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http://mathhelpforum.com/geometry/106345-true-false.html | # Thread:
1. ## True or false?
$x^2+y^2-2x-4y+6=0$ is the equation of a circle.
I answered true, but the answer in the book is false. So who is correct? This equation satisfies the general equation of a circle $x^2+y^2+2gx+2fy+c=0$
Thanks
2. Originally Posted by arze
$x^2+y^2-2x-4y+6=0$ is the equation of a circle.
I answered true, but the answer in the book is false. So who is correct? This equation satisfies the general equation of a circle $x^2+y^2+2gx+2fy+c=0$
Thanks
$x^2 - 2x + y^2 - 4y = -6$
complete the square for x and y ...
$x^2 - 2x + 1 + y^2 - 4y + 4 = -6 + 1 + 4$
$(x-1)^2 + (y-2)^2 = -1$
since the general circle equation is
$(x-h)^2 + (y-k)^2 = r^2$
$r^2 = -1$ ?
3. Originally Posted by arze
$x^2+y^2-2x-4y+6=0$ is the equation of a circle.
I answered true, but the answer in the book is false. So who is correct? This equation satisfies the general equation of a circle $x^2+y^2+2gx+2fy+c=0$
Thanks
Actually, the "general equation" of a circle is of the form
$(x - h)^2 + (y - k)^2 = r^2$
which is a circle of radius $r$, centred at $(h, k)$.
I would try to transform the equation you have been given into this form.
$x^2 + y^2 - 2x - 4y + 6 = 0$
$x^2 - 2x + y^2 - 4y = -6$
$x^2 - 2x + (-1)^2 + y^2 - 4y + (-2)^2 = -6 + (-1)^2 + (-2)^2$
$(x - 1)^2 + (y - 2)^2 = -1$.
Is $r^2 = -1$ possible?
4. Ah!!! How could I have missed that. Thanks! | {"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": 19, "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": 0.9505549669265747, "perplexity_flag": "middle"} |
http://mathoverflow.net/revisions/38445/list | ## Return to Answer
2 added 90 characters in body
I don't think you asked the question you intended. Even though this question is old, I'll say a bit:
For any fixed $n$ and $p$, there are a finite number of homomorphisms $F_2 \rightarrow SL_n(Z/pZ)$.
Hence, there are only a finite number of possibilities for the trace of a group element as a function of the representation. Therefore, there are many duplicates.
I think a better question is to ask whether there are pairs $(g_1, g_2)$ of non-conjugate elements of $F_2$ such that for all primes $p$ and all homomorphisms $F_2 \rightarrow SL_n(Z/pZ)$ the traces are equal. In this form, you can pass to a limit $p \rightarrow \infty$ and conclude the same trace identity would if nonconjugate elements can be true distinguished by traces for an infinite sequence of $SL_n(Z/pZ)$, they could be distinguished by a homomorphisms to $SL_n(\mathbb C)$. Conversely, if there are no trace identities in $SL_n(\mathbb C)$, then there are no trace identities true in all $SL_n(Z/pZ)$, by finding $Z/pZ$ quotients of the ring of coefficients.
Martin Kassabov has been interested in the latter question, and in discussions he and I have had, we've both come to the opinion that there are probably no trace identitites for $SL_n(C)$ when $n > 2$, but it's not easy to find a proof. One possible strategy is to first characterize all trace identities in $SL_2$, and then construct representations in $SL_3$ where they break down. This is interesting to me in any case because of its meaning in 2 and 3-manifold topology -- the trace identities give collections of distinct elements in $\pi_1$ that are forced to have the equal length in any hyperbolic structure.
A weaker question is whether the characteristic polynomials for representations in $SL_n$ can distinguish conjugacy classes in $F_2$.
1
I don't think you asked the question you intended. Even though this question is old, I'll say a bit:
For any fixed $n$ and $p$, there are a finite number of homomorphisms $F_2 \rightarrow SL_n(Z/pZ)$.
Hence, there are only a finite number of possibilities for the trace of a group element as a function of the representation. Therefore, there are many duplicates.
I think a better question is to ask whether there are pairs $(g_1, g_2)$ of non-conjugate elements of $F_2$ such that for all primes $p$ and all homomorphisms $F_2 \rightarrow SL_n(Z/pZ)$ the traces are equal. In this form, you can pass to a limit $p \rightarrow \infty$ and conclude the same trace identity would be true for homomorphisms to $SL_n(\mathbb C)$. Conversely, if there are no trace identities in $SL_n(\mathbb C)$, then there are no trace identities true in all $SL_n(Z/pZ)$, by finding $Z/pZ$ quotients of the ring of coefficients.
Martin Kassabov has been interested in the latter question, and in discussions he and I have had, we've both come to the opinion that there are probably no trace identitites for $SL_n(C)$ when $n > 2$, but it's not easy to find a proof. One possible strategy is to first characterize all trace identities in $SL_2$, and then construct representations in $SL_3$ where they break down. This is interesting to me in any case because of its meaning in 2 and 3-manifold topology -- the trace identities give collections of distinct elements in $\pi_1$ that are forced to have the equal length in any hyperbolic structure.
A weaker question is whether the characteristic polynomials for representations in $SL_n$ can distinguish conjugacy classes in $F_2$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9529118537902832, "perplexity_flag": "head"} |
http://mathhelpforum.com/math-topics/43732-magnitude-argument.html | # Thread:
1. ## magnitude and argument
I dont understand what this problem wants so can someone please help explain it... sqrt means square root of ...
what is the magnitude and argument of (-1+sqrt3i)^2
2. Originally Posted by cruxkitty
I dont understand what this problem wants so can someone please help explain it... sqrt means square root of ...
what is the magnitude and argument of (-1+sqrt3i)^2
As far as I understand the problem you are asked to simplify the term. Doing this you are supposed to know that
$\sqrt{i} = \frac12 \cdot \sqrt{2} + \frac12 \cdot \sqrt{2} \cdot i$ and
$\sqrt{3i} = \sqrt{3} \cdot \sqrt{i}$
Plug in these terms and calculate the square. Collect like terms.
3. Originally Posted by cruxkitty
I dont understand what this problem wants so can someone please help explain it... sqrt means square root of ...
what is the magnitude and argument of (-1+sqrt3i)^2
I assume you mean $(-1+i\sqrt{3})^2 =1 + (\sqrt{3}i)^2 - 2i\sqrt{3} = -2 - 2i\sqrt{3}$
MAGNITUDE:
Magnitude of a complex number $x+iy$ is denoted by $|x+iy|$ and is defined as $\sqrt{x^2 + y^2}$
Geometrically, that is, on the Argand plane, this denotes the distance of the complex number from the origin.
$|(-1+i\sqrt{3})^2| = |-2 - 2i\sqrt{3}| = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4+12} = \sqrt{16} = 4$
ARGUMENT:
Argument of a complex number $x+iy$ is denoted by $\text{arg}(x+iy)$ and is defined as
$\arg(x+iy)= \begin{matrix} \phi\cdot \text{sgn}(y) & x > 0 \\<br /> \frac{\pi}{2}\cdot \text{sgn}(y) & x = 0 \\<br /> (\pi - \phi)\cdot \text{sgn}(y) & x < 0 \\<br /> \end{matrix}$
Where $\phi = \tan^{-1} \left|\frac{y}{x}\right|$ and sgn is the signum function.
Geometrically, that is, on the Argand plane, this denotes the angle made by the line joining the origin and the complex number with the positive direction of the x axis.
$\text{arg}(-2 - 2i\sqrt{3}) = \tan^{-1}\left(\frac{2\sqrt{3}}{2}\right) - \pi =-\frac{2\pi}3$
Aliter:
If you know cube roots of unity and their positions in th argand plane.
Then observe that
$\omega = -\frac12 + \frac{i\sqrt{3}}2$
$2\omega = -1 + i\sqrt{3}$
$(2\omega)^2 = (-1 + i\sqrt{3})^2$
Thus the complex number is $4\omega^2$.
But we know that the cube roots of unity are on the unit circle and that they are distributed over the unit circle at $120^{\circ}$ to each other. 1 is on the real line. $\omega$ is at an angle $120^{\circ}$ to the x-axis. $\omega^2$ is at an angle $-120^{\circ}$ to the x-axis.
Mathematically they mean $|\omega^2| = 1, \arg{\omega^2} = -\frac{2\pi}{3}$
Lets use these facts:
$(-1 + i\sqrt{3})^2 = 4\omega^2$
$|4\omega^2| = |4||\omega^2| = 4$
$\arg(4\omega^2) = \arg(4) + \arg(\omega^2) = 0 -\frac{2\pi}{3} = -\frac{2\pi}{3}$
4. Originally Posted by Isomorphism
$(-1+i\sqrt{3})^2 =\color{red}-2 - 2i\sqrt{3}$
Note the correction above.
I am sure that the question deals with this number: $z = - 1 + \sqrt 3 i$.
“Multitude” must mean absolute value (distance from origin): $\left| {a + bi} \right| = \sqrt {a^2 + b^2 }$.
The argument is the angle the ray from the origin through the number makes with the positive real axis.
In doing this problem here are two useful facts: $\left| {z^2 } \right| = \left| z \right|^2 \,\& \,\arg (z^2 ) = 2\arg (z)$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 29, "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": 0.9317905902862549, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/187909/statistics-bertrands-box-paradox | # Statistics: Bertrand's Box Paradox [duplicate]
Possible Duplicate:
Probability problem
This is the Bertrand's Box Paradox I read about on Wikipedia:
Assume there is three boxes:
a box containing two gold coins,
a boxwith two silver coins
and a box with one of each.
After choosing a box at random and withdrawing one coin at random,
if that happens to be a gold coin,
the probability is actually 66% instead of 50%.
And the problem is equivalent to asking the question
"What is the probability that I will pick a box with two coins of the same color?".
No matter how hard I try, I just couldn't comprehend this..
How is the possibility of picking a gold coin the same as the probability of picking a box with two coins of the same color?
Does this imply there is a 66% chance of picking a gold coin and a 66% chance of picking a sliver coin?
If so, can we just say there is 50% chance of picking either one of them since both stand a 66% chance....?! and suddenly everything makes no sense..
[UPDATES] It is actually the probability of the remaining coin to be gold is 66% but not the probability of obtaining the gold coin is 66%.. I've misread it....
And everything makes sense now :D !
-
"the probability is". The probability of what? – Hurkyl Aug 28 '12 at 14:11
If we enumerate boxes $1,2,3$ and let probability to choose either of them be $\frac13$, then the probability of choosing a gold coin is $$\frac13\cdot 0+\frac13 \cdot\frac12+\frac13\cdot 1 = \frac12$$ by the law of total probability. There may be some trick with a sample space, though. – Ilya Aug 28 '12 at 14:13
– user38927 Aug 28 '12 at 14:17
@Ilya yeah.. And I just couldn't get the trick... – user38927 Aug 28 '12 at 14:19
– Ilya Aug 28 '12 at 14:19
show 2 more comments
## marked as duplicate by MJD, William, tomasz, Noah Snyder, Ayman HouriehOct 6 '12 at 0:43
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
## 1 Answer
The result is being incompletely quoted. Perform the experiment as described. Now suppose that you end up with a gold coin. The question is: what is the probability that the other coin in the same box is gold?
This probability is $\frac{2}{3}$. Let's do an informal computation. It will be imprecise, but could be made precise by using the notion of conditional probability.
Imagine repeating the experiment $3000$ times. Then each box will be picked roughly $1000$ times. We will get a gold coin about $1500$ times. Out of these $1500$ times that we get a gold, it will have come from the two-gold box $1000$ times.
So if we restrict attention to the $1500$ times that we get a gold, about $1000$ of these times it will come from the two-gold box. So given that we got a gold coin, the probability the other coin in the same box is gold should be around $\frac{1000}{1500}$.
-
Thanks for pointing out that! Now I understand why the problem is equivalent to asking the question "What is the probability that I will pick a box with two coins of the same color?" and I also understand why I spent 1hour on comprehending the logic behind the paradox but to no avail. I've misinterpreted the problem for so long.. – user38927 Aug 28 '12 at 14:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9428563714027405, "perplexity_flag": "middle"} |
http://mathoverflow.net/revisions/117511/list | ## Return to Answer
2 added 132 characters in body
One which I like much is $$\exp \left(\begin{bmatrix} . & . & . & . & .\\ 1 & . & . & . & . \\ . & 2 & . & . & . \\ . & . & 3 & . & . \\ . & . & . & 4 & . \\ \end{bmatrix} \right)= \begin{bmatrix} 1 & . & . & . & . \\ 1 & 1 & . & . & . \\ 1 & 2 & 1 & . & . \\ 1 & 3 & 3 & 1 & . \\ 1 & 4 & 6 & 4 & 1 \\ \end{bmatrix}$$ It is practically easier and a bit more iconic if we reduce it a bit - although for me it is not so pleasing, because the immediate remembering of the Pascal-triangle comes with the 1-4-6-4-1-row: $$\Large \exp \small \left(\begin{bmatrix} . & . & . & . \\ 1 & . & . & . \\ . & 2 & . & . \\ . & . & 3 & . \\ \end{bmatrix} \right)= \begin{bmatrix} 1 & . & . & . \\ 1 & 1 & . & . \\ 1 & 2 & 1 & . \\ 1 & 3 & 3 & 1 \\ \end{bmatrix}$$
With a bit explanation which might be useful for other guests http://go.helms-net.de/math/binomial/index-Dateien/image008.png
1 [made Community Wiki]
One which I like much is $$\exp \left(\begin{bmatrix} . & . & . & . & .\\ 1 & . & . & . & . \\ . & 2 & . & . & . \\ . & . & 3 & . & . \\ . & . & . & 4 & . \\ \end{bmatrix} \right)= \begin{bmatrix} 1 & . & . & . & . \\ 1 & 1 & . & . & . \\ 1 & 2 & 1 & . & . \\ 1 & 3 & 3 & 1 & . \\ 1 & 4 & 6 & 4 & 1 \\ \end{bmatrix}$$ It is practically easier and a bit more iconic if we reduce it a bit - although for me it is not so pleasing, because the immediate remembering of the Pascal-triangle comes with the 1-4-6-4-1-row: $$\Large \exp \small \left(\begin{bmatrix} . & . & . & . \\ 1 & . & . & . \\ . & 2 & . & . \\ . & . & 3 & . \\ \end{bmatrix} \right)= \begin{bmatrix} 1 & . & . & . \\ 1 & 1 & . & . \\ 1 & 2 & 1 & . \\ 1 & 3 & 3 & 1 \\ \end{bmatrix}$$ | {"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": 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": 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": 0.9062272906303406, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/34801?sort=oldest | ## Dirichlet’s regulator vs Beilinson’s regulator
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Consider a number field $F$ with ring of integers $O_F$. The Beilinson regulator can in this particular setting be viewed as a map from $K_n(O_F)$ to a suitable real vector space. Here $n$ is any positive odd integer. For $n \geq 3$, there is also another regulator map, defined by Borel, and Burgos has proved that the Borel regulator (suitably normalized) is twice Beilinson's regulator. For $n=1$, the Borel regulator is not defined (as far as I understand, correct me if I'm wrong), but we do have the original and most basic example of a regulator, which is that of Dirichlet.
Question: Is there some form of comparison theorem between the Beilinson regulator and the Dirichlet regulator?
-
By the Bass-Milnor-Serre theorem, $K_1(O_F)=O_F^*$; so the logarithmic embedding of $O_F^*$/(roots of unity in $F$) can be seen as a $K$-theory regulator map, generalized by Beilinson and Borel. – Robin Chapman Aug 7 2010 at 6:29
Hi Robin, sure, both the Beilinson regulator and the Dirichlet regulator are defined on $K_1(O_F)$, but the former is defined in terms of Gillet's general theory of Chern classes, and the latter in terms of the explicit logarithm formula found in any algebraic number theory textbook, and the question is if both definitions agree, or if they differ by some constant factor. – Andreas Holmstrom Aug 7 2010 at 15:05
## 1 Answer
As you comment, the Beilinson regulator is defined using Chern classes for Deligne cohomology. In particular, for $K_1(\mathbb C)$, the first Chern class induces the identity from $K_1(\mathbb C)\simeq \mathbb C^*$ to $H^1_{Deligne}(pt, \mathbb Z(1))\simeq \mathbb C^*$. One passes to $H^1_{Deligne}(pt, \mathbb R(1))\simeq \mathbb R$ by taking the logarithm. So the Belinson regulator for $K_1$ is induced by the map $\log |\; |:K_1(\mathbb C)\to \mathbb R$.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9245902299880981, "perplexity_flag": "head"} |
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## Einstein tensor with the cosmological constant present.
Quote by Messenger I need to read up more on the derivation of the LaGrangian that you showed.
I don't have a handy online reference, but MTW discusses it. See below for a quick summary.
Quote by Messenger I would think there would be some correlation between R and $\Lambda$ for them to show up together in it, similar to potential and kinetic energy.
No, there isn't. They have nothing to do with each other, other than appearing in the same Lagrangian.
The idea behind the Lagrangian is to construct the most general Lorentz scalar that is composed of no higher than second derivatives of the metric. There are only two such scalars possible: the scalar curvature $R$, composed of second derivatives of the metric, and a constant $\Lambda$, which is a "zeroth derivative" of the metric. So the most general Lagrangian is just the sum of those two terms (the factor of 2 in front of $\Lambda$ is there so that the field equation will just contain $\Lambda$ instead of $\Lambda / 2$). But there's no other connection between them; they are completely independent of each other as far as GR is concerned.
The $\sqrt{-g}$ is there to make a Lorentz invariant integration measure (since we're going to integrate the Lagrangian over all spacetime and then vary it with respect to the metric to obtain the field equation). The factor of $1 / 16 \pi$ is for convenience, to make certain formulas look the way physicists were used to having them look.
Quote by Mentz114 In this paper, Padmanabhan proposes an action for the gravitational field equations such that the cosmological constant arises as a constant of integration. It's here http://uk.arxiv.org/abs/gr-qc/0609012v2
I had not seen this, looks interesting.
This paper made me think quite a bit, oldie but a goodie:http://ajp.aapt.org/resource/1/ajpias/v39/i8/p901_s1
Quote by PeterDonis I don't have a handy online reference, but MTW discusses it. See below for a quick summary. No, there isn't. They have nothing to do with each other, other than appearing in the same Lagrangian. The idea behind the Lagrangian is to construct the most general Lorentz scalar that is composed of no higher than second derivatives of the metric. There are only two such scalars possible: the scalar curvature $R$, composed of second derivatives of the metric, and a constant $\Lambda$, which is a "zeroth derivative" of the metric. So the most general Lagrangian is just the sum of those two terms (the factor of 2 in front of $\Lambda$ is there so that the field equation will just contain $\Lambda$ instead of $\Lambda / 2$). But there's no other connection between them; they are completely independent of each other as far as GR is concerned. The $\sqrt{-g}$ is there to make a Lorentz invariant integration measure (since we're going to integrate the Lagrangian over all spacetime and then vary it with respect to the metric to obtain the field equation). The factor of $1 / 16 \pi$ is for convenience, to make certain formulas look the way physicists were used to having them look.
Gives me quite a bit to look into and think about. I really do appreciate you taking your time on this board Peter. Thanks.
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Quote by Messenger I really do appreciate you taking your time on this board Peter. Thanks.
You're welcome! Thanks for the appreciation.
Quote by PeterDonis There are only two such scalars possible: the scalar curvature $R$, composed of second derivatives of the metric, and a constant $\Lambda$, which is a "zeroth derivative" of the metric. So the most general Lagrangian is just the sum of those two terms (the factor of 2 in front of $\Lambda$ is there so that the field equation will just contain $\Lambda$ instead of $\Lambda / 2$). But there's no other connection between them; they are completely independent of each other as far as GR is concerned.
Hi Peter,
Looking into the "zeroth derivative". I take it from the Ricci scalar being composed of second derivatives, it is never possible for the action principle to = 0 with $2\Lambda=-R$ since this would mean that either the Ricci scalar or the cosmological constant would have to be the same order derivative of the metric. My inquiry stems from unimodular descriptions of the cosmological constant. If this were possible, and the definition of the Ricci scalar is $R=g^{\mu\nu}R_{\mu\nu}$, then this would imply that $\Lambda=g^{\mu\nu}\Lambda_{\mu\nu}$. I ask because for zero curvature it would seem feasible to then write
$\frac{1}{2}Rg_{\mu\nu}-R_{\mu\nu}=\frac{3}{4}\Lambda g_{\mu\nu}-\Lambda_{\mu\nu}=0$.
If I set $\frac{3}{4}\Lambda g_{\mu\nu}=0$, I don't see any way to tell that $\Lambda_{\mu\nu}=G_{\mu\nu}$ isn't true.
I would like to understand more in depth the proof that $\Lambda$ is only a zeroth derivative.
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http://mathforum.org/mathimages/index.php?title=Change_of_Coordinate_Systems&diff=4290&oldid=4278 | # Change of Coordinate Systems
### From Math Images
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| - | Often useful is mapping points from '''Cartesian Coordinates''' to '''[[Polar Coordinates]]'''. Such a mapping, as shown in this page's main image, can map a disk to a rectangle. Each origin-centered ring in the disk consists of points at constant distance from the origin and angles ranging from 0 to <math> 2\pi </math>. These points create a vertical line in Polar Coordinates. Each ring at different distance from the origin creates its own line in the polar system, and the collection of lines creates a rectangle. | + | The ellipse that is tilted relative to the coordinate axes is created by a combination of rotation and stretching, represented by the matrix |
| | | + | : <math>\begin{bmatrix} |
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| | | + | 2cos(\theta) & -sin(\theta)\\ |
| | | + | 2sin(\theta) & cos(\theta) \\ |
| | | + | \end{bmatrix}\vec{x} = \vec{x'}</math> |
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| | | + | Often useful is mapping points from '''Cartesian Coordinates''' to '''[[Polar Coordinates]]'''. Such a mapping, as shown in this page's main image, can map a disk to a rectangle. Each origin-centered ring in the disk consists of points at constant distance from the origin and angles ranging from 0 to <math> 2\pi </math>. These points create a vertical line in Polar Coordinates. Each ring at a different distance from the origin creates its own line in the polar system, and the collection of lines creates a rectangle. |
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## Revision as of 14:48, 11 June 2009
Change of Coordinates
Field: Algebra
Created By: Brendan John
Website: [ ]
Change of Coordinates
The same object, here a circle, can look completely different depending on which coordinate system is used.
# Basic Description
It is a common practice in mathematics to use different coordinate systems to solve different problems. An example of a switch between coordinate systems follows: suppose we take a set of points in regular x-y Cartesian Coordinates, represented by ordered pairs such as (1,2), then multiply their x-components by two, meaning (1,2) in the old coordinates is matched with (2,2) in the new coordinates.
Under this transformation, a set of points would become stretched in the horizontal x-direction since each point becomes further from the vertical y-axis (except for points originally on the y-axis, which remain on the axis). A set of points that was originally contained in a circle in the old coordinates would be contained by a stretched-out ellipse in the new coordinate system, as shown in this page's main image.
Many other such transformations exist and are useful throughout mathematics, such as mapping the points in a disk to a rectangle.
# A More Mathematical Explanation
[Click to view A More Mathematical Explanation]
Some of these mappings can be neatly represented in matrix notation, in the form
UNIQ649d189657a20 [...]
[Click to hide A More Mathematical Explanation]
Some of these mappings can be neatly represented in matrix notation, in the form
$A\vec{x}=\vec{x'}$
Where $\vec{x}$ is the coordinate vector of our point in the original coordinate system and $\vec{x'}$ is the coordinate vector of our point in the new coordinate system.
For example the transformation in the basic description, doubling the value of the x-coordinate, is represented in this notation by
$\begin{bmatrix} 2 & 0 \\ 0 & 1 \\ \end{bmatrix}\vec{x} = \vec{x'}$
As can be easily verified.
The ellipse that is tilted relative to the coordinate axes is created by a combination of rotation and stretching, represented by the matrix
$\begin{bmatrix} 2cos(\theta) & -sin(\theta)\\ 2sin(\theta) & cos(\theta) \\ \end{bmatrix}\vec{x} = \vec{x'}$
Often useful is mapping points from Cartesian Coordinates to Polar Coordinates. Such a mapping, as shown in this page's main image, can map a disk to a rectangle. Each origin-centered ring in the disk consists of points at constant distance from the origin and angles ranging from 0 to $2\pi$. These points create a vertical line in Polar Coordinates. Each ring at a different distance from the origin creates its own line in the polar system, and the collection of lines creates a rectangle.
# Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
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http://cstheory.stackexchange.com/questions/14379/the-motivation-for-using-karp-reductions-in-the-theory-of-mathcalnp-complet | # The motivation for using Karp-reductions in the theory of $\mathcal{NP}$-completeness
The notion of polynomial time reductions (Cook reductions) is an abstraction of a very intuitive concept: efficiently solving a problem by using an algorithm for a different problem.
However, in the theory of $\mathcal{NP}$-completeness, the notion of $\mathcal{NP}$-hardness is captured via mapping reductions (Karp reductions). This concept of "restricted" reductions is much-less intuitive (at least to me). It even seems a bit contrived, as it creates a somewhat less intuitive notion of hardness; by that I'm referring to the fact that $\mathcal{NP}$ does not trivially contains $co-\mathcal{NP}$. Although in complexity theory we are very used to the concept that being able to solve a problem such as $\mathsf{SAT}$ does not imply that we can solve $\overline{\mathsf{SAT}}$, in natural settings (which are captured by Cook reductions), assuming we have an algorithm for solving $\mathsf{SAT}$, we can solve $\overline{\mathsf{SAT}}$ just by running the algorithm for $\mathsf{SAT}$ and returning the opposite.
My question is why should we use Karp reductions for the theory of $\mathcal{NP}$-completeness? What intuitive notion does it capture? How does it relates to the way we understand "hardness of computation" in the real world?
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agreed that the basic defn of Cook and Karp reductions are not very transparent & subtle and not at all apparent in their distinction early on. you are not alone.. the wikipedia article on Ptime reduction is currently marked as "possibly confusing or unclear to readers" and many one reduction is not a lot better... on the other hand they do answer some of the basic questions similar to yours... – vzn Nov 19 '12 at 17:31
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– vzn Nov 19 '12 at 17:34
## 2 Answers
Like Turing reductions, many-one reductions came into complexity theory from computability/recursion theory literature. Cook and Karp reductions are natural complexity theoretic versions of similar existing reductions in computability.
There is a intuitive way of explaining many-one reductions: it is a restriction of Turing reductions where we can ask only a single question from the oracle and the oracle's answer will be our answer.
Now the question is why would we need to study this (and any other kind of reductions like truth-table, weak-truth-table, etc.)?
These reductions give a finer picture than Turing reductions. Turing reductions are too powerful to distinguish between many concepts. A very large portion of computability theory is devoted to the study of c.e./r.e. degrees. The notion of an c.e. set is central. We can have TM machine that can enumerate an infinite set, we may not be able to enumerate its complement. If you want to study c.e. sets then Turing reduction is too strong since c.e. sets are not closed under it. So many-one reductions are a (and maybe the) natural way of defining reductions for this purpose.
Other types of reductions are defined for similar reasons. If you are interested I would suggest checking Piergiorgio Odifreddi's "Classical Recursion Theory". It has a quite comprehensive chapter on different reductions and their relations.
Now for complexity theory the argument is similar. If you accept that $\mathsf{NP}$ is a extremely natural class of problems and you want to study $\mathsf{NP}$, then Cook reductions are too strong. The natural choice is a weaker reduction such that $\mathsf{NP}$ is closed under it and we can prove existence of a complete problem w.r.t. to those reduction for $\mathsf{NP}$. Karp reductions are the natural choice for this purpose.
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?? "cook reductions are too strong" to study NP? what do you mean by that? think it could be phrased a little clearer/better – vzn Nov 20 '12 at 19:11
there are several questions on this site related to Cook vs Karp reductions. have not seen a very clear description of this for the neophyte because its somewhat inherently subtle in many ways and its an active/open area of research. here are some refs that may be helpful to address it. as wikipedia summarizes, "Many-one reductions are valuable because most well-studied complexity classes are closed under some type of many-one reducibility, including P, NP, L, NL, co-NP, PSPACE, EXP, and many others. These classes are not closed under arbitrary many-one reductions, however."
it seems fair to say that even advanced theorists are actively pondering the exact distinction and differences as in below refs and the full story wont be available unless important open complexity class separations are resolved, ie these questions seem to cut to the center of the known vs unknown.
[1] Cook versus Karp-Levin: Separating Completeness Notions If NP Is Not Small (1992) Lutz, Mayordomo
[2] Are Cook and Karp Ever the Same? Beigel and Fortnow
[3] More NP-Complete Problems (PPT) see slides 9-14 on history & Cook vs Karp reduction distinctions
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http://mathoverflow.net/questions/84161?sort=oldest | Hopf algebras examples
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Following Richard Borcherds' questions 34110 and 61315, I'm looking for interesting examples of Hopf algebras for an introductory Hopf algebras graduate course.
Some of the examples I know are well-known:
• Examples related to groups and Lie algebras,
• Sweedler and Taft Hopf algebras,
• Drinfeld-Jimbo quantum groups.
(These examples can be found for example in Montgomery's book on Hopf algebras.) Other interesting examples are Lusztig small quantum groups and the generalized small quantum groups of Andruskiewitsch and Schneider.
Therefore I'm looking for interesting and non-standard examples maybe related to combinatorics or to other branches of mathematics.
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Since this isn't exactly what you asked, I'll leave it as a comment. Perhaps you will be interested in the crossed product algebras. Consider a Hopf algebra and some algebra acting on it, then you can form a Hopf algebra with underlying vector space of their tensor product. If you need a reference, I would suggest nlab. Sorry if this is off-topic. – B. Bischof Dec 24 2011 at 1:02
7 Answers
Symmetric functions, quasisymmetric functions, Connes-Kreimer algebra are all examples of combinatorial Hopf algebras. There are many more.
The original example is the cohomology ring of a Lie group.
Also Steenrod algebra and the analogue for any cohomology theory.
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Also the tensor and the shuffle Hopf algebra of a vector space (or, better, module). Besides, the Malvenuto-Reutenauer Hopf algebra $\mathrm{FQSym}$ (see math.tamu.edu/~maguiar/MR.pdf for definition and many properties) is a good example of something graded, neither commutative nor cocommutative but still easy to calculate in. There is also the Loday-Ronco Hopf algebra, but I am not really understanding its definition yet; it is definitely not that simple. But at least something made out of trees would be useful. I don't remember how the Hopf algebra in ... – darij grinberg Dec 23 2011 at 15:38
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... arxiv.org/abs/math/0408405 II.9.3 is called, but it is definitely elementary enough to be part of an introductory course. – darij grinberg Dec 23 2011 at 15:40
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Daril, you can see the Loday-Ronco Hopf algebra (also called ${\bf PBT}$) as the Hopf subalgebra of ${\bf FQSym}$ generated by all binary trees $T$ where the embedding is defined by $T := \sum_{\sigma \in \mathfrak{S}, {\tt bst}(\sigma) = T} \sigma$, where ${\tt bst}$ is the binary search tree obtained by inserting $\sigma$ (from left to right) following the binary search tree insertion algorithm. – Samuele Giraudo Mar 4 at 20:55
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.
I will elaborate on Bruce's examples and add a couple of my own.
Of course, the homology and cohomology of topological groups over a field are good examples.
For each prime prime $p$ the Steenrod algebra $\mathcal{A}_p$ which is the algebra of endomorphisms of the cohomology theory $H^*(-;\mathbb{F}_p)$. The cohomology of this Hopf algebra is the $E_2$ term of a spectral sequence, due to Adams, converging to the $p$-completed stable homotopy groups of spheres.
The functions on any affine algebraic groups over a field are another family of examples.
Formal group laws over a field $k$. You can read about these in Husemoller's book.
The rational homotopy groups of connected topological group or more generally an $H$-space is a Lie algebra. A nice result of Milnor-Moore shows that the universal enveloping algebra of this Lie algebra is isomorphic as Hopf algebras to the rational homology of the space.
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If you're interested in Hopf algebras in categories other than $\mathrm{Vect}$, you can look at the exterior algebra as a Hopf algebra in $\mathrm{SVect}$, the category of super vector spaces with degree-preserving morphisms. More precisely, let $V$ be a purely odd vector space (i.e. $V_0 = 0$ and $V_1 = V$) and form the exterior algebra $\Lambda(V)$ with its natural $\mathbb{Z}/2$ grading. This is a superalgebra, i.e. an algebra in $\mathrm{SVect}$. Then $\Lambda(V) \underline{\otimes} \Lambda(V)$ is an algebra in $\mathrm{SVect}$, where $\underline{\otimes}$ is the graded tensor product of graded algebras.
Now consider the map $\Delta : V \to \Lambda(V) \underline{\otimes} \Lambda(V)$ given by $$\Delta(v) = v \otimes 1 + 1 \otimes v.$$ With the sign conventions coming from the graded tensor product, you get $\Delta(v)^2 = 0$, and so according to the universal property of the exterior algebra, $\Delta$ extends to an algebra homomorphism $\Delta : \Lambda(V) \to \Lambda(V) \underline{\otimes} \Lambda(V)$. Coassociativity is clear. You can get the counit and antipode similarly using the universal property.
Another good example is the shuffle Hopf algebra, which is discussed in this question. Let $V$ be a vector space and $T(V)$ its tensor algebra. The shuffle Hopf algebra is a Hopf structure on $T(V)$ which uses neither the standard algebra nor coalgebra structures on the tensor algebra.
The comultiplication is given by deconcatenation: $$\Delta(v_1 \dots v_n) = \sum_{j=1}^{n+1} v_1 \dots v_{j-1} \otimes v_j \dots v_n,$$ while the multiplication is given by the shuffle product: $$(v_1 \dots v_k) \cdot (v_{k+1} \dots v_n) = \sum_{\sigma \in S_{k,n-k}} v_{\sigma^{-1}(1)} \dots v_{\sigma^{-1}(n)},$$ where $S_{k,n-k}$ is the set of $(k,n-k)$ shuffle permutations, i.e. $$\sigma(1) < \dots < \sigma(k)$$ and $$\sigma(k+1) < \dots < \sigma(n).$$ I haven't really worked much with the shuffle algebra myself, but the answers to the question linked above have some discussion of what it is good for.
