$\{f:G \to k\}$`. Thus `$H = k[U]$` for a set $U$, and closure under multiplication then makes $U$ a group.
The nice thing about finite-dimensional Hopf algebras over a field: You can turn them upside down. Vector space duality is an involution on finite-dimensional Hopf algebras. (But I think that this particular argument still works in the infinite-dimensional case: `$k[G]^*$` has a weak-* topology and `$H^*$` is a quotient by a closed ideal, so it should still create $U$.)
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Nice argument. Thanks – Ralph Oct 20 2010 at 11:44
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Moreover, finiteness of $G$ is irrelevant as Greg's proof works there as well. BTW, you don't need to go into $H^\ast$ if you are willing to use coalgebras: any subcoalgebra of $k[G]$ is $k[U]$ for some subset of $G$. $U$ must be a subgroup for the subco to be subHopf...
A more interesting question is to ask about forms of $k[G]$, i.e. Hopf algebras over a subfield $m$ such that $k\otimes_m H \cong k[G]$. There are a plenty of those which are not group subalgebras!!
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2
Forms of $k[G]$ are complementary to the work of Ahmad Chalabi, who studied the behavior of $k[G]$ under field extensions. So, better to look for those than for weapons of mass destruction. – Greg Kuperberg Oct 20 2010 at 11:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9342655539512634, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/1763/electron-transitions-in-an-infinite-square-well | Electron transitions in an infinite square well
Consider an electron in an infinite square well. The expectation values of momentum and angular momentum are all zero for energy eigenstates. An electron transition is accompanied by the emission or absorption of photons. And we know the momentum of a photon is hk and the angluar momentum (spin) is 1.
The momentum and angular momentum should be conserved in the transition process. Does that mean all electron transitions between energy levels are prohibited in such system?
If such transitions are prohibited, how do we explain light-emitting quantum dots?
===============================
Thanks for the answers and comments below. Now I realize a quantum dot is more like an atom than a hollow box. But if we consider the theoretical ideal square well, there seems to be no simultaneous eigenstates of both energy and momentum - at least I can't figure it out.
Consider a 2D quare well. The energey eigenstates can be shown as:
Energy eigenstates in a 2D Box
Apparently they are not momentum eigenstates. Since these states form a complete set, we can set a momentum eigenstate to be:
$|k_x\rangle = \sum_{n_x,n_y} C_{n_x,n_y} |n_x,n_y\rangle$
where C's are constants, and
$P_x |k_x\rangle = \hbar k_x |k_x\rangle$
Keep going on:
$P_x |k_x\rangle = -i\hbar\nabla_x \sum_{n_x,n_y} C_{n_x,n_y} |n_x,n_y\rangle = -i\hbar\nabla_x \left( C_{1,1} \sin(k_{1}x)\sin(k_{1}y)+\cdots\right)$ $= -i\hbar\left( C_{1,1} k_1 \cos(k_1 x)\sin(k_1 y)+\cdots\right)$
$=???\; \hbar k_x \left( C_{1,1} \sin(k_{1}x)\sin(k_{1}y)+\cdots\right) = \hbar k_x |k_x\rangle$
I don't know any way to transform cos*sin's to sin*sin's without messing them up.
Is there anything wrong with my calculation?
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2
A quantum dot (3D infinite square well) has states with nonzero l and m. – KennyTM Dec 9 '10 at 6:41
To elaborate on KennyTM's note: energy levels in multi-dimensional system can be hugely degenerated (because energy is no longer the only quantum number characterizing the system). For each such degenerated level you can choose more quantum numbers (this is not unique. It might be momentum, or angular momentum, or some function of the two; and for both you also have freedom in direction). If you choose angular momentum then it will in general have to distinguish between some of the degenerated states (so that it is a good quantum number). – Marek Dec 9 '10 at 12:43
To make the above a little bit more technical: if you found some basis of the Hilbert space on which some operator had only zero eigenvalues (which you claim for energy eigenstates and angular momentum operator) then it would imply that the operator itself is a zero operator. It should be obvious that this is not quite true. – Marek Dec 9 '10 at 12:46
So, what are the simultaneous eigenstates of enegey and momentum/angular momentum for such system? It's not spherical symmetric so that I don't see any easy connection to common n,l,m states in an atom. – skywaddler Dec 10 '10 at 5:07
skywaddler: I suppose that a particle in a box has cannot have a defined momentum due to the uncertainty principle $\Delta p \ge \hbar/2\Delta x$ – gigacyan Dec 10 '10 at 12:38
show 2 more comments
2 Answers
In an ideal 3-D square well the electron's movement along the three axes is independent. This is not true for a real quantum dot which has a spherical shape (close to it) and is thus more similar to hydrogen atom. As soon as there is a coupling between movement along $x$,$y$ and $z$ the electron will start orbiting around the quantum dot and will posess an angular momentum.
Now, a linear conjugated molecule can be considered as a 1-D square well. Linear carbon chains have been discovered in the interstellar space and also produced in the lab. To interact with a photon such molecule has to have a transition dipole moment which means that the electron cloud should move. If we don't consider $p$-orbitals (that would add a second dimension), only a non-symmetric molecule like $C_6N$ can have a nonzero transition moment and such molecules are in fact very strong light absorbers.
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If you're thinking about a 1-D square well, it doesn't really make sense to worry about angular momentum, since angular momentum necessarily requires multiple dimensions. When it comes to the linear momentum, there may be some subtlety regarding the meaning of expectation values. The expectation value of momentum ($\langle p \rangle$) is zero for a square-well state, but that does not mean that the particle is not moving. The expectation value of momentum squared ($\langle p^2 \rangle$) is non-zero. The conceptual explanation of this is that the particle is always in motion, but is equally likely to be found moving in either direction. Thus, when you average the momentum of many different realizations to get the expectation value, the momentum averages to zero, but when you remove the directional dependence by squaring the momentum, you get a non-zero value that corresponds to the speed of the particle.
This might allow enough wiggle room to let a photon absorption/emission conserve both energy and momentum, in that the momentum squared increases as you increase the energy state (as it must, the kinetic energy being $\frac{p^2}{2m}$). I'm not sure the numbers really work out, though.
I'll also note that when we talk about transitions in atoms at the same sort of level as used in talking about the 1-D infinite square well, we generally ignore the photon momentum-- we talk about electrons moving from one orbital to another, but the selection rules we find are based only on conservation of energy and conservation of angular momentum. The linear momentum carried by the photon is transferred to the atom as a whole, which is the basis for laser cooling, but this is generally neglected at the intro quantum physics level. So, it's not unprecedented for linear momentum to be ignored at this sort of crude level of approximation.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9236360192298889, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/69432/if-rm-is-free-and-n-is-free-is-m-otimes-n-neq-0 | # If $Rm$ is free and $N$ is free is $m \otimes n \neq 0$?
This post is a follow up to the counterexample presented in the following questions If $Rm$ is free, how do you show $m \otimes n \neq 0$?. The hope now is that we can eliminate the pathological behavior presented in the counterexamples by imposing the condition that one of the modules we are working with be free. In particular I hope the following question formulated below is sufficient to avoid the nice counter arguments cited above.
Let $R$ be a commutative ring with identity. Let $M, N$ be $R$-modules with $m \in M$ and $n\in N$ with $n \neq 0$.
If $Rm$ is free and $N$ is free is $m \otimes n \neq 0$?
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2
It suffices that N is flat. If you want a submodule of a tensor product to behave reasonably, then in general you need to tensor product with a flat module. – Jack Schmidt Oct 3 '11 at 3:51
## 1 Answer
Let $\{n_i:i\in I\}$ be a basis for $N$, and let $\{\phi_i:i\in I\}\subseteq\hom(N,R)$ be the dual basis.
Since $n\neq0$, there exists $i_0\in I$ such that $\phi_{i_0}(n)\neq0$. Now consider the composition $f:M\otimes N\to M$ of the map $\mathrm{id}_M\otimes\phi_{i_0}:M\otimes N\to M\otimes R$ with the canonical isomorphism $m\otimes r\in M\otimes R\mapsto mr\in R$.
Then $f(m\otimes n)=m\phi_{i_0}(n)\neq0$, so $m\otimes n\neq0$.
Of course, the same works if $N$ is only projective and $\{(n_i,\phi_i)\in N\times\hom(N,R):i\in I\}$ is a 'dual basis' on $N$.
More generally, the statememt works if $N$ is simply flat. Indeed, that $m$ be free in your sense means that the map $\mu:r\in R\mapsto rm\in M$ is injective so if $N$ is flat, then so is $\mu\otimes \mathrm{id}_N:R\otimes N\to M\otimes N$ is also injective. The element $1\otimes n$ of $R\otimes N$ is not zero: it follows that $m\otimes n=\mu(1\otimes n)$ is also not zero.
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@MaraianoSuarez-Alvarez Thanks for the answer. Is the map $f(m \otimes n) \neq 0$ imply $m \otimes n \neq 0$ because otherwise one of the maps in the composition would be zero? – user7980 Oct 3 '11 at 6:29
1
@user7980: if $f:X\to Y$ is a group homomorphism and $x\in X$ is such that $f(x)$ is not zero, then $x$ itself is not zero... – Mariano Suárez-Alvarez♦ Oct 3 '11 at 6:30
@MaraianoSuarez-Alvarez if $m = 0$ dosent $m \phi_{i_0} (n) = 0$ ? – user7980 Oct 3 '11 at 6:33
Is it possible to rule this case out becasuse $\{ 0 \}$ has no basis? – user7980 Oct 3 '11 at 6:38
@user7980: $\{0\}$ has a basis: the empty set... In any case, I don't understand what your two last comments come from. – Mariano Suárez-Alvarez♦ Oct 3 '11 at 6:45
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": 46, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8961090445518494, "perplexity_flag": "head"} |
http://mathhelpforum.com/advanced-statistics/171850-need-help-determing-number-test-questions-necessary.html | # Thread:
1. ## Need help determing number of test questions necessary
I need to devise a test that can state with 95% confidence that the test taker knows x percentage of a body of knowledge questions if they are tested on a sampling of the questions and score x percentage.
There are 20,000 questions. Obviously I cannot test my students on all of them, but I want to know what number of questions I need to test them on to determine with 95% confidence that they know the same percentage of the 20,000 questions as they do the questions they were tested with.
It would be great if you could show me how you determined the answer. I may need to change some of the criteria later.
2. I would say 377 questions would be best, but thats still a lot.
Look here Sample Size Calculator - Confidence Level, Confidence Interval, Sample Size, Population Size, Relevant Population - Creative Research Systems
Enter 20,000 and the desired error, I put is 5% which gives 95% $\pm$ 5% confidence.
3. Great. Good advice and helpful link. One question: How did you come up with 5% or 5 for the confidence interval? What does confidence interval really mean in this context? | {"extraction_info": {"found_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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9380115270614624, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/70556/irreducible-polynomial-of-mathrmgf216 | # Irreducible polynomial of $\mathrm{GF}(2^{16})$
I'm implementing some code for the Galois field $\mathrm{GF}(2^{16})$. Which irreducible polynomial do you recommend that I use?
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## 2 Answers
This table contains a list of primitive polynomials written in octal. The first entry in that table for $GF(2^{16})$ is $$210013_8=x^{16}+x^{12}+x^3+x+1.$$ IOW each octal digit gives 3 coefficients of the polynomial, the least significant bits correspond to lowest degree terms.
For many a hardware implementation a low weight polynomial is better - particularly for generating $m$-sequences, so you may want to support those. For reasons of compatibility / portability / independent verification on a different program you may also consider the primitive polynomial used by Matlab. I no longer have a copy on my laptop, so cannot tell, which polynomial they use.
If you decide to generate discrete logarithm tables as part of your program initialization, then you can support several primitive polynomials! The logic of the construction of the said table (and its inverse table, if judged useful) is independent from the exact form of the primitive polynomial.
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Of course, if you don't do discrete log tables, then there is no compelling reason to use a primitive polynomial. But I would recommend doing that. For implementing multiplication of two elements of the field a couple of table look-ups (and a modular addition of discrete logs) is a good approach. – Jyrki Lahtonen Oct 7 '11 at 11:35
2
– J. M. Oct 7 '11 at 11:46
I would recommend giving some thought to a different alternative. GF$(2^{16})$ is a degree-$2$ extension field of GF$(2^8)$, that is, each element of GF$(2^{16})$ can be represented as a polynomial $a_0 + a_1z$ where $a_0, a_1 \in \text{GF}(2^8)$, and so arithmetic operations in GF$(2^{16})$ become polynomial arithmetic operations in GF$(2^8)$. While the latter would take more time than the log table approach that can be used directly in GF$(2^{16})$, extensive subroutine libraries have been developed already for arithmetic in GF$(2^8)$ and you might be able to use them to save development effort and cost.
I will also suggest that the modular addition of discrete logarithms that @Jyrki has recommended be modified a little to avoid modular arithmetic, and also the unstated need to check if a multiplicand is $0$ before looking up its logarithm. By defining a suitable "value" for the logarithm of $0$, multiplication can be implemented via two table lookups, an addition, and another table lookup to get the product, without the need for testing the operands or for modular reduction following the addition. This is much faster. See here for ideas as to what can be done in this regard.
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+1 What a nice idea! (The doubled-up LUT) I know a dude who needs to do a little bit of coding sometime soon. When I needed this the most, I was battling against the 64 kByte barrier of a segment, so then this might not have been best. But that is a thing of the past, now. Of course, this won't necessarily do away with all the modular arithmetic, as sometimes one may want to compute powers using the log table - and doubling the range won't do then. But I often use a LUT of Frobenius automorphisms to aid that anyway, and then this trick will help very much. – Jyrki Lahtonen Oct 7 '11 at 13:02
And to get a multiplicative inverse you can take a suitable 2s complement of the logarithm, and offset that by a fixed amount (that I can't be bothered to compute right now...) – Jyrki Lahtonen Oct 7 '11 at 13:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9364281296730042, "perplexity_flag": "middle"} |
http://mathhelpforum.com/algebra/109476-logarithm-help-write-single-quantity.html | # Thread:
1. ## logarithm help! - write as single quantity
Hey guys! I need help with this question:
Write the expression as a logarithm of a single quantity
I know to make any of it computable in a calculator, the logs need to be in base10. but i'm not sure how to do it because of a, b & c.
2. Originally Posted by snypeshow
Hey guys! I need help with this question:
Write the expression as a logarithm of a single quantity
I know to make any of it computable in a calculator, the logs need to be in base10. but i'm not sure how to do it because of a, b & c.
Hi snypeshow,
Use these basic logarithm properties:
$\log_bx+\log_by=\log_b(xy)$
$\log_bx-\log_by=\log_b\left(\frac{x}{y}\right)$
$3\log_5a+4\log_5b-2\log_5c=(\log_5a^3+\log_5b^4)-\log_5c^2=$
$\log_5(a^3b^4)-\log_5c^2=\boxed{\log_5\left(\frac{a^3b^4}{c^2}\ri ght)}$ | {"extraction_info": {"found_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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9241526126861572, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/29000/volumes-of-sets-of-constant-width-in-high-dimensions | ## Background
The n dimensional Euclidean ball of radius 1/2 has width 1 in every direction. Namely, when you consider a pair of parallel tangent hyperplanes in any direction the distance between them is 1.
There are other sets of constant width 1. A famous one is the Reuleaux triangle in the plane. The isoperimetric inequality implies that among all sets of constant width 1 the ball has largest volume. Lets denote the Volume of the n-ball of radius 1/2 by $V_n$.
## The question
Is there some $\epsilon >0$ so that for every $n>1$ there exist a set $K_n$ of constant width 1 in dimension n whose volume satisfies $VOL(K_n) \le (1-\epsilon)^n V_n$.
This question was asked by Oded Schramm who also asked it for spherical sets of constant width r.
## A proposed construction
Here is a proposed construction (also by Schramm). It will be interesting to examine what is the asymptotic behavior of the volume. (And also what is the volume in small dimensions 3,4,...)
Start with $K_1=[-1/2,1/2]$. Given $K_n$ consider it embedded in the hyperplane of all points in $R^{n+1}$ whose (n+1)th coordinate is zero.
Let `$K^+_{n+1}$` be the set of all points $x$, with nonnegative (n+1)th coordinate, such that the ball of radius 1 with center at $x$ contains $K_n$.
Let `$K^-_{n+1}$` be the set of all points $x$, with nonpositive (n+1)th coordinate, such that $x$ belongs to the intersection of all balls of radius 1 containing $K_n$.
Let $K_{n+1}$ be the union of these two sets `$K^-_{n+1}$` and `$K^+_{n+1}$`.
## Motivation
Sets of constant width (other than the ball) are not lucky enough to serve as norms of Banach spaces and to attract the powerful Banach space theorist to study their asymptotic properties for large dimensions. But they are very exciting and this looks like a very basic question.
## References and additional motivation
In the paper: "On the volume of sets having constant width" Isr. J Math 63(1988) 178-182 Oded Schramm gives a lower bound on volumes of sets of constant width. Schramm wrote that a good way to present the volume of a set $K \subset R^n$ is to specify the radius of the ball having the same volume as $K$, called it the effective radius of the set $K$ and denote it by $er K$. Next he defined $r_n$ as the minimal effective radius of all sets having constant width two in $R^n$. Schramm proved that $r_n \ge \sqrt {3+2/(n+1)}-1$. He asked if the limit of $r_n$ exists and if $r_n$ is a monotone decreasing sequence.
Our question is essentially wheather $r_n$ tends to 1 as $n$ tends to infinity.
In the paper: O. Schramm, Illuminating sets of constant width. Mathematika 35 (1988), 180--189. Schramm proved a similar lower bound for the spherical case and deduced the best known upper bound for Borsuk's problem on covering sets with sets of smaller diameter.
-
2
Excellent question, Gil, not that I have any idea how to attack it. – Bill Johnson Jun 22 2010 at 5:48
3
The two 1988 papers of Oded Schramm mentioned in Gil's question were essentially Oded's MSc thesis written under the guidance of Gil. – Gideon Schechtman Jul 1 2010 at 10:10
3
I think Schramm's algorithm is the same as the one in lama.univ-savoie.fr/~lachand/pdfs/spheroforms.pdf ? Starting from the segment you generate Rouleaux, and then the Meissner body, and so on? According to springerlink.com/content/dn05ruk04k18l687/… Rouleaux is about 10% smaller than the disk, and the Meissner body is a bit shy of 20% smaller than the ball. From the numerics in the first paper I linked to, the 4-d one is about 23, 24% smaller than its sphere. – Willie Wong Jul 8 2010 at 15:02
3
Maybe I'm missing a word like "convex" in the statement of the problem. As written, I could dig out larger and larger holes from the middle of the radius-$1/2$ balls, and make the volumes decay however I want. – Theo Johnson-Freyd Jul 10 2010 at 21:10
1
@Theo If you made a hole, you could jam two parallel tangent planes in it that would of smaller width than the convex hull, so that wouldn't work. – Jonathan Fischoff Jul 14 2010 at 2:01
show 8 more comments | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9265073537826538, "perplexity_flag": "head"} |
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http://cms.math.ca/Reunions/hiver11/abs/st | Réunion d'hiver SMC 2011
Delta Chelsea Hotel, Toronto, 10 - 12 décembre 2011 www.smc.math.ca//Reunions/hiver11
Théorie des ensembles
Org: Ilijas Farah (York)
[PDF]
SAMUEL COSKEY, York University and The Fields Institute
Cardinal invariants and the Borel Tukey order [PDF]
Many proofs of inequalities between cardinal characteristics of the continuum are combinatorial in nature. These arguments can be carried out in any model of set theory, even a model of CH where the inequalities themselves are trivial. Thus, such arguments appear to establish a stronger relationship than a mere inequality. The Borel Tukey order was introduced by Blass in a 1996 article to address just this. Specifically, he observed that the combinatorial information linking two cardinal characteristics is often captured by a pair of Borel maps called a <em>Borel Tukey morphism</em>. The existence of a Borel Tukey morphisms between two cardinal invariants has since been found to have a couple of applications in other set-theoretic contexts. In this talk we will discuss a number of popular combinatorial cardinal invariants, and compare their traditional ordering of provable inequalities with the finer ordering given by the Borel Tukey morphisms.
SEAN COX, Universit\"at M\"unster
Catching antichains [PDF]
The notion of \emph{antichain catching} appeared in the Foreman-Magidor-Shelah paper on Martin's Maximum, and was used extensively in Woodin's proofs of the presaturation of various stationary tower forcings. For normal ideals $\mathcal{I}$ and $\mathcal{J}$, let us say that \emph{$\mathcal{J}$ catches $\mathcal{I}$} (and write $catch(\mathcal{J},\mathcal{I})$) iff $\mathcal{J}$ has sufficiently large support, the $\mathcal{J}$-positive sets project onto the $\mathcal{I}$-positive sets in a certain canonical manner (as ideals), and whenever $G \subset (\mathcal{J}^+, \subset)$ is generic then the projection of $G$ is generic for $(\mathcal{I}^+, \subset)$. Certain instances of $catch(\mathcal{J},\mathcal{I})$ are equivalent to saturation of $\mathcal{I}$ (namely when $\mathcal{J}$ is the \emph{conditional club filter relative to $\mathcal{I}$}; see Foreman's chapter in Handbook of Set Theory). But in general the statement:
\vspace{10pt}
there exists a $\mathcal{J}$ such that $catch(\mathcal{J},\mathcal{I})$''
\vspace{10pt}
is strictly weaker than saturation of $\mathcal{I}$ and strictly stronger than precipitousness of $\mathcal{I}$. I will discuss this result and others from some joint work with Martin Zeman; I will also discuss some joint work with Matteo Viale which made use of related notions.
JAMES CUMMINGS, Carnegie Mellon University
The number of normal measures [PDF]
(joint work with Arthur Apter and Katherine Thompson) We discuss some results on the set of normal measures on a measurable cardinal, some in ZFC and some in a choiceless setting.
JOHN KRUEGER, University of North Texas
Failure of Square Properties [PDF]
Square principles, which were introduced by Jensen in the context of the constructible universe L, are combinatorial properties which witness the uniform structure of canonical inner models. The construction of models in which square properties fail is an active and deep area of forcing and consistency results. We survey some of the more recent developments in this area, including the failure of partial square and good scales.
ALEKSANDRA KWIATKOWSKA, University of Illinois at Urbana-Champaign
Ample generics and the group of homeomorphisms of the Cantor set [PDF]
A topological group $G$ has ample generics if it has a comeager conjugacy class in every dimension, that is, if for every $m$ the diagonal conjugacy action of $G$ on $G^m$ has a comeager orbit. We discuss several examples and properties of groups with ample generics. Then, answering a question of Kechris and Rosendal, we show that the group of all homeomorphisms of the Cantor set has ample generics. The main tool we use is the projective Fraisse theory developed by Irwin and Solecki.
CLAUDE LAFLAMME, University of Calgary
Groups containing the automorphism group of the Rado graph [PDF]
We discuss groups of permutations containing the automorphism group of the Rado graph.
S. Thomas determined there are exactly five reducts (the closed ones), and P. Cameron and S. Tarzi investigated other natural overgroups. We review their results and answer some open questions.
This his is joint work with M. Pouzet, N. Sauer, and R. Woodrow.
JUSTIN MOORE, Cornell University
Transcending $\omega_1$-sequences of reals [PDF]
I will describe a procedure which, to each $\omega_1$-sequence of reals, assigns a sequence of tree orderings on $\omega_1$ which attempts to build a real number not in the range of the sequence. If this attempt fails, the result is a tree of countable closed subsets of $\omega_1$ which has no uncountable branch, is completely proper as a forcing notion (and remains so in any outer model with the same reals in which $T$ does not have an uncountable branch), and is self specializing'' in the sense that $$\{(s,t) \in T^2 : (\mathrm{ht}(s) = \mathrm{ht}(t)) \land (s \ne t)\}$$ can be decomposed into countably many antichains. In particular, this tree can have at most one branch in any outer model. This in particular shows that the forcing axiom for completely proper forcings is inconsistent with the Continuum Hypothesis, thus answering a longstanding problem of Shelah.
ITAY NEEMAN, UCLA
The tree property up to $\aleph_{\omega+1}$ [PDF]
The tree property is a combinatorial principle that resembles large cardinal reflection properties, but may hold at successor cardinals. We present some recent work showing, from $\omega$ supercompact cardinals, that the tree property can be forced to hold at all successor cardinals in the interval $[\aleph_2,\aleph_{\omega+1}]$. This is a further step in a general project of obtaining the tree property on increasingly large intervals of successor cardinals, and builds on work of Cummings--Foreman below $\aleph_\omega$, and work of Magidor--Shelah and Sinapova at $\aleph_{\omega+1}$.
ASSAF RINOT, Fields Institute and University of Toronto Mississauga
The extent of the failure of Ramsey's theorem at uncountable cardinals [PDF]
We study the principle $c_\kappa$, which asserts the existence of a coloring $c:[\kappa]^2\rightarrow\kappa$ with the property that $c[A\times B]=\kappa$ for every cofinal subsets $A,B$ of $\kappa$.
It is established that $c_\kappa$ holds whenever (1) $\kappa$ is the successor of a regular cardinal, (2) $\kappa$ is the successor of a singular cardinal and $\kappa\nrightarrow[\kappa]^2_\kappa$. This is joint work with Stevo Todorcevic.
GRIGOR SARGSYAN, Rutgers University
Square in Pmax extensions [PDF]
We will prove the consistency of the simultaneous failure of $\square_{\omega_2}$ and $\square(\omega_2)$ in a Pmax style extension. This considerably reduces the known large cardinal strength of the axiom needed to establish the above consistency.
DIMA SINAPOVA, UC Irvine
A hybrid Prikry type forcing [PDF]
Extender based Prikry type forcing is one of the most direct ways of violating the Singular Cardinal Hypothesis (SCH). We will describe a construction combining extender based Prikry and diagonal supercompact Prikry forcing. We will discuss its implications on singular cardinal arithmetic, and more precisely the relationship between SCH and combinatorial principles like Jensen's square and scales.
SLAWOMIR SOLECKI, University of Illinois
A proof of uniqueness of the Gurarii space [PDF]
A Gurarii space, constructed in 1965, is a separable Banach space that is universal for separable Banach spaces and is approximately homogeneous. Uniqueness up to isometry of the Gurarii space was proved by Lusky in 1976 using deep techniques developed by Lazar and Lindenstrauss. Subsequently, another proof of uniqueness was given by Henson using model theoretic methods of continuous logic. The question whether there is an elementary proof of uniqueness occurred to several mathematicians. This question was made current by recent increased interest in universal, homogeneous structures. In the talk, I will provide just such an elementary proof of isometric uniqueness of the Gurarii space.
This is a joint work with Wieslaw Kubis.
JURIS STEPRANS, York
Michael space in the Laver model [PDF]
This talk will discuss the status of Michael spaces in the Laver model.
ASGER TORNQUIST, University of Copenhagen
Conjugacy of measure preserving actions of $F_n$ [PDF]
It was recently shown by Foreman, Rudolph and Weiss that the conjugacy relation for ergodic probability measure preserving Z-actions is complete analytic. In this talk I will discuss some recent work with Inessa Epstein, where we show this also holds for ergodic p.m.p. actions of the non-amenable free groups $F_n$, $(n>1)$, and many other groups. We also obtain that orbit equivalence and von Neumann equivalence of such actions are analytic non-Borel relations.
MARTIN ZEMAN, University of California at Irvine
Universality of local core models [PDF]
Assume $M$ is a proper class inner model; denote the core model computed in $M$ by $K^M$. Under suitable anti-large cardinal hypothesis we show that if $\delta>\omega$ is a regular cardinal in $V$ such that $M$ correctly computes the its cardinal successor then $K^M||\delta$, the initial segment of $K^M$ of length $\delta$, is universal for all iterable premice in V of size strictly smaller than $\delta$. I will also discuss several variations and consequences of this fact. Perhaps the most interesting consequence is that the existence of $\delta$ as above implies that $M$ is $\Sigma^1_3$-correct. This is a joint work with Andres Caicedo.
## Commandites
Nous remercions chaleureusement ces commanditaires de leur soutien. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 60, "mathjax_display_tex": 1, "mathjax_asciimath": 2, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9076882600784302, "perplexity_flag": "head"} |
http://mathhelpforum.com/statistics/173321-balls-bag.html | # Thread:
1. ## Balls in a bag
There are 10 red balls, 10 green balls and 6 white balls.
Two balls picked at random - what is the probability that they are of different colors?
so probability of the first ball is 1/26 (if i'm right?), i don't get how to find t he probablity of the second one?
2. The probability of each ball being picked is $\frac{1}{26}$, which is a start.
The way I would solve this would be to find the probability of two red balls, the probability of two green balls, and the probability of two white balls. If you add these, you will find the probability of receiving two balls that are the same colour.
I'm going to assume (as you haven't specified) that the balls are not replaced, and that this is therefore conditional probability.
P(2 red) = $\frac{10}{26}\times\frac{9}{25} = \frac{9}{65}$. This is because there are $10$ red balls out of the $26$ total during the first selection, and then there are $9$ red balls left out of the $25$ remaining balls during the second selection.
P(2 green) is the same as p(2 red) because there are the same amount of red balls as green balls, so p(2 green) $=\frac{9}{65}$
P(2 white) = $\frac{6}{26}\times\frac{5}{25}=\frac{3}{65}$
Do you understand so far?
So what is the total probability of receiving two balls that are the same colour?
And, therefore, how can you work out from that the probability of receiving two balls which are not the same colour? Show your working if you get stuck.
3. Originally Posted by kandyfloss
There are 10 red balls, 10 green balls and 6 white balls.
Two balls picked at random - what is the probability that they are of different colors?
so probability of the first ball is 1/26 (if i'm right?), i don't get how to find t he probablity of the second one?
Draw a tree diagram.
4. So i did (1-P(sum of all of the probabilities)) and got 44/65. That seems right.
Thank you.
5. Originally Posted by kandyfloss
There are 10 red balls, 10 green balls and 6 white balls. Two balls picked at random - what is the probability that they are of different colors?
Probability both are white is $P(W_1\cap W_2)=\frac{6}{26}\cdot\frac{5}{25}$.
So find $P(R_1\cap R_2)+P(G_1\cap G_2)+P(W_1\cap W_2)$
6. Originally Posted by kandyfloss
So i did (1-P(sum of all of the probabilities)) and got 44/65. That seems right.
Thank you.
You could have tried...
What is the probability of getting
(1) a red with a green
or
(2) a red with a white
or
(3) a green with a white
There are $\binom{26}{2}$ ways to choose 2 balls. This is the denominator for your probability fraction.
There are 10(10)=100 ways to get a green with a red
There are 10(6)=60 ways to get a red with a white
There are 10(6)=60 ways to get a green with a white
The sum of the above three products is the numerator of your probability fraction.
7. I went through, both of your methods. Since I already did it using the first one, i'll stick to it.
But thanks you very much it really helped me understand better.
8. [1]Prob white : 6/26 = 3/13 ; [2]prob not white : 20/25 = 4/5 ; 3/13 * 4/5 = 12/65
[1]Prob not white: 20/26 = 10/13 ; [2]prob not same: 16/25 ; 10/13 * 16/25 = 32/65
12/65 + 32/65 = 44/65 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9572970867156982, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/84004/white-noise-in-lie-group | ## White noise in Lie group
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The matter is in the title.
Is there a means to define the white noise process in Lie group. A basic definition link text
Question: Can we replace $\mathbb{R}^n$ by a Lie group?
In fact, I would like for studying the stochastic differential equation
\begin{align} X_t^{x} &=x + \sum_{k=1}^m \int_0^t \mathcal{X}_k (X_s^{x}) dB^k_s & ( *) \end{align}
For $H > \frac{1}{2}$,
1. $\mathcal{X}_k$'s are $C^{\infty}$-bounded vector fields on $\mathbb{R}^m$ which I would like to remplace it by a lie group $G$.
2. $B$ is a $m$-dimensional fractional Brownian motion taking its values in $G$.
In Hida space $\mathcal{S}^*$, equation (*) becomes $$d X_t^{x} =\sum_{k=1}^m \mathcal{X}_k (X_t^{x})\diamondsuit B^k_t dt \qquad (**)$$ whith $X_0^x=x$. In order to exploit the underlying Lie algebra should be established white noise on the Lie group.
Thanks having taken bother to read this post.
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See Rogers and Williams, vol. 2 – Steve Huntsman Dec 21 2011 at 16:37
1
I'll second Steve Huntsman's comment. But, what is H? – George Lowther Dec 21 2011 at 17:34
$H\in ]\frac{1}{2}, 1[$ – Jonas Dec 21 2011 at 20:17
1
@alabair: ok, so that is the range of values that H can take. But, you never said what H means. Is it the Hurst index of fractional Brownian motion? So the question is a bit more complicated than just regular Brownian motion, and is not covered by Rogers and Williams. If you're asking about fractional Brownian motion on a Lie group, does that mean that you are already familiar with regular BM and stochastic differential equations on the group? Also, why specifically $H > 1/2$? Have I understood you correctly? I think the question should provide more information. – George Lowther Dec 21 2011 at 22:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.897767186164856, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/44501?sort=newest | ## What are the canonical and earliest references to trivial symmetries in gauge systems?
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I am trying to find canonical references and the history of trivial symmetries. The earliest text book reference I can find is on page 69 of Quantization of Gauge Systems by Henneaux and Teitelboim.
A trivial symmetry is a symmetry transformation of a classical mechanical or field theory system that reduces to the trivial transformation on-shell* (OS), i.e. in the classical mechanical system with action $S(q,\dot q, \dots)$ $$q \to q'=q'(q,\dot q,\dots) \quad\mathrm{st}\quad S(q',\dot q'\dots)=S(q,\dot q, \dots) \quad\mathrm{and}\quad q' \xrightarrow{OS} q$$ For infinitesimal symmetries $q \to q'=q+\delta q$ the above is written as (introducing indices $i$ for the coordinates $q$ and the summation convention) $$q^i \to q^i+\delta q^i \quad\mathrm{st}\quad \delta q^i \frac{\delta S}{\delta q^i}=0 \quad\mathrm{and}\quad \delta q^i \xrightarrow{OS} 0$$ Theorem (3.1) of Henneaux and Teitelboim says that such a transformation must be proportional* to the equation of motion $$\delta q^i = \varepsilon^{ij}\frac{\delta S}{\delta q^j}$$ where $\varepsilon^{ij}=-\varepsilon^{ji}$. This all generalises to field theories with both commuting and anticommuting fields. It is also proved in the article Symmetries and physical functions in general gauge theory by Gitman and Tyutin.
The above result means that infinitesimal trivial symmetries form an ideal in the algebra of gauge symmetries and can be basically ignored. In fact, they apparently weren't even really noticed in modern field theory until they turned up as the commutator of some non-trivial symmetries in some supergravity calculations. (This is stated without reference in Remarks on Gauge Invariance and First-Class Constraints.)
A more mathsy discussion can be found in (eg) Rigid Symmetries and Conservation Laws in Non-Lagrangian Field Theory.
So, to summarise, my questions are:
1. What is a canonical reference to be given when introducing trivial symmetries?
2. Where is theorem (3.1) of Henneaux and Teitelboimm first given?
3. What are the historical references for trivial gauge symmetries?*
EDIT - Footnotes:
• On-shell (OS) means that $q$ satisfies its equation of motion $\delta S/\delta q = 0$.
• Actually, it is normally written using a DeWitt-like condensed notation, so the index contraction actually includes an integration over time (or spacetime in field theories) - this is because there can also be terms with time derivatives of the equations of motion.
• Where were trivial symmetries first discussed in classical mechanics? Where were they first discussed in field theories? etc...
-
1
A slightly earlier, by 3 years, ref is from Henneaux slac.stanford.edu/spires/find/hep?key=2070154 I do not have it in hand; so, I do not know if it references an earlier paper. – Kelly Davis Nov 2 2010 at 10:48
@Kelly Thanks for the reference. Unfortunately, once again, the discussion seems to be given with no citations of earlier works. It might have been "common knowledge" in those communities but not put in the literature until about that time... – Simon Nov 2 2010 at 12:35
Yvette Kosmann-Schwarzbach has a book about the history of symmetries of Lagrangian systems and the Noether Theorems springer.com/mathematics/history+of+mathematics/… I don't know what it says about trivial symmetries but maybe you can email her and ask. – Michael Nov 2 2010 at 14:26
I like this question. I recommend, however, modifying the title to include the actual question, not just the topic. For example, "What are the canonical and earliest references to trivial symmetries in gauge systems?" will make it clearer to the readers (and to Google!) what question you mean to ask. (I tend to skim posts for question marks on first reading, and if I like the sentences ending with question marks, I read the post. The title counts. In that direction, you may also want to mildly rewrite your "looking for" list into entirely question form.) – Theo Johnson-Freyd Nov 2 2010 at 16:06
@Theo Thanks. I made the changes you suggested. – Simon Nov 2 2010 at 21:39
show 1 more comment
## 1 Answer
Remark: I think my answer should be ignored. (Apparently I did not understand the problem properly. Probably $S$ should be a linear functional on $C^\infty(M)$ (time-dependent?) etc. I still believe that the question ultimately boils down to something elementary once it is formulated in the right way. I just don't understand how the objects are defined.)
I think the question would be easier to answer for mathematicians if formulated in standard math language. I am not sure I am able to translate it, but let me try: You have some configuration space (probably a manifold, maybe infinite-dimensional) $M$, and a one-parameter group ${g_t}$ of $M$ (probably a diffeomorphism) and I guess that you assume that this is contained in some nice Lie group $G$ so that $g_t = \exp(tX)$ with $X \in \mathfrak g$, the Lie algebra of $G$. Now you have a (smooth?) function S on M and you want it to be invariant under ${g_t}$. This just means $XS = 0$. Now what does OS mean? Maybe you could explain this. It seems to me that you want to deduce something about $X$ and the derivative of $S$, but I am not quite sure I understand your notation there. I believe if you reformulate your question along these lines, more people can help.
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The action is a functional, not a function! – mathphysicist Nov 2 2010 at 10:37
Well, a functional is also a function, and the notation $S(q, \dot q, ...)$ seems to indicate that it depends only on input from the configuration space. – Tobias Hartnick Nov 2 2010 at 12:08
@Tobias Thanks for reminding me that on-shell (OS) is not standard maths terminology. I've added a footnote. I used this notation, because my question is a reference request, and as far as I know, most of the early work was done by physicists using this type of notation. (Also I wouldn't be entirely comfortable translating it) – Simon Nov 2 2010 at 12:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9383686780929565, "perplexity_flag": "middle"} |
http://en.wikipedia.org/wiki/Nuclear_magnetic_moment | # Nuclear magnetic moment
The nuclear magnetic moment is the magnetic moment of an atomic nucleus and arises from the spin of the protons and neutrons. It is mainly a magnetic dipole moment; the quadrupole moment does cause some small shifts in the hyperfine structure as well.
The nuclear magnetic moment varies from isotope to isotope of an element. It can only be zero if the numbers of protons and of neutrons are both even.
## Shell model
According to the shell model, protons or neutrons tend to form pairs of opposite total angular momentum. Therefore the magnetic moment of a nucleus with even numbers of both protons and neutrons is zero, while that of a nucleus with an odd number of protons and even number of neutrons (or vice versa) will have to be that of the "last", unpaired proton (or neutron). For a nucleus with odd numbers of both protons and neutrons, the total magnetic moment will be some combination of the magnetic moments of both of the "last", unpaired proton and neutron.
Nuclear magnetic moment is only partly predicted by simple versions of the shell model. The magnetic moment is calculated through j, l and s of the "last" nucleon, but nuclei are not in states of well defined l and s. Furthermore, for odd-odd nuclei, one has to consider the two "last" nucleons, as in deuterium. Therefore there are several possible answers for the nuclear magnetic moment, one for each possible combined l and s state, and the real state of the nucleus is a superposition of them. Thus the real (measured) nuclear magnetic moment is somewhere in between the possible answers.
## g-factors
The values of g(l) and g(s) are known as the g-factors of the nucleons.
The measured values of g(l) for the neutron and the proton are according to their electric charge. Thus, in units of nuclear magneton, g(l) = 0 for the neutron and g(l) = 1 for the proton.
The measured values of g(s) for the neutron and the proton are twice their magnetic moment (either the neutron magnetic moment or the proton magnetic moment). In nuclear magneton units, g(s) = -3.8263 for the neutron and g(s) = 5.5858 for the proton.
## Calculating the magnetic moment
In the shell model, the magnetic moment of a nucleon of total angular momentum j, orbital angular momentum l and spin s, is given by
$\mu=\langle(l,s),j,m_j=j|\mu_z|(l,s),j,m_j=j\rangle$
By projecting with the total angular momentum j we get $\mu=\langle(l,s),j,m_j=j|\overrightarrow{\mu}\cdot \overrightarrow{j}|(l,s),j,m_j=j\rangle \frac{\langle (l,s)j,m_j=j|j_z|(l,s)j,m_j=j\rangle}{\langle (l,s)j,m_j=j|\overrightarrow{j}\cdot \overrightarrow{j}|(l,s)j,m_j=j\rangle}$ $= {1\over (j+1)}\langle(l,s),j,m_j=j|\overrightarrow{\mu}\cdot \overrightarrow{j}|(l,s),j,m_j=j\rangle$
$\overrightarrow{\mu}$ has contributions both from the orbital angular momentum and the spin, with different coefficients g(l) and g(s):
$\overrightarrow{\mu} = g^{(l)}\overrightarrow{l} + g^{(s)}\overrightarrow{s}$
by substituting this back to the formula above and rewriting
$\overrightarrow{l}\cdot\overrightarrow{j} = {1\over 2} \left(\overrightarrow{j}\cdot \overrightarrow{j} + \overrightarrow{l}\cdot \overrightarrow{l} - \overrightarrow{s}\cdot \overrightarrow{s}\right)$
$\overrightarrow{s}\cdot\overrightarrow{j} = {1\over 2} \left(\overrightarrow{j}\cdot \overrightarrow{j} - \overrightarrow{l}\cdot \overrightarrow{l} + \overrightarrow{s}\cdot \overrightarrow{s}\right)$
$\mu = {1\over (j+1)}\langle(l,s),j,m_j=j|(g^{(l)}{1\over 2} \left(\overrightarrow{j}\cdot \overrightarrow{j} + \overrightarrow{l}\cdot \overrightarrow{l} - \overrightarrow{s}\cdot \overrightarrow{s}\right) + g^{(s)}{1\over 2} \left(\overrightarrow{j}\cdot \overrightarrow{j} - \overrightarrow{l}\cdot \overrightarrow{l} + \overrightarrow{s}\cdot \overrightarrow{s}\right)|(l,s),j,m_j=j\rangle$ $= {1\over (j+1)}\left(g^{(l)}{1\over 2} \left(j(j+1) + l(l+1) - s(s+1)\right) + g^{(s)}{1\over 2} \left(j(j+1) - l(l+1) + s(s+1)\right)\right)$
For a single nucleon $s =1/2$. For $j = l+1/2$ we get
$\mu_j = g^{(l)} l + {1\over 2}g^{(s)}$
and for $j = l-1/2$
$\mu_j = {j \over j+1} \left( g^{(l)} (l+1) - {1\over 2}g^{(s)} \right)$ | {"extraction_info": {"found_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": 14, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8823515176773071, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/92071/are-algebraic-geometry-error-correcting-codes-goppa-codes-good/92077 | ## Are algebraic geometry error correcting codes (Goppa codes) “good” ?
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Question (informal version): Are algebraic geometry error correcting codes (V.D. Goppa codes) "good" ?
Some details. There is certain construction of error-correcting codes by means of algebraic geometry, originating from pioneering work by Russian mathematician Valerii Denisovich Goppa (70-ies or early 80-ies ?).
I wonder what is known about these codes: a) are they "capacity-achieving" b) are there some "low-complexity" soft-decoders, like belief propogation which complexity is linear in the length of code c) are there some practical applications of these codes in error-correcting applications, if not - why ?
PS
It is known that they are involved in McEliece cryptosystem, but it is crypto-application, not error-correcting.
-
## 2 Answers
Given a $q$-ary code $C$ of length $n$ with minimum distance $d$, define its rate to be $(\log_q |C|)/n$, and its relative distance to be $d/n$. The Gilbert-Varshamov lower bound states that for any $q \ge 2$ and any $\delta \in (0,1)$ there is a $q$-ary code $C$ of relative distance $\ge \delta$ whose rate $r$ satisfies
$$r \ge 1 - H_q(\delta)$$
where $H_q(\delta) = \delta \log_q(q-1) - \delta \log_q(\delta) - (1-\delta)\log_q(1-\delta)$ is the $q$-ary entropy function.
The rate of an algebraic geometry Goppa code using a curve over $\mathbb{F}_q$ of genus $g$ satisfies
$$r \ge 1 - \delta - \frac{g-1}{n}.$$
This suggests that such codes could beat the Gilbert-Varshamov bound, and this was shown in 1982 by Tsfasman, Vladut and Zink. However I believe the best known improvement on the lower bound is very small, and so Goppa codes do not come close to meeting the Hamming bound $r \le 1-H_q(\delta/2)$.
In any case there are stronger generic bounds than the Hamming bound, for example, the Elias-Bassalygo bound, that show it is impossible to attain the channel capacity of a $q$-ary symmetric channel by hard nearest neighbour decoding in the Hamming setup.
I don't know much about decoding algebraic geometry Goppa codes. A quick web search found this paper from 1992. Roughly stated, the results in its introduction say that a Goppa code of length $n$ and minimum distance $d$ can be decoded in $O(n^3)$ time provided at most $d/2$ errors occur. There has been some more recent work on soft-decoding for Reed-Solomon codes (which are a special case of Goppa codes): see here, for example.
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Thank you very much for yours answer ! – Alexander Chervov Mar 25 2012 at 7:28
1
IIRC at least for some classes of Goppa codes $O(n^{7/3})$ decoding algorithms are known (search for Kötter, Feng-Rao, the Danes: Refslund, Hoholdt et al at least published something about it). A few years ago I asked Alex Vardy and Ralf Kötter, whether their soft decoding algorithm (a generalization of the list decoding algorithm due to Madhu Sudan) will generalize to one-point Goppa codes. Ralf Kötter in particular was convinced that it would. I don't know, whether they published anything on it, and sadly Kötter died a few year ago. – Jyrki Lahtonen Mar 29 2012 at 6:17
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AG codes for a given length n and alphabet q will beat corresponding turbo and LDPC but only over a channel with finite field alphabets (and over a complex channel if properly mapped and decoded properly).
Firstly hamming metric is not the true metric of gaussian channels which are the real channels (remember Trellis coded modulation which naively achieves coding gain by proper mapping of constellations that ordinary algebraic codes could not). There is no good way known to map algebraic codes to complex constellations (an analogy would be smaller distance code words should be mapped far apart in the complex constellations.. there is no subexponential way to do this). So even turbo and ldpc inspite of their bad minimum distance properties have an advantage.
Secondly even if you have a good map complex constellations (say gaussian distribution), there is no efficient soft decoder or hard decision ML decoder for AG codes and the notion of MAP decoding is hard (computationally). That is why even turbo and ldpc perform better inspite of having just a suboptimal MAP decoder over a complex channel.
My intuition is AG codes should achieve capacity faster (that is for a given rate could use much shorter codes) than inferior turbo and ldpc. However there is no proof for this. I believe the situation is due to the complexity involved in providing an argument through the first point (namely you have to show how many points are closer than distance d from each other and how should one map them over complex channels and how will the gains scale up.... these all seem to hit a wall due to the formidable computing complexity involved for any given n).
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Thank you very much for yours answer. Let me ask to clarify something. Do you mean that AG will beat turbo&LDPC for q>>2 ? (Actually I did not know about construction of turbo&LDPC for q>2). What if q=2 ? You write "There is no good way known to map algebraic codes to complex constellations" but why AG is worse than turbi&LDPC in this respect ? Would you be so kind to provide some references to the literature ? – Alexander Chervov Mar 25 2012 at 7:27
Hi Alexander: 1) I believe AG codes will beat any other construction that lies below or at the GV bound for all alphabets q>49 for some rates (if the other construction even meets the GV bound for q<49, they will be theoretically superior to AG codes since AG codes do not meet GV bound below q<49). 2) For the mapping question read TCM where if the underlying code words are close in binary, they will be mapped far apart in Euclidean distance. There is no easy way to do this for Block codes. – unknown (google) Mar 25 2012 at 7:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9186055064201355, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/191666/how-to-prove-o-log-n-is-true-for-a-binary-search-algorithm/195249 | # How to prove $O(\log n)$ is true for a binary search algorithm?
I have already looked at the answer here. I'm trying to understand how the poster got:
f(n) = O(1) = O(nlogba)
So far I have O(1) = T(n) - T(n/2). How is it that this became O(nlogba) ?
EDIT: After looking at the theorem, I'm also unsure how aT(n/b) = O(logb n). Is there is a proof for the limit as x->inf for (n/(b^x)) that equals logb n ?
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– Emmad Kareem Sep 6 '12 at 0:21
## 4 Answers
To do the specific case without the Master Theorem, the recurrence is $T(n)=T(n/2)+1$ because with one compare we can cut in half the number of places something can be. For simplicity, assume $n$ is a power of $2$. Then $T(1)=0$, because with one item we can find the right one without a compare. Similarly $T(2)=1$ because we have just one compare to do. Both of these satisfy $T(n)=\log_2 n$-the base case for our induction. Now to proceed by induction, assume $T(2^n)=n$. Then $T(2^{n+1})=T(2^n)+1=n+1=\log_2 (2^{n+1})$ and we have established it for all powers of $2$. The first equality used the recurrence, the second used the inductive assumption, and the third used the definition of $\log_2$.
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It's because in the answer you referenced to, $a=1$ and $b=2$ so $\log_ba=0$. This is done to match the form of $f$ in the Master Theorem
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The recurrence for binary search is $T(n)=T(n/2) + O(1)$. The general form for the Master Theorem is $T(n)=aT(n/b) + f(n)$. We take $a=1$, $b=2$ and $f(n)=c$, where $c$ is a constant. The key quantity is $\log_b a$, which in this case is $\log_2 1=0$.
If you look at the Wikipedia entry (through the link you posted) you will see that there are 3 main cases for the Master Theorem. Here we are in Case 2 since by taking $k=0$ we find that $n^{\log_b a}(\log n)^k=(n^0)(\log n)^0=1$; therefore $f(n)=c=\Theta(n^{\log_b a}\log^k n)$.
From Case 2 of the Master Theorem we know that $T(n)=\Theta(n^{\log_b a}(\log n)^{k+1}$ which in this case yields $T(n)=\Theta(n^0(\log n)^1)=\Theta(\log n)$.