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Hopf incidence algebras are not well understood and could be an important example as it deals with matrix multiplication. Fourier transform for these are not well understood.
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There's a (unique) semisimple noncocommutative Hopf algebra of dimension 8 that makes a nice example. (Unfortunately I don't remember where to find information on it at the moment, I learned about it in a survey paper of Susan Montgomery's.)
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Maybe you are misremembering and have the Sweedler 4-dimensional algebra in mind? I am pretty sure there are many 8-dimensional non-comm, non-cocomm Holp algebras. – Mariano Suárez-Alvarez Dec 23 2011 at 21:00
The classification for the pointed case in dimension 8 was done by [Caenepeel, S.; Dăscălescu, S. Pointed Hopf algebras of dimension $p\sp 3$. J. Algebra 209 (1998), no. 2, 622--634. MR1659887 (99k:16076)] and slightly later and differently by Nicolás Andruskiewitsch and Schneider (but maybe this second classification is for dimension $p^3$ with odd $p$...) – Mariano Suárez-Alvarez Dec 23 2011 at 21:12
Sorry, forgot to write semisimple. – Noah Snyder Dec 24 2011 at 4:36
A couple people mentioned the Steenrod algebra briefly, but you can do a few more topologically-related things:
• The subalgebras $\mathcal{A}(n)$ of the Steenrod algebra generated by $Sq^1, ..., Sq^n$ are neat. In particular, it is a good exercise in cohomology to compute $Ext_{\mathcal{A}(n)}(k,k)$. (One can do a minimal resolution and try to look for a pattern, and then prove that it works using a spectral sequence.)
• You can show that the Hopf algebras given by $\mathbb{Z}[c_1, c_2, ...]$ ($c_i$ living in degree 2i) and $\mathbb{Z}/2 [w_1, w_2, ...]$ ($w_i$ living in degree i) and comultiplications given by $y_n \mapsto \sum y_i \otimes y_j$ on the generators, are self-dual Hopf algebras and explicitly describe the relationship between itself and the dual. This is neat in and of itself, but then you can mention that these results lead to quick calculations of $H_*MU$ and $H_*MO$ as comodules over the dual of the Steenrod algebra, and thus allow for computations of cobordism groups via the Adams spectral sequence. This self-duality can also be used for a quick proof of the Bott periodicity theorem, though the only reference I know of for this is not yet published (by May), though it will be soon.
• If you're doing cohomology, it's always nice to do the cohomology of an exterior algebra; i.e. a Hopf algebra that's an exterior algebra on primitive generators. It's a very easy result, but you can use it to compute other things via spectral sequences.
I know I've forgotten several things I wanted to mention... if I remember them, I'll edit them in.
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There are lots of good combinatorial examples that are not hard to define: e.g., posets, matroids, quasisymmetric functions, graphs. A few places to look: W. Schmitt, Incidence Hopf Algebras, J. Pure Appl. Algebra 96 (1994), no. 3, 299–330. W. Schmitt, Hopf algebra methods in graph theory, J. Pure Appl. Algebra 101 (1995), no. 1, 77–90. M. Aguiar, N. Bergeron, and F. Sottile, Combinatorial Hopf algebras and generalized Dehn-Sommerville relations, Compos. Math. 142 (2006), no. 1, 1–30.
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http://mathhelpforum.com/discrete-math/18963-inductive-proof-size-power-sets-print.html | # inductive proof of size of power sets
Printable View
• September 14th 2007, 10:07 AM
polypus
inductive proof of size of power sets
hi everybody,
i have written an inductive proof of the statement that for any set $S$
$| \wp(S) | = 2 ^ {|S|}$
I am using my own notation for denoting set union with a disjoint set of cardinality 1, because it is such a convenience. is there some standard way of doing this? any tips/corrections, mathematical, stylistic, or notational will be highly appreciated.
Let $\psi(S) = |\wp(S)|$
and let
$S + 1$
denote $S \cup T$ where $T$ is some set disjoint from $S$ and where $|T| = 1$, so that we can write
| $S + 1| = |S| + 1$
The following holds trivially for $S = \emptyset$ and we will assume it to hold for any arbitrary set $S$ as well
$[\frac{\psi(S + 1)}{2} = 2 ^ {|S|}] \wedge [\psi(S) = 2 ^ {|S|}]$
thus the following also holds for any $S$
$\psi(S) = 2 ^ {|S|}$
we take as our inductive hypothesis that
$[\psi(S) = 2 ^ {|S|}] \rightarrow [\psi(S + 1) = 2^{|S + 1|}]$
recall that for any set $S$ we assume
$\frac{\psi(S + 1)}{2} = 2 ^ {|S|}$
which we can rearrange as follows
$\psi(S + 1) = 2 \times 2^{|S|}$
$\psi(S + 1) = 2^{|S| + 1}$
$\psi(S + 1) = 2^{|S + 1|}$
which proves the inductive step and the theorem.
thank again for any tips
• September 14th 2007, 10:12 AM
polypus
worries
i'm just worried that the fact that my consequent is essentially included in my antecedent invalidates the proof, correct?
• September 14th 2007, 11:01 AM
Plato
Sorry, but I have absolutely no idea what you have done there!
The usual way is to observe that $\wp (\{ 1,2,...,n,n + 1\} )$ contains $\wp (\{ 1,2,...,n\} )$ plus $\left\{ {X \cup \{ n = 1\} :X \in \wp (\{ 1,2,...,n\} )} \right\}$.
Hence $\wp (\{ 1,2,...,n, n+1\} )$ contains twice as many sets as $<br /> \wp (\{ 1,2,...,n\} )$.
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http://mathoverflow.net/questions/28776/does-the-ideal-class-of-the-different-of-a-number-field-have-a-canonical-square/33165 | ## Does (the ideal class of) the different of a number field have a canonical square root?
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A theorem of Hecke (discussed in this question) shows that if $L$ is a number field, then the image of the different $\mathcal D_L$ in the ideal class group of $L$ is a square.
Hecke's proof, and all other proofs that I know, establish this essentially by evaluating all quadratic ideal class characters on $\mathcal D_L$ and showing that the result is trivial; thus they show that the image of $\mathcal D_L$ is trivial in the ideal class group mod squares, but don't actually exhibit a square root of $\mathcal D_L$ in the ideal class group.
Is there any known construction (in general, or in some interesting cases) of an ideal whose square can be shown to be equivalent (in the ideal class group) to $\mathcal D_L$.
Note: One can ask an analogous question when one replaces the rings of integers by Hecke algebras acting on spaces of modular forms, and then in some situations I know that the answer is yes. (See this paper.) This gives me some hope that there might be a construction in this arithmetic context too. (The parallel between Hecke's context (i.e. the number field setting) and the Hecke algebra setting is something I learnt from Dick Gross.)
Added: Unknown's very interesting comment below seems to show that the answer is "no", if one interprets "canonical" in a reasonable way. In light of this, I am going to ask another question which is a tightening of this one.
On second thought: Perhaps I will ask a follow-up question at some point, but I think I need more time to reflect on it. In the meantime, I wonder if there is more that one can say about this question, if not in general, then in some interesting cases.
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If this is possible, it likely will have to be very arithmetic. In a paper by Frohlich, Serre, and Tate (A different with an odd class, Crelle 209 (1962), 6–-7) they give a non-arithmetic example of an extension of Dedekind domains where the different is not a square in the class group. By comparison, if you look at discriminant ideals instead of different ideals then there is an algebraic construction of a square root: the Steinitz class. That is, if S/R is your Dedekind domain extension, decompose S as an R-module as S = R^{n-1}+M (direct sum) for an ideal M in R. Then [M]^2 = [disc(S/R)]. – KConrad Jun 19 2010 at 20:36
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Comments on this paper at end of volume II of Serre's collected works give two proofs for fn fields over (quasi)finite fields. The first is Hecke's, and the second (given away from char. 2) is geometric: constructs divisor class on the curve (not just over alg. closure) which doubles to the canonical class. Since the different in fn field case for a sep'ble cover is difference of canonical classes (base pulled back to source), one gets construction in the fn field case. Probably no useful analogue for number fields or else Serre would have said something. Though could email Serre to ask him! – BCnrd Jun 19 2010 at 20:55
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I'd guess the answer has to be "no", if one wants a canonical construction. In the fn field case, there's a moduli space $S_g$ of "spin curves" (curves endowed with a sqrt of can. bundle), naturally a cover of $M_g$. Any natural construction of a sqrt of can. bundle would lead to a global section of the cover $S_g / M_g$. But this cover is provably nontrivial (for $g \geq 2$?). By analogy, I'd doubt the existence of a natural construction in the number field case as well. – Marty Jun 20 2010 at 16:25
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There exists a quadratic extension of Q with class group cyclic of order 4, the different the element of order 2, and such that the automorphism group acts non-trivially on the class group. It follows that there is no natural square root of the dfferent. (I had found such an example in response to a question of Fulton about the arithmetic Riemann-Roch theorem around 20 years ago by looking up some tables but unfortunately do not recall all the details.) – ulrich Jun 21 2010 at 10:40
2
Unknown, there is a mistake in your comment: the different ideal in a quadratic field is always principal. More generally, if the ring of integers has the form Z[a] then the different ideal is (f'(a)) for f(x) the min. poly. of a over Q. In any quadratic field the ring of integers has the form Z[a] for some a. Thus whatever example you may have found couldn't possibly be quadratic over Q. Was your base field the quadratic field instead of Q (and some extension of it had a class group of order 4, etc.)? – KConrad Jun 21 2010 at 21:51
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## 6 Answers
The following example shows that, in its strongest form, the answer to Professor Emerton's question is no. This answer is essentially an elaboration on what is already in the comments.
Let $p \equiv q \equiv 5 \pmod 8$. Let $K/\mathbb{Q}$ be a cyclic extension of degree four totally ramified at $p$ and $q$ and unramified everywhere else (it exists). To make life easier, suppose that the $2$-part of the class group of $K$ is cyclic. The Galois group of $K$ is $G = \mathbb{Z}/4 \mathbb{Z}$. Let $C$ denote the class group of $K$. I claim that $C^G$ is cyclic of order two. Since the $2$-part of $C$ is cyclic, this is equivalent to showing that $C_G$ is cyclic of order two. By class field theory, $C_G$ corresponds to a Galois extension $L/\mathbb{Q}$ unramified everywhere over $K$ such that there is an exact sequence
$$1 \rightarrow C_G \rightarrow \mathrm{Gal}(L/\mathbb{Q}) \rightarrow \mathbb{Z}/4 \mathbb{Z} \rightarrow 1.$$
If $\Gamma$ is any finite group with center $Z(\Gamma)$, then an easy exercise shows that $\Gamma/Z(\Gamma)$ is cyclic only if it is trivial. We deduce that $L$ is the genus field of $K$. There is a degree four extension $M/L$ contained inside the cyclotomic field $\mathbb{Q}(\zeta_p,\zeta_q)$ that is unramified over $K$ at all finite primes. However, the congruence conditions on $p$ and $q$ force $K$ to be (totally) real and $M$ (totally) complex. Thus $M/K$ is ramified at the infinite primes, and $C_G = \mathbb{Z}/2\mathbb{Z}$, and the claim is established.
We note, in passing, that $L = K(\sqrt{p}) = K(\sqrt{q})$.
Suppose there exists a canonical element $\theta \in C$ such that $\theta^2 = \delta_K$, where $\delta_K$ is the different of $K$. The different $\delta_K$ is invariant under $G$. If $\theta$ is canonical in the strongest sense then it must also be invariant under $G$. In particular, the element $\theta \in C^G$ must have order dividing two, and hence $\theta^2 = \delta_K$ must be trivial in $C$. We conclude that if $\delta_K$ is not principal, no such $\theta$ exists.
It remains to show that there exists primes $p$ and $q$ such that $\delta_K$ is not principal and the $2$-part of $C$ is cyclic. A computation shows this is so for $p = 13$ and $q = 53$. For those playing at home, $K$ can be taken to be the splitting field of $$x^4 + 66 x^3 + 600 x^2 + 1088 x - 1024,$$ where $C = \mathbb{Z}/8 \mathbb{Z}$ and $\delta_K = [4]$. $C$ is generated by (any) prime $\mathfrak{p}$ dividing $2$, and $G$ acts on $C$ via the quotient $\mathbb{Z}/2\mathbb{Z}$, sending $\mathfrak{p}$ to $\mathfrak{p}^3$.
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Thankyou very much for working out this concrete example! – Emerton Jul 24 2010 at 2:53
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Let $N/K$ be a finite Galois extension of number fields, with Galois group $G$, then Hilbert's formula for the valuation of the different at a prime ideal $P$ of $N$ states that: $$v_P(D)=\sum_{i\geq 0}(|G_i(P)|-1)$$ where $G_i(P)$ is the $i$-th ramification group (in lower notation) at $P$ in $N/K$, and $|G_i(P)|$ is its order. The sum is finite since $G_i(P)={1}$ for large $i$; it is an even integer if $G$ is of odd order, since this forces its subgroups $G_i(P)$ to be of odd order as well for any $i$, any $P$.
It follows that, in any odd degree Galois extension of number fields, there exists a fractional ideal whose square equals the inverse different. This fractional ideal, known as the "square root of the inverse different", has rich properties as a Galois module and as an hermitian module, which were first revealed and studied by Boas Erez, see e.g.
MR1128708 (92g:11108) Erez, B. The Galois structure of the square root of the inverse different. Math. Z. 208 (1991), no. 2, 239–255.
In conclusion we get a bit more than what was asked in the odd degree Galois extension case (an ideal whose square equals the inverse different, not only with the same class). But this does not say anything in the even degree or non-Galois case.
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Let me point out that I would like to interpret "canonical" in a less strict sense. In order to explain what I mean, consider the following question:
Does there exist a canonical square root of $-1$ modulo primes $p = 4n+1$?
If canonical means that the result should be independent of any automorphism of the residue class group modulo $p$, then the answer is no: of $i$ is one answer, then the automorphism sending every residue class to its inverse will send $i$ to the other root $-i$.
Yet I would accept $i \equiv (\frac{p-1}2)! \bmod p$ as an answer to the question.
Edit 2 (28.07.10) I'm still thinking about what canonical should mean in this context. The ideal class of the square root is defined up to classes of order $2$; doesn't this mean that a "canonical" choice of this ideal class should be an element in the group $Cl(K)/Cl(K)[2]$?
Edit. Trying to generalize frictionless jellyfish's example I ended up with the following results (which so far I have only partially proved).
Let $p$ and $q$ be two prime numbers with $p \equiv q \equiv 1 \bmod 4$. There is a unique cyclic quartic extension $K/\mathbb Q$ with conductor $pq$ and discriminant $p^3q^3$. Let $\mathfrak p$ and $\mathfrak q$ denote the prime ideals in $K$ above $p$ and $q$. Then $diff(K/\mathbb Q) = {\mathfrak p}^3 {\mathfrak q}^3$. Moreover, ${\mathfrak p}^2 {\mathfrak q}^2 = (\sqrt{pq}\,)$ is principal, so the ideal class of the different is either trivial or has order $2$.
Theorem If $(p/q) = -1$, then $Cl_2(K) \simeq [2]$ if $p \equiv q \equiv 5 \bmod 8,$ and $Cl_2(K) \simeq [4]$ otherwise. In both cases, the $2$-class field of $K$ is abelian over $\mathbb Q$, hence is equal to its genus class field.
The ideal classes of each of the prime ideals above $\mathfrak p$ and $\mathfrak q$ generates the $2$-class group.
Taking $p=5$ and $q = 17$ gives an example of a "non-canonical" square root. I have not yet found a criterion that would tell me when the different is principal and when its class has order $2$. Both cases do occur.
The case where $p$ and $q$ are quadratic residues of each other is more involved, more interesting (and more conjectural):
Theorem Assume that $(p/q) = +1$, and that $p \equiv q \equiv 5 \bmod 8$. Then $Cl_2(K) \simeq [2,2]$ if $(p/q)_4 (q/p)_4 = -1$, $Cl_2(K) \simeq [4]$ if $(p/q)_4 = (q/p)_4 = -1$, and $Cl_2(K) \simeq [2^n], \ n \ge 3$ if $(p/q)_4 = (q/p)_4 = +1$.
If $(p/q)_4 = (q/p)_4 = -1$, then the ideal classes $[\mathfrak p]$ and $[\mathfrak q]$ are squares, but not fourth powers; in particular, the different is principal.
If $(p/q)_4 = (q/p)_4 = +1$, there are two cases:
1. $h_2(K) = h_2(k)= 2^n$; then $[\mathfrak p]$ and $[\mathfrak q]$ both have order $2$, and the different is principal.
2. $h_2(K) = 2h_2(k)= 2^n$; then either $\mathfrak p$ or $\mathfrak q$ is principal, whereas the other ideal generates a class of order $2$. In particular, the ideal class of the different has order $2$.
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Dear Franz, Thanks for this thoughtful remark, and please do let us know if you find something interesting! Regards, Matthew – Emerton Jul 25 2010 at 20:49
I thought about trying to do something like this, but then decided to leave it to someone who was a master of using CFT on extensions of small degree - such as yourself. Minor quibble, there are exactly two cyclic quartic extensions K with conductor pq and discriminant p^3 q^3. – Lavender Honey Jul 26 2010 at 19:18
Thanks for your "minor quibble" - I have proved the first theorem now, and without having computed a single example I bet that among the two cyclic quartic extensions with class group [4], the different is principal in one of them and generates a class of order $2$ in the other. The distinction will come from the value of the quadratic residue symbol [a+bi / c+di], where p = a^2 + b^2 and q = c^2 + d^2. – Franz Lemmermeyer Jul 27 2010 at 13:13
Since you asked for special cases where something was known, I'd like to mention this paper which I happened to come across :
MR2019023 (2004j:11135) Vinatier (Stéphane), Sur la racine carrée de la codifférente. Les XXIIèmes Journées Arithmetiques (Lille, 2001). J. Théor. Nombres Bordeaux 15 (2003), no. 1, 393--410
which begins by stating that the codifferent of an odd-degree galoisian extension $N|{\bf Q}$ has a canonical square root.
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Dear Chandan, Thank you very much; I'm looking forward to reading it! – Emerton Sep 29 2010 at 15:56
Recently the following was explained to me by Melanie Wood. (What follows is a summary of some results from her preprint here; see in particular the discussion at the top of page 3.)
If $f$ is a primitive irreducible binary $n$-ic form with coefficients in $\mathbb Z$, then we can consider the locus $S_f$ in $\mathbb P^1_{\mathbb Z}$ that $f$ cuts out. This will be finite over $\mathbb Z$, and hence is equal to Spec $R$, where $R = H^0(S_f, \mathcal O)$. The finite $\mathbb Z$-algebra $R$ will be an order in a number field of degree $n$. Not all orders in all number fields arise this way (if $n >3$), but certainly some do!
Now it turns out that restriction of $\mathcal O(n-2)$ to $S_f$ (which is an invertible sheaf on $S_f$, and hence gives an ideal class of $R$) is canonically isomorphic to the inverse different of $R$, and hence if $n$ is even then we can consider the restriction to $S_f$ of $\mathcal O((n-2)/2)$, and so obtain a square root of the inverse different of $R$.
As is explained in the above linked preprint, $R$ alone does not suffice to determine $f$; rather $f$ is determined by $R$ together with some extra ideal-theoretic strucure. So (as far as I know) this construction of the square root does not depend on $R$ alone, but on extra data. Nevertheless, it gives a very interesting answer to my question in those cases where it applies.
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In the interesting case n=4, the extra data attached to R is a "cubic resolvent ring." When R is the ring of integers of a quartic number field (which I think is the case you have in mind in your original question) the cubic resolvent is unique, so it isn't really extra data. But, as you say, not all quartic maximal orders arise this way -- only those R whose cubic resolvents are monogenic (generated by a single element over Z.) The case n=3 is intriguing -- could there be a "parametrization" in anything like Bhargava's sense for cubic rings together with a square root of inverse different? – JSE Jan 22 2011 at 6:05
7
prepint - is that what one drinks in Australia before going to the pub? – Lavender Honey Jan 22 2011 at 6:57
1
"Prepint" now corrected to "preprint"; apologies to those who thought we were going down to the pub. – Emerton Jan 22 2011 at 14:55
ADDED: As pointed out in the exchange of comments by unknown and KConrad, unknown's example doesn't work as stated. Nevertheless, it suggests a concrete test for one interpretation of the question, which might be interesting to investigate.
In light of unknown's very interesting comment above, it seems that the question asked has the answer "no" in general, at least under one very reasonable interpretation of "canonical".
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As pointed out by KConrad in his comment to the question, my "example" cannot possibly be correct as stated. Since I do not remember the details, I have to withdraw my claim that a canonical square root of the different does not exist. – ulrich Jun 22 2010 at 10:17 | {"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": 177, "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": false}, "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": 0.9401608109474182, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/14758/flux-homomorphism-for-manifolds-with-boundary | Flux homomorphism for manifolds with boundary
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Hi all,
I am wondering whether someone has considered the definition of the flux homomorphism for manifolds with boundary. More specifically, I am looking at the annulus and I want the diffeomorphism $\phi(r, \theta) = (r, \theta + \alpha)$ for $\alpha$ constant to have non-zero flux.
I was thinking of extending the flux homomorphism using the relative homology like this:
$flux: Symp_0(A) \times H_1(A, \partial A) \to R$
where the value of $flux$ on the pair $(\phi, C)$ is given by choosing a 2-cycle $S$ such that $\partial S = C - \phi(C)$ and integrating the symplectic form over this cycle. Now, I think $H_1(A, \partial A)$ is generated by a circle homologous to the boundary and an arc connecting the two, and if you compute the value of $flux$ on the rotiational diffeomorphism and the latter generator, you get a non-zero result.
My questions are
1. Is this anywhere near correct, and if so, has this been done before?
2. Is there a way to define the flux homomorphism in exactly the same way as for the case of manifolds without boundary, so that the standard results are still valid?
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1
Shouldn't your map be circle-valued? The ambiguity is the area of the annulus. – Tim Perutz Feb 10 2010 at 0:46
Thanks for pointing this out. It took me a while to get it, but you're right: when choosing the 2-cycle S you can add any multiple of the fundamental class. I guess I need to add the condition that $[S] = 0$ as an element of $H_2$. Hmm, this is a good point -- I'm not sure how this is addressed in the usual definition of the flux homomorphism. – jvkersch Feb 11 2010 at 12:11
Of course, in the usual setup the flux homomorphism is circle valued -- that's the whole point. Sorry for being so slow on the uptake, I realized it the moment I added my comment. – jvkersch Feb 11 2010 at 12:13
1 Answer
1) $H_1(A,\partial A)$ is just one-dimensional, it is generated by a path that joins two sides of $A$.
2) The definition that you gave works for the annulus, and for surfaces with a boundary as well. This will also work for manifolds with boundary $M^n$, in the case when you consider fluxes of volume-preserving maps (i.e. you work with $\Omega^n$ and $H_{n-1} (M^n, \partial M^n$). I have not seen this definition before, but it is so natural, that it would be strange if no one considered it.
3) It does not look that this definition will work for higher-dimensional symplectic manfiolds, if you want to study fluxes of symplectomorphisms (and you work with $\omega$ and $H_1(M^{2n},\partial M^{2n})$), because the restriction of $\omega$ to $\partial$ will be non-zero.
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1) You're absolutely right: the other "generator" that I mentioned in my question is of course homologous to the boundary, hence zero. This also puts an end to much of my confusion about relative homology, so thanks again! 2-3) You're right about volume vs. symplectic manifolds. I'm trying to formalize some construction in Arnold's book on Topological Fluid Dynamics, so luckily volume-preserving maps are what I need. Putting it in the context of manifolds with a volume form gives me a pointer of where to look for references -- I remember seeing a flux homomorphism for volume manifolds... – jvkersch Feb 10 2010 at 10:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "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": 0.9351047277450562, "perplexity_flag": "head"} |
http://mathhelpforum.com/algebra/119177-difficulty-writing-summation-notation.html | # Thread:
1. ## Difficulty writing the summation notation..
I'm having some difficulty writing the geometric summation notation of this question:
3. a) write the series 10 + 5 + 5/2 + 5/4 + 5/8 + 5/16 + 5/32 using summation notation (sigma).
I discovered each term is divided by 2 so in this case r= 1/2
The formula Sn = [a(r^n-1)] / (r-1) should be this:
S1 = [10 (1/2^2 - 1)] / (1/2 - 1)
However, when I then try to verify this formula I get 10 when n=1 but 15 when n=2.
I must be doing something wrong, but I don't know what. I will give this another crack and see if I can find my errors. Any help is appreciated.
2. Hi
You made one simple mistake only:
The geometric sum is indeed $S_n = \frac{a(r^n-1)}{r-1}$ but this only applies when |r| > 1
When |r| < 1, it's actually $S_n = \frac{a(1-r^n)}{1-r}$
I can show you proof of this if you like. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 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": 0.8995950818061829, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showthread.php?t=388482 | Physics Forums
## against natural tendencies?
1. The problem statement, all variables and given/known data
CONSIDER A SITUATION
AN ELECTRON SAY MOVING ON THE SCREEN OF UR COMPUTER FROM BOTTOM TO TOP . SUDDENLY A MAGNETIC FIELD IS SWITCHED ON OUTWARDS TOWARDS YOUR FACE FROM THE SCREEN . NOW VISUALISE THE MOTION OF ELECTRON SURELY A ANTI CLOCKWISE CIRCLE { AN ELECTRON IS NEGATIVELY CHARGED }
NOW CONSIDER THIS REVOLVING ELECTRON AS AN EFFECTIVE CURRENT LOOP IN THE SAME MAGNETIC FIELD . SO A LOOP WITH CLOCKWISE CURRENT & ABOVE SAID FIELD. NOW VISUALISE THE MAGNETIC MOMENT VECTOR OF THE LOOP & TO THE SURPRISE IT IS AT AN ANGLE OF 180(deegre) TO THE FIELD THE CONDITION OF UNSTABLE EQUILIBRIUM WITH MAXIMUM POTENTIAL ENERGY.
2. Relevant equations
3. The attempt at a solution
MY DOUBT IS AS THE SYSTEM HAS MOVED INTO THE CONDITION ITSELF THEN WHY IT MOVED TO MAX. ENERGY CONFIGURATION AND NOT TO MINIMUM ENERGY ONE . DISPUTING WITH ALL WE STUDY ABOUT THE NATURALBEHAVIOR OF SYSTEMS.
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Quote by DV10 1. The problem statement, all variables and given/known data CONSIDER A SITUATION AN ELECTRON SAY MOVING ON THE SCREEN OF UR COMPUTER FROM BOTTOM TO TOP . SUDDENLY A MAGNETIC FIELD IS SWITCHED ON OUTWARDS TOWARDS YOUR FACE FROM THE SCREEN . NOW VISUALISE THE MOTION OF ELECTRON SURELY A ANTI CLOCKWISE CIRCLE { AN ELECTRON IS NEGATIVELY CHARGED } NOW CONSIDER THIS REVOLVING ELECTRON AS AN EFFECTIVE CURRENT LOOP IN THE SAME MAGNETIC FIELD . SO A LOOP WITH CLOCKWISE CURRENT & ABOVE SAID FIELD. NOW VISUALISE THE MAGNETIC MOMENT VECTOR OF THE LOOP & TO THE SURPRISE IT IS AT AN ANGLE OF 180(deegre) TO THE FIELD THE CONDITION OF UNSTABLE EQUILIBRIUM WITH MAXIMUM POTENTIAL ENERGY. 2. Relevant equations 3. The attempt at a solution MY DOUBT IS AS THE SYSTEM HAS MOVED INTO THE CONDITION ITSELF THEN WHY IT MOVED TO MAX. ENERGY CONFIGURATION AND NOT TO MINIMUM ENERGY ONE . DISPUTING WITH ALL WE STUDY ABOUT THE NATURALBEHAVIOR OF SYSTEMS.
What makes you so certain that the system has moved to a higher energy configuration? When you speak of "the system", are you referring to just the moving electron, or are you also including the two magnetic fields present? Remember, the fields themselves can be thought of as carrying energy and are capable of transferring it to a dipole, much like a rocket engine transfers energy to a rocket.
In any case, a single electron moving subject to a magnetic field is a poor example. If you apply the Lorentz force Law to the electron, you will find the force on it will always be perpendicular to its motion, and hence no work will be done on it, and it's energy will not change.
as far as i've read,potential energy of a current carrying loop in a uniform magnnetic field is U=-p.B where p,B are magnetic dipole moment vector of the loop and the magnetic field vector(the field already present) the tendency of a current carrying loop should have been to align itself WITH along the magnetic field already present..that would've been the min energy configuration..
## against natural tendencies?
Quote by gabbagabbahey Remember, the fields themselves can be thought of as carrying energy and are capable of transferring it to a dipole, much like a rocket engine transfers energy to a rocket.
could you elaborate a bit on this?
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Quote by DV10 as far as i've read,potential energy of a current carrying loop in a uniform magnnetic field is U=-p.B where p,B are magnetic dipole moment vector of the loop and the magnetic field vector
There are two important reasons why you can't apply this formula blindly to a single electron moving in a circle subject to a magnetic field:
(1) The formula you give is only valid for current loops that are small enough to be approximately considered as pure magnetic dipoles.
(2) Current carrying loops typically contain a very large number of charged particles. Each of these particles will be accelerating (the direction of their velocity changes as they traverse the loop), and accelerating charges produce time-varying magnetic fields, which in turn induce electric fields (Faraday's Law). It is a straight-forward calculation to show that the magnetic fields never directly do any work on a charged particle, and hence the work in this case must come from the intermediary electric fields. When their are many moving charges, then each individual charge will produce an electric field that will do work on all other charges, but (according to Newton's 3rd Law) not on itself. When you add up all the forces from all the electric fields created by all the accelerating charges, you will find that $\textbf{F}=\nabla(\textbf{p}\cdot\textbf{B})$ and hence $U=-\textbf{p}\cdot\textbf{B}$, provided that the loop is closely approximated by a magnetic dipole. (This is the underlying mechanism as to how that equation is derived via the Lorentz Force Law)
For a single moving electron, their are no electric fields to do work on it. The only electric field present is the one it produces, which according to Newton's 3rd Law will do no work on it. (In fact, classical electrodynamics does predict that accelerating charges will exert a force on themselves. But, neglecting this so-called radiation reaction force, which violates Newton's 3rd Law, this is a valid argument).
For this reason, a single electron moving in a circle is a poor example, as its potential energy doesn't change.
the tendency of a current carrying loop should have been to align itself WITH along the magnetic field already present..that would've been the min energy configuration..
A magnetic dipole would have a lower energy if it were aligned with the external field. However, this doesn't violate the laws of nature one bit.
A rocket ship has a lower energy sitting on the surface of Earth than it does in high orbit, does this mean that rocket ships violate the laws of nature? Of course not; the rockets engines burn fuel which provides additional energy to the rocket and raise it to a higher altitude.
So, a situation where you have a magnetic dipole anti-aligned with an external magnetic field must be similar to the rocket ship. If the dipole were created by the field, then its energy must have come from the field. Your particular example fails because a single electron in circular motion is not an ideal magnetic dipole, but if you were to devise some other scenario where an external field created a dipole that was opposed to the field, then you would have to conclude its energy came from the field (or rather, the power source that created the field) and was transferred to the charges that composed the dipole via intermediary induced electric fields.
If you are curious as to how to find the energy stored "in a magnetic field", look up Poynting's theorem.
Quote by gabbagabbahey (1) The formula you give is only valid for current loops that are small enough to be approximately considered as pure magnetic dipoles. Your particular example fails because a single electron in circular motion is not an ideal magnetic dipole, .
how do you define an idealistic situation here?i mean,lets say instead of a single electron, i have a beam of electrons..doesnt that work as a better approxiamtion?
in that case doesnt your 2nd arguement fail?
Quote by gabbagabbahey (2) Current carrying loops typically contain a very large number of charged particles. Each of these particles will be accelerating (the direction of their velocity changes as they traverse the loop), and accelerating charges produce time-varying magnetic fields, which in turn induce electric fields (Faraday's Law). It is a straight-forward calculation to show that the magnetic fields never directly do any work on a charged particle, and hence the work in this case must come from the intermediary electric fields. When their are many moving charges, then each individual charge will produce an electric field that will do work on all other charges, but (according to Newton's 3rd Law) not on itself. When you add up all the forces from all the electric fields created by all the accelerating charges, you will find that and hence , provided that the loop is closely approximated by a magnetic dipole. (This is the underlying mechanism as to how that equation is derived via the Lorentz Force Law) For a single moving electron, their are no electric fields to do work on it. The only electric field present is the one it produces, which according to Newton's 3rd Law will do no work on it. (In fact, classical electrodynamics does predict that accelerating charges will exert a force on themselves. But, neglecting this so-called radiation reaction force, which violates Newton's 3rd Law, this is a valid argument)..
as much as i've infered from this discussion is, that basically are you trying to say that in the scenario i've presented,an external agency has to do work on the system in order to sustain the uniformity of the magnetic field?(by uniformity i mean keeping the configuration of the field from changing)(this is the only seemingly valid arguement here)
ps-i appreciate your time here,thanks for trying to clear things up.. but im still a bit confused here..
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General Math 13 | {"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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9209950566291809, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/112772/how-to-find-co-ordinates-of-a-point-from-four-points-that-can-translate-or-rotat | # How to find co-ordinates of a point from four points that can translate or rotate? All these points form a rigid body.
I have a rigid body that translates and/or rotates about an axis perpendicular to the screen. I have co-ordinates of four points on the rigid body. How can I get the co-ordinates of a unknown point on the rigid body?
I have untransformed data for all five points. I have the transformed data for four points. I wish to find the transformed co-ordinates of fifth point.