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refering to the final answer
$T(n) = O(n^{\log_b a} \log_{2} n)) = O(\log_{2} n)$
The answer is very simple because the $O(\log_{2} n)$ is much more rapid than $O(n^{\log_b a}$ we igonore the other parametr,
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9472498893737793, "perplexity_flag": "head"} |
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Antimatter | # All Science Fair Projects
## Science Fair Project Encyclopedia for Schools!
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# Science Fair Project Encyclopedia
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# Antimatter
Antimatter is matter that is composed of the antiparticles of those that constitute normal matter. In 1929-31, Paul Dirac put forward a theory that for each type of particle, there is an antiparticle for which each additive quantum number has the negative of the value it has for the normal matter particle. The sign reversal applies only to quantum numbers (properties) which are additive, such as charge, but not to mass, for example. So, the antiparticle of the normal electron is called the positron, as it has a positive charge, but the same mass as the electron. An atom of antihydrogen, for instance, is composed of a negatively-charged antiproton being orbited by a positively-charged positron. Paul Dirac's theory has been experimentally verified and today a wide range of antiparticles have been detected. This is one of the few examples of a fundamental particle being predicted in theory and later discovered by experiment.
If a particle/antiparticle pair comes in contact with each other, the two annihilate and produce a burst of energy, which may manifest itself in the form of other particles and antiparticles or electromagnetic radiation. In these reactions, rest mass is not conserved, although (as in any other reaction), mass-energy is conserved.
Scientists in 1995 succeeded in producing anti-atoms of hydrogen, and also anti-deuteron nuclei, made out of an antiproton and an antineutron, but not yet more complex antimatter. In principle, sufficiently large quantities of antimatter could produce anti-nuclei of other elements, which would have exactly the same properties as their positive-matter counterparts. However, such a "periodic table of anti-elements" is thought to be, at best, highly unlikely, as the quantities of antimatter required would be, quite literally, astronomical.
Antiparticles are created elsewhere in the universe where there are high-energy particle collisions, such as in the center of our galaxy ,but none have been detected that are residual from the Big Bang, as most normal matter is [1]. The unequal distribution between matter and antimatter in the universe has long been a mystery. The solution likely lies in the violation of CP-symmetry by the laws of nature [2].
Positrons and antiprotons can individually be stored in a device called a Penning trap, which uses a combination of magnetic field and electric fields to hold charged particles in a vacuum. Two international collaborations (ATRAP and ATHENA) used these devices to produce thousands of slowly moving antihydrogen atoms in 2002. It is the goal of these collaborations to probe the energy level structure of antihydrogen to compare it with that of hydrogen as a test of the CPT theorem. One way to do this is to confine the anti-atoms in an inhomogenous magnetic field (one cannot use electric fields since the anti-atom is neutral) and interrogate them with lasers. If the anti-atoms have too much kinetic energy they will be able to escape the magnetic trap, and it is therefore essential that the anti-atoms are produced with as little energy as possible. This is the key difference between the antihydrogen that ATRAP and ATHENA produced, which was made at very low temperatures, and the antihydrogen produced in 1995 which was moving at a speed close to the speed of light.
Antimatter/matter reactions have practical applications in medical imaging, see Positron emission tomography (PET). In some kinds of beta decay, a nuclide loses surplus positive charge by emitting a positron (in the same event, a proton becomes a neutron, and neutrinos are also given off). Nuclides with surplus positive charge are easily made in a cyclotron and are widely generated for medical use.
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## Notation
Physicists need a notation to distinguish particles from antiparticles. One way is to denote an antiparticle by adding a bar (or macron) over the symbol for the particle. For example, the proton and antiproton are denoted as p and $\bar{\mathrm{p}}$, respectively.
Another convention is to distinguish particles by their electric charge. Thus, the electron and positron are denoted simply as e− and e+. Adding a bar over the e+ symbol would be redundant and is not done.
## Antimatter as fuel
In antimatter-matter collisions, the entire rest mass of the particles is converted to energy. The energy per unit mass is about 10 orders of magnitude greater than chemical energy, and about 2 orders of magnitude greater than nuclear energy that can be liberated today using chemical reactions or nuclear fission/fusion respectively. The reaction of 1 kg of antimatter with 1 kg of matter would produce 1.8×1017 J of energy (by the equation E=mc²). In contrast, burning a kilogram of gasoline produces 4.2×107 J, and nuclear fusion of a kilogram of hydrogen would produce 2.6×1015 J. Not all of that energy can be utilized by any realistic technology, because as much as 50% of energy produced in reactions between nucleons and antinucleons is carried away by neutrinos, so, for all intents and purposes, it can be considered lost. [3]
The scarcity of antimatter means that it is not readily available to be used as fuel, although it could be used in antimatter catalyzed nuclear pulse propulsion. Generating a single antiproton is immensely difficult and requires particle accelerators and vast amounts of energy—millions of times more than is released after it is annihilated with ordinary matter, due to inefficiencies in the process. Known methods of producing antimatter from energy also produce an equal amount of normal matter, so the theoretical limit is that half of the input energy is converted to antimatter. Counterbalancing this, when antimatter annihilates with ordinary matter energy equal to twice the mass of the antimatter is liberated—so energy storage in the form of antimatter could (in theory) be 100% efficient. Antimatter production is currently very limited, but has been growing at a nearly geometric rate since the discovery of the first antiproton in 1955[4]. The current antimatter production rate is between 1 and 10 nanograms per year, and this is expected to increase dramatically with new facilities at CERN and Fermilab. With current technology, it is considered possible to attain antimatter for \$25 billion per gram (roughly 1,000 times more costly than current space shuttle propellants) by optimizing the collision and collection parameters, given current electricity generation costs. Antimatter production costs, in mass production, are almost linearly tied in with electricity costs, so economical pure-antimatter thrust applications are unlikely to come online without the advent of such technologies as deuterium-deuterium fusion power.
Since the energy density is vastly higher than these other forms, the thrust to weight equation used in antimatter rocketry and spacecraft would be very different. In fact, the energy in a few grams of antimatter is enough to transport a small ship to the moon. It is hoped that antimatter could be used as fuel for interplanetary travel or possibly interstellar travel, but it is also feared that if humanity ever gets the capabilities to do so, there could be the construction of antimatter weapons.
## Antimatter in Popular Culture
The most famous fictional example of this kind of power source in action is in the science fiction franchise, Star Trek, where antimatter is a common energy source for starships. Antimatter as a weapon is also explored in the fictional work by author Dan Brown in the novel 'Angels and Demons'.
## External links
• Fermi National Accelerator Laboratory (Fermilab)—Among other things, an Antimatter research center.
• CERN
03-10-2013 05:06:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9344792366027832, "perplexity_flag": "middle"} |
http://en.wikipedia.org/wiki/Scientific_law | # Scientific law
This article discusses the philosophy of scientific laws. For the scientific and mathematical details, see laws of science.
A scientific law is a statement based on repeated experimental observations that describes some aspect of the world. A scientific law always applies under the same conditions, and implies that there is a causal relationship involving its elements. Factual and well-confirmed statements like "Mercury is liquid at standard temperature and pressure" are considered to be too specific to qualify as scientific laws. A central problem in the philosophy of science, going back to David Hume, is that of distinguishing causal relationships (such as those implied by laws) from principles that arise due to constant conjunction.[1]
Laws differ from scientific theories in that they do not posit a mechanism or explanation of phenomena: they are merely distillations of the results of repeated observation. As such, a law is limited in applicability to circumstances resembling those already observed, and may be found to be false when extrapolated. Ohm's law only applies to linear networks, Newton's law of universal gravitation only applies in weak gravitational fields, the early laws of aerodynamics such as Bernoulli's principle do not apply in case of compressible flow such as occurs in transonic and supersonic flight, Hooke's law only applies to strain below the elastic limit, etc. These laws remain useful, but only under the conditions where they apply.
Many laws take mathematical forms, and thus can be stated as an equation; for example, the Law of Conservation of Energy can be written as $\Delta E = 0$, where E is the total amount of energy in the universe. Similarly, the First Law of Thermodynamics can be written as $\mathrm{d}U=\delta Q-\delta W\,$.
The term "scientific law" is traditionally associated with the natural sciences, though the social sciences also contain laws.[2] An example of a scientific law in social sciences is Zipf's law.
Like theories and hypotheses, laws make predictions (specifically, they predict that new observations will conform to the law), and can be falsified if they are found in contradiction with new data.
## Introduction
A scientific law or scientific principle is a concise verbal or mathematical statement of a relation that expresses a fundamental principle of science, like Newton's law of universal gravitation. A scientific law must always apply under the same conditions, and implies a causal relationship between its elements. The law must be confirmed and broadly agreed upon through the process of inductive reasoning. As well, factual and well-confirmed statements like "Mercury is liquid at standard temperature and pressure" are considered to be too specific to qualify as scientific laws. A central problem in the philosophy of science, going back to David Hume, is that of distinguishing scientific laws from principles that arise merely accidentally because of the constant conjunction of one thing and another.[3]
A law differs from a scientific theory in that it does not posit a mechanism or explanation of phenomena: it is merely a distillation of the results of repeated observation. As such, a law is limited in applicability to circumstances resembling those already observed, and is often found to be false when extrapolated. Ohm's law only applies to constant currents, Newton's law of universal gravitation only applies in weak gravitational fields, the early laws of aerodynamics such as Bernoulli's principle do not apply in case of compressible flow such as occurs in transonic and supersonic flight, Hooke's law only applies to strain below the elastic limit, etc.
## References
1. Honderich, Bike, ed. (1995), "Laws, natural or scientific", Oxford Companion to Philosophy, Oxford: Oxford University Press, pp. 474–476, ISBN 0-19-866132-0
2. Andrew S. C. Ehrenberg (1993), "Even the Social Sciences Have Laws", Nature, 365:6445 (30), page 385.(subscription required)
## Further reading
• Dilworth, Craig (2007). "Appendix IV. On the nature of scientific laws and theories". Scientific progress : a study concerning the nature of the relation between successive scientific theories (4th ed.). Dordrecht: Springer Verlag. ISBN 978-1-4020-6353-4.
• Hanzel, Igor (1999). The concept of scientific law in the philosophy of science and epistemology : a study of theoretical reason. Dordrecht [u.a.]: Kluwer. ISBN 978-0-7923-5852-7.
• Nagel, Ernest (1984). "5. Experimental laws and theories". The structure of science problems in the logic of scientific explanation (2nd ed.). Indianapolis: Hackett. ISBN 978-0-915144-71-6.
• R. Penrose (2007). . Vintage books. ISBN 0-679-77631-1.
• Swartz, Norman (20 February 2009). "Laws of Nature". Internet encyclopedia of philosophy. Retrieved 7 May 2012. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8893486261367798, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/68676/traceable-representation-of-reductive-group-over-a-p-adic-field | ## Traceable representation of reductive group over a p-Adic field.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I have a question on how to define a trace of a unitary representation. If $G$ is a reductive group over a p-adic field it is known ( I do not know who prove it) that is of type I. Knowing this we have that every representation is traceable. However I think that there are more than one way to get a trace of a representation but only one canonical. For instance the regular representation has $f(1)$.
What will be the appropriate trace for the induced representation of the trivial representation of a subgroup $N$? i.e. what is correct trace for $c-ind_N^G1_n$?
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## 2 Answers
If $G$ is a $p$-adic reductive group, it is a result of Jacquet that any irreducible smooth representation $(\pi ,V)$ of $G$ is admissible : for any compact open subgroup $K$ of $G$, the fixed vector space $V^K$ is finite dimensional. From this it follows that for all locally constant function $f$ on $G$ with compact support, the operator $\pi (f) \in {\rm End}(V)$ has finite range, and therefore has a trace.
If $N$ is not cocompact in $G$, the compactly induced representation ${\rm c-Ind}_N^G {\mathbf 1}$ is not admissible in general. Moreover it is not semisimple in general neither of finite length.
For more information you may read Bushnell's paper "Induced representations of locally profinite groups".
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+1 Nice reference. – Marc Palm Jun 24 2011 at 11:17
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
There is a notion of a CRR group in Dixmier's book on $C^*$ algebras, which is a group such that for every unitary irreducible representation $\pi$, the integrated representation $\int \pi$, a representation of $C_0(G)$, is trace class, if you restrict to compactly supported function.
The expression $tr \int \pi(f)$ is the trace of the representation here.
Lie groups, abelian locally compact groups, reductive p-adic groups all are CRR groups.
I do not have the book with me, but you should find it in the index of the mentioned book.
Now to your main question. I assume for simplicity that the subgroup $N$ is cocompact in $G$, then $c-ind$ and $Ind$ are isomorphic functors, then the chapter Selberg trace formula in Deitmar-Echterhoff "principles of harmonic analysis" proves this fact.
If you look at arbitrary subgroups, I am not sure, if $c-ind_N^G 1$ has a nice decomposition into irreducibles with finite multiplicity, but I am not an expert, so you might want to wait until somebody knows more.
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Lie groups are not CCR, only reductive ones are. – Marc Palm Mar 5 at 6:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.923005223274231, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/1869/what-reasons-are-there-to-suspect-string-theory-may-be-an-incorrect-theory/3772 | # What reasons are there to suspect string theory may be an incorrect theory?
It's a truism that in science it is at least as important to state the reasons why a theory or idea might be wrong, as to state the reasons why it is might be correct. For example, early renormalization methods made experimental predictions that were very accurate, but the inventor's famously worried for decades about the conceptual problems of divergences; and many of their papers/articles/books on the subject include a long discussion about how acutely worried they were that their method for handling divergences was deeply flawed somehow.
My intention with this post is not to engage any user in a debate about whether string theory is true or not. I think anyone who takes string theory seriously should be happy to acknowledge as clearly as possible all the potential theoretical issues/problems with the theory.
Therefore I would like to ask, what are the major theoretical problems which would lead a reasonable person to be worried that string theory is not an accurate theory? I am excluding the fact that string theory has made no (major) predictions that have been confirmed, I am only interested in conceptual and theoretical ideas which might discredit the theory.
Please keep all answers to one issue per post.
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Have you looked at Peter Woit's "Not even wrong" blog? – Sean Tilson Dec 13 '10 at 5:50
@Sean: I haven't read Woit's blog in awhile, but I used to read it everyday. @kall43: It's hard for me to answer your question, I am trying to be a broad and general as possible in my formulation; so basically if there is anything at all that bothers you about string theory, but not that it has never made a prediction that was confirmed, then it's acceptable for an answer in my opinion. – Matt Dec 14 '10 at 4:25
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I think you may be missing a bunch of interesting answers because of the phrasing of your question. Based on your analogy, you may be interested in the ways string theory is incomplete, and the various ideas people have on how to remedy that. None of this necessarily leads you to suspect there is something wrong (as in the QFT analogy), just that something is still missing. – user566 Jan 25 '11 at 16:03
## 8 Answers
I don't know any reason to suspect string theory may be incorrect, but the closest analogy to renormalisation infinities is the fact that we only know string theory as a perturbation series from a fixed background, plus a collection of non-perturbative dualities that relate different versions. By rights there should be a better formulation from which all the perturbative string theories can be derived directly. M-theory is a name sometimes used in this connection but we still don't know what it is.
Even if string theory were not a theory of physics there would be good reason from the mathematics to think that such a formulation exists because otherwise all the relationships between the different string theories would be just creepy coincidences. There really has to be some underlying framework that explains them.
I think this is comparable in some ways with the original view of renormalisation which made people question the original perturbative formulations of quantum field theories. Once non-perturbative formulations such as lattice gauge theory emerged they were able to understand renormalisation as a scaling behaviour related to critical points in the theory and things were better understood.
String theory has other problems such as the need to understand how a vacuum can be selected that explains low energy particle physics and the cosmological constant. We also need to understand better what happens to spacetime in string theory at Planckian energies. These problems are often highlighted because they are the reason why string theory cannot yet make any testable predictions, but I don't think they can be resolved properly until the underlying principles of the theory are known. So that is the key issue to resolve first.
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Philip wrote: "Once non-perturbative formulations such as lattice gauge theory emerged they were able to understand renormalisation as a scaling behaviour related to critical points in the theory and things were better understood." What critical points are you talking about? Is the lattice step has physical meaning? I thought it could be made as small as you liked, similarly to arbitrariness of the cutoff L in the before-lattice analytical calculations. – Vladimir Kalitvianski Jan 25 '11 at 16:51
In this case the critical point is in the limit where lattice spacing goes to zero. – Philip Gibbs Jan 26 '11 at 7:35
So there is no better understanding but better illusion of understanding. – Vladimir Kalitvianski Jan 26 '11 at 9:36
There are two sides to this coin. String/M theory is more than a theory, it is more in a way a framework of theories. It is a vast territory, at least from a mathematical perspective. A lot of it is not likely correct, or should we say does not manifest itself in anything physical. On the other hand there appears to be universality to some stringy structure, where it is appearing in condensed matter physics. Some stringy or AdS structure is making an appearance in heavy ion physics. So the other side of the coin is that stringy/M structures are too rich for the whole thing to be completely false.
When looking at string physics it is probably best to hold more to basic physical ideas, such as how Susskind approaches things, and to remain close to mathematics which has some foundational aspect to it. Highly rococo mathematics that deviates from “basics,” and theory which is increasingly complex might be held up to greater suspicion. Where the dividing line lies here is a bit hard to define.
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If I understand what you're looking for correctly:
One issue is that string theory predicts the number of spacetime dimensions to be 10 (or 11 in M-theory), larger than the 4 we observe. Of course, it's possible to explain that away by postulating that 6 (or 7) of the dimensions form a compact manifold with characteristic scale much smaller than we are able to detect, but as far as I know, nobody knows whether that's a realistic configuration. The predictions of the theory would depend on the particular topology of the compact manifold, and it's not clear whether there is any choice of topology that makes the theory correspond to reality.
And even if there is, it'd still be an open question why nature has "chosen" that particular topology, out of all the possibilities.
N.B. I haven't studied this in full detail so perhaps I've gotten some of the details wrong, in which case hopefully someone more knowledgeable will correct me.
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Can I ask an additional question to this: is it true that the conclusion of " exactly 11 dimentions" comes from non-divergence of certain quantity, while this quantity is just a term in perturbation theory? – Kostya Dec 13 '10 at 6:31
@Kostya It's been a while since I've looked at string theory, but if I remember correctly, the restriction on the dimensionality of spacetime comes from a computation of the Poisson Brackets of the classical constaints of the theory. Unless the dimensionality of spacetime is a certain number (that is different for different string theories-10 for superstrings, 11 for M-theory, 26 for bosonic strings), then the Poisson brackets of a subset of the constraints will be equal to something nonzero, meaning that the constraints can't be solved and the theory is inconsistent. But its been a while. – Jerry Schirmer Dec 13 '10 at 20:41
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Vafa and Brandenberger argued that spacetime will not expand unless wrapped strings are able to combine and annihilate (lest their energy grow with the size of the universe) -- two-dimensional worldsheets will only meet generically in no more than four dimensions. This is a mostly heuristic argument, I guess, but one which tries to address the observed dimensionality of large spacetime dimensions. As for the total number of dimensions, this is a kind of technical point in string theory. The number 10 comes from preserving Lorentz invariance, in one derivation. (M-theory adds the 11th.) – Eric Zaslow Dec 14 '10 at 3:01
11=7+4 appears in traditional compactification, because the antisymmetric tensor $A_{pqr}$ from 11D SUGRA allows for this kind of solutions, Freund-Robin. It only fails to be "realistic" because the 4D part happens to be anti-deSitter. – arivero Jan 24 '11 at 19:23
This question is a little confusing. There are really no reasons to "suspect it may be wrong" (and there are surely many reasons to suspect it may be right!) in the sense you're probably thinking of. It seems you're more likely asking the very different question: "what things would we like string theory to do, but do not know how to do in it."
In that case, of course, the list is very long because we would like to do everything with it!
More concisely, the obvious point is that we'd like to be able to explicitly construct a model in string theory which reduces, at low energy, to the Standard Model plus classical gravity, and we do not know how to do this.
One could potentially wonder, as in the case of the old Kaluza-Klein theories, whether there is an obstruction to getting the correct structure. In Kaluza-Klein, everything looks promising until you (or, Ed Witten) prove that while the theory does permit the correct gauge groups, it does not permit the correct representations (you don't get chiral theories).
However, in this case, we know that we can get both the right groups and representations out. As far as I know, there are no good reasons to suspect that any particular property of the Standard Model is not allowed in string theory. In fact, many models very similar to the Standard Model can be constructed, so in a sense, we can qualitatively get the right kinds of answers.
We also have things like the AdS/CFT correspondence, which allows us to explicitly construct stringy duals of some ordinary field theories, so string theory is at least as right as those field theories are!
So I would say there are no theoretical reasons to expect string theory to be "incorrect" in any reasonable way, and there are only marginally good ones to suspect that it could be unable to give us the Standard Model.
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Wow, I'm so happy that all of a sudden lots of people are answering this old post. Since I seem to have confused you Jeremy, let me just explain that I am a hardcore disbeliever in string theory, and have been for about five years. I don't really want to get into a debate in this thread, because I promised not to in the OP, but basically my intention behind the question was to compile a list of problems with string theory, without offering any of my own ideas/opinions. Actually I think this thread can have a lot of value for the folks who like string theory, as well as those who dont. – Matt Jan 27 '11 at 7:54
Contrary to D=11 Kaluza Klein theory, where you have a few small groups as SO(8) or SU(3)xSU(2)xS(2), in string theory you can produce very huge gauge groups down from E8xE8 or from SO(32), or even perhaps more complicated structures down from M-theory. This is suspicious but it could be more a a wrong interpretation that an incorrect theory. After all, there are some limiting procedures going to D=11 supergravity -and its multiplets-, which has a more adequate size.
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beacuse it is not falsable :D :D .. and this is the biggest problem of this theory ,
it is like saying 'God create the universe' you can not prove me neither wrong or right :) however i can give no proofs of my sentence.
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I think that QFT and the string theory have the following flaw in common: they treat perturbatively coupling the things coupled permanently. This is a conceptual flaw that not always can be "repaired".
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it is not clear at all what you are trying to say and what bearing it has on the question. Can you add more detail? Thanks. – user346 Jan 24 '11 at 15:17
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OK, let us consider one electron and the quantized electromagnetic field. They are permanently coupled. When you push the real electron, you excite the photon oscillators. This means the electron is a part of these oscillators and vice versa. This is not implemented as such in QED. In QED we have a free electron and free oscillators as the initial approximation. Then we couple them perturbatively. In the first Born approximation you can push the electron and have no radiation at all (elastic processes). Meanwhile the (soft) radiation is a process with unity probability - it happens always. – Vladimir Kalitvianski Jan 24 '11 at 16:13
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– Vladimir Kalitvianski Jan 24 '11 at 16:18
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Seeing so many downvotes, I wonder if anybody can give any reasonable objection to my argument? Can anybody treat the electron and proton coupling in Hydrogen atom perturbatively? No? Why do you think you can treat quarks perturbatively? Do you think you know what quarks are? No? Do you know what strings are then? – Vladimir Kalitvianski Jan 24 '11 at 22:21
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Dear downvoters, tell me where I am wrong, please. I would like to learn. – Vladimir Kalitvianski Jan 31 '11 at 16:23
String theory considers the Lorentz symmetry and existence of extradimensions between others postulates. The problem is, these extradimensions will manifest itself just with violation of Lorentz symmetry, which makes string theory a fringe theory at the rudimentary logic level - until it finds a way, how to demonstrate the existence of extradimensions without violation of Lorentz symmetry.
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Neither my downvote, nor my expertise, but this seems like a pretty hefty claim. Care to elaborate? – wsc Nov 7 '11 at 16:13
You can understand this problem for example with water surface model of space-time. Until the wave spreading is not affected with underwater, it remains perfectly background independent in accordance to relativity theory. The only way, how to observe the additional extradimension of 2D surface is the dispersion of surface ripples in 3D underwater. Unfortunately, just this dispersion will break the background invariance of surface ripple spreading. This explains, why string theory can never get some definite solution, being stuck at the landscape of 10E+500 possible solutions. – Zephir Nov 7 '11 at 18:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9537771940231323, "perplexity_flag": "middle"} |
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Given the following function Collatz: ... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8797028660774231, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/50876?sort=votes | ## Background/Motivation
My interest in this problem traces back to an 11 year old girl who really took to one-way path counting problems. After doing several configurations of streets, she decided to come up with a problem of her own. She presented a $3 \times 3$ gridwork of two-way streets (forming 4 blocks in a $2 \times 2$ arrangement) and added the condition that a street could be traversed at most once. She asked how many such paths are there from upper left to lower right? (Answer: 16.)
Stirred by her enthusiasm, we tried generalizing in various directions. If you have 2 long horizontal streets with $N$ verticals, and let $a_N$ be the number of edge-disjoint paths from upper left to lower right and $b_N$ be the number of edge-disjoint paths from upper left to upper right, then $a_{N+1} = b_{N+1} = a_N + b_N$ for $N > 1$ and $a_1 = b_1 = 1$.
The $3 \times N$ case is trickier, but the number of edge-disjoint paths from upper left to lower right still satisfies a finite linear recurrence relation.
Naturally, I turned to OEIS and found sequences A013991-A013997, where Dan Hoey gives the number of edge-disjoint paths between opposite corners of $K \times N$ grids for $K = 3, 4, 5, ..., 9$ and small $N$. He also provides the first few values for the $N \times N$ cases (sequence A013990). (Note, his numbering counts blocks, not streets.) For $K=3$, he provides a generating function. In a recent communication, he explained the computer algorithm he used to compute the values but indicates that he did not find a recurrence relation for these sequences, so as far as I know, there is no known way to determine the answer to the title question for large $N$.
I've also spoken with Gregg Musiker, Bjorn Poonen, and Tim Chow about this problem. Although none knew how to do the $4 \times N$ case, Gregg simplified my recurrence relations for the $3 \times N$ case, Bjorn suggested many related questions and suggested an asymptotic formula for the $N \times N$ case, and Tim suggested looking at the related literature on self-avoiding walks, such as the book by Neal Madras and Gordon Slade, though it's not clear to me how related edge-disjoint and self-avoiding are with respect to counting them.
Because there are finite linear recurrence relations for the $2 \times N$ and $3 \times N$ cases, it seems natural to also ask:
Is there a finite linear recurrence relation for the number of edge-disjoint paths between opposite corners of a $4 \times N$ gridwork of streets?
Are these problems intractable?
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3
An edge-disjoint path on a graph is the same thing as a self-avoiding walk on its line graph. – Qiaochu Yuan Jan 1 2011 at 19:40
If I remember correctly a talk I heard from C. Krattenthaler, the number of self-avoiding graph is the value of some explicit Pfaffian. Can anyyone confirm that ? – Denis Serre Jan 1 2011 at 21:39
For a related problem see kcollins.web.wesleyan.edu/publications/grid-r.pdf. – Richard Stanley Jan 2 2011 at 2:50
Thank you for these comments and the reference. – Ken Fan Jan 2 2011 at 4:47
Isn't www-fourier.ujf-grenoble.fr/%7Erivoal/journeehyp/… related? "A chess Rook can move any number of squares horizontally or vertically on an $N\times N$ chess board. Assuming that the Rook moves right or up at every step, how many different paths can the Rook travel in moving from the lower-left corner to the upper-right corner on the board?" – Wadim Zudilin Jan 2 2011 at 12:11
show 3 more comments
## 2 Answers
To amplify on Christian's answer, the problem on a $K \times N$ grid for fixed $K$ and varying $N$ admits a finite-state transition model, so in particular is given by a linear recurrence.
The key is to find the right set of states. If you take an edge-disjoint path on a $K \times N$ grid and slice it on a vertical line through the middle passing through a set of horizontal edges, you'll see the path crossing along some odd number of these edges ($\le K$). On both the left and right we'll see a collection of paths with these endpoints. There's another constraint, that we end up with a single, connected path without disjoint loops; to take that in to account, also record a matching: which endpoints are paired up on the right hand side. All but one of the endpoints are paired up in this way. (You could also choose the left, and end up with a slightly different matrix.)
For instance, in the $3 \times N$ case, there are $6$ states. If we record an occupied edge by $\times$ and an unoccupied edge by $\circ$ and turn everything on its side, the states are $$\times\circ\circ\quad\circ\times\circ\quad\circ\circ\times\quad\times_1\times_1\times \quad\times_1\times\times_1\quad\times\times_1\times_1$$ where the subscript indicates the matching. (In this case, there is at most one matched pair.)
Next consider the transitions. If you consider two adjacent vertical slicings of a path, you'll see two possibly different states. The set of edges that are occupied in the middle is determined by which edges are occupied in the two different states. There is sometimes a choice about how the strands are connected up. However, some of these choices will be ruled out by the constraints on the connectivity; usually you will end up with just $0$ or $1$ possibilities.
For instance, in the $3 \times N$ case, with the states in the order above, I get the following matrix of possibilities: ```$$
M =\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 0 \\
1 & 1 & 1 & 1 & 0 & 1\\
1 & 1 & 1 & 0 & 1 & 1\\
0 & 1 & 1 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
1 & 1 & 0 & 0 & 0 & 1
\end{pmatrix}
$$```
For the ultimate answer, you want to look at paths that start at the upper-left and go to the lower-right. You can incorporate that nicely by adding an extra slice to the left of the entire diagram, with only the top slot occupied, and another to the right of the diagram, with only tho lower slot occupied.
Concretely, in the $3 \times N$ case, the number of paths is given by the $(1,3)$ entry of $M^N$.
For the $4 \times N$ case, you would get a $16 \times 16$ matrix, which is straightforward but somewhat tedious to work out. As a result, the answer will satisfy a linear recurrence of order $16$.
An interesting variation is to consider only crossingless paths. In this case, the matching must be crossingless, so we only get 5 states in the $3 \times N$ case and $12$ in the $4 \times N$ case.
Update Jan 7: The matrix above is wrong: it should be ```$$
M =\begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 0 \\
1 & 1 & 1 & 2 & 2 & 2\\
1 & 1 & 1 & 0 & 1 & 1\\
0 & 1 & 1 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
1 & 1 & 0 & 0 & 0 & 1
\end{pmatrix}
$$```
Update 2: And here's an image illustrating what is actually being counted: I permuted the entries slightly, but they're labelled along the sides. The dotted paths are there to help in the counting: the non-allowed configurations would form a closed loop.
-
Thanks for including the matrix graphic! I understand now that you're counting physically different path snippets (which includes the information of pairings on the right) without regard to orientation. In retrospect, Christian does seem to have this notion (with pairings on the left), but I was thrown by the first matrix with all zeroes and ones. I think as far as the question I posed here goes, the problem is settled and I think it should be so noted. I wish I could accept both answers. I've marked both up, but will go ahead and accept this one because of the pretty picture... – Ken Fan Jan 8 2011 at 17:30
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Let $v_1$, $\ldots$, $v_N$ be the vertical streets and let $h_{1,j}$, $\ldots$, $h_{4,j}$ be the horizontal edges between $v_{j-1}$ and $v_j$. An admissible path $\gamma$ induces a coloring of the horizontal edges as follows: Consider a vertical street $v_j$. The path $\gamma$ uses either one or three of the incoming edges $h_{k,j}$ $(1\le k\le 4)$. If $\gamma$ uses one edge, color it black and the three other edges white. If $\gamma$ uses three edges, two of them are linked to each other by a part of $\gamma$ extending only to the left of $v_j$. Color these two edges red, the third edge black and the unused edge white. In all, there are 16 possible colorings $c:\lbrace 1,2,3,4\rbrace\to\lbrace b, r, w\rbrace$ that can result in this way. There is a $16\times 16$ transition matrix $T$ that encodes the possible matchings between the coloring $c$ of the edges $h_{k,j}$ and the coloring $c'$ of the edges $h_{k,j+1}$ (e.g., circles must be avoided). This matrix $T$ has to be determined "the hard way", i.e., by listing for each $c$ the possible $c'$. The number of admissible paths $\gamma$ is then obtained by applying $T^{N-1}$ to a suitable starting vector; so there is indeed a linear recurrence for the number of these paths.
An example: If $c$ contains one black and two red edges, then using the vertical edges on $v_j$ in an admissible way one may
(a) continue the black and the two red edges into the next column individually, maybe at a different level,
or
(b) connect the black end of $c$ to either one of the red ends by a vertical segment creating a $\supset$ and continue the other red edge of $c$ into the next column, but as a black edge,
and, if room on $v_j$ permits, one may
(c) throw in two red edges beginning on $v_j$ which are connected by a vertical segment creating a $\subset$.
Here is a pictorial list (hopefully complete) of the possible transitions $c\to c'$:
http://www.math.ethz.ch/~blatter/grid.pdf
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2
Is it obvious that admissible colorings correspond exactly to edge-dispoint paths? – Qiaochu Yuan Jan 2 2011 at 16:04
@Qiaochu, admissible colorings correspond not to paths, but to possible states of partial paths. – Dylan Thurston Jan 3 2011 at 15:54
@all commenters: Thanks Dylan and Christian for these detailed answers. I've been thinking about them and here's what I've come up with so far. As described in both answers, it isn't quite right because there is more than one way to string up a given admissible covering into an edge disjoint path. However, in the 3 by N case, if one adds an order to the two horizontal left-to-right edges indicating which is traversed first, then you get a 9 by 9 matrix and this seems to work! For the 4 by N case, this would mean using a 28 by 28 matrix instead of 16 by 16. – Ken Fan Jan 4 2011 at 1:42
cntd.: However, I think that even with order of traversal information, there is a problem in the 4 by N case because it seems there can be more than one path per admissible covering. (I haven't actually checked it in the 3 by N case, but the agreement with the data seems to suggest that it is ok in this case.) But in the 4 by N case, using a variant of Dylan's notation with U=up and D=down, if o U1 D U2 is stacked on top of U1 D U2 o, there are two ways to string these together. (The numbers indicate the order the streets are traversed.) – Ken Fan Jan 4 2011 at 1:55
Here's the 9 by 9 matrix for the 3 by N case: [1 1 1 0 0 0 1 0 1; 1 1 1 1 1 1 1 1 1; 1 1 1 1 0 1 0 0 0; 1 1 0 1 0 0 0 0 0; 1 1 0 0 1 0 0 0 0; 0 1 0 0 0 1 0 0 0; 0 1 0 0 0 0 1 0 0; 0 1 1 0 0 0 0 1 0; 0 1 1 0 0 0 0 0 1]. Take powers of this and, just as Dylan says, the (1,3) entry seems to count the number of paths. – Ken Fan Jan 4 2011 at 2:06
show 3 more comments | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 82, "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}, "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}, "english_confidence": 0.94251549243927, "perplexity_flag": "head"} |
http://mathhelpforum.com/pre-calculus/187100-solving-x-log-x-type-equations.html | # Thread:
1. ## Solving x = log(x) type equations.
It looks like a simple problem, but for some reason, it's giving me trouble. How do you find the solution algebraically:
$x = 8\ log _2 x$
By drawing the 2 functions, I can see that there will be two solutions. But that's about all I can say right now. How do you find the values of these two roots?
2. ## Re: Logarithmic functions
You can not solve it elementary.
You could use numerical methods. | {"extraction_info": {"found_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}, "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}, "english_confidence": 0.9226473569869995, "perplexity_flag": "middle"} |
http://mathhelpforum.com/differential-geometry/142053-conformal-map-upper-half-plane-print.html | # conformal map, upper half-plane
Printable View
• April 28th 2010, 08:43 PM
eskimo343
conformal map, upper half-plane
Find a conformal map $w(z)$ of the right half-disk $\{ \text{Re}(z), |z|<1 \}$ onto the upper half-plane that maps $-i$ to $0$, $+i$ to $\infty$, and $0$ to $-1$. What is $w(1)$?
The back of the book says that $w=\frac{-(z+i)^2}{(z-i)^2}$, $w(1)=1$. However, I still do not see how they get $w=\frac{-(z+i)^2}{(z-i)^2}$. For the second part we can plug in $1$ to get $1$ as the output. I just don't see how to get $w=\frac{-(z+i)^2}{(z-i)^2}$. Thanks in advance.
All times are GMT -8. The time now is 04:02 AM. | {"extraction_info": {"found_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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.917969286441803, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/43408/is-this-formulation-of-the-singular-value-decomposition-standard/43410 | ## Is this formulation of the Singular Value Decomposition standard?
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In customary formulations of the Singular Value Decomposition or SVD that I have seen, (e.g., Wikipedia or Gil Strang's textbooks) it is always stated in terms of writing an $m \times n$ matrix $M$ (say of rank $r$) as a product $U \Lambda V$, where $U$ and $V$ are orthogonal $m \times m$ and $n \times n$ matrices and $\Lambda$ is a diagonal $m \times n$ matrix with non-negative "singular values" on the diagonal, $s_1 \ge s_2 \ge \ldots \ge s_r >0$ and the rest zero. Looking back over the many times I have taught linear algebra to both undergraduates and graduate students, I realized that I have not once covered the SVD, and even though I consider myself pretty knowledgeable about linear algebra, I have never felt comfortable with the statement of SVD or felt that I understood it in a more than formal way. (I might add that many theoretical linear algebra texts do not mention the SVD and many of my more theoretically minded colleagues do not even recognize the term.) But a few days ago, a social scientist friend of mine asked me about
a problem he was interested in; one that involved the SVD in an essential way. After thinking about it for a while, I realized that SVD can be reformulated as a statement about linear transformations that, to me at least, seems a lot more conceptual and geometric:
If $T$ is a linear map, say of rank $r$, between finite dimensional inner-product spaces $V$ and $W$, then there are orthonormal bases $v_1, \ldots, v_m$ for $V$ and $w_1, \ldots, w_n$ for $W$ and $r$ positive numbers $s_1 \ge s_2 \ge \ldots \ge s_r$, such that $T v_i$ equals $s_i w_i$ if $i \le r$ and equals zero if $i > r$.
I certainly realize that this is a pretty obvious reformulation of SVD, once you see it (and
those poor misguided souls who prefer matrices to linear transformations may even see it as a step backwards :-), but my question is whether there is some standard source for this reformulation that I can reference.
-
By a strange coincidence, I'm about to teach it precisely this way this afternoon ... – Andrew Stacey Oct 25 2010 at 7:43
It seems that the more "applied" you get, the more occasion you have to think about the SVD. The "purest" mathematicians seem to be the ones "do not even recognize [that] term". In some contexts, it also seems as if the "pure" assume that matrices are square. Once I wrote down the matrix $A(A^T A)^{-1} A^T$ and a highly respected "pure" mathematician told me that's the identity matrix. So it is, if $A$ has as many columns as rows. – Michael Hardy Nov 10 2010 at 17:52
Michael, since you're a statistician, do you think there might be sample bias in your observations about the iniquities of pure mathematicians (here and elsewhere)? – Yemon Choi Nov 10 2010 at 18:24
Well, I wouldn't have actually called this an "iniquity". – Michael Hardy Nov 10 2010 at 19:20
## 5 Answers
I just looked in Wikipedia (http://en.wikipedia.org/wiki/Singular_value_decomposition). There is a very thorough discussion, including a section "Geometric meaning" in which your interpretation is clearly explained. Well, there $K^n$, where $K = \mathbb R$ or $K = \mathbb C$ is used, instead of arbitrary inner product spaces, but I suppose you'll agree that the difference is not essential.
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1
OUCH ! You are absolutely correct. Somehow I missed that section---or rather I guess that I never got down that far in the Wikipedia article, perhaps because it comes well after the explanation in terms of matrices and a long section on applications. Many thanks. – Dick Palais Oct 24 2010 at 19:41
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The singular values do have a sound geometrical meaning. The first one is nothing but the operator norm of $T$, that is the maximum dilation coefficient $$\frac{\|Tx\|}{\|x\|}.$$ The next ones can be seen also as maximum of dilation coefficients, provided you replace lines by subspaces. For instance $$s_k=\max_{\dim F=k}\min\left\{\frac{\|Tx\|}{\|x\|};x\in F,x\ne0\right\}.$$ In terms of the exterior algebra over $\mathbb C^k$ ($k=m$ or $n$), you have $$s_1\cdots s_k=\sup\left\{\frac{\|Tx_1\wedge\cdots\wedge Tx_k\|}{\|x_1\wedge\cdots\wedge x_k\|};x_1,\ldots,x_k\in F\right\}.$$
Notice also that there is a $p$-adic version of the SVD decomposition. See K. S. Kedlaya. $p$-adic differential equations. Cambridge Studies in Advanced Mathematics, 125. Cambridge University Press, Cambridge, 2010
Edit. I should have also given the formula $$s_1+\cdots+s_k=\max\{{\rm Tr}(PTQ); P\hbox{ and } Q \hbox{ are unitary projectors of rank } k\}.$$ This has the interesting consequence that $T\mapsto s_1+\cdots+s_k$ is a convex function.
Edit (bis). Actually, $T\mapsto s_1\cdots s_k$ is rank-one convex. This means that it is convex along every line $A+{\mathbb R} B$ for which $B-A$ is a rank-one matrix.
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1
I recognized the definition of $s_k$ as Courant-Fischer, but I was not aware of the formula for the product $s_1\cdots s_k$. This is really fascinating! Are there any interesting applications of that formula that would be easily readable? – Thierry Zell Nov 10 2010 at 13:02
@Thierry. See my edits. – Denis Serre Nov 10 2010 at 14:22
2
@Denis: very interesting. You should write a book about this :) – Piero D'Ancona Nov 10 2010 at 14:28
Also worth noting here is that the trace-norm $s_1+\cdots+s_k$ is the convex function "closest" to the numerically intractable matrix rank function. – S. Sra Nov 10 2010 at 14:46
@Piero. You don't miss any occasion of variation to your favorite joke. More seriously, see exercises 108, 109 in umpa.ens-lyon.fr/~serre/DPF/exobis.pdf . – Denis Serre Nov 10 2010 at 15:13
The book "Numerical Linear Algebra" by Trefethen and Bau, also introduces the SVD in this way. The relevant chapters (4 and 5) seem to be complete in Google books. The exposition is not tainted by the 'numerical' in the title.
I don't understand why the SVN isn't given more emphasis in standard introductions to linear algebra. It seems to give a good intuitive decomposition of a linear operator. I would expect it to be a useful theoretical tool even if numerical issues are not being considered.
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The first five chapters of the book are available online: people.maths.ox.ac.uk/~trefethen/text.html – Ramsay Oct 24 2010 at 21:15
3
A week ago I might have demurred from that second statement, but now I think I have to agree. My clouded eyes have been opened. :-) BTW, I really like the way that Trefethan and Bau express the key point (page 32) : "The SVD makes it possible for us to say that every matrix is diagonal---if only we choose the proper bases for the domain and range spaces." – Dick Palais Oct 24 2010 at 21:21
2
That quote by itself is not the best summary of the SVD. That quote by itself is the best summary of the rank theorem (by which I mean that every linear transformation can have matrix [I_k 0; 0 0]). SVD says that the rank theorem still holds if we insist that all our isomorphisms are isometries, providing we replace the identity submatrix by a diagonal one. – Andrew Stacey Oct 25 2010 at 7:43
1
Trivial, but at least doable! My main gripe with things like SVD is that so often they are presented as things that can actually be computed (Jordan canonical form is another example) whereas to actually do it involves finding eigenvalues which is, as the Norwegians know very well, impossible. Of course, the fact that it exists is tremendously important but for reasons other than actually being able to compute it. – Andrew Stacey Oct 26 2010 at 7:15
3
That all depends on what you mean by "computed". It can certainly be computed in a practical sense, in that there are programs that can compute the SVD of a matrix (over C^n). More theoretically, I also expect that it's possible to construct an algorithm that provably computes the SVD to any accuracy. It is not clear to me what the relevance is of the fact that you cannot write down the SVD in closed form. The Jordan canonical form is a very different beast because it is discontinuous. – Jitse Niesen Nov 10 2010 at 10:17
show 1 more comment
This formulation, or something very close to it (I don't have the book with me) is in Axler's Linear Algebra Done Right.
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1
You can also find it in Treil's Linear Algebra Done Wrong, section 3.3 (pdf of book: math.brown.edu/~treil/papers/LADW/LADW.pdf ). – alex Oct 24 2010 at 20:06
1
So it seems that whether you do things right or wrong, the linear transformation formulation I mentioned IS the way to go. :-) BTW, when you say it is in Treil, are you referring to the formula (3.1) on page 168? – Dick Palais Oct 24 2010 at 21:05
Right - I actually meant Proposition 3.6 just above that. – alex Oct 24 2010 at 23:40
It's in my favourite linear algebra book Advanced Linear Algebra by Steven Roman, chapter 17 (called Singular Values and the Moore-Penrose inverse).
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http://mathhelpforum.com/pre-calculus/177341-inequality-print.html | # Inequality
Printable View
• April 9th 2011, 06:13 AM
aonin
Inequality
Hey guys I'm confused whether I doing these right, I used 2 different methods to solve this problem, and they seem to be giving different answers
$\text{solve} \dfrac{1}{x-3}> 4$
1st method:
$\dfrac{1}{x-3}> 4$
$\dfrac{1}{(x-3)^2}>16$
$1>16(x-3)^2$
$1>16x^2-96x+144$
$16x^2-96x+143<0$
$x=\dfrac{96\pm\sqrt{96^2-4(16)(143)}}{32}$
$\dfrac{88}{32}<x<\dfrac{13}{4}$
2nd method
$\dfrac{1}{x-3}> 4$
$\dfrac{1}{x-3}-4>0$
$\text{multiply both sides by}\hspace{5mm}(x-3)^2$
$(x-3)-4(x-3)^2>0$
$(x-3)(1-4(x-3))>0$
$(x-3)(13-4x)>0$
$3<x<\dfrac{13}{4}$
they give different answers...? Pls let me know if i've done anything wrong (Doh)
Thanks
• April 9th 2011, 06:22 AM
Prove It
Whenever you square an equation or inequation to aid in solving it, you are likely to bring in extraneous solutions.
What I would do is invert both sides, noting that inverting both sides of an inequation changes the direction of the inequality sign. The rest should be obvious.
• April 9th 2011, 06:23 AM
TheEmptySet
First we need to get a common denominator on both sides
$\displaystyle \frac{1}{x-3}> 4 \iff \frac{1}{x-3}>\frac{4(x-3)}{x-3}$
simplify to get
$\displaystyle \frac{1}{x-3}>\frac{4x-12}{x-3} \iff 0> \frac{4x-13}{x-3}$
So the two critical numbers(where the inequality can chage) are $\displaystyle x=3,\quad x=\frac{13}{4}$
If you test some points you will get
$\displaystyle \left(3,\frac{13}{4}\right)$
• April 9th 2011, 06:28 AM
Prove It
Quote:
Originally Posted by TheEmptySet
First we need to get a common denominator on both sides
$\displaystyle \frac{1}{x-3}> 4 \iff \frac{1}{x-3}>\frac{4(x-3)}{x-3}$
simplify to get
$\displaystyle \frac{1}{x-3}>\frac{4x-12}{x-3} \iff 0> \frac{4x-13}{x-3}$
So the two critical numbers(where the inequality can chage) are $\displaystyle x=3,quad x=\frac{13}{4}$
If you test some points you will get
$\displaystyle (-\infty,3) \cup \left(\frac{13}{4},\infty \right)$
Sorry but for any $\displaystyle x$ in $\displaystyle (-\infty, 3)$, $\displaystyle \frac{1}{x - 3}$ gives something negative, which is clearly NOT greater than $\displaystyle 4$...
• April 9th 2011, 06:35 AM
HallsofIvy
The way I would have done it is this: $\frac{1}{x-3}> 4$ is the same as $0< 4- \frac{1}{x- 3}= \frac{4x- 13}{x- 3}$. Now, that can change from "> 0" to "< 0" only where the numerator, 4x- 13, is equal to 0 (so the fraction is 0) or where x- 3= 0 (where the fraction is "discontinuous"). The sign is constant on the intervals $(-\infty, 3)$, $(3, 13/4)$, and $(13/4, \infty)$. If x= 0, which is less than 3, $\frac{1}{0-3}= -\frac{1}{3}< 4$. If x= 4, which is larger than 13/4, $\frac{1}{4-3}= 1< 4$. If x= 3.1, which is between 3 and $\frac{13}{4}= 3\frac{1}{4}$, $\frac{1}{3.1- 3}= \frac{1}{0.1}= 10> 4$.
Therefore, the inequality is true on the interval $\left(3, \frac{13}{4}\right)$.
• April 9th 2011, 06:35 AM
TheEmptySet
Quote:
Originally Posted by Prove It
Sorry but for any $\displaystyle x$ in $\displaystyle (-\infty, 3)$, $\displaystyle \frac{1}{x - 3}$ gives something negative, which is clearly NOT greater than $\displaystyle 4$...
You are correct I cant add this morning! thanks
• April 9th 2011, 06:41 AM
veileen
"If you test some points you will get" Absolutely wrong. In general, you learn math for not testing things - anybody can do that.
The second method is wrong. You don't know if $\frac{(x-3)^2}{x-3}=x-3$ or $3-x$.
First method is wrong too. You can square only after you say that x must be bigger then 3.
• April 9th 2011, 06:47 AM
Prove It
Quote:
Originally Posted by HallsofIvy
The way I would have done it is this: $\frac{1}{x-3}> 4$ is the same as $0< 4- \frac{1}{x- 3}= \frac{4x- 13}{x- 3}$. Now, that can change from "> 0" to "< 0" only where the numerator, 4x- 13, is equal to 0 (so the fraction is 0) or where x- 3= 0 (where the fraction is "discontinuous"). The sign is constant on the intervals $(-\infty, 3)$, $(3, 13/4)$, and $(13/4, \infty)$. If x= 0, which is less than 3, $\frac{1}{0-3}= -\frac{1}{3}< 4$. If x= 4, which is larger than 13/4, $\frac{1}{4-3}= 1< 4$. If x= 3.1, which is between 3 and $\frac{13}{4}= 3\frac{1}{4}$, $\frac{1}{3.1- 3}= \frac{1}{0.1}= 10> 4$.
Therefore, the inequality is true on the interval $\left(3, \frac{13}{4}\right)$.
Actually $\displaystyle \frac{1}{x - 3} > 4$ is the same as $\displaystyle 0 > 4 - \frac{1}{x - 3}$...
• April 9th 2011, 07:16 AM
veileen
A way this kind of inequalities can be solved:
$\frac{1}{x-3}>4 \Leftrightarrow 4-\frac{1}{x-3}<0 \Leftrightarrow \frac{4x-13}{x-3}<0$
http://img193.imageshack.us/img193/5701/frlu.jpg
You want that fraction be negativ, so $x \in \left ( 3, \frac{13}{4} \right )$.
• August 14th 2011, 03:37 AM
aonin
Re: Inequality
Quote:
Originally Posted by veileen
The second method is wrong. You don't know if $\frac{(x-3)^2}{x-3}=x-3$ or $3-x$.
sorry for bringing up an old thread, but I thought about this inequality thing again, looking over what you wrote isn't $\frac{(x-3)^2}{x-3}=x-3$ always?, isn't this true for all real numbers x, for example sub in x=5 then the answer would be 2, which is 5-3, if subbing in lets say a negative number -1, then it would be 16/-4=-1-3 which is true again, the same is true for x=0 or a positive number smaller than 3, eg 1
• August 14th 2011, 03:53 AM
Siron
Re: Inequality
The statement:
$\frac{(x-3)^2}{x-3}=3-x$ is not true because, that would mean $(3-x)\cdot(x-3)=(x-3)^2 \Leftrightarrow -x^2+6x-9=x^2-6x+9$ which is not true for all $x$.
• August 14th 2011, 12:23 PM
HallsofIvy
Re: Inequality
Quote:
Originally Posted by veileen
"If you test some points you will get" Absolutely wrong. In general, you learn math for not testing things - anybody can do that.
There are a lot of things that anybody can do- that doesn't mean it is wrong.
The only places an inequality can change form ">" to "<" is where the function is 0 or is not continuous. For fractions that is where the numerator is 0 or where the denominator is 0. Once you have found those points, checking one point in each interval tells you which. What TheEmptySet said was perfectly valid.
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http://physics.stackexchange.com/questions/5829/why-are-some-solitons-formed-from-bosonic-fields-fermionic/5854 | # Why are some solitons formed from bosonic fields fermionic?
Some topological solitons formed from bosonic fields have fermionic statistics. Why?
-
## 3 Answers
A short answer: The soliton in bosonic field theory can be fermionic because the model secretly contains massive fermions at high energies.
This is because in order to define an bosonic field theory, we need to non-perturbatively regulate the field theory. So let us put the bosonic field theory on a lattice to non-perturbatively regulate the theory (otherwise, the field theory is not even well defined at non-perturbative level). The claim is that to produce the needed topological term that makes the soliton an fermion, the lattice model must contain fermions with finite energy gap.
A more precise claim: Any gapped bosonic model on lattice that have unique ground state on closed space of any topology do not contain fermionic quasiparticles.
Basically, "gapped bosonic model on lattice that have unique ground state on closed space" implies that the model has no topological order. The only way to have emergent fermion from a bosonic lattice model is to have a non-trivial topological order. Also see a related discussion, where I claim that, in lattice bosonic model, the emergent fermion must appear together with emergent gauge theory at low energies. Skyrme-model contain no low energy gauge theory. This is why I claim that the Skyrme-model secretly contains massive fermions at high energies.
-
There are many examples. Moshe focused on lower-dimensional theories; for example, in 1+1 dimensions, bosonic and fermionic conformal field theories are typically totally equivalent to each other (and one may get fermions as kinks of the bosons, and bosons as bilinear currents of the fermions). I will focus on higher-dimensional theories.
I don't really know how to get fermions in the $D=26$ bosonic string theory. But in other, $D=10$ string theories that normally contain bosons only, fermions may be obtained - see e.g. this paper by Justin David, Shiraz Minwalla, and Carlos Nunez (whose writing I remember pretty well, back from Santa Barbara 2001 etc.)
http://arxiv.org/abs/hep-th/0107165
The basic trick, known already in 3+1-dimensional bosonic gauge theory (by Jackiw-Rebbi-Hasenfratz-’t Hooft), is "spin from isospin".
The soliton is only invariant under the "diagonal group" of the rotations and some internal isospin group. Noether's theorem dictates that the right definition of the spin of the excitations is associated with whatever symmetry we have. In this case we have the diagonal symmetry, so the new conserved angular momentum will be the diagonal one. Because half-integer-isospin excitations are possible, they will look like half-integer-spin excitations of the soliton.
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The fermionic statistics has to do with a minus sign acquired when two objects are interchanged. Usually this is unambiguously attributed to statistics, as opposed to interaction between the objects, since any interaction falls off rapidly when they are far apart, but the relative phase acquired due to interchange is the same whether they are close or far.
However, in low dimensional systems the interaction strength between well-separated objects does not always fall off very fast (or not at all). So, one can get fermionic (or more generally anyonic) behaviour from interaction. For some solitons the interactions is such that they acquire a phase when interchanged, no matter how far apart they are.
This arguments only shows it can happen in low dimensional systems, more detailed arguments will show you how it happens (for example google sine-Gordon and Thirring models, or bosonization). The why questions I am never sure how to answer.
-
4
Additional comment is that even in 4d, the WZW term can make a soliton constructed from scalar fields alone fermionic; the skyrmion becoming a baryon when the number of color is odd is the typical example. In a sense the Lagrangian in this case has a term which produces long-range phase. – Yuji Feb 25 '11 at 6:39
Great point, thanks! – user566 Feb 25 '11 at 6:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8971505761146545, "perplexity_flag": "middle"} |
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1
### Keep the light beam in a closed room, is it possible? [duplicate]
No mirror can be perfectly reflective due to quantum tunneling so that already answers your question. But even if it could be done, you would never be able to check the situation because when you look inside, the light almost instantly leaves through the peephole. This also poses a problem for your initiation method, which John M already touched on: you need ...
0
### Keep the light beam in a closed room, is it possible? [duplicate]
Short answer: No. You have to remember that light is very fast. The law of specular reflection states that an angle made with a mirror and incoming beam will be the same angle the outgoing beam and the mirror make. If you were able to put light in completely perpendicular to the surface of lets say a half-a-meter radius mirror(insides are mirrors) sphere ...
2
### Infinite reflection of light and the conservation of energy / momentum
First, of course there's no perfect mirror. But let's assume there was one. Next, the question is: Is the bouncing off the mirrors elastic or inelastic. If the photon is absorbed and re-emitted with the same frequency, then the bouncing is elastic and no energy is lost by the photon. It would then go on forever and ever. But what if it does lose energy ...
1
### Why do focal lengths affect magnification?
It's like leverage. The longer the distance from the objective lens to the virtual image, the larger the virtual image. Imagine there's a piece of frosted glass at the focal point. It will show the virtual image. Now the eyepiece looks at that virtual image with a magnifying glass. That also makes it look bigger.
1
### The weight of a cavity of radiation
You're not being very careful about your terminology. Instead of weight I think you're really talking about the total mass-energy of the sealed cavity. If you want to discuss weight the it would be better to think of that as the force on the total mass-energy in a gravitational field. Thinking about it in terms of mass-energy makes the question much ...
0
### Why does the quantum eraser seem to violate energy and momentum conservation?
There is a misunderstanding here. If the beam(s) are not changed at the origin, they deliver the same energy on the screen whether one sees interference or not. It just has a different spatial distribution. Conservation of energy says that the energy of the beam(s) should equal the energy absorbed by the screen in total plus the energy reflected. It has ...
3
### Selectively visible laser beam with a controllable means
There is no magic here. To see the laser beam, the beam has to use light in the visible frequency and some of the light needs to enter your eyes. Unless the beam is pointing at your eyes you won't see it. Remember, in coherent light all the photons are traveling in essentially the same direction, with the same frequency, in the same phase. If you want to ...
7
### Can a photon exhibit multiple frequencies?
Yes. Consider quantizing electromagnetic fields in a box. This corresponds to photons being trapped inside of said box since photons are just the mode quanta of the EM fields. The Hilbert space (called Fock space in this case) of the quantized radiation is found to be spanned by states |\mathbf k_1, \mu_1; \dots, ; \mathbf k_N, \mu_N\rangle, \qquad ...
1
### Selectively visible laser beam with a controllable means
A piece of white paper is the most common device to show a laser beam. The beam is visible on both sides of the paper. Fix it on a electronically controlled motor shaft you can rotate the paper into the laser beam. If precise control is necessary: just align an iris aperture to the center of the laser beam.
1
### Light Ray Reflection from concave mirror
Assuming mirror to be spherical section. C is the center of sphere. See, Using trigonometry. $$x=d \times \sin(2\theta)$$ $$x=R\times\sin\theta$$ Eliminate $\theta$ and get $d$ : distance from Center of curvature as a function of $x$. Verify for small theta where $\sin\theta\approx\theta$ If you just want to see that which side ray bents then see. ...
1
### Light Ray Reflection from concave mirror
A parabolic mirror is a special case of a concave mirror. The rays at the rim are refracted to the focal point. This is true in simplification of geometric optics and perfect manufactured mirror. However in modern techiques like injection molding there are more imperfections at the rims of mirrors. The question of a general concave mirror can be answered ...
0
### Is light red shifted in optical tweezers?
In theory, yes, the light will be redshifted. In practice, it sounds like the glass bead is too large for any measurable red shift. This is actually used in Mossbauer spectroscopy. What happens is that if your $\gamma$-ray source is a free atom, the recoil of the atom will cause the resulting radiation to be red-shifted relative to the natural frequency of ...
1
### Is light red shifted in optical tweezers?
Sure - the relativistic doppler effect means that light which is scattered off a moving object can be redshifted or blueshifted. And there can be more redshifted photons than blueshifted photons, or vice-versa, depending on where the object is, and how it's moving, relative to the center of the trap. But since the object is moving much much much less than ...
1
### Confusing mirror problem
You find $s$ in the same way you found it for concave mirror, just keep in mind that for the convex mirror the image is always virtual(as explained nicely in the Wikipedia article on curved mirrors, it cannot be projected on a surface, unlike the real image), so in this case (by convention) $s'=-35cm$ and $f=-53cm$.
3
### Three polarizers, 45° apart
This link: http://alienryderflex.com/polarizer/ has an excellent explanation; much better than anything I could write here. Essentially, it says that this occurs because the 45 degree filter outputs a projection of the vertical rays at 45 degrees. This, in turn, has a horizontal component, which the final filter projects in its output.
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### Young's double slit experiment viva question
The intensity distribution will remain same as fraunhofer one only , however intensity will increase at each point . Means supposedly earlier there was Amplitude 2A , at a point of maxima , now it will become 4A , since 2 waves will come from the new fraunhofer slit also as it is not very distant from the previous slit , hence , you can assume the maximas ...
2
### Can the choice of reflection angle for light can be derived from a minimality condition?
This follows from from Fermat's principle. A derivation for the reflection is e.g. found here.
2
### How can a Jones vector give linear polarization along an axis?
The component of the Jones vector are not defined as two pure imaginary exponentials but as two complex numbers $z_1$ and $z_2$ whose polar decomposition is $z_i=|z_i|e^{i\phi_i}$ for $i=1,2$. To describe linearly polarized light along the $x$ axis $$|H \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$ we just take $z_1 \in \mathbb{R}$ ($\phi_1=0$) and ...
Top 50 recent answers are included | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9216422438621521, "perplexity_flag": "middle"} |
http://crypto.stackexchange.com/questions/3123/probability-of-repeated-encrypted-block-in-output-feedback-ofb-mode/3125 | # Probability of Repeated Encrypted Block In Output Feedback (OFB) Mode
Is there any probability/chances of repeated encrypted block in Output Feedback (OFB) mode? Is there any existing documentation/studies about that? If it is existing, can you please provide me a link for that documentation, I want to analyze it carefully, I can't find some documentation about the probability of repeated encrypted block in OFB, maybe it is existing but it is hard to find.
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## 1 Answer
Well, if the block cipher is modeled as a random $N$ bit permutation (that is, each permutation from the set of $2^N$ bit patterns to itself is equally probable), then the answer is really quite easy (and this answer is exact): the probability that we will repeat a block within $M$ outputs is precisely $(M-1) 2^{-N}$ (for $0 < M \le 2^N+1$).
The reasoning behind this is remarkably simple. First, when we consider the sequence $X, P(X), P(P((X)) = P^2(X), P^3(X), ...$, we see that the first pair of repeated elements must include the starting point $X$. Here's why: if we have the pair $P^i(X) = P^j(X)$ for $i, j \ne 0$, then we have $P^{i-1}(X) = P^{j-1}(X)$ (remember, $P$ is a permutation, that is, if $P(A)=P(B)$, then $A=B$), and so if we have a pair $i,j \ne 0$, that can't be the first pair.
Now, let us compute the conditional probability that, if we haven't had a repeat after $M-1$ outputs, we get a repeat after $M$ outputs. Now, after we have cycled through $M-1$ outputs, there are $2^N - (M-1)$ outputs we haven't generated before and $1$ output that we have that might be generated in the next step (the other outputs we have generated are impossible, by the reasoning in the above paragraph). If we've specified $(M-1)$ outputs of a permutation, the rest of the possible outputs are equiprobable, and hence the probability that we're generate the one output that would generate a repeat is $1 / (1 + 2^N - (M-1)) = 1/( 2^N-M)$
Now, the probability that we'll generate the first repeat at step $M$ is the probability that we'll not generate a repeat after $M-1$ steps, and then repeat at step $M$. This is:
$Prob(M) = (2^N-1)/2^N \cdot (2^N-2)/(2^N-1) \cdot ... \cdot (2^N-M) / (2^N-M+1) \cdot 1 / (2^N-M) = 1/2^N$
That is, the probability that we'll get a repeat after exactly $M$ steps is independent of the value of $M$ (as long as it is in range).
And hence, the probability we'll get a repeat after $M$ steps (or earlier) is just the sum of the probabilities that we'll generate the first repeat after $k$ steps for $1 < k \le M$, and that's $(M-1)/2^N$.
If you insist on a link, you can look at this one; this uses different logic to come up with the same result.
If the block cipher cannot be modeled as a random permutation (or a random even permutation; that answer differs only in the $M=2^N$ case), well, I don't know on any specific study; that would probably depend on the specific block cipher, and how it differs from a random permutation.
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Nice answer, although I did understand the question a bit different ("what is the probability that ciphertext blocks repeat", which is harder, as it depends on the plaintext) - but your interpretation makes more sense. – Paŭlo Ebermann♦ Jul 2 '12 at 17:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9493228793144226, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/39307/question-on-sakurais-treatment-of-the-harmonic-oscillator | Question on Sakurai's treatment of the Harmonic Oscillator:
In Section 2.3 of the second edition of Modern Quantum Mechanics (which discusses the harmonic oscillator), Sakurai derives the relation $$Na\left|n\right> = (n-1)a\left|n\right>,$$ and states that
this implies that $a\left|n\right>$ and $\left|n-1\right>$ are the same up to a multiplicative constant.
To my sensibilities, this is only implied if the $\lambda$-eigenspace of the number operator $N:=a^{\dagger}a$ corresponding to $\lambda=n-1$ is one-dimensional. If it is multidimensional, then we cannot say that $a\left|n\right>$ and $\left|n-1\right>$ are proportional. So (unless I've made some fundamental error) how do we know that the $\lambda$ eigenspaces of $N$ are one dimensional?
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1 Answer
OP wrote(v1):
So (unless I've made some fundamental error) how do we know that the $\lambda$-eigenspaces of $N$ are one dimensional?
Yes, OP is right. In general, we cannot know. There exist reducible unitary representations of the Heisenberg algebra $[a,a^\dagger]=1$, where the eigenvalues of $N$ are degenerated.
However, if one assumes that the ket Hilbert space is a non-trivial irreducible unitary representation of the Heisenberg algebra, then one may show that the eigenvalues of $N$ must be non-degenerated. See e.g. this Phys.SE answer.
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Much appreciated. Thanks for the link. – Nick Thompson Oct 8 '12 at 3:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "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}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8651791214942932, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/tagged/torsion | ## Tagged Questions
0answers
148 views
### Torsion product Tor^R_1(,) [closed]
We know that: Torsion product Tor$^R_1(,)$, which maps two modules, $M_1$ over $R$ and $M_2$ over $R$, to a third module $M_3$ over $R$: $M_3$ = Tor$^R_1(M_1,M_2)$. See for examp …
4answers
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### Why is it important that partial derivatives commute?
I am asking this in the context of differential geometry (specifically Riemannian). When the Levi-Civita Connection is defined, we require that the torsion tensor is 0, which in l …
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### About the parallel transport and choice of connection
Thought Experiment Consider a 2-sphere, $S^2$, and let $p$ be a point at the equator. Case 1 Let us parallel transport a vector, $V$ from $p$ using the recipe: Move one unit o …
13answers
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### What is torsion in differential geometry intuitively?
Hi, given a connection on the tangent space of a manifold, one can define its torsion: $$T(X,Y):=\triangledown_X Y - \triangledown_Y X - [X,Y]$$ What is the geometric picture behin …
2answers
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### Interpretation of Curvature and Torsion
Dear all, When dealing with General Relativity one uses the Levi-Civita connection with is torsion-free. Thus the commutator of the covariant derivatives yields \$[\nabla_\mu,\nab …
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### Rolling without slipping interpretation of torsion
This is, in a sense, a follow up to this question. Hehl and Obukhov try to give an intuitive description of torsion. I am confused about their description. I am looking at the fol …
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### Torsion-free tensor powers
Does there exist an integral domain $R$ and an $R$-module $M$ that is not flat over $R$ such that every tensor power of $M$ over $R$ is nonzero and $R$-torsion-free? (If such a do …
0answers
221 views
### connection between non-orientable manifolds and torsion in 1D (co) homology
I'm interested in understanding the probability that given a prime $p$, $p$ divides the order of the torsional part of $H^k(X,Z)$, where $X$ is a finite simplicial complex. Lets s … | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 24, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8598427176475525, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/70377/meaning-of-regular-neighborhood-for-homology-basis-curves-in-s-g-2 | ## Meaning of Regular Neighborhood for Homology Basis Curves in $S_{g,2}$
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi All:
I have been trying to understand the meaning of the expression "regular neighborhood" in the context described below, but I'm stuck:
We have a collection of curves ci; i=1,2,..,n. embedded in Sg, the orientable genus-g surface, with 2 boundary components , satisfying these properties:
i) ci intersects ci+1 transversely in a single point, and the algebraic intersection ci∩ci+1 is +1
ii)ci∩cj is empty for |i-j|>1, and
iii) The homology classes of the ci are linearly-independent.
The claim is then made that if n is even, the regular 'hood (neighborhood) has genus
$\frac{n−1}{2}$ ,and two boundary components, while if n is odd, the 'hood has $\frac{n}{2}$ and one boundary component if n is even.
To add some context, this paper is an effort to show that (here for the case of $S_{g,2}$ where g is the genus and 2 is the number of boundary components )the Torelli group is finitely-generated by "bounding pairs" (BP's) , meaning Dehn twists in opposite directions about the bounding curves.
My understanding of regular neighborhoods is limited; I have had trouble finding a precise definition for them; there is one for simplicial complexes, and I have also seen , I think, descriptions of regular neighborhoods which see to come down to being tubular 'hoods, but neither of these seems to apply. The case of the algebraic intersection has to see with a choice of orientation for the (tangent spaces at intersection points of the )ci's , so that I do not see how this would help.
I think the author (D.Johnson) is choosing the ci′s in a way that consecutive ones are "perpendicular", in that planes containing them would be perpendicular. I think also the cj′s are supposed to be some variant of a symplectic basis (def. as being a basis {x_i,y_i} for H1(Sg) so that xi intersects yj exactly once when i=j, and the xi do not intersect the yj otherwise.
Edit: I hope this is not against MO protocol, but I thought I may tie-in another question: what is the spine of a surface? I think this is a generalization of the simplicial skeleton, but it would be nice to have something more precise.
Thanks in Advance.
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I would interpret that as the union of regular neighbourhoods of the single curves: basically a union of bands $S^1\times (-1,1)$, one for each $c_i$, such that the intersection of any two is either empty or a square (i.e. the product of an interval in $S^1$ with $(-1,1)$, with the identification reversing the two factors). The genus count works well in this case. – Marco Golla Jul 14 2011 at 23:22
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I'm voting to close as this is a duplicate of your question on math.stackexchange. See: math.stackexchange.com/questions/51484/… – Ryan Budney Jul 15 2011 at 2:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.951111912727356, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/203923/help-with-relation-on-infimum | # Help with relation on infimum
I'm trying to show: $\inf(x_k+y_k) \leq \inf(x_k)+\inf(y_k)$. So I did:
Let $r>0$, there is a $k_1$ such that $x_{k_1}\leq \inf(x_k)+r/2$ and there is a $k_2$ such that $y_{k_2}\leq \inf(y_k)+r/2$.
Then: $x_{k_1}+y_{k_2} \leq \inf(x_k)+\inf(y_k)+r$
I'd like to take the inf and make $r \rightarrow 0$, but since $k_1$ and $k_2$ may be different I can't conclude. Please somebody help!
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"Inf" is super-additive, while "Sup" is subadditive. – Siminore Sep 28 '12 at 11:17
## 1 Answer
Let $x_k=(-1)^k$, $y_k=(-1)^{k+1}$. Then $x_k+y_k=0$ for all $k$, hence the $\inf$ is $0$ as well. However, $\inf x_k=\inf y_k=-1$.
- | {"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": 17, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9759030342102051, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/269128/compact-maps-problem-in-lax/277252 | # Compact maps problem in Lax
In Functional Analysis of Peter Lax there are the following exercise
Show that if $\bf C$ is compact and $\{{\bf M}_n \}$ tends strongly to $\bf M$, then $\bf CM_n$ tends uniformly to $\bf CM$.
Assumptions are that ${\bf C}: {\bf X} \rightarrow {\bf X},{\bf M_n}: {\bf X} \rightarrow {\bf X}$ where $\bf X$ is a Banach space
I was thinking that one could use that if let $x_i$ be such that $|({\bf M_i-M})x_i|\ge||{\bf M_i-M}|| - \epsilon$. Then from compactness of $\bf C$ we have that there is a finite sequence $j=1\ldots,n$ of ${\bf C}({\bf M_j-M})x_j$ s.t $\min_j ||{\bf C}({\bf M_j-M})x_j - {\bf C}({\bf M_i-M})x_i||<\epsilon$ for all i. But I dont get anywhere.
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Maybe try to approximate $C$ by suitable finite rank operators? (Assuming you work in Hilbert space) It might also be useful to note also that a strongly convergent sequence must be norm-bounded by uniform boundedness. – Martin Jan 2 at 13:43
@Davide from X->X, where X is Banach. – Jonas Wallin Jan 2 at 13:58
@Martin not in a Hilbert space. Yes, norm-bounded might be needed. – Jonas Wallin Jan 2 at 14:01
## 2 Answers
I believe that the problem is wrong. Here is an counter example:
Assume that ${\bf X}=l^2$, ${\bf C}x=(x,e_1)e_1$ and ${\bf M}_nx=(x,e_n)e_1$.
${\bf M_n} x\rightarrow 0$ for all $x$, so ${\bf M_n} \rightarrow {\bf 0}$.
But $||{\bf C M}_n - {\bf C 0}|| = ||{\bf C M}_n|| \geq ||{\bf C M}_ne_n||=||(e_n,e_n)(e_1,e_1)e_1||=1.$
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As Jonas Wallin points out in his answer, it's not necessarily true that $CM_n-MC\to 0$ uniformly even in the context of Hilbert spaces. It's however true in finite dimensional vector spaces, as strong convergence is the same thing as uniform convergence in this particular case.
However, maybe Lax meant to show that $M_nC-MC\to 0$ uniformly. In this case, we can follow the following steps.
1. We assume that $M=0$, otherwise consider $M_n-M$.
2. Using the principle of uniform boundedness, we obtain that $R:=\sup_{n\in\Bbb N}M_n$ is finite.
3. Let $\delta$ such that $\delta\leqslant\limsup_{n\to +\infty}\lVert M_nC\rVert$. Let $\{n_k\}$ a strictly increasing sequence of integers such that $\delta\leqslant \lVert M_{n_k}C\rVert$, and $x_k$ of norm $1$ such that $\lVert M_{n_k}C\rVert\leqslant k^{-1}+\lVert M_{n_k}Cx_k\rVert$.
4. The sequence $\{Cx_k\}$ lies in a compact set, hence we extract a converging subsequence $\{x_{k'}\}$ to some $y$.
5. We have for all $k'$ that $$\delta\leqslant k'^{-1}+\lVert M_{n_{k'}}(Cx_{k'}-y)+M_{n_{k'}}y\rVert\leqslant k'^{-1}+R\lVert Cx_{k'}-y\rVert+\lVert M_{n_{k'}}y\rVert.$$
6. Taking the limit $k'\to +\infty$ in the later inequality, we get that $\delta=0$.
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1
Actually Lax's question was two show it both ways that is both $M_nC-MC \rightarrow 0$ and $CM_n-CM \rightarrow 0$. – Jonas Wallin Jan 17 at 12:42
So compact operators correct strong convergence only "by the right side". – Davide Giraudo Jan 17 at 14:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9464272260665894, "perplexity_flag": "head"} |
http://mathhelpforum.com/differential-geometry/160568-convexity-epigraph-question.html | # Thread:
1. ## convexity and epigraph question
Prove that if the epigraph is a convex set then the function below the epigraph is convex.
Here's what I have so far.
The epigraph is going to be in $\mathbb{R}^{n+1}$ and the supporting function in $\mathbb{R}^n$.
In the epigraph we are going to have a hyperplane of n+1 dimensions and a hyperplane of n dimensions for $f(x_1,x_2,..x_n)$. So what I"m thinking is to find find the maximum slope of of the function $f$ and rotating the hyperplane in the epigraph so that both are going to be parallel (this is the part I have no idea on how to do, as we can possibly in 4+ dimensions. I have no idea how a rotation matrix will look like). At which point we simply lower the hyperplane from the epigraph to hit the function $f(x_1,x_2,..x_n)$ but still remain above $f(x_1,x_2,..x_n)$. Now every value from $f(x_1,x_2,..x_n)$ will be below the hyperplane (from the epigraph) showing that $f(x_1,x_2,..x_n)$ is convex. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9448419809341431, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/69946?sort=oldest | ## Why is this theorem (about $L(P(\omega_1))^V$ and $L(P(\omega_1))^{V[G]}$) nice?
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I was recently told that the following (due to M. Viale) is a nice theorem:
Suppose there are arbitrarily large supercompacts, and $\mathrm{MM}$ holds in $V$. Let $G$ be generic for a proper forcing and $V[G] \vDash \mathrm{MM}$. Then $L(P(\omega_1))^V$ is elementarily equivalent to $L(P(\omega_1))^{V[G]}$.
My question (borne of ignorance, not skepticism) is:
Why is this theorem nice, and how does it fit into the bigger picture?
Some slightly-more-specific questions that refine my main question are: Do the hypotheses of this theorem often come up in natural settings? What's the upshot of the conclusion? Is it that proper forcing which preserves $\mathrm{MM}$, leaves the theory of a small but not-that-small chunk of the universe unchanged?
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My view is that questions about highly technical results such as this needn't be community wiki, even when they are asking for intuition or soft information about them. – Joel David Hamkins Jul 11 2011 at 0:43
It happens that Viale just spoke on his theorem today at a conference here in Singapore (www2.ims.nus.edu.sg/Programs/011aiic), and I had a chance to talk to him about it. The theorem, a little different than stated in the question, asserts that if MM++ holds and there are a proper class of inaccessible limits of supercompact cardinals, then the theory of the $\omega_1$-Chang model $L([\text{Ord}]^{\lt\omega_1})$ cannot be changed by stationary-preserving forcing that preserves MM++. The axiom MM++ strengthens MM by requiring that names for stationary sets have stationary values. – Joel David Hamkins Jul 21 2011 at 10:40
Oops, I meant $L([\text{Ord}]^{\lt\omega_2})$ as Matteo says. – Joel David Hamkins Jul 22 2011 at 4:06
## 2 Answers
By your tags you've asked for a soft answer, and so let me try to provide one.
The theorem is indeed very nice and engages with and reinforces a number of philosophical views in set theory.
First, there is the idea that large cardinal axioms are leading us towards the final, true set-theory, and so set-theorists are keenly interested when the existence of large cardinals causes a fragment of set-theoretic truth to stabilize in the lower part of the universe. One could take this to mean that that stabilized fragment is a part of the final theory we seek. Although Gödel's hopes that large cardinals would settle CH were dashed by the Levy-Solovay theorem, nevertheless the phenomenon does exist. Increasingly large large cardinal assumptions, for example, imply that increasingly large portions of the projective hierarchy have extremely nice properties (Lebesgue measurable, property of Baire, determinacy, etc.). If there are infinitely many strong cardinals, then it is consistent that projective truth is invariant by forcing. When there is a proper class of Woodin cardinals, then the theory of $L(\mathbb{R})$ is invariant by set forcing. This paper by Neeman and Zapletal shows that if there is a weakly compact Woodin cardinal, then for any proper forcing extension $V[G]$, there is an elementary embedding $j:L(\mathbb{R})^V\to L(\mathbb{R})^{V[G]}$. The idea is that once we have sufficient large cardinals, then one cannot change the universe too much by this kind of forcing.
Viale's theorem essentially extends this from $L(\mathbb{R})$ to $L(P(\omega_1))$, which is impressive.
Second is the philosophical idea that much of the indeterminism of set-theory is due to the anomalous effects of forcing. That is, the extreme flexibility of forcing causes so much set-theoretic chaos---we can turn the continuum hypothesis on and off like a lightswitch---and so when we come to know that a statement or class of statements is invariant by forcing, or by a huge natural class of forcing such as proper forcing, then this is really significant. The theorem and the others I have mentioned are all instances where the chaotic nature of forcing is controlled or restricted by the existence of large cardinals.
Third, there is the view of Martin's Maximum MM as expressing a fundamentally important truth about the universe. Justin Moore has emphasized its attractive nature as a fundamental axiom generalizing the Baire category theorem. Viale's theorem essentially says that when there are sufficient large cardinals, then Martin's maximum completely smoothes out the chaotic effects of (proper) forcing, since under statements in $L(P(\omega_1))$, which is a huge fragment of the universe, cannot be affected by proper forcing preserving MM. This result therefore underscores the idea that MM and large cardinals cause a measure of stability in the set-theoretic universe. (Note that the theorem is definitely false without the MM preservation, since proper forcing can switch the CH lightswitch, and CH is expressible in $L(P(\omega_1))$.)
Meanwhile, I expect that other set-theorists can add important insights.
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In the Neeman/Zapletal result, I should have said "for any small proper forcing," meaning the size of the forcing notion is below the weakly compact Woodin cardinal. – Joel David Hamkins Jul 11 2011 at 13:58
Thanks, this is great, as always! – Amit Kumar Gupta Jul 16 2011 at 2:54
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Let me give a brief answer as the author of the mentioned theorem. I shall first say that (as you might imagine) I was very pleased to see that my theorem raised your interest. Now, as mentioned by Joel in his latest post, your statement of the theorem is not correct. However there is a typo also in Joel's post, namely the conclusion of the theorem holds for the Chang model $L([Ord]^{<\omega_2})$ and not for the Chang model $L([Ord]^{<\omega_1})$ as posted by Joel. For this Chang model the result is due to Woodin and was already known, it appears for example in Chapter 3 of Larson's book on the stationary tower forcing. To complement the very good answer Joel has already given you, I invite you to read the many survey papers related to the philosophical position subsumed by this theorem, here is a sample list: Woodin's two short papers for the "notices of the AMS" where he exposes the basic ideas behind the generic absoluteness results for $L(\mathbb{R})$:
• Woodin, W. Hugh (2001a). "The Continuum Hypothesis, Part I" (PDF). Notices of the AMS 48 (6): 567–576. http://www.ams.org/notices/200106/fea-woodin.pdf.
• Woodin, W. Hugh (2001b). "The Continuum Hypothesis, Part II" (PDF). Notices of the AMS 48 (7): 681–690. http://www.ams.org/notices/200107/fea-woodin.pdf.
Several of Peter Koellner's papers available at his webpage:
http://www.people.fas.harvard.edu/~koellner/
Most of them contains a long introductory part which motivates and explains very carefully and plainly the ideas at the heart of the $\Omega$-logic approach to absoluteness result.
It has to be noted that the kind of solution to the continuum problem prospected by my theorem follows Woodin's view as exposed in the papers on AMS notices. Currently Woodin has pursued a different approach towards the solution of the continuum problem that leads him to prospect a view of the universe (Ultimate $L$) which is radically different from the one given by MM.
Finally in case you are interested there are in my webpage several slides of talks I gave on this and related subjects, as well as a proof of the theorem you have mentioned in your question: http://www2.dm.unito.it/paginepersonali/viale/index.html (unfortunately my university is currently changing the websites locations, so for some time you may have troubles to consult it....)
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Matteo, welcome to MathOverflow! – Joel David Hamkins Jul 22 2011 at 4:12
Thanks! Yes, I've seen Koellner's great paper as well as Woodin's articles and the slideshows on your site, I really appreciate seeing the clear explanations of how these results fit into the broader perspective and their philosophical consequences for questions of foundations. Cheers! – Amit Kumar Gupta Jul 25 2011 at 22:25 | {"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": 24, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9556078910827637, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/85595?sort=votes | ## Relation between combinatorial manifolds and PL manifolds
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In
W. W. Boone, W. Haken, and V. Poenaru, On Recursively Unsolvable Problems in Topology and Their Classification, Contributions to Mathematical Logic (H. Arnold Schmidt, K. Schütte, and H. J. Thiele, eds.), North-Holland, Amsterdam, 1968.
a combinatorial manifold is defined as a simplicial complex with the property that the star of every vertex is combinatorially equivalent to the standard $n$-simplex. (two simplicial complexes are combinatorially equivalent if they possess linear subdivisions, the associated abstract simplicial complexes of which are isomorphic) This is equivalent to the condition that the link of every vertex be a combinatorial $(n-1)$-sphere (=boundary of the standard $n$-simplex) if the underlying manifold has no boudnary.
However, in
A. Ranicki (ed.), The Hauptvermutung Book, K-Monographs in Mathematics, vol. 1, Kluwer Academic Publishers, Dordrecht, Boston, London, 2010.
on page 4, this condition is used to define the term
combinatorial manifold (or PL manifold)''.
I find this very weird; a PL manifold should be defined as a topological manifold with a maximal atlas of homeomorphisms with PL coordinate changes (and I know a lot of authors who use this definition).
The obvious question now is: Is a simplicial complex, the vertices of which have $S^{n-1}$ as link, the same as a topological manifold with a maximal atlas of homeomorphisms piecewise linear coordinate changes? Of course, Ranickis nomenclature implies that it does.
Obviously, the condition on the links can be used to construct such an atlas. However, the converse puzzles me, as it seems to be equivalent to the question if every manifold with a maximal PL atlas admits a triangulation.
If anyone could point me to an article where this problem is addressed, I would be thrilled.
Best regards, Malte
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Perhaps I'm missing something but this seems clear to me. take a locally finite PL atlas and triangulate each chart linearly on some compact polyhedral domains that still cover M. Since the transition maps are PL and the cover is locally finite you can subdivide the triangulations to make the transition maps linear on simplexes. The resulting triangulation will obviously have spheres as links of vertices. – Vitali Kapovitch Jan 13 2012 at 18:54
This looks like a fine idea. I was wondering for a moment why this wouldn't work with an arbitrary atlas, but this is required as the intersection of two simplices $f_0: \Delta_0 \rightarrow M$ and $f_1: \Delta_1 \rightarrow M$ might otherwise be ``non-linear'' (I guess this is what you mean by ``make the transition maps linear''). Anyway, you're a very sweet man, thanks a lot! – Malte Jan 14 2012 at 20:01
## 2 Answers
It is claimed here that "A PL manifold is easily shown to be PL homeomorphic to a simplicial complex that is a so-called combinatorial manifold [37]", [37] being Hudson's Piecewise Linear Topology. I think the whole thing is worked out in Chapter 3 of that book.
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Take a look at MR0271956 (42 #6837) Siebenmann, L. C. Are nontriangulable manifolds triangulable? 1970 Topology of Manifolds (Proc. Inst., Univ. of Georgia, Athens, Ga., 1969) pp. 77–84 Markham, Chicago, Ill.57.01
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Thank you for pointing out this article, I didn't know about it. However, it doesn't seem to answer the question. – Malte Jan 14 2012 at 20:01
I agree. It only reinforces it. I thought it might have some relevant information, but I am not surprised that it does not. In the meanwhile, Vitali's response has me fairly convinced that the two definitions are equivalent, but I have not thought it through. – Matt Brin Jan 15 2012 at 5:25 | {"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": 6, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9213438034057617, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/212183-volume-revolution-problem.html | 1Thanks
• 1 Post By Prove It
# Thread:
1. ## Volume of revolution problem
Hi everyone, I have the following problem: What is the solid formed by rotating the region bounded by y=ln(x+1),x=0,y=2 around the y axis? The answer would be its pi/2(e^4-4e^2+7). How would one solve this problem? Would one find the intersection of y=2 and y=ln(x+1) which i got to be ((e^2)-1,ln(e^2)) and then take the integral of ln(x+1) from o to (e^2)-1 away from the area under y=2 from o to (e^2)-1 to get formula the cross section and then to find the volume integrate the formula for the cross section squared (radius) from y=0 to 2 and then times by pi?
Is this correct?
Thanks
2. ## Re: Volume of revolution problem
First of all, surely you can simplify $\displaystyle \begin{align*} \ln{\left( e^2 \right)} \end{align*}$...
3. ## Re: Volume of revolution problem
So my point of intersection is ((e^2)-1,2) but is the process correct?
Thanks for the reply
4. ## Re: Volume of revolution problem
First, graph the region to be revolved:
Next, compute the volume of an arbitrary disk:
$dV=\pi r^2\,dy$
$r=x=e^y-1$
Hence:
$dV=\pi (e^y-1)^2\,dy$
Finally sum up the disks by integrating:
$V=\pi\int_0^2(e^y-1)^2\,dy$
5. ## Re: Volume of revolution problem
How did you get the radius = e^y-1?
6. ## Re: Volume of revolution problem
Because the function is $\displaystyle \begin{align*} y = \ln{(x + 1)} \end{align*}$ and the radius of your solid is given by how far horizonally you extend, so $\displaystyle \begin{align*} x \end{align*}$. Surely you can see that if $\displaystyle \begin{align*} y = \ln{(x + 1)} \end{align*}$ then $\displaystyle \begin{align*} x = e^y - 1 \end{align*}$...
7. ## Re: Volume of revolution problem
Im sorry i cannot see that could you please explain? You extend to the x coordinate of intersection dont you which is (e^2)-1.
8. ## Re: Volume of revolution problem
No, for an arbitrary disk, the radius is found by extending from the y-axis to the x-coordinate on the logarithmic curve.
Look at the graph I provided and picture an arbitrary horizontal line in the shaded area serving as the radius of some disk.
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http://en.wikibooks.org/wiki/Statistics/Distributions/Binomial | # Statistics/Distributions/Binomial
A Wikibookian suggests that R Programming/Probability Functions/Binomial be merged into this book or chapter. Discuss whether or not this merger should happen on the discussion page.
## Contents
### Binomial Distribution
Notation
Probability mass function
Cumulative distribution function
B(n, p)
n ∈ N0 — number of trials p ∈ [0,1] — success probability in each trial
k ∈ { 0, …, n } — number of successes
$\textstyle {n \choose k}\, p^k (1-p)^{n-k}$
$\textstyle I_{1-p}(n - k, 1 + k)$
np
⌊np⌋ or ⌈np⌉
⌊(n + 1)p⌋ or ⌊(n + 1)p⌋ − 1
np(1 − p)
$\frac{1-2p}{\sqrt{np(1-p)}}$
$\frac{1-6p(1-p)}{np(1-p)}$
$\frac12 \log_2 \big( 2\pi e\, np(1-p) \big) + O \left( \frac{1}{n} \right)$
$(1-p + pe^t)^n \!$
$(1-p + pe^{it})^n \!$
$G(z) = \left[(1-p) + pz\right]^n.$
$g(p,n) = \frac{n}{p(1-p)}$ (continuous parameter only)
Where the Bernoulli Distribution asks the question of "Will this single event succeed?" the Binomial is associated with the question "Out of a given number of trials, how many will succeed?" Some example questions that are modeled with a Binomial distribution are:
• Out of ten tosses, how many times will this coin land heads?
• From the children born in a given hospital on a given day, how many of them will be girls?
• How many students in a given classroom will have green eyes?
• How many mosquitos, out of a swarm, will die when sprayed with insecticide?
The relation between the Bernoulli and Binomial distributions is intuitive: The Binomial distribution is composed of multiple Bernoulli trials. We conduct $n$ repeated experiments where the probability of success is given by the parameter $p$ and add up the number of successes. This number of successes is represented by the random variable X. The value of X is then between 0 and $n$.
When a random variable X has a Binomial Distribution with parameters $p$ and $n$ we write it as X ~ Bin(n,p) or X ~ B(n,p) and the probability mass function is given by the equation:
$P\left[X = k\right] = \begin{cases} {n \choose k} p^k \left(1-p\right)^{n-k}\ & 0 \le k \le n \\ 0 & \mbox{otherwise} \end{cases} \quad 0 \leq p \leq 1, \quad n \in \mathbb{N}$
where ${n \choose k}={n! \over k!(n-k)!}$
For a refresher on factorials (n!), go back to the Refresher Course earlier in this wiki book.
#### An example
Let's walk through a simple example of the Binomial distribution. We're going to use some pretty small numbers because factorials can be hard to compute. (Few basic calculators even feature them!) We are going to ask five random people if they believe there is life on other planets. We are going to assume in this example that we know 30% of people believe this to be true. We want to ask the question: "How many people will say they believe in extraterrestrial life?" Actually, we want to be more specific than that: "What is the probability that exactly 2 people will say they believe in extraterrestrial life?"
We know all the values that we need to plug into the equation. The number of people asked, n=5. The probability of any given person answering "yes", p=0.3. (Remember, I said that 30% of people believe in life on other planets!) Finally, we're asking for the probability that exactly 2 people answer "yes" so k=2. This yields the equation:
$P \left[X = 2 \right] = {5 \choose 2} \cdot {{0.3^2 \cdot} {\left( 1 - 0.3 \right)^{3}}} = {10} \cdot {{0.3^2} \cdot {\left( 1-0.3 \right)^{3}}} = 0.3087$ since ${5 \choose 2}={5! \over 2! \cdot 3!}={5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \over (2 \cdot 1) \cdot (3 \cdot 2 \cdot 1)}={120 \over 12}=10$
Here are the probabilities for all the possible values of X. You can get these values by replacing the k=2 in the above equation with all values from 0 to 5.
Value for k Probability f(k)
0 0.16807
1 0.36015
2 0.30870
3 0.13230
4 0.02835
5 0.00243
What can we learn from these results? Well, first of all we'll see that it's just a little more likely that only one person will confess to believing in life on other planets. There's a distinct chance (about 17%) that nobody will believe it, and there's only a 0.24% (a little over 2 in 1000) that all five people will be believers.
#### Explanation of the equation
Take the above example. Let's consider each of the five people one by one.
The probability that any one person believes in extraterrestrial life is 30%, or 0.3. So the probability that any two people both believe in extraterrestrial life is 0.3 squared. Similarly, the probability that any one person does not believe in extraterrestrial life is 70%, or 0.7, so the probability that any three people do not believe in extraterrestrial life is 0.7 cubed.
Now, for two out of five people to believe in extraterrestrial life, two conditions must be satisfied: two people believe in extraterrestrial life, and three do not. The probability of two out of five people believing in extraterrestrial life would thus appear to be 0.3 squared (two believers) times 0.7 cubed (three non-believers), or 0.03087.
However, in doing this, we are only considering the case whereby the first two selected people are believers. How do we consider cases such as that in which the third and fifth people are believers, which would also mean a total of two believers out of five?
The answer lies in combinatorics. Bearing in mind that the probability that the first two out of five people believe in extraterrestrial life is 0.03087, we note that there are C(5,2), or 10, ways of selecting a set of two people from out of a set of five, i.e. there are ten ways of considering two people out of the five to be the "first two". This is why we multiply by C(n,k). The probability of having any two of the five people be believers is ten times 0.03087, or 0.3087.
#### Mean
The mean can be derived as follow.
$\operatorname{E}[X] = \sum_i f(x_i) \cdot x_i = \sum_{x=0}^n {n \choose x} p^x \left(1-p\right)^{n-x} \cdot x$
$\operatorname{E}[X] = \sum_{x=0}^n {n! \over x!(n-x)!} p^x \left(1-p\right)^{n-x} x$
$\operatorname{E}[X] = {n! \over 0!(n-0)!} p^0 \left(1-p\right)^{n-0} \cdot 0 + \sum_{x=1}^n {n! \over x!(n-x)!} p^x \left(1-p\right)^{n-x} x$
$\operatorname{E}[X] = 0 + \sum_{x=1}^n {n(n-1)! \over x(x-1)!(n-x)!} p \cdot p^{x-1} \left(1-p\right)^{n-x} x$
$\operatorname{E}[X] = np\sum_{x=1}^n {(n-1)! \over (x-1)!(n-x)!} p^{x-1} \left(1-p\right)^{n-x}$
Now let w=x-1 and m=n-1. We see that m-w=n-x. We can now rewrite the summation as
$\operatorname{E}[X] = np \left[\sum_{w=0}^m {m! \over w!(m-w)!} p^{w} \left(1-p\right)^{m-w}\right]$
We now see that the summation is the sum over the complete pmf of a binomial random variable distributed Bin(m, p). This is equal to 1 (and can be easily verified using the Binomial theorem). Therefore, we have
$\operatorname{E}[X] = np \left[1\right]=np$
#### Variance
We derive the variance using the following formula:
$\operatorname{Var}[X] = \operatorname{E}[X^2] - (\operatorname{E}[X])^2.$
We have already calculated E[X] above, so now we will calculate E[X2] and then return to this variance formula:
$\operatorname{E}[X^2] = \sum_i f(x_i) \cdot x^2 = \sum_{x=0}^n x^2 \cdot {n\choose x}p^x(1-p)^{n-x}.$
We can use our experience gained above in deriving the mean. We use the same definitions of m and w.
$\operatorname{E}[X^2] = \sum_{x=0}^n {n! \over x!(n-x)!} p^x \left(1-p\right)^{n-x} x^2$
$\operatorname{E}[X^2] = 0 + \sum_{x=1}^n {n! \over x!(n-x)!} p^x \left(1-p\right)^{n-x} x^2$
$\operatorname{E}[X^2] = np\sum_{x=1}^n {(n-1)! \over (x-1)!(n-x)!} p^{x-1} \left(1-p\right)^{n-x}x$
$\operatorname{E}[X^2] = np \sum_{w=0}^m {m\choose w} p^{w} \left(1-p\right)^{m-w}(w+1)$
$\operatorname{E}[X^2] = np \left[\sum_{w=0}^m {m\choose w} p^{w} \left(1-p\right)^{m-w}w+\sum_{w=0}^m {m\choose w} p^{w} \left(1-p\right)^{m-w} \right]$
The first sum is identical in form to the one we calculated in the Mean (above). It sums to mp. The second sum is 1.
$\operatorname{E}[X^2] = np \cdot ( mp + 1) = np((n-1)p + 1) = np(np - p + 1).$
Using this result in the expression for the variance, along with the Mean (E(X) = np), we get
$\operatorname{Var}(X) = \operatorname{E}[X^2] - (\operatorname{E}[X])^2 = np(np - p + 1) - (np)^2 = np(1-p).$ | {"extraction_info": {"found_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": 34, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9215989112854004, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showthread.php?p=4064098 | Physics Forums
## Why are we so sure the universe existed in a point 13.7 billion years ago?
I've given this a lot of thought the last couple days. I look at the Hubble equation $v=H_0 d$, and I don't see how this necessitates everything existing in a single point at some time in the past. Certainly, everything was really close together 13.7 billion years ago, but not in an infinitely small volume.
$1/H_0$ (Hubble constant is in km/s/Mpc) does not give you the age of the universe (assuming the Hubble constant is constant through time). It gives you the amount of time it takes to travel a Megaparsec going at $H_0$ km/s. If you calculate the amount of time it takes to travel a Megaparsec going 71 km/s, it is indeed around 13.78 billion years. However, any object currently a Megaparsec from the Earth was not always receding at 71 km/s relative to the Earth, as you can see from the Hubble equation. As you go back in time and the object gets closer to the Earth, the velocity decreases. The way I see it, d=0 acts like an asymptote. The object approaches Earth the further back in time you go, but the distance is always non-zero, no matter how far back you go.
Another way to look at it is to see what happens when you apply the case of everything existing in a single point to the Hubble equation. The distance between any object is 0, so the velocity is also 0. If the velocity is 0, nothing moves, and the distance remains 0 for all time. Nothing ever expands! Something magical must have happened for these objects to separate before the expanding could begin.
It seems more logical to conclude that the universe has been expanding for an infinite amount of time. It's just that before 13.7 billion years ago, everything was so condensed that matter could not exist as we know it exists now.
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Quote by PhyPsy I don't see how this necessitates everything existing in a single point at some time in the past.
Uh ... who fed you that nonsense? It was never a point it was just smaller than it is today, and VERY much hotter and more dense.
Mentor
Quote by PhyPsy I've given this a lot of thought the last couple days. I look at the Hubble equation $v=H_0 d$, and I don't see how this necessitates everything existing in a single point at some time in the past. Certainly, everything was really close together 13.7 billion years ago, but not in an infinitely small volume.
To put it more politely than the previous poster, this is not true, and is in fact a common misconception. The standard cosmological model, or 'big bang' model, simply states that the universe was once much hotter, denser and smaller than it is today, and that is has expanded from such a state.
If you extrapolate this backwards you arrive with what's called a 'singularity': a point of infinite density and zero volume. However, this is simply a mathematical issue, telling us that our model is not valid at such a time. This is no surprise, since the cosmological model is governed by general relativity, yet when the universe gets very small and hot, we would expect quantum mechanics to be important. When we have a theory of quantum gravity, we expect it to deal with this singularity.
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## Why are we so sure the universe existed in a point 13.7 billion years ago?
Quote by PhyPsy I've given this a lot of thought the last couple days. I look at the Hubble equation $v=H_0 d$, and I don't see how this necessitates everything existing in a single point at some time in the past. Certainly, everything was really close together 13.7 billion years ago, but not in an infinitely small volume. ...
Ordinary mainstream cosmologists do not think of the universe being in an "infinitely small volume". That is, I think, more what you hear on television, or is suggested by flashy visuals being fed to the public.
So you are right, in a sense, to challenge the "infinitely small volume" idea. The professionals would agree with you that it is a misconception.
But we don't know what conditions were at the very start of expansion. There are several ideas that have been proposed. They need to be tested observationally. It would be premature to jump to the conclusion that one or the other is right. I would guess that among the various ideas for cosmic models that avoid the infinite-density failure (i.e. that have no "singularity" breakdown) currently more people are studying "bounce" cosmologies than any other model. that does not mean it's right, just that it is currently the most-studied alternative.
The Hubble expansion rate has changed enormously over the history of expansion.
You are right, if you question its constancy! It has definitely not been a constant, over time.
The present value of round 70.4 km/s per Mpc corresponds to distances taking
139 million years to grow by 1%.
So if you watch some distance, between widely separated galaxies, for a million years you can expect to see it increase by 1/139 of one percent.
that fractional or percentage growth rate has been much larger in the past. The 1/139 number is expected to continue declining and level out at around 1/163.
I've read about theories that state not only did matter as we understand it not exist before 13.7 billion years ago, but spacetime itself did not exist before then. I understand that GR posits a singularity at that time in history, but does this mean spacetime itself did not exist before then? I find it easier to believe that spacetime existed before then and matter simply expanded enough to finally be able to become how it is known now.
Quote by phinds Uh ... who fed you that nonsense? It was never a point it was just smaller than it is today, and VERY much hotter and more dense.