The point(center) about which the body is rotated is not known.
Can I form a linear equation with four known points to get the fifth unknown point? How to accommodate translation and rotation in the equation?
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## 1 Answer
This is way overdetermined. You've got $8$ coordinates to determine $3$ parameters, one rotation angle and two translation parameters. You can use any two of the points to find the transformation (assuming that you have the untransformed coordinates of all five points, which you didn't state but presumably intended to imply?).
You can use a system of equations, but you can also just do this: Find the angle at which the second point you're using appears from the first, $\operatorname{atan2}(\Delta y,\Delta x)$, in both the transformed and the untransformed coordinates. The difference is the rotation angle. Then write the transformation as $TR$, where $R$ is a rotation around the origin and $T$ is a translation, apply $R$ with the calculated rotation angle to either of the untransformed points, and determine the translation parameters as the differences between its transformed coordinates and the coordinates of the rotated untransformed point.
(Note that operations are composed right-to-left, that is, $TR$ means first apply the rotation, then apply the translation. This notation is used so that $TRp$ can be interpreted as $T(R(p))$.)
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Thanks for the answer. But what if the point about which the body is rotated is not known. How to determine the point? – Arun Feb 24 '12 at 9:49
@Arun: The answer doesn't refer to a point about which the body is rotated. This is because any orientation-preserving isometry of the plane can be written as $TR$, with $T$ a translation and $R$ a rotation about an arbitrary point, for convenience the origin. Try rotating the body around two different points through the same angle; you'll see that the difference between the two results is a translation. – joriki Feb 24 '12 at 10:18 | {"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": 10, "end_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": 0.9324072003364563, "perplexity_flag": "head"} |
http://mathhelpforum.com/algebra/37757-adding-fractions.html | # Thread:
1. ## adding fractions
Show how to solve 5 3/4+ 1 2/3 in two different ways. In each case, express your answer as a mixed number. Explain why both of the methods are legitimate.
Thank you
The numbers above are fractiones
2. Hi
how to solve 5 3/4+ 1 2/3 in two different ways
What are you supposed to solve ? (there is no $=$ sign, no unknown...)
3. Its in form of a fraction so its
3 2
5-- + 1---
4 3
They want the answer to be expressed as a mixed number.
4. Originally Posted by flyingsquirrel
Hi
What are you supposed to solve ? (there is no $=$ sign, no unknown...)
The OP meant simplify.
In this case....
$5 \frac{3}{4} + 1 \frac{2}{3}$
Simplifies to:
$\frac{20+3}{4} + \frac{3+2}{3}$
$\frac{23}{4} + \frac{5}{3}$
LCD is 12
$\frac{69}{12} + \frac{20}{12}$
$\frac{89}{12}$
$7 \frac{5}{12}$
Another way to do this is to find the LCD when they are mixed numbers, and then simplify them to normal fractions.
5. 5 3/4 + 1 2/3 = (5 + 1) + (3/4 + 2/3) = 6 + (9/12 + 8/12) = 6 + 17/12 = 7 + 5/12 = 7 5/12. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9295995831489563, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/24094/thermodynamics-and-cross-entropy?answertab=votes | # Thermodynamics and cross entropy
I am facing with the concept of cross entropy. I would like to know the thermodynamic and statistical meaning of cross entropy (if exists)?
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What is cross entropy? Can you define it or provide a reference? – Vijay Murthy Apr 20 '12 at 20:16
– Emanuele Luzio Apr 21 '12 at 9:42
## 1 Answer
Thank you for your interesting question, which lead me to quite a nice little result. I couldn't find a nice thermodynamic meaning for the cross entropy, but I did find one for a related quantity called the Kullback-Leibler divergence. As far as I know, this is a new result.
The cross entropy between two probability distributions is defined as $$H(P, Q) = -\sum_i p_i \log q_i.$$ These two probability distributions should both refer to the same set of underlying states. Normally in thermodynamics we think of a system only having one probability distribution, which represents (roughly) the range of possible states the system might be in at the present time. But systems can change over time. So let's imagine have a system (with constant volume) that's initially in equilibrium with a heat bath at a temperature $T_1$. According to the usual principles of statistical mechanics, its state can be represented by the probability distribtion $$p_i = \frac{1}{Z_1} e ^{-\beta_1 u_i},$$ where $\beta_1=1/T_1$ (I've set Boltzmann's constant equal to 1 for clarity) and $Z_1$ is a normalisation factor called the partition function. The $u_i$ are the energy levels of the system's permitted microscopic states.
Now let's imagine we pick up our system and put it in contact with a heat bath at a different temperature, $T_2$, and let it come to equilibrium again. Since no work has been done, all the $u_i$ values will be unchanged and we'll have a new distribtion that looks like this $$q_i = \frac{1}{Z_2} e ^{-\beta_2 u_i},$$ where $\beta_2 = 1/T_2$.
Now we can do a bit of algebra to find the cross-entropy: $$H(P,Q) = \frac{1}{Z_1}\sum_i e^{-\beta_1u_i}(\beta_2u_i + \log(Z_2)) = \log(Z_2) + \frac{1}{Z_1}\sum_i e^{-\beta_1 u_i}\beta_2 u_i$$ $$= \log(Z_2) + U_1 \beta_2.$$
It's a standard result from statistical mechanics that $$S_2 = H(Q) = \log(Z_2) + U_2\beta_2.$$ Solving this for $\log(Z_2)$ and substituting into the cross entropy formula we have $$H(P,Q) = S_2 - \beta_2(U_2 - U_1) = S_2 - \frac{U_2 - U_1}{T_2}.$$ Physicists are, generally speaking, afraid of any quantity that has entropy units, and if they see one they like to multiply it by a temperature in order to make it look like an energy. If we multiply this by $T_2 = 1/\beta_2$ we get $$T_2H(P,Q) = T_2S_2 - \Delta U.$$ It's possible that this might have a nice thermodynamic interpretation in terms of something like the maximum amount of work that we can extract from doing this transformation under a particular set of circumstances --- but if it does then I haven't seen it just yet. The expression looks tantalisingly like a change in free energy ($\Delta U - T\Delta S$), but it's not quite the same.
However, we can get a much more interesting result if we note that in information theory, the Kullback-Leibler divergence (aka information gain) is often seen as more fundamental than the cross entropy. The KL-divergence is defined as $$D_{KL}(P\|Q) = \sum_i p_i \log\frac{p_i}{q_i} = H(P,Q)-H(P),$$ which in our case is equal to $$S_2 - S_1 - \frac{U_2-U_1}{T_2} = \Delta S - \Delta U/T_2.$$ This is much more interesting than the result for the cross-entropy, because it does have a clear thermodynamic interpretation. When we put the system in contact with the second heat bath, its entropy changes by $\Delta S$, and the entropy of the heat bath changes by $-\Delta U/T_2$. (This is because entropy is heat divided by temperature - an amount $\Delta U$ leaves the system, so $-\Delta U$ enters the heat bath.) So the KL-divergence is just the total change in entropy after we put the system in contact with the new heat bath. I'm quite excited about this because I didn't know it before, and I don't think anyone else did either!
We can even take this a bit further. Let's imagine putting a heat engine in between the system and the second heat reservoir. So we'll try to extract some useful work from the flow of heat that takes place as the system and the heat bath equilibriate. If we do this the total change of entropy becomes $\Delta S + (-\Delta U - W)/T_2$. This has to be greater than 0, which means that $W\le T_2\Delta S - \Delta U$.
Now, if we do that physicist thing of multiplying $D_{KL}$ by $T_2$, it becomes $T_2\Delta S - \Delta U$, which is the value for the maximum work that we just calculated. So while the thermodynamic meaning of the cross-entropy isn't clear to me, the KL-divergence does seem to have a nice interpretation in terms of work.
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Great. Is exactly the answer that i was looking for. – Emanuele Luzio Apr 21 '12 at 14:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 10, "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": false}, "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": 0.9582698345184326, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/24881/list | ## Return to Answer
2 added 187 characters in body
Just as in any area of mathematics, different parts of the subject have different strengths. The theory of Reverse Mathematics investigates these strengths by determining for a large number of mathematical theorems, including theorems arising in the development of first order logic, the weakest axioms that are able to prove them. In order to do this, one reverses the usual direction of mathematical proof, by proving the axioms from the theorems(, over an extremely weak base theory)theory.
The remarkable fact of Reverse Mathematics is that many/most many theorems of classical mathematics fall into five large equivalence classes, consisting of provably equivalent theorems. Many instances of this are listed on the Wikipedia page I link to above. I believe
It appears that all the most trivial parts of logical syntax can be developed in the weak base system $RCA_0$. Beyond RCA_0\$, and this , here are some may be the answer to your question. For more powerful parts of the less trivial examplestheory, one needs stronger axioms.
• A weak version of the Goedel Incompleteness Theorem (for countable theory, already closed under conseqeunce) is also provable already in the base theory $RCA_0$. (As is the Baire Category theorem and the existence, but not uniqueness, of an algebraic closure for any countable field.)
• The Completeness Theorem for countable languages is equivalent to $WKL_0$. (As is the Heine-Borel theorem, the Jordan curve theorem, and many others.)
• the sequential completeness of the reals and various forms of Ramsey's theorem are equivalent to $ACA_0$.
• Comparability of well-orders and Open Determinacy are equivalent to $ATR_0$.
• the Cantor-Bendixon theorem is equivalent to $\Pi^1_1-CA_0$.
1
Just as in any area of mathematics, different parts of the subject have different strengths. The theory of Reverse Mathematics investigates these strengths by determining for a large number of mathematical theorems, including theorems arising in the development of first order logic, the weakest axioms that are able to prove them. In order to do this, one reverses the usual direction of mathematical proof, by proving the axioms from the theorems (over an extremely weak base theory).
The remarkable fact of Reverse Mathematics is that many/most theorems of classical mathematics fall into five large equivalence classes, consisting of provably equivalent theorems. Many instances of this are listed on the Wikipedia page I link to above. I believe that all the most trivial parts of logical syntax can be developed in the weak base system $RCA_0$. Beyond this, here are some of the less trivial examples.
• A weak version of the Goedel Incompleteness Theorem (for countable theory, already closed under conseqeunce) is provable in the base theory $RCA_0$.
• The Completeness Theorem for countable languages is equivalent to $WKL_0$. (As is the Heine-Borel theorem, the Jordan curve theorem, and many others.)
• the sequential completeness of the reals and various forms of Ramsey's theorem are equivalent to $ACA_0$.
• Comparability of well-orders and Open Determinacy are equivalent to $ATR_0$.
• the Cantor-Bendixon theorem is equivalent to $\Pi^1_1-CA_0$. | {"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": 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": 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": 0.9290969967842102, "perplexity_flag": "head"} |
http://www.haskell.org/haskellwiki/index.php?title=User:Michiexile/MATH198/Lecture_10&diff=32078&oldid=32053 | # User:Michiexile/MATH198/Lecture 10
### From HaskellWiki
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| | No homework at this point. However, if you want something to think about, a few questions and exercises: | | No homework at this point. However, if you want something to think about, a few questions and exercises: |
| | | | |
| - | # | + | # Prove the relations showing that <math>\leq_1</math> is indeed a partial order on <math>[U\to\Omega]</math>. |
| | | + | # Prove the universal quantifier theorem. |
| | | + | # The ''extension'' of a formula <math>\phi</math> over a list of variables <math>x</math> is the sub-object of the product of domains <math>A_1\times\dots\times A_n</math> for the variables <math>x_1,\dots,x_n=x</math> classified by the interpretation of <math>\phi</math> as a morphism <math>A_1\times\dots\times A_n\to\Omega</math>. A formula is ''true'' if it classifies the entire product. A ''sequent'', written <math>\Gamma:\phi</math> is the statement that using the set of formulae <math>\Gamma</math> we may prove <math>\phi</math>, or in other words that the intersection of the extensions of the formulae in <math>\Gamma</math> is contained in the extension of <math>\phi</math>. If a sequent <math>\Gamma:\phi</math> is true, we say that <math>\Gamma</math> ''entails'' <math>\phi</math>. (some of the questions below are almost embarrassingly immediate from the definitions given above. I include them anyway, so that a ''catalogue'' of sorts of topoidal logic inferences is included here) |
| | | + | ## Prove the following entailments: |
| | | + | ### Trivial sequent: <math>\phi:\phi</math> |
| | | + | ### True: <math>:true</math> (note that true classifies the entire object) |
| | | + | ### False: <math>false:\phi</math> (note that false classifies the global minimum in thepreorder of subobjects) |
| | | + | ## Prove the following inference rules: |
| | | + | ### Implication: <math>\Gamma,\phi:\psi</math> is equivalent to <math>\Gamma:\phi\Rightarrow\psi</math>. |
| | | + | ### Thinning: <math>\Gamma:\phi</math> implies <math>\Gamma,\psi:\phi</math> |
| | | + | ### Cut: <math>\Gamma,\psi:\phi</math> and <math>\Gamma:\psi</math> imply <math>\Gamma:\phi</math> if every variable free in <math>\psi</math> is free in <math>\Gamma</math> or in <math>\phi</math>. |
| | | + | ### Negation: <math>\Gamma, \phi: false</math> is equivalent (implications both ways) to <math>\Gamma: \neg\phi</math>. |
| | | + | ### Conjunction: <math>\Gamma:\phi</math> and <math>\Gamma:\psi</math> together are equivalent to <math>\Gamma:\phi\wedge\psi</math>. |
| | | + | ### Disjunction: <math>\Gamma,\phi:\theta</math> and <math>\Gamma,\psi:\theta</math> together imply <math>\Gamma, \phi\vee\psi: \theta</math>. |
| | | + | ### Universal: <math>\Gamma:\phi</math> is equivalent to <math>\Gamma:\forall x.\phi</math> if <math>x</math> is not free in <math>\Gamma</math>. |
| | | + | ### Existential: <math>\Gamma,\phi: \psi</math> is equivalent to <math>\Gamma,\exists x.\phi:\psi</math> if <math>x</math> is not free in <math>\Gamma</math> or <math>\psi</math>. |
| | | + | ### Equality: <math>:q=q</math>. |
| | | + | ### Biconditional: <math>(v\Rightarrow w)\wedge(w\Rightarrow v):v=w</math>. We usually write <math>v\Leftrightarrow w</math> for <math>v=w</math> if <math>v,w:A\to\Omega</math>. |
| | | + | ### Product: <math>p_1u = p_1u', p_2u = p_2u' : u = u'</math> for <math>u,u'\in A\times B</math>. |
| | | + | ### Product revisited: <math>:(p_1(s\times s')=s)\wedge(p_2(s\times s')=s')</math>. |
| | | + | ### Extensionality: <math>\forall x\in A. f(x) = g(x) : f = g</math> for <math>f,g\in[A\to B]</math>. |
| | | + | ### Comprehension: <math>(\lambda x\in A. s)x = s</math> for <math>x\in A</math>. |
| | | + | ## Prove the following results from the above entailments and inferences -- or directly from the topoidal logic mindset: |
| | | + | ### <math>:\neg(\phi\wedge\neg\phi)</math>. |
| | | + | ### <math>:\phi\Rightarrow\neg\neg\phi</math>. |
| | | + | ### <math>:\neg(\phi\vee\psi)\Rightarrow(\neg\phi\wedge\neg\psi)</math>. |
| | | + | ### <math>:(\neg\phi\wedge\neg\psi)\Rightarrow\neg(\phi\wedge\psi)</math>. |
| | | + | ### <math>:(\neg\phi\vee\neg\psi)\Rightarrow\neg(\phi\vee\psi)</math>. |
| | | + | ### <math>\phi\wedge(\theta\vee\psi)</math> is equivalent to <math>(\phi\wedge\theta)\vee(\phi\wedge\psi)</math>. |
| | | + | ### <math>\forall x.\neg\phi</math> is equivalent to <math>\neg\exists x.\phi</math>. |
| | | + | ### <math>\exists x\phi\Rightarrow\neg\forall x.\neg\phi</math>. |
| | | + | ### <math>\exists x\neg\phi\Rightarrow\neg\forall x.\phi</math>. |
| | | + | ### <math>\forall x\phi\Rightarrow\neg\exists x.\neg\phi</math>. |
| | | + | ### <math>\phi:\psi</math> implies <math>\neg\psi:\neg\phi</math>. |
| | | + | ### <math>\phi:\psi\Rightarrow\phi</math>. |
| | | + | ### <math>\phi\Rightarrow\not\phi:\not\phi</math>. |
| | | + | ### <math>\not\phi\vee\psi:\phi\Rightarrow\psi</math> (but not the converse!). |
| | | + | ### <math>\neg\neg\neg\phi</math> is equivalent to <math>\neg\phi</math>. |
| | | + | ### <math>(\phi\wedge\psi)\Rightarrow\theta</math> is equivalent to <math>\phi\Rightarrow(\psi\Rightarrow\theta)</math> (currying!). |
| | | + | ## Using the Boolean negation rule: <math>\Gamma,\neg\phi:false</math> is equivalent to <math>\Gamma:\phi</math>, prove the following additional results: |
| | | + | ### <math>\neg\neg\phi:\phi</math>. |
| | | + | ### <math>:\phi\vee\neg\phi</math>. |
| | | + | ## Show that either of the three rules above, together with the original negation rule, implies the Boolean negation rule. |
| | | + | ### The converses of the three existential/universal/negation implications above. |
| | | + | ## The restrictions introduced for the cut rule above block the deduction of an entailment <math>:\forall x.\phi\Rightarrow\exists x.\phi</math>. The issue at hand is that <math>A</math> might not actually have members; so choosing one is not a sound move. Show that this entailment can be deduced from the premise <math>\exists x\in A. x=x</math>. |
| | | + | ## Show that if we extend our ruleset by the quantifier negation rule <math>\forall x\Leftrightarrow \neg\exists x.\neg</math>, then we can derive the entailment <math>:\forall w: w=t \vee w = false</math>. From this derive <math>:\phi\vee\neg\phi</math> and hence conclude that this extension gets us Boolean logic again. |
| | | + | # A ''topology'' on a topos <math>E</math> is an arrow <math>j:\Omega\to\Omega</math> such that <math>j\circ true=true</math>, <math>j\circ j=j<math> and <math>j\circ\wedge = \wedge\circ j\times j</math>. For a subobject <math>S\subseteq A</math> with characteristic arrow <math>\chi_S:A\to\Omega</math>, we define its <math>j</math>-closure as the subobject <math>\bar S\subseteq A</math> classified by <math>j\circ\chi_S</math>. |
| | | + | ## Prove: |
| | | + | ### <math>S\subseteq\bar S</math>. |
| | | + | ### <math>\bar S = \bar{\bar S}</math>. |
| | | + | ### <math>\bar{S\cap T} = \bar S\cap\bar T</math>. |
| | | + | ### <math>S\subseteq T</math> implies <math>\bar S\subseteq\bar T</math>. |
| | | + | ### <math>\bar{f^{-1}(S)} = f^{-1}(\bar S)</math>. |
| | | + | ## We define <math>S</math> to be <math>j</math>-closed if <math>S=\bar S</math>. It is <math>j</math>-dense if <math>\bar S=A</math>. These terms are chosen due to correspondences to classical pointset topology for the topos of sheaves over some space. For a logical standpoint, it is more helpful to look at <math>j</math> as a modality operator: "''it is <math>j</math>-locally true that''" Given any <math>u:1\to\Omega</math>, prove that the following are topologies: |
| | | + | ### <math>(u\to -): \Omega\to\Omega</math> (the ''open topology'', where such a <math>u</math> in a sheaf topos ends up corresponding to an open subset of the underlying space, and the formulae picked out are true on at least all of that subset). |
| | | + | ### <math>u\vee -): \Omega\to\Omega</math> (the closed topology, where a formula is true if its disjunction with <math>u</math> is true -- corresponding to formulae holding over at least the closed set complementing the subset picked out) |
| | | + | ### <math>\neg\neg: \Omega\to\Omega</math>. This may, depending on the topos, end up being interpreted as ''true so far as global elements are concerned'', or ''not false on any open set'', or other interpretations. |
| | | + | ### <math>1_\Omega</math>. |
| | | + | ## For a topos <math>E</math> with a topology <math>j</math>, we define an object <math>A</math> to be a ''sheaf'' iff for every <math>X</math> and every <math>j</math>-dense subobject <math>S\subseteq X</math> and every <math>f:S\to A</math> there is a unique <math>g:X\to A</math> with <math>f=g\circ s</math>. In other words, <math>A</math> is an object that cannot see the difference between <math>j</math>-dense subobjects and objects. We write <math>E_j</math> for the full subcategory of <math>j</math>-sheaves. |
| | | + | ### Prove that any object is a sheaf for <math>1_\Omega</math>. |
| | | + | ### Prove that a subobject is dense for <math>\neg\neg</math> iff its negation is empty. Show that <math>true+false:1+1\to\Omega</math> is dense for this topology. Conclude that <math>1+1</math> is dense in <math>\Omega_{\neg\neg}</math> and thus that <math>E_{\neg\neg}</math> is Boolean. |
## Revision as of 18:20, 2 December 2009
IMPORTANT NOTE: THESE NOTES ARE STILL UNDER DEVELOPMENT. PLEASE WAIT UNTIL AFTER THE LECTURE WITH HANDING ANYTHING IN, OR TREATING THE NOTES AS READY TO READ.
This lecture will be shallow, and leave many things undefined, hinted at, and is mostly meant as an appetizer, enticing the audience to go forth and seek out the literature on topos theory for further studies.
## Contents
### 1 Subobject classifier
One very useful property of the category Set is that the powerset of a given set is still a set; we have an internal concept of object of all subobjects. Certainly, for any category (small enough) C, we have a contravariant functor $Sub(-): C\to Set$ taking an object to the set of all equivalence classes of monomorphisms into that object; with the image Sub(f) given by the pullback diagram
If the functor Sub( − ) is representable - meaning that there is some object $X\in C_0$ such that Sub( − ) = hom( − ,X) - then the theory surrounding representable functors, connected to the Yoneda lemma - give us a number of good properties.
One of them is that every representable functor has a universal element; a generalization of the kind of universal mapping properties we've seen in definitions over and over again during this course; all the definitions that posit the unique existence of some arrow in some diagram given all other arrows.
Thus, in a category with a representable subobject functor, we can pick a representing object $\Omega\in C_0$, such that Sub(X) = hom(X,Ω). Furthermore, picking a universal element corresponds to picking a subobject $\Omega_0\hookrightarrow\Omega$ such that for any object A and subobject $A_0\hookrightarrow A$, there is a unique arrow $\chi: A\to\Omega$ such that there is a pullback diagram
One can prove that Ω0 is terminal in C, and we shall call Ω the subobject classifier, and this arrow $\Omega_0=1\to\Omega$ true. The arrow χ is called the characteristic arrow of the subobject.
In Set, all this takes on a familiar tone: the subobject classifier is a 2-element set, with a true element distinguished; and a characteristic function of a subset takes on the true value for every element in the subset, and the other (false) value for every element not in the subset.
### 2 Defining topoi
Definition A topos is a cartesian closed category with all finite limits and with a subobject classifier.
It is worth noting that this is a far stronger condition than anything we can even hope to fulfill for the category of Haskell types and functions. The functional programming relevance will take a back seat in this lecture, in favour of usefulness in logic and set theory replacements.
### 3 Properties of topoi
The meat is in the properties we can prove about topoi, and in the things that turn out to be topoi.
Theorem Let E be a topos.
• E has finite colimits.
#### 3.1 Power object
Since a topos is closed, we can take exponentials. Specifically, we can consider $[A\to\Omega]$. This is an object such that $hom(B,[A\to\Omega]) = hom(A\times B, \Omega) = Sub(A\times B)$. Hence, we get an internal version of the subobject functor. (pick B to be the terminal object to get a sense for how global elements of $[A\to\Omega]$ correspond to subobjects of A)
#### 3.2 Internal logic
We can use the properties of a topos to develop a logic theory - mimicking the development of logic by considering operations on subsets in a given universe:
Classically, in Set, and predicate logic, we would say that a predicate is some function from a universe to a set of truth values. So a predicate takes some sort of objects, and returns either True or False.
Furthermore, we allow the definition of sets using predicates:
$\{x\in U: P(x)\}$
Looking back, though, there is no essential difference between this, and defining the predicate as the subset of the universe directly; the predicate-as-function appears, then, as the characteristic function of the subset. And types are added as easily - we specify each variable, each object, to have a set it belongs to.
This way, predicates really are subsets. Type annotations decide which set the predicate lives in. And we have everything set up in a way that opens up for the topos language above.
We'd define, for predicates P,Q acting on the same type:
$\{x\in A : \top\} = A$
$\{x\in A : \bot\} = \emptyset$
$\{x : (P \wedge Q)(x)\} = \{x : P(x)\} \cap \{x : Q(x)\}$
$\{x : (P \vee Q)(x)\} = \{x : P(x)\} \cup \{x : Q(x)\}$
$\{x\in A : (\neg P)(x) \} = A \setminus \{x\in A : P(x)\}$
We could then start to define primitive logic connectives as set operations; the intersection of two sets is the set on which both the corresponding predicates hold true, so $\wedge = \cap$. Similarily, the union of two sets is the set on which either of the corresponding predicates holds true, so $\vee = \cup$. The complement of a set, in the universe, is the negation of the predicate, and all other propositional connectives (implication, equivalence, ...) can be built with conjunction (and), disjunction (or) and negation (not).
So we can mimic all these in a given topos:
We say that a universe U is just an object in a given topos. (Note that by admitting several universes, we arrive at a typed predicate logic, with basically no extra work.)
A predicate is a subobject of the universe.
We can now proceed to define all the familiar logic connectives one by one, using the topos setting. While doing this, we shall introduce the notation $t_A: A\to\Omega$ for the morphism $t_A = A\to 1\to^{true}\Omega$ that takes on the value true on all of A. We note that with this convention, $\chi_{A_0}$, the characteristic morphism of a subobject, is the arrow such that $\chi_{A_0}\circ i = t_{A_0}$.
Conjunction: Given predicates P,Q, we need to define the conjunction $P\wedge Q$ as some $P\wedge Q: U\to\Omega$ that corresponds to both P and Q simultaneously.
We may define $true\times true: 1\to\Omega\times\Omega$, a subobject of $\Omega\times\Omega$. Being a subobject, this has a characteristic arrow $\wedge:\Omega\times\Omega\to\Omega$, that we call the conjunction arrow.
Now, we may define $\chi_P\times\chi_Q:U\to\Omega\times\Omega$ for subobjects $P,Q\subseteq U$ - and we take their composition $\wedge\circ\chi_P\times\chi_Q$ to be the characteristic arrow of the subobject $P\wedge Q$.
And, indeed, this results in a topoidal version of intersection of subobjects.
Implication: Next we define $\leq_1: \Omega_1\to\Omega\times\Omega$ to be the equalizer of $\wedge$ and proj1. Given $v, w: U\to\Omega$, we write $v\leq_1 w$ if $v\times w$ factors through $\leq_1$.
Using the definition of an equalizer we arrive at $v\leq_1 w$ iff $v = v\wedge w$. From this, we can deduce
$u\leq_1 true$
$u\leq_1 u$
If $u\leq_1 v$ and $v\leq_1 w$ then $u\leq_1 w$.
If $u\leq_1 v$ and $v\leq_1 u$ then u = v
and thus, $\leq_1$ is a partial order on $[U\to\Omega]$. Intuitively, $u\leq_1 v$ if v is at least as true as u.
This relation corresponds to inclusion on subobjects. Note that $\leq_1:\Omega_1\to\Omega\times\Omega$, given from the equalizer, gives us Ω1 as a relation on Ω - a subobject of $\Omega\times\Omega$. Specifically, it has a classifying arrow $\Rightarrow:\Omega\times\Omega\to\Omega$. We write $h\Rightarrow k = (\Rightarrow)\circ h\times k$. And for subobjects $P,Q\subseteq A$, we write $P\Rightarrow Q$ for the subobject classified by $\chi_P\Rightarrow\chi_Q$.
It turns out, without much work, that this $P\Rightarrow Q$ behaves just like classical implication in its relationship to $\wedge$.
Membership: We can internalize the notion of membership as a subobject $\in^U\subseteq U\times\Omega^U$, and thus get the membership relation from a pullback:
For elements $x\times h: 1\to U\times\Omega^U$, we write $x\in^U h$ for $x\times h\in\in^U$. Yielding a subset of the product $U\times\Omega^U$, this is readily interpretable as a relation relating things in U with subsets of U, so that for any x,h we can answer whether $x\in^Uh$. Both notations indicate $ev_A\circ h\times x = true$.
Universal quantification: For any object U, the maximal subobject of U is U itself, embedded with 1U into itself. There is an arrow $\tau_U:1\to\Omega^U$ represents this subobject. Being a map from 1, it is specifically monic, so it has a classifying arrow $\forall_U:\Omega^U\to\Omega$ that takes a given subobject of U to true precisely if it is in fact the maximal subobject.
Now, with a relation $r:R\to B\times A$, we define $\forall a. R$ by the following pullback:
where λχR comes from the universal property of the exponential.
Theorem For any $s:S\to B$ monic, $S\subseteq \forall a.R$ iff $S\times A\subseteq R$.
This theorem tells us that the subobject given by $\forall a.R$ is the largest subobject of B that is related by R to all of A.
Falsum: We can define the false truth value using these tools as $\forall w\in\Omega.w$. This might be familiar to the more advanced Haskell type hackers - as the type
x :: forall a. a
which has to be able to give us an element of any type, regardless of the type itself. And in Haskell, the only element that inhabits all types is
undefined
.
From a logical perspective, we use a few basic inference rules:
and connect them up to derive
for any φ not involving w - and we can always adjust any φ to avoid w.
Thus, the formula $\forall w.w$ has the property that it implies everything - and thus is a good candidate for the false truth value; since the inference
is the defining introduction rule for false.
Negation: We define negation the same way as in classical logic: $\neg \phi = \phi \Rightarrow false$.
Disjunction: We can define
$P\vee Q = \forall w. ((\phi\Rightarrow w)\wedge(\psi\Rightarrow w))\Rightarrow w$
Note that this definition uses one of our primary inference rules:
as the defining property for the disjunction, and we may derive any properties we like from these.
Existential quantifier: Finally, the existential quantifier is derived similarly to the disjunction - by figuring out a rule we want it to obey, and using that as a definition for it:
$\exists x.\phi = \forall w. (\forall x. \phi \Rightarrow w)\Rightarrow w$
Here, the rule we use as defining property is
Before we leave this exploration of logic, some properties worth knowing about: While we can prove $\neg(\phi\wedge\neg\phi)$ and $\phi\Rightarrow\neg\neg\phi$, we cannot, in just topos logic, prove things like
$\neg(\phi\wedge\psi)\Rightarrow(\neg\phi\vee\neg\psi)$
$\neg\neg\phi\Rightarrow\phi$
nor any statements like
$\neg(\forall x.\neg\phi)\Rightarrow(\exists x.\phi)$
$\neg(\forall x.\phi)\Rightarrow(\exists x.\neg\phi)$
$\neg(\exists x.\neg\phi)\Rightarrow(\forall x.\phi)$
We can, though, prove
$\neg(\exists x.\phi)\Rightarrow(\forall x.\neg\phi)$
If we include, extra, an additional inference rule (called the Boolean negation rule) given by
then suddenly we're back in classical logic, and can prove $\neg\neg\phi\Rightarrow\phi$ and $\phi\or\neg\phi$.
#### 3.3 Examples: Sheaves, topology and time sheaves
The first interesting example of a topos is the category of (small enough) sets; in some sense clear already since we've been modelling our axioms and workflows pretty closely on the theory of sets.
Generating logic and set theory in the topos of sets, we get a theory that captures several properties of intuitionistic logic; such as the lack of Boolean negation, of exclusion of the third, and of double negation rules.
For the more interesting examples, however, we shall introduce the concepts of topology and of sheaf:
Definition A (set-valued) presheaf on a category C is a contravariant functor $E: C^{op}\to Set$.
Presheaves occur all over the place in geometry and topology - and occasionally in computer science too: There is a construction in which a functor $A\to Set$ for a discrete small category A identified with its underlying set of objects as a set, corresponds to the data type of bags of elements from A - for $a\in A$, the image F(a) denotes the multiplicity of a in the bag.
Theorem The category of all presheaves (with natural transformations as the morphisms) on a category C form a topos.
Example Pick a category on the shape
A contravariant functor on this category is given by a pair of sets G0,G1 and a pair of function $source, target: G_1\to G_0$. Identities are sent to identities.
The category of presheaves on this category, thus, is the category of graphs. Thus graphs form a topos.
The subobject classifier in the category of graphs is a graph with two nodes: in and out, and five arrows:
$in \to^{all} in$
$in \to^{both} in$
$in \to^{source} out$
$out \to^{target} in$
$out \to^{neither} out$
Now, given a subgraph $H \leq G$, we define a function $\chi_H:G\to\Omega$ by sending nodes to in or out dependent on their membership. For an arrow a, we send it to all if the arrow is in H, and otherwise we send it to both/source/target/neither according to where its source and target reside.
To really get into sheaves, though, we introduce more structure - specifically, we define what we mean by a topology:
Definition Suppose P is a partially ordered set. We call P a complete Heyting algebra if
• There is a top element 1 such that $x\leq 1 \forall x\in P$.
• Any two elements x,y have an infimum (greatest lower bound) $x\wedge y$.
• Every subset $Q\subseteq P$ has a supremum (least upper bound) $\bigvee_{p\in P} p$.
• $x\wedge(\bigvee y_i) = \bigvee x\wedge y_i$
Note that for the partial order by inclusion of a family of subsets of a given set, being a complete Heyting algebra is the same as being a topology in the classical sense - you can take finite unions and any intersections of open sets and still get an open set.
If {xi} is a subset with supremum x, and E is a presheaf, we get functions $e_i:E(x)\to E(x_i)$ from functoriality. We can summarize all these ei into $e = \prod_i e_i: E(x)\to\prod_i E(x_i)$.