I suppose I ignored other things that would lead one to think the universe existed in a point at one time, such as the singularity predicted by GR. However, there are definitely people who have stated the universe existed in an infinitely small volume. Hawking, for example, said that the big bang singularity is a point of no volume from which everything begins to expand. Well, Hubble's law seems to contradict there ever being a single point in which everything existed.
Recognitions: Gold Member Science Advisor GR (formulated in 1915) is rather old classical (pre-quantum) theory and the fact that it develops a singularity is generally viewed as a symptom that it's wrong or has something missing. A quantum version might fix that. Several quantum theories of how geometry behaves are being studied and in at least one what's been found is that gravity in effect becomes repellent at extreme density. the theory predicts a bounce. IOW collapse CAN'T be total and complete. When a region collapses to sufficiently high density gravity stops attracting and starts repelling. Quantized geometry resists infinite density. In the recent MG13 meeting at Stockholm in July there was a special session devoted to "Nonsingular" cosmology models in general, mostly they are of bounce type. And there were also 2 or 3 other sessions devoted specifically to Loop, the general theory and the cosmology, which is one of those where you get a that kind of cosmology. With such quantized geometry, if you follow our universe's expansion back far enough, where you reach high enough density, then the equations show a prior contracting phase. So that's one possibility: extrapolating back in time with a model of gravity that bounces rather than suffering a singularity, you see an ordinary universe like ours, with ordinary matter etc., that is collapsing instead of expanding---and when a critical density is reached, bounces and begins the expansion that we do in fact witness. I'll get a link for the Stockholm MG meeting. It is one of the biggest international conferences where they discuss GR , quantum GR, cosmology, and things like that. You can check out the relevant sessions. It may not mean much but you can get a general idea of which areas of research are active and what questions people find interesting. Here's the list of sessions: http://www.icra.it/mg/mg13/parallel_sessions.htm Ones to click on, to see the titles of the talks are CM3 QG1 A QG1 B QG4 A QG4 B If you find a talk title interesting you can click on it and often get a brief abstract or summary of the talk.
Recognitions: Gold Member It's easy to be misled when you see comments like "The universe started off no larger than the size of a pea".....when in fact such references allude to a VISIBLE size, not the overall size. In other words, if observers existed back then all of them, perhaps many lights years apart, would have made the same observation. In fact the universe was much larger then, perhaps even infinite, just as today we can only observe what is likely an infinitesimally small part of the the total universe.
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| | Astrophysics | 3 | | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9625531435012817, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/285276/differential-eqn-by-separation-of-variables | # Differential Eqn by separation of variables.
I have the following differential equation:$$\frac{dx}{dt} = k(a-x)(b-x), 0 < a < b$$.
This is easy enough to solve using a partial fraction decomposition and using logs: I get to $$\left[-\frac{1}{b-a} \ln|a-x| + \frac{1}{a-b}\ln |b-x|\right]_0^x =kt,$$ which can be expressed as $$\frac{1}{a-b} \ln\left(|(a-x)(b-x)|\right) = kt$$ Manipulating this further gives $$|(a-x)(b-x)| = e^{(a-b)kt}$$ How would I isolate $x$ now to get an explicit function $x = x(t)?$
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1
Your partial fraction decomposition is not correct. Check it again. – Christopher A. Wong Jan 23 at 19:55
You also didn't include your constant of integration, which is important toward the end. – JohnD Jan 23 at 20:04
@ChristopherA.Wong What is wrong with it? I checked it again. I used the fact that $(a-b) = -(b-a)$ and it works? – CAF Jan 23 at 20:11
@JohnD I have done a definite integral. My notation is a bit sloppy but I have the integral from $0$ to $x$ $d\hat{x}$ say and the integral from $0$ to $t$ $d\hat{t}$. I can use this because I have an initial condition (which I forgot to put in the question - sorry) – CAF Jan 23 at 20:14
You have a minus sign error. When corrected, it will leave you with a linear equation for $x$. – André Nicolas Jan 23 at 20:21
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http://math.stackexchange.com/questions/281337/bounds-for-solutions-of-xex-n-0-for-n-geq-3?answertab=votes | # Bounds for solutions of $xe^{x} -n =0$ for $n\geq 3$
$\alpha _{ n }$ is a solution for $f_{ n }=xe^{ x }-n =0$. How can I prove that $\forall n\geq 3$, $\ln(n)-\ln(\ln(n))\le \alpha _{ n }\le \ln(n)$
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## 3 Answers
For $x\ge-1$, $\frac{\mathrm{d}}{\mathrm{d}x}xe^x=(1+x)e^x\ge0$. Thus, $f(x)=xe^x$ is monotonic increasing for $x\ge-1$. Therefore, for $x\ge e$, $$\begin{align} f(\log(x)) &=\log(x)e^{\log(x)}\\ &=x\log(x)\\ &\ge x \end{align}$$ Furthermore, $$\begin{align} f(\log(x)-\log(\log(x))) &=(\log(x)-\log(\log(x)))e^{\log(x)-\log(\log(x))}\\ &=\frac{\log(x)-\log(\log(x))}{\log(x)}x\\ &=\left(1-\frac{\log(\log(x))}{\log(x)}\right)x\\ &\le x \end{align}$$ Therefore, for $x\ge e$, $$\log(x)-\log(\log(x))\le f^{-1}(x)\le\log(x)$$
Lambert W
$f^{-1}$ is commonly known as the Lambert W function.
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Note that $f_n(x)=0$ iff $g(x)=xe^x=n$.
$g(x)$ is monotone increasing, so $(g(x)>n) \Leftrightarrow (x>\alpha_n)$ and $(g(x)<n) \Leftrightarrow (x<\alpha_n)$
The upper bound: $g(\ln n) = n\ln n > n$, so as written above, $\alpha_n < \ln$.
The lower bound:
$$g(\ln n - \ln(\ln n)) = (\ln n - \ln(\ln n))e^{\ln n - \ln(\ln n)}$$ $$= (\ln n - \ln(\ln n))n\cdot \frac{1}{\ln n}$$ $$= \left(1 - \frac{\ln(\ln n)}{\ln n}\right)n < n$$
So again because $g(x)$ is monotone increasing, $\alpha_n>\ln n - \ln(\ln n)$.
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Let $\alpha_n$ be a solution for some $n\geq 3$. Then $$\alpha_n e^{\alpha_n}=n \iff \alpha_n+\log(\alpha_n)=\log(n),$$ and so if $\log(\alpha_n)>0$ or equivalently $\alpha_n>1$ (which is true since $n\geq 3$), then $\alpha_n\leq \log(n)$ by the above.
Now, since $\alpha_n\leq \log(n)$ and $x\mapsto \log(x)$ is increasing, then $\log(\alpha_n)\leq\log(\log(n))$ and so we obtain the lower bound: $$\alpha_n=\log(n)-\log(\alpha_n)\geq \log(n)-\log(\log(n)).$$
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http://mathoverflow.net/questions/9484?sort=votes | ## Number of permutations with a specified number of fixed points
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Let F(k,n) be the number of permutations of an n-element set that keep k elements fixed.
We know:
1. F(n,n) = 1
2. F(n-1,n) = 0
3. F(n-2,n) = $\binom {n} {2}$
...
4. F(0,n) = n! $\cdot \sum_{k=0}^n \frac {(-1)^k}{k!}$ (the subfactorial)
The summation formula is obviously
$\displaystyle\sum_{k=0}^n F(k,n) = n!$
A recursive definition of F(k,n) is (my claim):
$F(k,n) = \binom {n} {k} \cdot ( k! - \displaystyle\sum_{i=0}^{k-1} F(i,k) )$
Question 1: Is there a common name for the "generalized factorial" F(k,n)?
Question 2: Does anyone know a closed form for F(k,n) or have an idea how to get it from the recursive definition? (generating function?)
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## 3 Answers
The "semi-exponential" generating function for these is
$\sum_{n=0}^\infty \sum_{k=0}^n {F(k,n) z^n u^k \over n!} = {\exp((u-1)z) \over 1-z}$
which follows from the exponential formula.
These numbers are apparently called the rencontres numbers although I'm not sure how standard that name is.
Now, how do we get a formula for these numbers out of this? First note that
$$exp((u-1)z) = 1 + (u-1)z + {(u-1)^2 \over 2!} z^2 + {(u-1)^3 \over 3!} z^3 + \cdots$$
and therefore the "coefficient" (actually a polynomial in $u$) of $z^n$ in $exp((u-1)z)/(1-z)$ is
$$P_n(u) = 1 + (u-1) + {(u-1)^2 \over 2!} + \cdots + {(u-1)^n \over n!} = \sum_{j=0}^n {{(u-1)^j } \over j!}$$
since division of a generating function by $1-z$ has the effect of taking partial sums of the coefficients.
The coefficient of $u^k$ in $P_n(u)$ (which I'll denote $[u^k] P_n(u)$, where $[u^k]$ denotes taking the $u^k$-coefficient) is then
$$[u^k] P_n(u) = \sum_{j=0}^n [u^k] {(u-1)^j \over j!}$$
But we only need to do the sum for $j = k, \ldots, n$; the lower terms are zero, since they are the $u^k$-coefficient of a polynomial of degree less than $k$. So
$$[u^k] P_n(u) = \sum_{j=k}^n [u^k] {(u-1)^j \over j!}$$
and by the binomial theorem,
$$[u^k] P_n(u) = \sum_{j=k}^n {(-1)^{j-k} \over k! (j-k)!}$$
Finally, $F(k,n) = n! [u^k] P_n(u)$, and so we have
$$F(k,n) = n! \sum_{j=k}^n {(-1)^{j-k} \over k!(j-k)!}$$
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Thanks. Am I - as an MO user - supposed to know how to get the closed form for F(k,n) from this "semi-exponential" generating function? – Hans Stricker Dec 21 2009 at 16:35
Not necessarily. It's not hard, though; I'll edit the solution to explain that. – Michael Lugo Dec 21 2009 at 16:57
The name "recontres numbers" is standard in the following ways: (1) EIS, (2) canonical name in Wikipedia, (3) 10 hits in Google Scholar. Although that last one is not a huge number, if you take all three together, it's plenty standard enough for a relatively obscure concept. – Greg Kuperberg Dec 21 2009 at 17:38
1
My claim that the name was nonstandard was entirely subjective; basically, this is something that I felt I should have known a name for, and the name was new to me. – Michael Lugo Dec 21 2009 at 17:50
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
A permutation of {1, ..., n} with k fixed points is determined by choosing which k elements of {1, ..., n} it fixes and choosing a derangement of the remaining n-k elements. So,
$F(k, n) = {n \choose k} F(0, n-k)$.
(This formula is also on the page Michael Lugo linked to.) You have already given one formula for the number of derangements on n letters. Another one is F(0, n) = the nearest integer to n!/e.
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http://mathhelpforum.com/statistics/93643-simple-question-about-propabilites.html | # Thread:
1. ## Simple question about propabilites
Hi everybody,
Combinatorics always confusing me. I ask you to give me the type of a classic issue in propability.
We have N boxes. Each box contains P0...Ps elements (so means s to count).
If we pick up an element from each box:
- How many will be the UNIQUE comhinations of the elements?
if you provide and an analytic solution, it would be greatful.
2. Originally Posted by gdmath
We have N boxes. Each box contains P0...Ps elements (so means s to count).
If we pick up an element from each box:
- How many will be the UNIQUE comhinations of the elements?
if you provide and an analytic solution, it would be greatful.
Not a good description of the problem.
Are the elements in the same box identical?
If not how many of each kind?
Can two boxes have any common elements?
If so, how many?
I really think that you should try to rewrite the to make all the condition on the boxes and on the elements as clear as possible.
3. Ok,
We have N boxes.
Assume each box contains elements that can be described as a number.
So each box contains numbers from 1 to s (1,2,3,...,s). So these elements are deferent from each other.
Next I say "each box". This mean that the first box has exactly the same elements with any other box.
So if we pick one element from each box (total N elements) how many unique combinations will have?
(Unique means that if i have the combination 123456...N i cannot accept the combination 213456..N)
Hope that i am clear.
4. MUCH better.
Let’s give an example where there are sill some questions.
Suppose we have six boxes each looks this: $\left\{ {0,1,2,3,4,5,6} \right\}$.
Choosing one element from each box we could have this sample: $\left\{ {0,1,3,4,5,6} \right\}$. Clearly six different elements.
But we could also have $\left\{ {0,1,1,4,0,0} \right\}$ with some repetitions.
Now as I understand you mean this is the same as $\left\{ {0,0,0,1,1,4} \right\}$.
Is that correct?
Under this understanding the answer is $\binom{6+7-1}{6}$.
That is the number of ways to select six items from a variety of seven.
$\binom{N+S-1}{N}$ is the number of ways to select N items from a variety of S.
If this is not what you mean, let’s try again.
5. Yes exactly this i need.
I knew it was something with binomial, but not exactly what...
First of all thank you
Now look something interesting,
The way i was trying to realize the problem,
when there are two boxes the probabilty is (s elements in each box):
$\sum_{i=1}^{s+1} i = \frac{1}{2} s(s+1)=\binom{s+2-1}{2}$
When we have three Boxes:
$\sum_{j=1}^{j+2} ( \sum_{i=1}^{j+1} i ) = \frac{1}{6} s(s+1)(s+2)=\binom{s+3-1}{3}$
And goes on recursively at sums...
Thank you again | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9343288540840149, "perplexity_flag": "middle"} |
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### Has the black hole information loss paradox been settled?
This question was triggered by a comment of Peter Shor's (he is a skeptic, it seems.) I thought that the holographic principle and AdS/CFT dealt with that, and was enough for Hawking to give John ... | {"extraction_info": {"found_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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.926850438117981, "perplexity_flag": "middle"} |
http://fliptomato.wordpress.com/2007/07/11/braneworld-and-the-hierarchy-in-rs1/ | ### Braneworld and the Hierarchy in RS1
11Jul07
Earlier this year I wrote a Part III essay on bulk neutrinos in the Randall-Sundrum I (RS1) model. After scouring the introductory braneworld literature, I noticed that there seems to be set of standard figures used. Somewhat uninspired by these, I went ahead and made my own graphic. The resulting image, below, was nifty enough that I feel compelled to write a post around it.
Fig 1. Scientists are searching for nature’s answers while stuck on the brane, represented by Gauguin’s Tahitian panting, “Where Do We Come From? What Are We? Where Are We Going?” (1897, image from the Artchive). The red line depicts the warp factor, by which mass scales are exponentially shrank on our brane. The $\phi$ axis runs from $-\pi$ to 0.
So what’s going on in the funky picture above (fig 1)? What’s heuristically represented is the RS1 `toy model’ of the universe in which our observed (3+1)-dimensional spacetime is embedded as a hypersurface (brane) in a 4+1 dimensional bulk spacetime. The fifth dimension is an interval, which is actually an $S^1/\mathbb Z^2$ orbifold. Our brane, the visible brane, is located at one endpoint of the interval while another brane, the hidden brane, is located at the other end. In the figure above, this hidden brane is represented by the yellow translucent sheet. All Standard Model particles are confined to live on the brane while gravity is allowed to propagate freely in the bulk.
As an ansatz, we can write our bulk metric as:
(1) $ds^2 = e^{-2\sigma(\phi)}(\eta_{\mu\nu}dx^\mu dx^\nu) + r^2d\phi^2$.
One can see heuristically that the warp factor $e^{-2\sigma(\phi)}$ conformally transforms 3+1-dimensional cross sections of the bulk spacetime relative to one another. This will be the mechanism by which the RS1 model can naturally generate large hierarchies (more on this shortly).
Solving the 55-component of the Einstein equation does two things for us: first it fixes the bulk cosmological constant, and then it fixes $\sigma$. The cosmological constant turns out to be negative so that the bulk space is Anti deSitter (AdS$_5$). The heuristic form of the warp factor is depicted in Fig. 1 as a red line. Explicitly, one ends up with:
(2) $\sigma = kr|\phi|$.
Where $k = -\sqrt{\Lambda/12M^3}$ is naturally on the order of the inverse `fundamental’ Planck scale. The combination $kr$ is dimensionless and naturally order 1.
Just as a rubber sheet can be stretched, the branes are allowed to have tension, i.e. a constant energy density localized on the brane. To sentient life confined to live on the brane, this constant energy density looks like a 4D cosmological constant. These brane tensions compensate for the curvature of the bulk space so that the 4D branes are flat. (Sanity check: the cosmological constant indeed supplies the `missing’ energy density so that the universe is flat or very nearly so.)
After some manipulations (see references below), this construction yields something fantastic. First of all, the 4D Planck scale $M_{Pl}$ is roughly the same order as the 5D `fundamental’ Planck scale. The exact relation ends up as:
(3) $M_{Pl}^2 = \frac{M^3}{k}(1-e^{-2\pi kr})$
This comes from perturbing the 4D part of the metric (i.e. introducing 4D gravitons) and playing with the corresponding gravitaton terms in the action. One can see that with $k \sim M$ and $kr \sim \mathcal{O}(1)$, the right hand side is indeed $\sim M^2$.
But what about the Standard Model (SM)? The SM `lives’ on the visible brane. Effectively, this means that in the 5D action the SM lagrangian terms are accompanied by delta functions which fix these fields to the visible brane. But the extra dimension still influences the SM lagrangian in two ways: (1) since we are dealing with gravity, there is an overall factor of the square root of the metric multiplying every term in the action, and (2) contractions of vector indices depend on the metric. Since the metric is nonfactorisable (i.e. the 4D block depends explicitly on the 5th coordinate), the metric is different depending on where in the 5th dimension it is being evaluated. In particular, at the visible brane we get factors of $e^{-2\pi kr}$ floating around — the very same factors that were floating around in the previous paragraph.
However, we are effectively 4D beings that only care about 4D effective actions. We want to massage the Standard Model terms in the action into a form we are used to. Namely, we perform the trivial integration over the extra dimension (using the brane delta function) and we want to canonically normalise our fields relative to the 4D theory. In the Higgs sector, the effective Lagrangian looks like:
(4) $e^{-4\pi kr} \left(e^{2\pi kr}|D_\mu H|^2 + \lambda(|H|^2-v_0^2)^2\right)$.
The exponential factors come from the metric as discussed above. We canonically normalise the Higgs field by rescaling $H \rightarrow H' = e^{-\pi kr} H$. Thus, we discover that the effective Lagrangian (properly normalised) takes the form:
(5) $|D_\mu H'|^2 + \lambda(|H'|^2-e^{-2\pi kr} v_0^2)^2$.
That is to say that the effective Higgs vev $v=e^{-\pi kr}v_0$ is equal to the `fundamental’ scale $v_0$ multiplied by the warp factor. The natural value for $v_0$ is the fundamental Planck scale, $M$. For values of $kr \sim 30$, we are able to reproduce the observed Standard Model Higgs vev. Meanwhile, the effective lagrangian (with this definition for $v$) is exactly the same as the usual Standard Model lagrangian.
Aha! Did you catch it? Something fantastic has happened. Because of the warp factor, we have been able to explain the hierarchy between the electroweak and Planck scales using only natural parameters, namely the warping of spacetime turns naturally Planck-scale quantities into [naturally] electroweak scale quantities. That is to say that this scenario `solves’ the hierarchy problem (if it really is a problem)! Ok, so the cost is that we’ve had to introduce a new parameter here and there, but the point is that these parameters all take natural values.
A good analogy (by Rattazzi in his Cargese lectures, see references) is that of redshifting. One can think of the mass (energy) of a particle in terms of its frequency. Just as photons are redshifted near a strong gravitational field, the idea of warped extra dimensions is to redshift the mass-scale of our visible brane such that the fundamental Planck scale is `warped’ to the electroweak scale.
Now back to Gauguin. Where does the brane come from? What is it? Where is it going? One may feel awkward that we’ve made a rather arbitrary set up to get a nice model. To some extent, this is true — but this is model-building, after all. It turns, though, out that this scenario comes out naturally from Horava-Witten models in M-theory. However, we shall take the stance that the mechanism that generated the brane is related to some high-energy theory that is otherwise decoupled from our low-energy universe. All that is relevant is that it is reasonable that a high-energy theory (which would turn out to have extra dimensions) may have, a Randall-Sundrum braneworld set up as a lower-energy limit.
Extra Dimensions Review Articles I Liked
An excellent starting point is Bee’s literature review. Seriously, if more bloggers wrote similar lit reviews with lots of useful and beginner-friendly references, then the world would be a better place for grad students.
Because I hope to cater to colleagues thoroughly rooted in the 21st century, here are some nice multimedia references… they’re great for serious studying, background noise, or even for an iPod while study-jogging (does anyone else do this?):
For some great background reading, I suggest looking at several TASI lectures on the subject that have been given in recent years:
• Sundrum: To the Fifth Dimension and Back
• Csaki: Lectures on XD and Branes
• Dienes: New Directions for New Dimensions (not readily available on the arXiv… but it’s online if you look for it, otherwise you can find it in the 2002 TASI book)
• Kribs: Lectures on the Pheno. of XD
For the RS1 set up, Csaki’s reference above is very nice, as well as the original paper by Randall and Sundrum. Some good background can be found in an earlier papers by Sundrum: EFT for a 3-Brane Universe, Compactification for a 3-Brane Universe.
I’ll specialise even a bit deeper and suggest a very nice paper by Grossman and Neubert regarding bulk neutrinos in a warped extra dimension. This paper provides the mathematical framework for placing fermions in the bulk, something which became a bit of a model-building trend.
Hopefully these will provide enough of a framework to make the subject easily accessible to a first year postgraduate.
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#### 15 Responses to “Braneworld and the Hierarchy in RS1”
1. 1 evankeane on July 12, 2007 said:
nice post there flip – well explained.
This is a nice model – I especially like the idea of the 4D cosmological constant compensating for the negative cosmological constant of the bulk to keep the 4D brane flat.
I will have to trust you that “this scenario comes out naturally from Horava-Witten models in M-theory”. How serious do you treat this model then? I figured it was simply a nice toy model until I read that … Could tests be thought up? Would the small differences in Planck scales be noticeable to some super experiment of the future?
Evan
2. 2 evankeane on July 12, 2007 said:
oh ya forgot to say – the figure is very nice!
3. 3 on July 12, 2007 said:
Thanks, Evan. Indeed the basic RS1 model is more of a toy model, but I think it has many of the necessary `moving parts’ for people to build upon it. There are a few questions that I didn’t bring up, for example radius stabilisation: 5D graviton excitations should also make the radius of the extra dimension vary (i.e. we have a “radion field”); but in the toy model we assume the radius is constant. Thus model-builders have to play with radius-stabilisation mechanisms (the standard one is called Goldberger-Wise).
The signature of these models, as is generic for models with compactified extra dimensions, are Kaluza-Klein excitations of bulk fields. Since we only have gravity propagating in the bulk, this would mean we would expect a tower of Kaluza-Klein gravitons. Think of the particle in a box question in ordinary quantum mechanics. If we impose periodic boundary conditions (the same thing as turning the interval into a circle), then we have a standing wave solution and all higher harmonics. Thus the signature would be a family of gravitons with masses corresponding to each harmonic. As we are stuck on the visible brane, I don’t think there’s a straightforward way to probe the small difference between the effective 4D Planck scale and the fundamental 5D Planck scale… but if we see the Kaluza-Klein modes, we know that we should be thinking about extra dimensional models. (That is to say *if* we can disambiguate the KK modes from, say, supersymmetric partners… see the paper “Getting Fooled at the LHC” by Cheng et al.)
As far as how serious should we treat the model, this is a good question. The appropriate context is that the past 30 years has been marked by the hegemony of the Standard Model in nearly all experimental tests. Model-builders had all this time to think of creative new physics, with very little experimental hints about which direction to proceed. The fact that there is some sector of string theory that accomodates braneworld scenarios is somewhat comforting from a motivational point of view… but it’s important to stress that braneworld does not imply string theory and string theory does not imply braneworld.
The two most popular models for new TeV-scale physics are supersymmetry and extra dimensions. Supersymmetry is the unique extension of the 4D Poincare group under the Coleman-Mandula theorem. Extra dimensions extends the 4D Poincare group by turning it into the (4+N)D Poincare group. From this point of view we are trying to squeeze in new physics by circumventing well-accepted restrictions.
4. 4 evankeane on July 12, 2007 said:
i could be cynical (me?) and play on your comment on string theory and braneworld –> string theory implies nothing, nothing implies string theory (oh look i did say it!)
interesting stuff though – also i am heartened that even though in the past few decades, as you say, there has been no experimental pointers as to where the modelling should go, particle physicists have tried to find holes, ambigueities, possible problems, etc. in their theories! cf: the Cheng et al paper you mentioned! this is healthy and scientific!
Q. what are the motivations for supersymmetry and extra dimensions, besides string theory, if any?
NB. “because it works” IS an acceptable answer by the way. in fact this is often given as the ‘derivation’ of the Schrodinger Equation if I recall correctly and everyone is happy with that!!!
Evan
5. 5 on July 12, 2007 said:
Motivation for SUSY and XD! Excellent question. Maybe I’ll write a brief post on these some time… but these are rather standard in most SUSY literature. Here’s a short list: (By SUSY, by the way, I mean TeV-scale SUSY)
1) The Hierarchy Problem (why is the electroweak scale so much smaller than the Planck scale) … SUSY and XD provide ways to avoid the fine tuning. In SUSY there are new particles that cancel the radiative corrections to the Higgs mass. In XD one either reparameterises the Hierarchy in terms of fine-tuning the size of the XD (the Large XD scenario) or one says that there’s no fine-tuning because all parameters are natural (the RS1 scenario above).
2) Dark Matter. We know DM exists. We have good reason to believe that it is weakly interacting. SUSY with R-parity (which also prevents unfavourable interactions that lead to proton decay/flavour changing neutral currents) provides a natural DM candidate, the lightest supersymmetric particle (LSP). R-parity prevents the LSP from decaying into anything else.
3) Grand unification. Based on results at LEP, running the renormalisation group equations in the Standard Model would not lead to gauge coupling unification. However, the minimally supersymmetric standard model provides the framework for the gauge couplings (as measured by LEP at the 100 GeV scale) to unify at a higher scale.
4) Theoretical beauty. SUSY is the only extension of the Poincare Group under the Coleman-Mandula theorem. If SUSY did NOT exist, we would be left wondering why Nature didn’t make full use of this symmetry.
5) Other stuff… there are some other nonperturbative effects that are relevant (BPS states), but unfortunately I don’t understand these well.
6) It’s a necessary ingredient for String Theory. (Some people accept this, some people don’t.) Though this doesn’t mean we need to have low-energy SUSY (TeV scale).
7) Anthropic principle: if we didn’t have low energy SUSY, then a LOT of people will have essentially wasted their time with their research in the past 30 years.
8) Strong Anthropic principle: I’m interested in studying SUSY. So the universe should accomodate my interest.
6. 6 Alejandro Rivero on July 13, 2007 said:
One should not say that string theory is a motivation for SUSY. On the contrary, SUSY plus Kaluza Klein plus other math technicalities (Leech lattice, Exceptional Groups, …) are motivations for string theory. This was an argument attributed to Polchinski: that if one follows the research on top-down particle theory, then it does not matter the road you travel (I say: GUTs, octonions, Susy, etc) you arrive to string theory at the end.
7. 7 Alejandro Rivero on July 13, 2007 said:
Now, an amateur can find more colateral motivations for SUSY. One of them is that the representations you use in GUT theories are generated by fermionic creation operators. This was observed time ago in a paper from Wilczek and Zee, and you can still find their |++—> notation both in Zee book and in Wilczek opencourseware lecture notes (QFT III at MIT). Well, the case is that if we are speaking fermions here in the representation theory, we can hope a GUT setup will integrate smoothly with supersymmetry and its powers-of-two. And remember than neutrino oscillations have strongly motivated the GUT way again, as now we need the right handed neutrino.
Another motivation I revisited recently (in http://groups.google.com/group/sci.physics.research/browse_thread/thread/769d65ce12474189 but I had noticed it already in hep-ph/0512065v1 ) is a coincidence with the number of different open QCD strings (or Reggee traj.) when you terminate them wither with quarks or antiquarks. They happen to be, charge-by-charge, the same number that the number of degrees of freedom of the standard model fermions. This is because you the mass of the top is larger than the QCD scale and then you can not attach the top to a QCD string. In the spr thread I stress that if we took this equality between strings and fermions as postulate, then forcefully num generations >= 3, and if we extend it to the leptons, then =3.
Still another motivation, of course, is the use of SUSY symmetry in other areas of physics. Generically, graded Lie superalgebras and supergroups can be useful in nuclear physics, and who knows if also in solid state physics.
8. 8 evankeane on July 14, 2007 said:
hey flip – i actually looked up some SUSY books (while in Heffers) re: the motivations for SUSY and XD and i’m happy enough with them! so with TeV SUSY (ie. this is the scale it breaks at) the heirarchy problem is solved, the couplings unify and there is a dark matter candidate. fair enough! so that’s your first 3 reasons and that’s enough for me. Mars wil be happy with your 4th reason i will take your word for it on number 5. 6 and 7 – yes well string theory and the anthropic principle … no comment. your final reason is also cool
9. 9 evankeane on July 14, 2007 said:
ok i will comment on string theory …
i agree with Alejandro Rivero – string theory is not a motivation for SUSY and XD but SUSY and XD (which are themselves motivated by what they produce as you mentiones) are motivations for string theory. this makes sense to me!
10. 10 Alejandro Rivero on July 14, 2007 said:
hey evan! I like your abreviation for eXtra Dimensions. In emoticon language, it would translate as “dead while laughing” ( X for the eyes, dead shut, and D for the mouth, laughing loud). Btw, let me to apologise by the poor grammar in previous postings.
About what motivates XD, I would think that its sources of motivation are: 1) the representation theory of Clifford Algebras 2) SuGra or Susy, and 3) The paper of Witten on Kaluza Klein. This paper actually motivates, I think, M-theory, moving the dimension up to 11. A 4th motivation can be Connes’s model, where dim=6 mod 8. The exact number of extra dimensions is burred because Clifford (n,m) = Clifford (n+1,m+1) So dimension 25+1 is as well motivated as 24+0. And here a 5th) motivation: Leech lattice.
11. 11 on July 18, 2007 said:
Great post! I’d just like to add that the energy of the KK-excitations in the RS scenario is not equally spaced as it is in scenarios that compactify on a torus. (if one solves the wave-equation in the above background geometry, they turn out to be at the zeros of one of the Bessel functions, forgot which). It’s not theoretically a big issue, but makes a difference for the phenomenology of that scenario.
@Alejandro: When we founded our group on large extra dimensions in 2000 back in Germany, we were looking for a good abbreviation. the choice LED didn’t sound too appealing since we didn’t plan to emit very much light. It was my husband who suggested ‘LXD’, what we used from then on in our papers (and also on the homepage… gee, I just saw that it still exists!). I still like XD and LXD better than ED and LED, but most people seem to use the latter.
12. 12 alex on July 23, 2007 said:
Another motivation for SUSY is REWSB (radiative electroweak symmetry breaking).
13. 13 Alejandro Rivero on July 23, 2007 said:
Alex, right, but only for Minimal Susy in the orthodox way, and with the mass of the top a lot greater than the mass of the other fermions. Then the mass term of the higgs field (not to be confused with the mass of the higgs) changes sign in the adequate scale.
Which remembers me another different view of the hierarchy: the expectation of that the electromagnetic mass of the electron, with cut-off at the Planck mass, should be of the order of the total mass of the electron. This is old tale, but for instance Polchiski revisits it when speaking of hierarchies in the second volume of his book.
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• ## Flip Tomato
I'm a postgraduate, expatriate physics student. RSS Feed.
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http://math.stackexchange.com/questions/127436/continuity-of-a-bivariate-function?answertab=votes | Continuity of a bivariate function
Let $f : \mathbb{R}^2 \to \mathbb{R}$ be defined by $$f(x,y) = \begin{cases} \frac{x^2 y^2}{x^4 + y^4}, & \text{ }\text{(x,y)} \neq (0,0) \\ 0, & \text{ }\text{(x,y)} = (0,0) \end{cases} .$$
Show that $\frac{df}{dx} (0,0)$ exists, and $f$ is not continuous at $(0,0)$.
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Write down the limit defining the partial derivative. What do you get? Compute the limit of $f$ along the path $x=y$ and compare with that value of $f(0,0)$. What do you get? – David Mitra Apr 2 '12 at 23:27
1 Answer
For existence of the partial derivative:
By definition, $f_x(0,0)$ is the limit $$\lim_{h\rightarrow 0}{f(h,0)-f(0,0)\over h }.$$ Show that this limit exists for your particular function.
For continuity:
Note that,
$$\lim_{x\rightarrow 0} f(x,x) =\lim_{x\rightarrow 0}{x^4\over2x^4}={1\over2}.$$ Keeping this in mind and the fact that $f(0,0)=0$, what can you say about the continuity of $f$ at $(0,0)$?
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http://mathoverflow.net/revisions/45337/list | Return to Answer
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Let `$T = [1,1 ; \begin{pmatrix} 1 & 1 \\ 0 , & 1 ] \end{pmatrix} \in \Gamma_{\infty}$. Gamma_{\infty}$`. There is a natural map $V_{k+1} \to H^1(\Gamma_\infty,V_k)$ sending a polynomial $P$ to the cocycle $c_P$ determined by $c_P(T) = P(X+1)-P(X)$. It is easy to check that this map induces an isomorphism $\psi : V_{k+1}/V_k \cong H^1(\Gamma_\infty,V_k)$, so that the latter space is one-dimensional.
\bigl(\frac{1}{2\pi left(\frac{1}{2\pi i} \frac{d}{dz}\bigr)^{k-1} frac{d}{dz}\right)^{k-1} \widetilde{f}(z) = f(z).(\frac{d}{dz})^{k-1} \left(\frac{d}{dz} \right)^{k-1} (F |_{2-k} g) = (\frac{d^{k-1} F}{dz^{k-1}}) \left(\frac{d^{k-1} F}{dz^{k-1}} \right) |_k gLet $c_f \in Z^1(\Gamma,V_k)$ be the cocycle associated to this choice of $\widetilde{f}$. Let us compute the value of $c_f$ on $T$ and `$S=[0,-1;1,0]$. S= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$`. First as explained above, we have
c_f(S) = \frac{(2\pi i)^{k-1}}{(k-2)!} \int_0^{\infty} \left(f(z)-\frac{a_0}{z^k}-a_0 \right) (f(z)-\frac{a_0}{z^k}-a_0)(z-X)^{k-2} z-X)^{k-2} dz\frac{(k-2)!}{(2\pi i)^{k-1}} c_f(\gamma) = &= \int_{z_0}^{\infty} (f(z)-a_0)(z-X)^{k-2} dz + \int_{\gamma^{-1} \infty}^{z_0} (f(z) \left(f(z) -\frac{a_0}{(cz+d)^k}) \frac{a_0}{(cz+d)^k} \right) (z-X)^{k-2} dz \end{equation}& + \frac{a_0}{k-1} ((X-z_0)^{k-1}-(X-\gamma \left((X-z_0)^{k-1}-(X-\gamma z_0)^{k-1} |{2-k} _{2-k} \gamma + X^{k-1} |{2-k} _{2-k} (\gamma-1))where $z_0 \in \mathcal{H}$ is arbitrary and `$\gamma= [a,b;c,d]$.\begin{pmatrix} a & b \\ c & d \end{pmatrix}$`.
5 added 36 characters in body
The result you're looking for is contained in the following article :
Haberland, Klaus. Perioden von Modulformen einer Variabler and Gruppencohomologie I (German) [Periods of modular forms of one variable and group cohomology I], Math. Nachr. 112 (1983), 245-282.
Let $S_k$ (resp. $M_k$) be the space of holomorphic cusp forms (resp. holomorphic modular forms) for $\Gamma = SL_2(\mathbf{Z})$. Let $\Gamma_{\infty}$ be the stabilizer of $\infty$ in $\Gamma$. Let $V_k$ be the space of polynomials of degree $\leq k-2$ with complex coefficients. Haberland proves an exact sequence
\begin{equation} (*) \qquad 0 \to S_k \oplus \overline{S_k} \to H^1(\Gamma,V_k) \to H^1(\Gamma_\infty,V_k) \to 0. \end{equation} Let $T = [1,1 ; 0, 1] \in \Gamma_{\infty}$. There is a natural map $V_{k+1} \to H^1(\Gamma_\infty,V_k)$ sending a polynomial $P$ to the cocycle $c_P$ determined by $c_P(T) = P(X+1)-P(X)$. It is easy to check that this map induces an isomorphism $\psi : V_{k+1}/V_k \cong H^1(\Gamma_\infty,V_k)$, so that the latter space is one-dimensional.
The "Eisenstein cocycle" you're looking for is a natural map $\delta : M_k \to H^1(\Gamma,V_k)$ which Haberland constructs the following way (actually I learnt this construction and many other properties of $\delta$ during Zagier's 2002-2003 lectures at the Collège de France).
Let $f \in M_k$. Let $\widetilde{f}$ be an Eichler integral of $f$, that is any holomorphic function on $\mathcal{H}$ such that
\begin{equation} \bigl(\frac{1}{2\pi i} \frac{d}{dz}\bigr)^{k-1} \widetilde{f}(z) = f(z). \end{equation} Note that $\widetilde{f}$ is unique up to adding some element of $V_k$. Since we integrate $k-1$ times, the function $\widetilde{f}$ should be thought of as a function of "weight" $k-2\cdot (k-1) = 2-k$ (of course this isn't true in the strict sense). Let us make this more precise.
For any $n \in \mathbf{Z}$, let $|_n$ denote the weight $n$ action of $SL_2(\mathbf{R})$ on the space of complex-valued functions on $\mathcal{H}$ (so that any $f \in M_k$ is a fixed vector of the weight $k$ action of $\Gamma$). Note also the weight $2-k$ action gives the usual action of $\Gamma$ on $V_k$. The crucial fact is that we have
\begin{equation} \widetilde{f} |_{2-k} (\gamma-1) \in V_k \qquad (\gamma \in \Gamma). \end{equation} This can be proved using Bol's identity
\begin{equation} (\frac{d}{dz})^{k-1} (F |_{2-k} g) = (\frac{d^{k-1} F}{dz^{k-1}}) |_k g \end{equation} which holds for any holomorphic function $F$ on $\mathcal{H}$ and any $g \in SL_2(\mathbf{R})$.
Since $\gamma \mapsto \widetilde{f} |_{2-k} (\gamma-1)$ is obviously a coboundary in the space of functions on $\mathcal{H}$, it defines a cocycle in the space $V_k$. Therefore we get $\delta(f) \in H^1(\Gamma,V_k)$ and this element doesn't depend on the choice of $\widetilde{f}$. Thus we have constructed $\delta : M_k \to H^1(\Gamma,V_k)$.
It is not difficult to check that if $f =\sum_{n \geq 0} a_n e^{2i\pi nz}$ then the image of $\delta(f)$ in $H^1(\Gamma_\infty,V_k)$ is the image of the polynomial $\frac{a_0 \cdot (2\pi i)^{k-1}}{(k-1)!} \cdot X^{k-1} \in V_{k+1}$ under the isomorphism $\psi$ above. In particular $\delta$ is injective, and the exact sequence $(*)$ gives the isomorphism you want.
Note that there is a distinguished choice of $\widetilde{f}$, namely
\begin{equation} \widetilde{f} = \sum_{n \geq 1} \frac{a_n}{n^{k-1}} e^{2i\pi nz} + \frac{a_0 \cdot (2\pi i)^{k-1}}{(k-1)!} z^{k-1}. \end{equation} Let $c_f \in Z^1(\Gamma,V_k)$ be the cocycle associated to this choice of $\widetilde{f}$. Let us compute the value of $c_f$ on $T$ and $S=[0,-1;1,0]$. First as explained above, we have
\begin{equation} c_f(T)=\frac{a_0 \cdot (2\pi i)^{k-1}}{(k-1)!} ((X+1)^{k-1}-X^{k-1}). \end{equation} To compute $c_f(S)$, Haberland uses the natural integral representation of $\widetilde{f}$ in terms of $f-a_0$, and gets
\begin{equation} c_f(S) = \frac{(2\pi i)^{k-1}}{(k-2)!} \int_0^{\infty} (f(z)-\frac{a_0}{z^k}-a_0)(z-X)^{k-2} dz \end{equation} (there is a similar but more complicated formula for $c_f(\gamma)$ for any $\gamma \in \Gamma$, see below). Then $c_f(S)$ can be expressed in terms of the special values of $L(f,s) := \sum_{n=1}^\infty a_n/n^s$ at integers $s = 1,\ldots,k-1$. It is then a good exercise to compute $c_f(S)$ when $f$ is the Eisenstein series $E_k$, in terms of Bernoulli numbers and of $\zeta(k-1)$ (this is Satz 3 in Haberland's article, Kapitel 1).
Please tell me if something isn't clear in my explanation.
EDIT : I found the following expression for $c_f(\gamma)$ where $\gamma \in \Gamma$. It is quite complicated (maybe it could be somewhat simplified) :
\begin{equation} \frac{(k-2)!}{(2\pi i)^{k-1}} c_f(\gamma) = \int_{z_0}^{\infty} (f(z)-a_0)(z-X)^{k-2} dz + \int_{\gamma^{-1} \infty}^{z_0} (f(z) -\frac{a_0}{(cz+d)^k}) (z-X)^{k-2} dz \end{equation} \begin{equation} + \frac{a_0}{k-1} ((X-z_0)^{k-1}-(X-\gamma z_0)^{k-1}|z_0)^{k-1} |{2-k} \gamma + X^{k-1} |{2-k} (\gamma-1)) \end{equation} where $z_0 \in \mathcal{H}$ is arbitrary and $\gamma= [a,b;c,d]$.
4 added 567 characters in body; deleted 14 characters in body
Let $c_f \in Z^1(\Gamma,V_k)$ be the cocycle associated to this choice of $\widetilde{f}$. Let us compute the value of $c_f$ on $T$ and $S$. S=[0,-1;1,0]\$. First as explained above, we have
(there should be is a similar but more complicated formula for $c_f(\gamma)$ for any $\gamma \in \Gamma$). Gamma$, see below). Then$c_f(S)$can be expressed in terms of the special values of$L(f,s) := \sum_{n=1}^\infty a_n/n^s$at integers$s \in {1,\ldots,k-1}$. = 1,\ldots,k-1$. It is then a good exercise to compute $c_f(S)$ when $f$ is the Eisenstein series $E_k$, in terms of Bernoulli numbers and of $\zeta(k-1)$ (this is Satz 3 in Haberland's article, Kapitel 1).
EDIT : I found the following expression for $c_f(\gamma)$ where $\gamma \in \Gamma$. It is quite complicated (maybe it could be somewhat simplified) :
\begin{equation}\frac{(k-2)!}{(2\pi i)^{k-1}} c_f(\gamma) = \int_{z_0}^{\infty} (f(z)-a_0)(z-X)^{k-2} dz + \int_{\gamma^{-1} \infty}^{z_0} (f(z) -\frac{a_0}{(cz+d)^k}) (z-X)^{k-2} dz + \frac{a_0}{k-1} ((X-z_0)^{k-1}-(X-\gamma z_0)^{k-1}|{2-k} \gamma + X^{k-1} |{2-k} (\gamma-1))where $z_0 \in \mathcal{H}$ is arbitrary and $\gamma= [a,b;c,d]$.
3 Added the computation of the value of the cocycle at S
Let $T = \begin{pmatrix} 1 & 1 \ [1,1 ; 0& , 1\end{pmatrix} ] \in \Gamma_{\infty}$. There is a natural map $V_{k+1} \to H^1(\Gamma_\infty,V_k)$ sending a polynomial $P$ to the cocycle $c_P$ determined by $c_P(T) = P(X+1)-P(X)$. It is easy to check that this map induces an isomorphism $\psi : V_{k+1}/V_k \cong H^1(\Gamma_\infty,V_k)$, so that the latter space is one-dimensional.
Note that $\widetilde{f}$ is unique up to adding some element of $V_k$. Since we integrate $k-1$ times, the function $\widetilde{f}$ should be thought of as a function of "weight" $k-2\cdot (k-1) = 2-k$ (of course this isn't true in the strict sense). Let us make this more precise.
For any $n \in \mathbf{Z}$, let $|_n$ denote the weight $n$ action of $\Gamma$ SL_2(\mathbf{R})$on the space of complex-valued functions on$\mathcal{H}$(so that any$f \in M_k$is a fixed vector of the weight$k$action). action of$\Gamma$). Note also the weight$2-k$action gives the usual action of$\Gamma$on$V_k\$. The crucial fact is that we have
Since $\gamma \mapsto \widetilde{f} |_{2-k} (\gamma-1)$ is obviously a coboundary in the space of functions on $\mathcal{H}$, it defines a cocycle $r_f$ in the space $V_k$, V_k$. Therefore we get$\delta(f) \in H^1(\Gamma,V_k)$and therefore an this element doesn't depend on the choice of$H^1(\Gamma,V_k)$. \widetilde{f}$. Thus we have constructed $\delta : M_k \to H^1(\Gamma,V_k)$.This map is compatible with the Eichler-Shimura isomorphism
It is not difficult to check that if $f =\sum_{n \geq 0} a_n e^{2i\pi nz}$ then the image of $\delta(f)$ in $H^1(\Gamma_\infty,V_k)$ is the image of the polynomial $\frac{a_0}{k-1} \frac{a_0 \cdot (2\pi i)^{k-1}}{(k-1)!} \cdot X^{k-1} \in V_{k+1}$ under the isomorphism $\psi$ above. In particular $\delta$ is injective, and the exact sequence $(*)$ gives the isomorphism you want.
Note : that there is also a formula for the value distinguished choice of $\widetilde{f}$, namely
\begin{equation}\widetilde{f} = \sum_{n \geq 1} \frac{a_n}{n^{k-1}} e^{2i\pi nz} + \frac{a_0 \cdot (2\pi i)^{k-1}}{(k-1)!} z^{k-1}.Let $c_f \in Z^1(\Gamma,V_k)$ be the cocycle associated to this choice of $r_f$ at \widetilde{f}$. Let us compute the matrix value of$S=\begin{pmatrix} 0 & -1 c_f$on$T$and$S\$. First as explained above, we have
\1 & 0 begin{equation}c_f(T)=\frac{a_0 \end{pmatrix}$, cdot (2\pi i)^{k-1}}{(k-1)!} ((X+1)^{k-1}-X^{k-1}).To compute$c_f(S)$, Haberland uses the natural integral representation of$\widetilde{f}$in terms of$f-a_0\$, and gets
\begin{equation}c_f(S) = \frac{(2\pi i)^{k-1}}{(k-2)!} \int_0^{\infty} (f(z)-\frac{a_0}{z^k}-a_0)(z-X)^{k-2} dz(there should be a similar formula for $c_f(\gamma)$ for any $\gamma \in \Gamma$). Then $c_f(S)$ can be expressed in terms of the special values $L(f,s)$ for of $1 L(f,s) := \leq sum_{n=1}^\infty a_n/n^s$ at integers $s \leq k-1$, which was also explained in Zagier's lectures{1,\ldots,k-1}$. It is then a good exercise to compute$c_f(S)$when$f$is the Eisenstein series$E_k$, in terms of Bernoulli numbers and of$\zeta(k-1)\$ (this is Satz 3 in Haberland's article, Kapitel 1).
2 Improved formatting
The result you're looking for is contained in the following article :
Haberland, Klaus. Perioden von Modulformen einer Variabler and Gruppencohomologie I (German) [Periods of modular forms of one variable and group cohomology I], Math. Nachr. 112 (1983), 245-282.
Let $S_k$ (resp. $M_k$) be the space of holomorphic cusp forms (resp. holomorphic modular forms) for $\Gamma = SL_2(\mathbf{Z})$. Let $\Gamma_{\infty}$ be the stabilizer of $\infty$ in $\Gamma$. Let $V_k$ be the space of polynomials of degree $\leq k-2$ with complex coefficients. Haberland proves an exact sequence
\begin{equation} (*) \qquad 0 \to S_k \oplus \overline{S_k} \to H^1(\Gamma,V_k) \to H^1(\Gamma_\infty,V_k) \to 0. \end{equation} Let $T =\begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix} \in \Gamma_{\infty}$. There is a natural map $V_{k+1} \to H^1(\Gamma_\infty,V_k)$ sending a polynomial $P$ to the cocycle $c_P$ determined by $c_P(T) = P(X+1)-P(X)$. It is easy to check that this map induces an isomorphism $\psi : V_{k+1}/V_k \cong H^1(\Gamma_\infty,V_k)$, so that the latter space is one-dimensional.
The "Eisenstein cocycle" you're looking for is a natural map $\delta : M_k \to H^1(\Gamma,V_k)$ which Haberland constructs the following way (actually I learnt this construction and many other properties of $\delta$ during Zagier's 2002-2003 lectures at the Collège de France).
Let $f \in M_k$. Let $\widetilde{f}$ be an Eichler integral of $f$, that is any holomorphic function on $\mathcal{H}$ such that
\begin{equation} \bigl(\frac{1}{2\pi i} \frac{d}{dz}\bigr)^{k-1} \widetilde{f}(z) = f(z). \end{equation} Note that $\widetilde{f}$ is unique up to adding some element of $V_k$.
For any $n \in \mathbf{Z}$, let $|_n$ denote the weight $n$ action of $\Gamma$ on the space of complex-valued functions on $\mathcal{H}$ (so that any $f \in M_k$ is a fixed vector of the weight $|k$)k$action). The crucial fact is that for any$\gamma \in \Gamma$, we have$\widetilde{f}
\begin{equation} \widetilde{f} |{2-k} _{2-k} (\gamma-1) \in V_k\$. V_k \qquad (\gamma \in \Gamma). \end{equation} This can be proved using Bol's identity
\begin{equation} (\frac{d}{dz})^{k-1} (F |_{2-k} g) = (\frac{d^{k-1} F}{dz^{k-1}}) |_k g \end{equation} which holds for any holomorphic function $F$ on $\mathcal{H}$ and any $g \in SL_2(\mathbf{R})$.
Since $\gamma \mapsto \widetilde{f} |_{2-k} (\gamma-1)$ is obviously a coboundary in the space of functions on $\mathcal{H}$, it defines a cocycle $r_f$ in the space $V_k$, and therefore an element of $H^1(\Gamma,V_k)$. Thus we have constructed $\delta : M_k \to H^1(\Gamma,V_k)$. This map is compatible with the Eichler-Shimura isomorphism
It is not difficult to check that if $f =\sum_{n \geq 0} a_n e^{2i\pi nz}$ then the image of $\delta(f)$ in $H^1(\Gamma_\infty,V_k)$ is the image of the polynomial $\frac{a_0}{k-1} \cdot X^{k-1} \in V_{k-1}$ V_{k+1}$under the isomorphism$\psi$above. In particular$\delta$is injective, and the exact sequence$(*)\$ gives the isomorphism you want.
Note : there is also a formula for the value of the cocycle $r_f$ at the matrix $S=\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$, in terms of the special values $L(f,s)$ for $1 \leq s \leq k-1$, which was also explained in Zagier's lectures.
Please tell me if something isn't clear in my explanation.
1
The result you're looking for is contained in the following article :
Haberland, Klaus. Perioden von Modulformen einer Variabler and Gruppencohomologie I (German) [Periods of modular forms of one variable and group cohomology I], Math. Nachr. 112 (1983), 245-282.
Let $S_k$ (resp. $M_k$) be the space of holomorphic cusp forms (resp. holomorphic modular forms) for $\Gamma = SL_2(\mathbf{Z})$. Let $\Gamma_{\infty}$ be the stabilizer of $\infty$ in $\Gamma$. Let $V_k$ be the space of polynomials of degree $\leq k-2$ with complex coefficients. Haberland proves an exact sequence
\begin{equation} (*) \qquad 0 \to S_k \oplus \overline{S_k} \to H^1(\Gamma,V_k) \to H^1(\Gamma_\infty,V_k) \to 0. \end{equation} Let $T =\begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix} \in \Gamma_{\infty}$. There is a natural map $V_{k+1} \to H^1(\Gamma_\infty,V_k)$ sending a polynomial $P$ to the cocycle $c_P$ determined by $c_P(T) = P(X+1)-P(X)$. It is easy to check that this map induces an isomorphism $\psi : V_{k+1}/V_k \cong H^1(\Gamma_\infty,V_k)$, so that the latter space is one-dimensional.
The "Eisenstein cocycle" you're looking for is a natural map $\delta : M_k \to H^1(\Gamma,V_k)$ which Haberland constructs the following way (actually I learnt this construction and many other properties of $\delta$ during Zagier's 2002-2003 lectures at the Collège de France).
Let $f \in M_k$. Let $\widetilde{f}$ be an Eichler integral of $f$, that is any holomorphic function on $\mathcal{H}$ such that
\begin{equation} \bigl(\frac{1}{2\pi i} \frac{d}{dz}\bigr)^{k-1} \widetilde{f}(z) = f(z). \end{equation} Note that $\widetilde{f}$ is unique up to adding some element of $V_k$.
For any $n \in \mathbf{Z}$, let $|_n$ denote the weight $n$ action of $\Gamma$ on the space of complex-valued functions on $\mathcal{H}$ (so that any $f \in M_k$ is a fixed vector of $|k$). The crucial fact is that for any $\gamma \in \Gamma$, we have $\widetilde{f} |{2-k} (\gamma-1) \in V_k$. This can be proved using Bol's identity
\begin{equation} (\frac{d}{dz})^{k-1} (F |_{2-k} g) = (\frac{d^{k-1} F}{dz^{k-1}}) |_k g \end{equation} which holds for any holomorphic function $F$ on $\mathcal{H}$ and any $g \in SL_2(\mathbf{R})$.
Since $\gamma \mapsto \widetilde{f} |_{2-k} (\gamma-1)$ is obviously a coboundary in the space of functions on $\mathcal{H}$, it defines a cocycle $r_f$ in the space $V_k$, and therefore an element of $H^1(\Gamma,V_k)$. Thus we have constructed $\delta : M_k \to H^1(\Gamma,V_k)$. This map is compatible with the Eichler-Shimura isomorphism
It is not difficult to check that if $f =\sum_{n \geq 0} a_n e^{2i\pi nz}$ then the image of $\delta(f)$ in $H^1(\Gamma_\infty,V_k)$ is the image of the polynomial $\frac{a_0}{k-1} \cdot X^{k-1} \in V_{k-1}$ under the isomorphism $\psi$ above. In particular $\delta$ is injective, and the exact sequence $(*)$ gives the isomorphism you want.
Note : there is also a formula for the value of the cocycle $r_f$ at the matrix $S=\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$, in terms of the special values $L(f,s)$ for $1 \leq s \leq k-1$, which was also explained in Zagier's lectures.
Please tell me if something isn't clear in my explanation. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 261, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8172773122787476, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/52452/topology-of-black-holes/52468 | ## Topology of black holes
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I've asked this question of some physicist friends of mine and I've never gotten a satisfactory answer: What is topologically possible for a neighborhood of a black hole? To clarify, I'm curious about the topology as a 4-manifold, although I'd also be interested to hear about time-like and space-like slices as well. I've heard that a time-like slice of the event horizon can be a torus or sphere, but this isn't really what I'd like to know, although I imagine that there is a close connection between the topology of the event horizon (as a 3-manifold) and the topology of a neighborhood of the black hole. Please ask if any clarifications are needed.
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What do you want to use as a definition of a black hole? For example if your space time is a compact 3-manifold x a real line, wouldn't the whole of space-time satisfy Hawking and Ellis's definition? – Ryan Budney Jan 19 2011 at 2:36
## 4 Answers
So far J Verma and RBega provided two succint descriptions on topology of the apparent horizon itself (for any space-time admitted trapped regions), and so by association, the topology of the event horizon in a stationary asymptotically flat black-hole space-time.
I'll try to provide an answer to an interpretation of the question you asked for: namely that of the topology of a neighborhood of the event horizon. That I have to interpret the question is because you have not actually provide a description of what you mean by a neighborhood. Using the Hawking topology theorem you can easily manage that the topology of the event horizon is $\mathbb{S}^2\times \mathbb{R}$, and is an embedded null hypersurface. So a tubular neighborhood of it necessarily has the topology $\mathbb{S}^2\times \mathbb{R}^2$, which is, for one thing, simply connected. But there is absolutely nothing to prevent you from choosing a neighborhood to be some arbitrary open set containing the event horizon with complicated topology. Indeed, you can easily imagine removing a four dimensional tube disjoint from the event horizon from the $\mathbb{S}^2\times\mathbb{R}^2$ set to get something that is not simply connected.
Because of this freedom to choose subsets, the naive reading of you question leads to the answer that "pretty much as bad as you want".
But that answer is rather physically useless: it doesn't capture anything essential about black holes. In fact, the answer given above is identical to the answer to the following question: let $U$ be an open connected subset of $\mathbb{R}^4$, what kinds of topology can $U$ admit?
A more useful question to ask is: given an isolated gravitating body (such as a black hole), what is the topology of the space-time outside of it? And that question is one admitting a good answer. The content is the topological censorship theorem. In physicist's language, to quote Friedman, Schleich, and Witt,
general relativity does not allow an observer to probe the topology of spacetime: Any topological structure collapses too quickly to allow light to traverse it.
An early version of this is due to Gannon, who showed that Cauchy hypersurface with non-trivial topology will necessarily generate a development which is null geodesically incomplete. The FSW paper showed that under some restrictions, all the non-trivial topology must be hidden behind the event horizon.
A stronger generalisation of the topological censorship theorem is due to Galloway, who later, with Schleich, Witt, and Woolgar, extended the result from asymptotically flat space-times to also asymptotically anti-deSitter ones. One interesting crucial assumption of these theorems is the requirement for null or time-like "Scri".
A somewhat related paper is this one by Schleich and Witt which I didn't read in detail so cannot say more about.
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great answer.+1 – J Verma Jan 19 2011 at 16:42
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Hawking's Theorem of Black Hole topology asserts that the in case of $4$d asymptotically flat stationary black holes satisfying the suitable energy condition (dominant energy condition), the cross sections of the evernt horizon are spherical.
Galloway and Schoen extended this theorem to higher dimensions; they showed that the cross sections of event horizon (stationary case) and the outer (apparent) horizon (general case) are of positive Yamabe type. This paper can be found at Galloway's webpage www.math.miami.edu/~galloway/papers/220_2006_19_OnlinePDF.pdf
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The previous answers dealt with the physically relevant case of $d=4$ spacetime dimensions. One of the surprising discoveries in recent years is that in higher dimensions the possible topologies are much richer. I believe this started with the discovery by Emparan and Reall of a black hole in $d=5$ with horizon topology $S^2 \times S^1$ (hep-th/0110260). The recent paper arXiv:1002.0490 by Hollands et. al. surveys the situation and discusses restrictions on the possible topologies of the horizon for $d=5$ black holes.
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thanks for pointing this.The review article "Black Holes in Higher dimensions" by Reall and Emparan in Living Reviews also has a section on Black hole topology. – J Verma Jan 19 2011 at 15:05
1
Actually, in $d=5$, the most relevant question is "which of the ones allowed by Galloway-Schoen is attained?" Since in 5 space-time dimensions, the horizon cross section is 3 dimensional, and all compact 3 manifolds in positive Yamabe class is known, the question becomes "which of those actually corresponds to stationary asymptotically flat solutions". A futher interesting question is the consideration of black-saturn or di-ring type horizons: cases with more than one connected component of the horizon. Such stationary solutions are not expected to exist in 4d. – Willie Wong Jan 19 2011 at 15:18
Willie, thanks for the clarification. Black-saturn solutions in $d=5$ can be found in arXiv.org/pdf/hep-th/0701035. – Jeff Harvey Jan 19 2011 at 15:33
Some di-ring solutions are also known. The first example that Google returns is arxiv.org/abs/hep-th/0511007 ; I mentioned those two by name because there are ones for which I know have been more extensively studied, and they are the known examples of the more general cases of disconnected horizons for which the field is mostly wide open. – Willie Wong Jan 19 2011 at 16:56
[While I was writing, J. Verma answered the question more succinctly but I already wrote most of my answer so figured I would post it]
First of all I am not a physicist nor do I work in GR, but I have taken a number of courses over the year so I will attempt to give some idea of an answer (I'm sure an expert can expand).
Basically one wants to restrict attention to asymptotically flat spacetimes (in 4d) i.e complete spacelike slices of the spacetime are asymptotically euclidean (i.e. in a neighborhood of infinity the metric looks like the euclidean metric in a definite way). This physically models an isolated gravitating system. I think one probably also wants a global non-vanishing timelike vectorfield (so no timelike closed loops) and also that the the dominant energy condition holds (this is too technical to state but physically means no energy enters the past light cone of a point from outside the past light cone).
Given this setup the event horizon is still not an easy concept to grasp (it depends on the global hyperbolic structure and so requires one to understand the entire spacetime all at once... I'm not sure what can be said about it honestly other than for highly symmetric spacetimes). An easier thing to grab hold of are MOTS (marginally outer trapped surfaces). To find these one takes a space-like complete slice and looks for surfaces whose area is unchanged under the flow by null vectors (this is marginally trapped) and that there is no surface strictly outside of this surface with this property (this is outermost). The existence of one of these implies the formation of a singularity (due to Penrose?) so are natural stand-ins for black holes (I'm not sure how much more is known about the full relationship between MOTS and black holes though I believe understanding it completely is about as hard as showing cosmic censorship). Anyway, it can be shown that these guys (since they are "stable") must be either spheres or tori (this is analogous to theorems about minimal surfaces) and the latter has a rigidity result attached to it which I believe makes it unphysical (due to Galloway and Schoen).
I hope this makes sense and I haven't made any glaring errors.
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The chronology assumption is not necessary for the topology theorem of Hawking/Galloway-Schoen. The main argument works essentially on an initial data set satisfying the constraint equations (there will be constraints if you take a 1+3 hyperbolic-elliptic decomposition of the Einstein equations). You should think of it as a statement about a 3 manifold and its embedding into a 4 manifold: you only need Einstein's equation to be satisfied on a small neighborhood the embedded 3 manifold (in fact, I think just pointwise along the 3 manifold may be enough). – Willie Wong Jan 19 2011 at 13:15
Likewise, the asymptotic flatness assumption is also unnecessary, except for physical interpretation to attach meaning to the words "inside the horizon" and "outside the horizon". The result is essentially local. Of course, these facts are connected to what you wrote in the second to last paragraph: the statement of Galloway-Schoen is a statement about apparent horizons (MOTS), and not a statement about event horizons (which is connected to the global 4-dimensional geometry of the space-time). MOTS can be defined purely locally, and hence can be studied without regard to global structures. – Willie Wong Jan 19 2011 at 13:20
Thanks for the comments! I see I was being too conservative in my assumptions. – Rbega Jan 19 2011 at 15:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9371412992477417, "perplexity_flag": "middle"} |
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Dark energy sound speed
Author Message
Antony Lewis
Joined: 23 Sep 2004
Posts: 673
Affiliation: University of Sussex
Posted: December 15 2004
Are there any theoretical bounds on what sound speed (meaning $c_s^2 \equiv \delta p / \delta \rho$) the dark energy could have? In particular $c_s^2 <0$ or $c_s^2 > 1$ are often not consisdered, but I don't see any immediate contradictions from the cosmological perturbation equations?
Anze Slosar
Joined: 24 Sep 2004
Posts: 198
Affiliation: Brookhaven National Laboratory
Posted: December 15 2004
Well, isn't the sound speed just what is says on the can, i.e. the speed at which small perturbations move? I would expecte that cs2>1 would violate causality, while cs2<0 would just give you evanescent waves, i.e. a dissipative media. So cs2<0 might actually be an interesting option after all.
Adam Amara
Joined: 25 Sep 2004
Posts: 18
Affiliation: ETH
Posted: December 16 2004
I'm about to show my ignorance but I have no idea what an evanescent wave is. Cs2<0 doesn’t seem to make much sense to me but it would be really interested to hear more if it does. Adam
Adam Amara
Joined: 25 Sep 2004
Posts: 18
Affiliation: ETH
Posted: December 16 2004
Sorry, I've just been reading around. Just in case anyone else was as clueless as i was: "An electromagnetic wave observed in total internal reflection, undersized waveguides, and in periodic dielectric heterostructures. While wave solutions have real wavenumbers k, k for an evanescent mode is purely imaginary. Evanescent modes are characterized by an exponential attenuation and lack of a phase shift." Again, sounds interesting, but how does this apply to cosmology? Adam
Antony Lewis
Joined: 23 Sep 2004
Posts: 673
Affiliation: University of Sussex
Posted: December 16 2004
How it applies to cosmology is really the question. If you give me a can of dark energy, then indeed $c_s^2$ is just what you think of as the sound speed: for a constant equation of state $\delta'' + k^2 c_s^2 \delta = 0$ (if the sound speed is negative, you immediately see that the solution is exponential rather than oscillating). The question I'm asking is about dark energy in the universe. Here the full equations are relativistic and coupled to other matter by GR: $\delta' + 3H(\hat{c}_{s}^2-w)(\delta +3H(1+w)v/k)+ (1+w)kv + 3H w' v/k= -3(1+w)h'$ $v' + H(1-3\hat{c}_{s}^2)v + kA = k \hat{c}_{s}^2 \delta/(1+w)$ Much more complicated! (h' is a source from background expansion fluctuations, v is the velocity, A is the acceleration (zero in synchronous gauge), H is the conformal hubble rate, and derivatives are conformal time). The sound speed is defined in the rest frame of the dark energy as $c_s^2 \equiv \delta p/ \delta \rho$, which is not by definition positive despite the notation. In general the equation of state w is a function of time, but the sound speed can be a function of time and wavenumber k. So whether or not there are physical constraints on it - especially on super-horizon scales - is unclear to me. If the dark energy is some effective stuff, say a GR representation of some brane-world effect, I don't think you can argue constraints on its sound speed as though it were a simple fluid that could exist independent of cosmology?
Anze Slosar
Joined: 24 Sep 2004
Posts: 198
Affiliation: Brookhaven National Laboratory
Posted: December 16 2004
Antony Lewis wrote:
In general the equation of state w is a function of time, but the sound speed can be a function of time and wavenumber k. So whether or not there are physical constraints on it - especially on super-horizon scales - is unclear to me. If the dark energy is some effective stuff, say a GR representation of some brane-world effect, I don't think you can argue constraints on its sound speed as though it were a simple fluid that could exist independent of cosmology?
Yes, I agree with you. My point was that locally the dark energy doesn't know about cosmology and hence $c_s^2$ cannot be greater than one, at least for large k. If you take $c_s^2$ to be a function of k, then it can be anything on superhorizon scales... Do you buy this?
Your complicated equations are perfect for beard stroking at 5pm with a cup of tea!
Anze Slosar
Joined: 24 Sep 2004
Posts: 198
Affiliation: Brookhaven National Laboratory
Posted: December 16 2004
Antony Lewis wrote:
Much more complicated! (h' is a source from background expansion fluctuations, v is the velocity, A is the acceleration (zero in synchronous gauge), H is the conformal hubble rate, and derivatives are conformal time). The sound speed is defined in the rest frame of the dark energy as $c_s^2 \equiv \delta p/ \delta \rho$, which is not by definition positive despite the notation.
Following Amara's suit, I will now ask a stupid question. Isn't it that dp / dρ = w by definition of w (i.e. 1/3 for standard waves, etc...) How can you then define δp / δρ for a given k mode and how does it relate to w?
Antony Lewis
Joined: 23 Sep 2004
Posts: 673
Affiliation: University of Sussex
Posted: December 16 2004
The equation of state is defined as w≡p' / ρ'. This is not the same as the definition of $c_s^2$ so in general they are different (indeed for quintessence $c_s^2=1$ always). They are only equal if you apply adiabaticity: not the case for quintessence. Some refs are astro-ph/0307100, astro-ph/0307104, astro-ph/0410680.
Pier Stefano Corasaniti
Joined: 11 Nov 2004
Posts: 42
Affiliation: LUTH, Observatoire de Paris-Meudon
Posted: December 16 2004
Antony Lewis wrote:
not the case for quintessence.
This statement is too strong, whether scalar field perturbations are adiabatic or isocurvature depends on the initial conditions, which we think are set by inflation.
If the quintessence field is a decay product of the inflaton the perturbations in the quintessence will be adiabatic, which means that not only the perturbations of the quintessence field relative to the other fluids
(Sq f =0) initially vanish, but also the scalar field intrinsic entropy perturbation (Gamma_q =0) is initially zero.
If these conditions are verified at an initial time the conservation of the energy momentum tensor (which provide the equation of motion for the perturbations) impose that they are verified at any other time (no matter the gauge, actually everything can be formulated in gauge invariant language as S and Gamma are gauge invariant variables).
In other words adiabaticity is preserved by the flow equations and no matter what is the background scalar field evolution or the choice of the gauge, the scalar field perturbations will remain adiabatic (Gamma_q=0 always) as those in the other fluids.
This can be inferred from General Relativity arguments in the context of the Separate Universe Approach (Wands, Malik, Lyth and Liddle, astro-ph/0003278).
More specifically we have shown this by directly looking at the flow equations in the gauge invariant formulation (Bartolo, Corasaniti, Liddle, Malquarti, astro-ph/0311503).
Also we have found that if the initial conditions are isocurvature ones, which could be the case if the quintessence or k-essence or whatever is not a decay product of the inflaton, then the evolution of the isocurvature quintessence mode depends on the dynamics of the background. In particular during kination regimes the isocurvature mode is amplified, while it decays during tracker ones. This implies that only those quintessence scenarios with initial isocurvature perturbations and with the field undergoing a long kination phase followed by a short period of tracking can give rise to a cosmologically relevant isocurvature perturbation that can survive at present time. (The specific time evolution of this isocurvature mode can be different in the case of k-essence, since the two scenarios (Q <-> K) can be mapped one into the other only at the homogeneus level).
I tend to accept the fact that cs2, as a free parameter for an effective fluid description of scalar field perturbations, is a simple way of parametrizing something which depends on initial conditions, however from a pure physical point of view I find this effective description rather misleading, so one should have always in mind the microscopic description of the scalar field whatsoever is its lagrangian.
Pier-Stefano
Antony Lewis
Joined: 23 Sep 2004
Posts: 673
Affiliation: University of Sussex
Posted: December 16 2004
I should probably have said internally adiabatic (following e.g. astro-ph/0410680). The rest frame sound speed of a quintessence field is always identically 1 independent of the initial conditions (a simple proof is given in e.g. astro-ph/0307104). Hence for w≠ - 1 it is never internally adiabatic. I hope this is correct? I take you point to be that for models in which $c_s^2$ is not fixed, it may depend on the initial conditions. This is an interesting point I'd not considered.
Pier Stefano Corasaniti
Joined: 11 Nov 2004
Posts: 42
Affiliation: LUTH, Observatoire de Paris-Meudon
Posted: December 17 2004
Antony Lewis wrote:
I should probably have said internally adiabatic (following e.g. astro-ph/0410680). The rest frame sound speed of a quintessence field is always identically 1 independent of the initial conditions (a simple proof is given in e.g. astro-ph/0307104). Hence for w≠ - 1 it is never internally adiabatic. I hope this is correct? I take you point to be that for models in which $c_s^2$ is not fixed, it may depend on the initial conditions. This is an interesting point I'd not considered.
If by internally adiabatic you mean that the intrinsic entropy perturbation Γq = 0, this is true as long as the initial conditions are adiabatic and on the large scales approximation. In fact inside the horizon the modes get mixed up even if you start from initial adiabatic conditions, as can be seen from Eq.(33) and (34) in astro-ph/0311503.
Perhaps it is just a different language, $c_s^2=1$ it is used to define adiabatic scalar field perturbations in the effective fluid description, while $c_s^2\ne 1$ would correspond to isocurvature ones.
But I find this wording misleading anyway since as you say in the scalar field rest frame $c_s^2=1$ independently of the initial conditions, while adiabaticity has to do with the initial conditions independently of the gauge. And personally I prefer the microscopic description rather than the effective one, although the former requires to specify a lagrangian.
Fabio Finelli
Joined: 12 Nov 2004
Posts: 2
Affiliation: INAF-CNR/IASF, Sezione di Bologna
Posted: January 19 2005
Dear all,
I take this chance to communicate my ideas on the topic, which have generated
astro-ph/0211626, astro-ph/0304325, astro-ph/0405041.
My motivation was to understand if you can discriminate within different Dark Energy models with a different speed of sound: this was suggested by K-essence authors, since quintessence has always cx2=1. By cX2 I mean the quantity in the pressure
perturbations
δ px = cx2 δ ρx + f(η,wx) Θx
where the second term denotes the intrinsic non-adiabatic pressure perturbations,
θx is the velocity potential for DE, wx is the equation of state of wx and η is the conformal time.
Within scalar field theories (where f(η) ≠ 0)
it is difficult to discriminate between cx2=1 and generic cx (unless cx is drastically different from 1) as it can be seen in astro-ph/0405041.
For perfect fluid of model of DE (where f(η) = 0) , cx2 is locked in some way to wx
and there is a chance to have stronger effects on CMB and LSS (as can be seen in
astro-ph/0211626, astro-ph/0304325). For instance, the Chaplygin gas is ruled out at more than 3 σ as a dark energy model (see astro-ph/0304325).
Cheers,
Fabio
Antony Lewis wrote:
The equation of state is defined as w≡p' / ρ'. This is not the same as the definition of $c_s^2$ so in general they are different (indeed for quintessence $c_s^2=1$ always). They are only equal if you apply adiabaticity: not the case for quintessence. Some refs are astro-ph/0307100, astro-ph/0307104, astro-ph/0410680.
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You cannot vote in polls in this forum | {"extraction_info": {"found_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": 20, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9177214503288269, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/238220/basis-for-this-mathbbp-3-subspace | # Basis for this $\mathbb{P}_3$ subspace.
Just had an exam where the last question was:
Find a basis for the subset of $\mathbb{P}_3$ where $p(1) = 0$ for all $p$.
I answered $\{t,t^2-1,t^3-1\}$, but I'm not entirely confident in the answer. Did I think about the question in the wrong way?
-
1
Your first polynomial, $t$, isn’t in the subspace: if if $p(t)=t$, then $p(1)=1$, not $0$. Did you mean $t-1$? That would work. – Brian M. Scott Nov 15 '12 at 20:58
Since you have only told us your answer, and haven't told us how you thought about the question, it is impossible to answer your query as to whether you thought about the question the wrong way. – Gerry Myerson Nov 16 '12 at 1:51
## 2 Answers
Another way to get to the answer:
$P_3=\{{ax^3+bx^2+cx+d:a,b,c,d{\rm\ in\ }{\bf R}\}}$. For $p(x)=ax^3+bx^2+cx+d$ in $P_3$, $p(1)=0$ is $$a+b+c+d=0$$ So, you have a "system" of one linear equation in 4 unknowns. Presumably, you have learned how to find a basis for the vector space of all solutions to such a system, or, to put it another way, a basis for the nullspace of the matrix $$\pmatrix{1&1&1&1\cr}$$ One such basis is $$\{{(1,-1,0,0),(1,0,-1,0),(1,0,0,-1)\}}$$ which corresponds to the answer $$\{{x^3-x^2,x^3-x,x^3-1\}}$$ one of an infinity of correct answers to the question.
-
I'm assuming by $\mathbb{P}_3$ you mean the vector space of polynomials of degree 3 or less. It has dimension $4$, and since you have one condition (namely $p(1)=0$), we expect the subspace to have dimension 3, one less.
So it is enough to find three linearly independent basis vectors. In this example, that means three linearly independent polynomials, all with $p(1)=0$.
In your answer ($\{t, t^2-1, t^3-1\}$), the first one must be wrong, since $t(1)=1$. If you replace $t$ by $t-1$, then you have three linearly independent polynomials, all satisfying $p(1)=0$.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9587767720222473, "perplexity_flag": "head"} |
http://mathhelpforum.com/advanced-algebra/112438-order-element.html | # Thread:
1. ## Order of element
Let G be a finite group. Prove the order of any element v (which v is an element of G) divides the order |G|of G.
2. Originally Posted by apple2009
Let G be a finite group. Prove the order of any element v (which v is an element of G) divides the order |G|of G.
Hint: $\langle v\rangle$ is a subgroup of $G$. Now apply Lagrange's theorem.
3. am I need to prove v is a subgroup of G? and how?
4. Originally Posted by apple2009
am I need to prove v is a subgroup of G? and how?
You already know that $\langle v\rangle$ is a subgroup of $G$ (or you should). Now use the fact that $|v|=|\langle v\rangle|$ and Lagranges theorem. | {"extraction_info": {"found_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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9006344676017761, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/87669?sort=newest | ## $\psi$ class in $\overline{M}_{0,n}$
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Basic question, but I found no reference.
Is the $\psi$ class the only one which is not a boundary class in the PIcard group of the Deligne-Mumford compactification of $\mathcal{M}_{0,n}$? Or can it be expressed in terms of boundary divisors? If yes what is its expression?
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## 1 Answer
Unless you mean something else when you write $\psi$ class, it is expressible in terms of boundary divisors.
That is, if $\psi_i$ is the $i$-th cotangent bundle, then you can write it in terms of boundary divisors. One reference for this is the tome "Mirror Symmetry" by Hori, Katz, Klemm, et al. on p. 513, the comparison lemma.
The idea is that you can consider the forgetful maps $\pi$ from $M_{0,n}$ to $M_{0,n-1}$ and look at the divisor $$\psi_i - \pi^*\psi_i$$ (where the $\psi$ classes are, by abuse of notation, living on different spaces). This is expressible in terms of boundary divisors, and so we can inductively write out the $\psi$ classes on any $M_{0,n}$.
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Thank you Simon, excellent and precise answer. – IMeasy Feb 7 2012 at 22:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8747188448905945, "perplexity_flag": "head"} |
http://en.wikipedia.org/wiki/Finitary_relation | # Finitary relation
For other uses, see Relation (disambiguation).
In set theory and logic, a relation is a property that assigns truth values to $k$-tuples of individuals. Typically, the property describes a possible connection between the components of a $k$-tuple. For a given set of $k$-tuples, a truth value is assigned to each $k$-tuple according to whether the property does or does not hold.
An example of a ternary relation (i.e., between three individuals) is: "$X$ was introduced to $Y$ by $Z$", where $\left(X, Y, Z\right)$ is a 3-tuple of persons; for example, "Beatrice Wood was introduced to Henri-Pierre Roché by Marcel Duchamp" is true, while "Karl Marx was introduced to Friedrich Engels by Queen Victoria" is false.
The variable $k$ giving the number of "places" in the relation, 3 for the above example, is a non-negative integer, called the relation's arity, adicity, or dimension. A relation with $k$ places is variously called a $k$-ary, a $k$-adic, or a $k$-dimensional relation. Relations with a finite number of places are called finite-place or finitary relations. It is possible to generalize the concept to include infinitary relations between infinitudes of individuals, for example infinite sequences; however, in this article only finitary relations are discussed, which will from now on simply be called relations.
Since there is only one 0-tuple, the so-called empty tuple ( ), there are only two zero-place relations: the one that always holds, and the one that never holds. They are sometimes useful for constructing the base case of an induction argument. One-place relations are called unary relations. For instance, any set (such as the collection of Nobel laureates) can be viewed as a collection of individuals having some property (such as that of having been awarded the Nobel prize). Two-place relations are called binary relations or, in the past, dyadic relations. Binary relations are very common, given the ubiquity of relations such as:
• Equality and inequality, denoted by signs such as '$=$' and '$<$' in statements like '$5 < 12$';
• Being a divisor of, denoted by the sign '$\mid$' in statements like '$13 \mid 143$';
• Set membership, denoted by the sign '$\in$' in statements like '$1 \in \mathbb{N}$'.
A $k$-ary relation is a straightforward generalization of a binary relation.
## Informal introduction
Relation is formally defined in the next section. In this section we introduce the concept of a relation with a familiar everyday example. Consider the relation involving three roles that people might play, expressed in a statement of the form "X thinks that Y likes Z ". The facts of a concrete situation could be organized in a Table like the following:
Relation S : X thinks that Y likes Z
Person X Person Y Person Z
Alice Bob Denise
Charles Alice Bob
Charles Charles Alice
Denise Denise Denise
Each row of the Table records a fact or makes an assertion of the form "X thinks that Y likes Z ". For instance, the first row says, in effect, "Alice thinks that Bob likes Denise". The Table represents a relation S over the set P of people under discussion:
P = {Alice, Bob, Charles, Denise}.
The data of the Table are equivalent to the following set of ordered triples:
S = {(Alice, Bob, Denise), (Charles, Alice, Bob), (Charles, Charles, Alice), (Denise, Denise, Denise)}.
By a slight abuse of notation, it is usual to write S(Alice, Bob, Denise) to say the same thing as the first row of the Table. The relation S is a ternary relation, since there are three items involved in each row. The relation itself is a mathematical object defined in terms of concepts from set theory (i.e., the relation is a subset of the Cartesian product on {Person X, Person Y, Person Z}), that carries all of the information from the Table in one neat package. Mathematically, then, a relation is simply an "ordered set".
The Table for relation S is an extremely simple example of a relational database. The theoretical aspects of databases are the specialty of one branch of computer science, while their practical impacts have become all too familiar in our everyday lives. Computer scientists, logicians, and mathematicians, however, tend to see different things when they look at these concrete examples and samples of the more general concept of a relation.
For one thing, databases are designed to deal with empirical data, and experience is always finite, whereas mathematics at the very least concerns itself with potential infinity. This difference in perspective brings up a number of ideas that may be usefully introduced at this point, if by no means covered in depth.
## Formal definitions
When two objects, qualities, classes, or attributes, viewed together by the mind, are seen under some connexion, that connexion is called a relation.
The simpler of the two definitions of k-place relations encountered in mathematics is:
Definition 1. A relation L over the sets X1, …, Xk is a subset of their Cartesian product, written L ⊆ X1 × … × Xk.
Relations are classified according to the number of sets in the defining Cartesian product, in other words, according to the number of terms following L. Hence:
• Lu denotes a unary relation or property;
• Luv or uLv denote a binary relation;
• Luvw denotes a ternary relation;
• Luvwx denotes a quaternary relation.
Relations with more than four terms are usually referred to as k-ary or n-ary, for example, "a 5-ary relation". A k-ary relation is simply a set of k-tuples.
The second definition makes use of an idiom that is common in mathematics, stipulating that "such and such is an n-tuple" in order to ensure that such and such a mathematical object is determined by the specification of n component mathematical objects. In the case of a relation L over k sets, there are k + 1 things to specify, namely, the k sets plus a subset of their Cartesian product. In the idiom, this is expressed by saying that L is a (k + 1)-tuple.
Definition 2. A relation L over the sets X1, …, Xk is a (k + 1)-tuple L = (X1, …, Xk, G(L)), where G(L) is a subset of the Cartesian product X1 × … × Xk. G(L) is called the graph of L.
Elements of a relation are more briefly denoted by using boldface characters, for example, the constant element $\mathbf{a}$ = (a1, …, ak) or the variable element $\mathbf{x}$ = (x1, …, xk).
A statement of the form "$\mathbf{a}$ is in the relation L " is taken to mean that $\mathbf{a}$ is in L under the first definition and that $\mathbf{a}$ is in G(L) under the second definition.
The following considerations apply under either definition:
• The sets Xj for j = 1 to k are called the domains of the relation. Under the first definition, the relation does not uniquely determine a given sequence of domains.
• If all of the domains Xj are the same set X, then it is simpler to refer to L as a k-ary relation over X.
• If any of the domains Xj is empty, then the defining Cartesian product is empty, and the only relation over such a sequence of domains is the empty relation L = $\varnothing$. Hence it is commonly stipulated that all of the domains be nonempty.
As a rule, whatever definition best fits the application at hand will be chosen for that purpose, and anything that falls under it will be called a relation for the duration of that discussion. If it becomes necessary to distinguish the two definitions, an entity satisfying the second definition may be called an embedded or included relation.
If L is a relation over the domains X1, …, Xk, it is conventional to consider a sequence of terms called variables, x1, …, xk, that are said to range over the respective domains.
Let a Boolean domain B be a two-element set, say, B = {0, 1}, whose elements can be interpreted as logical values, typically 0 = false and 1 = true. The characteristic function of the relation L, written ƒL or χ(L), is the Boolean-valued function ƒL : X1 × … × Xk → B, defined in such a way that ƒL($\mathbf{x}$) = 1 just in case the k-tuple $\mathbf{x}$ is in the relation L. Such a function can also be called an indicator function, particularly in probability and statistics, to avoid confusion with the notion of a characteristic function in probability theory.
It is conventional in applied mathematics, computer science, and statistics to refer to a Boolean-valued function like ƒL as a k-place predicate. From the more abstract viewpoint of formal logic and model theory, the relation L constitutes a logical model or a relational structure that serves as one of many possible interpretations of some k-place predicate symbol.
Because relations arise in many scientific disciplines as well as in many branches of mathematics and logic, there is considerable variation in terminology. This article treats a relation as the set-theoretic extension of a relational concept or term. A variant usage reserves the term "relation" to the corresponding logical entity, either the logical comprehension, which is the totality of intensions or abstract properties that all of the elements of the relation in extension have in common, or else the symbols that are taken to denote these elements and intensions. Further, some writers of the latter persuasion introduce terms with more concrete connotations, like "relational structure", for the set-theoretic extension of a given relational concept.
## Transitive Relations
Transitive relations are binary relations R on a single set X where for all a, b, c in X, aRb and bRc implies aRc. Transitive relations fall into two broad classes, equivalence relations and order relations. Equivalence relations are also symmetric and reflexive, while order relations are antisymmetric (complete order) or asymmetric (partial order) and may be reflexive (inclusive order) or anti-reflexive (strict order). The algebraic structure of equivalence relations builds on transformation groups; that of order relations builds on lattice theory.
## Analogy with functions
A binary relation R on sets X and Y may be considered to associate, with each member of X, zero or more members of Y. (In the case of a relation T on more than two sets, X or Y or both can be cross products of any of the sets on which T is defined.) X is then referred to a the domain of R. Y is called the range or codomain of R. The subset of Y associated with a member x of X, is called the image of x, written as R(x). The subset of Y associated with a subset ξ of X is the union of the images of all the x in ξ and is called the image of ξ, written as R(ξ).
R is fully defined or total at X, if for every member x of X, there is at least one member y of Y where xRy. R is uniquely defined or tubular at X, if for every member x of X, there is at most one member y of Y where xRy. R is surjective or total at Y, if for every member y of Y, there is at least one member x of X where xRy. R is injective or tubular at Y, if for every member y of Y, there is at most one member x of X where xRy. If R is both fully defined and uniquely defined then R is well defined or 1-regular at X (for every member x of X, there is one and only one member y of Y where xRy). If R is both surjective and injective then R is bijective or 1-regular at Y. If R is both uniquely defined and injective then R is one-to-one.
A function is a well defined relation. A uniquely defined relation is a partial function. A surjective function is a surjection. An injective function is an injection. A bijective function is a bijection.
Relations generalize functions. Just as there is composition of functions, there is composition of relations.
Every binary relation R has a transpose relation R−1, which is related to the inverse function. For a relation R that is both fully defined and injective, the transpose relation R−1 is a true inverse in that R−1 faithfully restores any element x or subset ξ: R−1(R(ξ)) = ξ.
## Examples
This section discusses, by way of example, the arithmetical binary relation of divisibility.
### Divisibility
A more typical example of a 2-place relation in mathematics is the relation of divisibility between two positive integers n and m that is expressed in statements like "n divides m" or "n goes into m." This is a relation that comes up so often that a special symbol "|" is reserved to express it, allowing one to write "n|m" for "n divides m."
To express the binary relation of divisibility in terms of sets, we have the set P of positive integers, P = {1, 2, 3, …}, and we have the binary relation D on P such that the ordered pair (n, m) is in the relation D just in case n|m. In other turns of phrase that are frequently used, one says that the number n is related by D to the number m just in case n is a factor of m, that is, just in case n divides m with no remainder. The relation D, regarded as a set of ordered pairs, consists of all pairs of numbers (n, m) such that n divides m.
For example, 2 is a factor of 4, and 6 is a factor of 72, which can be written either as 2|4 and 6|72 or as D(2, 4) and D(6, 72).
## Suggested reading
The logician Augustus De Morgan, in work published around 1860, was the first to articulate the notion of relation in anything like its present sense. He also stated the first formal results in the theory of relations (on De Morgan and relations, see Merrill 1990). Charles Sanders Peirce restated and extended De Morgan's results. Bertrand Russell (1938; 1st ed. 1903) was historically important, in that it brought together in one place many 19th century results on relations, especially orders, by Peirce, Gottlob Frege, Georg Cantor, Richard Dedekind, and others. Russell and A. N. Whitehead made free use of these results in their epochal Principia Mathematica. For a systematic treatise on the theory of relations see R. Fraïssé, Theory of Relations (North Holland; 2000).
## Notes
1. De Morgan, A. (1858) "On the syllogism, part 3" in Heath, P., ed. (1966) On the syllogism and other logical writings. Routledge. P. 119,
## References
• Peirce, C.S. (1870), "Description of a Notation for the Logic of Relatives, Resulting from an Amplification of the Conceptions of Boole's Calculus of Logic", Memoirs of the American Academy of Arts and Sciences 9, 317–78, 1870. Reprinted, Collected Papers CP 3.45–149, Chronological Edition CE 2, 359–429.
• Ulam, S.M. and Bednarek, A.R. (1990), "On the Theory of Relational Structures and Schemata for Parallel Computation", pp. 477–508 in A.R. Bednarek and Françoise Ulam (eds.), Analogies Between Analogies: The Mathematical Reports of S.M. Ulam and His Los Alamos Collaborators, University of California Press, Berkeley, CA.
## Bibliography
• Bourbaki, N. (1994) Elements of the History of Mathematics, John Meldrum, trans. Springer-Verlag.
• Carnap, Rudolf (1958) Introduction to Symbolic Logic with Applications. Dover Publications.
• Halmos, P.R. (1960) Naive Set Theory. Princeton NJ: D. Van Nostrand Company.
• Lawvere, F.W., and R. Rosebrugh (2003) Sets for Mathematics, Cambridge Univ. Press.
• Lucas, J. R. (1999) Conceptual Roots of Mathematics. Routledge.
• Maddux, R.D. (2006) Relation Algebras, vol. 150 in 'Studies in Logic and the Foundations of Mathematics'. Elsevier Science.
• Merrill, Dan D. (1990) Augustus De Morgan and the logic of relations. Kluwer.
• Peirce, C.S. (1984) Writings of Charles S. Peirce: A Chronological Edition, Volume 2, 1867-1871. Peirce Edition Project, eds. Indiana University Press.
• Russell, Bertrand (1903/1938) The Principles of Mathematics, 2nd ed. Cambridge Univ. Press.
• Suppes, Patrick (1960/1972) Axiomatic Set Theory. Dover Publications.
• Tarski, A. (1956/1983) Logic, Semantics, Metamathematics, Papers from 1923 to 1938, J.H. Woodger, trans. 1st edition, Oxford University Press. 2nd edition, J. Corcoran, ed. Indianapolis IN: Hackett Publishing.
• Ulam, S.M. (1990) Analogies Between Analogies: The Mathematical Reports of S.M. Ulam and His Los Alamos Collaborators in A.R. Bednarek and Françoise Ulam, eds., University of California Press. | {"extraction_info": {"found_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": 29, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9237171411514282, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/229173/limit-of-dirac-delta-and-test-functions | # limit of dirac delta and test functions
I am trying to understand dirac delta better. We know that $\int_{\mathbb{R}^n} \delta(x) \phi(x)=\phi(0)$ where $\phi(x)$ is a test function and $\delta_{0}(x)$ represents dirac distribution. I have a function $\lim \limits_{l\rightarrow 0 }f(l)=c(constant)$. So, I think $\lim \limits_{l\rightarrow 0 }\int_{\mathbb{R}^n} \delta(x) f(l) \phi(x)=c\phi(0)$. What does happen, if I have such a function $\lim \limits_{l\rightarrow 0 }f_{l}(x)=\infty$, i.e like $f(x)=\frac{x}{l^2}$
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What is $\delta(\cdot)$? – Davide Giraudo Nov 4 '12 at 21:11
@thanks for your remark, I edited my question. – pcepkin Nov 4 '12 at 21:23
I asked the question because it's known that Dirac distribution cannot be represented by a locally integrable function. – Davide Giraudo Nov 4 '12 at 21:25
@Davide Giraudo then I should have defined as a measure and rewrited $\int_{\mathbb {R}}\phi(x)\delta(dx)=\phi(0)$ or didn't I get again? – pcepkin Nov 4 '12 at 21:40
Yes. I don't understand what you do next: what is $f_l(x)$ (why do you allow $f$ to depend on a parameter)? – Davide Giraudo Nov 4 '12 at 21:41
show 8 more comments
## 1 Answer
I think your notation is at the very least misleading: $\delta (x)$ makes little sense, since it's a distribution, i.e. a functional on $\mathcal{D} ( \mathbb{R}^n)$. Although it's indeed quite common notation, handy for some manipulations, I'd rather write $\delta ( \phi) = 0$ or $\langle \delta_0, \phi \rangle_{\mathcal{D}' ( \mathbb{R}^n)} = \phi ( 0)$ if you like verbosity. Now, if your function $f_l$ is in $\mathcal{D} ( \mathbb{R}^n) = C^{\infty}_0 ( \mathbb{R}^n)$, you can define for any distribution $T$ the product $f_l T$ via $\langle f_l T, \phi \rangle := \langle T, f_l \phi \rangle$ and it obviously follows $( f_l \delta) ( \phi) = ( f_l \phi) ( 0)$. If $f_l \phi$ is not in $C^{\infty}_0$ then it cannot be an argument to $\delta \in \mathcal{D}'$ (although you can extend the domain of definition of $\delta$, of course).
In your example, and for some fixed test function, you can consider the delta as the limit of bumps it is, and apply each of these to the product $f_l \phi$ then see what happens in the limit, which may not even exist depending on $f_l$ and $\phi$.
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thank you answering my stupid question, I finally got it. – pcepkin Nov 4 '12 at 22:19
1
You are welcome. Also, I don't think not understanding something is stupid (I'd be totally depressed by now if I did...). I must cite here my favorite maths quotation: "Young man, in mathematics you don't understand things. You just get used to them." (Von Neumann) – Miguel Nov 4 '12 at 22:26
Also: shouldn't you accept the answer, to mark this question as resolved? Not that I'm reputation-hungry or anything... ;) – Miguel Nov 4 '12 at 22:31
sorry, I was excited to understand something:)I did it with some delay. – pcepkin Nov 4 '12 at 22:35
Actually, it's usually better to wait a while before accepting an answer. Question with an accepted answer often get less attention, and people may be discouraged to post alternative answers to such questions. – mrf Nov 4 '12 at 22:53
show 2 more comments | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 28, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9581876397132874, "perplexity_flag": "middle"} |
http://terrytao.wordpress.com/tag/topological-entropy/ | What’s new
Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao
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## The Chowla conjecture and the Sarnak conjecture
14 October, 2012 in expository, math.NT, math.PR | Tags: Chowla conjecture, Mobius function, Peter Sarnak, pseudorandomness, Sarnak conjecture, topological entropy | by Terence Tao | 12 comments
One of the basic general problems in analytic number theory is to understand as much as possible the fluctuations of the Möbius function ${\mu(n)}$, defined as ${(-1)^k}$ when ${n}$ is the product of ${k}$ distinct primes, and zero otherwise. For instance, as ${\mu}$ takes values in ${\{-1,0,1\}}$, we have the trivial bound
$\displaystyle |\sum_{n \leq x} \mu(n)| \leq x$
and the seemingly slight improvement
$\displaystyle \sum_{n \leq x} \mu(n) = o(x) \ \ \ \ \ (1)$
is already equivalent to the prime number theorem, as observed by Landau (see e.g. this previous blog post for a proof), while the much stronger (and still open) improvement
$\displaystyle \sum_{n \leq x} \mu(n) = O(x^{1/2+o(1)})$
is equivalent to the notorious Riemann hypothesis.
There is a general Möbius pseudorandomness heuristic that suggests that the sign pattern ${\mu}$ behaves so randomly (or pseudorandomly) that one should expect a substantial amount of cancellation in sums that involve the sign fluctuation of the Möbius function in a nontrivial fashion, with the amount of cancellation present comparable to the amount that an analogous random sum would provide; cf. the probabilistic heuristic discussed in this recent blog post. There are a number of ways to make this heuristic precise. As already mentioned, the Riemann hypothesis can be considered one such manifestation of the heuristic. Another manifestation is the following old conjecture of Chowla:
Conjecture 1 (Chowla’s conjecture) For any fixed integer ${m}$ and exponents ${a_1,a_2,\ldots,a_m \geq 0}$, with at least one of the ${a_i}$ odd (so as not to completely destroy the sign cancellation), we have
$\displaystyle \sum_{n \leq x} \mu(n+1)^{a_1} \ldots \mu(n+m)^{a_m} = o_{x \rightarrow \infty;m}(x).$
Note that as ${\mu^a = \mu^{a+2}}$ for any ${a \geq 1}$, we can reduce to the case when the ${a_i}$ take values in ${0,1,2}$ here. When only one of the ${a_i}$ are odd, this is essentially the prime number theorem in arithmetic progressions (after some elementary sieving), but with two or more of the ${a_i}$ are odd, the problem becomes completely open. For instance, the estimate
$\displaystyle \sum_{n \leq x} \mu(n) \mu(n+2) = o(x)$
is morally very close to the conjectured asymptotic
$\displaystyle \sum_{n \leq x} \Lambda(n) \Lambda(n+2) = 2\Pi_2 x + o(x)$
for the von Mangoldt function ${\Lambda}$, where ${\Pi_2 := \prod_{p > 2} (1 - \frac{1}{(p-1)^2}) = 0.66016\ldots}$ is the twin prime constant; this asymptotic in turn implies the twin prime conjecture. (To formally deduce estimates for von Mangoldt from estimates for Möbius, though, typically requires some better control on the error terms than ${o()}$, in particular gains of some power of ${\log x}$ are usually needed. See this previous blog post for more discussion.)
Remark 1 The Chowla conjecture resembles an assertion that, for ${n}$ chosen randomly and uniformly from ${1}$ to ${x}$, the random variables ${\mu(n+1),\ldots,\mu(n+k)}$ become asymptotically independent of each other (in the probabilistic sense) as ${x \rightarrow \infty}$. However, this is not quite accurate, because some moments (namely those with all exponents ${a_i}$ even) have the “wrong” asymptotic value, leading to some unwanted correlation between the two variables. For instance, the events ${\mu(n)=0}$ and ${\mu(n+4)=0}$ have a strong correlation with each other, basically because they are both strongly correlated with the event of ${n}$ being divisible by ${4}$. A more accurate interpretation of the Chowla conjecture is that the random variables ${\mu(n+1),\ldots,\mu(n+k)}$ are asymptotically conditionally independent of each other, after conditioning on the zero pattern ${\mu(n+1)^2,\ldots,\mu(n+k)^2}$; thus, it is the sign of the Möbius function that fluctuates like random noise, rather than the zero pattern. (The situation is a bit cleaner if one works instead with the Liouville function ${\lambda}$ instead of the Möbius function ${\mu}$, as this function never vanishes, but we will stick to the traditional Möbius function formalism here.)
A more recent formulation of the Möbius randomness heuristic is the following conjecture of Sarnak. Given a bounded sequence ${f: {\bf N} \rightarrow {\bf C}}$, define the topological entropy of the sequence to be the least exponent ${\sigma}$ with the property that for any fixed ${\epsilon > 0}$, and for ${m}$ going to infinity the set ${\{ (f(n+1),\ldots,f(n+m)): n \in {\bf N} \} \subset {\bf C}^m}$ of ${f}$ can be covered by ${O( \exp( \sigma m + o(m) ) )}$ balls of radius ${\epsilon}$. (If ${f}$ arises from a minimal topological dynamical system ${(X,T)}$ by ${f(n) := f(T^n x)}$, the above notion is equivalent to the usual notion of the topological entropy of a dynamical system.) For instance, if the sequence is a bit sequence (i.e. it takes values in ${\{0,1\}}$), then there are only ${\exp(\sigma m + o(m))}$ ${m}$-bit patterns that can appear as blocks of ${m}$ consecutive bits in this sequence. As a special case, a Turing machine with bounded memory that had access to a random number generator at the rate of one random bit produced every ${T}$ units of time, but otherwise evolved deterministically, would have an output sequence that had a topological entropy of at most ${\frac{1}{T} \log 2}$. A bounded sequence is said to be deterministic if its topological entropy is zero. A typical example is a polynomial sequence such as ${f(n) := e^{2\pi i \alpha n^2}}$ for some fixed ${\sigma}$; the ${m}$-blocks of such polynomials sequence have covering numbers that only grow polynomially in ${m}$, rather than exponentially, thus yielding the zero entropy. Unipotent flows, such as the horocycle flow on a compact hyperbolic surface, are another good source of deterministic sequences.
Conjecture 2 (Sarnak’s conjecture) Let ${f: {\bf N} \rightarrow {\bf C}}$ be a deterministic bounded sequence. Then
$\displaystyle \sum_{n \leq x} \mu(n) f(n) = o_{x \rightarrow \infty;f}(x).$
This conjecture in general is still quite far from being solved. However, special cases are known:
• For constant sequences, this is essentially the prime number theorem (1).
• For periodic sequences, this is essentially the prime number theorem in arithmetic progressions.
• For quasiperiodic sequences such as ${f(n) = F(\alpha n \hbox{ mod } 1)}$ for some continuous ${F}$, this follows from the work of Davenport.
• For nilsequences, this is a result of Ben Green and myself.
• For horocycle flows, this is a result of Bourgain, Sarnak, and Ziegler.