Furthermore, functoriality gives us families of functions $c_{ij}: E(x_i)\to E(x_i\wedge x_j)$ and $d_{ij}: E(x_j)\to E(x_i\wedge x_j)$. These can be collected into $c: \prod_i E(x_i)\to\prod_{ij}E(x_i\wedge x_j)$ and $d:\prod_j E(x_j)\to\prod_{ij}E(x_i\wedge x_j)$.
Definition A presheaf E on a Heyting algebra is called a sheaf if it satisfies:
$x = \bigvee x_i$
implies that
is an equalizer. If you have seen sheaves before, you may recognize this as the covering axiom.
In other words, E is a sheaf if whenever $x=\bigvee x_i$ and c(α) = d(α), then there is some $\bar\alpha$ such that $\alpha = e(\bar\alpha)$.
Theorem The category of sheaves on a Heyting algebra is a topos.
For context, we can think of sheaves over Heyting algebras as sets in a logic with an expanded notion of truth. Our Heyting algebra is the collection of truth values, and the sheaves are the fuzzy sets with fuzziness introduced by the Heyting algebra.
Recalling that subsets and predicates are viewed as the same thing, we can view the set E(p) as the part of the fuzzy set E that is at least p true.
As it turns out, to really make sense of this approach, we realize that equality is a predicate as well - and thus can hold or not depending on the truth value we use.
Definition Let P be a complete Heyting algebra. A P-valued set is a pair (S,σ) of a set S and a function $\sigma: S\to P$. A category of fuzzy sets is a category of P-valued sets. A morphism $f:(S,\sigma)\to(T,\tau)$ of P-valued sets is a function $f:S\to T$ such that $\tau\circ f = \sigma$.
From these definitions emerges a fuzzy set theory where all components of it being a kind of set theory emerges from the topoidal approach above. Thus, say, subsets in a fuzzy sense are just monics, thus are injective on the set part, and such that the valuation, on the image of the injection, increases from the previous valuation: $(T,\tau)\subseteq(S,\sigma)$ if $T\subseteq S$ and σ | T = τ.
To get to topoi, though, there are a few matters we need to consider. First, we may well have several versions of the empty set - either a bona fide empty set, or just a set where every element is never actually there. This issue is minor. Much more significant though, is that while we can easily make (S,σ) give rise to a presheaf, by defining
$E(x) = \{s\in S: \sigma(s)\geq x\}$
this definition will not yield a sheaf. The reason for this boils down to $E(0) = S \neq 1$. We can fix this, though, by adjoining another element - $\bot$ - to P giving P + . The new element $\bot$ is imbued with two properties: it is smaller, in P + , than any other element, and it is mapped, by E to 1.
Theorem The construction above gives a fuzzy set (S,σ) the structure of a sheaf on the augmented Heyting algebra.
Corollary The category of fuzzy sets for a Heyting algebra P forms a topos.
Final note While this construction allows us to make membership a fuzzy concept, we're not really done fuzzy-izing sets. There are two fundamental predicates on sets: equality and membership. While fuzzy set theory, classically, only allows us to make one of these fuzzy, topos theory allows us - rather easily - to make both these predicates fuzzy. Not only that, but membership reduces - with the power object construction - to equality testing, by which the fuzzy set theory ends up somewhat inconsistent in its treatment of the predicates.
### 4 Literature
At this point, I would warmly recommend the interested reader to pick up one, or more, of:
• Steve Awodey: Category Theory
• Michael Barr & Charles Wells: Categories for Computing Science
• Colin McLarty: Elementary Categories, Elementary Toposes
or for more chewy books
• Peter T. Johnstone: Sketches of an Elephant: a Topos Theory compendium
• Michael Barr & Charles Wells: Toposes, Triples and Theories
### 5 Exercises
No homework at this point. However, if you want something to think about, a few questions and exercises:
1. Prove the relations showing that $\leq_1$ is indeed a partial order on $[U\to\Omega]$.
2. Prove the universal quantifier theorem.
3. The extension of a formula φ over a list of variables x is the sub-object of the product of domains $A_1\times\dots\times A_n$ for the variables $x_1,\dots,x_n=x$ classified by the interpretation of φ as a morphism $A_1\times\dots\times A_n\to\Omega$. A formula is true if it classifies the entire product. A sequent, written Γ:φ is the statement that using the set of formulae Γ we may prove φ, or in other words that the intersection of the extensions of the formulae in Γ is contained in the extension of φ. If a sequent Γ:φ is true, we say that Γ entails φ. (some of the questions below are almost embarrassingly immediate from the definitions given above. I include them anyway, so that a catalogue of sorts of topoidal logic inferences is included here)
1. Prove the following entailments:
1. Trivial sequent: φ:φ
2. True: :true (note that true classifies the entire object)
3. False: false:φ (note that false classifies the global minimum in thepreorder of subobjects)
2. Prove the following inference rules:
1. Implication: Γ,φ:ψ is equivalent to $\Gamma:\phi\Rightarrow\psi$.
2. Thinning: Γ:φ implies Γ,ψ:φ
3. Cut: Γ,ψ:φ and Γ:ψ imply Γ:φ if every variable free in ψ is free in Γ or in φ.
4. Negation: Γ,φ:false is equivalent (implications both ways) to $\Gamma: \neg\phi$.
5. Conjunction: Γ:φ and Γ:ψ together are equivalent to $\Gamma:\phi\wedge\psi$.
6. Disjunction: Γ,φ:θ and Γ,ψ:θ together imply $\Gamma, \phi\vee\psi: \theta$.
7. Universal: Γ:φ is equivalent to $\Gamma:\forall x.\phi$ if x is not free in Γ.
8. Existential: Γ,φ:ψ is equivalent to $\Gamma,\exists x.\phi:\psi$ if x is not free in Γ or ψ.
9. Equality: :q = q.
10. Biconditional: $(v\Rightarrow w)\wedge(w\Rightarrow v):v=w$. We usually write $v\Leftrightarrow w$ for v = w if $v,w:A\to\Omega$.
11. Product: p1u = p1u',p2u = p2u':u = u' for $u,u'\in A\times B$.
12. Product revisited: $:(p_1(s\times s')=s)\wedge(p_2(s\times s')=s')$.
13. Extensionality: $\forall x\in A. f(x) = g(x) : f = g$ for $f,g\in[A\to B]$.
14. Comprehension: $(\lambda x\in A. s)x = s$ for $x\in A$.
3. Prove the following results from the above entailments and inferences -- or directly from the topoidal logic mindset:
1. $:\neg(\phi\wedge\neg\phi)$.
2. $:\phi\Rightarrow\neg\neg\phi$.
3. $:\neg(\phi\vee\psi)\Rightarrow(\neg\phi\wedge\neg\psi)$.
4. $:(\neg\phi\wedge\neg\psi)\Rightarrow\neg(\phi\wedge\psi)$.
5. $:(\neg\phi\vee\neg\psi)\Rightarrow\neg(\phi\vee\psi)$.
6. $\phi\wedge(\theta\vee\psi)$ is equivalent to $(\phi\wedge\theta)\vee(\phi\wedge\psi)$.
7. $\forall x.\neg\phi$ is equivalent to $\neg\exists x.\phi$.
8. $\exists x\phi\Rightarrow\neg\forall x.\neg\phi$.
9. $\exists x\neg\phi\Rightarrow\neg\forall x.\phi$.
10. $\forall x\phi\Rightarrow\neg\exists x.\neg\phi$.
11. φ:ψ implies $\neg\psi:\neg\phi$.
12. $\phi:\psi\Rightarrow\phi$.
13. $\phi\Rightarrow\not\phi:\not\phi$.
14. $\not\phi\vee\psi:\phi\Rightarrow\psi$ (but not the converse!).
15. $\neg\neg\neg\phi$ is equivalent to $\neg\phi$.
16. $(\phi\wedge\psi)\Rightarrow\theta$ is equivalent to $\phi\Rightarrow(\psi\Rightarrow\theta)$ (currying!).
4. Using the Boolean negation rule: $\Gamma,\neg\phi:false$ is equivalent to Γ:φ, prove the following additional results:
1. $\neg\neg\phi:\phi$.
2. $:\phi\vee\neg\phi$.
5. Show that either of the three rules above, together with the original negation rule, implies the Boolean negation rule.
1. The converses of the three existential/universal/negation implications above.
6. The restrictions introduced for the cut rule above block the deduction of an entailment $:\forall x.\phi\Rightarrow\exists x.\phi$. The issue at hand is that A might not actually have members; so choosing one is not a sound move. Show that this entailment can be deduced from the premise $\exists x\in A. x=x$.
7. Show that if we extend our ruleset by the quantifier negation rule $\forall x\Leftrightarrow \neg\exists x.\neg$, then we can derive the entailment $:\forall w: w=t \vee w = false$. From this derive $:\phi\vee\neg\phi$ and hence conclude that this extension gets us Boolean logic again.
4. A topology on a topos E is an arrow $j:\Omega\to\Omega$ such that $j\circ true=true$, $j\circ j=j<math> and <math>j\circ\wedge = \wedge\circ j\times j$. For a subobject $S\subseteq A$ with characteristic arrow $\chi_S:A\to\Omega$, we define its j-closure as the subobject $\bar S\subseteq A$ classified by $j\circ\chi_S$.
1. Prove:
1. $S\subseteq\bar S$.
2. $\bar S = \bar{\bar S}$.
3. $\bar{S\cap T} = \bar S\cap\bar T$.
4. $S\subseteq T$ implies $\bar S\subseteq\bar T$.
5. $\bar{f^{-1}(S)} = f^{-1}(\bar S)$.
2. We define S to be j-closed if $S=\bar S$. It is j-dense if $\bar S=A$. These terms are chosen due to correspondences to classical pointset topology for the topos of sheaves over some space. For a logical standpoint, it is more helpful to look at j as a modality operator: "it is j-locally true that" Given any $u:1\to\Omega$, prove that the following are topologies:
1. $(u\to -): \Omega\to\Omega$ (the open topology, where such a u in a sheaf topos ends up corresponding to an open subset of the underlying space, and the formulae picked out are true on at least all of that subset).
2. $u\vee -): \Omega\to\Omega$ (the closed topology, where a formula is true if its disjunction with u is true -- corresponding to formulae holding over at least the closed set complementing the subset picked out)
3. $\neg\neg: \Omega\to\Omega$. This may, depending on the topos, end up being interpreted as true so far as global elements are concerned, or not false on any open set, or other interpretations.
4. 1Ω.
3. For a topos E with a topology j, we define an object A to be a sheaf iff for every X and every j-dense subobject $S\subseteq X$ and every $f:S\to A$ there is a unique $g:X\to A$ with $f=g\circ s$. In other words, A is an object that cannot see the difference between j-dense subobjects and objects. We write Ej for the full subcategory of j-sheaves.
1. Prove that any object is a sheaf for 1Ω.
2. Prove that a subobject is dense for $\neg\neg$ iff its negation is empty. Show that $true+false:1+1\to\Omega$ is dense for this topology. Conclude that 1 + 1 is dense in $\Omega_{\neg\neg}$ and thus that $E_{\neg\neg}$ is Boolean. | {"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": 199, "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": 0.8355367183685303, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/90234/positive-weighted-directed-graphs | ## positive weighted directed graphs
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Consider a directed graph with real-number weights on the edges. I'll call the graph "positive" if the sum of weights along every circuit is positive. (It's easy to check for positivity: just make sure the circuit sum is positive for all circuits that don't pass through the same vertex twice.)
Given a directed graph, call two weightings "equivalent" if they produce the same sums on circuits.
Given a weighting, it's easy to produce an equivalent one: choose a vertex, choose a real number m, then add m to each incoming edge weight and subtract m from each outgoing edge weight.
Question #1: is the set of equivalent weightings generated by that operation?
Question #2: given a positive directed graph, does there always exist an equivalent weighting in which each edge weight is positive?
-
## 2 Answers
I assume that when you say the weight of any circuit is positive, you mean that for every directed cycle $C$, $\sum_{e \in E(C)} w(e) > 0$. There's a similar model of group labeled graphs where you calculate the weight of a circuit by either adding or subtracting $w(e)$ depending on whether you traverse the edge according to the direction or not.
The answer to 2. is "yes" if we assume the graph has every edge contained in a cycle. First, observe that it suffices to prove it for rational weights - just subtract a tiny epsilon from each edge to make it's weight rational without altering the fact that each cycle has positive weight.
Your operation for finding an equivalent weight function on the edges can be generalized: if we take any edge cut $\delta(U)$ for a subset $U \subseteq V(G)$ and a value $m \in \mathbb{R}$, then if we add $m$ to all the edges of $\delta^{in}(U)$ and subtract $m$ from $\delta^{out}(U)$, we won't change the weight of any cycle. Call this operation $resigning~on~a~cut$. (In fact, resigning on a cut $\delta(U)$ for a subset of vertices $U$ is the same as repeatedly resigning by the same value on each of vertices in $U$). Say two weight functions are $flip~equivalent$ if one can be obtained from the other by repeatedly re-signing on a cut.
A weight function is $non-negative$ if the weight of any cycle is not negative. We claim that if $G$ has the property that every edge is contained in a cycle, then any non-negative rational weight function is flip equivalent to one where every edge has non-negative weight. Assume the claim is false. It suffices to prove the claim for integer weight functions by re-scaling. Pick a counterexample on a minimal number of vertices, and subject to that, to minimize $\sum_{e \in E(G): w(e) \ge 0} w(e)$. If there exists a cycle $C$ such that $w(C) = 0$, then there exists a weight function $\bar{w}$ which is flip-equivalent to $w$ such that $\bar{w}(e) = 0$ for all edges $e \in E(C )$. To see this, number the edges of $C$ $e_1, \dots, e_k$ so that they occur in that order on $C$. We can force $e_i$ for $1 \le i \le k-1$ to have weight 0 by sequentially resigning on $\delta(v_{i+1})$. After doing so, the weight on $e_k$ will be the weight of the cycle, namely 0.
Consider the graph $G'$ obtained by contracting the cycle $C$ to a single vertex and deleting any loops which arise. Note that this preserves the property that every edge is contained in a cycle, and it also holds that each loop must have positive weight in $\bar{w}$. Let $w'$ be the weight function obtained by restricting $\bar{w}$ to the edges of $G'$. Then $G'$ has no negative weight circuit, since any such circuit could be rerouted through $C$ to give a negative weight cycle in $G$. Thus, $w'$ on $G'$ can be made non-negative by repeatedly resigning on cuts. Since each cut of $G'$ is a cut of $G$ as well, it follows that $\bar{w}$ can be made non-negative by repeatedly resigning on cuts.
Thus, we may assume that every cycle has strictly positive weight, and consequently, weight at least 1. It follows that there must exist an edge $f$ with $w(f) \ge 1$. Fix such an edge $f$, and let $w''$ be the weight function with $w''(e) = w(e)$ for all $e \neq f$ and let $w''(f) = w(f) - 1$. By construction, $w''$ is a non-negative since the weight of any cycle decreases by at most $1$. Moreover, $\sum_{e \in E(G): w''(e) \ge 0} w''(e)$ has strictly decreased, so by our choice of counterexample, there exists a weight function which is flip equivalent to $w''$ where every edge has non-negative weight. By resigning on the same series of cuts, we find a weight function which is flip equivalent to $w$ where every edge has non-negative weight, contradicting our choice of counterexample.
Now to see that the same result holds for rational weight functions where every cycle has strictly positive weight, let $z$ be such a weight function. By above there exists a weight function $z'$ which is flip-equivalent to $z$ such that every edge has $z'(e) \ge 0$. Pick such a $z'$ to have has few edges of weight zero as possible. The function $z'$ does not have any cycles where every edge has weight 0 in $z'$. Thus, if there are any edges $z'(e) = 0$, it follows that there is a vertex $v$ such that $v$ has some out edge of weight zero and no in edge of weight zero. If we let $\epsilon= min_{e \in \delta(v)} z'(e)$, then adding $\epsilon/2$ to each of the out edges of $v$ and subtracting $\epsilon/2$ from all the in edges will maintain the property of being a non-negative weighting and strictly decrease the number of edges of weight zero, a contradiction.
-
Thanks! -- that's just what I was looking for. – David Hillman Mar 9 2012 at 19:44
Great - glad it works. If you need the result for general graphs (i.e. not strongly connected), the proof can be extended. If there is a directed cut, inductively re-sign both halves of the cut. Performing the same operations on the original graph will leave every edge positive, except for possibly edges of the cut we decomposed on. But since it is a directed cut, we can add some arbitrarily large number to every edge of the cut and the resulting graph has every edge with positive weight. – Paul Wollan Mar 9 2012 at 23:36
Slightly simpler proof (based heavily on yours): suppose every edge is in a cycle and the weight function is nonnegative (positive). Choose cycle with minimal weight w and n vertices, n>1, with no vertex repeated. In the usual way transform the cycle so each edge has weight w/n. Claims: 1) every edge beginning and ending at vertices in the cycle is nonnegative (positive). 2) if you contract the cycle to a point, the weight function on the resulting graph is nonnegative (positive). 3) any valid transformation of the contracted graph can be done on the corresponding edges of the original graph. – David Hillman Mar 17 2012 at 18:27
Proof of (1): consider $e: u\to v$ having weight $z$. If $u=v$ then $e$ determines a cycle so done. If not, there is a path in the cycle containing $k$ edges having weight $kw/n$ with $k>0$, and $z>=kw/n$ or else that cycle didn't have minimum weight. Done. – David Hillman Mar 17 2012 at 18:39
Proof of (2): consider cycle $c$ in contracted graph with weight $z$. If $c$ doesn't contain contracted vertex, clearly $z$ is nonnegative (positive). Otherwise $z$ is the weight of a path $p:u\to v$ in the original graph with $u$ and $v$ in the original cycle. Now use argument of (1) with $e$ replaced by $p$. – David Hillman Mar 17 2012 at 19:00
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### 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.
The set of weightings whose sum vanishes along each cycle corresponds exactly to gradients of functions on vertices (in other words, the weight of an edge is $f(h) - f(t),$ where $h, t$ are head and tail respectively. This is easy to show. I assume this is equivalent to what you are suggesting for Question 1. For Question 2, I am quite sure that the answer is NO, and there is a very round-about geometric way of seeing it (basically, conditions on cycle sums of exterior dihedral angles together with positivity gives a necessary and sufficient conditions for a realization of a polyhedron as a convex ideal polyhedron, and some combinatorial types are not so realizable, but can be realized as non-convex ideal polyhedra, which means that you can't keep the cycle sums the same while making the edge weights positive. A complete explanation is too long, but you can check out my papers (sorry for the self-promotion, but these are the ones I know best :)):
For Q1: Arxiv preprint "Walks on free groups and other stories". For Q2: "On the geometry of convex ideal polyhedra in hyperbolic 3 space" (Annals of Math '94) and "Combinatorial optimization in geometry" (advances in applied math, 2002(?), there is also an arxiv version.
I am sure there is a much simpler way to construct a counterexample to Q2, but nothing leaps to mind.
-
1
@Igor: You must have some extra conditions in mind for Q1. Consider an acyclic digraph. – Brendan McKay Mar 5 2012 at 2:16
@Brendan: I think I am thinking of a somewhat different setting than the one in the question... Hmm... – Igor Rivin Mar 5 2012 at 13:15
The gradient thing is equivalent to what I was saying. Clearly it fails (e.g. consider a graph with two vertices and two parallel edges between them: we should be able to make the weights arbitrary, but can't). – David Hillman Mar 5 2012 at 21:29
I think the condition that makes it work is that the graph is strongly connected. Construct outgoing and incoming spanning trees with root at $v$. Let $p_w$ be path from $v$ to $w$ on outgoing tree and $q_w$ be path from $w$ to $v$ on incoming tree. Zero edges on outgoing tree by subtracting the gradient where $f(w)$ is the sum of weights on $p_w$. Then the weight on edge $e:w\to x$ is the sum of weights on circuit $p_w e q_x$ minus the sum of weights on circuit $p_x q_x$. So this is a canonical form achieved by the desired moves. – David Hillman Mar 5 2012 at 21:38
@David: Right, and to take this a tiny bit further, a characterization of when Q1 has answer "yes" is that the digraph is a disjoint union of strongly connected digraphs (which is equivalent to every edge lies on a cycle). – Brendan McKay Mar 6 2012 at 13:09 | {"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": 103, "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": 0.9461051225662231, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/16850/is-0-0-0-an-undefined-vector?answertab=votes | # Is (0,0,0) an undefined vector?
I'm not sure what to make of the direction of a vector with components (0,0,0). Is it an undefined vector?
-
– Qmechanic♦ Nov 11 '11 at 22:01
## 2 Answers
There's absolutely nothing wrong with a vector having all zero components. But a vector like that does have an undefined direction. So the definition of a vector as "something which has a magnitude and a direction" can throw you off here a little.
-
In mathematics that is the identity element 0 for vector addition, and is a key object in the definition of a vector space.
Physically, if $(0, 0, 0)$ is the position vector of some particle, then that just means it is located at the origin.
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http://mathhelpforum.com/math-topics/166961-mechanics.html | Thread:
1. Mechanics
Which is the right answer of R?
Rcos27-85g=0
R= 85g/cos27
R= 935N
or
R-85gcos27=0
R=85gcos27
R= 742N
Im bit confused. In M1, I was told to use the Second method, which is right for M1.
But Im doing M2 atm, and the solution uses the first method.
Can someone explain?
Attached Thumbnails
2. $R = mg\cos{\theta}$
3. So the second one is right?
4. Originally Posted by BabyMilo
So the second one is right?
if $85g$ is the weight of the mass, then $R = 85g\cos(27^\circ)$
5. If you ever get confused, just remember that the normal force on a ramp cannot exceed the weight (unless something else is pushing it against the ramp).
6. Originally Posted by skeeter
if $85g$ is the weight of the mass, then $R = 85g\cos(27^\circ)$
what if there is a horizontal force of mw^2r?
would R still be 85gcos27 ?
thanks
7. Originally Posted by BabyMilo
what if there is a horizontal force of mw^2r?
would R still be 85gcos27 ?
thanks
You are talking about centripedal force right? In that case 85g/cos27.
Remember, if the block is sliding down the ramp, the force is pointing down (parallel) to the ramp. In this case, the force components perpendicular to the ramp should sum to zero. If you draw the free body diagram, you get R = 85gcos(27).
If it is the angle of a ramp in a turn, then the force is horizontal (pointed directly towards the center of the circle). In this case, the vertical force components should sum to zero. If you draw the free body diagram, you get R = 85g/cos(27).
Also, usually R means radius and not normal force. I would suggest changing the notation to avoid confusion.
8. well we use r for radius instead of R.
but thanks for you help.
9. next time, post the entire problem.
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http://physics.stackexchange.com/questions/37864/longitudinal-and-transverse-part-of-vector-field-components | # Longitudinal and transverse part of vector field components
I was reviewing a paper of coupling to vector field and tensor field. I have got stuck with the term $$A_k \varepsilon^{kmn}\partial_mV_n=V^{T}.(\nabla\times A^{T})-\nabla.(A^{T}\times V^{T})+V^{T}.(\nabla\times A^{L})-\nabla.(A^{L}\times V^{T})$$ Where $A^T$ and $V^T$ are the transverse part of vector $A$ and $V$ respectively. And $A^L$ is the longitudinal part of $A$ . According to the paper, after calculation the only term that survives is this one-$$A_k \varepsilon^{kmn}\partial_mV_n=V^{T}.(\nabla\times A^{T})$$ other vanishes. Well we can use $$\nabla .V^T=0=\nabla \times V^L$$. That vanishes the term $V^T.(\nabla\times A^L)$. But still there are 2 more term left. How can I solve this? Does integration by parts can do any help?
-
I see no reason why those terms should be zero for general $A$ and $V$. Are you sure there are no other relationships between $A$ and $V$? Could you post a link to the paper please. – Mistake Ink Sep 20 '12 at 22:49
– aries0152 Sep 20 '12 at 23:36
It comes when I tried to re-write equation (4) to equation (20) – aries0152 Sep 21 '12 at 0:01 | {"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": 11, "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": 0.9488101601600647, "perplexity_flag": "head"} |
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15 30 50 per page | {"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": 22, "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": 0.18, "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": 0.893458902835846, "perplexity_flag": "middle"} |
http://mathhelpforum.com/business-math/189858-european-call-option.html | # Thread:
1. ## European call option
Consider a binomial model with one period and with parameters r(=interest rate) = 1/5, S_0 =
1, d = 3/4, u = 5/4. Compute the price of a European call option with strike price
K = 1, and find a replicating portfolio.
u(=upfactor) and d(=downfactor) are the following parameters: $u=\frac{S_0(H)}{S_0}$
$d=\frac{S_0(T)}{S_0}$
$S_1(H)=\frac{5}{4}$ $S_1(T)=\frac{3}{4}$
How can I determine the price at S_0 ?
2. ## Re: European call option
if you are working from first principles you can find the replicating porfolio first. the price is the cost of the replicating portfolio.
do you know how to find the replicating portfolio?
3. ## Re: European call option
Thank you very much for you answer, but I already solved the exercise.
I have an other exercise, where I have some problems, may you can help me with that:
Consider a model of a financial market consisting of one stock and one bond
with risk-free interest rate r > 0. Assume that the stock price at time 0 is a constant
S_0 > 0, and at time 1 can have any of the three values d,m and u , each with strictly positive probability, where we assume that 0 < d < m < u. On what conditions is this model free of arbitrage?
This model is known as a trinomial, and I have the following attempt:
S_0 is the start price and after a specific time I get three different results, each with different properies
$p_1 => S_0u$
$p_2 => S_0m$
$p_3 => S_0d$
and $p_1+p_2+p_3=1$
What are some examples for conditions that this is a model free of arbitrage?
4. ## Re: European call option
Does somebody has an idea?
5. ## Re: European call option
When the probability of an upward move (u) is greater than 50% by the amount that the probability of a downward move (d) is less than 50%, and a sideways move (m) is 50%. I assume you are doing this in class and not to trade, so write it on a blackboard as a time series and it should become very clear. | {"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": 0.9260149598121643, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/52863/number-of-ways-to-multiply-elements-with-a-non-associative-operation?answertab=votes | # Number of ways to multiply elements with a non-associative operation
The question comes from considerations in algebra but it's probably more related to combinatorics.
Suppose I have a set of $n$ elements $(a, b, c, ...)$ and a possibly non-associative operation $*$ between these elements.
What I want to do is to count the number of elements I could obtain with this operation without interchanging the order of the elements.
Let me illustrate this. Here below the parentheses mean that the operation has higher priority (usual notation).
• For $a$ and $b$, there is only 1 way to multiply them: $a * b$.
• For $a$, $b$ and $c$, we can have 2: $a*(b*c)$ or $(a*b)*c$.
• For $a$, $b$, $c$ and $d$, there are 5: $(a*b)*(c*d)$, $((a*b)*c)*d$, $(a*(b*c))*d$, $a*((b*c)*d)$ or $a*(b*(c*d))$.
It seems that for $5$ elements we have $14$ possibilities unless I've made a mistake by enumerating them all by hand.
Would anyone have an idea on how many could we obtain for $n$ elements? I fail to detect a pattern here.
-
## 1 Answer
The number of ways of inserting parentheses into $n$ objects is given by the $n-1$'th Catalan number (the relationship is mentioned here): $$C_{n-1}=\frac{1}{n}\binom{2n-2}{n-1}.$$ For example, $C_4=\frac{1}{5}\binom{8}{4}=14$.
-
1
It's the $n-1$th Catalan number. $C_4=14$, $C_5=\frac{1}{6}\binom{10}{5}=42$. – Chris Eagle Jul 21 '11 at 11:04
@Chris: Thanks for the correction, I answered too hastily :) – Zev Chonoles♦ Jul 21 '11 at 11:05
Thank you very much! I surely haven't heard of these before. – johnny Jul 21 '11 at 11:13
1
– Jyrki Lahtonen Jul 21 '11 at 14:14 | {"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": 29, "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": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9304543137550354, "perplexity_flag": "middle"} |
http://en.wikipedia.org/wiki/Geographical_distance | # Geographical distance
Geodesy
Fundamentals
Concepts
Technologies
Standards
History
Geographical distance is the distance measured along the surface of the earth. The formulae in this article calculate distances between points which are defined by geographical coordinates in terms of latitude and longitude. This distance is an element in solving the second (inverse) geodetic problem.
## An abstraction
Calculating the distance between geographical coordinates is based on some level of abstraction; it does not provide an exact distance, which is unattainable if one attempted to account for every irregularity in the surface of the earth.[1] Common abstractions for the surface between two geographic points are:
• Flat surface;
• Spherical surface;
• Ellipsoidal surface.
All abstractions above ignore changes in elevation. Calculation of distances which account for changes in elevation relative to the idealized surface are not discussed in this article.
## Nomenclature
Distance, $D,\,\!$ is calculated between two points, $P_1\,\!$ and $P_2\,\!$. The geographical coordinates of the two points, as (latitude, longitude) pairs, are $(\phi_1,\lambda_1)\,\!$ and $(\phi_2,\lambda_2),\,\!$ respectively. Which of the two points is designated as $P_1\,\!$ is not important for the calculation of distance.
Latitude and longitude coordinates on maps are usually expressed in degrees. In the given forms of the formulae below, one or more values must be expressed in the specified units to obtain the correct result. Where geographic coordinates are used as the argument of a trigonometric function, the values may be expressed in any angular units compatible with the method used to determine the value of the trigonometric function. Many electronic calculators allow calculations of trigonometric functions in either degrees or radians. The calculator mode must be compatible with the units used for geometric coordinates.
Differences in latitude and longitude are labeled and calculated as follows:
$\begin{align} \Delta\phi&=\phi_2-\phi_1;\\ \Delta\lambda&=\lambda_2-\lambda_1. \end{align} \,\!$
It is not important whether the result is positive or negative when used in the formulae below.
"Mean latitude" is labeled and calculated as follows:
$\phi_m=\frac{\phi_1+\phi_2}{2}.\,\!$
Colatitude is labeled and calculated as follows:
For latitudes expressed in radians:
$\theta=\frac{\pi}{2}-\phi;\,\!$
For latitudes expressed in degrees:
$\theta=90^\circ-\phi.\,\!$
Unless specified otherwise, the radius of the earth for the calculations below is:
$R\,\!$ = 6,371.009 kilometers = 3,958.761 statute miles = 3,440.069 nautical miles.
$D_\,\!$ = Distance between the two points, as measured along the surface of the earth and in the same units as the value used for radius unless specified otherwise.
## Singularities and discontinuity of latitude/longitude
Longitude has singularities at the Poles (longitude is undefined) and a discontinuity at the ±180° meridian. Also, planar projections of the circles of constant latitude are highly curved near the Poles. Hence, the above equations for delta latitude/longitude ($\Delta\phi\!$, $\Delta\lambda\!$) and mean latitude ($\phi_m\!$) may not give the expected answer for positions near the Poles or the ±180° meridian. Consider e.g. the value of $\Delta\lambda\!$ (“east displacement”) when $\lambda_1\!$ and $\lambda_2\!$ are on either side of the ±180° meridian, or the value of $\phi_m\!$ (“mean latitude”) for the two positions ($\phi_1\!$=89°, $\lambda_1\!$=45°) and ($\phi_2\!$=89°, $\lambda_2\!$=−135°).
If a calculation based on latitude/longitude should be valid for all Earth positions, it should be verified that the discontinuity and the Poles are handled correctly. Another solution is to use n-vector instead of latitude/longitude, since this representation does not have discontinuities or singularities.
## Flat-surface formulae
A planar approximation for the surface of the earth may be useful over small distances. The accuracy of distance calculations using this approximation become increasingly inaccurate as:
• The separation between the points becomes greater;
• A point becomes closer to a geographic pole.
The shortest distance between two points in plane is a straight line. The Pythagorean theorem is used to calculate the distance between points in a plane.
Even over short distances, the accuracy of geographic distance calculations which assume a flat Earth depend on the method by which the latitude and longitude coordinates have been projected onto the plane. The projection of global latitude and longitude coordinates onto a plane is the realm of cartography.
The formulae presented in this section provide varying degrees of accuracy.
### Spherical Earth projected to a plane
This formula takes into account the variation in distance between meridians with latitude:
$D=R\sqrt{(\Delta\phi)^2+(\cos(\phi_m)\Delta\lambda)^2}{\color{white}\frac{\big|}{.}}\,\!$
where:
$\Delta\phi\,\!$ and $\Delta\lambda\,\!$ are in radians;
$\phi_m\,\!$ must be in units compatible with the method used for determining $\cos(\phi_m).\,\!$
To convert latitude or longitude to radians use
$1^\circ = (\pi/180)\,\mathrm{radians}.$
Note: This approximation is very fast and produces fairly accurate result for small distances[citation needed]. Also, when ordering locations by distance, such as in a database query, it is much faster to order by squared distance, eliminating the need for computing the square root.
### Ellipsoidal Earth projected to a plane
The FCC prescribes essentially the following formulae in 47 CFR 73.208 for distances not exceeding 475 km /295 miles:[2]
$D=\sqrt{(K_1\Delta\phi)^2+(K_2\Delta\lambda)^2};{\color{white}\frac{\big|}{.}}\,\!$
where
$D\,\!$ = Distance in kilometers;
$\Delta\phi\,\!$ and $\Delta\lambda\,\!$ are in degrees;
$\phi_m\,\!$ must be in units compatible with the method used for determining $\cos(\phi_m);\,\!$
$\begin{align} K_1&=111.13209-0.56605\cos(2\phi_m)+0.00120\cos(4\phi_m);\\ K_2&=111.41513\cos(\phi_m)-0.09455\cos(3\phi_m)+0.00012\cos(5\phi_m).\end{align}\,\!$
It may be interesting to note that:
$K_1=M\frac{\pi}{180}\,\!$ = kilometers per degree of latitude difference;
$K_2=\cos(\phi_m)N\frac{\pi}{180}\,\!$ = kilometers per degree of longitude difference;
where $M\,\!$ and $N\,\!$ are the meridional and its perpendicular, or "normal", radii of curvature (the expressions in the FCC formula are derived from the binomial series expansion form of $M\,\!$ and $N\,\!$, set to the Clarke 1866 reference ellipsoid).