• For the Thue-Morse sequence, this is a result of Dartyge-Tenenbaum (with a stronger error term obtained by Maduit-Rivat). A subsequent result of Bourgain handles all bounded rank one sequences (though the Thue-Morse sequence is actually of rank two), and a related result of Green establishes asymptotic orthogonality of the Möbius function to bounded depth circuits, although such functions are not necessarily deterministic in nature.
• For the Rudin-Shapiro sequence, I sketched out an argument at this MathOverflow post.
• The Möbius function is known to itself be non-deterministic, because its square ${\mu^2(n)}$ (i.e. the indicator of the square-free functions) is known to be non-deterministic (indeed, its topological entropy is ${\frac{6}{\pi^2}\log 2}$). (The corresponding question for the Liouville function ${\lambda(n)}$, however, remains open, as the square ${\lambda^2(n)=1}$ has zero entropy.)
• In the converse direction, it is easy to construct sequences of arbitrarily small positive entropy that correlate with the Möbius function (a rather silly example is ${\mu(n) 1_{k|n}}$ for some fixed large (squarefree) ${k}$, which has topological entropy at most ${\log 2/k}$ but clearly correlates with ${\mu}$).
See this survey of Sarnak for further discussion of this and related topics.
In this post I wanted to give a very nice argument of Sarnak that links the above two conjectures:
Proposition 3 The Chowla conjecture implies the Sarnak conjecture.
The argument does not use any number-theoretic properties of the Möbius function; one could replace ${\mu}$ in both conjectures by any other function from the natural numbers to ${\{-1,0,+1\}}$ and obtain the same implication. The argument consists of the following ingredients:
1. To show that ${\sum_{n<x} \mu(n) f(n) = o(x)}$, it suffices to show that the expectation of the random variable ${\frac{1}{m} (\mu(n+1)f(n+1)+\ldots+\mu(n+m)f(n+m))}$, where ${n}$ is drawn uniformly at random from ${1}$ to ${x}$, can be made arbitrary small by making ${m}$ large (and ${n}$ even larger).
2. By the union bound and the zero topological entropy of ${f}$, it suffices to show that for any bounded deterministic coefficients ${c_1,\ldots,c_m}$, the random variable ${\frac{1}{m}(c_1 \mu(n+1) + \ldots + c_m \mu(n+m))}$ concentrates with exponentially high probability.
3. Finally, this exponentially high concentration can be achieved by the moment method, using a slight variant of the moment method proof of the large deviation estimates such as the Chernoff inequality or Hoeffding inequality (as discussed in this blog post).
As is often the case, though, while the “top-down” order of steps presented above is perhaps the clearest way to think conceptually about the argument, in order to present the argument formally it is more convenient to present the arguments in the reverse (or “bottom-up”) order. This is the approach taken below the fold.
Read the rest of this entry »
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http://mathhelpforum.com/discrete-math/130009-differences-sets.html | # Thread:
1. ## Differences of sets
If the difference of sets A and B is the set containing elements that are in A but not in B and B is an all-inclusive set then is A-B={null set} or just A-B=null set?
I was thinking that it would be A-B={null set} since the definition of the difference of sets is that it is, "the set containing..."
Thanks!
2. Originally Posted by buddyp450
If the difference of sets A and B is the set containing elements that are in A but not in B and B is an all-inclusive set then is A-B={null set} or just A-B=null set?
I was thinking that it would be A-B={null set} since the definition of the difference of sets is that it is, "the set containing..."
I, for one, have no idea what you point is. What are you asking?
Here is a fact: $A \subseteq B\quad \Leftrightarrow \quad A\backslash B = \emptyset$.
3. I'm sorry, let me try to be more clear as I'm new to discrete mathematics.
A = {1, 2, 3, 4, 5}
B = {1, 2, 3, 4, 5}
so does
A-B = $\emptyset$
or
A-B = { $\emptyset$}
and could you please explain why?
4. ## Notation
Originally Posted by buddyp450
I'm sorry, let me try to be more clear as I'm new to discrete mathematics.
A = {1, 2, 3, 4, 5}
B = {1, 2, 3, 4, 5}
so does
A-B = $\emptyset$
or
A-B = { $\emptyset$}
and could you please explain why?
$\emptyset$ is equivalent to {}. This is standard notation. E.g., { $\emptyset$} is the set containing the empty set. So, A-B is $\emptyset$ or {}, not { $\emptyset$}.
5. Originally Posted by Plato
I, for one, have no idea what you point is. What are you asking?
Here is a fact: $A \subseteq B\quad \Leftrightarrow \quad B\backslash A = \emptyset$.
Is that a typo, or am I misunderstanding the notation? $\{1\}\subset\{1,2\}$ but $\{1,2\}-\{1\}=\{2\}$
6. I think that by "B is an all inclusive set" you mean that B is the universal set. In that case, everything is in B so "are in A but not in B" does not apply to anything since there is nothing that is "not in B". A- B is the empty set. | {"extraction_info": {"found_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": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9636600017547607, "perplexity_flag": "middle"} |
http://nrich.maths.org/895/solution | ### All in the Mind
Imagine you are suspending a cube from one vertex (corner) and allowing it to hang freely. Now imagine you are lowering it into water until it is exactly half submerged. What shape does the surface of the water make around the cube?
### Qqq..cubed
It is known that the area of the largest equilateral triangular section of a cube is 140sq cm. What is the side length of the cube? The distances between the centres of two adjacent faces of another cube is 8cms. What is the side length of this cube? Another cube has an edge length of 12cm. At each vertex a tetrahedron with three mutually perpendicular edges of length 4cm is sliced away. What is the surface area and volume of the remaining solid?
### Painting Cubes
Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours?
# Marbles in a Box
##### Stage: 3 and 4 Challenge Level:
Joel and Sarah from Sawston Village College found an alternative method for working out the number of different winning lines in a $3 \times 3 \times 3$ cube. Here is the description of their method. Congratulations! Finding all the lines once and none of them twice is a real challenge.
The second part of the problem asked if it was possible to adapt the methods described for the $3 \times 3 \times 3$ cube to give a general formula for any size cube.
Jamal from King Edward VII School, Sheffield managed to adapt Caroline's method to a
$4 \times 4 \times 4$ cube.
Remember that on a $3 \times 3 \times 3$ cube all winning lines must pass either:
• along an edge of the cube
• through the middle of a face
• through the centre of the cube
but these must be changed to make it work for a $4 \times 4 \times 4$ cube.
There are still 12 edges on a $4 \times 4 \times 4$ cube so there are 12 winning lines along edges.
Since on the larger cube there is no 'middle' of a face, I counted all the lines on each face that aren't along the edge instead, each one passes through the 'middle' four pieces of a face. There are 2 diagonals, 2 horizontal and 2 vertical lines on each of the 6 faces, so that's 36 more lines.
Again the $4 \times 4 \times 4$ doesn't have a 'centre' piece so I counted all the lines through the central $2 \times 2 \times 2$ block. I found there is exactly one winning line through any two of those eight pieces in the block. There are (8x7)/2 =28 pairs because I can choose from eight pieces for the first one, then from the remaining seven for the second, but I divide by two because it doesn't matter in what order I choose at the pair.
Adding these up I have 12 + 36 + 28 = 76 winning lines.
Niharika used similar reasoning to find the number of winning lines in an $n \times n \times n$ cube.
Rachael from Lancaster Girls' Grammar School used James' method to find the same result, and then generalised the answer to $n \times n \times n$ cubes.
Considering the non-diagonal winning lines first:
There are 16 from front to back.
There are 16 from left to right.
There are 16 from top to bottom.
Considering the diagonal winning lines:
On each layer there are 2 diagonal winning lines so:
There are 8 from front to back.
There are 8 from top to bottom.
There are 8 from left to right.
There are 4 lines from a vertex to a diagonally opposite vertex.
In total, there are 48+24+4 = 76 winning lines.
If I now take a cube of size nxnxn, there are n2 non-diagonal lines from front to back, and left to right, and top to bottom. That makes 3n2 lines. Counting the diagonal ones, there are 2 in each of the n layers, in each of the 3 dimensions; so 6n lines. Then there's the four lines from a vertex to a diagonally opposite vertex. In total there are 3n2 + 6n +4 lines.
Llewellyn from Cowbridge Comprehensive School saw that Alison's method generalises easily, jumping straight to the nxnxn formula.
A cube has 8 vertices, 12 edges and 6 faces. Correspondingly there are three kinds of pieces: vertex, edge, and face. An $n \times n \times n$ cube has 8 vertex peices, 12(n-2) edge pieces and 6(n-2)2 face pieces.
From a vertex piece there are 7 other vertex pieces that you can join to in order to make a winning line. 7×8 = 56 lines, but this counts each line from both ends, so there are 28 vertex winning lines.
From an edge piece there are 3 other edge pieces that you can join to in order to make a winning line. Remembering that this counts each line from both ends there are 3×12(n-2)/2 = 18n - 36 edge winning lines.
From a face piece there is one winning line, joining to the opposite face, so there are 6(n-2)2/2 = 3n2 - 12n + 12 face winning lines.
So in total, there are 28 + 18n - 36 + 3n2 - 12n + 12 = 3n2 + 6n + 4 winning lines. That means there are 76 winning lines in a 4x4x4 cube and 364 winning lines in a 10x10x10 cube.
Leah from St Paul's Girls' School saw how to adapt Grae's method to a solution for 4x4x4 cubes.
The winning lines may be counted by looking at lines:
• in each horizontal plane
• in each vertical plane from left to right
• in each vertical plane from front to back
• in the diagonal planes
On a non-diagonal plane there are 10 winning lines (4 horizontal, 4 vertical and 2 diagonal lines).
In the cube, there are 4 horizontal planes, so 10×4=40 winning lines.
There are also 4 vertical planes going from left to right, but now with only 6 new winning lines per plane, as the 4 horizontal lines have already been counted. So 6×4=24 winning lines.
On the 4 vertical planes going from front to back, we now only have 2 new (diagonal) winning lines per plane. So 2×4 = 8 winning lines.
Finally, there are also diagonal planes to consider. There are 4 winning lines going from corner to diagonally opposite corner.
In total, there are 40 + 24 + 8 + 4 = 76 winning lines.
It is possible to adapt Grae's and Caroline's methods to the general formula and you might like to try using Leah's or Jamal's work to help you get started. Of course it's easier now you know what formula you're looking for!
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9420955181121826, "perplexity_flag": "middle"} |
http://mathhelpforum.com/differential-geometry/183537-absolute-value-no-longer-sqrt-x-2-complex-numbers.html | # Thread:
1. ## Absolute value no longer sqrt(x^2) for complex numbers
Why is |a+bi| defined as: $\sqrt{a^2+b^2$
Why is it not defined as $\sqrt{(a+bi)^2$
This would yield $\sqrt{a^2-b^2+2abi$
I understand that it comes from applying pythagorean's theorem to the complex plane but since proofs of pythagorean's theorem obviously involve only real numbers I guess it's just a convenient definition so that other results come out the way we want? Is that the idea of even defining the imaginary plane to begin with?
2. ## Re: Absolute value no longer sqrt(x^2) for complex numbers
The point of an absolute value is that |x| is the distance from x to 0. Of course, a "distance" must be a non-negative real number. For a complex number, we can represent the number x+ iy by the point (x,y) in the "complex plane". The distance form (x, y) to (0, 0), by the Pythagorean theorem: $\sqrt{x^2+ y^2}$.
That is not $\sqrt{x^2}$ but it is $\sqrt{z\overline{z}}$ where $\overline{z}$ is the "complex conjugate" of z: the complex conjugate of z= x+ iy is $\overline{z}= x- iy$ which, in the case that z is real, z= x+ 0i, reduces to x so that $\sqrt{z\overline{z}}= \sqrt{x^2}$.
3. ## Re: Absolute value no longer sqrt(x^2) for complex numbers
Well it's our choice which interpretation of absolute value we want to stay with us when we expand to the complex numbers. Why do we choose that particular interpretation?
4. ## Re: Absolute value no longer sqrt(x^2) for complex numbers
Originally Posted by lamp23
Well it's our choice which interpretation of absolute value we want to stay with us when we expand to the complex numbers. Why do we choose that particular interpretation?
No, sorry in this case it is not your choice.
In mathematics absolute value is a metric (i.e. a distance).
You may chose to redefine a distance function but it must conform with the axioms of a metric. If it does not then it is a new definition and therefore needs a new name.
5. ## Re: Absolute value no longer sqrt(x^2) for complex numbers
Originally Posted by lamp23
Well it's our choice which interpretation of absolute value we want to stay with us when we expand to the complex numbers. Why do we choose that particular interpretation?
You're right in that we're certainly free to define modulus anyway we like. The problem is, you're definition has no use!
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http://math.stackexchange.com/questions/145862/visualizing-solids-of-revolution | # Visualizing solids of revolution
Can someone shed a little bit of light on the problem of volumes of revolution about the $x$-axis and the $y$-axis of the same shapes? Take for example $f(x)=x^2$. If we want to find the volume bounded by the parabola and the $x$-axis with axis of revolution at $y=0$ we would use the standard method of disks and get the volume $\pi/5$ for $x=[0,1]$. Now if we want to find the volume by rotating the curve around the $y$-axis instead, using the cylindrical shells method, we get the volume to be $\pi/2$.
Now, maybe it's just my flawed intuition, but since we are basically rotating the same shape/area by 360$^\circ$ (but in different "directions"), I would guess the two volumes to be the same. Anyone knows some good graphics/animation/resources to help me visualize this problem?
-
Note also that for rotating about y, you could delete the whole area to the left of the axis and not change the volume. Rotating about x you cannot. – Ross Millikan May 16 '12 at 15:36
@RossMillikan the area to the left that you speak of, is as if it was "deleted" when integrating for $x=[0,1]$, isn't it? Or am I missing your point? – Milosz Wielondek May 16 '12 at 15:45
my point was that it is another way of seeing the disconnect between area and rotated volume. You could have it or not when rotating about y and get the same volume, but the area changes by a factor 2. – Ross Millikan May 16 '12 at 16:00
## 1 Answer
When you rotate about $y$, most of the area is farther from the axis than if you rotate around $x$. If you think about a small area of size $dxdy$ when rotated around $x$ it sweeps out a volume element $ydxdy$ and when rotated around $y$ is sweeps out a volume $xdxdy$. The methods of discs and shells just do one dimension of the integral for you.
You could think of a rectangle $[0,10] \times [0,0.1]$. If you rotate it around $x$ you get a cylinder of radius $0.1$ and height $10$, for volume $0.1\pi$. If you rotate it around $y$ the radius is $10$ and the height is $0.1$ for a volume of $10\pi$
-
Naturally, it does make perfect sense mathematically. But am I the only one to think it's somewhat counter-intuitive? – Milosz Wielondek May 16 '12 at 16:25 | {"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": 27, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.938591718673706, "perplexity_flag": "head"} |
http://mathhelpforum.com/advanced-algebra/156553-beginning-abstract-algebra-problem.html | Thread:
1. Beginning Abstract Algebra Problem
Prove this is true for every group G, or give a counter example to show this is false in some groups:
If x^2 = a^2, then x=a
2. Originally Posted by jzellt
Prove this is true for every group G, or give a counter example to show this is false in some groups:
If x^2 = a^2, then x=a
Obviously, you need a counterexample to $x^2=a^2\Rightarrow x=a$. - What about $\mathbb{R}\backslash\{0\}$ with multiplication?
3. Another, simpler, example is the set {1, -1} with multiplication as operation. Yet, another, the Klein 4-group. In fact, any finite group whose order is even! | {"extraction_info": {"found_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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8904101252555847, "perplexity_flag": "middle"} |
http://nrich.maths.org/6677&part= | Roots and Coefficients
If xyz = 1 and x+y+z =1/x + 1/y + 1/z show that at least one of these numbers must be 1. Now for the complexity! When are the other numbers real and when are they complex?
Target Six
Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.
8 Methods for Three by One
This problem in geometry has been solved in no less than EIGHT ways by a pair of students. How would you solve it? How many of their solutions can you follow? How are they the same or different? Which do you like best?
Impedance Can Be Complex!
Stage: 5 Challenge Level:
An AC generator of rms voltage E is connected in series with two reactive impedances,
$Z_1$ and $Z_2$, as shown below:
$\theta$ is the argument of $Z_1$ and $\phi$ is the argument of $Z_2$.
Find the power dissipated in $Z_2$
If the phase of $Z_2$ ($\phi$) is fixed whilst allowing the magnitude to vary, find the relationship between $Z_1$ and $Z_2$ when we achieve maximum power transfer to $Z_2$.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "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}, "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}, "english_confidence": 0.9448577165603638, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/208163/probability-of-winning-a-game-from-raw-data?answertab=votes | # Probability of winning a game from raw data.
I have a dataset made of couples $(n_i,v_i)$ where $n_i$ denotes the number of times a game has been played, and $v_i$ the number of victories at the $i-$th day.
What is the best way to evaluate the probability $P$ of winning the game? (We can assume that winning the game does not depend on time).
My first thought is to evaluate $P$ as the total number of victories over the number of games, i.e. $$P = \frac{\sum_{i=1}^t v_i}{\sum_{i=1}^t n_i}$$
I then had the doubt though that I could evaluate $p_i = v_i / n_i$ and define $P$ as the mean of the individual probabilities: $$P = \frac{1}{t} \sum_{i=1}^t p_i$$.
Somehow I feel this second approach is wrong, but can't entirely understand why.
How would you evaluate $P$ and why? Can you give me some links explaining how to evaluate reliable statistics?
-
## 1 Answer
The answer may depend on whart you know beforehand about that probability $p$. What we can says, is: If the probability is $p$ then the probability of observing $v=\sum v_i$ victories within $n=\sum n_i$ games is $${n\choose v}p^v(1-p^{n-v})$$ and this expression is maximal when $p=\frac v n$. Thus if we have no a priori knowledga about $p$ (i.e. consider each value $\in[0,1]$ equally likely), then $p=\frac vn$ is the best guess.
You can also do this on a day.by.day basis, but then your adjustment after day $i$ must take into account that you do have some a priori knowledge. You need to apply Bayes theory and this will in the end lead to exactly the same result.
-
Thanks Hagen, but I can't really follow your reasoning (it is a long time since I applied probability last time). Can you please send me some reference? Why would maximizing p be the best guess? Anyway yes, I assume that p is uniform in [0,1]. I think this is quite standard job for statisticians, and I am looking at the standard way to do this. Also I don't understand how a priori knowledge would influence the result, since $p_i$ doesn't depend on previous times... – Luca Cerone Oct 6 '12 at 12:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9634770750999451, "perplexity_flag": "head"} |
http://mathhelpforum.com/pre-calculus/9762-even-odd-functions.html | # Thread:
1. ## even odd functions
1. For each of the following functions, decide whether it is even, odd, or neither. Enter E for an EVEN function, O for an ODD function and N for a function which is NEITHER even nor odd.
A. f(x)= x^3+x^5+x^7
B. f(x)= (-5x^2)-(3x^4)-2
C. f(x)= (x^2)+(3x^4)-2
D. f(x)= x^-2
2. Originally Posted by qbkr21
1. For each of the following functions, decide whether it is even, odd, or neither. Enter E for an EVEN function, O for an ODD function and N for a function which is NEITHER even nor odd.
A. f(x)= x^3+x^5+x^7
B. f(x)= (-5x^2)-(3x^4)-2
C. f(x)= (x^2)+(3x^4)-2
D. f(x)= x^-2
The definition of an even function is that f(-x) = f(x), and an odd function has the property f(-x) = -f(x).
So for A: $f(-x) = (-x)^3 + (-x)^5 + (-x)^7 = -x^3 - x^5 - x^7 = -f(x)$. So A is an odd function.
B: $f(-x) = -5(-x)^2 - 3(-x)^4 - 2 = -5x^2 - 3x^4 -2 = f(x)$ so this function is even.
You should be able to do C and D by yourself. (They are both even.)
-Dan
3. Sir I think that one of these is incorrect could you please recheck the problem. Thanks
4. Originally Posted by qbkr21
Sir I think that one of these is incorrect could you please recheck the problem. Thanks
(Shrugs) Which one do you think is wrong?
Originally Posted by topsquark
The definition of an even function is that f(-x) = f(x), and an odd function has the property f(-x) = -f(x).
So for A: $f(-x) = (-x)^3 + (-x)^5 + (-x)^7 = -x^3 - x^5 - x^7 = -f(x)$. So A is an odd function.
B: $f(-x) = -5(-x)^2 - 3(-x)^4 - 2 = -5x^2 - 3x^4 -2 = f(x)$ so this function is even.
You should be able to do C and D by yourself. (They are both even.)
-Dan
C: $f(-x) = (-x)^2 + 3(-x)^4 -2 = x^2 + 3x^4 - 2 = f(x)$ so this is even.
D: $f(-x) = \frac{1}{(-x)^2} = \frac{1}{x^2} = f(x)$ so this is even.
I note that B and C are fairly similar. Is there possibly a typo?
-Dan
5. My Fault it, here I will try typing it in through Latex...
A.
$f(x)=x^3+x^5+x^7$
B.
$f(x)=-5x^2-3x^4-2$
C.
$f(x)=x^2+3x^4+2x^7$
D.
f(x)=x^-2
Sorry Again, hope this makes a bit easier.
6. Originally Posted by qbkr21
My Fault it, here I will try typing it in through Latex...
A.
$f(x)=x^3+x^5+x^7$
B.
$f(x)=-5x^2-3x^4-2$
C.
$x^2+3x^4+2x^7$
D.
X^-2
Sorry Again, hope this makes a bit easier.
Okay, so the problem is with C. Let's try this again:
$f(-x) = (-x)^2 + 3(-x)^4 + 2(-x)^7 = x^2 + 3x^4 - 2x^7 \neq f(x), -f(x)$
Since f(-x) is equal to neither f(x) nor -f(x), this function is a "neither." (Or in more advanced language, is not a "parity eigenstate.")
-Dan
7. Thanks Dan I appreciate the help. Things are worked out now
8. Originally Posted by qbkr21
Thanks Dan I appreciate the help. Things are worked out now
Always pleased to be of service!
-Dan | {"extraction_info": {"found_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": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9142468571662903, "perplexity_flag": "middle"} |
http://www.nag.com/numeric/CL/nagdoc_cl23/html/G08/g08intro.html | # NAG Library Chapter Introductiong08 – Nonparametric Statistics
## 1 Scope of the Chapter
The functions in this chapter perform nonparametric statistical tests which are based on distribution-free methods of analysis. For convenience, the chapter contents are divided into five types of test: tests of location, tests of dispersion, tests of distribution, tests of association and correlation, and tests of randomness. There are also functions to fit linear regression models using the ranks of the observations.
The emphasis in this chapter is on testing; if you wish to compute nonparametric correlations you are referred to Chapter g02, which contains several functions for that purpose.
There are a large number of nonparametric tests available. A selection of some of the more commonly used tests are included in this chapter.
## 2 Background to the Problems
### 2.1 Parametric and Nonparametric Hypothesis Testing
Classical techniques of statistical inference often make numerous or stringent assumptions about the nature of the population or populations from which the observations have been drawn. For instance, a testing procedure might assume that the set of data was obtained from Normally distributed populations. It might be further assumed that the populations involved have equal variances, or that there is a known relationship between the variances. In the Normal case, the test statistic derived would usually be a function of the sample means and variances, since a Normal distribution is completely characterised by its mean and variance. Alternatively, it might be assumed that the set of data was obtained from other distributions of known form, such as the gamma or the exponential. Again, a testing procedure would be devised based upon the arguments characterising such a distribution.
The type of hypothesis testing just described is usually termed parametric inference. Distributional assumptions are made which imply that the arguments of the chosen distribution, as estimated from the data, are sufficient to characterise the difference in distribution between the populations.
However, problems arise with parametric methods of inference when these assumptions cannot be made, either because they are contrary to the known nature of the mechanism generating a population, or because the data obviously do not satisfy the assumptions. Some parametric procedures become unreliable under relatively minor departures from the hypothesised distributional form. In the Normal case for example, tests on variances are extremely sensitive to departures from Normality in the underlying distribution.
There are also common situations, particularly in the behavioural sciences, where much more basic assumptions than that of Normality cannot be made. Data values are not always measured on continuous or even numerical scales. They may be simply categorical in nature, relating to such quantities as voting intentions or food preferences.
Techniques of inference are therefore required which do not involve making detailed assumptions about the underlying mechanism generating the observations. The functions in this chapter perform such distribution-free tests, evaluating from a set of data the value of a test statistic, together with an estimate of its significance.
For a comparison of some distribution-based and distribution-free tests, the interested reader is referred to Chapter 31 of Kendall and Stuart (1973). For a briefer and less mathematical account, see Conover (1980) or Siegel (1956).
### 2.2 Types of Nonparametric Test
This introduction is concerned with explaining the basic concepts of hypothesis testing, and some familiarity with the subject is assumed. Chapter 22 of Kendall and Stuart (1973) contains a detailed account, and the outline given in Conover (1980) or Siegel (1956) should be sufficient to understand this section.
Nonparametric tests may be grouped into three categories:
1. Tests of location
3. Tests of randomness
Tests can also be categorised by the design that they can be applied to:
1. One sample
2. Two related (paired) samples
3. Two independent samples
4. $k\left(>2\right)$ related (matched) samples
5. $k\left(>2\right)$ independent samples
A third classification of a test relates to the type of data to which it may be applied. Variables are recorded on four scales of measurement: nominal (categorical), ordinal, interval, and ratio.
The nominal scale is used only to categorise data; for each category a name, perhaps numeric, is assigned so that two different categories will be identified by distinct names. The ordinal scale, as well as categorising the observations, orders the categories. Each is assigned a distinct identifying symbol, in such a way that the order of the symbols corresponds to the order of the categories. (The most common system for ordinal variables is to assign numerical identifiers to the categories, though if they have previously been assigned alphabetic characters, these may be transformed to a numerical system by any convenient method which preserves the ordering of the categories.) The interval scale not only categorises and orders the observations, but also quantifies the comparison between categories; this necessitates a common unit of measurement and an arbitrary zero-point. Finally, the ratio scale is similar to the interval scale, except that it has an absolute (as opposed to arbitrary) zero-point.
It is apparent that there are many possible combinations of these three characteristics of a problem, and many nonparametric tests have been derived to meet the different experimental situations. However, it is not usually a difficult matter to choose an appropriate test given the nature of the data and the type of test which one wishes to perform.
### 2.3 Principles of Nonparametric Tests
In this section, each type of test is considered in turn, and remarks are made on the design principles on which each is based.
#### 2.3.1 Location tests
These tests are primarily concerned with inferences about differences in the location of the population distributions. In some cases, however, the tests are only concerned with inferences about the population distributions unless added assumptions are made which allow the hypotheses to be stated in terms of the location arguments.
For most of these tests, data must be measured numerically on at least an ordinal scale, in order that a measure of location may be devised. Ordinal measurement implies that pairs of values may be compared and numerically ordered. A vector of $n$ values may therefore be ranked from smallest to largest using the ordering operation. The resultant ranks contain all the information in the original data, but have the advantage that tests may be derived easily based on them, and no testing bias is introduced by the use of ordinal values as though they were measured on an interval scale. Note that the requirement of the measurement scale being ordinal does not imply that all tests of this type involve the actual ranking of the original data.
For the one-sample or matched pairs case, test statistics may be derived based on the number of observations (or differences) lying either side of zero (or some other fixed value), as in the sign test, for example. Under the hypothesis that the median of the single population is zero or the difference in the medians of the paired populations is zero the number of positive and negative values should be similar. The Wilcoxon signed rank test goes further than the sign test by taking into account the magnitude of the single sample values or of the differences.
For the two-sample case, if median equality is hypothesised, the distribution of the ranks of each sample in the total pooled sample should be similar. Test statistics, such as the Mann–Whitney U statistic, which are based on the ranks of each sample and summarize the differences in rank sums for each sample, may be computed. These statistics are referred to their expected distributions under the null hypothesis. The above hypothesis can also be tested using the median test. Its test statistic is based on the number of values in each sample which are greater than or less than the pooled median of the two samples, rather than the ranks of each sample.
If median equality is hypothesised for several samples, the distribution ranks of the members of each sample in the total pooled sample should be ‘homogeneous’. Test statistics can be derived which summarize the differences in rank sums for the various samples, and again referred to their expected distributions under the null hypothesis.
#### 2.3.2 Tests of fit
In the one-sample case, these are tests which investigate whether or not a sample of observations can be considered to follow a specified distribution. In the two-sample case, a test of fit investigates whether the two samples can be considered to have arisen from a common probability distribution.
For the one-sample problem, the null hypothesis may specify only the distributional form, for example $\mathrm{Normal}\left(\mu ,{\sigma }^{2}\right)$, or it may incorporate actual argument values, for example Poisson with mean $10$.
Some tests of this type proceed by forming the sample cumulative distribution function of the observations and computing a statistic whose value measures the departure of the sample cumulative distribution function from that of the null distribution. In the two-sample case, a statistic is computed which provides a measure of the difference between the sample cumulative distribution function of each sample. These tests are known as one- or two-sample Kolmogorov–Smirnov tests.
The significance for these test statistics can be computed directly for moderate sample sizes but for larger sample sizes asymptotic results are often used.
Another goodness-of-fit test is the ${\chi }^{2}$ test. For this test, the data is first grouped into intervals and then the difference between the observed number of observations in each interval and the number expected, if the null hypothesis is true, is computed. A statistic based on these differences has asymptotically a ${\chi }^{2}$-distribution.
#### 2.3.3 Tests of randomness
These tests are designed to investigate sequences of observations and attempt to identify any deviations from randomness. There are clearly many ways in which a sequence may deviate from randomness. The tests provided here primarily detect some form of dependency between the observations in the sequence.
The most common application of this type of tests is in the area of random number generation. The tests are used as empirical tests on a sample of output from a generator to establish local randomness. Theoretical tests are necessary and useful for testing global randomness. Some of the more common empirical tests are discussed below.
A runs-up or runs-down test investigates whether runs of different lengths are occurring with greater or lesser frequency than would be expected under the null hypothesis of randomness. A run up is defined as a sequence of observations in which each observation is larger than the previous observation. The run up ends when an observation is smaller than the previous observation. A test statistic, modified to take into account the dependency between successive run lengths, is computed. The test statistic has an asymptotic ${\chi }^{2}$-distribution.
The pairs test investigates the condition that, under the null hypothesis of randomness, the non-overlapping $2$-tuples (pairs) of a sequence of observations from the interval $\left[0,1\right]$ should be uniformly distributed over the unit square (${\left[0,1\right]}^{2}$). The triplets test follows the same idea but considers $3$-tuples and checks for uniformity over the unit cube (${\left[0,1\right]}^{3}$). In each test, a test statistic, based on differences between the observed and expected distribution of the 2- or $3$-tuples, is computed which has an asymptotic ${\chi }^{2}$-distribution.
The gaps test considers the ‘gaps’ between successive occurrences of observations in the sequence lying in a specified range. Under the null hypothesis of randomness, the gap length should follow a geometric distribution with an argument based on the length of the specified range, relative to the overall length of the interval containing all possible observations. The expected number of ‘gaps’, of a certain length, under the null hypothesis may thus be computed together with a test statistic based on the differences between the observed and expected numbers of ‘gaps’ of different length. Again the test statistic has an asymptotic ${\chi }^{2}$-distribution.
Other empirical tests such as the ${\chi }^{2}$ goodness-of-fit test and the one-sample Kolmogorov–Smirnov test may be used to investigate a sequence for non-uniformity.
### 2.4 Regression using ranks
If you wish to fit a regression model but is unsure about what transformation to take for the observed response to obtain a linear model, then one strategy is to replace response observations by their ranks. Estimates for regression arguments can be found by maximizing a likelihood function based on the ranks and the proposed regression model. The present functions give approximate estimates which are adequate when the signal-to-noise ratio is small, which is often the case with data from the medical and social sciences. Approximate standard errors of estimated regression coefficients are found. Also ${\chi }^{2}$ statistics can be used to test the null hypothesis of no regression effect.
## 3 Recommendations on Choice and Use of Available Functions
### 3.1 Location Tests
#### 3.1.1 One-sample or matched-pairs case
Note that a random sample of matched pairs, (${x}_{i},{y}_{i}$), may be reduced to a single sample by considering the differences, ${d}_{i}={x}_{i}-{y}_{i}$ say, of each pair. The matched pair may be thought of as a single observation on a bivariate random variable.
nag_sign_test (g08aac) performs the sign test on two paired samples. Each pair is classified as a $+$ or $-$ depending on the sign of the difference between the two data values within the pair. Under the assumptions that the ${d}_{i}$ are mutually independent and that the observations are measured on at least an ordinal scale, the sign test tests the hypothesis that for any pair sampled from the population distribution, $\text{Probability}\left(+\right)=\text{Probability}\left(-\right)$. The hypothesis may be stated in terms of the equality of the location arguments but the test is no longer regarded as unbiased and consistent unless further assumptions are made. If you wish to test the hypothesis that the location arguments differ by a fixed amount then that amount must be added or subtracted from one of the samples as required before calling nag_sign_test (g08aac).
nag_wilcoxon_test (g08agc) performs the one-sample Wilcoxon signed-rank test. The test may be used to test if the median of the population from which the random sample was taken is equal to some specified value (commonly used to test if the median is zero). In this test not only is the sign of the difference between the data values and the hypothesised median value important but also the magnitude of this difference. Thus, where the magnitude of the differences (or the data values themselves if the hypothesised median value is zero) is important this test is preferred to the sign test because it is more powerful. The test may easily be used to test whether the medians of two related populations are equal by taking the differences between the paired sample values and then testing the hypothesis that the median of the differences is zero, using the single sample of differences. The significance of the test statistic may be computed exactly for a moderate sample size but for a larger sample a Normal approximation is used. The exact method allows for ties in the differences.
#### 3.1.2 Two independent samples
nag_median_test (g08acc) performs the median test and nag_mann_whitney (g08amc) performs the Mann–Whitney $U$ test.
For both tests the two samples are assumed to be random samples from their respective populations and mutually independent. The measurement scale must be at least ordinal.
Note that, although the median test may be generalized to more than two samples, nag_median_test (g08acc) only deals with the two-sample case. For the median test, each observation is classified as being above or below the pooled median of the two samples. It may be used to test the hypothesis that the two population medians are equal; under the assumption that if the two population medians are equal then the probability of an observation exceeding the pooled median is the same for both populations.
The Mann–Whitney $U$ test involves the ranking of the pooled sample. The Mann–Whitney test thus attaches importance to the position of each observation relative to the others and not just its position relative to the median of the pooled sample as in the median test. Thus when the magnitude of the differences between the observations is meaningful the Mann–Whitney $U$ test is preferred as it is more powerful than the median test. The test tests whether the two population distributions are the same or not. If it is assumed that any difference between the two population distributions is a difference in the location then the test is testing whether the population means are the same or not.
In nag_mann_whitney (g08amc), the significance of the $U$ test statistic is computed either using a Normal approximation or the exact probability.
#### 3.1.3 More than two related samples
nag_friedman_test (g08aec) performs the Friedman two-way analysis of variance. This test may in some ways be regarded as an extension of the sign test to the case of $k$ ($k>2$) related samples. The data is in the form of a number of multivariate observations which are assumed to be mutually independent. This test also assumes that the measurement within each observation across the $k$ variates is at least ordinal so that the observation for each variate may be ranked according to some criteria.
#### 3.1.4 More than two independent samples
nag_kruskal_wallis_test (g08afc) performs the Kruskal–Wallis one-way analysis of variance. The test assumes that each sample is a random sample from its respective distributions and in addition that there is both independence within the samples and mutual independence among the various samples. The test requires that the measurement scale is at least ordinal so that the pooled sample may be ranked.
### 3.2 Tests of Fit
nag_1_sample_ks_test (g08cbc) performs the one-sample Kolmogorov–Smirnov distribution test. This test is used to test the null hypothesis that the random sample arises from a specified null distribution against one of three possible alternatives.
With nag_1_sample_ks_test (g08cbc) you may choose a null distribution from one of the following: the uniform, Normal, gamma, beta, binomial, exponential, and Poisson. The argument values may either be specified by you or estimated from the data by the function.
nag_2_sample_ks_test (g08cdc) performs the two-sample Kolmogorov–Smirnov test which tests the null hypothesis that the two samples may be considered to have arisen from the same population distribution against one of three possible alternative hypotheses, again corresponding to one-sided and two-sided tests. The distribution of the test statistic is computed using an exact method for moderate sample sizes, but for larger samples approximations based on asymptotic results are used.
Note that nag_prob_1_sample_ks (g01eyc) and nag_prob_2_sample_ks (g01ezc) are available for computing the distributions of the one-sample and two-sample Kolmogorov–Smirnov statistics respectively.
nag_chi_sq_goodness_of_fit_test (g08cgc) performs the ${\chi }^{2}$ goodness-of-fit test on a single sample which again tests the null hypothesis that the sample arises from a specified null distribution. You may choose a null distribution from one of the following: the Normal, uniform, exponential, ${\chi }^{2}$, and gamma; or may define the distribution by specifying the probability that an observation lies in a certain interval for a range of intervals covering the support of the null distribution. The significance of this test is computed using the ${\chi }^{2}$-distribution as an approximation to the distribution of the test statistic.
Several functions are available to perform Anderson–Darling goodness-of-fit tests:
• nag_anderson_darling_uniform_prob (g08cjc) for the case of uniformly distributed data;
• nag_anderson_darling_normal_prob (g08ckc) for a full-unspecified Normal distribution;
• nag_anderson_darling_exp_prob (g08clc) for an unspecified exponential distribution.
Note that data from a fully-specified distribution can be transformed to standard uniform by applying its cumulative distribution function. In all other cases use nag_anderson_darling_stat (g08chc) to calculate the Anderson–Darling statistic, ${A}^{2}$, and simulate its probability.
Tests of Normality may also be carried out using functions in Chapter g01.
### 3.3 Tests of Randomness
nag_runs_test (g08eac) performs the runs-up test on a sequence of observations. The runs-down test may be performed by multiplying each observation by $-1$ before calling the function. All runs whose length is greater than or equal to a certain chosen length will be treated as a single group.
nag_pairs_test (g08ebc) performs the pairs (serial) test on a sequence of observations from the interval $\left[0,1\right]$. The number of equal subintervals into which the interval $\left[0,1\right]$ is to be divided must be specified.
nag_triplets_test (g08ecc) performs the triplets test on a sequence of observations from the interval $\left[0,1\right]$. The number of equal subintervals into which the interval $\left[0,1\right]$ is to be divided must be specified.
nag_gaps_test (g08edc) performs the gaps test on a sequence of observations. The total length of the interval containing all possible values the observations could take must be specified together with the interval being used to define the ‘gaps’. All ‘gaps’ whose length is greater than or equal to a certain chosen length will be treated as a single group.
### 3.4 Regression Using Ranks
nag_rank_regsn (g08rac) fits a multiple linear regression model in which the observations on the response variable are replaced by their ranks.
nag_rank_regsn_censored (g08rbc) performs the same function but takes into account observations which may be right-censored.
### 3.5 Related Functions
Tests of location and distribution may be based on scores which are estimates of the expected values of the order statistics. nag_ranks_and_scores (g01dhc) may be used to compute Normal scores, an approximation to the Normal scores (Blom, Tukey or van der Waerden scores) or Savage (exponential) scores. For more accurate Normal scores nag_normal_scores_exact (g01dac) may be used. Other functions in this chapter may be used to test for Normality.
## 4 Functionality Index
Regression using ranks:
right-censored data nag_rank_regsn_censored (g08rbc)
uncensored data nag_rank_regsn (g08rac)
Tests of fit:
A2 and its probability of for a fully-unspecified Normal distribution nag_anderson_darling_normal_prob (g08ckc)
A2 and its probability of for an unspecified exponential distribution nag_anderson_darling_exp_prob (g08clc)
A2 and its probability of for uniformly distributed data nag_anderson_darling_uniform_prob (g08cjc)
Anderson–Darling test statistic A2 nag_anderson_darling_stat (g08chc)
Kolmogorov–Smirnov one-sample distribution test:
for standard distributions nag_1_sample_ks_test (g08cbc)
Kolmogorov–Smirnov two-sample distribution test nag_2_sample_ks_test (g08cdc)
Mann–Whitney two-sample test nag_mann_whitney (g08amc)
χ2 goodness-of-fit test nag_chi_sq_goodness_of_fit_test (g08cgc)
Tests of location:
Friedman two-way analysis of variance on k matched samples nag_friedman_test (g08aec)
Kruskal–Wallis one-way analysis of variance on k samples of unequal size nag_kruskal_wallis_test (g08afc)
Median test on two samples of unequal size nag_median_test (g08acc)
Sign test on two paired samples nag_sign_test (g08aac)
Wilcoxon one sample signed rank test nag_wilcoxon_test (g08agc)
Tests of randomness:
Gaps test nag_gaps_test (g08edc)
Pairs (serial) test nag_pairs_test (g08ebc)
Runs up or runs down test nag_runs_test (g08eac)
Triplets test nag_triplets_test (g08ecc)
None.
## 6 References
Conover W J (1980) Practical Nonparametric Statistics Wiley
Kendall M G and Stuart A (1973) The Advanced Theory of Statistics (Volume 2) (3rd Edition) Griffin
Siegel S (1956) Non-parametric Statistics for the Behavioral Sciences McGraw–Hill | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 40, "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}, "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}, "english_confidence": 0.8767683506011963, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/55114/how-to-make-a-nongrounded-conductor-have-equipotential | # How to make a nongrounded conductor have equipotential?
I'm studying the Method of Images and I seemed to have come to a conundrum. Method of Images takes advantage of grounded objects, (I am currently studying spheres), to set boundary conditions. However, how would one use the idea of MoI to set the potential of a conducting sphere as constant?
Since $E = \nabla V$, a constant potential would be the electric field inside would be constant? Therefore, a density of charge inside the sphere?
-
I edited your question, to fix $\nabla E = V$, which should be $E=\nabla V$. but perhaps that typo is the source of your equation. – askewchan Feb 26 at 2:04
1
Also note that conundrum is a noun and not a verb. – Emilio Pisanty Feb 26 at 16:20
## 1 Answer
Regardless of whether it is grounded, if there were a nonzero electric field inside a conductor, it would push the charge carriers around until there were no longer forces on them.
Thus: A perfect conductor, grounded or isolated, will have a surface (and volume) of equal potential, and the electric field inside will be zero.
If the conductor is not grounded, then there will be some net charge (possibly zero) on the conductor. If it's a symmetric shape (a sphere) then it will have a uniform surface density, namely the total charge over the total surface area ($Q / 4\pi r^2$).
You cannot assume to know $Q$ on the sphere. Whether or not $Q=0$, there will be $E=0$ inside the sphere (Gauss Law), and outside the sphere there will be $E\neq 0$ if $Q\neq0$.
Adding an image charge at the origin would actually create a nonzero $E$ inside the sphere, which you cannot have inside a conductor (I'm assuming the sphere is solid). The charge on the surface of the sphere already cancels its own $E$ field inside the sphere. If you want to cancel the $E$ field outside the sphere, then an image at the origin would do that.
-
thank you for your edit. It was a mistype when I was typing the question. I understand that a perfect conductor would have a surface of equal potential, but what if I want to make a non grounded sphere have equi potential inside? Method of Images tells me that if I introduce a charge to correct boundary conditions, it will give me that specific case. I can't seem to figure out where though.. – julesverne Feb 26 at 16:00
@julesverne see my edit – askewchan Feb 26 at 16:17
Can I interpret the conductor as "neutrally charged"? If there was a net charge in the sphere, there would be an electric field. Therefore, the second charge that must be placed to create an equipotential throughout the non grounded sphere, would be at the origin, and would have a charge of -Q? As that would cancel out the charge on the surface of the sphere, creating no electric field, giving a constant potential. – julesverne Feb 26 at 17:12
@julesverne: I edited the answer again. – askewchan Feb 26 at 17:18
So if I were to find the potential outside of a conducting sphere that is not grounded, the potential inside would have to be constant. Therefore that second image charge would have to be placed at the origin to mimic if it were grounded. – julesverne Feb 26 at 17:53
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": 12, "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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9572772979736328, "perplexity_flag": "middle"} |
http://mathhelpforum.com/differential-geometry/81362-homology-3-torus.html | # Thread:
1. ## Homology of the 3-Torus
I'm learning the basics of homology theory. I'm trying to give the 3-Torus a CW structure and then compute its homology.
I think I understand the approach to giving the 2-torus a CW complex. Then you have a diagram with sides identified to work from. Can I create a 3-d box with side identifications and do something similar? Even if I can, that approach is unsatisfying, as generalizing to the n-torus would be difficult.
2. Originally Posted by robeuler
I think I understand the approach to giving the 2-torus a CW complex. Then you have a diagram with sides identified to work from. Can I create a 3-d box with side identifications and do something similar? Even if I can, that approach is unsatisfying, as generalizing to the n-torus would be difficult.
I think you can. Giving the 2-torus a CW structure is based on the observation that $T^2=S^1\times S^1 = \mathbb{R}/\mathbb{Z} \times \mathbb{R}/\mathbb{Z}$.
Now we can add another factor to get the 3-torus:
$T^3=S^1\times S^1 \times S^1= \mathbb{R}/\mathbb{Z} \times \mathbb{R}/\mathbb{Z} \times \mathbb{R}/\mathbb{Z}$.
This shows how to identify sides on a cube to get a 3-torus. And that should lead to a CW-structure on $T^3$.
Further generalizations for higher dimensions should be analogous. | {"extraction_info": {"found_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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9343420267105103, "perplexity_flag": "head"} |
http://en.wikibooks.org/wiki/OpenGL_Programming/Modern_OpenGL_Tutorial_Arcball | OpenGL Programming/Modern OpenGL Tutorial Arcball
The arcball is a tool to rotate objects with the mouse naturally.
The concept
Arcball angle during a rotation
Imagine a virtual ball that is just behind the screen. By clicking it with your mouse, you would pinch it, and by moving your mouse you would make the ball spin around its center. And the same rotation would be applied to an object in the OpenGL scene!
The diagram on the right shows the virtual ball in top view. The black line at the bottom is the screen, and x1 and x2 are two successive mouse positions during the drag. This is shown in 2D but the same principle applies in 3D.
The goal is to compute the α angle and the rotation axis. This is the point where you understand that maths are really necessary as soon as you get serious with OpenGL ;) In particular we'll need the Pythagorean theorem, the vector dot product and cross product.
We'll:
Wikipedia has related information at Pythagorean theorem
1. Convert the screen coordinates (in pixels) to camera coordinates (in [-1, 1])
2. Compute the vectors OP1 and OP2, the points at the surface of the ball that match our mouse click
• x and y coordinates are directly taken from the click in camera coordinates
• z coordinate is computed using the classical Pythagorean theorem $\scriptstyle b \;=\; \sqrt{c^2 \,-\, a^2}. \,$
• If P1 or P2 is too far away from the sphere ($\scriptstyle \left|\mathbf{OP}\right| \;>\; 1$), we normalize it to get the nearest point on the surface of the ball
3. We have $\scriptstyle \mathbf{a} \cdot \mathbf{b} \;=\; \left\|\mathbf{a}\right\| \, \left\|\mathbf{b}\right\| \cos (\theta)$, and the ball's size is 1 ($\scriptstyle \left\|\mathbf{a}\right\| \;=\; \left\|\mathbf{b}\right\| \;=\; 1$), so we get the angle using $\scriptstyle \arccos\left(\mathbf{OP1}\cdot\mathbf{OP2}\right)$.
4. Get the rotation axis in 3D, we compute $\scriptstyle\mathbf{OP1}\times\mathbf{OP2}$, which will give a unit perpendicular vector
The code
Capturing mouse events
GLUT offers a way to get the mouse clicks and drag events:
`glutMouseFunc(onMouse)` will call `onMouse(int button, int state, int x, int y)` for each mouse click, where:
• button is GLUT_LEFT_BUTTON, GLUT_MIDDLE_BUTTON or GLUT_RIGHT_BUTTON
• state is GLUT_DOWN or GLUT_UP
• x and y are the screen coordinates, starting from the top-left corner (y is reversed compared to OpenGL coordinates!)
`glutMotionFunc(onMotion)` will call `onMotion(int x, int y)` for each mouse move when any button is pressed down, where x and y are the screen coordinates.
You also have `glutPassiveMotionFunc(...)` which works similarly for mouse moves when no button is pressed at all.
So we add two functions to keep track of the mouse moves when the left button is pressed:
```/* Global */
int last_mx = 0, last_my = 0, cur_mx = 0, cur_my = 0;
int arcball_on = false;
```
```/* main() */
glutMouseFunc(onMouse);
glutMotionFunc(onMotion);
```
```void onMouse(int button, int state, int x, int y) {
if (button == GLUT_LEFT_BUTTON && state == GLUT_DOWN) {
arcball_on = true;
last_mx = cur_mx = x;
last_my = cur_my = y;
} else {
arcball_on = false;
}
}
void onMotion(int x, int y) {
if (arcball_on) { // if left button is pressed
cur_mx = x;
cur_my = y;
}
}
```
Compute OP1 and OP2
We add a new function to compute the arcball surface point:
```/**
* Get a normalized vector from the center of the virtual ball O to a
* point P on the virtual ball surface, such that P is aligned on
* screen's (X,Y) coordinates. If (X,Y) is too far away from the
* sphere, return the nearest point on the virtual ball surface.
*/
glm::vec3 get_arcball_vector(int x, int y) {
glm::vec3 P = glm::vec3(1.0*x/screen_width*2 - 1.0,
1.0*y/screen_height*2 - 1.0,
0);
P.y = -P.y;
float OP_squared = P.x * P.x + P.y * P.y;
if (OP_squared <= 1*1)
P.z = sqrt(1*1 - OP_squared); // Pythagore
else
P = glm::normalize(P); // nearest point
return P;
}
```
We first convert the x,y screen coordinates to [-1,1] coordinates (and reverse y coordinates). Then we use the Pythagorean theorem to check the length of the OP vector and compute the z coordinate, as explained above.
Compute the angle and axis
``` /* onIdle() */
if (cur_mx != last_mx || cur_my != last_my) {
glm::vec3 va = get_arcball_vector(last_mx, last_my);
glm::vec3 vb = get_arcball_vector( cur_mx, cur_my);
float angle = acos(min(1.0f, glm::dot(va, vb)));
glm::vec3 axis_in_camera_coord = glm::cross(va, vb);
glm::mat3 camera2object = glm::inverse(glm::mat3(transforms[MODE_CAMERA]) * glm::mat3(mesh.object2world));
glm::vec3 axis_in_object_coord = camera2object * axis_in_camera_coord;
mesh.object2world = glm::rotate(mesh.object2world, glm::degrees(angle), axis_in_object_coord);
last_mx = cur_mx;
last_my = cur_my;
}
```
Once we have OP1 and OP2 (here named `va` and `vb`), we can compute the angle with `acos(dot(va,vb))`.