### Polar coordinate flat-Earth formula
$D=R\sqrt{\theta^2_1\;\boldsymbol{+}\;\theta^2_2\;\mathbf{-}\;2\theta_1\theta_2\cos(\Delta\lambda)};{\color{white}\frac{\big|}{.}}\,\!$
where the colatitude values are in radians. For a latitude measured in degrees, the colatitude in radians may be calculated as follows: $\theta=\frac{\pi}{180}(90^\circ-\phi).\,\!$
## Spherical-surface formulae
If we are willing to accept a possible error of 0.5%, we can use formulas of spherical trigonometry on the sphere that best approximates the surface of the earth.
The shortest distance along the surface of a sphere between two points on the surface is along the great-circle which contains the two points.
### Tunnel distance
A tunnel between points on Earth is defined by a line through three-dimensional space between the points of interest. For a spherical Earth, this line is also the chord of the great circle between the points. For points near each other, the tunnel distance is only slightly less than the great-circle distance.[3]
The great circle chord length may be calculated as follows for the corresponding unit sphere:
$\begin{align} &\Delta{X}=\cos(\theta_2)\cos(\lambda_2) - \cos(\theta_1)\cos(\lambda_1);\\ &\Delta{Y}=\cos(\theta_2)\sin(\lambda_2) - \cos(\theta_1)\sin(\lambda_1);\\ &\Delta{Z}=\sin(\theta_2) - \sin(\theta_1);\\ &C_h=\sqrt{(\Delta{X})^2 + (\Delta{Y})^2 + (\Delta{Z})^2}.\end{align}\,\!$
The tunnel distance between points on the surface of a spherical Earth is: $D = R C_h$
### Great-circle distance
The great-circle distance article gives the formula for calculating the distance along a great-circle on a sphere about the size of the Earth. The article includes an example of the calculation.
## Ellipsoidal-surface formulae
An ellipsoidal approximation for the surface of the earth may be useful over great distances. The shortest distance along the surface of an ellipsoid between two points on the surface is along the geodesic.
### Exact method
Clairaut[4] derived a conserved quantity for geodesics on an ellipsoid of revolution, Clairaut's relation. This enabled Legendre[5] to transfer the ellipsoidal geodesics onto an "auxiliary sphere". On this sphere, the geographic latitude is replaced by the reduced latitude, the azimuth of the geodesic is preserved, and the distance along the geodesic and the geographic longitude are related to the spherical arc length and spherical longitude by simple one dimensional integrals. A succinct derivation of these results is given by Bessel.[6] The problem was further elaborated by Helmert,[7] and Rapp[8] provides a good modern summary of the method.
Vincenty[9][10] formulated an algorithm based the solution using the auxiliary sphere. This is accurate to third order in the flattening of the ellipsoid, i.e., about 0.5 mm; however, the algorithm fails to converge for points that are nearly anti-podal. For details, see Vincenty's formulae.
Karney[11] tackled the inverse geodesic problem using Newton's method to accelerate the convergence. In addition, he employs series which are accurate to sixth order in the flattening. This results in an algorithm which is accurate to full double precision and which converges for arbitrary pairs of points on the earth. This algorithm is implemented in GeographicLib.[12]
### Approximate methods
The exact methods described above are feasible when carrying out calculations on a computer. Before the advent of computers, several authors derived approximate formulas suitable for hand calculations.
If we limit distance to 100–150 km we can get millimeter accuracy with equations given by Bowring.[13][14] In the linked pdf "e' 2" is the second eccentricity squared, a constant for the chosen spheroid
$e'^2 = \frac {2r - 1}{(r - 1)^2}$
where r is the reciprocal of the flattening (r = 298.257223563 for the WGS84 spheroid). $\scriptstyle \phi_1$ and $\scriptstyle \phi_2$ are the latitudes of the two points, $\scriptstyle \lambda_1$ and $\scriptstyle \lambda_2$ are the longitudes; $\scriptstyle \Delta \phi$ is the difference in latitude (which at one point must be in radians). Calculate A, B, C and w on the first page of the pdf, then skip to "Inverse Problem" on the second page.
Lambert's formulae[15] give accuracy on the order of 10 meters over thousands of kilometers. First convert the latitudes $\scriptstyle \phi_1$ , $\scriptstyle \phi_2$ of the two points to reduced latitudes $\scriptstyle \psi_1$ , $\scriptstyle \psi_2$
$\tan \psi \; = \; \left ( \frac {r - 1}{r} \right ) \tan \phi.$
Then calculate the central angle $\sigma$ in radians between two points $(\psi_1 , \; \lambda_1)$ and $(\psi_2 , \; \lambda_2)$ on a sphere in the usual way (law of cosines or haversine formula), with longitudes $\lambda_1 \;$ and $\lambda_2 \;$ being the same on the sphere as on the spheroid.
$P = \frac { \psi_1 + \psi_2 }{2}$
$Q = \frac {\psi_2 - \psi_1}{2}$
$X = ( \sigma - \sin \sigma) \frac {\sin^2 P \cos^2 Q}{ \cos^2 \frac { \sigma}{2}}$
$Y = ( \sigma + \sin \sigma) \frac {\cos^2 P \sin^2 Q}{ \sin^2 \frac { \sigma}{2}}$
$distance = a \left ( \sigma - \frac {X + Y}{2r} \right )$ where $a$ is the equatorial radius of the chosen spheroid.
On the GRS 80 spheroid Lambert's formula is off by
0 North 0 West to 40 North 120 West, 12.6 meters
0N 0W to 40N 60W, 6.6 meters
40N 0W to 40N 60W, 0.85 meter
## References
1. Clairaut, A. C. (1735). "Détermination géometrique de la perpendiculaire à la méridienne tracée par M. Cassini" [Geometrical determination of the perpendicular to the meridian drawn by Jacques Cassini]. Mém. de l'Acad. Roy. des Sciences de Paris 1733 (in French): 406–416.
2. Legendre, A. M. (1806). "Analyse des triangles tracės sur la surface d'un sphėroïde". Mém. de l'Inst. Nat. de France (1st sem.): 130–161. Retrieved 2011-07-30.
3. Bessel, F. W. (2010). "The calculation of longitude and latitude from geodesic measurements (1825)". Astron. Nachr. 331 (8): 852–861. arXiv:0908.1824. doi:10.1002/asna.201011352. English translation of Astron. Nachr. 4, 241–254 (1825).
4. Helmert, F. R. (1964). Mathematical and Physical Theories of Higher Geodesy, Part 1 (1880). St. Louis: Aeronautical Chart and Information Center. Retrieved 2011-07-30. English translation of Die Mathematischen und Physikalischen Theorieen der Höheren Geodäsie, Vol. 1 (Teubner, Leipzig, 1880).
5. Rapp, R. H. (March 1993). Geometric Geodesy, Part II. Ohio State University. Retrieved 2011-08-01.
6. Vincenty, T. (April 1975a). "Direct and Inverse Solutions of Geodesics on the Ellipsoid with application of nested equations". Survey Review. XXIII (misprinted as XXII) (176): 88–93. Retrieved 2009-07-11.
7. Vincenty, T. (April 1976). "Correspondence". Survey Review. XXIII (180): 294.
8. Karney, C. F. F. (2013). "Algorithms for geodesics". J. Geodesy 87 (1): 43–55. doi:10.1007/s00190-012-0578-z. Retrieved 2012-07-14.
9.
10. Bowring, B. R. (1981). "The direct and inverse problems for short geodesics lines on the ellipsoid". Surveying and Mapping 41 (2): 135–141.
11. Bowring's formulae at the Wayback Machine (archived March 19, 2012)
12. Lambert, W. D (1942). "The distance between two widely separated points on the surface of the earth". J. Washington Academy of Sciences 32 (5): 125–130. | {"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": 68, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8277013897895813, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/111592/prove-a-rightarrow-b-rightarrow-a-rightarrow-a | # Prove: $((A \rightarrow B) \rightarrow A) \rightarrow A )$
How could I derive the following proposition:
$$((A \rightarrow B) \rightarrow A) \rightarrow A )$$
using any of the following axioms:
1) $A→(B→A)$
2) $(A→(B→C))→((A→B)→(A→C))$
3) $((¬B)→(¬A))→(A→B)$
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1
You have supplied only axioms, and no rules of inference. – André Nicolas Feb 21 '12 at 7:14
Sorry. Modus ponens should be used. – user25376 Feb 21 '12 at 7:16
This is Peirce's law, which is equivalent to (3) in the presence of (1) and (2). – Zhen Lin Feb 21 '12 at 7:22
Thanks. I'm still not sure how to prove this. :| I'm pretty new to propositional calculus. – user25376 Feb 21 '12 at 8:16
3
@Arjang: It's fairly standard in logic to use $\to$ for the implication symbol in the object language and reserve ⟹ for implication in the metalanguage. – Chris Eagle Feb 21 '12 at 12:49
show 3 more comments
## 2 Answers
By the deduction theorem, it's enough to show that $((A \to B) \to A) \vdash A$. This can be done by contradiction, i.e. we assume $((A \to B) \to A)$ and $\lnot A$ and derive a contradiction. From $\lnot A$, we can use axiom 1 and modus ponens to get $(\lnot B \to \lnot A)$, and then use axiom 3 and MP to get $(A \to B)$. Then using MP with $(A \to B)$ and $((A \to B) \to A)$ gives us $A$, contradicting $\lnot A$ as desired.
If you haven't seen the proof-by-contradiction inference rule before, here's how to derive it: we suppose $\Gamma$ (a set of sentences) and $\lnot A$ together lead to a contradiction, and want to show that $\Gamma \vdash A$. Since $\Gamma$ and $\lnot A$ are contradictory, they prove anything, so $\Gamma, \lnot A \vdash \lnot (A \to (A \to A))$ say (here $(A \to (A \to A))$ could be replaced by any tautology). Then by the deduction theorem, we have $\Gamma \vdash (\lnot A \to \lnot (A \to (A \to A)))$, and then by axiom 3 and MP, $\Gamma \vdash ((A \to (A \to A)) \to A)$. Since $(A \to (A \to A))$ is a tautology, using MP again gives $\Gamma \vdash A$, as desired.
joriki points out that my proof of proof-by-contradiction uses the fact that a contradiction implies anything, that is, the inference rule $A, \lnot A \vdash B$. As it happens, I did most of the work of proving this above, when I inferred $(A \to B)$ from $\lnot A$. One more use of MP gives the result.
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I think to complete this you'd have to show how "Since $\Gamma$ and $\neg A$ are contradictory, they prove anything" can be derived in this system. – joriki Feb 21 '12 at 13:07
Here's an unraveled version of Chris' proof, with proof by contradiction and ex falso quodlibet unraveled but the deduction theorem still being used:
$$\begin{array}{lrcl} 1.&\neg A &\vdash& \neg C\to\neg A &&\text{axiom 1, modus ponens} \\ 2.& \neg A &\vdash& A\to C &&\text{1., axiom 3, modus ponens} \\ 3.& (A\to B)\to A,\neg A &\vdash& A &&\text{2.} [C\mapsto B]\text{, modus ponens} \\ 4.& (A\to B)\to A,\neg A &\vdash& A\to\neg(A\to(A\to A)) &&\text{2.} [C\mapsto \neg(A\to(A\to A))] \\ 5.& (A\to B)\to A,\neg A &\vdash& \neg(A\to(A\to A)) &&\text{3., 4., modus ponens} \\ 6.& (A\to B)\to A &\vdash& \neg A\to\neg(A\to(A\to A)) &&\text{5., deduction theorem} \\ 7.& (A\to B)\to A &\vdash& (A\to(A\to A))\to A &&\text{6., axiom 3, modus ponens} \\ 8.& (A\to B)\to A &\vdash& A &&\text{7., axiom 1, modus ponens} \\ 9.& &\vdash& ((A\to B)\to A)\to A &&\text{8., deduction theorem} \end{array}$$
If you want to unravel it further to use only the axioms and modus ponens, no deduction theorem, you can transform it as described here. The basic idea is that if you have a proof using hypothesis $A$, you conditionalize on $A$ by tacking $A$ on as a premise in front of every statement in the proof, including $A$ itself; then you transform every application of modus ponens that derives $C$ from $B$ and $B\to C$ by using axiom 2 and applying modus ponens twice (using the conditionalized inputs $A\to(B\to C)$ and $A\to B$ of the modus ponens step to derive its conditionalized output $A\to C$); then it just remains to prove the conditionalized hypothesis itself, $A\to A$, which is shown in the Wikipedia article:
$$\begin{array}{lrl} 1.& (A→((A→A)→A))→((A→(A→A))→(A→A)) &\text{axiom 2} \\ 2.& A→((A→A)→A) & \text{axiom 1} \\ 3.&(A→(A→A))→(A→A)&\text{1., 2., modus ponens} \\ 4.&A→(A→A)&\text{axiom 1} \\ 5.&A→A&\text{3., 4., modus ponens} \end{array}$$
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How did you produce firs step? Modus pones says $(A \implies B) \land A \vdash B$ but in your case A was not any axiom and it was not infered so it looks like you could not produce B. It looks like you used deduction theorem not modus ponens. – Trismegistos Feb 21 '12 at 17:59
@Trismegistos: I'm assuming that you mean the first step in the first of the two proofs. As indicated in the right-hand column, this is modus ponens applied to an instance of axiom $1$. The substitution for the $A\to(B\to A)$ axiom schema that produces this is $A\mapsto\neg A,B\mapsto\neg C$, yielding $\neg A\to(\neg C\to\neg A)$. The premise $\neg A$ is noted as a hypothesis on the left-hand side, and applying modus ponens yields $\neg C\to\neg A$. The deduction theorem moves premises from the left of the turnstyle to the right; modus ponens moves them from the right to the left. – joriki Feb 21 '12 at 18:00
but if you marked $\lnot A$ as hypotesis you showed that sentence you wanted to prove is true if your hypothesis $\lnot A$ is true. Isn't it? – Trismegistos Feb 21 '12 at 18:06
@Trismegistos: I'm not sure why you talk about truth. To my mind this is about proof. Apart from that, yes, I showed that the sentence $\neg C\to\neg A$ can be proved if the hypothesis $\neg A$ is assumed. That's what the deduction theorem is all about -- you assume hypotheses, prove something from them, then turn the hypotheses into antecedents of the proved statement by applying the deduction theorem. In this case, the hypothesis $\neg A$ is used in step 6 in applying the deduction theorem. – joriki Feb 21 '12 at 18:10
I undestood your proof after reading Chris Eagle message. – Trismegistos Feb 21 '12 at 18:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 54, "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": 0.9321098923683167, "perplexity_flag": "head"} |
http://planetmath.org/grothendieckgroup | # Grothendieck group
Let $S$ be an abelian semigroup. The Grothendieck group of $S$ is $K(S)=S\times S/\mathord{\sim}$, where $\sim$ is the equivalence relation: $(s,t)\sim(u,v)$ if there exists $r\in S$ such that $s+v+r=t+u+r$. This is indeed an abelian group with zero element $(s,s)$ (any $s\in S$), inverse $-(s,t)=(t,s)$ and addition given by $(s,t)+(u,v)=(s+u,t+v)$. It is common to use the suggestive notation $t-s$ for $(t,s)$.
The Grothendieck group construction is a functor from the category of abelian semigroups to the category of abelian groups. A morphism $f\colon S\to T$ induces a morphism $K(f)\colon K(S)\to K(T)$ which sends an element $(s^{+},s^{-})\in K(S)$ to $(f(s^{+}),f(s^{-}))\in K(T)$.
###### Example 1
Let $(\mathbb{N},+)$ be the semigroup of natural numbers with composition given by addition. Then, $K(\mathbb{N},+)=\mathbb{Z}$.
###### Example 2
Let $(\mathbb{Z}-\{0\},\times)$ be the semigroup of non-zero integers with composition given by multiplication. Then, $K(\mathbb{Z}-\{0\},\times)=(\mathbb{Q}-\{0\},\times)$.
###### Example 3
Let $G$ be an abelian group, then $K(G)\cong G$ via the identification $(g,h)\leftrightarrow g-h$ (or $(g,h)\leftrightarrow gh^{{-1}}$ if $G$ is multiplicative).
Let $C$ be a (essentially small) symmetric monoidal category. Its Grothendieck group is $K([C])$, i.e. the Grothendieck group of the isomorphism classes of objects of $C$.
Type of Math Object:
Definition
Major Section:
Reference
## Mathematics Subject Classification
16E20 no label found13D15 no label found18F30 Grothendieck groups
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http://mathoverflow.net/questions/15202?sort=oldest | ## Different interpretations of moduli stacks
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I'm taking my first steps in the language of stacks, and would like something cleared up. The intuitive idea of moduli spaces is that each point corresponds to an object of what we're trying to classify (smooth curves of genus g over ℂ, for example). Fine moduli spaces are defined to be the objects that represent the functor that takes an object and gives you the [set, for schemes; groupoid, as I understand it, for stacks] of ways that that object parametrizes families of the object we want to classify. Now, for schemes - this makes sense in the following way:
Let that functor be F, and let it be represented by M. Then F(Spec ℂ) are the families of (desired) objects parametrized by Spec ℂ (a point), so it corresponds to all desired objects (the ones we want to classify). But F(Spec ℂ) is also Hom(Spec ℂ, M), and so corresponds to the closed points of M. Thus M really does, intuitively, have as points the objects it wishes to classify.
Does this idea go through to moduli stacks? Of course, it probably does, and this is all probably trivial - but I feel like I need someone to assure me that I'm not crazy. So let me put the question like this: Can you formulate how to think of a fine moduli stack (as an object that represents an F as above; also: how would you define this F in the category of stacks?) in a way that makes it clear that it parametrizes the desired objects?
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2
Thanks for the question. It put clearly what my own problem with the algebraic geometric version of stacks. I know the non-Abelian cohomology motivation much better. – Tim Porter Feb 13 2010 at 21:39
## 2 Answers
I'll assure you that you're not crazy. Not only does the idea go through for stacks, but it's impossible (or at least very hard) to make sense of stacks without that idea.
If you're trying to parameterize wigits, you can build a functor F(T)={flat families of wigits over T}. If there is a space M that deserves to be called the moduli space of wigits, it should represent F. It's not just that the points of M must correspond to isomorphism classes of wigits, so we must have F(Spec ℂ)=Hom(Spec ℂ,M). The points are also connected up in the right way. For example, a family of widits over a curve should correspond to a choice of wigit for every point in the curve in a continuous way, so it should correspond to a morphism from the curve to M.
It happens that if wigits have automorphisms, there's no hope of finding a geometric object M so that maps to M are the same thing as flat (read "continuous") families of wigits. The reason is that any geometric object should have the property that maps to it can be determined locally. That is, if U and V cover T, specifying a map from T to M is the same as specifying maps from U and V to M which agree on U∩V. The jargon for this is "representable functors are sheaves." If a wigit X has an automorphism, then you can imagine a family of wigits over a circle so that all the fibers are X, but as you move around the circle, it gets "twisted" by the automorphism (if you want to think purely algebro-geometrically, use a circular chain of ℙ1s instead of a circle). Locally, you have a trivial family of wigits, so the map should correspond to a constant map to the moduli space M, but that would correspond to the trivial family globally, which this isn't. Oh dear!
Instead of giving up hope entirely, the trick is to replace the functor F by a "groupoid-valued functor" (fibered category), so the automorphisms of objects are recorded. Now of course there won't be a space representing F, since any space represents a set-valued functor, but it turns out that this sometimes revives the hope that F is represented by some mythical "geometric object" M in the sense that objects in F(T) (which should correspond to maps to M) can be determined locally. If this is true, we say that "F is a stack" or that "M is a stack." Part of what makes your question tricky is that as things get stacky, the line between M and F becomes more blurred. M isn't really anything other than F. We just call it M and treat it as a geometric object because it satisfies this gluing condition. We usually want M to be more geometric than that; we want it to have a cover (in some precise sense) by an honest space. If it does, then we say "M (or F) is an algebraic stack" and it turns out you can do real geometry on it.
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I'm sure Anton knows this, but to the questioner, in general, a fibered category is not fibered in groupoids. A general fibered category is merely "fibered in categories". Another way to say this (although it requires the axiom of choice) is that a fibered category is a pseudofunctor $\mathcal{F}:\mathcal{C}\to \mathcal{C}at$, and a category fibered in groupoids is a pseudofunctor $\mathcal{F}:\mathcal{C}\to \mathcal{G}pd$. – Harry Gindi Feb 13 2010 at 20:50
@fpqc: thanks for pointing that out. I was trying hard to drop the relevant terms you'd want to know while presenting the intuitive picture. I hope there aren't too many other technical lies I told in the process. – Anton Geraschenko♦ Feb 13 2010 at 20:58
Well, by pseudofunctor I meant contravariant pseudofunctor, so it should in fact be from $\mathcal{C}^{op}$. Sorry for the confusion if the OP was confused. The whole point of a fibered category is that it's a direct generalization of a presheaf, which is a contravariant functor into sets. – Harry Gindi Feb 13 2010 at 21:00
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Yeah, you've got the right idea! Say you've got some moduli problem -- for instance you have some class of objects X over C you're interested in -- and you want to say what the moduli space of X's is, as a variety (or stack or whatever) over C. The idea is that you can describe a variety by how other varieties map to it -- Yoneda's Lemma. So the moduli space M will be determined if you know what to call Hom(T,M) for every variety T. Certainly if T=Spec(C) is the point, this should be the original set (groupoid, whatever) of objects X over C; but in general Hom(T,M) should be the set of families of X's continuously parametrized by T, which usually is nicely encoded by some flat family over T whose fibers over C-points are X's.
So, to sum up, the notion of a moduli space as a functor exactly encodes the idea that the points of a moduli space should be the objects you're trying to parametrize. One just needs to use "points" in the more general context of maps from an arbitrary variety T, and be able to say what it means to give an object of the type you're interested in over an arbitrary such T. This is not formally determined by the case T=Spec(C) a point, but there is usually a natural extension to make.
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Sorry, I just reread your question, and it looks like your concern was exactly with the change from varieties to stacks, i.e., from sets to groupoids, which I handled with a couple of dismissive "whatever"s. I'm sorry for my misinterpretation and resulting uncouthness! What my "whatever"s were trying to convey, however clumsily, is that the formalism and ideas are exactly the same whether one requires sets or groupoids. – Dustin Clausen Feb 13 2010 at 20:07
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The reason one still only has to define one's functor on varieties, and not general stacks, is the same as why one only ever needs to define it on affine varieties, and not all varieties: by definition stacks can be "glued" from varieties, just like varieties can be "glued" from affines. Formally, any stack is a colimits of varieties, or affine varieties, and this implies that the functor of points restricted even to affines recovers the space. – Dustin Clausen Feb 13 2010 at 20:10
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While stacks can be glued from affines, there is no way to define a general algebraic stack without first defining an algebraic space. This is simply because the requirement that a sheaf of groupoids in the fppf topology be an algebraic stack is that its diagonal is representable by an algebraic space. There are other constructions you can use, but I think that most of them are equivalent to defining an algebraic space first. – Harry Gindi Feb 13 2010 at 20:38
What I meant was this: intuitively, we want stacks to be glued from affine varieties -- and this is exactly what justifies the formal definition of stack as a kind of sheaf on affine varieties, since the category of sheaves on a site is exactly what you get when you formally adjoin colimits to the site (colimits = gluing) subject to the restriction that the base X of a cover {U_i} --> X should be glued from the Cech nerve of the cover. – Dustin Clausen Feb 13 2010 at 20:48
Note that this has nothing to do with potential "algebraicity" of stacks, and is independent of what definition one takes there. – Dustin Clausen Feb 13 2010 at 20:51
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": 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": 10, "end_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": 0.952399730682373, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/238906/how-many-distinct-n-letter-words-can-be-formed-from-a-set-of-k-letters-where-s/267670 | # How many distinct n-letter “words” can be formed from a set of k letters where some of the letters are repeated?
How many distinct n-letter "words" can be formed from a set of k letters where some of the letters are repeated?
Examples:
__
How many 6-letter words can be formed from the letters: ABBCCC?
This is elementary. There are 6! arrangements counting repeats. Then we just divide by 2!3! to account for the repeats caused by the 2!3! identical arrangements of the Bs and Cs
How many 5-letter words can be formed from the letters: ABBCCC?
Excluding A we have $\frac{5!}{2!3!}$
Excluding either B $\frac{5!}{3!}$
Excluding any of the Cs $\frac{5!}{2!2!}$
Then take the sum.
How many 4-letter words can be formed from the letters: ABBCCC?
At this point I find it difficult to procede without far too many cases.
Is there a general approach?
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Interesting question, perhaps difficult to give a nice answer to. In the particular instance, we remove AB or AC or BB or CC or BC, so the number of cases is manageable. – André Nicolas Nov 16 '12 at 22:11
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– Jair Taylor Nov 17 '12 at 20:35
Thank you Jair! – a little don Nov 29 '12 at 23:42
## 2 Answers
Let's say you want the word of letters to be $M$-long, with $N$ choices for letters in each space on the word. Then draw something like this:
to enumerate all possible words. In the picture $M=3$ and $N=2$. Let's say you have a deficit in $A$s of magnitude $D_A$ from being able to make a full, $M$-long word of $A$s.
First, assuming none of the deficits overlap (that is, there are no words that violate 2 or more restrictions in the number of letters), find the number of nodes that are less than or equal to $M-D_A$ away from $(A,A,...A)$. This turns out to be $\sum_{k=0}^{k=M-D_A}\binom{M}{k}{(N-1)}^{M-k}$.
For the proof of this, think of a 3-letter word with choices of letter $A$, $B$ or $C$. There are $\binom{3}{0}2^0$ ways of making words with 3 $A$s, $\binom{3}{1}2^1$ ways of making a word with 2 $A$s, etc.
So evidently you need to subtract each sum taking into account the restrictions of each letter from the total number of unrestricted words.
However, it's much more complicated when the sum of two deficits is more than $M$ so that because of the overlap the subtraction will have been too great. Basically the $(N-1)$ in the sum will be some lower number for higher values of $k$ when other letter's restrictions are overstepped, but I can't see a nice formula for it for a given $D_B, D_C...$.
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It is solved here. In general, you have to account for the duplicates as detailed in the link.
$\frac{_nP_r}{n_1!n_2!...n_k!}$.
where the denominator denotes factorial for indistinguishable or repeated letters.
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Thanks, but, everything at that link is equivalent to my first example only. That formula will not work when n < k. – a little don Nov 16 '12 at 22:02 | {"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": 26, "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": false}, "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": 0.9386323690414429, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/98764/simplicity-of-cuntz-algebra/99478 | # Simplicity of Cuntz algebra
It is well known that the Cuntz algebra $\mathcal{O}_N$ (for fixed $N$) is simple. Is there any easy way to exhibit it (as a Banach algebra only) as a quotient of a Banach algebra modulo some maximal ideal? Of course, I am not interested in the quotient of the form $\mathcal{O}_N/ \{0\}$.
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## 1 Answer
Cuntz introduced the algebras $\mathcal O_n$ in the paper "Simple C*-algebras generated by isometries", 1977. Section 3 of the paper is called "Extensions of $\mathcal O_n$." It is proved there that if $n$ is finite, and if $V_1,\ldots V_n$ are isometries on Hilbert space such that $\sum\limits_{k=1}^n V_k V_k^*\leq I$ and $\sum\limits_{k=1}^n V_k V_k^*\neq I$ (i.e., if $V_1,\ldots, V_n$ are isometries with orthogonal ranges that do not add up to the whole space), then $\mathcal O_n$ is a quotient of $C^*(V_1,\ldots,V_n)$ by the ideal generated by the projection $I-\sum\limits_{k=1}^n V_k V_k^*$, and this ideal is isomorphic to the algebra of compact operators.
Since it is easy to give examples of such isometries, this might be the easy way you're looking for. Here is one particular construction that may not be the easiest, but that has an important role in the subject, and appears in D.E. Evans's "On $O_n$", 1980. Let $H$ be an $n$-dimensional Hilbert space with orthonormal basis $\{e_1,\ldots,e_n\}$. Let $F(H)=\mathbb C\oplus H\oplus (H\otimes H)\oplus (H\otimes H\otimes H) \oplus \cdots=\sum\limits_{k=0}^\infty H^{\otimes k}$ be the full Fock space of $H$, and for $i\in\{1,2,\ldots,n\}$, let $V_i\in B(F(H))$ be defined by $V_i x=e_i\otimes x$. Then each $V_i$ is an isometry, $\sum\limits_{k=1}^n V_kV_k^*\leq I$, and $I-\sum\limits_{k=1}^n V_kV_k^*$ is the projection onto the first summand of $F(H)$. In this case the ideal in question is precisely the set of compact operators on $F(H)$, $K(F(H))$, and $C^*(V_1,\ldots,V_n)/K(F(H))\cong \mathcal O_n$.
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http://mathoverflow.net/questions/1151/sheaf-cohomology-and-injective-resolutions/1153 | ## Sheaf cohomology and injective resolutions
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In defining sheaf cohomology (say in Hartshorne), a common approach seems to be defining the cohomology functors as derived functors. Is there any conceptual reason for injective resolution to come into play? It is very confusing and awkward to me that why taking injective stuff into consideration would allow you to "extend" a left exact functor.
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## 8 Answers
Since everybody else is throwing derived categories at you, let me take another approach and give a more lowbrow explanation of how you might have come up with the idea of using injectives. I'll take for granted that you want to associate to each object (sheaf) $F$ a bunch of abelian groups $H^i(F)$ with $H^0(F)=\Gamma(F)$, and that you want a short exact sequence of objects to yield a long exact sequence in cohomology.
I also want one more assumption, which I hope you find reasonable: if $F$ is an object such that for any short exact sequence $0\to F\to G\to H\to 0$ the sequence $0\to \Gamma(F)\to \Gamma(G)\to \Gamma(H)\to 0$ is exact, then $H^{i}(F)=0$ for $i>0$. This roughly says that $H^{i}$ is zero unless it's forced to be non-zero by a long exact sequence (you might be able to run this argument only using this for $i=1$, but I'm not sure). Note that this implies that injective objects have trivial $H^{i}$ since any short exact sequence with $F$ injective splits.
Now suppose I come across an object $F$ that I'd like to compute the cohomology of. I already know that $H^{0}(F)=\Gamma(F)$, but how can I compute any higher cohomology groups? I can embed $F$ into an injective object $I^{0}$, giving me the exact sequence $0\to F\to I^{0}\to K^{1}\to 0$. The long exact sequence in cohomology gives me the exact sequence $$0\to \Gamma(F)\to \Gamma(I^{0})\to \Gamma(K^{1})\to H^{1}(F)\to 0 = H^1(I^{0})$$
That's pretty good; it tells us that $H^{1}(F)= \Gamma(K^{1})/\mathrm{im}(\Gamma(I^{0}))$, so we've computed $H^{1}(F)$ using only global sections of some other sheaves. We'll come back to this, but let's make some other observations first.
The other thing you learn from the long exact sequence associated to the short exact sequence $0\to F\to I^{0}\to K^{1}\to 0$ is that for $i>0$, you have $$H^{i}(I^{0}) = 0\to H^{i}(K^{1})\to H^{i+1}(F)\to 0 = H^{i+1}(I^{0})$$
This is great! It tells you that $H^{i+1}(F)=H^{i}(K^{1})$. So if you've already figured out how to compute $i$-th cohomology groups, you can compute $(i+1)$-th cohomology groups! So we can proceed by induction to calculate all the cohomology groups of $F$.
Concretely, to compute $H^{2}(F)$, you'd have to compute $H^{1}(K^{1})$. How do you do that? You choose an embedding into an injective object $I^{1}$ and consider the long exact sequence associated to the short exact sequence $0\to K^{1}\to I^{1}\to K^{2}\to 0$ and repeat the argument in the third paragraph.
Notice that when you proceed inductively, you construct the injective resolution $$0\to F\to I^{0}\to I^{1}\to I^{2}\to\cdots$$ so that the cokernel of the map $I^{i-1}\to I^{i}$ (which is equal to the kernel of the map $I^{i}\to I^{i+1}$) is $K^{i}$. If you like, you can define $K^{0}=F$. Now by induction you get that $$H^{i}(F) = H^{i-1}(K^{1}) = H^{i-2}(K^{2}) = \cdots = H^{1}(K^{i-1}) = \Gamma(K^{i})/\mathrm{im}(\Gamma(I^{i-1})).$$
Since $\Gamma$ is left exact and the sequence $0\to K^{i}\to I^{i}\to I^{i+1}$ is exact, you have that $\Gamma(K^{i})$ is equal to the kernel of the map $\Gamma(I^{i})\to \Gamma(I^{i+1})$. That is, we've shown that $$H^{i}(F) = \ker[\Gamma(I^{i})\to \Gamma(I^{i+1})]/\mathrm{im}[\Gamma(I^{i-1})\to \Gamma(I^{i})].$$
Whew! That was kind of long, but we've shown that if you make a few reasonable assumptions, some easy observations, and then follow your nose, you come up with injective resolutions as a way to compute cohomology.
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It is probably worth making explicit that this same argument shows why you can use acyclics for the given functor you are interested in to resolve by and compute derived functors. The splitting property of injectives just ensures that they are suitable for any left exact additive functor. – Greg Stevenson Oct 19 2009 at 7:02
I'd also like to mention that the reader interested in following this line of thought might read Cartan and Eilenberg's book, which studies homological algebra from pretty basic principles. – Tyler Lawson Oct 19 2009 at 12:53
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This exact development is found in Gunter Harder's book "Lectures on Algebraic Geometry 1". It is also implicit in Hartshorne's treatment but I never knew that "universal delta functors" were as simple as this before reading through Harder's book. I have no idea why this line of thought is not usually presented upfront in presentation of derived functors (Ideally AFTER you have gotten a feeling for some cohomology with your bare hands, such as building $H^1$ from the ground up in group cohomology). Too many people seem content to use the definition without any motivation... – Steven Gubkin Jun 9 2010 at 13:41
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Even though I'm far from a historian, it seems to me the initial reason for considering injectives is
-prior to derived categories;
-but comes later than the task of extending left exact functors.
For the latter, in particular, there are apparently many ways of doing it. But:
injective resolutions are almost ideal for comparing different definitions of cohomology.
It is important to note that cohomology was around before injective resolutions, appropriate to different situations, and then the question came up of comparing several of them when they all made sense. As a concrete example, you might consider Cech and De Rham cohomology on a smooth manifold, both with real coefficients. It's also obvious that cohomology was not initially thought of in terms of the failure of exactness, which, in any case, can be good or bad depending on the situation.
Typically, you have two complexes A and O with rather different constitutions. How do we then compare their cohomology? The standard method is to find a third one C that admits natural maps A->C and O->C. Then we proceed to show that both of these induce isomorphisms on cohomology. A very basic form of this argument occurs even when showing that the cohomology computed using triangulations is independent of the triangulation. There, C is the complex associated to a common refinement.
Even in other situations, it makes sense to consider C as a 'common refinement' of some sort. The point then, is that an injective complex gives an ultimate common refinement in a wide variety of situations. This is because injectives, by their very definition, receive maps (more precisely, map extensions) very easily, so that you don't need to cook up a separate C for each pair of A and O. Once the injective definition is around, the different comparisons can be made in one fell stroke with the theorem that all acyclic resolutions compute the same cohomology as the injective one. Of course, `acyclicity' here can only be defined in terms of the fixed definition using injectives, and checking for it can be tricky and situation-dependent. For example, checking that the Cech resolution is acyclic on a variety requires that the covering consist of affines, and then knowledge of some property of affines. Checking that the smooth differential forms on a manifold form an acyclic requires some technicalities on partitions of unity and extensions of C-infinity functions, and so on. (Of course such verification processes are routine, once you're used to them. But whenever they are examined afresh, they always strike me as quite technical in interesting ways.) In the end, however, it's clear that this approach brings considerable conceptual unity to the ubiquitous problem of comparing cohomology. It's my best guess at the real initial motivation behind the definition.
You might even say that the definition of an injective object incarnates purely wishful thinking with regard to the problem of comparisons. What could be more naive than thinking there is one thing that frees you at once from all future specific consideration of C? The deep insight is that objects realizing such wishful thinking do exist often enough in nature. If I may add a bit of philosophical reflection, the definition of injectives illustrates quite well the sense of child-like innocence that Grothendieck regarded as fundamental to his mathematical nature. Several people have commented on the meaning of Grothendieck's self-evaluation, especially in view of the apparent sophistication of the resultant technology. It is interesting to identify the precise mathematical locus of such innocence, if only to gain some sense of when innocence is likely to yield fruit.
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I particularly like this line: "it's clear that this approach brings considerable conceptual unity to the ubiquitous problem of comparing cohomology." – David Zureick-Brown♦ Oct 19 2009 at 18:08
it's a wonderful answer! regarding the final paragraph, what do you mean by the "resultant technology"? do you mean "resulting technology"? Or is there some technology about the resultant?? – natura Apr 3 2010 at 1:13
I'm glad you liked it. I just meant 'resulting technology.' – Minhyong Kim Apr 5 2010 at 9:31
"It is interesting to identify the precise mathematical locus of such innocence, if only to gain some sense of when innocence is likely to yield fruit." This is a very nice sentence - I think innocence bearing fruit is one of more satisfying feelings of math. – Peter Samuelson Sep 23 at 6:25
Here is a funny aspect of the story:
When learning (co)homology one usually encounters first projective resolutions, projective objects are better to imagine in most people's opinion. But in sheaf categories there tend to be few projectives for two reasons:
In usual cohomology for modules, the projectivity is a requirement on the module structure only; set theoretically the map required in the definition of "projective" always exists. This requirement on module structure has of course to be met for sheaves of modules as well.
But secondly, even if you drop the requirement to be a homomorphism the map of sheaves does not need to exist: By definition of "projective" any epimorphism into a projective object has a section. For sets this is the statement of the Axiom of Choice ("every set is projective"). But in a sheaf category (which is a topos) the internal logic is in general intuitionistic, in particular the axiom of choice need not be valid and one can fail to find the section already on the level of sheaves of sets, even before looking for sections that are homomorphisms...
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This is an interesting discussion to someone raised in the 60's. It illustrates how lack of motivation creeps into books unnoticed. Back when Hartshorne was being written everyone was steeped in the then standard derived functor formalism of Cartan Eilenberg and Grothendieck, axiomatizing constructions of cohomology via complexes. So the pattern of this formalism explained so nicely by Anton and Andrew above was taken for granted. As the subject evolved, it was understood that acyclic resolutions could replace the more categorically natural injective ones. This is the gist of Evan's answer. Perhaps also there is a tradition in mathematics books of giving definitions without historical background. I have always had difficulty with any unmotivated definitions, such as abstract "residues" and modern Riemann Roch theorems, so I keep stressing to my students the value of learning the original version of Riemann, even if their goal is to understand cohomological and arithmetic versions. Anyway, excellent question.
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If you're willing to take for granted that (bounded-below) chain complexes and quasi-isomorphisms are good things to study, then left exact functors have the defect that they do not preserve quasi-isomorphisms between chain complexes. Derived functors are a way to correct this defect. The idea is that left exact functors do preserve quasi-isomorphisms between complexes of injective modules, and every complex is quasi-isomorphic to a complex of injectives. So whenever we want to apply a left exact functor to a complex, we should first replace the complex by a quasi-isomorphic complex of injectives--it doesn't matter which replacement we pick--and then apply our functor. This process goes by some name like the total right derived functor. In summary, derived functors are a way to force functors to preserve quasi-isomorphisms of chain complexes.
Often in homological algebra we start with a single object M rather than a complex, view it as a complex concentrated in degree 0, apply the right derived functor of F, and then take the ith cohomology group of the resulting complex--we write this as RFi(M).
There is a massive generalization of these ideas due to Quillen--the theory of model categories.
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One conceptual reason is, in technical terms, that "the derived category of (bounded-below) complexes is isomorphic to the category of (bounded-below) injective complexes." In less fancy language:
First, when working with a single sheaf A, we can make it into a complex A• with one A and everything else 0:
...→0→A→0→0→0→....
Then an "injective resolution" of A is really a complex of injectives I• with a quasi-isomorphism A•→I• (a map of complexes which induces an isomorphism on cohomology).
Now, say A• and B• are any (bounded below) complexes of sheaves which have the same (cochain) cohomology. There may not be a map from one to the other giving rise to the isomorphism of their cohomologies (kernels mod image), i.e. a quasi-isomorphism (qis).
However, you can find complexes of injectives I• and J•, and maps a: A• → I•, b: B• → J•, and f: I• → J• such that a,b are qis, and f is a homotopy equivalence (in particular a qis). So, "as far as cohomology is concerned", you can replace A• by I• and B• by J•.
The "big picture" reason for this is that injectives are "flexible" in terms of extending maps into them (that's how they're defined), which is what allows the construction of the maps in the previous paragraph, and having maps between things is good, because maps transform nicely under the application of functors.
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In my opinion, all answers go a little too far. In this (non-topological!) setting I think about this as follows: The aim is to analyze the lack of right-exactness of a left-exact functor $\Gamma:C \rightarrow D$ between abelian categories. One functorial notion that captures this inexactness is a delta-functor $H = \lbrace H^n, n \in \mathbb{N}$ with $H^0 = \Gamma$. This is what I call a cohomology theory for $\Gamma$. Now, there may exist a lot of them, and so it makes sense to look for universal cohomology theories for $\Gamma$, where universal means, that given any delta functor $K = \lbrace K^n, n \in \mathbb{N} \rbrace$ (not necessarily with $K^0 = \Gamma$ !) and a morphism $f^0:H^0 \rightarrow K^0$ in degree zero, then there exists a unique extension of $f^0$ to a morphism $f:H \rightarrow K$ of delta-functors. Up to canonical isomorphism there exists only one universal cohomology theory for $\Gamma$ which is then called the right derivative of $\Gamma$. The question is of course if such a universal cohomology theory for $\Gamma$ exists. And here it comes: If the category $C$ has enough injectives, then $\Gamma$ has a right derivative which can be computed by injective resolutions. (You can confer Lang's Algebra book for all the above notions, it's actually pretty nice).
The universality condition forces a strong connection between $\Gamma$ and $H$. This allows you for example to extend constructions in degree 0 to arbitrary degree. An example from group cohomology: Let U be a subgroup of a group G and let $g \in G$. Then conjugation is an isomorphism of functors $(-)^U \rightarrow (-)^{gUg^{-1}}$, where $(-)^U$ is the U-invariant functor on G-modules. Now, $H^n(U,-)$ is a universal cohomology theory for $(-)^U$ and $H^n(gUg^{-1},-)$ is a universal cohomology theory for $(-)^{gUg^{-1}}$. Hence, the exists a unique isomorphism of delta-functors $H^n(U,-) \rightarrow H^n(gUg^{-1},-)$ and in this way we have extended the conjugation to all cohomology groups. This is very inexplicit of course, but this is a powerful property for more advanced constructions. And verifying each time that constructions are compatible with the deltas is not a nice job (normally people simply say, it works).
In Lang's book "Topics in the Cohomology of Groups" you will find a lot more of such arguments.
If you now want to get rid of all choices, you can move to the derived category.
Hope this helps.
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1
Sorry if this question sounds stupid, but why is universal object good/special? In some very concrete cases, like free groups or tensor products, it is very clear to me why the universal properties in those cases are of interest. However in the situation of delta functors, I don't even know why we should care about the universal property of unique extension of degree 0 morphism. – Ho Chung Siu Oct 19 2009 at 9:25
Not a stupid question! I have edited my above post. – S1 Oct 19 2009 at 11:45
This answer is the same as Anton's, just with more machinery under the hood. In particular if you spell out what a delta functor is, what it means for one to be universal, and why injective resolutions give you one, you will have Anton's answer. – Steven Gubkin Jun 9 2010 at 13:46
The reasons injective (and dually, projective) resolutions are important in the construction of derived functor cohomology are primarily technical in nature. You can think of an injective resolution as a "nice" replacement for an object (or more generally, for a chain complex) in the homotopy category of chain complexes in the same way that we think of a CW complex as a "nice" homotopical replacement for an arbitrary topological space. A very important feature is that these replacements are functorial in the homotopy category. In other words, the resolutions are unique up to chain homotopy equivalence, and given a map between two objects, this map extends to a map between their injective resolutions uniquely (again up to chain homotopy).
However, once you actually show the existence of a good notion of derived functor, you can actually use more broad classes of resolutions for computations. It's generally fairly impractical to take injective or projective resolutions. Instead of injective sheaves, one can take resolutions of acyclic objects, which are objects that themselves have no higher cohomology. There are various classes of acyclic sheaves that are often used in various contexts, such as soft sheaves, flasque (or flabby) sheaves, soft sheaves, and fine sheaves. Finding some sort of acyclic resolution is often much easier than finding an injective resolution, so understanding when various types of sheaves are acyclic is very useful.
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http://physics.stackexchange.com/questions/6867/how-does-slow-anti-hydrogen-annihilate-with-normal-matter-in-the-lab | # How does slow anti-hydrogen annihilate with normal matter in the lab?
In a recent article:
Physical Review A 83, 032903 (2011), A.Yu. Voronin, P.Froelich, V.V. Nesvizhevsky, Antihydrogen Gravitational Quantum States
the authors claim that anti-hydrogen has a lifetime of 0.1 seconds when placed in a container with metal walls. They say that the atom is kept from annihilating with the walls because of the Casimir-Polder effect. This raises in my mind the question of how it is that the atoms annihilate. My first guess is that the positron would annihilate against an electron, and then the proton would be attracted into the metal by the image charge effect. But this would seem to be in contradiction to the Casimir-Polder effect mentioned in the paper.
So how does slow anti-hydrogen decay in the presence of normal matter?
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– voix Mar 14 '11 at 18:59
## 2 Answers
Dear Carl, the correct paper to derive the 0.1-second lifetime of the anti-Hydrogen atom in the gravitational field is described after the very sentence you quoted.
There is a "[20]" symbol which means that the sentence is justified in the reference number 20 in the list of literature at the end of the paper you quoted. So the correct paper that answers your question is
Quantum reflection of ultracold antihydrogen from a solid surface
A Yu Voronin and P Froelich 2005 J. Phys. B: At. Mol. Opt. Phys. 38 L301, doi: 10.1088/0953-4075/38/18/L02
http://iopscience.iop.org/0953-4075/38/18/L02
http://iopscience.iop.org/0953-4075/38/18/L02/pdf/0953-4075_38_18_L02.pdf
which is fully available online - click the last link for the PDF file. At distances longer than 15 nanometers from the metal, the Casimir-Polder potential has the form $-74/z^4$ in atomic units. At shorter distances between 1 Bohr radius or so to 15 nanometers, the potential becomes $-0.25/z^3$ in atomic units - a van der Waals form.
Only at distances shorter than 1 Bohr radius or so, the anti-Hydrogen behaves differently than the Hydrogen, and is annihilated. This portion of the potential is not probed in the 0.1-second case: note that to get annihilation, the positron must approach the electron at a shorter distance than the Bohr radius - comparable to the Compton wavelength of the electron (and even shorter, nuclear distances if we want to annihilate the hadrons). The authors calculate the reflection probability as a function of the kinetic energy of the anti-atom. The lower energy/temperature they have, the more likely they will bounce off. For low enough kinetic energy, the reflection probability is as high as 0.99987 (a table).
In those cases, one doesn't probe the short-distance portion of the potential. So the answer to your question why there's no annihilation is the same as the answer to the question why you don't get fusion if you fill a tank with the Hydrogen gas.
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1
As usual your comment is useful, and as usual I'll upvote it when I get my votes back in 4 hours (I always use them up). But what I'm asking is "how does anti-hydrogen annihilate". Is it the case that the positron goes first and then the proton is attracted into the metal and eventually hits a nucleus? Does that mean that the anti-hydrogen can escape the metal container with a fairly high probability (if it misses the nuclei)? – Carl Brannen Mar 14 '11 at 19:03
Dear Carl, thanks for your niceties. The antiproton is negatively charged so it is electrically attracted to the normal nuclei. As it accelerates, it emits and loses energy, and eventually falls to one of those normal nuclei and they annihilate. The electrons and positrons play the role of the shield that protects the hadrons from close encounter - additional potential energy - but the shield only works when the distances are atomic or longer. I don't know who annihilates "first" (proton or electron) and whether the question makes sense at all. e- e+ will probably form a positronium for a time – Luboš Motl Mar 20 '11 at 7:21
Let me say that antihydrogen is never " slow" in the process of annihilation, when it comes to matter it is already accelerated by Casimir-Polder potential to the energies of few eV, which corresponds to temperature of dozens of thousands of degrees. The attraction forces between particles and antiparticles tear antihydrogen into antiproton, which is captured by matter atom, and positron, which most likely forms pair with electron. Antiproton falls down on nuclei, ejecting electrons from the atom. Then it Usually destroys nuclei and annihilates. This process usually takes time of order of 10^-9 s or so, electron-positron in best case could survive for 10^-7 s.
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Where the gravitational pull of Earth exist up to? What distance from Earth it will be zero? How do the skydivers fly at a same altitude? Won't they feel gravitational pull? What is the Earth'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": 16, "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": 0.9427828788757324, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/248617/bounds-for-maximal-blowup-contained-in-graph/248657 | # Bounds for maximal blowup contained in graph
In my homework, I'm asked to prove the following: By denoting $b_n(r,\epsilon)$ - the largest integer $b$ so that any graph with $(1-\frac{1}{r} +\epsilon)\frac{n^2}{2}$ edges, contains a $b$-blowup of $K_{r+1}$ (meaning, that it contains a complete $r+1$-partite graph, with $b$ vertices in each partition). I need to show that $b_n(1,\epsilon)=\Theta\left(\frac{\log n}{\log(1/\epsilon)}\right)$ and that $b_n(2,\epsilon)=O\left(\frac{\log n}{\log(1/\epsilon)}\right)$.
First, I'm not sure how to show either the lower bound, or the upper bound. For the first part, I know that $ex(n; K_{b,b}) \geq c\binom{n}{2}^{1-\frac{1}{t+1}}$ and $ex(n; K_{b,b}) \leq cb^{\frac{1}{b}}n^{2-\frac{1}{b}}$ and I think that for one of the bounds I need to combine one of those inequalities, with the fact that for $\epsilon\binom{n}{2}$ there is a bipartite graph.
I'll appreciate any light you can shed on the subject.
Thanks in advance.
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## 2 Answers
I have a feeling we are in the same class. If I understand correctly, we defined a b-blowup to be an addition of b vertices to each original vertex, and each of the new ones is connected only to the neighbors of the original vertex. So a b blowup of K_(2) is the graph K_(1,b) and not K_(b,b).
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Seems I was wrong about that. – user38310 Dec 1 '12 at 17:01
Highly possible. After all, there are not that much extremal graph theory courses, commencing simultaneously around the world. Now, about the definition - the one that I have is the one I wrote. From the searches I made, this definition also coincides with the theory. – Pavel Dec 1 '12 at 17:05
Your definition is correct and I managed to sovle the problem now using the bounds you posted. Just take them and "ask" for which values of b are the bounds achieved, i.e when is ex(n;Kb,b)≥c(n^2)1−1t+1 > ϵn^22.
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Well, it's a good idea and I've kind of thought of it already, but it gives the lower bound only - $b > \frac{logn}{log(1/\epsilon)}$. And even then, it requires some fine tuning on the constants. – Pavel Dec 1 '12 at 19:05 | {"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": 10, "end_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": 0.9513505697250366, "perplexity_flag": "head"} |
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Walter Lewin's best performance was the pendulum demonstration, and I copy the transcript now: Would the period come out to be the same or not? [students respond] Some of you think it'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": 51, "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": 0.9355021715164185, "perplexity_flag": "middle"} |
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Maximum_flow_problem | # All Science Fair Projects
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# Max flow min cut theorem
(Redirected from Maximum flow problem)
The max flow min cut theorem is a statement in optimization theory about optimal flows in networks.
Suppose G is a finite directed graph and every edge e has a capacity w(e), a non-negative real number. Further assume two vertices s and t have been distinguished. Think of G as a network of pipes; we want to pump as much stuff as possible from the source s to the sink t, never exceeding any edge's capacity.
A flow f on G from s to t assigns to each edge e a real number f(e) with 0 ≤ f(e) ≤ w(e) such that for every vertex x of G different from s and t, we have
$\sum_{e \; \in \; I(x)} f(e) = \sum_{e \; \in \; O(x)} f(e)$
where I(x) is the set of all incoming edges of x and O(x) is the set of all outgoing edges of x ("whatever goes in must come out"). For such a flow, we automatically have
$\sum_{e \; \in \; O(s)} f(e) \; - \sum_{e \; \in \; I(s)} f(e) \quad = \quad \sum_{e \; \in \; I(t)} f(e) \; - \sum_{e \; \in \; O(t)} f(e)$
this number is called the amount of the flow. Intuitively, it is the net total amount of stuff we are pumping from s to t. We are interested in flows with maximal amount.
A cut C on G between s and t is a set of edges such that every directed path from s to t passes through at least one edge in C. The capacity of the cut C is the sum of all its edge capacities.
The statement of the theorem is:
The maximal amount of a flow is equal to the capacity of a minimal cut.
Note that there may be several flows which attain the maximum amount, and there may be several cuts which attain the minimal weight.
There is a partial correspondence between maximal flows and minimal cuts: if C is a minimal capacity cut, then there exists at least one maximal value flow f such that f(e) = w(e) for all e in C. Conversely, if f is a maximal flow, then there is a minimal cut C such that f(e) = w(e) for all e in C.
The theorem was proved by A. Feinstein and C.E. Shannon in 1956, and independently also by L.R. Ford, Jr. and D.R. Fulkerson in the same year. Determining maximum flows is a special kind of linear programming problem, and the max flow min cut theorem can be seen as a special case of the duality theorem for linear programming.
Maximum flows can be computed with the Ford-Fulkerson method, which is most commonly implemented according to the Edmonds-Karp algorithm. This implies that the maximum flow of a graph with integer capacities has an integer value as well. Another way to prove this is to notice that one of the linear programming formulations of the max flow problem has a totally unimodular constraint matrix, and so has integer solutions.
## External link
A review of current literature on computing maximum flows
03-10-2013 05:06:04
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http://gowers.wordpress.com/2012/03/21/endre-szemeredi-wins-the-2012-abel-prize/ | # Gowers's Weblog
Mathematics related discussions
## Endre Szemeredi wins the 2012 Abel Prize
If you were looking for a clue about this year’s winner, you could perhaps have paid attention to the curious incident of my recent Mathoverflow question.
“But you haven’t asked any Mathoverflow questions recently.”
That was the curious incident.
Anyhow, it was wonderful to be told that Endre Szemerédi was to be this year’s winner. I won’t say any more in this post, but instead refer you to the Abel Prize website and to the written version of the talk I gave, which was intended for non-mathematicians.
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### 65 Responses to “Endre Szemeredi wins the 2012 Abel Prize”
1. Wednesday Highlights | Pseudo-Polymath Says:
March 21, 2012 at 1:46 pm | Reply
[...] A prize and for what explained. [...]
2. Endre Szemerédi laureado com Prémio Abel 2012 « problemas | teoremas Says:
March 21, 2012 at 2:14 pm | Reply
[...] “The Work of Endre Szemeredi” de Timothy Gowers encontra-se disponível a partir deste artigo do seu blogue. Partilhar isto:TwitterGostar disto:GostoBe the first to like this . Deixe um [...]
3. Konstantinos Says:
March 21, 2012 at 3:28 pm | Reply
Reblogged this on Room 196, Hilbert's Hotel.
4. Hakuna Matata Says:
March 21, 2012 at 5:41 pm | Reply
Well done Norwegian Academy.
5. Attila Says:
March 22, 2012 at 8:14 am | Reply
Why is it a clue that you asked no question on Mathoverflow?
• gowers Says:
March 22, 2012 at 8:29 am
That’s a reasonable question. The comment was a joke that only a small proportion of readers could be expected to get, so I’ll now explain it. Last year, I was talking about John Milnor, and I couldn’t understand why his famous result about exotic spheres wasn’t obviously false. So I asked a related question on Mathoverflow, without revealing why. So this year my failure to ask a question could have been taken as a sign that I was more familiar with the work of the winner, as was indeed very much the case. But it would have been a silly argument, as I wouldn’t have dared to ask about an unfamiliar topic. So there were in reality no circumstances in which I could have used Mathoverflow to help me this year.
6. Attila Says:
March 22, 2012 at 8:43 am | Reply
Thanks for the explanation, Tim.
And I’m very happy to learn that you are a Holmes aficionado …
7. Ionica Smeets (@ionicasmeets) Says:
March 22, 2012 at 9:57 am | Reply
I predicted the winner correctly (for the first time in ten years)! My clue was that Terence Tao told me a few years ago that Endre Szeméredi is his favorite mathematician. The interview is here (in Dutch): http://www.wiskundemeisjes.nl/20070426/de-favoriete-nog-levende-wiskundige-van%E2%80%A6-11/
Your talk was excellent yesterday, I liked it even better than the written version.
8. Endre Szemerédi wins the 2012 Abel Prize | OU Math Club Says:
March 22, 2012 at 5:23 pm | Reply
[...] Fellow math blogger, Tim Gowers, was in charge of giving a talk for non-mathematicians (i.e. journalists and such) about Dr. Szemerédi’s research. A tough challenge which Dr. Gowers adroitly pulls off. You can read the text on his blog here. [...]
9. plm Says:
March 22, 2012 at 7:53 pm | Reply
Before the announcement I looked at your recent questions, noticed there were none, thought that perhaps Terence Tao was picked (I know little combinatorics so I could not value Szemerédi’s achievements myself), then thought more probable that you had decided not to ask publicly, knowing that you would reveal you were giving the popular talk. This made sense especially if you were going to be a regular presenter, to avoid a guessing game. Then I made the Tarski joke because I could not think of a logician for the Abel prize -now I think perhaps Hrushovsky, Shelah, a computer scientist, a set theorist. The big question is: why “mereological sum”?
Also I would be curious to know how many nonmathematicians looked at the public lecture, what they think about it and how it influenced them.
• gowers Says:
March 23, 2012 at 10:06 am
The mereological sum business was another thing not to be taken too seriously. I wanted to be very careful not to give anything away, including the number of winners (which can be greater than 1). My initial impulse was to refer to a set of mathematicians, but that isn’t correct: the Abel Prize is not awarded to a set. So I wanted to do something like the following operation: take a set of mathematicians and remove the curly brackets. So my question was this. If I take some mathematicians, form a set out of them, and remove the curly brackets, then what is the relationship between what I now have and the mathematicians I started with? I think the answer is that I have their mereological sum, though I wouldn’t be too surprised if a philosopher read this and told me I was wrong.
• plm Says:
March 23, 2012 at 6:23 pm
Thanks alot for the explanation. I really should have understood that by myself. I guess philosophy requires too much rigor for me.
Thank you.
• plm Says:
March 23, 2012 at 6:39 pm
On a second thought, I think “union of mathematicians” would be the appropriate counterpart of “mereological sum of mathematicians”, because “set of mathematicians” implies that you take the set of recipients (say using the axiom of separation), not the union of the recipients (the set made of all elements in at least one of them, whatever that may mean). And in mereology the counterpart of “set of…” would actually also be “set of…”, as a mereological object.
• Richard Baron Says:
March 23, 2012 at 10:07 pm
Perhaps the difficulty of deciding how to describe the situation in a way that allows for an award to one mathematician, and also an award to a greater number of mathematicians, arises out of a decision to regard the prize as an abstract entity.
The prize money is straightforward. It consists of a pile of kroner, and we can think of each mathematician taking a part of the pile (an improper part when there is only one winner).
The prize itself is different. If we think of an entity, the prize, being awarded to more than one mathematician, we have to worry about curly brackets and mereological sums. But we could avoid this if we did not think in terms of an entity which was won by other entities, and thought only in terms of properties of the winner or winners. That is, we could describe people as Abel-Prize-winners, without thinking that there had to be an Abel Prize that they, or a set of them, or a mereological sum of them, had won.
This approach is not free of philosophical difficulties. Nothing is, not even nothing. And it would contradict clause 2 of the statutes of the prize: ” … en internasjonal pris … “.
10. iancreid Says:
March 23, 2012 at 4:38 am | Reply
Tim
Thanks very much for your written talk – I enjoyed it very much.
I think there may be a small correction:
“pair up the rocks in L with the rocks in R” should be “pair up the rocks in L with the rocks in H” I think?
• gowers Says:
March 23, 2012 at 10:07 am
No prizes for working out how I made that mistake …
Anyway, thanks for pointing it out. I’m not sure whether I can change the version on the Abel Prize website, but with a bit of effort I can change the version linked to from this blog. Maybe I’ll wait a bit to see whether any other errors get picked up.
11. Budapest, Seattle, New Haven | Combinatorics and more Says:
March 23, 2012 at 7:56 am | Reply
[...] yesterday the Abel prize Laureate 2012. http://www.abelprize.no/ Congratulations, Srac! (See also this post on Gowers’s [...]
12. Alon Amit Says:
March 24, 2012 at 6:18 am | Reply
A couple of additional typos in the written version:
- “Some ‘elementary’ proof are amongst the hardest…” (should be “proofs”)
- “I’m just going to repeat that is elementary in the technical sense…” (missing “it”)
- “…computer program that was able learn from experience” (missing “to”)
Thanks for sharing the talk!
13. Anonymous Says:
March 31, 2012 at 5:02 am | Reply
Tim, do you agree with the following interpretation of your essay
“Gowers speaks about a possible scenario of the development of mathematics in his essay “Rough structure and classification” (published in a special issue of “Geometric and Functional Analysis” in 2000 or 2001). See Section 2 entitled “Will mathematics exists in 2099?” He outlines a scenario in which mathematicians are gradually replaced by computers. This essay cannot be separated from his more famous essay “The two cultures in mathematics”, published nearly simultaneously. All supporting examples for his scenario belong to his “second culture”, which is more or less coinciding with the Hungarian style combinatorics. His attempts to approach some issues of the first culture along the lines outlined by him failed abysmally. A Wigner-shift to the Hungarian combinatorics would make his scenario a much more probable one. Indeed, it would be not very surprising if a Szemeredi-Gowers-like mathematician could be surpassed by computers (assuming that Gowers’ own description of his style in “Two cultures…” is correct), very much like Kasparov was surpassed by a very primitive by the current standards computer. But it is hard to imagine that Serre or Milnor can be ever approached by a computer.”
Taken from a comment thread here
http://www.blogger.com/comment.g?blogID=4049018105272921172&postID=8690732893167971234
• gowers Says:
March 31, 2012 at 1:11 pm
I agree with some of it, disagree with some of it, and am not sure about some of it. The main point of agreement is that I am very interested indeed in the possibility of getting computers to do a lot of what can currently only be done by mathematicians. I do not know what “His attempts to approach some issues of the first culture along the lines outlined by him failed abysmally” is referring to. I don’t recall having made any serious attempt in this direction. I disagree with some of the disparaging remarks about Hungarian-style combinatorics that you did not reproduce here. Also, the writer seems to have a picture of me as a wheeler-dealer that I don’t recognise.
But the most interesting question is whether Serre or Milnor can be approached by a computer. The first thing I’d say is that that’s setting the bar very high: by and large Serre and Milnor can’t be approached by humans either, except in very rare cases. So the real question (for me) is whether there is something about first-culture mathematics that means that computers would necessarily be terrible at it, as opposed to merely not the best in the world. The difficulty I have here is that I am not sufficiently trained in that kind of mathematics to be able to think about this question by analysing carefully how I do it. I’d love to be able to do more, but in the end I think my efforts are better spent on thinking about how computers would do the kind of maths I myself do.
I’d be very interested indeed if somebody who works in more of a first-culture sort of area took the view that computers ought to be able to do that too. (Actually, Gromov strongly believes that computers can take over, and he has plenty of first-culture experience.)
• Anonymous Says:
April 3, 2012 at 9:00 am
Computers obviously cannot replace human beings, Godel and Turing prohibit this. But of course it is intersting to which extent computers may help mathematician to solve their problems, how low is the complexity of the “real-life” mathematical questions. So, Timothy really has a point here, and I don’t see why one who loves “first” culture (may be it is second actually?!), and who obviously hasn’t succeeded in building any theories (great mathematicians never go down for condescending way of talking to others), would start casuistry about “computers replacing people”.
• gowers Says:
April 3, 2012 at 10:28 am
I’d query your use of the word “obviously”, since only a small minority of mathematicians and philosophers agree with Penrose that Gödel and Turing’s arguments demonstrate that there is something that humans can do that computers can’t.
• sowa Says:
April 3, 2012 at 12:32 pm
To Anonymous:
Here is a quote from Gowers’ GAFA Visions essay.
“In the end, the work of the mathematician would be simply to learn how to use theorem-proving machines effectively and to find interesting applications for them, This would be a valuable skill, but it would hardly be pure mathematics as we know it today.”
If this is not “computers replacing mathematicians”, then what it is?
• sowa Says:
April 3, 2012 at 12:59 pm
To Timothy Gowers:
The quoted by Anonymous comment requires some clarifications after being transferred here. You understood me partially correctly, partially not, and partially attributed to me something that I never said or even thought (namely, a picture of you as a wheeler-dealer). Unfortunately, a proper reply will hardly fit in this narrow column (sorry, the pun is not intended). It is already two standard pages long. I will try to find a suitable place to post it. Here I will say just a few words.
What you called the main point of agreement at least now hides the main point of disagreement. In the above quote from your GAFA Visions paper you seem to agree that using computers is not (pure) mathematics. But it is hard to understand how such a mathematician as you may be interested in eliminating mathematics. Anyhow, some people (including me) love mathematics in the original sense of this word as opposed to assisting computers.
I agree that the most interesting question is the one about Serre and Milnor. Are they humans or some kind of superhumans? But it leads nowhere, we have to assume that they are humans, as we do with respect to all similarly looking creatures.
Finally, after being charmed for several years by your essay about two cultures and studying some part of that presumed second culture as a result, I come to the conclusion that there is no two cultures, there is only one (pure) mathematics.
• gowers Says:
April 3, 2012 at 9:51 pm
Let me try to answer you when you say, “But it is hard to understand how such a mathematician as you may be interested in eliminating mathematics.” I’ll start from the premise that such an elimination (that is, an elimination resulting from a computer program that can do mathematics as well as we can and much more quickly) is possible. Thus, I’m trying to explain why if it is possible, then I am interested in it. Whether or not it is possible is a separate question.
The main reason I think it is worth trying to do an elimination of this kind is that the challenges along the way seem to me to be fascinating. It will be impossible to program a computer to prove theorems without a deep understanding of the theorem-proving process, and that, it seems to me, is an understanding that is worth striving for: I rebel against the idea of accepting that we rely on mysterious and irreducible flashes of characteristically human genius, and I want to know what is going on.
If it did turn out to be possible to get computers to do mathematics, then something I value would of course be lost. But there have been many such losses as technology has improved, and nobody would argue that those losses outweigh the gains. A recent example has occurred in the humanities: it used to be a mark of a good scholar in the humanities that they knew a huge amount and could therefore make connections that other people wouldn’t spot. That ability is still useful of course, but the vast amount of data that is now available online and searchable has made it far less important than it used to be. I can well imagine a recently retired clever-connection spotter thinking ruefully that what gave him/her such pleasure throughout an entire career would now be something that one could do just by typing a few words into Google. But we wouldn’t want to give up Google for that kind of reason.
Similarly, if it turns out that computers can do maths, then I think that things like the activity of sitting around for two weeks struggling to prove a lemma and only then realizing that there is a counterexample would come to seem as quaint as writing a PhD thesis by hand and paying somebody to type it up. And a generation of people would grow up knowing that computers could answer their mathematical questions (not all of them obviously, but the ones that humans might have been able to answer) and wouldn’t miss the hard slog that had previously been necessary.