Since we're using `float` variables, there may be precision issues: `dot` may return a value slightly greater than 1, and `acos` will return `nan`, which means an invalid float. The consequence is that our rotation matrix will be all messed, and usually our object will just disappear from the screen! To remedy this, we cap the value with a maximum of `1.0`.
An extra trick is converting the rotation axis from camera coordinates to object coordinates. It's useful when the camera and object are placed differently. For instace, if you rotate the object by 90° on the Y axis ("turn its head" to the right), then perform a vertical move with your mouse, you make a rotation on the camera X axis, but it should become a rotation on the Z axis (plane barrel roll) for the object. By converting the axis in object coordinates, the rotation will respect that the user work in camera coordinates (WYSIWYG). To transform from camera to object coordinates, we take the inverse of the MV matrix (from the MVP matrix triplet).
And last we can apply our transformation using `glm::rotate` as usual :)
Exercises
• Is the rotation angle proportional to the mouse move? Try to make a move near the border of the virtual ball.
• The virtual ball will stop rolling when the mouse is too far away. Other mouse controls are possible. For instance, study how dragging with the middle button works in the Blender 3D modeler.
• Try different roll speeds, by multiplying the rotation angle.
< OpenGL Programming
Browse & download complete code | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8775720000267029, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/92807/finding-the-intersection-points-of-two-circles-where-one-circle-has-a-diff-x-a | # Finding the intersection points of two circles - where one circle has a diff x and y coord than the other
I now understand how to calculate the offset of the radical line from circle_a (a)
However:
````Let two circles of radii and and centered at and intersect in a region shaped like an asymmetric lens. The equations of the two circles are
x^2 + y^2 = R^2
(x-d)^2 + y^2 = R^2
````
So this methods assumes that the y coordinates of both circles are the same?
How do I calculate the intersection points where both the x and y coordinates are not the same then?
Thanks.
-
translating so that the center of one circle is at the origin and rotating so that the $x$-coordinate of the center of the other is zero reduces the general case to this one. – yoyo Dec 19 '11 at 20:53
2
– Joseph O'Rourke Dec 19 '11 at 22:02
## 2 Answers
You can always assume that the y-coordinates of both circles are the same because you can rotate both circles in a way that they lie on a line that are parallel to the x-axis. Then you can translate both circles so the first one is in the origin.
Using that rotation the $d$ in your formula will be the distance of both centers and your formulas will look a lot better. To undo this you have to undo the translation and then the rotation. Either you use rotation matrices or you use a complete solution like for example this one.
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I see, thanks a lot :) – xcrypt Dec 19 '11 at 21:03
No problem :-). – Listing Dec 19 '11 at 21:04
to give an example, for instance $(x-1)^2+(y-1)^2=1, (x-2)^2+(y-2)^2=1$. change variables, $(x,y)\mapsto(x+1,y+1)$ to get a new set of equations $x^2+y^2=1, (x-1)^2+(y-1)^2=1$. the center of the first circle is now at the origin and the other is centered at $(1,1)$. now the angle from the positive $x$-axis to $(1,1)$ is $\pi/4$ (in general you would have to find some inverse tangent; i picked an easy one). so you want to rotate the whole plane by $-\pi/4$ ie $$(x,y)\mapsto \left( \begin{array}{cc} 1/\sqrt{2}&1/\sqrt{2}\\ -1/\sqrt{2}&1/\sqrt{2}\\ \end{array} \right) \left( \begin{array}{c} x\\ y\\ \end{array} \right) =\Bigg(\frac{(x+y)}{\sqrt{2}},\frac{(y-x)}{\sqrt{2}}\Bigg)$$ now you have the equations $x^2+y^2=1, (x-\sqrt{2})^2+y^2=1$.
- | {"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": 11, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9238340854644775, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/35881/easy-or-formal-proof-of-hawking-effect-that-anyone-can-understand | # easy or formal proof of Hawking effect … that anyone can understand?
where could i find an explanation (appart from Wikipedia) of how the Hawking's effect is obtained from quantum field theory GR and thermondynamic :D
-
## 1 Answer
You can try this approach which is only superficially different from Wikipedia:
Consider a black hole metric
$$(1 - {2m\over r} ) dt^2 + {dr^2\over 1 - {2m\over r}} + r^2 d\Omega^2$$
Continued to imaginary time (I flipped the sign of $dt^2$). Now write $r=2m + u$, and expand the metric to leading order:
$${ u \over 2m} dt^2 + {2m\over u} du^2 + 2m^2 d\Omega^2$$
and notice that after a coordinate change ${u = 8m v^2}$, the du part turns into $dv^2$ and the $dt$ part becomes the angular form of a polar metric
$${ v^2 {dt^2\over 16m^2} + dv^2 }$$
Which means that ${t\over 4m}$ needs to be periodic with period $2\pi$, just like $\theta$ in polar coordinates, or else you have a curvature singularity at $v=0$. The variable t is periodic at infinity with the same period, and this means that the inverse temperature for a path-integral on this background is $8\pi m$. That tells you the temperature of doing quantum field theory on this background is the Hawking temperature, assuming there is no singularity on the Euclidean horizon. This is Hawking's argument of the late 1970s.
The physical justification for all this formal stuff is just Wikipedia's argument--- taking a near horizon limit turns this into Rindler space. Assuming nothing funky happens on the horizon is equivalent to using the equivalence principle to get Unruh radiation in the Rindler space, and making the periodicity of t constant over the whole space is just continuing the radiation with Einstein redshift.
Hawking's original calculation of 1974-1976 performs a Bogoliubov transformation on the outgoing and incoming modes, assuming that the horizon-crossing modes look like vacuum. The calculation uses an eikonal approximation in the near horizon limit, and this means Hawking is really considering only the rate of geodesic separation near the horizon. So you might as well do it only in the near-horizon limit, and this is tantamount to just doing Unruh's calculation. The rest of Hawking's calculation just matches Unruh's result to the far-away modes by consistently redshifting them, and this you might as well do using the local Unruh temperature.
Further, Hawkings assumption that the infalling modes are not occupied, that they are vacuum, is equivalent to the statement that an infalling observer sees nothing special, and this is equivalent to the statement that the near-horizon radiation is Unruh. So all these superficially different calculations are exactly the same in physical content and in assumptions.
I am just going through this to convince you that the Wikipedia presentation is exactly the same as all others, except made so that everyone can understand it. There is nothing more in any of the pre-1980 demonstrations. There are some more recent ones involving tunneling of particles, associated with Wilczek,, which I never went through.
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http://math.stackexchange.com/questions/187374/partitioning-a-triangulated-2-sphere-into-two-triangulated-discs | # Partitioning a triangulated 2-sphere into two triangulated discs
Take a triangulation of the 2-sphere, $S^2$. Let the triangulation be denoted $T$. The Euler characteristic tells you that the number of triangles in $T$ is even. Since triangulations of the 2-sphere are shellable, this means you can write $T$ as the union of two triangulated discs along their common boundary, moreover you can ensure the number of triangles in each disc is half the number of triangles in $T$.
Do you know an efficient algorithm to find a partition of $T$ into two discs with equal number of triangles? I imagine there are some fairly efficient algorithms out there, but I'm not sure what to search for in the literature.
I'm particularly interested in partitions into two discs that are somehow optimal. For example, one where the triangulated discs have boundaries with the least number of edges.
edit: Joriki's comment shows the first part of my question has a relatively simple answer. In my words: take a triangulated 2-sphere. Remove a triangle to get a triangulated disc (if you're not using simplicial triangulations this will require a judicious choice). That triangulated disc is shellable, so you can remove one of the triangles that contains a boundary edge (or two) to get a combinatorial disc. etc. By shellability of triangulated discs, this process can be continued until you have a triangulated disc with half the number of triangles.
So the only part of my question remaining is finding ideal such partitions. I suppose the greedy way to attempt to do this would be to prefer to remove triangles with two boundary edges over one. Is that enough to ensure you get a combinatorial disc with the minimum number of boundary edges?
-
Why can't you just start from some triangle and add adjacent triangles until you have half of them? You'd just need to make sure that you don't add ones that contain vertices you've already included elsewhere, since that would separate the remaining ones, but it seems that you shouldn't run out of suitable triangles before reaching half? – joriki Aug 27 '12 at 7:55
That algorithm isn't fast, since it requires a lot of back-tracking to correct for poor choices of triangles. Think of something like a near "space-filling" combinatorial disc in a triangulated 2-sphere. – Ryan Budney Aug 27 '12 at 7:57
1
I don't understand -- wouldn't such a disk still always allow adding to it some suitable triangles (near the "bends")? Or is the problem that there might be few of these and it might be inefficient to have to find them? But it seems one should be able to keep a list of suitable triangles and keep it up to date in constant time per triangle added, which would lead to linear time overall. (Also it seems it shouldn't be too difficult to avoid such a situation by a judicious choice of triangles to add, e.g. always adding the one closest to a fixed point on the sphere.) – joriki Aug 27 '12 at 8:11
– David Speyer Aug 27 '12 at 14:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "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}, "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}, "english_confidence": 0.9605152606964111, "perplexity_flag": "head"} |
http://mathhelpforum.com/algebra/77425-solved-recurring-decimal-help-0-1575757-a.html | # Thread:
1. ## [SOLVED] Recurring Decimal Help - 0.1575757...
I need an explanation of how to do this question
0.1(57) Where the brackets mark recurring signs.
I know you start it by saying let x = 0.157575757... just not quite sure where to go from here.
2. Originally Posted by Domoz
I need an explanation of how to do this question
0.1(57) Where the brackets mark recurring signs.
I know you start it by saying let x = 0.157575757... just not quite sure where to go from here.
Well, if we let $x=0.157575757\ldots$ then $100x=15.757575757\ldots.$ Now subtract.
$\begin{tabular}{crcr@{.}l}<br /> &$100x$&$=$&15&757575757\ldots\\<br /> $-$&$x$&$=$&0&157575757\ldots\\\hline<br /> &$99x$&=&15&6<br /> \end{tabular}$
So $x=\frac{15.6}{99}=\frac{156}{990}=\frac{26}{165}.$
3. Thankyou! I got confused because I got a decimal over a fraction and didn't know what to do.
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": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8932052850723267, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/122668?sort=newest | ## Triangulation of the surface determined by sampling two of its cross-sections
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I have a data set that essentially looks like the picture below, i.e., it's given by sets of points in $\mathbb{R}^3$ that sample the cross-sections of a certain surface that in principle I do not know. In fact, I would like to recover this surface or, more precisely, a triangulation/mesh of it.
In this particular example, it is not difficult to see that we have a section of a cone. In this sense, I do not want to compute the convex hull of these points, but instead I just want to find its boundary (if this makes any sense; I actually don't know whether it does), i.e., I want to find the surface that one could obtain by simply joining the "corresponding" points in both "slices". (I've put "corresponding [points]" in quotes because I believe it's not straightforward to perform this matching exercise, i.e., find the point on the adjacent slice that isn't necessarily the closest one in the $\mathbb{R}^3$ metric, but would correspond to the closest one if we projected one slice into another and considered only the $\mathbb{R}^2$ distance between them, as seen below.)
-
1
I'm not sure I understand what your question is, but the matching method you've described seems reasonable. If the cross sections are in parallel planes project one sample to the other sample's plane and pair each point in one sample to one of its nearest neighbors in the other sample. You can make triangles out of the triples of (1) a point from the first sample (2) one of its closest neighbors in the other sample (in the shared plane) and (3) one of its nearest neighbors in the same sample. – Aaron Golden Feb 23 at 0:15
@Aaron: this is precisely what i had in mind when i wrote the question last night — sorry if i wasn't clear enough (at that point of the day i couldn't think straight anymore :-P). And thank you for your time and attention, i really appreciate it: i've been working with this problem for a couple of months now, and progress has been à la "two steps forwards, one backwards". ;-) – Daniel Feb 23 at 18:47
## 1 Answer
This is a problem that has been studied for a long time. I showed with my students Carol Gitlin and Vinita Subramanian, that there does not always exist a polyhedron that connects two arbitrary polygons in parallel slices. In other words, Daniel's hope to "simply `[`join`]` the 'corresponding' points in both 'slices' " cannot always be realized:
C. Gitlin, J. O'Rourke, V. Subramanian. "On reconstruction of polyhedra from parallel slices," International Journal of Computational Geometry & Applications, 6(1) 1996, 103-122.
These two polygons constitute a counterexample:
There is a very nice summary of the early work, and a practical algorithm, in:
Gill Barequet, Daniel Shapiro, Ayellet Tal. "History Consideration in Reconstructing Polyhedral Surfaces from Parallel Slices." Proceedings of IEEE Visualization, 1996. (CiteSeer link)
Maybe look at this more recent work?:
Samir Akkouche, Eric Galin. "Implicit surface reconstruction from contours." The Visual Computer. August 2004, Volume 20, Issue 6, pp 392-401. (Springer link)
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@O'Rourke: Professor, thank you much for your help, and for the references as well — i really appreciate them, and i'm sure they'll lead me forward. As i mentioned above, i wasn't very clear-headed when i wrote up this question, but i believe that you capture the essence of my doubts (which i have a hard time putting in words, being fresh to this problem and field and all). In hindsight, i should've asked earlier… ;-) – Daniel Feb 23 at 18:58
1
@Daniel: I should have been clearer myself: "I want to find the surface that one could obtain by simply joining the 'corresponding' points in both 'slices.'" What my paper shows is that sometimes it is impossible to find such a correspondence. But this negative result is balance by many practical algorithms. – Joseph O'Rourke Feb 23 at 23:30
@O'Rourke: Professor, i haven't yet had a chance to check the references you posted, so maybe i'm asking something that you already answered, but there you go: Do you know where i can find out more about such algorithms you mentioned? (I'm fairly convinced that my problem will fall into this "impossible" category, given how feature-rich each slice is, but maybe i can work around it somehow…) – Daniel Feb 24 at 0:45
1
@Daniel: I would start with the Barequet et al. paper. It is a very good paper, very clear. And they were well aware of my negative result. (The link I provided leads to the conference version PDF, rather than the journal version.) – Joseph O'Rourke Feb 24 at 0:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9548132419586182, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/tagged/heat+crystals | Tagged Questions
1answer
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Why do phonons cause excellent heat conduction in diamonds?
Phonons are the quantum of lattice vibrations in crystals and are not to be confused with photons, the gauge bosons of the electromagnetic force. Apparently, they contribute to heat conduction, but I ...
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Does a thermally expanding torus experience internal stress?
I'm trying to learn continuum mechanics and thermo-mechanics. As we know, heating an object increases the mean atomic distance $a_0$ of the atoms in a rigid body. Let's assume it is a linear elastic ... | {"extraction_info": {"found_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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.892582893371582, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/4713/if-the-order-divides-a-prime-p-then-the-order-is-p-or-1 | # If the order divides a prime P then the order is P (or 1)
I've just come up with this question as I'm studying for a number theory midterm. If $p$ and $q$ are different prime numbers, and it's known that $2^p \equiv 1 \bmod{q}$, then $q\equiv 1 \bmod{p}$. I've tried a few cases and it seems to be true. How can it be proved?
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6
Let $k$ be minimal natural number, such that $2^k\equiv 1\mod{q}$. Then its easy to see that $k|p$ and $k|(q-1)$(from the little theorem of Fermat). But, because $p$ is prime, there just 2 cases: 1)$k=1$ - impossible. 2)$k=p\to p|(q-1)$ – Nurdin Takenov Sep 15 '10 at 13:54
1
That was neat. болшое спасибо, Такенов. – Weltschmerz Sep 15 '10 at 14:04
## 1 Answer
$\rm\displaystyle\;\ 2^p,\; 2^{q-1} \equiv 1 \;\Rightarrow\; 2^{gcd(p,q-1)} \equiv 1\ (mod\ q)\;\Rightarrow\; gcd(p,q-1) = p\;$ (not $1$ else $\rm \:q|2^1-1\:$)
Said in group theory language: if an element $\ne 1$ has order dividing a prime $\:\rm p\;$ then it must have order $\rm p\:$. In ring theory language: if a principal ideal$\;\ne 1$ contains an irreducible element then that element generates the ideal (which is why the minimal polynomial is sometimes called the "irreducible polynomial"). Here the group / ideal is simply the so-called order ideal $\rm\; \{n : x^n = 1 \}$ which, being nonempty and closed under subtraction, comprises a subgroup / ideal of $\:\mathbb Z\:$. The Euclidean algorithm implies that ideals in $\mathbb Z$ are principal, so every element of a nonzero ideal is a multiple of the least positive element. For an order ideal this simply says that every "possible" order is a multiple of the least possible order $\rm m\:$, $\:$ i.e. $\rm\; x^n = 1 \;\Rightarrow\; m\:|\:n\;$. Compare this ring-theoretic proof to the ubiquitous group-theoretic proof using Lagrange's theorem.
Here is a simple additive example: the order ideal of a fraction is the denominator ideal $\rm\; \{n : n\: x \in \mathbb Z \:\}\:$. Here the analogous result is that if a proper fraction can be written with a prime denominator then it is the least possible denominator (in the multiplicative sense), i.e. it divides ever other possible denominator.
Denominator ideals and fractional ideals can yield significant number-theoretical simplifications - in much the same way that fractional arithmetic often serves to simplify integer arithmetic in elementary number theory. For example, see my post on irrationality proofs using the denominator ideal or, more generally, see my proof employing Dedekind's conductor ideal - a sort of universal denominator for an entire extension ring. Follow the links there to my sci.math posts for much further discussion on this and related topics.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9092785716056824, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/11104/constructing-a-maximally-entangled-qutrit-state-from-n-bell-states | # Constructing a maximally entangled qutrit state from $n$ Bell states
I've read that maximally entangled qubit states are a good "unit" of bipartite entanglement since it is possible to create any other entangled state from them using local operations and classical communication (LOCC) provided sufficiently many copies are available. What would the protocol be to construct a maximally entangled qutrit ($\vert \psi \rangle_{AB} = \frac{1}{\sqrt{3}}(\vert 00 \rangle + \vert 11 \rangle + \vert 22 \rangle)$) between two space separated parties from a set of $n$ Bell states ($\vert \Phi^+ \rangle^{\otimes n}_{AB} = 2^{-n/2} (\vert 00 \rangle + \vert 11 \rangle)^{\otimes n}$) initially shared by those parties using only LOCC?
If you can, please include in your answer why local operations alone would not be sufficient.
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## 1 Answer
Basically, when you have $n$ Bell-states, you already have a maximally entangled $2^n$ dimensional state. For example, when $n=2$,
$$\begin{align} \lvert\Phi^+\rangle_{AB}^{\otimes 3} &= \frac12 (\lvert0\rangle_A\lvert0\rangle_B + \lvert1\rangle_A\lvert1\rangle_B)(\lvert0\rangle_A\lvert0\rangle_B + \lvert1\rangle_A\lvert1\rangle_B)\\ &=\frac12 (\lvert00\rangle_A\lvert00\rangle_B + \lvert01\rangle_A\lvert01\rangle_B + \lvert10\rangle_A\lvert10\rangle_B + \lvert11\rangle_A\lvert11\rangle_B\\ &=\frac12 (\lvert0\rangle_A\lvert0\rangle_B + \lvert1\rangle_A\lvert1\rangle_B + \lvert2\rangle_A\lvert2\rangle_B + \lvert3\rangle_A\lvert3\rangle_B ) \end{align}$$
If you really want to confert thes ququarts to qutrits, then you have to do this probabilistically. If you are not to picky about the success probability, you just have to measure whether the sate written above is in the $\lvert3\rangle$ state or not. If not (which happens with probability 3/4, you have converted 2 qubit maximally entangled pairs into 1 maximally entangled qutrit pairs. You need therefore on average $\frac2{3/4}=8/3=2.667$ qubit pairs per produced qutrit pairs.
Of course, processing more pairs simultaneously allows you to approach the asymptotic rate of $\log_2 3=1.585$ qubit pair per qutrit pairs.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8849135637283325, "perplexity_flag": "head"} |
http://mathhelpforum.com/pre-calculus/203159-complex-number-plane.html | # Thread:
1. ## complex number plane
(a+bi)^2=i What is the squareroot of i solve geometrically in a complex number plane
2. ## Re: complex number plane
Originally Posted by kalemale
(a+bi)^2=i What is the squareroot of i solve geometrically in a complex number plane
There are two square roots of $i$: $\exp \left( {\frac{{i\pi }}{4}} \right)\& \;\exp \left( {\frac{{ - 3i\pi }}{4}} \right)$
3. ## Re: complex number plane
I do not really understand how do you get to that?
4. ## Re: complex number plane
Originally Posted by kalemale
(a+bi)^2=i What is the squareroot of i solve geometrically in a complex number plane
One method:
$\displaystyle \begin{align*} (a + b\,i)^2 &= i \\ a^2 - b^2 + 2ab\,i &= 0 + 1i \\ a^2 - b^2 = 0 \textrm{ and } 2ab &= 1 \end{align*}$
From the second of these equations we have $\displaystyle \begin{align*} b = \frac{1}{2a} \end{align*}$. Substituting into the first equation we have
$\displaystyle \begin{align*} a^2 - \left(\frac{1}{2a}\right)^2 &= 0 \\ a^2 - \frac{1}{4a^2} &= 0 \\ a^2 &= \frac{1}{4a^2} \\ 4a^4 &= 1 \\ a^4 &= \frac{1}{4} \\ a &= \pm \frac{\sqrt{2}}{2} \end{align*}$
Back-substituting gives
$\displaystyle \begin{align*} b &= \frac{1}{2\left( \pm \frac{\sqrt{2}}{2}\right) } \\ &= \frac{1}{ \pm \sqrt{2}} \\ &= \pm \frac{\sqrt{2}}{2} \end{align*}$
Therefore the solutions to your equation are $\displaystyle \begin{align*} \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \end{align*}$ and $\displaystyle \begin{align*} -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i \end{align*}$.
These are consistent with the solutions given by Plato, who probably used the Exponential form of each number and Index Laws as an alternative method.
5. ## Re: complex number plane
Yeah but I need to solve in a complex number plane. Sorry if I am slow but your solutions is equational?
6. ## Re: complex number plane
Originally Posted by kalemale
I do not really understand how do you get to that?
Do you understand that $\exp(i\theta)=\cos(\theta)+i\sin(\theta)~?$
Do you know how to do $\left(\cos(\theta)+i\sin(\theta)\right)^2~?$
7. ## Re: complex number plane
Originally Posted by kalemale
Yeah but I need to solve in a complex number plane. Sorry if I am slow but your solutions is equational?
I don't understand what the problem is. You will always need to use some algebra to at least find one solution. If you wish to use more geometry after that, you can use the fact that there are as many "roots" as its power (i.e. square roots have 2 solutions, cube roots have 3 solutions, fourth roots have 4 solutions, etc) and they are all evenly spaced around a circle centred at the origin.
8. ## Re: complex number plane
Thanks for all the help! | {"extraction_info": {"found_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": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9113660454750061, "perplexity_flag": "middle"} |
http://mathhelpforum.com/advanced-algebra/195817-showing-non-empty-subset-m-affine-subset-r-4-a.html | # Thread:
1. ## Showing that a non-empty subset M is an affine subset of R^4
Hi, I have to do a project on affine subsets and affine mappings, but I have no clue what they are... We are given only one clue and I can't find many notes on google. I would really appreciate it if someone could help me with this first problem (and if you could also give me a link to some good notes on the topic). Thanks a lot!
The problem says:
"To illustrate this concept, show that
$M=\{x=(x_1, ..., x_4) \epsilon \mathbb{R}^4 : 2x_1-x_2+x_3=1$ and $x_1+4x_3-2x_4=3\}$
is an affine subset of $\mathbb{R}^4$."
And the only hint we get is
"Throughout Part A of this project, V will be a real vector space and, for a non-empty subset S of V and a $\epsilon$ V, the set { $x+a : x \epsilon S$} will be denoted by S+a.
An affine subset of V is a non-empty subset M of V with the property that $\lambda *x+(1- \lambda )*y \epsilon M$ whenever $x,y \epsilon M$ and $\lambda \epsilon \mathbb{R}$."
2. ## Re: Showing that a non-empty subset M is an affine subset of R^4
OK, I have managed to solve the problem... I just have to substitute lambda*x+(1-lambda)*y into both of the conditions given. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9434089660644531, "perplexity_flag": "head"} |
http://euler.slu.edu/escher/index.php/Tessellations_by_Squares,_Rectangles_and_other_Polygons | # Tessellations by Squares, Rectangles and other Polygons
### From EscherMath
K-12: Materials at high school level.
Oberkapfenberg Castle, Syria.
## Some Basic Tessellations
Recall that a polygon is just a simple geometric shape. The polygons we will be talking about are squares, rectangles, parallelograms, the rhombus and maybe a couple of other ones.
If you haven't thought about polygons for a couple of weeks it may be useful to quickly sneek a peek at some of the examples we talked about earlier: Squares, Rectangles, Parallelograms and Other Polygons
In this section we will be looking at the following:
Question: Which polygons tile the plane?
Another way of saying this is: which polygons can be used to tile say a floor or a wall? This means we are looking for shapes that fit together nicely, without any gaps or overlaps to create a pattern. Some people call these patterns tilings, while others call them tessellations. Both words are correct. We may use them both in this text.
The most common and simplest tessellation uses a square. You may not have thought about it, but you will ahve seen titlings by squares before. A lot of bathrooms have square tiles on the floor. A lot of classsrooms will have squares on the floor and there may even be squares in the ceiling.
Squares easily form horizontal strips:
Stacks of these strips cover a rectangular region and the pattern can clearly be extended to cover the entire plane. This easily gives us the result that:
All squares tessellate.
The same technique works with parallelograms. You can put parallelograms side by side and create these strips. If you stack the trips you will have a tiling by parallelograms, and so:
All parallelograms tessellate.
Looking for other tessellating polygons is a complex problem, so we will organize the question by the number of sides in the polygon. The simplest polygons have three sides, so we begin with triangles:
All triangles tessellate.
To see this, take an arbitrary triangle and rotate it about the midpoint of one of its sides. The resulting parallelogram tessellates:
The picture works because all three corners (A, B, and C) of the triangle come together to make a 180° angle - a straight line. This property of triangles will be the foundation of our study of polygon tessellations, so we state it here:
The sum of angles of any triangle is 180°.
Moving up from triangles, we turn to four sided polygons, the quadrilaterals. Before continuing, try the Quadrilateral Tessellation Exploration.
## Tessellations by Quadrilaterals
Recall that a quadrilateral is a polygon with four sides.
The sum of angles in any quadrilateral is 360°
To prove, divide a quadrilateral into two triangles as shown:
Since the angle sum of any triangle is 180°, and there are two triangles, the angle sum of the quadrilateral is 180° + 180° = 360°. Taking a little more care with the argument, we have:
$\alpha_1 + \delta_1 + \gamma = 180^\circ$ and $\alpha_2 + \delta_2 + \beta = 180^\circ$.
Then
$360^\circ = \alpha_1 + \delta_1 + \gamma + \alpha_2 + \delta_2 + \beta = \alpha_1 + \alpha_2 + \beta + \gamma + \delta_1 + \delta_2 = \angle A + \angle B + \angle C + \angle D$.
This division into triangles does not calculate the angle sum of the quadrilateral.
The point of all the letters is that the angles of the triangles make the angles of the quadrilateral, which would not work if the quadrilateral was divided as shown on the right.
We now turn to the main result of this section:
All quadrilaterals tessellate.
Begin with an arbitrary quadrilateral ABCD. Rotate by 180° about the midpoint of one of its sides, and then repeat using the midpoints of other sides to build up a tessellation.
The angles around each vertex are exactly the four angles of the original quadrilateral. Since the angle sum of the quadrilateral is 360°, the angles close up, the pattern has no gaps or overlaps, and the quadrilateral tessellates.
Recall from Fundamental Concepts that a convex shape has no dents. All triangles are convex, but there are non-convex quadrilaterals. The technique for tessellating with quadrilaterals works just as well for non-convex quadrilaterals:
It is worth noting that the general quadrilateral tessellation results in a wallpaper pattern with p2 symmetry group.
## Tessellations by Convex Polygons
Every shape of triangle can be used to tessellate the plane. Every shape of quadrilateral can be used to tessellate the plane. In both cases, the angle sum of the shape plays a key role. Since triangles have angle sum 180° and quadrilaterals have angle sum 360°, copies of one tile can fill out the 360° surrounding a vertex of the tessellation.
The next simplest shape after the three and four sided polygon is the five sided polygon: the pentagon. The angle sum of any pentagon is 540°, because we can divide the pentagon into three triangles:
Since each triangle has angle sum 180° the angle sum of the pentagon is 180° + 180° + 180° = 540°.
Rather than repeat the angle sum calculation for every possible number of sides, we look for a pattern. The angle sum of a triangle (3-gon) is 180°, the angle sum of a quadrilateral (4-gon) is 2x180°, and the angle sum of a pentagon is 3x180°. A general polygon with n sides can be cut into n − 2 triangles and so we have:
The sum of the angles of an n-gon is $(n-2)\times 180^\circ$.
Unlike the triangle and quadrilateral case, the pentagon's angle sum of 540° is not helpful when trying to fit a bunch of pentagons around a vertex. In fact, there are pentagons which do not tessellate the plane.
For example, the regular pentagon has five equal angles summing to 540°, so each angle of the regular pentagon is $\frac{540^\circ}{5} = 108^\circ$. Attempting to fit regular polygons together leads to one of the two pictures below:
Both situations have wedge shaped gaps that are too narrow to fit another regular pentagon. Thus, not every pentagon tessellates. On the other hand, some pentagons do tessellate, for example this house shaped pentagon:
The house pentagon has two right angles. Because those two angles sum to 180° they can fit along a line, and the other three angles sum to 360° (= 540° - 180°) and fit around a vertex.
Thus, some pentagons tessellate and some do not. The situation is the same for hexagons, but for polygons with more than six sides there is the following:
No convex polygon with seven or more sides can tessellate.
This remarkable fact is difficult to prove, but just within the scope of this book. However, the proof must wait until we develop a counting formula called the Euler characteristic, which will arise in our chapter on Non-Euclidean Geometry.
Nobody has seriously attempted to classify non-convex polygons which tessellate, because the list is quite likely to be too long and messy to describe by hand. However, there has been quite a lot of work towards classifying convex polygons which tessellate. Because we understand triangles and quadrilaterals, and know that above six sides there is no hope, the classification of convex polygons which tessellate comes down to two questions:
• Which convex pentagons tessellate?
• Which convex hexagons tessellate?
Question 2 was completely answered in 1918 by K. Reinhardt.[1] Reinhardt showed that there are only three types of convex hexagons which tessellate:
File:Convex-hexagons.svg
Reinhardt also addressed Question 1 and gave five types of pentagon which tessellate. In 1968, R. Kershner[2] found three new types, and claimed a proof that the eight known types were the complete list. A 1975 article by Martin Gardner[3] in Scientific American popularized the topic, and led to a surprising turn of events.
In fact Kershner's "proof" was incorrect. After reading the Scientific American article, a computer scientist, Richard James III, found a ninth type of convex pentagon that tessellates. Not long after that, Marjorie Rice, a San Diego homemaker with only a high school mathematics background, discovered four more types, and then a German mathematics student, Rolf Stein, discovered a fourteenth type in 1985.
Since 1985, no new types have been discovered, and many mathematicians believe that the list is finally complete. However, there is no well accepted proof of the classification, so it remains possible that there is a fifteenth or even many more types of convex pentagons that tessellate. Today, question 1 is an open problem, a problem whose solution is unknown.
Summary of Polygon Tessellations
Sides Angle Sum Tessellates?
3 180° Yes. All triangles tessellate.
4 360° Yes. All quadrilaterals tessellate.
5 540° Sometimes. There is a list of 14 types of convex pentagons which tessellate, but nobody knows if the list is complete.
6 720° Sometimes. There is a list of 3 types of convex hexagons which tessellate, and these are the only three.
7 & up 900°, 1080°, ... No convex n-gon tessellates for $n \geq 7$
## Tessellations by Regular Polygons
Recall that a regular polygon is a polygon whose sides are all the same length and whose angles all have the same measure. A regular n-gon has n equal angles that sum to $(n-2)180^\circ$, so:
The corner angle of a regular n-gon is $\frac{(n-2)180^\circ}{n}$.
The table shows the corner angles for the first few regular polygons:
| | | | | | | | | Number of Sides
| Corner angle
|
|-----|-----|------|------|---------|------|------|------|---------------------|------------------|
| 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| 60° | 90° | 108° | 120° | ~128.6° | 135° | 140° | 144° | ~147.3° | 150° |
### Regular Tessellations
Regular tessellation
A tessellation using one regular polygon tile, arranged so that edges match up.
Corners of the tiles need to fit together around a point, which means the corner angle of the regular polygon must evenly divide 360°. Since
$6 \times 60^\circ = 360^\circ$,
there is a regular tessellation using six triangles around each vertex. Since
$4 \times 90^\circ = 360^\circ$,
there is a regular tessellation using four squares around each vertex. And since
$3 \times 120^\circ = 360^\circ$
there is a regular tessellation using three hexagons around each vertex. We have already seen that the regular pentagon does not tessellate. A regular polygon with more than six sides has a corner angle larger than 120° (which is 360°/3) and smaller than 180° (which is 360°/2) so it cannot evenly divide 360°. We conclude:
There are three regular tessellations of the plane: by triangles, by squares, by hexagons.
A major goal of this book is to classify all possible regular tessellations. Apparently, the list of three regular tessellations of the plane is the complete answer. However, these three regular tessellations fit nicely into a much richer picture that only appears later when we study Non-Euclidean Geometry.
Tessellations using different kinds of regular polygon tiles are fascinating, and lend themselves to puzzles, games, and certainly tile flooring. Try the Pattern Block Exploration.
### Archimedean tessellations - * Optional
An Archimedean tessellation (also known as a semi-regular tessellation) is a tessellation made from more that one type of regular polygon so that the same polygons surround each vertex.
| | | | |
|-----------|-------------|-----------|-------------|
| | | | |
| (3,6,3,6) | (3,4,6,4) | (3,12,12) | (3,3,3,4,4) |
| | | | |
| (4,8,8) | (3,3,4,3,4) | (4,6,12) | (3,3,3,3,6) |
We can use some notation to clarify the requirement that the vertex configuration be the same at every vertex. We can list the types of polygons as they come together at the vertex. For instance in the top row we see on the left a semi-regular tessellation with at every vertex a (3,6,3,6) configuration. We see a 3-gon, a 6-gon, a 3-gon and a 6-gon.
The other tessellations on the top row have a (3,4,6,4), a (3,12,12), and a (3,3,3,4,4) configuration. These configurations are unique up to cyclic reordering (and possibly reversing the order). For example (3,12,12) can also be written as (12,12,3) or (12,3,12). In the bottom row we have (4,8,8), (3,3,4,3,4), (4,6,12) and (3,3,3,3,6) configurations.
Recall that creating a tessellation requires the angle sums of the polygons to add up to 360° around a vertex. We know from a previous section that the angle of a regular n-gon is $\frac{(n-2)180^\circ}{n}$. If we have a collection of n-gons: n1,n2,...,nk , then the angles of those polygons need to add up to 360°:
$\frac{(n_1-2)180^\circ}{n_1}+ \frac{(n_2-2)180^\circ}{n_2} + ... + \frac{(n_k-2)180^\circ}{n_k} = 360^\circ$
Dividing both sides by 180° gives us the following equation:
$\frac{n_1-2}{n_1}+ \frac{n_2-2}{n_2} + ... + \frac{n_k-2}{n_k} = 2$.
For example, suppose that n1 = 3,n2 = 3,n3 = 3,n4 = 4,n5 = 5, then we have: $\frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{2}{4} + \frac{2}{4} = 2$. This means that 3 triangles and 2 squares will give us a vertex type. In this case we can arrange these polygons around the vertex in two different ways: (3,3,3,4,4) and (3,3,4,3,4). Both of these will give rise to a semi-regular tessellation.
Recall that a semi-regular tessellation satisfies the following requirements:
• All the tiles are regular polygons
• It is a tessellation, hence no gaps or overlaps
• All the vertices are of the same type.
There are only 21 combinations of regular polygons that will fit around a vertex. And of these 21 there are there are only 11 that will actually extend to a tessellation. Below are the different vertex types. An asterisk indicates that this vertex type cannot be extended to a tessellation.
type N=3 N=4 N=5 N=6 N=7 N=8 N=9 N=10 N=12 N=15 N=18 N=20 N=24 N=42 vertex configuration(s)
1 6 (3,3,3,3,3,3)
2 4 1 (3,3,3,3,6)
3 3 2 (3,3,3,4,4)
4 3 2 (3,3,4,3,4)
5 2 1 1 (3,3,4,12) (*)
6 2 1 1 (3,4,3,12) (*)
7 2 2 (3,3,6,6) (*)
8 2 2 (3,6,3,6)
9 1 2 1 (3,4,4,6) (*)
10 1 2 1 (3,4,6,4)
11 1 1 1 (3,7,42) (*)
12 1 1 1 (3,8,24) (*)
13 1 1 1 (3,9,18) (*)
14 1 1 1 (3,10,15) (*)
15 1 2 (3,12,12)
16 4 (4,4,4,4)
17 1 1 1 (4,5,20) (*)
18 1 1 1 (4,6,12)
19 1 2 (4,8,8)
20 2 1 (5,5,10) (*)
21 3 (6,6,6)
From the table we can see that:
Theorem 1: There are 3 regular tessellations of the plane.
Theorem 2: There are 8 semi-regular (Archimedean) tessellations of the plane.
## Relevant examples from Escher's work
• Fundamental forms of regular division of the plane, Visions of Symmetry pg. 33
• Sketch #A7 (Regular division with triangles)
• Tessellation by triangles, sketch (2) from the abstract motif notebook, Visions of Symmetry pg. 83.
• Sketch #131-134 (Pentagon tessellations), and Tiled Column, New Lyceum, Baarn
• Tessellation by hexagons, Visions of Symmetry pg. 90
## Related Sites
• Java applet for quadrilateral tessellations. By Shodor Interactive.
• Topics in Tiling by Eric Weisstein at Wolfram Research.
### Convex Pentagons
• The 14 known types of convex pentagons that tessellate. By http://www.mathpuzzle.com
• A brief history of convex pentagon tessellations. By Doris Schattschneider.
• Intriguing Tessellations. By Marjorie Rice.
• Tessellations by Hexagons. By David King.
## Notes
1. ↑ K. Reinhardt, Über die Zerlegung der Ebene in Polygone. (Inaugural-Disstertation, Univ. Frankfurt a.M.) R. Noske, Borna and Leipzig, 1918.
2. ↑ R. Kershner, On Paving the Plane, The American Mathematical Monthly 75, October 1968, pg. 839-844
3. ↑ Martin Gardner, Time Travel and Other Mathematical Bewilderments Ch. 13. W.H. Freeman, 1988
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http://mathhelpforum.com/geometry/48183-help-geometry.html | # Thread:
1. ## Help! Geometry
The question is:
The oval athletic field is the union of a square and semicircles at opposites ends, as shown in Figure. If the area of the field is 1300 square yards, find the dimensions of the square.
Basically the figure is a football field with two semicircles at the end. I can do this problem forward, but going backward is not going so well.
Kelli
2. So you have a circle and a square. I think there has to be more info!
3. Originally Posted by wynkky
The question is:
The oval athletic field is the union of a square and semicircles at opposites ends, as shown in Figure. If the area of the field is 1300 square yards, find the dimensions of the square.
Basically the figure is a football field with two semicircles at the end. I can do this problem forward, but going backward is not going so well.
Kelli
Let us suppose radius of the semicircular ends = r.
So, side of square = diameter of semicircle = 2r
Now area of field = Area of two semicircles + Area of square.
$1300 = 2 \times {\frac{\pi r^2}{2}} + (2r)^2$
$1300 = \pi r^2 + 4r^2$
$1300 = r^2(\pi + 4)$
$r^2 = \frac {1300}{\pi + 4}$
$r= \sqrt {182.032}$
r= 13.49 yards
Side of square = 2r = $2 \times 13.49$ = 26.98 yards.= 27 yards | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9531502723693848, "perplexity_flag": "middle"} |
http://cms.math.ca/10.4153/CMB-2003-001-2 | Canadian Mathematical Society
www.cms.math.ca
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| Site map | | | CMS store | |
location: Publications → journals → CMB
Abstract view
# Condensed Domains
Read article
[PDF: 154KB]
Published:2003-03-01
Printed: Mar 2003
• D. D. Anderson
• Tiberiu Dumitrescu
Features coming soon:
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Format: HTML LaTeX MathJax PDF PostScript
## Abstract
An integral domain $D$ with identity is condensed (resp., strongly condensed) if for each pair of ideals $I$, $J$ of $D$, $IJ=\{ij; i\in I, j\in J\}$ (resp., $IJ=iJ$ for some $i\in I$ or $IJ =Ij$ for some $j\in J$). We show that for a Noetherian domain $D$, $D$ is condensed if and only if $\Pic(D)=0$ and $D$ is locally condensed, while a local domain is strongly condensed if and only if it has the two-generator property. An integrally closed domain $D$ is strongly condensed if and only if $D$ is a B\'{e}zout generalized Dedekind domain with at most one maximal ideal of height greater than one. We give a number of equivalencies for a local domain with finite integral closure to be strongly condensed. Finally, we show that for a field extension $k\subseteq K$, the domain $D=k+XK[[X]]$ is condensed if and only if $[K:k]\leq 2$ or $[K:k]=3$ and each degree-two polynomial in $k[X]$ splits over $k$, while $D$ is strongly condensed if and only if $[K:k] \leq 2$.
MSC Classifications: 13A15 - Ideals; multiplicative ideal theory 13B22 - Integral closure of rings and ideals [See also 13A35]; integrally closed rings, related rings (Japanese, etc.) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8244125843048096, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/6400/are-tidal-power-plants-slowing-down-earths-rotation?answertab=active | Are tidal power plants slowing down Earth's rotation?
Are tidal power plants slowing down Earth's rotation to the speed of the orbiting moon? (1 rotation per 28 cca days)
Are they vice versa increasing the speed of moon orbiting by generating some waves in gravitation field?
If yes, can you calculate how much energy must be produced by how many tidal power plants (compare it to average nuclear plant please) to slow down the Earth's rotation to 25 hours / day?
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11
– Ami Mar 6 '11 at 5:53
4 Answers
In principle, yes, the ultimate source of energy for a tidal power plant is Earth's rotational energy, so these plants are slowing down the Earth's rotation. By conservation of angular momentum, that means they are pushing the Moon further away as well, although I wouldn't phrase it as being due to "waves in the gravitational field," as that expression suggests a different phenomenon.
The Earth's rotational kinetic energy is about $10^{29}$ J, and the world uses something like $10^{22}$ J/year, so you could power the entire world for millions of years before you'd run out of rotational energy.
To answer your numerical question, you should work out the rotational kinetic energy of the Earth now, and also when the day is 25 hours long. The difference between those is the total energy required. The way to figure out the rotational kinetic energy is ${1\over 2}I\omega^2$. Here $I$ is the Earth's moment of inertia, which is about $0.4MR^2$ where $M$ and $R$ are Earth's mass and radius. $\omega$ is the Earth's rotation rate in radians per second -- that is, $2\pi$ over the time for one rotation.
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But are tidal plants slowing down the rotation more than normal tides would without the plants there? – Mark Eichenlaub Mar 5 '11 at 20:12
@Ted Thanks for the energy consumption calculation. Are you saying the earth will stop rotating one day completely? That will happen even without the power plants, right? – daniel.sedlacek Mar 5 '11 at 22:10
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@Mark -- I think they must be in principle. I haven't done the calculation to see which of the two (very small) effects is smaller. – Ted Bunn Mar 5 '11 at 22:36
2
@daniel.sedlacek -- It'll stop rotating relative to the Moon. That is, eventually the rotation rate will be such that the Earth always keeps the same face toward the Moon. The Earth will still rotate with respect to the Sun and the stars, though. The Moon has already done this, by the way -- that's why it always keeps the same face towards the Earth. This sort of "tidal locking" turns out to be pretty common in the solar system: quite a few moons have done it. (Mercury has done something similar, but in kind of a weird way.) – Ted Bunn Mar 5 '11 at 22:39
1
Also, note that the full system (earth and moon ignoring the sun for now) conserve angular momentum. The loss of angular momentum by the earth is counteracted by the increase in angular momentum represented by the lunar orbit. The oribital energy of the moon is increased and that term must be deducted from the loss of rotational kinetic energy of the earth before determining how much energy must be dissipated by a unit transference of angular momentum between the two bodies. – Omega Centauri Mar 6 '11 at 2:01
show 7 more comments
I wonder if hydroelectric or fossil fuel plants slow rotation more?
Hydroelectric concentrates large masses of water at slightly high elevations, while fossil fuel puts much larger masses of CO2 + H2O much higher in the atmosphere
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"Hydroelectric power" does not concentrate anything, sun sea evaporation, clouds rise the water, rain and rivers bring it down. In case of an artificial hydroelectric lake in the mountains, the water is temporarily "hold up" that is all. If the same lake would be created by a landslide, this were the same. – Georg Mar 18 '11 at 19:38
I'm not 'blaming' hydroelectric power but there is a measurable difference in earth rotation in the winter as snow accumulates in the northern hemisphere. Something like a pumped storage scheme must have a tiny effect on rotation – Martin Beckett Mar 18 '11 at 20:17
But that this effects are only temporarily. When the snow/water is down, everything is as before. – Georg Mar 18 '11 at 21:44
I am a bit puzzled by the answers.
The earth moon system has a specific distance at a given moment and the gravitational force on the earth a specific magnitude that gives a "spin" to the earth. Does the composition of the earth play a role to the amount of the spin? i.e. water versus land?
I may be completely wrong but the bulge of the moon goes at the same rate over land and sea, irrespective of friction. I would expect the angular momentum transfer to exist even if the earth were rock solid and no bulge possible, in the same magnitude, the same way the moon's angular momentum changes by the interaction with the earth's gravitation.
Thus my answer is that the problem with the power plants would be a matter of earth angular momentum conservation that will not be transferred to the moon, and thus the power plants would make no difference because angular momentum of the earth will be conserved whether they are there or not.
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For the Earth to transfer angular momentum to the Moon, work has to be done (otherwise the conservation of energy won't be satisfied). So if the Earth was rock solid with no tides, there would be no work done, and the moon could not gain angular momentum. I don't know whether tidal power plants increase the amount of work being done, or just divert it into energy usable by people. I don't understand the microscopic details of how this works, either. – Peter Shor Mar 6 '11 at 11:22
@Peter Shor Could not the energy go to an increase of the spin of the earth? Like a giant top? I suppose that the asymmetry comes because the earth is not a point gravitational mass but is a sphere, which will be interacting gravitationally incrementally to the distance from the moon. In any case the only interaction between the moon and the earth is gravitational and I do not see how , lets say continental drifts as an example, will affect the moon as long as the gravitational mass does not change. – anna v Mar 6 '11 at 11:59
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If you look at the Earth-moon system, there are two parameters, the speed of the Earth's rotation, and the distance to the moon. There are also two conserved quantities: energy and angular momentum. If you want to keep both energy and angular momentum fixed, you can't change the parameters. Currently, the energy of the Earth-moon system is decreasing because it's going into powering the tides, which is only possible because the Earth is not quite a rigid body. – Peter Shor Mar 6 '11 at 12:23
@Anna, of course the "rock solid" earth dissipates angular momentum too! An example is the moon, whose momentum was consumed eons ago. Any "earth" which is not an ideally elastic body will "crunch" angular momentum into heat. Maybe this is far more than the energy used up in the oceans? I dont know. – Georg Mar 6 '11 at 12:32
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Some misconceptions above. Angular momentum cannot be consumed, only transfered in this case between rotational and orbital. If both earth and the moon were perfectly rigid, there would be no angular momentum transfer. Likewise if both bodies were perfectly elastic (they don't dissapate mechanical energy) there also couldn't be angular momentum transfer. So one (or both) bodies must be both defomable, and inelastic (at least some mechnical energy is transformed to heat) for the angular momentum to be transferred. – Omega Centauri Mar 6 '11 at 15:47
show 4 more comments
In order to slow down the rotation of a body angular momentum must transferred off that body. In the case of Earth and the moon this occurs from the difference in gravity across the Earth, or tidal force. A tidal power generating system simply converts a tiny fraction of energy in the tidal bulge of the Earth, mostly in the oceans, as it moves around the globe into mechanical or electrical power. The question is whether that induces a torque on the Earth.
These systems might serve to reduce the tidal bulge of the oceans a very tiny amount. So from the systems perspective the flow of water is impeded, the effective viscosity increased, friction increased and the tidal bulge reduced. This represents a tiny amount of energy reduced on the Earth in this form. However, this is a near infinitesimal amount of the Earth’s rotational kinetic energy.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "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}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9293119311332703, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/161749/prove-that-int-01-fracx4-log-xx2-1-le-frac18 | # Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$
Here is another interesting integral inequality :
$$\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$$
According to W|A the difference between RS and LS is extremely small, namely 0.00241056. I don't know what would work here since the difference is so small.
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If there will be badges for those who asks interesting question, surely you will get it :) – Norbert Jun 22 '12 at 18:38
Thank you all for the solutions you gave! (+1) for all – Chris's wise sister Jun 22 '12 at 19:59
@Norbert: well, it's enough for me if people here enjoy my questions. :-) – Chris's wise sister Jun 22 '12 at 20:02
## 4 Answers
You can actually just evaluate the integral explicitly. You can divide $x^2 -1$ into $x^4$ and get $$\frac{x^4}{x^2 - 1} = x^2 + 1 + \frac {1}{x^2 - 1}$$ So the integral is the same as $$\int_0^1 (x^2 + 1)\log(x)\,dx + \int_0^1 \frac{\log(x)}{x^2 - 1}\,dx$$ The second integral is related to the famous dilogarithm integral, and as explained in Peter Tamaroff's answer can be evaluated to $\frac{\pi^2}{8}$. For the first term, just integrate by parts; you get $$({x^3 \over 3} + x)\log(x)\big|_{x = 0}^{x =1} - \int_0^1 ({x^2 \over 3} + 1)\,dx$$ The first term vanishes, while the second term is $-{10 \over 9}$. So the answer is just ${\pi^2 \over 8} - {10 \over 9}$ which is less than ${1 \over 8}$.
A way of doing the whole integral in one fell swoop occurs to me. Note that ${\displaystyle {1 \over 1 - x^2} = \sum_{n=0}^{\infty} x^{2n}}$. So the integral is $$-\sum_{n = 0}^{\infty} \int_0^1 x^{2n + 4}\log(x)\,dx$$ $$= -\sum_{m = 2}^{\infty} \int_0^1 x^{2m}\log(x)\,dx$$ Integrating this by parts this becomes $$\sum_{m = 2}^{\infty} \int_0^1 {x^{2m} \over 2m + 1}$$ $$= \sum_{m = 2}^{\infty} {1 \over (2m + 1)^2}$$ This is the sum of the reciprocals of the odd squares starting with $5$. The sum of the reciprocals of all odd squares is ${\pi^2 \over 8}$, so one subtracts off $1 + {1 \over 9} = {10 \over 9}$. Hence the result is ${\pi^2 \over 8} - {10 \over 9}$.