There’s a very important qualification here, concerning how the programs actually work. If they are something like Deep Blue, using huge searches as a substitute for human understanding (which I actually don’t think is possible, so this is a rather hypothetical qualification), then something important would definitely be lost. So my requirement of a computer program that I would be happy (if a bit wistful) to see taking over is that it should be able to explain its thought processes in a way that humans can understand — at least when such explanations exist.
• plm Says:
April 3, 2012 at 10:19 pm
Dear Tim,
You say “nobody would argue that those losses outweigh the gains”. I think I understand what you mean and am a bit surprised: I see alot (I couldn’t easily give numbers) of people arguing against technology making humans obsolete, in several ways claiming “it’s not the same result”, or “it was nice to do it by hand” (say chorizo), or “it was happier times back when humans did that”, or generally when humans used less technology.
I personally like all technology, without having thought about the issue thoroughly, but I think those are complicated and grave matters -so I plan to think about them more.
• sowa Says:
April 4, 2012 at 5:53 am
To Timothy Gowers:
Your explanation of this particular point is understandable, but not without serious qualifications. First of all, if this is your motivation, you are not mathematician anymore. You are a scientist, interested in a particular phenomenon. Mathematics is only partially a science; to a big extent it is an art. If your project succeeds, this art will disappear, in accordance with what you said in your GAFA Visions paper.
If you agree that you are not a mathematician anymore, then that my question is answered, but there will be other questions.
Second, your project is doubtful even for a scientist. Not all scientific and/or technological projects should be pursued, and not all of them are beneficial for the humanity. Imagine that A-bomb and H-bomb were created without the justification of the WWII and later of the Cold War, just out of scientific curiosity (many people consider this to be very objectionable even with these justifications), and then, of course, were tested, as the scientific method requires. Or imagine somebody working on a global weapon capable to turn the whole Earth into dust, and then testing it (for the sake of the argument, let us assume that it is impossible to deliver this weapon to another planet).
I would be surprised if you will not find such a research objectionable. Your goals are more moderate, of course: only to eliminate some kind of human activity. Still, your project is an experiment on human subjects, and now there are very strict rules for such experiments.
The Deep Blue example is quite relevant. The development of the chess-playing software started with exactly the same motivation you suggested: to understand how humans play chess. The success of Deep Blue told us nothing about this; it plays chess in a completely different manner than humans. Note that chess is, as G. Hardy mentioned in his famous book, a part of mathematics, only a very uninteresting one from the mathematical point of view. It is only natural to expect that if your project succeeds, the result will be almost the same. We will learn nothing about how humans prove theorems. But while chess survives as an entertainment, mathematics will not survive, since it cannot support itself as a form of entertainment.
• gowers Says:
April 4, 2012 at 7:55 am
I don’t have the space to explain why in detail, but I do not believe that anything like a Deep Blue approach to mathematics can be successful. I therefore think that the only way to program a computer to prove theorems is to get it to mimic closely how humans prove theorems. Therefore, I couldn’t disagree more when you say, “If your project succeeds, the result will be almost the same. We will learn nothing about how humans prove theorems.” One of my main reasons for interest in this is to learn about how humans prove theorems.
• sowa Says:
April 4, 2012 at 11:18 am
To Timothy Gowers:
Certainly, your explanations would be of interest not only for me. I hope you will have an occasion to write them down. Your claim contradicts all the experience of the humanity: successful technologies do not imitate the ways humans do things. Steam power, electricity, cars, planes, phones, computers, computers playing chess do not imitate humans. In any case, it is hard to accept this as a justification for your dangerous experiment: even in your own opinion, it can be justified only if your will get a particular outcome. The people who build the Deep Blue were interested in creating a tool to be used (of course, the chess was only a toy problem from the very beginning). The works in way different from humans, but it does not matter for the tool-designer. These issues are discussed in the book of Feng-hsiung Hsu, the system architect of Deep Blue. But as you say now, you are interested only in demonstrating that the most advanced intellectual human activity is not really human (Deep Blue did not demonstrated this for chess, and this wasn’t a goal).
After this short exchange of comments the question if you still consider yourself to be a mathematician emerged as the most intriguing one.
• gowers Says:
April 4, 2012 at 7:17 pm
I’m not sure it’s as intriguing as all that. I consider myself a mathematician because I try to solve mathematical problems. But probably the question you’re really asking is whether the activity of trying to work out how to program a computer to solve mathematical problems is itself mathematics. That is, would I still be a mathematician if I was devoting 100% of my time to trying to develop such a program? One could ask the same question of the process of trying to understand how somebody thought of a proof, or could have thought of it. Both activities involve thinking hard about how humans do mathematics, which is not itself mathematics. However, it is an activity that cannot be done well without significant mathematical experience (at least if my thesis that the Deep Blue approach won’t work is correct), and I don’t rule out that it might be possible to develop a formal model of “discoverable mathematical proof” that would turn the not fully mathematical question “How did anyone think of that?” into “Exhibit a discoverable [in the formal sense] proof of that.” I don’t claim to have such a model right now.
• sowa Says:
April 4, 2012 at 11:24 pm
To Timothy Gowers:
Of course, I am aware that you are working on some mathematical problems, and not only try, but actually solve some. In this sense you are a mathematician; and there is no much sense in discussing this triviality. The question is little bit more complicated than form you gave to it, but let start with your version. If you devote all your time to your program, then, as you say, you will exclude the Deep Blue approach, and will use your mathematical experience. I think that using the past mathematical experience cannot qualify you as a mathematician in this hypothetical situation. If some mathematician starts working on the Wall Street, she/he inevitably will use his past mathematical experience. Still, we usually say about such people “he/she quit mathematics and now works for Goldman-Sachs”. As all other people, mathematician can and do change their profession to some other sometimes.
But my question was not this one; I asked exactly what I wanted. It seems that your program of eliminating mathematicians is of very high value for you. In your explanations above with respect to the question “how such mathematician as you can be interested in eliminating mathematics” you presented yourself as a scientist, not as a mathematician. It seems that this scientific (presumably, it may be not) problem outweighs your recent work in mathematics. It seems that for you mathematics is a sort of an entertainment you are good at (like some people play chess or climb mountains), but for you the *real* problem is “can we replace mathematicians by computers, and if we can, how we do this”. For me this is hardly compatible with being a mathematician
• Scott Carnahan Says:
April 5, 2012 at 4:36 am
sowa: Your argument rests on a definition of “mathematician” that is not universally accepted. Some people talk about “quitting mathematics” when they refer to leaving the academic mathematical world, but professional research mathematics is far from the only place where mathematics is done. While the word “mathematician” is sometimes used in an exclusive sense you seem to indicate (but have not precisely defined), one may also use the word “mathematician” to mean anyone who is skilled or educated in mathematics, or someone who uses mathematics as a fundamental part of day-to-day life.
At any rate, why are you wasting your time telling mathematicians that they aren’t mathematicians on the internet, when you could be using your antagonistic skills for far greater pursuits, like running for political office?
• sowa Says:
April 5, 2012 at 5:44 am
To Scott Carnahan:
I am not sure why are you wasting your time trying to tell some person on the internet how he should use his presumable skills. Anyhow, the answer to your question is very simple: I love mathematics, I am interested in the continuation of its existence, and I am not interested at all in holding a political office.
You don’t know my definition of a mathematician; I pointed out in my very first comment that the proper reply is quite long and does not fit in the comments here. If not the insinuations of the first Anonymous, I would try to write down my ideas carefully and to find a proper place to post them. Now many of them are already mentioned in the comments.
I do not think that one can be considered a mathematician only if he or she belongs to the academic world, although it seems that earning a living in some other way is not compatible nowadays with doing research in mathematics (it is well know that it was compatible sometimes at some places).
In any case, that is the point of arguing about what definition is the right one? There are some qualities which are necessary for being a mathematician in “my sense”. It is not my personal definition, it is the one a learned from other people either directly or through their writings. It seems to me now that I was influenced most by my Ph.D. thesis advisor and by writings of A. Weil and J. Dieudonne (who once usggested another definition, more suitable for his goals). Answering the question if somebody is a mathematician in this sense could be enlightening even if you adhere to another one.
The issue is not if the definition I use is universally accepted (in any case, such issues are not decided by a vote, this is not politics). The issues at hand are what is mathematics, is it valuable for the human race, would it be good to sacrifice it for the sake of scientific curiosity alone, etc.
14. JuanPi Says:
April 2, 2012 at 4:02 pm | Reply
Good text!
“to” missing in the sentence
[...] a computer program that was able learn from experience [...]
15. Another anonymous Says:
April 4, 2012 at 11:34 am | Reply
“Serre and Milnor can’t be approached by humans either, except in very rare cases.”
Really? Both are renowned for the clarity of their mathematics, besides anything else. With Milnor this is even official; he has a Steele Prize for exposition.
“it used to be a mark of a good scholar in the humanities that they knew a huge amount and could therefore make connections that other people wouldn’t spot. That ability is still useful of course, but the vast amount of data that is now available online and searchable has made it far less important than it used to be. I can well imagine a recently retired clever-connection spotter thinking ruefully that what gave him/her such pleasure throughout an entire career would now be something that one could do just by typing a few words into Google.”
I do not doubt that connection-spotting in mathematics is currently valued highly. In my own low-level experience (and the principle of mediocrity suggests that this is wide-spread; I am not Gel’fand) this is a matter of knowing one thing, or bunch of things, seeing another, and then recognizing a similarity, or parallel, or analogy. But “knowing” very often means “having some hazy acquaintance with or memory of”; how is this kind of knowledge to be uploaded to a database, when the person knowing it scarcely knows it themselves? And often the useful output is just as vague as the input: “Hmm, that reminds me, but of what exactly?”. Processing this imprecision is, currently, a crucial part of doing mathematics; when do you expect databases to be able to handle it as well as we do?
• gowers Says:
April 4, 2012 at 7:10 pm
That’s a very good question that I have certainly thought about but will not have a good answer to without thinking about it a lot more (and not necessarily even then).
I think you misinterpreted my remark about Serre and Milnor. I meant that only very few humans can do mathematics of the depth and originality of that of Serre and Milnor. I did not mean that very few humans can understand what they did.
• sowa Says:
April 4, 2012 at 11:52 pm
To Another Anonymous and Timothy Gowers:
I think that can add some perspective to the Serre-Milnor question. By the way, Serre also got a Steele Prize for expository writing (for “Arithmetics”). The quality of writing is indeed relevant here. Both of them wrote (and still write!) very clearly. The way they wrote their papers and books helps tremendously to understand not only why the results are true, but also that the results (not always their own) are natural and inevitable. Later on both of them wrote extremely interesting accounts of how they were lead to discover their Fields medal results. So, they are “approachable” not only in the sense that we can easily read their remarkable books and papers. We can also get a lot of insight into how their results were discovered.
So, the answer to my question above is that Serre and Milnor are humans. It was to some extent a rhetorical question. As I said, the answer “yes, they are humans” is true by much more general reasons than outlined here. The choice of these two mathematicians was not my; they were mentioned in the original post of avzel on blogspot as mathematicians with whom E. Szemeredi cannot be even compared. I indeed had some candidates for being superhuman (it does not matter if Serre or Milnor were among them), and even had discussed this issue online (but this was a private conversation). My current position is that they are all humans both by the general reasons, and that this also follows form an analysis of their work, their historical context, etc., even if they did not provided as with such good clues as Serre and Milnor did. Such an analysis is very rarely done. For example, Galois work if sometimes presented as coming out of sky, but this is not true (despite an irreducible revelations is present, but it is usually present in the works of much lower level too.)
16. Another anonymous Says:
April 4, 2012 at 11:07 pm | Reply
About Serre and Milnor: if I misinterpreted you, it was because the word “approached” is ambiguous.
More substantively, my question of imparting vague knowledge to a computer or database is one on which the AI people have been stuck for 55 years. This is not a criticism of AI; the problem is worth being stuck on. And I am no luddite; were connection-spotting to be mechanized that would be a huge intellectual advance. But *why* do you think progress might be possible here when it has been elusive elsewhere?
• gowers Says:
April 5, 2012 at 11:57 am
Apologies — I didn’t mean to suggest that the misinterpretation was your fault.
As for the more substantive question, it isn’t one that I can answer briefly, partly because I don’t have a satisfactory answer at all, and partly because the answer I do have is not all that brief. But let me try to give an unsatisfactory summary of my unsatisfactory answer.
Basically, what you are asking is why I think that this formidable AI problem might be less formidable in the particular context of automatic theorem proving. Part of the reason is the general one that I think that all formidable AI problems become a little less frightening in this context, because to do mathematics we don’t have to handle large amounts of messy real-world data. How might this play out for the particular problem of spotting vague (but real) connections? Here is one idea. It is too simple to do everything — let me make that very clear before I start — but I think that it is powerful enough that it could be used to spot at least some vague connections, including ones that are of genuine mathematical interest.
I’ll illustrate the idea with two examples. The first is very simple indeed. Suppose we have just been told the definition of a group and are given as an exercise to prove the cancellation law, that $gh=g'h$ implies $g=g'$. Somehow we have to think of multiplying both sides on the right by $h^{-1}$ and then applying associativity, which is problematic (from the point of view of coming up with a general method for solving problems) because it makes the two sides of the equation more complicated before it simplifies them. And if we’re allowed to do that, then why are we not allowed to do all sorts of other things that make problems more complicated?
The reason the approach feels natural to a human is almost certainly that they have seen cancellation laws before. The very notation for groups reminds us of multiplication of real numbers (or other numbers if you prefer), so we might think, “This looks like the multiplicative cancellation law for numbers, so perhaps we can imitate the proof of that. But how can one `divide both sides by $h$‘? Ah, we could think of the identity as a bit like 1, and then the inverse is a bit like a reciprocal. So maybe multiplying both sides by $h^{-1}$ would work.” At that point, being completely unfamiliar with the group axioms, we still need to check that our approach works, but that is a much easier problem. So our spotting of an analogy has been hugely helpful.
The key to finding this analogy was to recognise the more general concept of a cancellation law and then to search one’s memory for other examples of cancellation laws. At that point, the problem is reduced to smaller questions like, “What is the analogue of division?” that can be answered more automatically. (For instance, the “reciprocal” of $h$ “ought” to be the $x$ that satisfies $xh=e$. And this turns out to exist and to be a good answer.)
This comment is getting long, so I’ll continue in a new comment.
• gowers Says:
April 5, 2012 at 1:40 pm
One might well object to the example I’ve just given by saying that it isn’t an example of a vague connection, since it is in fact completely precise. All I’ve done is take a familiar cancellation law — the multiplicative law for the non-zero reals — and generalized it to groups. I think that a lot of interesting mathematics can be done by generalizing, or by finding common generalizations, but sooner or later we will end up in situations where we exploit vague resemblances that we do not know how to turn into common generalizations. How might searching for those be automated?
Let me attempt a partial answer to that, by discussing another example. A useful lemma in additive combinatorics is the following. A dissociated set in an Abelian group is a subset $K$ such that if $x_1,\dots,x_r\in K$, then no two of the sums $\sum_i\epsilon_ix_i$, where each $\epsilon_i$ is 0 or 1, are equal. (Other definitions are possible, but this one will do for my purposes here.) Suppose now that you have a subset $A$ of an Abelian group $G$ and $A$ does not contain any dissociated set of size greater than $k$. What can be said about $A$?
A natural reaction to this question is to spot that the definition of a dissociated set resembles that of a linearly independent set in a vector space. It is therefore reasonable to make the vague conjecture that $A$ ought in some sense to be $k$-dimensional. And once one has had that thought it is natural to speculate that a maximal dissociated subset of $A$ ought in some sense to “span” $A$. And it is in fact easy to show that this is true in the following sense. If $u_1,\dots,u_k$ is a maximal dissociated subset of $A$, then every element of $A$ can be written in the form $\sum_i\eta_iu_i$, where each $\eta_i$ is 0, 1 or -1. The proof is essentially the same as the corresponding proof is for vector spaces: if $a$ is not spanned in this sense, then one checks that it can be added to the dissociated set without losing the dissociativity.
Before writing this, I realized that one can with a bit of effort come up with a slightly artificial statement that simultaneously generalizes this lemma and the statement that a maximal linearly independent subset of a vector space is also a spanning set. However, that isn’t really the point, since it is not via that common generalization that one spots the connection. So what is going on?
I suggest that when we spot connections that are not wholly precise, what we are doing is finding common generalizations that can be written in a precise language (and therefore in principle handled by a computer) even if they are not precisely interpreted or universally true. For example, the common generalization here might be something like, “If you’ve got a maximal subset for which all combinations are distinct, then that subset generates the whole set.” There is then some work to be done: what is a combination? what does “generates” mean? etc. However, that is lower-level work, and should be easier than the preliminary spotting of the connection.
So what I would suggest is that a program for spotting connections would take precise mathematical statements and rewrite them in vaguer and vaguer language. My hope would be that many statements that start out distinct would become identical after a few stages of this process, and that is how vague connections would be spotted.
The big question for me is how far a method like that can take us. I’m convinced that it can do a lot, but I do not have a good reason to think that it can do everything. Indeed, I would expect many people to react at this point by saying, “That’s all very well for those simple examples, but what about X?” where X is a more sophisticated example. I’d be very interested in examples of vague connections that do not look as though they could be spotted in this way — or rather, I’d be interested in simple examples of such connections. If they involve a lot of sophisticated mathematical machinery, then it is not clear that the difficulty is not principally in the sophistication of the machinery.
17. André Joyal Says:
April 6, 2012 at 9:54 pm | Reply
I would like to add my grain of salt to the ongoing discussion on Gowers’s idea that computers may eventually replace mathematicians in some future. I dont know what computers will be in the future and they maybe able to do miraculous things, as in science fiction. I love science fiction, but I do not want to discuss it here. I do not want to speculate about quantum computing either. I assume that future computers will remain Von Neuman machines for a long time to come. They will be faster, have a bigger memory and they will be programmed differently.
Experience shows that mathematical thinking is depending on the alternation of two kinds of activities. One is about developing a mental picture of the problem, an intuition of its nature. The other is about making a calculation, as in algebra or formal logic. The two activities are performed alternatively as in a dance. A calculation is guided by an intuition, and the intuition is confirmed or rejected by a calculation. Few mathematical progresses can be made by using intuitions or calculations alone. Computers are very good at computing and they can help us doing mathematics, like a pen and a piece of paper. But they have no intuition, no consciousness. They are mindless mechanical machines.
I claim that we do not understand the nature of consciousness. Of course, we are making progress in studying the brain. But artificial intelligence seems to show that the brain does not need to be conscious to perform its tasks. So why should the brain produces consciousness? Consciousness is the blind spot of modern science.
• gowers Says:
April 7, 2012 at 6:57 am
My view is that the brain is just as much of a “mindless mechanical machine” (the apparent contradiction in terms is deliberate), and yet it somehow produces consciousness. That makes me optimistic (at least in the long term) about what computers might be able to do.
18. Pete Says:
April 7, 2012 at 6:35 pm | Reply
Maybe one needs to be a bit more clear about what ‘long-term’ and ‘possible’ are supposed to mean in order for this discussion to get further?
I think no-one who understands basic computer science would seriously claim that it is impossible for a computer to do something arbitrarily closely resembling human thought (whether you believe that is equivalent to being human, or genuinely self-conscious, or simply clever simulation seems to me to be a religious matter, and I’m not). If nothing else, one could in principle simulate the entire workings of a brain (noting that with current technology we cannot determine these workings well enough, but there isn’t any reason why it should stay impossible). So in principle we can replace Serre with a computer simulation, and it would no doubt show incredible intuitive abilities. But in practice, we do not have anything like the hardware to do this, nor are we likely to get it any time soon (i.e. here long-term could mean centuries)
On the other hand, it seems unlikely that simulating one particular computer which happens to run a desired algorithm is the best way to run the algorithm. So one could hope to somehow abstract the algorithm and run it on something more like current technology. In principle there is no reason why this shouldn’t work, and there is probably no reason why we should not be able to do it in a year’s time: it’s likely the solution will be simple (as compared to say the Windows code-base). But it’s probably also true that solutions to most major mathematical problems are similarly simple. I think it’s fair to compare the situation to the P=NP conjecture: nothing we have tried comes close, the only results we really have are of the form ‘this approach cannot work’. We still might solve it next year, but we probably won’t.
As some kind of middle ground, it’s possible we could develop a program which produces the kind of results we want but without our understanding how it works; evolutionary algorithms aren’t new, some of them really solve things in unexpected ways (I remember a decade or so back there was an FPGA circuit which solved a problem in a way which the researchers didn’t understand until they tried the same circuit with a different component layout: this didn’t work, and eventually it was realised that two neighbouring but unconnected components on the FPGA had a capacitance which was critical to the function but wasn’t ‘supposed’ to exist). Since these things are basically massively parallel maybe one could run an evolutionary algorithm in the style of SETI@home and get something useful. But it’s neither clear how one would design a scoring function for the algorithms being evolved, nor how long it would take.
19. Hans Says:
April 7, 2012 at 6:36 pm | Reply
To sowa:
If computers would be able to solve important mathematical problems and develop important mathematical structures/tools, then this wouldn’t prohibit you from doing mathematics nor would it destroy mathematics as an art. Quite the opposite, it would just put mathematics on the same scale with other arts – you have to do it out of your free time, and you only get payed when you do something very beautiful. Anyone who really loves mathematics, would still do mathematics. There would be even extra joy coming from the facts that 1) you have a mighty collaborator 2) much more (also beautiful) mathematics gets done 3) mathematics is potentially much more widely used.
So in fact, your accusations towards prof. Gowers of not being a real mathematician could using a Tolstoian argument be very easily turned against yourself.
• sowa Says:
April 8, 2012 at 9:27 am
To Hans:
First of all, since than claiming that somebody is not a mathematician, but “only” a scientist is an “accusation”? Is it now a sort of crime not to be a mathematician?
Pure mathematics exists as other arts anyhow. You are paid only when you did something other people like a lot. If you do not manage to produce something like this in your early years, you are out of the profession. The experience shows that it is nowadays impossible to do pure mathematics in your free time, as it was possible, say, to Fermat. Other jobs are too demanding, and mathematics requires prolonged concentration, which is not compatible with regular jobs.
In fact, most of traditional arts had disappeared during the last century. The modern paintings are essentially a kind of financial instruments. Still, a lot of people may appreciate visual arts and may pay for them for the pleasure and not as an investment. The situation in sports is similar. A lot of people appreciate chess even if they only know the rules. There is quite a lot or rich people among of them, and they played an essential role in supporting chess.
The situation with mathematics differs dramatically. The good current mathematics can be appreciated only by other professional mathematicians. You can do something incredibly beautiful, but it will be understood and appreciated first by a half-dozen of your closest colleagues. Even later, when a result finds its way in research monographs and then advanced level textbooks, the appreciation remains inside of the mathematical community. So, there is no outside source of support for this kind of beauty, and the mathematical community relays on at least potential usefulness of its production, as opposed to its artistic value.
Next, about the joy. There will be a joy of doing something better than a computer did. This kind of you exists without computers also; one may prove a theorem better than it was done the first time. But this joy is not comparable with the joy of doing something first time. What to do? Keep your ideas in secret from assistants to theorem-proving machines? They will spoil the joy anyhow, reproducing your result within a day.
I fail to see how a computer can be a mighty (or not) collaborator in the sort of mathematics I like. Computer may help to prove some theorem, like to (in)famous 4 colors conjecture, but there is nothing beautiful there.
I also fail to see why we (the human race) may need more mathematics. We already have too much. And why this computer mathematics will be beautiful. All beauty I ever seen in the output of computer was injected into it by humans before the computer started working.
The idea that pure mathematics is useful and is actually used is a fortunate (for mathematicians) misconception. The heart of pure mathematics, the proofs, is not needed for any applications. A heuristic argument together with an experimental verification is always sufficient.
20. André Joyal Says:
April 7, 2012 at 8:09 pm | Reply
I thank you Tim Gowers for expressing your position. I agree that the humain brain can be “mindless” in a rethorical sense (I am often mindless myself). You wrote that the brain can “somehow produces consciousness”. I would like to understand what you mean. Let me propose a thought experiment. One can imagine a universe parallel to ours, ruled by the same physical laws, with a planet supporting life like ours and everything exactly the same except for one thing: all animals and humans on this planet would be totally unconscious. This bizarre hypthesis is not absurd, since artificial intelligence is teaching us that consciousness is not essential to intelligence. Also, the hypothesis does not seem to contradict the laws of physics that I know. It follows from this thought experiment, that the presence of conciousness in our universe cannot be deduced from the law of physics as we know them. At this point, there are many possible positions. One would say that the idea of consciousness is an illusion of language. This is tantamout to saying that we are zombies. Another would say (as I do) that consciousness is real but we dont know what it is.
What do you mean when you write that the brain “somehow produces consciousness”?
• gowers Says:
April 7, 2012 at 10:27 pm
The zombie argument is a well-known argument in the philosophy of mind, and so is any response I am likely to think of. Like many philosophers (Daniel Dennett being a well-known example), I take the view that if a quasi-human on another planet had a brain that worked according to the same physical laws (complete with neurons firing etc.) then it would be conscious. In other words, I believe that consciousness is an emergent phenomenon and not some mysterious non-physical thing that can be added or subtracted.
In case I’m misunderstood, I don’t think consciousness is a black and white phenomenon. So if you had a sequence of ever more sophisticated computer programs, starting with today’s programs and ending with something that had software more or less identical to our brain’s software, I’d say that these programs would start out with virtually nothing that deserved to be called consciousness and would end up as conscious as we are. In between, they would have something intermediate.
• Nyaya Says:
April 8, 2012 at 1:26 pm
I thought the following are far from being settled:
1) The brain is indeed a computational device
2) The hard problem of consciousness is no more a problem and as Dennett says there is no such thing as Qualia.
Also, if we were to simulate ‘doing mathematics’, would it be like simulating the weather or like simulating addition?
• gowers Says:
April 8, 2012 at 4:16 pm
@Nyaya It is true that there are many who do not accept 1) or 2). However, there is a strong case for saying that 1) is true in a trivial sense if you believe in current physics — the brain is a physical system and physical systems can in principle be simulated computationally. The question then becomes whether it is a computational device in a less trivial sense: roughly, that we can hope to simulate it well enough on a computer to reproduce its outward behaviour — to which some, including me, would add the requirement that the software should be basically the same. (That is, I wouldn’t be happy with brute-force methods that just happened to give the same output, not that I believe that is remotely practical.)
As for 2), my view is that very strong arguments have been put forward against qualia by Dennett and others, and nothing I have ever read has come close to countering them. So I agree that it is not settled, but it seems to me that it ought to have been settled. (I feel the same way about climate change, though my respect for people who believe in qualia is infinitely greater than my respect for people who don’t believe in man-made global warming.)
21. André Joyal Says:
April 8, 2012 at 3:09 am | Reply
I thank you Tim Gowers for your reply. I am not a professionnal philosopher and I was not aware that my argument is standard. Anyway, it is a very natural argument and I am glad you have given an answer. I agree that consciousness is not a black and white phenomenon and that it is in some sense emerging. But I find the notion of emergence too vague and universal to be the basis of a real scientific explanation of the phenomenon. In any cases, if we succeed in constructing robots that behave intelligently, they better be unconscious. Because they will likely be used as servants, slaves, guards, soldiers and what? They will be reponsible of all the bad work. I would hate to know that my computer is suffering because it is computing for me days and nights.
• sowa Says:
April 8, 2012 at 9:45 am
To André Joyal:
Being conscious and being able to suffer (or to love, which are two sides of the same coin) seem to be completely independent phenomena.
In general, it is fairly amusing to learn that modern thinkers had only repackaged some century-old ideas of a well known political leader, V.I. Lenin. The metaphor of zombies may be new, but the idea of consciousness as an emergent phenomenon is worked out in his writings; it did not appeared to be so natural at the time.
• André Joyal Says:
April 8, 2012 at 9:59 pm
I thank you Sowa for the reference to Lenin and for expressin your view. You wrote that “Being conscious and being able to suffer seem to be completely independent phenomena”. Your are surely not saying that pain can exist without the consciousness of that pain? Are you saying that a conscious person may be devoided of emotion? A psychologist would probably diagnose this person as a psychopath. Of course, we may imagine that this person is a nice guy like Spock in Star Trek, but this is pure fiction. In your reply to Hans, you wrote that pure mathematics may be regarded as a form of art. I completely agree with you. Good mathematics allways carries an aesthetic emotion. Mathematicians are not purely rational minds. They are more like artists exploring the beauty of pure reason. It seems foolish to think that mathematical beauty can be fully rationalised. This is why we may never be able to replace mathematicians by intelligent machines. Why should we try?
• sowa Says:
April 8, 2012 at 10:57 pm
To André Joyal:
To be more precise, I am quite sure that one can conscious person devoid of emotions. Yes, such persons are usually classified as psychopaths (if they are not smart enough to hide this quality). By some estimates, about 5% of the population is such (I have no idea how accurate is this estimate, but at least it agrees with my own experience). But they are considered as humans, of course.
In the other direction the question is more subtle. Is it possible to suffer without being conscious? I believe that this is possible, at least to some extent. Let us look, for example, at sufficiently distant (in the biological sense) from us animals. It seems obvious that they can suffer, can be attracted to their mate or to a human, etc. But it doesn’t seem to be clear that they are conscious and if they are, to what degree. With the “emerging phenomenon” theory, if one wants a coherent point of view, one has to accept that *everything* is conscious, only to different degrees. Even stones, an even an electron should have a rudiment of consciousness. This is one of the issues Lenin recognized and dealt with. If we take such a position, the question disappears.
If one takes some other position, then it is natural to think that consciousness is needed for experiencing emotions. This only shifts the question. What is consciousness? I must admit that I don’t know what exactly the (post)modern philosophers understand by the consciousness. But nowhere had I seen a serious discussion of the following issue: is consciousness is just a receiver, a passive entity getting information from outside of it? Or it is also a transmitter, or, to put it better, is it active and creative? Personally, I believe that one cannot suffer without a receiver, but one can without a transmitter.
Starting with the words “In your reply to Hans”, I agree with every word you said, and don’t see any need for any qualifications or clarification. Here we are in complete agreement, including the last phrase “Why should we try?”
22. André Joyal Says:
April 9, 2012 at 2:00 am | Reply
@Sowa: In your reply to Hans, you wrote: “I also fail to see why we (the human race) may need more mathematics. We already have too much.”
I dont share your pessimism. Possibly because I am optimistic by nature,
It is true that mathematics is now too vast for one human been to know it all. It is expanding at an exponential rate (I would like to know the rate). More mathematics is produced every year (maybe every week) than what I can learn in my lifetime. The quality of the average mathematical paper seems to be going down. The number of mathematicians may double during the next 25 years, largely due to the contributions of developing countries like India and China (25 years is a rough estimate) . These developements will affect the mathematical culture (they are surely influencing it already). Mathematical knowledge appears increasely fragmented the barriers between the fields higher.
The traditional culture inherited from the age of enlightenment is under enormous stress, it maybe gone already. But something new may emerge from the ashes of the old culture. Can we figure it out? What we should do?
@Gowers: I am very interested in knowing your opinion on this. But my question could be out of the context you have created for this discussion. Please, let me know.
23. sowa Says:
April 9, 2012 at 3:16 am | Reply
To André Joyal:
Well, you quite nicely detailed what I meant: why there is “too much” mathematics.
I may add that the quality of mathematical papers is going down independently of developing countries. Take any top journal, like “Annals of Mathematics” or any other. During the last 20 years, the number of pages per year in “Annals” increased 3-fold, I think. There is no noticeable presence in “Annals” of mathematicians from China or India (I mean working there, not ethnicity). Most of the authors work in US, UK, rarely in France, and sometimes in other European countries. The number of positions did not increased, it is actually decreased. Inevitably, the level of “Annals” went down quite noticeably.
I don’t think that the quoted remark qualifies me as a pessimist (may be I am, but not because of this opinion). My point is there is no inherent good in producing more mathematics. It is not needed for applications (I already said this here: proofs are not needed, even in physics). Given the situation you outlined, more mathematics will serve no good for any individual mathematician. Today there is no way to learn in 100 years any sizable fraction of the already existing mathematics I would like to know (in particular, know about).
Craig Smorynski suggested about 25-30 years ago that mathematicians need to slow down the production new results for a while, and to put in order the things presumably done already. There are huge gaps in the literature, and many papers are hardly understandable. There are many examples from many branches of mathematics. The most distressing is the fact the some results or proofs are apparently lost despite their discoverers are still alive and well. They are just not interested in their old (and, occasionally, even new) results anymore.
I would like to stress that I am not speaking about the production of an average mathematician, I speak about superhuman insights of some of our contemporaries. Also, let me repeat, I am interested in the beauty of these insights and do not care much if a particular statement is true or not (probably, the generalized Riemann hypothesis is an exception, may be the only one).
• André Joyal Says:
April 9, 2012 at 4:34 am
@Sowa: You wrote that “Smorynski suggested about 25-30 years ago that mathematicians need to slow down the production new results for a while, and to put in order the things presumably done already” I feel exactly the same. Not that we should entirely stop producing new results, but that we should spend a lot of time reorganising what we already know, simplifying it, explaining it to others, learning other fields, reading old papers, writing introductory books, mastering the applications, etc. I am tempted call it SLOW MATHEMATICS, the opposite of FAST MATHEMATICS (like SLOW FOOD is the opposite of FAST FOOD). A fast mathematicians must write his papers quickly because he is in a rat race. His goal is to publish as much papers in a year as he can. His position and career are depending on that. Many mathematicians I know would love to slow down but they can’t. The value of slow mathematics is hardly recognized. We may be approaching the point where fast mathematics will destroy mathematics by turning it into a meaningless game, a pure rat race. Could slow mathematics save mathematics and mathematicians?
• sowa Says:
April 9, 2012 at 5:29 am
@André Joyal: I cannot agree more. I appreciate a lot when mathematicians do these slow things, and try to such things when possible. I like a lot learning other fields, and very happy when somebody writes an introduction aimed to mathematicians (and advanced graduate students). I found reading old papers be extremely illuminating even if the material is already in textbooks. I even had a couple of projects for introductory books (based on my graduate courses), but during last few months I realized that the mathematical books publishing may disappear sooner than I finish any of my projects.