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Haha, I just went straight to an inequality! Silly me! – Peter Tamaroff Jun 22 '12 at 19:52
@Zarrax: that's a really interesting approach! – Chris's wise sister Jun 22 '12 at 20:01
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Practically this solution uses @PeterTamaroff work, without it you could as well write right away $\frac{\pi^2}{8} - \frac{10}{9} \le \frac{1}{8}$ :/ – qoqosz Jun 22 '12 at 20:07
@qoqosz Of course what he did there was the bulk of the effort... the point is, once you have that you can just do the original integral. – Zarrax Jun 22 '12 at 20:12
$+\left\lceil \frac 12 \right\rceil$ for the second part. – Sam Jun 28 '12 at 21:34
Representing natural logarithm as an integral and changing the order of integration we obtain: $$\ldots = \int_0^1 \frac{x^4}{x^2 - 1} \, dx \int_1^x \frac{dt}{t} = \int_0^1 \frac{dt}{t} \int_0^t \frac{x^4}{x^2 - 1} \, dx \\= \int_0^1 \frac{t + \frac{1}{3} t^3 - \tanh^{-1} t}{t}\, dt = \frac{10}{9} - \int_0^1 \frac{\tanh^{-1} t}{t} \, dt$$ So we now want to show that: $$\frac{71}{72} = \frac{10}{9} - \frac{1}{8} \le \int_0^1 \frac{\tanh^{-1} t}{t} \, dt$$ Using Maclauirn series we have: $$\frac{\tanh^{-1}}{t} = 1 + \frac{t^2}{3} + \frac{t^4}{5} + \ldots$$ Integrating first three terms yields: $\frac{259}{225} > \frac{71}{72}$.
Actually it turns out that $\frac{\tanh^{-1} t}{t} \approx 1$ is enough.
Added
Alternatively we can write: $$\int_0^1 \frac{\tanh^{-1} t}{t} dt = \int_0^1 \frac{dt}{t} \int_0^t \frac{dx}{1-x^2} \ge \int_0^1 \frac{dt}{t} \int_0^t dx = \int_0^1 dt = 1 \ge \frac{71}{72}$$
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Maybe this is easier:
$$\eqalign{ & \int\limits_0^1 {\frac{{{x^4}\log x}}{{{x^2} - 1}}dx} = \int\limits_0^1 {\frac{{\left( {{x^4} - 1} \right)\log x}}{{{x^2} - 1}}dx} + \int\limits_0^1 {\frac{{\log x}}{{{x^2} - 1}}dx} \cr & = \int\limits_0^1 {\left( {{x^2} + 1} \right)\log xdx} + \int\limits_0^1 {\frac{{\log x}}{{{x^2} - 1}}dx} \cr}$$
$$= \int\limits_0^1 {\left( {{x^2} + 1} \right)\log xdx} = \left[ {\left( {\frac{{{x^3}}}{3} + x} \right)\log x} \right]_0^1 - \int\limits_0^1 {\left( {\frac{{{x^2}}}{3} + 1} \right)dx} = - \frac{{10}}{9}$$
$$\int_0^1 \frac{\log x}{x^2-1}dx=\frac 1 2 \int_0^1 \frac{\log x}{x-1}dx-\frac 1 2 \int_0^1 \frac{\log x}{x+1}dx$$
Now those two evaluate in terms of $\pi$, since we get
$$-\int_0^1 \frac{\log x}{1-x}dx=-\int_0^1 \log x \sum_{k=0}^\infty x^kdx=$$
$$=-\sum_{k=0}^\infty\int_0^1 x^k\log x dx=\sum_{k=0}\frac{1}{(k+1)^2}=\frac{\pi^2}{6}$$
Since
$$\int_0^1 x^k\log x dx=-\frac{1}{(k+1)^2}$$
And for the other, we get the similar:
$$\int_0^1 \frac{\log x}{1+x}dx=\int_0^1 \log x \sum_{k=0}^\infty (-1)^kx^kdx=$$
$$=\sum_{k=0}^\infty\int_0^1 \log x (-1)^kx^kdx=\sum_{k=0}^\infty(-1)^k\int_0^1 x^k\log x dx=\sum_{k=0}\frac{(-1)^{k+1}}{(k+1)^2}=-\frac{\pi^2}{12}$$
So we have that
$$\int_0^1 \frac{\log x}{x^2-1}dx=\frac 1 2 \left( \frac{\pi^2}{6}+\frac{\pi^2}{12} \right)$$
$$\int_0^1 \frac{\log x}{x^2-1}dx=\frac{\pi^2}{8}$$
and finally
$$I=\frac{\pi^2}{8}-\frac{10}{9}\approx 0.1225 < 0.125$$
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it's a bit longer proof, but it's a detailed answer. That's a good point! Thanks. – Chris's wise sister Jun 22 '12 at 20:20
You can do the $\int_0^1 {\log(x) \over x^2 - 1}$ integral the same way you did the two integrals there, expand ${1 \over 1 - x^2} = \sum_n x^{2n}$. You end out with the sum of the reciprocals of the odd squares, which gives $\frac {\pi^2}{8}$. – Zarrax Jun 22 '12 at 20:51
@Zarrax Hmhm. True. – Peter Tamaroff Jun 22 '12 at 21:07
@Peter Tamaroff: i will delete the above posts. ;) – Chris's wise sister Jun 28 '12 at 21:19
According to Maple, your integral is $\dfrac{\pi^2}{8} - \dfrac{10}{9}$, so your inequality becomes $\pi < \sqrt{89}/{3}$. In fact, an antiderivative is $$F(x) = \dfrac{x^3 \ln(x)}{3} - \dfrac{x^3}{9} + x \ln(x) - x - \dfrac{\ln(x) \ln(x+1)}{2} - \dfrac{\text{dilog}(x)+\text{dilog}(x+1)}{2}$$
More generally, for $p > -1$ $$\int_0^1 \dfrac{x^p \ln(x)}{x^2-1}\ dx = \dfrac{\Psi(1,(p+1)/2)}{4}$$ where for even integers $p=2n$, $$\Psi(1,n+1/2) = \sum_{k=n+1}^\infty \dfrac{4}{(2k-1)^2}$$ while for odd integers $p=2n-1$, $$\Psi(1,n) = \sum_{k=n}^{\infty} \dfrac{1}{k^2}$$
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hmmm. That's interesting. – Chris's wise sister Jun 22 '12 at 18:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 27, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9353296160697937, "perplexity_flag": "middle"} |
http://nrich.maths.org/8292 | ### Cubes
Investigate the number of faces you can see when you arrange three cubes in different ways.
### I'm Eight
Find a great variety of ways of asking questions which make 8.
### Let's Investigate Triangles
Vincent and Tara are making triangles with the class construction set. They have a pile of strips of different lengths. How many different triangles can they make?
# Doing and Undoing
##### Stage: 1 Challenge Level:
Maybe you are used to making knots and sometimes you undo them!
Let's look at undoing some maths that's been done.
Suppose we have a starting number and then we doubled it.
That's the bit we'll call "doing".
To "undo" the maths we start with just the answer and see if we can get back to our starting number.
When I doubled my starting number, I got $6$. What do I have to do, to "undo" the $6$ and get back to my starting number?
Suppose that I did it again with a new starting number so I doubled and got to $16$. What would you have to do to "undo" and get back to my new starting number?
Can you think what you would have to do to "undo" these three children's maths?
Danesh says
"I took $4$ away, what should I do to get back to my starting number?"
Meg says
" I added $8$, what should I do to get back to my starting number?"
Chris says
"I halved, what should I do to get back to my starting number?"
Now, if they all finished with a $12$ what were their starting numbers?
Photograph acknowledgements;
http://www.instructables.com/image/FPRA3T8FZ8J4A39/How-to-tie-various
-knots.jpg
http://2.bp.blogspot.com/-Jg0te1K_F5g/TpFcEY8xT6I/AAAAAAAAHy4/vgkke
-0QQ98/s1600/knot.jpg
http://www.instructables.com/image/F0PMDJ8FZ8J4A36/Figure-8-Knot.jpg
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9402989149093628, "perplexity_flag": "middle"} |
http://stats.stackexchange.com/questions/24203/is-it-ever-good-to-increase-significance-level/24208 | # Is it ever good to increase significance level?
Suppose you are using $\alpha = 0.01$ as the significance level. IF you are not getting significant results, is it better to increase the significance level to $\alpha = 0.05$?
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Absolutely. And if your results aren't significant at 0.05, you can always go to 0.10 and even 0.20: whatever it takes to get the answer you need, right? :-) – whuber♦ Mar 6 '12 at 20:16
4
A farmer had lots of targets painted on the side of his barn with bullet holes perfectly centered in the bullseye. As he explained, "I shoot first and paint the target afterwards." – Dilip Sarwate Mar 6 '12 at 20:33
1
– chl♦ Mar 6 '12 at 23:36
Like it or not, this practice is widespread and even if you know its wrong, you're often pressured to do it. People want validation of their hunches and will tailor methods to do so; asides from refusing (and getting replaced) how do you stand up to that? – user4673 Mar 7 '12 at 18:31
See if you can find a good reference for the culture of "Null Hypothesis Significance Testing" - There isn't many. In fact you will find that there are many more articles which criticise this practise. I'm pretty sure its a blend of Neyman-Pearson hypothesis testing with Fishers p-values. I'm also pretty sure niether said their method should be used this way. – probabilityislogic Oct 7 '12 at 17:14
## 3 Answers
Expanding on @EpiGrad 's answer (which is a good answer):
1. There are many reasons to ignore p-values altogether: Principally, they answer a question we are very rarely interested in.
2. If you are going to use p-values, using them as cutoffs often makes little sense.
3. If you are going to use them as cut-offs, you should decide on the cutoff before the analysis
4. Making a more stringent cutoff for type I error means lower power (more type II error). Typical values are .05 for type I and .20 for type II (power = .8). But there is no reason why type II errors are necessarily less bad than type I errors. Suppose you develop a drug that treats a disease that is terminal and rapidly so (e.g. something like Ebola). You test it.
Type I error - you say the drug does something when it doesn't, and then give a useless drug to dying people.
Type II error - you say the drug does nothing when it does something, and you fail to give a beneficial drug to dying people.
Which is worse? Type II, by my book.
To quote Prof. David Cox
There are no routine statistical questions, only questionable statistical routines
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Except, the type I scenario could also end up in significant patient harm if the patient doesn't have a terminal condition and the (falsely positive) treatment has significant side effects. Both type I and type II are bad, depending on the context. – pmgjones Mar 6 '12 at 20:43
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@pmgjones: Yes, I believe that's the point that Peter Flom is making. He wrote that "there is no reason why type II errors are necessarily less bad than type I errors" (emphasis mine), then gave one example situation where he would consider type II errors to be worse. – ruakh Mar 6 '12 at 22:06
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True - I had missed that part of his message. Sorry Peter! – pmgjones Mar 6 '12 at 22:17
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+1 for many reasons to ignore them completely. If I could give it another +1, I would for the emphasis on Type II errors still being errors. – EpiGrad Mar 7 '12 at 0:14
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This is reminiscence of the trade off between Precision and Recall. – Itamar Mar 12 '12 at 19:16
show 1 more comment
Just if you're not getting significant results? No. Fiddling with the significance level after the experiment is conducted and the results are known is never good practice.
There are circumstances where you might chose a more relaxed p-value, but doing it post hoc is just a bad idea.
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+1 for the restraint... – gung Mar 6 '12 at 20:14
On one hand it is somewhat artificial to discount a variable because its p value is higher than 0.01 (that's an unusually stringent criteria). How you get there may be more important than what is the ultimate significance level. A variable that is well grounded on logic or causal links with an acceptable p value may be much more meaningful than a variable with a lower p balue but with no meaningful logic supporting it.
If you are dealing with hypothesis testing watch out that the statistical significance is in good part just a function of your sample size. A large sample size will translate into a low standard error and higher statistical significance. And, this process is somewhat artificial as large samples will render immaterial differences statistically significant. If you are dealing within such a domain I recommend you move towards an Effect Size method where the unit of statistical distance is not Standard Error but instead Standard Deviation. And, the latter can't be manipulated by sample size.
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0.01 is only small by conventional standards. If you use BIC to assess significance, the equivalent cutoff can be much smaller. For example a sample size of 10,000 means using a p-value of something like 0.001. – probabilityislogic Oct 7 '12 at 17:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9459157586097717, "perplexity_flag": "middle"} |
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# The Fundamental Theorem of Calculus
Course Topic(s): One-Variable Calculus | Fundamental Theorem
This web page includes a brief statement of the second form of the fundamental theorem of calculus and a focused Java applet that allows an instructor to demonstrate an example of an accumulation or area function, where the area under the graph of $$f(x) = \sin 2x$$ (from $$0$$ to $$4\pi$$) is graphed on a second plot. It is a nice example, since you can see the relative extrema of the accumulation function occuring where the function $$f(x)$$ is zero, etc.
Resource URL: http://www.dean.usma.edu/math/research/mathtech/java/FTCProject/ftcprojectpage.html
### To rate this resource on a 1-5 scheme, click on the appropriate icosahedron:
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Subject classification(s): Integration | Single Variable Calculus | Calculus
Creator(s): Garrett Heath
Contributor(s): Garrett Heath
This resource was cataloged by Paul Seeburger
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United States Military Academy West Point
This review was published on March 05, 2011
### Comments
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http://math.stackexchange.com/questions/160328/what-is-the-usefulness-of-matrices/160561 | # What is the usefulness of matrices?
I have matrices for my syllabus but I don't know where do they find their use. I even asked my teacher but she also has no answer. Can anyone please tell me where are they used? And please also give me an example of how they are used?
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A matrix is a way to express a linear map between finite-dimensional vector spaces, given a choice of bases for said spaces. Since linear maps are useful, matrices are obviously useful. – gspr Jun 19 '12 at 14:19
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Also, if the teacher who couldn't answer your question is the same one who is teaching you about matrices, I would be very skeptical of her and the course. – gspr Jun 19 '12 at 14:21
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Services like Netflix store information in matrices. When it is deciding what movies to recommend to you, it compares your movie matrix with ones 'similar' to yours, and recommends movies that other users with your preferences also enjoyed. Determining 'orthoganality' is one of the topics you will likely cover. – Joshua Shane Liberman Jun 19 '12 at 16:18
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Matrices were invented to have an endless supply of homework for the poor students. – Peter Sheldrick Jun 19 '12 at 19:21
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Pretty much everything in computer graphics uses matrices. Without matrices, there'd be no video games :) – BlueRaja - Danny Pflughoeft Jun 19 '12 at 21:12
show 3 more comments
## 11 Answers
I work in the field of applied math, so I will give you the point of view of an applied mathematician.
I do numerical PDEs. Basically, I take a differential equation (an equation whose solution is not a number, but a function, and that involves the functions and its derivatives) and, instead of finding an analytical solution, I try to find an approximation of the value of the solution at some points (think of a grid of points). It's a bit more deep than this, but it's not the point here. The point is that eventually I find myself having to solve a linear system of equations which usually is of huge size (order of millions). It is a pretty huge number of equations to solve, I would say.
Where do matrices come into play? Well, as you know (or maybe not, I don't know) a linear system can be seen in matrix-vector form as
$$\text{A}\underline{x}=\underline{b}$$ where $\underline{x}$ contains the unknowns, A the coefficients of the equations and $\underline{b}$ contains the values of the right hand sides of the equations. For instance for the system
$$\begin{cases}2x_1+x_2=3\\x_1-x_2=1\end{cases}$$ we have
$$\text{A}=\left[ \begin{array}{cc} 2 & 1\\ 1 & -1 \end{array} \right],\qquad \underline{x}= \left[\begin{array}{c} x_1\\ x_2 \end{array} \right]\qquad \underline{b}= \left[\begin{array}{c} 3\\ 1 \end{array} \right]$$
For what I said so far, in this context matrices look just like a fancy and compact way to write down a system of equations, mere tables of numbers.
However, in order to solve this system fast is not enough to use a calculator with a big RAM and/or a high clock rate (CPU). Of course, the more powerful the calculator is, the faster you will get the solution. But sometimes, faster might still mean days (or more) if you tackle the problem in the wrong way, even if you are on a Blue Gene.
So, to reduce computational cost, you have to come up with a good algorithm, a smart idea. But in order to do so, you need to exploit some property or some structure of your linear system. These properties are encoded somehow in the coefficients of the matrix A. Therefore, studying matrices and their properties is of crucial importance in trying to improve linear solvers efficiency. Recognizing that the matrix enjoys a particular property might be crucial to develop a fast algorithm or even to prove that a solution exists, or that the solution has some nice property.
For instance, consider the linear system
$$\left[\begin{array}{cccc} 2 & -1 & 0 & 0\\ -1 & 2 & -1 & 0\\ 0 & -1 & 2 & -1\\ 0 & 0 & -1 & 2 \end{array} \right] \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ x_4 \end{array} \right]= \left[ \begin{array}{c} 1\\ 1\\ 1\\ 1 \end{array} \right]$$ which corresponds (in equation form) to
$$\begin{cases} 2x_1-x_2=1\\ -x_1+2x_2-x_3=1\\ -x_2+2x_3-x_4=1\\ -x_3+2x_4=1 \end{cases}$$
Just giving a quick look to the matrix, I can claim that this system has a solution and, moreover, the solution is non-negative (meaning that all the components of the solution are non-negative). I'm pretty sure you wouldn't be able to draw this conclusion just looking at the system without trying to solve it. I can also claim that to solve this system you need only 25 operations (one operation being a single addition/subtraction/division/multiplication). If you construct a larger system with the same pattern (2 on the diagonal, -1 on the upper and lower diagonal) and put a right hand side with only positive entries, I can still claim that the solution exists and it's positive and the number of operations needed to solve it is only $8n-7$, where $n$ is the size of the system.
Moreover, people already pointed out other fields where matrices are important bricks and plays an important role. I hope this thread gave you an idea of why it is worth it to study matrices. =)
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Matrices are a useful way to represent, manipulate and study linear maps between finite dimensional vector spaces (if you have chosen basis).
Matrices can also represent quadratic forms (it's useful, for example, in analysis to study hessian matrices, which help us to study the behavior of critical points).
So, it's a useful tool of linear algebra.
Moreover, linear algebra is a crucial tool in math.
To convince yourself, there are a lot of linear problems you can study with little knowledge in math. For examples, system of linear equations, some error-correcting codes (linear codes), linear differential equations, linear recurrence sequences...
I also think that linear algebra is a natural framework of quantum mechanics.
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"Linear algebra is useful" is a bit of an understatement. To the extent of my knowledge, linear algebra is a crucial tool in literally every branch of science, engineering, and mathematics. – Qiaochu Yuan Jun 19 '12 at 14:36
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Indeed, but I'm not good enough in English to clarify some nuances. I edited the answer with your words ;-). – JBC Jun 19 '12 at 14:43
Graph Theory --loosely, the study of connect-the-dot figures-- uses matrices to encode adjacency and incidence structures. More than simply bookkeeping, however, the matrices have computational uses. From powers of the adjacency matrix, for a simple example, one can read the number of available paths between any two dots.
"Spectral" Graph Theory derives graph-theoretical information from matrix-theoretical results (specifically, "eigenvalues" and "eigenvectors" --by the way, the set of eigenvalues is the "spectrum" of a matrix, hence "spectral"-- which come from the linear map interpretation of matrices). My own work generates coordinates for "symmetric" geometric realizations of graphs --think Platonic and Archimedean solids-- from this kind of analysis of their adjacency matrices.
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Matrices are a useful tool for studying finite groups. Every finite group has a representation as a set of invertible matrices; the study of such representations is called, well, Representation Theory.
One of the major theorems of all time in finite group theory is the classification of all finite simple groups. These are the building blocks of group theory, the group-theoretic version of prime numbers. The "proof" took scores of mathematicians many decades, and could not have been completed without viewing these groups as groups of matrices. One just has to open the ATLAS of Finite Groups, or wonder what a group of Lie type is, to get my point!
(Of course, linear algebra is exceptionally useful etc. etc. but that is a topic better covered by an engineer...)
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@PranitBauva: I do not think my answer should be the accepted answer. It was intended to illustrate a different use for matrices. The usual/main use is in linear algebra. Indeed, this is a rich topic, and can be extended to the study of Lie groups, manifolds, etc. (or functional analysis, if you want to leave matrices behind and go down the route of linear transformations). I am in no way qualified to give an account of this extended subject, but I would advise you to "unaccept" my answer, in the hope that someone comes along and gives an account of this lovely area. – user1729 Jun 19 '12 at 15:59
Matrices can represent Markov Chains. They provide a means for the tabular representation of data. Their utility in mathematics and computing is huge.
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Matrices are used in engineering, physics, computer science, and other applications of mathematics.
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In the most general sense, matrices (and a very important special case of matrices, vectors) provide a way to generalize from single variable equations to equations with arbitrarily many variables. Some of the rules change along the way, hence the importance of learning about matrices - more precisely, learning Linear Algebra, or the algebra of matrices.
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Matrices are used very often in 3D geometry (e.g. computer graphics) and are very powerful. A simple 4x4 matrix can represent a lot of transformations at once (translation, rotation, scaling, perspective/orthogonal projection). You can then multiply a 3D position vector (x, y, z, 1) by this matrix to obtain a new position with all the trasformations applied. Notice that this vector is also a 1x4 matrix (although the position is in 3D, the fourth component is added to make the multiplication possible and allow for the projection transformation, if you want to know more read about homogeneous coordinates). Similar ideas can be used in 2D or even in higher dimensions like 4D.
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I know and use matrices for two things: systems of equations and holding data in programming.
1. As @bartgol said, matrices in math are useful for solving systems of equations. You arrange all the equations in standard form and make a matrix of their coefficients, making sure to use 0s as placeholders (like if there isn't an x term). We call this matrix A. Then make a second matrix of the constants and call it B. It will be one term wide (long). Plug these into your calculator and then evaluate (A^-1)*(B). The resulting matrix, if there is a solution, will solve for each variable. The first row is x, the second y and so on.
2. I've also used two-dimensional arrays in programming (C++, Java) to help store information that just makes sense to be in matrix-form. For example: one program generated magic squares which were stored in a matrix. Another used a 3-by-3 to keep track of spaces on a tic-tac-toe board.
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We are forever indebted to the mathematicians that made matrix algebra possible...
Real world applications of matrices make them extremely important and include the some of following I've had some experience with... (there are so many more!)
Robotics / Kinematics - matrices allow rotations, translations through planes to be easily calculated
Betting - matrices allow complex dutching/betting combinations without separate formulae such as multiple complex simultaneous equations.
Data Mining - Most data mining software use matrices to calculate the algorithms as it's fundamental to this field of mathematics both in the theory and the handling of data.
Graphics/Gaming - Anything from particle collision to ray tracing use matrices
All posts I've read so far have valid uses of matrices and so many more that I couldn't even comprehend...
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I never fully got matrices until I left university. I am glad I understand now.
A good quality camera will save the captured image uncorrected, along with a 3x3 colour correction matrix. Your computer will multiply this with the colour correction matrix of your display, and then by every pixel in the image before putting in on your display. The computer will use a different display matrix for the printer (as it is a different display).
Look at several real world examples. Experiment with colour or 2D/3D transformations, they are fun and visual (if you are a visual person). 2D is easiest and most visual.
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## protected by Qiaochu YuanJun 23 '12 at 21:46
This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9401232600212097, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/268314/how-to-bound-int-pi-pid-nt-dt-from-above?answertab=active | # How to bound $\int^{\pi}_{-\pi}|D_{n}(t)|\,dt$ from above?
Rudin asked me to bound $$\int_{-\pi}^{\pi} |D_{n}(t)|\,dt$$ from above. I need a bound at the level of $o(\log(n))$.
The background is:
If $s_{n}$ is the $n$-th partial sum of the Pourier series of a function $f\in C(T)$, prove that $$\frac{s_{n}}{\log[n]}\rightarrow 0$$ uniformly. That is, prove that $$\lim_{n\rightarrow \infty}\frac{|s_{n}|_{\infty}}{\log(n)}=0$$On the other hand, if $\lambda_{n}/\log[n]\rightarrow 0$, prove that there exist an $f\in C^(T)$ such that sequence $s_{n}(f,0)/\lambda_{n}$ is unbounded.
Update: a numerical evaluation for $n=10^{800}$ is inconclusive.
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Is not the integral of the Dirichlet kernel equal to $2\pi$, according to its representation as a trigonometric sum? Or did you mean the integral of $|D_n|$? – user53153 Dec 31 '12 at 18:55
The Dirichlet kernel. – user32240 Dec 31 '12 at 19:03
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Be sure to rewrite your question to make this clear. As for the estimate, my suggestion is to focus on the intervals where $|\sin (n+1/2)t|\ge 1/2$ and use the bound $|\sin t/2|\le t/2$ there. This way you get a sum of integrals of $\int t^{-1}dt$ over a bunch of intervals, which should lead to the desired bound. – user53153 Dec 31 '12 at 19:27
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– user53153 Dec 31 '12 at 20:17
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OK, this makes things clear. You are referring to Real and Complex Rudin. Exercise 19 does not say that $\|D_n\|_1=o(\log n)$; this particular statement is yours, not Rudin's. And, as noted above, a false one. – user53153 Dec 31 '12 at 20:38
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## 2 Answers
I'll replace $|\sin t/2|$ by $|t|$ since they are comparable: $$\frac{|t|}{\pi}\le \left|\sin\frac{t}{2}\right| \le \frac{|t|}{2} \ \text{ for }t\in [-\pi,\pi]$$ Claim: for all $\lambda\ge 1$ $$\frac{1}{3}\log \lambda \le \int_0^\pi \frac{|\sin \lambda t|}{t}\,dt \le \log \lambda +\log \pi +1.$$
Proof. For the upper bound, split the integral into "small $t$" part and the rest: $$\int_0^{\lambda^{-1}} \frac{|\sin \lambda t|}{t}\,dt \le \int_0^{\lambda^{-1}} \frac{\lambda t}{t}\,dt =1$$ and $$\int_{\lambda^{-1}}^\pi \frac{|\sin \lambda t|}{t}\,dt \le \int_{\lambda^{-1}}^\pi \frac{1}{t}\,dt = \log\lambda + \log \pi$$
The lower bound needs a bit more work. Since the integrand is nonnegative, we can restrict the region of integration to the set $|\sin \lambda t|\ge 1/2$. This set contains the intervals $I_k=[\pi \lambda^{-1} (k+1/6), \pi \lambda^{-1} (k+5/6)]$ for all integers $k$ such that $0\le k \le \lambda-1$. The integral over $I_k$ is at least $$\int_{I_k} \frac{1/2}{t}\,dt \ge |I_k| \frac{1/2}{\pi \lambda^{-1} (k+1)} = \frac{1/3}{k+1}$$ Therefore, the integral is bounded from below by $$\frac13 \sum_{k=0}^{\lfloor \lambda-1\rfloor }\frac{1}{k+1} \ge \frac{1}{3} \log \lambda.$$
Remark. If $\|D_n\|_{L^1}=o(\log n)$ were true, the estimate in #19 would hold for the Fourier series of any finite measure on $[-\pi,\pi]$. This is not the case. The fact that $s_n$ comes from a continuous function should be used.
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This is not really needed, because Rudin already showed this in his book. – user32240 Dec 31 '12 at 23:55
@user32240 You are welcome. – user53153 Dec 31 '12 at 23:56
But thanks for the help nevertheless; I need to think about the problem myself for a while. Since it is in 3rd edition I assume Rudin must had been serious about it. – user32240 Dec 31 '12 at 23:58
Well, you can get the value of the integral very explicitely. As you already noted $$\int_{-\pi}^\pi \frac{\sin[(n+\frac12)t]}{\sin\frac t2}\,dt = \int_{-\pi}^\pi \frac{\sin(nt)}{\sin\frac t2}\cos\frac t2dt = \int_{-\pi/2}^{\pi/2} \frac{\sin(2nt)}{\sin t}2\cos t\,dt\,.$$ Now the fraction can be transformed into $$\frac{\sin(2nt)}{\sin t} = \frac{e^{2nti}-e^{-2nti}}{e^{ti}-e^{-ti}} = \frac{e^{4nti}-1}{e^{2ti}-1} = 1+e^{2ti}+e^{4ti}+\cdots+e^{(4n-2)ti}$$ We multiply this with $2\cos t=e^{-ti}+e^{ti}$ and get for the integrand $$\frac{\sin(2nt)}{\sin t}2\cos t = e^{-ti}+2e^{ti}+2e^{3ti}+\cdots+2e^{(4n-5)ti}+2e^{(4n-3)ti}+e^{(4n-1)ti}.$$ Furthermore, $$\int_{-\pi/2}^{\pi/2} e^{kti}dt = \frac2k\sin\frac{k\pi}2,$$ and hence we finally get $$\int_{-\pi}^\pi \frac{\sin[(n+\frac12)t]}{\sin\frac t2}\,dt =2+4-\frac43+\frac45-\frac47\pm\cdots+\frac{4}{4n-3}-\frac{2}{4n-1}$$ which is obviously convergent by the Leibniz criterion and hence bounded by a constant.
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Hi, thanks for the effort - but I have to integral the absolute value instead. – user32240 Dec 31 '12 at 20:06
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+1 and please don't delete the answer: combining it with one of my comments, one can get a proof of $\pi = 4-4/3+4/5-4/7+\dots$ which is neat by itself. – user53153 Dec 31 '12 at 20:10
True. Nice. Unfortunately, I need to go now. Maybe I'll solve the problem another time. – Ralph Tandetzky Dec 31 '12 at 20:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "mathjax_display_tex": 14, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9512279033660889, "perplexity_flag": "head"} |
http://en.wikipedia.org/wiki/Perturbation_(astronomy) | # Perturbation (astronomy)
In astronomy, perturbation is the complex motion of a massive body subject to forces other than the gravitational attraction of a single other massive body.[1] The other forces can include a third (fourth, fifth, etc.) body, resistance, as from an atmosphere, and the off-center attraction of an oblate or otherwise misshapen body.[2]
The perturbing forces of the Sun on the Moon at two places in its orbit. The blue arrows represent the direction and magnitude of the gravitational force on the Earth. Applying this to both the Earth's and the Moon's position does not disturb the positions relative to each other. When it is subtracted from the force on the Moon (black arrows), what is left is the perturbing force (red arrows) on the Moon relative to the Earth. Because the perturbing force is different in direction and magnitude on opposite sides of the orbit, it produces a change in the shape of the orbit.
## Introduction
The study of perturbations began with the first attempts to predict planetary motions in the sky, although in ancient times the causes remained a mystery. Newton, at the time he formulated his laws of motion and of gravitation, applied them to the first analysis of perturbations,[2] recognizing the complex difficulties of their calculation.[3] Many of the great mathematicians since then have given attention to the various problems involved; throughout the 18th and 19th centuries there was demand for accurate tables of the position of the Moon and planets for purposes of navigation at sea.
The complex motions of gravitational perturbations can be broken down. The hypothetical motion that the body follows under the gravitational effect of one other body only is typically a conic section, and can be readily described with the methods of geometry. This is called a two-body problem, or an unperturbed Keplerian orbit. The differences between that and the actual motion of the body are perturbations due to the additional gravitational effects of the remaining body or bodies. If there is only one other significant body then the perturbed motion is a three-body problem; if there are multiple other bodies it is an n-body problem. Analytical solutions (mathematical expressions to predict the positions and motions at any future time) for the two-body and three-body problems exist; none has been found for the n-body problem except for certain special cases. Even the two-body problem becomes insoluble if one of the bodies is irregular in shape.[4]
Mercury's orbital longitude and latitude, as perturbed by Venus, Jupiter and all of the planets of the Solar System, at intervals of 2.5 days. Mercury would remain centered on the crosshairs if there were no perturbations.
Most systems that involve multiple gravitational attractions present one primary body which is dominant in its effects (for example, a star, in the case of the star and its planet, or a planet, in the case of the planet and its satellite). The gravitational effects of the other bodies can be treated as perturbations of the hypothetical unperturbed motion of the planet or satellite around its primary body.
## Mathematical analysis
### General perturbations
In methods of general perturbations, general differential equations, either of motion or of change in the orbital elements, are solved analytically, usually by series expansions. The result is usually expressed in terms of algebraic and trigonometric functions of the orbital elements of the body in question and the perturbing bodies. This can be applied generally to many different sets of conditions, and is not specific to any particular set of gravitating objects.[5] Historically, general perturbations were investigated first. The classical methods are known as variation of the elements, variation of parameters or variation of the constants of integration. In these methods, it is considered that the body is always moving in a conic section, however the conic section is constantly changing due to the perturbations. If all perturbations were to cease at any particular instant, the body would continue in this (now unchanging) conic section indefinitely; this conic is known as the osculating orbit and its orbital elements at any particular time are what are sought by the methods of general perturbations.[2]
General perturbations takes advantage of the fact that in many problems of celestial mechanics, the two-body orbit changes rather slowly due to the perturbations; the two-body orbit is a good first approximation. General perturbations is applicable only if the perturbing forces are about one order of magnitude smaller, or less, than the gravitational force of the primary body.[4] In the Solar System, this is usually the case; Jupiter, the second largest body, has a mass of about 1/1000 that of the Sun.
General perturbation methods are preferred for some types of problems, as the source of certain observed motions are readily found. This is not necessarily so for special perturbations; the motions would be predicted with similar accuracy, but no information on the configurations of the perturbing bodies (for instance, an orbital resonance) which caused them would be available.[4]
### Special perturbations
In methods of special perturbations, numerical datasets, representing values for the positions, velocities and accelerative forces on the bodies of interest, are made the basis of numerical integration of the differential equations of motion.[6] In effect, the positions and velocities are perturbed directly, and no attempt is made to calculate the curves of the orbits or the orbital elements.[2] Special perturbations can be applied to any problem in celestial mechanics, as it is not limited to cases where the perturbing forces are small.[4] Once applied only to comets and minor planets, special perturbation methods are now the basis of the most accurate machine-generated planetary ephemerides of the great astronomical almanacs.[2][7] Special perturbations are also used for modeling an orbit with computers.
#### Cowell's method
Cowell's method. Forces from all perturbing bodies (black and gray) are summed to form the total force on body i (red), and this is numerically integrated starting from the initial position (the epoch of osculation).
is perhaps the simplest of the special perturbation methods;[8] mathematically, for $n$ mutually interacting bodies, Newtonian forces on body $i$ from the other bodies $j$ are simply summed thus,
$\mathbf{\ddot{r}}_i = \sum_{\underset{j \ne i}{j=1}}^n {Gm_j (\mathbf{r}_j-\mathbf{r}_i) \over r_{ij}^3}$
where $\mathbf{\ddot{r}}_i$ is the acceleration vector of body $i$, $G$ is the gravitational constant, $m_j$ is the mass of body $j$, $\mathbf{r}_i$ and $\mathbf{r}_j$ are the position vectors of objects $i$ and $j$ and $r_{ij}$ is the distance from object $i$ to object $j$, all vectors being referred to the barycenter of the system. This equation is resolved into components in $x$, $y$, $z$ and these are integrated numerically to form the new velocity and position vectors. This process is repeated as many times as necessary. The advantage of Cowell's method is ease of application and programming. A disadvantage is that when perturbations become large in magnitude (as when an object makes a close approach to another) the errors of the method also become large.[9] For many problems in celestial mechanics, this is never the case. Another disadvantage is that in systems with a dominant central body, such as the Sun, it is necessary to carry many significant digits in the arithmetic because of the large difference in the forces of the central body and the perturbing bodies, although with modern computers this is not nearly the limitation it once was.[10]
#### Encke's method
Encke's method. Greatly exaggerated here, the small difference δr (blue) between the osculating, unperturbed orbit (black) and the perturbed orbit (red), is numerically integrated starting from the initial position (the epoch of osculation).
begins with the osculating orbit as a reference and integrates numerically to solve for the variation from the reference as a function of time.[11] Its advantages are that perturbations are generally small in magnitude, so the integration can proceed in larger steps (with resulting lesser errors), and the method is much less affected by extreme perturbations. Its disadvantage is complexity; it cannot be used indefinitely without occasionally updating the osculating orbit and continuing from there, a process known as rectification.[9] Encke's method is similar to the general perturbation method of variation of the elements, except the rectification is performed at discrete intervals rather than continuously.[12]
Letting $\boldsymbol{\rho}$ be the radius vector of the osculating orbit, $\mathbf{r}$ the radius vector of the perturbed orbit, and $\delta \mathbf{r}$ the variation from the osculating orbit,
$\delta \mathbf{r} = \mathbf{r} - \boldsymbol{\rho}$, and the equation of motion of $\delta \mathbf{r}$ is simply
()
$\ddot{\delta \mathbf{r}} = \mathbf{\ddot{r}} - \boldsymbol{\ddot{\rho}}$.
()
$\mathbf{\ddot{r}}$ and $\boldsymbol{\ddot{\rho}}$ are just the equations of motion of $\mathbf{r}$ and $\boldsymbol{\rho}$,
$\mathbf{\ddot{r}} = \mathbf{a}_{\text{per}} - {\mu \over r^3} \mathbf{r}$ for the perturbed orbit and
()
$\boldsymbol{\ddot{\rho}} = - {\mu \over \rho^3} \boldsymbol{\rho}$ for the unperturbed orbit,
()
where $\mu = G(M+m)$ is the gravitational parameter with $M$ and $m$ the masses of the central body and the perturbed body, $\mathbf{a}_{\text{per}}$ is the perturbing acceleration, and $r$ and $\rho$ are the magnitudes of $\mathbf{r}$ and $\boldsymbol{\rho}$.
Substituting from equations (3) and (4) into equation (2),
$\ddot{\delta \mathbf{r}} = \mathbf{a}_{\text{per}} + \mu \left( {\boldsymbol{\rho} \over \rho^3} - {\mathbf{r} \over r^3} \right)$,
()
which, in theory, could be integrated twice to find $\delta \mathbf{r}$. Since the osculating orbit is easily calculated by two-body methods, $\boldsymbol{\rho}$ and $\delta \mathbf{r}$ are accounted for and $\mathbf{r}$ can be solved. In practice, the quantity in the brackets, ${\boldsymbol{\rho} \over \rho^3} - {\mathbf{r} \over r^3}$, is the difference of two nearly equal vectors, and further manipulation is necessary to avoid the need for extra significant digits.[13][14] Encke's method was more widely used before the advent of modern computers, when much orbit computation was performed on mechanical calculating machines.
## Periodic nature
Gravity Simulator plot of the changing orbital eccentricity of Mercury, Venus, Earth, and Mars over the next 50,000 years. The 0 point on this plot is the year 2007.
In the Solar System, many of the disturbances of one planet by another are periodic, consisting of small impulses each time a planet passes another in its orbit. This causes the bodies to follow motions that are periodic or quasi-periodic – such as the Moon in its strongly perturbed orbit, which is the subject of lunar theory. This periodic nature led to the discovery of Neptune in 1846 as a result of its perturbations of the orbit of Uranus.
On-going mutual perturbations of the planets cause long-term quasi-periodic variations in their orbital elements, most apparent when two planets' orbital periods are nearly in sync. For instance, five orbits of Jupiter (59.31 years) is nearly equal to two of Saturn (58.91 years). This causes large perturbations of both, with a period of 918 years, the time required for the small difference in their positions at conjunction to make one complete circle, first discovered by Laplace.[2] Venus currently has the orbit with the least eccentricity, i.e. it is the closest to circular, of all the planetary orbits. In 25,000 years' time, Earth will have a more circular (less eccentric) orbit than Venus. It has been shown that long-term periodic disturbances within the Solar System can become chaotic over very long time scales; under some circumstances one or more planets can cross the orbit of another, leading to collisions.[15]
The orbits of many of the minor bodies of the Solar System, such as comets, are often heavily perturbed, particularly by the gravitational fields of the gas giants. While many of these perturbations are periodic, others are not, and these in particular may represent aspects of chaotic motion. For example, in April 1996, Jupiter's gravitational influence caused the period of Comet Hale–Bopp's orbit to decrease from 4,206 to 2,380 years, a change that will not revert on any periodic basis.[16]
## See also
• Nereid one of the outer moons of Neptune with a high orbital eccentricity of ~0.75 and is frequently perturbed
• Osculating orbit
• Orbital resonance
• Stability of the Solar System
• Formation and evolution of the Solar System
• Orbit Modeling | {"extraction_info": {"found_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": 45, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9354372620582581, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/289622/limit-of-sum-variable-in-summand-and-limit?answertab=votes | # Limit of sum (variable in summand and limit)
Can someone help evaluate this limit: $$\lim_{n\rightarrow \infty} \sum_{k=0}^n\frac{k^k(n-k)^{n-k}}{k!(n-k)!}n!n^{-n-1/2}$$
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What have you tried? And if this is homework, please add the `homework` tag. – Stefan Hansen Jan 29 at 10:03
## 3 Answers
Stirling's approximation gives us that $$n!=\sqrt{2\pi n}\ n^n\,e^{-n}(1+O(1/n))\tag{1}$$ As it is simply the $j=k$ term in $$\sum_{j=0}^n\binom{n}{j}\left(\frac{k}{n}\right)^{j}\left(1-\frac{k}{n}\right)^{n-j}=1\tag{2}$$ we can deduce that $$\binom{n}{k}\left(\frac{k}{n}\right)^{k}\left(1-\frac{k}{n}\right)^{n-k}\le1\tag{3}$$ Therefore, $$\begin{align} &\sum_{k=0}^n\frac{k^k(n-k)^{n-k}}{k!(n-k)!}n!n^{-n-1/2}\\ &=\sum_{k< m}\binom{n}{k}\left(\frac{k}{n}\right)^{k}\left(1-\frac{k}{n}\right)^{n-k}n^{-1/2}\\ &+\sum_{k=m}^{n-m}\binom{n}{k}\left(\frac{k}{n}\right)^{k}\left(1-\frac{k}{n}\right)^{n-k}n^{-1/2}\\ &+\sum_{k>n-m}\binom{n}{k}\left(\frac{k}{n}\right)^{k}\left(1-\frac{k}{n}\right)^{n-k}n^{-1/2}\\ &=O(mn^{-1/2})+\frac1{\sqrt{2\pi}}\sum_{k=m}^{n-m}\frac1{\sqrt{k}\sqrt{n-k}}(1+O(1/m))+O(mn^{-1/2})\\ &=O(mn^{-1/2})+\frac1{\sqrt{2\pi}}\sum_{k=0}^n\frac1{\sqrt{k/n}\sqrt{1-k/n}}\frac1n(1+O(1/m))\tag{4} \end{align}$$ We can add the tails to the last sum in $(4)$ since each of the $2m$ terms is $\le n^{-1/2}$ and absorb the difference into $O(mn^{-1/2})$.
Approximating with Riemann sums, $(4)$ becomes $$\lim_{n\to\infty}\sum_{k=0}^n\frac{k^k(n-k)^{n-k}}{k!(n-k)!}n!n^{-n-1/2} =\frac1{\sqrt{2\pi}}\int_0^1\frac{\mathrm{d}x}{\sqrt{x-x^2}}(1+O(1/m))\tag{5}$$ Since $m$ was arbitrary, we get $$\begin{align} \lim_{n\to\infty}\sum_{k=0}^n\frac{k^k(n-k)^{n-k}}{k!(n-k)!}n!n^{-n-1/2} &=\frac1{\sqrt{2\pi}}\int_0^1\frac{\mathrm{d}x}{\sqrt{x-x^2}}\\ &=\frac1{\sqrt{2\pi}}\int_{-1}^1\frac{\mathrm{d}x}{\sqrt{1-x^2}}\\ &=\sqrt{\frac\pi2}\tag{6} \end{align}$$
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Stirling's approximation shows that the limit can be rewritten $$\lim_{n\rightarrow \infty} \sum_{k=0}^n\frac{k^k(n-k)^{n-k}}{k!(n-k)!}\sqrt{2\pi}e^{-n}$$, but I don't know if that makes it easier.
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No, it cant be written like that based on what you know, the limit of the product of two expressions only equals the product of each individual limit, if both limits exist. – Ethan Jan 29 at 11:05
Are you sure the binomial theorem works? Are you suggesting $$\sum_{k=0}^n \frac{n!}{k!(n-k)!}k^k(n-k)^{n-k} = (k+n-k)^n$$? $k$ is the dummy variable used in the sum though; it is not constant.
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It doesn't sorry – Ethan Jan 29 at 11:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9318613409996033, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/104450/road-to-solovays-land/104462 | ## Road to Solovay’s Land.
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In the first semester of 2012 I took a course in General Topology and Set Theory, at undergraduate level. For topology, I was instructed to use Engelking's General Topology; albeit I had a great difficult to approach it, I got used to the text and did (and I'm still doing) some exercises (but none of the problems until now). For the Set Theory course we used Jech & Hrbacek's Introduction to Set Theory, which I think was suitable for my level back there. In these courses, I heard about Boolean Algebras, Forcing, Independence Proofs, Models, Topological Games, Kunen's book (which I just bought a copy), and others interesting things that caused me to change my favorite mathematical area (in fact I was a physics undergrad student when this year began).
In this semester, I enrolled myself in Measure and Integration course, also at undergraduate level, where I discovered about Solovay's model, which completely drove me to madness.
I'm looking for advice about my background and the path that I have to follow to reach these mentioned topics; is too early to begin? do my background is sufficiently enough to start? And where to begin with ? what books do I have to read?
P.S.: I had no background in mathematical logic, the only thing I can do is some proofs with truth-tables.
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## 2 Answers
With no background in logic, it's a bit of a long road to Solovay's model; the good news is that every step of it is incredibly interesting!
(Some of this you may already know - I'm just listing a complete roadmap to Solovay.)
To start with, you need a good understanding of what models of ZFC look like. The last couple chapters of Hrbacek/Jech cover this; alternatively, it's at the beginning of Kunen's book.
Then comes forcing. Forcing is basically a way of building models of ZFC "to specification." This is a big deal, since ZFC is a really complicated theory, unlike, say, the theory of rings: while it's very easy to build lots and lots of (models of the theory of) rings, it's incredibly hard to build models of ZFC, and forcing accomplishes this.
The picture of forcing in ZFC is reasonably straightforward (although the details, of course, take a lot of work): you take a model V of ZFC to start with, look at some poset P in V, and the machinery of forcing gives you a* model V[G] containing V with properties that can be discovered reasonably easily by looking at P, and conversely, there are natural strategies for building a P such that the resulting V[G] will have properties you want it to. Playing around with Martin's Axiom might make forcing make a lot more sense; it certainly did for me!
(*OK, actually forcing gives you lots of different models, one for each "generic filter" G of P over V, but for almost all intents and purposes the precise generic filter doesn't matter, and all the information is contained in the poset P alone.)
Now we can prove lots of nice properties about forcing over models of ZFC, including one which for our purposes is actually a bad property: any V[G] is also a model of ZFC. The reason this is bad for us is that Solovay's model is definitely not a model of Choice, so we have to add another layer of complexity: the symmetric submodel construction. By doing some complicated shenanigans** with automorphisms of P, we can build intermediate models W of ZF set theory, containing V and contained in V[G]. Solovay's model is built in this fashion.
(**Specifically, elements of the extension V[G] have "names" in V; the symmetric submodel construction is a way of defining "hereditarily symmetric" names, which are basically names fixed by "a lot" of automorphisms of P (the precise choice of definition of "a lot" determines the properties of the symmetric submodel), and models W consist of the elements of V[G] with hereditarily symmetric names.)
So there's really four different steps in getting to Solovay's model: understanding the ZFC picture of the universe (Hrbacek/Jech's final chapters, or Kunen's intro chapter, do this well); understanding forcing over models of ZFC (Kunen covers this well, as does Jech's gigantic set theory tome); understanding symmetric submodels (this is covered in Jech's big tome, but not Kunen; so it might be a good idea to use Jech throughout); and finally, understanding the details behind Solovay's particular construction (covered in a bunch of sources, including Jech's book). Basically, Jech's giant tome of set theory - "Set Theory," Third Millennium Edition - has everything you need. It's pretty expensive, though.
Good luck!
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Having some recursion theory knowledge would also be very helpful to understand Solovay's construction. – Liang Yu Jan 13 at 0:33
To elaborate on Yu's point: recursion theory (AKA computability theory) is definitely not a prerequisite for Solovay's construction. The relevance is that the intuition behind the symmetric submodel construction is (at least for me) a "limited knowledge" sort of thing: I imagine the model of set theory I'm building as a character who is only aware of some (specific) portion of the larger model $V[G]$. Recursion theory is all about this sort of picture, so a familiarity with recursion theory can really make Solovay's construction "click." – Noah S Jan 13 at 21:33
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Let me give an alternative ending to Noah's road map. The splitting point is at symmetric models.
After you've understood the basics of forcing well, you can switch to Kanamori's The Higher Infinite. In the chapter about the real numbers and forcing he again reviews forcing (and if you're new to this - such review is always good) and constructs Solovay's model in a very clear approach.
He avoids [1] talking about symmetric models (which can be a rather complicated tool) by using the "external" construction: we add some sort of generic set to $V$ then we consider an inner model of $V[G]$ which is $HOD(\mathbb R)$ or $L(\mathbb R)$, the latter being thrown around a lot in discussions about models of set theory without choice.
In Kanamori you can find a good introduction to large cardinals (if you haven't run into them in previous steps) which also play a role in Solovay's construction, although that appears in another chapter of the book.
I want to add that studying the construction of symmetric extensions is a good idea. This is an extremely illuminating construction which sheds a lot of light on how set theory works, at least this is how I felt in the past year. However for this particular case I think that using the approach of relative constructibility is better.
Footnotes:
1. This is not entirely true that Kanamori avoids the symmetric models because as it turns out all symmetric models are $HOD(A)$ (whatever that means) of some generic set $A$. In the case of Solovay's model it is just much simpler to use this sort of construction rather going through the complication of symmetric forcing.
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Let me second Asaf's statement that, in this case, inner models are clearer than symmetric submodels. My preference for inner models comes purely from the fact that I really like other things they do - constructing amorphous sets, for example - where they are very clear, so I like them for their own sake. – Noah S Aug 11 at 17:32
Noah, I suppose you meant your preference for symmetric models and not inner models. Do note that Grigorieff proved that every symmetric extension is actually this sort of an inner model. In particular those that have amorphous sets etc.; however I agree completely that for general negations of AC symmetric models are preferable because they allow you better control, in some sense. – Asaf Karagila Aug 11 at 17:55
Yeah, that was a typo. Re: Grigorieff, I'm aware of that; I just meant (as you said) that many results make more sense to me when phrased in terms of symmetric submodel constructions than as inner models. – Noah S Aug 12 at 2:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9571540355682373, "perplexity_flag": "middle"} |
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