I do not really understand why do we have this rat race. It seems that it is a fairly recent phenomenon. Personally, not very long ago I had the luxury to develop a little theory in the course of seven years after publishing only an announcement, devoted to this theory only partially. (I did publish papers about other things.) The final result is a very short monograph; a rat race would force me to publish it as nearly 10 papers, which would be overlapping and interconnected in a complicated way.
Could it be the case that we do compete for a smaller and smaller number of positions in pure mathematics? But what about people with tenure? What forces them to continue this race?
• André Joyal Says:
April 9, 2012 at 9:43 pm
Sowa wrote: “Could it be the case that we do compete for a smaller and smaller number of positions in pure mathematics?”
Probably so.
Higher education has expanded a lot during the second part of last century, starting after WWII. But the expansion seems to have slowed down recently. Let me discuss the general context. It is quite clear that we are now living in an era of triomphant capitalism, despite the lasting recession created by the financial system. I would like to give you a small example. In Canada (where I live) the grant agency supporting mathematical research is the NSERC (the Natural Sciences and Engineering Research Council). The freshly reelected Harper government (conservative) stated in his last budget, that “from now the NSERC will concentrate its energy to serve exclusively the priorities of the enterprises”. Wow!
Happily, the popularity of the Harper goverment is rapidly decreasing.
I would like to make it clear that I am not against capitalism. Surely, capitalism can be good, since it encourages initiatives and inovations via competion. But unreined capitalism is dangerous, it may destroy everything including itself. This is why democratic societies must impose stringent rules on corporations (like anti-monopolistic laws).
You wrote: “But what about people with tenure? What forces them to continue this race?”
I guess by self interest. Again, I would like to make it clear that self interest and competion can be good for academia. The problem here is that the rules governing academic research have been fixed a long time ago according to a pattern which is now partly outdated. The explosion of mathematical research does not translate into a broadening of the mathematical culture, except possibly for a very small number of peoples, if any. I fear that mathematics may eventually collapse on itself if it does not broaden its intellectual base. I feel strange when I meet someone who knows everything about nothing and nothing about everything. Hopefully, the danger will be recognised on time and the rules will be changed. Peoples contributing to the developpement of general mathematical culture should be better rewarded by the system.
Some efforts have been made in the past to unify and broaden the mathematical culture of the time. The Bourbaki group is famous. Despite their rigor, the mathematics of Bourbaki are poorly motivated and the applications are absent from the books. The Soviet Encyclopedia of Mathematics edited by Vinogradov is probably a better tentative, but I know it less. I guess that it has contributed to the dominance of russian mathematics during the last 20 years. The “Princeton Companion to Mathematics” edited by Tim Gowers is the latest example I know. Thank you Tim, your book is beautifully! I hope that Tim will not mind if I criticise his book a bit. I believe the book should have paid more attention to category theory, since it is possibly the most important tool for unifying mathematics.
Modestly, I would like to formulate a dream that many mathematicians have today. A new collective effort to present and unify mathematics should be undertaken. It should dawrf all previous efforts and it should be open ended. It should use the internet.
I dont know how such a collective effort may start. I would love to contribute to it with my modest means.
• sowa Says:
April 10, 2012 at 2:02 am
@André Joyal:
Instead of venturing into the political philosophy of capitalism, I would prefer to stay closer to mathematics. Things similar to what you wrote about NSERC do happen in its US equivalent, NSF, and even on the level of the Division of Mathematical Science. But I don’t see how this may be related to capitalism, triumphant or not, or to what party is in the power. Both agencies are purely socialist institutions and function in a way characteristic for socialist institutions. For example, exactly the same words as you quoted could be said by some USSR Communist Party apparatchik overseeing sciences, and, I believe, they were said many times.
While young non-tenured people may be forced to enter this rat race by external to mathematics circumstances, like the shortage of the new positions, the rat race of tenured people is a phenomenon internal to mathematics and we have nobody to blame except ourselves. Who is rewarded is determined by us by a very trivial reason: no administrator or an NSF officer can distinguish good mathematics from bad and even an expository work from purely original research (a good expository work requires a lot of creativity, in fact). Well, perhaps there are or at least were some exceptions in NSF, usually some former mathematicians (but usually even former mathematicians are not exceptions). Anyhow, any grant award, any promotion decision, and any salary rise eventually depends on the peer review. I think that there is no need to change the rules. Instead of this, we should change our own priorities, what we do recognized and what we do not.
My defense of Bourbaki turned out to be fairly long, and I posted it at another place. Also, I would like to defer my critisism of Encyclopedias and of the “Princeton Companion to Mathematics” for another occasion.
I do not have any big expectations for big collective online projects. To date, Bourbaki is the only example of a successful expository collaboration of several people. The Bourbaki enterprise is not reproducible, at least without some dramatic changes in the attitudes of the mathematical community, and even with an appropriate change, only very rarely.
What needs to be done collectively is exactly the change of attitude. We should appreciate the expository work much more than now, write enthusiastically about such work in letters of reference, say to the authors more often that they are doing very valuable work, etc. Then much more individuals or two-three authors together will start to write expositions, and, may be some bigger online projects will eventually mature.
As of your idea of dwarfing all previous efforts, I must say that I subscribe to Freeman Dyson’s maxim: “Small is beautiful”.
• André Joyal Says:
April 10, 2012 at 3:55 pm
@Sowa. I agree with what you said. I have streched the connection between academic life and globalisation too far, at least for the sake of the present discussion. I am beginning to worry that we may be abusing the hospitality of our host. We could discuss elsewhere. You are welcome to to contact me. You know my real name.
• sowa Says:
April 10, 2012 at 9:18 pm
@André Joyal:
Yes, I already suspected that we are abusing the hospitality of our host. We may move to one of my Google places:
http://owl-sowa.blogspot.com .
The top post is for your comments. I suspect that you missed the reference to my previous post there with my comments about Bourbaki. Here the links are not very visible unless given in the http:// form.
I am planning to use that blog for a discussion of other issues related to this post of T. Gowers.
My experience showes that a discussion in comment is more convenient than an exchange of e-mails, since the whole thread is at one place. I plan to write you later (right now I have to go, anyhow). I am not announcing my identity here by completely unrelated with the discussion issues. I am prepared to stay by my words under my real name.
24. John Says:
April 13, 2012 at 5:30 pm | Reply
Tim
Thank you very much for your written talk – I enjoyed a lot.
25. André Joyal Says:
April 16, 2012 at 5:22 pm | Reply
I would like to correct a factual error I made in my reply to Sowa April 9, 2012. I wrote that the Harper government has decided that This was wrong! The
government actually decided that I have confused the NSERC with the NRC!
http://www.nrc-cnrc.gc.ca/eng/index.html
http://www.nserc-crsng.gc.ca/index_eng.asp
I apologise. The former is supporting pure mathematics, not the latter. The new policy of the government does not seem to be as bad as I thought. There is something to worry however, since the NRC is supporting a wide variety of scientific researchs, including research in biology and environment.
26. Matthew Kennedy awarded CMS 2012 Doctoral Prize « Noncommutative Analysis Says:
December 13, 2012 at 10:34 am | Reply
[...] and Terry Tao have set a fine example in their expositions of the works of Fields Medalists or Abel Prize laureates. These are among the most interesting and important posts out there, I [...]
27. vznvzn Says:
April 2, 2013 at 5:11 pm | Reply
hi all, found this post belatedly. sowa pointed to the dialog on computer-based mathematics. there are many related areas here including computer assisted vs automatic theorem proving, the role of human vs computer in mathematics, etcetera. have been studying this subject for many years. a few thoughts.
it seems to me there is a lot of debate around here about competitive approaches eg “the two cultures” and “mathematicians vs computers” which at extremes border on adversarial. this shows up a lot in sowa’s writing. lets all take a deep breath. think about symbiosis & cooperation. its the natural/higher/global order between the society of mathematicians and between them and computers. its a feedback loop.
if you read gowers original paper on “rough structure and classification”, the interesting dialog with the computer is particularly *collaborative*. it is sowa that is emphasizing that computers could make human mathematicians obsolete and putting words into the mouth of gowers [this pattern continues on newer blogs on this site].
computers will never replace mathematics just as the field of chess has been *strengthened* with the advance of computer power.
it appears to me there will always be a tension between very difficult proofs that are *objectively* “true” but inscrutable and it will always take humans to rearrange and reorient the same proof in different ways, also called psychological “chunking”, that *subjectively* are more understandable and yes, *aesthetic*.
in this way there is a strong similarity to architecture and refactoring code from the field of software development. think this core analogy [between proofs and coding] will continue to become more prominent and strengthened in the future even with major advances in automated thm proving.
over the years Ive been looking at a particular model for human vs computer theorem proving. its the collatz conjecture. in some ways a toy problem, but in other ways, possibly the precursor to a new style of computer-assisted mathematics in a remarkable style.
along these lines: have some preliminary, promising results on the collatz conjecture related to computational analysis & am looking for volunteer(s) for a project that would bring about this new order into reality. needed: mathematical background, programming ability, and enthusiasm for pushing the [extreme?] boundaries of mathematical and scientific knowledge. the idea is to apply very deep new technical principles to a “toy” problem but only as a start, a prototype or “proof of principle” on the way to grander plateaus…
plz reply on my blog if interested!
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http://math.stackexchange.com/questions/177029/on-the-relationship-between-fermats-last-theorem-and-elliptic-curves/177044 | # On the relationship between Fermats Last Theorem and Elliptic Curves
I have to give a presentation on elliptic curves in general. It does not have to be very in depth. I have a very basic understanding of elliptic curves (The most I understand is the concept of ranks). I was wondering if anyone could explain to me simply what is the connection between the equation $x^n +y^n=z^n$ and elliptic curves.
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2
– William Jul 31 '12 at 3:31
2
I may be wrong, but I do not think there is a simple explanation. – Alex Becker Jul 31 '12 at 3:33
## 2 Answers
Given non-zero integers $A,$ $B$, and $C$, such that $A + B = C$, we can form the so-called Frey curve (named after the mathematician Frey, who first considered elliptic curves in the context of FLT)
$$E: y^2 = x(x-A)(x+B),$$
which has discriminant (up to some power of $2$ which one can compute precisely, but which I will ignore here) equal to $ABC$.
Suppose now that $A = a^p$, $B = b^p$, and $C = c^p$ (so that we have a solution to the Fermat equation of exponent $p$). Then the elliptic curve $E$ has a discriminant which (up to the power of $2$ that I am ignoring) is a perfect $p$th power.
This means that the group of $p$-torsion points $E[p]$ on $E$ (which is a two-dimensional vector space over the field $\mathbb F_p$ of $p$-elements, equipped with an action of the Galois group of $\overline{\mathbb Q}$ over $\mathbb Q$) has very special properties --- in algebraic number theory terms, it is very close to being unramified. (More specifically, but more technically, it is unramified except possibly at $2$ and $p$, and at $p$ the ramification is very mild --- it is finite flat.)
Now the Shimura--Taniyama conjecture, which is what Wiles (together with Taylor) proved, shows that $E$, and so $E[p]$, arises from a weight two modular form. Ribet's earlier results on Serre's epsilon conjecture imply that this modular form must actually be of level $2$. (This is where we use the above information about the ramification.) But there are no non-zero cuspforms of weight $2$ and level $2$, and we get a contradiction.
Although it is much harder (in that the only way we know to rule out the existence of $E[p]$ is by the --- quite difficult --- Shimura--Taniyama conjecture, or else by related more recent results such as Khare and Wintenberger's work on Serre's conjecture), one can think of the non-existence of $E[p]$ as being analogous to Minkowski's theorem in algebraic number theory, which says that an everywhere-unramified extension of $\mathbb Q$ cannot exist.
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There are certainly people on this site that are much more capable than I am of explaining the connection competently, so I'll stick to providing a reference. The last chapter of Diamond-Shurman's A First Course in Modular Forms gives an overview of how this story goes, at the level of a graduate-level textbook (of course, given the scope of the result, they have to black-box many, many things). They explain how to construct the Frey curve, which is an elliptic curve associated to a hypothetical solution to Fermat's equation, and they show how the theory of modular forms (in particular the Taylor-Wiles result that all elliptic curves are modular in the sense that they have an "associated" modular form) allows one to show that no such Frey curves can exist (very briefly, if a Frey curve existed, its associated modular form would have to live in a particular space of forms which one can show directly to be empty).
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Connective Real $$K$$-Theory of Finite Groups
Robert R. Bruner, Wayne State University, Detroit, MI, and J. P. C. Greenlees, University of Sheffield, UK
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Mathematical Surveys and Monographs
2010; 318 pp; hardcover
Volume: 169
ISBN-10: 0-8218-5189-6
ISBN-13: 978-0-8218-5189-0
List Price: US\$92
Member Price: US\$73.60
Order Code: SURV/169
See also:
The Connective K-Theory of Finite Groups - R R Bruner and J P C Greenlees
Parametrized Homotopy Theory - J P May and J Sigurdsson
This book is about equivariant real and complex topological $$K$$-theory for finite groups. Its main focus is on the study of real connective $$K$$-theory including $$ko^*(BG)$$ as a ring and $$ko_*(BG)$$ as a module over it. In the course of their study the authors define equivariant versions of connective $$KO$$-theory and connective $$K$$-theory with reality, in the sense of Atiyah, which give well-behaved, Noetherian, uncompleted versions of the theory. They prove local cohomology and completion theorems for these theories, giving a means of calculation as well as establishing their formal credentials. In passing from the complex to the real theories in the connective case, the authors describe the known failure of descent and explain how the $$\eta$$-Bockstein spectral sequence provides an effective substitute.
This formal framework allows the authors to give a systematic calculation scheme to quantify the expectation that $$ko^*(BG)$$ should be a mixture of representation theory and group cohomology. It is characteristic that this starts with $$ku^*(BG)$$ and then uses the local cohomology theorem and the Bockstein spectral sequence to calculate $$ku_*(BG)$$, $$ko^*(BG)$$, and $$ko_*(BG)$$. To give the skeleton of the answer, the authors provide a theory of $$ko$$-characteristic classes for representations, with the Pontrjagin classes of quaternionic representations being the most important.
Building on the general results, and their previous calculations, the authors spend the bulk of the book giving a large number of detailed calculations for specific groups (cyclic, quaternion, dihedral, $$A_4$$, and elementary abelian 2-groups). The calculations illustrate the richness of the theory and suggest many further lines of investigation. They have been applied in the verification of the Gromov-Lawson-Rosenberg conjecture for several new classes of finite groups.
Readership
Graduate students and research mathematicians interested in connective $$K$$-theory.
Reviews
"The book is very carefully written, including many diagrams and tables, and also a thorough review of the authors' previous work on the complex case."
-- Donald M. Davis, Mathematical Reviews
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http://mathoverflow.net/questions/67509/extending-a-bi-invariant-metric-from-a-set-of-generators-to-the-whole-group | ## Extending a bi-invariant metric from a set of generators to the whole group.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $S$ be a symmetric set of generators of the (finite) group $G$. Having a bi-invariant metric $d$ on $S$ (meaning that whenever $s,t,g,gs,gt,sg,tg\in S$, then $d(s,t)=d(sg,tg)=d(gs,gt)$), is it always possible to extend $d$ to a bi-invariant metric on $G$?
Update: It has been shown below by Lucasz that the answer is negative. Let me try to add a condition on $S$ which hopefully makes the question more interesting. Suppose that $S$ verifies the following condition: whenever there are $(s_1,t_1),(s_2,t_2)\in S\times S$ with $s_1\neq t_1$ and $g\in G$ such that either $gs_1=s_2, gt_1=t_2$ or $s_1g=s_2,t_1g=t_2$, then $g=1$.
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1
The new condition implies that S consists of just one element: if there are two different elements a and b in S then take pairs (a,a), (b,b) and g=ba^{-1}. Together with the assumption that S is symmetric it actually implies that G is the cyclic group of order 2. Maybe assuming $s_1 \neq t_1$ will make it more interesting. – Łukasz Grabowski Jun 11 2011 at 18:33
yes, of course. Thanks! – Valerio Capraro Jun 12 2011 at 10:30
However, with the new condition the original condition is empty, i.e. there are no elements $s,t,g$, $g\neq1$, such that $s,t, gs, gt \in S$. Therefore any metric on $S$ is bi-invariant - you sure you want it this way? Then the question could be more clearly phrased as: given a finite group $G$ and a set of generators $S$ fulfilling ..., is it true that any metric on $S$ can be extended to a bi-invariant metric on $G$? – Łukasz Grabowski Jun 12 2011 at 11:03
I'm doing a mess! Here is more or less my situation. Let $U(r)$ be the unitary group of rank $r$. I have a symmetric family $g_1,...g_n\in U(r)$ and I want to embed it (multiplicatively) into a finite group equipped with a bi-invariant metric which agrees with the metric induced by the 2-norm in $U(r)$. The idea would be easy: $g_1,...g_n$ generates a residually finite group and then they are multiplicatively embeddable into a finite group $G$. Inside $G$ they anyway carry the (pushforward of the) metric induced by the 2-norm and then I was trying to extend this metric to the whole $G$... – Valerio Capraro Jun 12 2011 at 15:34
I can't see a way to control a priori that $S$ is fulfilling for the $G$ constructed in this way.. – Valerio Capraro Jun 12 2011 at 15:40
show 2 more comments
## 1 Answer
No, it's not always possible. Take $G$ to be the cyclic group of order $20$, let $g$ be its generator. Let $S =$ { $g,g^3,g^{17},g^{19}$ }. Define $d$ on $S$ by putting $d(g,g^3)=2$ and all the other distances equal to $1$.
Now, the pair $(g,g^3)$ is not in the same "$(S\times S)$-orbit" as $(g^{17},g^{19})$, and therefore the above defines a bi-invariant metric on $S$, but it is in the same $G\times G$-orbit, so it cannot be extended to a bi-invariant metric on $S$.
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### Iterative Function Mapping [duplicate]
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### Using pure functions in Table
I need a table with the elements made of pure functions and list elements. This is a simplified example: I need a list as: ... | {"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": 18, "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": 0.9044606685638428, "perplexity_flag": "middle"} |
http://nrich.maths.org/7547 | ### More Mods
What is the units digit for the number 123^(456) ?
### Days and Dates
Investigate how you can work out what day of the week your birthday will be on next year, and the year after...
### Mod 3
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
# Filling the Gaps
##### Stage: 4 Challenge Level:
Charlie has been thinking about which numbers can be written as a sum of two square numbers. He took a $10\times10$ grid, and shaded the square numbers in blue and the sums of two squares in yellow.
He hoped to find a pattern, but couldn't see anything obvious.
Vicky suggested changing the number of columns in the grid, so they reduced it by one:
"There seems to be a diagonal pattern."
"If the rows were one shorter, then those diagonals would line up into vertical columns, wouldn't they?"
"Let's try it..."
What do you notice about the positions of the square numbers?
What do you notice about the positions of the sums of two square numbers?
Can you make any conjectures about the columns in which squares, and sums of two squares, would appear if the grid continued beyond 96?
Can you prove any of your conjectures?
You might like to look back at the nine-column grid and ask yourself the same questions.
Charlie couldn't write every number as a sum of two squares. He wondered what would happen if he allowed himself three squares.
Will any of the numbers in the seventh column be a sum of three squares?
Can you prove it?
"We must be able to write every number if we are allowed to include sums of four squares!"
"Yes, but it's not easy to prove. Several great mathematicians worked on it over a long period before Lagrange gave the first proof in 1770."
With thanks to Vicky Neale who created this task in collaboration with NRICH.
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": 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": 0.9454561471939087, "perplexity_flag": "middle"} |
http://unapologetic.wordpress.com/2010/02/23/proving-the-classification-theorem-ii/ | # The Unapologetic Mathematician
## Proving the Classification Theorem II
We continue with the proof of the classification theorem that we started yesterday.
1. No more than three edges (counting multiplicities) can be incident on any given vertex of $\Gamma$.
2. Consider some vector $\epsilon\in\mathfrak{D}$, and let $\eta_1,\dots,\eta_k$ be the vectors whose vertices are connected to $\epsilon$ by at least one edge. Since by step 3 the graph $\Gamma$ cannot contain any cycles, we cannot connect any $\eta_i$ and $\eta_j$, and so we have $\langle\eta_i,\eta_j\rangle=0$ for $i\neq j$.
Since the collection $\{\epsilon,\eta_1,\dots,\eta_k\}$ is linearly independent, there must be some unit vector $\eta_0$ in the span of these vectors which is perpendicular to each of the $\eta_i$. This vector must satisfy $\langle\epsilon,\eta_0\rangle\neq0$. Then the collection $\{\eta_0,\eta_1,\dots,\eta_k\}$ is an orthonormal basis of this subspace, and we can thus write
$\displaystyle\epsilon=\sum\limits_{i=0}^k\langle\epsilon,\eta_i\rangle\eta_i$
and thus
$\displaystyle\begin{aligned}1=\langle\epsilon,\epsilon\rangle&=\left\langle\sum\limits_{i=0}^k\langle\epsilon,\eta_i\rangle\eta_i,\sum\limits_{j=0}^k\langle\epsilon,\eta_j\rangle\eta_j\right\rangle\\&=\sum\limits_{i=0}^k\sum\limits_{j=0}^k\langle\epsilon,\eta_i\rangle\langle\epsilon,\eta_j\rangle\langle\eta_i,\eta_j\rangle\\&=\sum\limits_{i=0}^k\sum\limits_{j=0}^k\langle\epsilon,\eta_i\rangle\langle\epsilon,\eta_j\rangle\delta_{i,j}\\&=\sum\limits_{i=0}^k\langle\epsilon,\eta_i\rangle^2\end{aligned}$
Now since we can’t have $\langle\epsilon,\eta_0\rangle=0$, we must find
$\displaystyle\sum\limits_{i=1}^k\langle\epsilon,\eta_i\rangle^2<1$
and thus
$\displaystyle\sum\limits_{i=1}^k4\langle\epsilon,\eta_i\rangle^2<4$
But $4\langle\epsilon,\eta_i\rangle^2$ is the number of edges (counting multiplicities) between $\epsilon$ and $\eta_i$, and so this sum is the total number of edges incident on $\epsilon$.
3. The only connected Coxeter graph $\Gamma$ containing a triple edge is $G_2$
4. Indeed, step 4 tells us that no vertex can support more than three incident edges, counting multiplicities. Once we put the triple edge between two vertices, neither of them has any more room for another edge to connect to any other vertices. Thus the connected component cannot grow larger than this.
### Like this:
Posted by John Armstrong | Geometry, Root Systems
## 3 Comments »
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2. [...] Theorem IV We continue proving the classification theorem. The first three parts are here, here, and [...]
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3. [...] we conclude the proof of the classification theorem. The first four parts of the proof are here, here, here, and [...]
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 25, "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": 0.9296424984931946, "perplexity_flag": "head"} |
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# Introduction to turbulence/Statistical analysis/Estimation from a finite number of realizations
### From CFD-Wiki
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| | == Estimators for averaged quantities == | | == Estimators for averaged quantities == |
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| - | Thus ''the variability of the stimator depends inversely on the number of independent realizations, <math>N</math>, and linearly on the relative fluctuation level of the random variable itself <math>\sigma_{x}/ X</math>''. Obviously if the relative fluctuation level is zero (either because there the quantity being measured is constant and there are no measurement errors), then a single measurement will suffice. On the other hand, as soon as there is any fluctuation in the <math>x</math> itself, the greater the fluctuation ( relative to the mean of <math>x</math>, <math>\left\langle x \right\rangle = X</math> ), then the more independent samples it will take to achieve a specified accuracy. | + | Thus ''the variability of the estimator depends inversely on the number of independent realizations, <math>N</math>, and linearly on the relative fluctuation level of the random variable itself <math>\sigma_{x}/ X</math>''. Obviously if the relative fluctuation level is zero (either because there the quantity being measured is constant and there are no measurement errors), then a single measurement will suffice. On the other hand, as soon as there is any fluctuation in the <math>x</math> itself, the greater the fluctuation ( relative to the mean of <math>x</math>, <math>\left\langle x \right\rangle = X</math> ), then the more independent samples it will take to achieve a specified accuracy. |
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| | '''Example:''' In a given ensemble the relative fluctuation level is 12% (i.e. <math>\sigma_{x}/ X = 0.12</math>). What is the fewest number of independent samples that must be acquired to measure the mean value to within 1%? | | '''Example:''' In a given ensemble the relative fluctuation level is 12% (i.e. <math>\sigma_{x}/ X = 0.12</math>). What is the fewest number of independent samples that must be acquired to measure the mean value to within 1%? |
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| | or <math>N \geq 144</math>. | | or <math>N \geq 144</math>. |
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| - | |- | | |
| - | | [[Statistical analysis in turbulence|Up to statistical analysis]] | [[Multivariate random variables|Back to multivariate random variables]] | [[Generalization to the estimator of any quantity|Forward to generalization to the estimator of any quantity]] | | |
| - | |} | | |
| | | | |
| | {{Turbulence credit wkgeorge}} | | {{Turbulence credit wkgeorge}} |
| | | | |
| - | [[Category: Turbulence]] | + | {{Chapter navigation|Multivariate random variables|Generalization to the estimator of any quantity}} |
## Latest revision as of 16:42, 31 August 2007
Nature of turbulence
Statistical analysis
Ensemble average Probability Multivariate random var ... Estimation from a finite ... Generalization to the esti ...
Reynolds averaged equation
Turbulence kinetic energy
Stationarity and homogeneity
Homogeneous turbulence
Wall bounded turbulent flows
Study questions ... template not finished yet!
## Estimators for averaged quantities
Since there can never an infinite number of realizations from which ensemble averages (and probability densities) can be computed, it is essential to ask: How many realizations are enough? The answer to this question must be sought by looking at the statistical properties of estimators based on a finite number of realization. There are two questions which must be answered. The first one is:
• Is the expected value (or mean value) of the estimator equal to the true ensemble mean? Or in other words, is yje estimator unbiased?
The second question is
• Does the difference between the and that of the true mean decrease as the number of realizations increases? Or in other words, does the estimator converge in a statistical sense (or converge in probability). Figure 2.9 illustrates the problems which can arise.
## Bias and convergence of estimators
A procedure for answering these questions will be illustrated by considerind a simple estimator for the mean, the arithmetic mean considered above, $X_{N}$. For $N$ independent realizations $x_{n}, n=1,2,...,N$ where $N$ is finite, $X_{N}$ is given by:
$X_{N}=\frac{1}{N}\sum^{N}_{n=1} x_{n}$
Figure 2.9 not uploaded yet
Now, as we observed in our simple coin-flipping experiment, since the $x_{n}$ are random, so must be the value of the estimator $X_{N}$. For the estimator to be unbiased, the mean value of $X_{N}$ must be true ensemble mean, $X$, i.e.
$\lim_{N\rightarrow\infty} X_{N} = X$
It is easy to see that since the operations of averaging adding commute,
$\begin{matrix} \left\langle X_{N} \right\rangle & = & \left\langle \frac{1}{N} \sum^{N}_{n=1} x_{n} \right\rangle \\ & = & \frac{1}{N} \sum^{N}_{n=1} \left\langle x_{n} \right\rangle \\ & = & \frac{1}{N} NX = X \\ \end{matrix}$
(Note that the expected value of each $x_{n}$ is just $X$ since the $x_{n}$ are assumed identically distributed). Thus $x_{N}$ is, in fact, an unbiased estimator for the mean.
The question of convergence of the estimator can be addressed by defining the square of variability of the estimator, say $\epsilon^{2}_{X_{N}}$, to be:
$\epsilon^{2}_{X_{N}}\equiv \frac{var \left\{ X_{N} \right\} }{X^{2}} = \frac{\left\langle \left( X_{N}- X \right)^{2} \right\rangle }{X^{2}}$
Now we want to examine what happens to $\epsilon_{X_{N}}$ as the number of realizations increases. For the estimator to converge it is clear that $\epsilon_{x}$ should decrease as the number of sample increases. Obviously, we need to examine the variance of $X_{N}$ first. It is given by:
$\begin{matrix} var \left\{ X_{N} \right\} & = & \left\langle X_{N} - X^{2} \right\rangle \\ & = & \left\langle \left[ \lim_{N\rightarrow\infty} \frac{1}{N} \sum^{N}_{n=1} \left( x_{n} - X \right) \right]^{2} \right\rangle - X^{2}\\ \end{matrix}$
since $\left\langle X_{N} \right\rangle = X$ from the equation for $\langle X_{N} \rangle$ above. Using the fact that operations of averaging and summation commute, the squared summation can be expanded as follows:
$\begin{matrix} \left\langle \left[ \lim_{N\rightarrow\infty} \sum^{N}_{n=1} \left( x_{n} - X \right) \right]^{2} \right\rangle & = & \lim_{N\rightarrow\infty}\frac{1}{N^{2}} \sum^{N}_{n=1} \sum^{N}_{m=1} \left\langle \left( x_{n} - X \right) \left( x_{m} - X \right) \right\rangle \\ & = & \lim_{N\rightarrow\infty}\frac{1}{N^{2}}\sum^{N}_{n=1}\left\langle \left( x_{n} - X \right)^{2} \right\rangle \\ & = & \frac{1}{N} var \left\{ x \right\} \\ \end{matrix}$
where the next to last step follows from the fact that the $x_{n}$ are assumed to be statistically independent samples (and hence uncorrelated), and the last step from the definition of the variance. It follows immediately by substitution into the equation for $\epsilon^{2}_{X_{N}}$ above that the square of the variability of the estimator, $X_{N}$, is given by:
$\begin{matrix} \epsilon^{2}_{X_{N}}& =& \frac{1}{N}\frac{var\left\{x\right\}}{X^{2}} \\ & = & \frac{1}{N} \left[ \frac{\sigma_{x}}{X} \right]^{2} \\ \end{matrix}$
Thus the variability of the estimator depends inversely on the number of independent realizations, $N$, and linearly on the relative fluctuation level of the random variable itself $\sigma_{x}/ X$. Obviously if the relative fluctuation level is zero (either because there the quantity being measured is constant and there are no measurement errors), then a single measurement will suffice. On the other hand, as soon as there is any fluctuation in the $x$ itself, the greater the fluctuation ( relative to the mean of $x$, $\left\langle x \right\rangle = X$ ), then the more independent samples it will take to achieve a specified accuracy.
Example: In a given ensemble the relative fluctuation level is 12% (i.e. $\sigma_{x}/ X = 0.12$). What is the fewest number of independent samples that must be acquired to measure the mean value to within 1%?
AnswerUsing the equation for $\epsilon^{2}_{X_{N}}$ above, and taking $\epsilon_{X_{N}}=0.01$, it follows that:
$\left(0.01 \right)^{2} = \frac{1}{N}\left(0.12 \right)^{2}$
or $N \geq 144$.
## Credits
This text was based on "Lectures in Turbulence for the 21st Century" by Professor William K. George, Professor of Turbulence, Chalmers University of Technology, Gothenburg, Sweden. | {"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": 39, "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": 0.8918430209159851, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/38360/distribution-of-transformed-multinomial-variable | ## Distribution of transformed multinomial variable?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose we have a uniform multinomial distribution over $2^d$ outcomes. Multinomial coefficients give distribution of vector valued variable $v$ where $v$ is the vector of observed counts.
Is there a combinatorial expression for the distribution of $D v$ where $D$ is a matrix obtained by $d$-fold Kronecker product of $A$, a $(-1,0,1)$ valued 2-by-2 matrix?
Three particular choices of $A$ I'm looking at are
$$A_1=\left(\begin{matrix}1&1\\1&-1\end{matrix}\right)$$ $$A_2=\left(\begin{matrix}1&1\\0&1\end{matrix}\right)$$ $$A_3=\left(\begin{matrix}1&0\\-1&1\end{matrix}\right)$$
Motivation: distribution of $Dv$ corresponds to distribution of feature vectors for three commonly feature bases for exponential families (Walsh, Amari's $\eta$ and Amari's $\theta$ bases respectively)
```d = 3;
walsh = KroneckerProduct @@ Table[{{1, 1}, {1, -1}}, {d}];
amariEta = KroneckerProduct @@ Table[{{1, 1}, {0, 1}}, {d}];
amariTheta = KroneckerProduct @@ Table[{{1, 0}, {-1, 1}}, {d}];
```
Edit: all 3 matrices are invertible, so we just need to multiply by appropriate inverse to get back vector of counts and corresponding multinomial coefficient. Inverse of $D$ is just the Kronecker product of inverses of corresponding base matrices
$$A_1^{-1}=\left(\begin{matrix}1/2&1/2\\1/2&-1/2\end{matrix}\right)$$ $$A_2^{-1}=\left(\begin{matrix}1&-1\\0&1\end{matrix}\right)$$ $$A_3^{-1}=\left(\begin{matrix}1&0\\1&1\end{matrix}\right)$$
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I'm voting to close as no longer relevant – Yaroslav Bulatov Sep 13 2010 at 17:55 | {"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": 6, "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": 0.8836541771888733, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/32248/extraordinary-cohomology-as-a-derived-functor | ## Extraordinary cohomology as a derived functor?
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The purpose of this question is to find out whether one can view the Atiyah-Hirzebruch spectral sequence as a particular case of the "composition of derived functors" spectral sequence.
The Leray spectral sequence of a continuous map $f:X\to Y$ of topological spaces can be constructed as follows. Let $a_X:X\to pt$ and $a_Y:Y\to pt$ be the maps from $X$ and $Y$ respectively to the one point space $pt$; we obviously have $a_X=a_Y\circ f$. So for any sheaf $L$ on $X$ we have $(a_X)_\ast L=(a_Y)_*f_\ast L$. But $(a_X)_\ast$ is just the functor of the global sections and so is $(a_Y)_*$. Recall that if $A,B$ and $C$ are abelian categories and $F:A\to B,G:B\to C$ are left exact functors, then (under mild hypotheses) $R_\ast(G\circ F)$ is isomorphic to $R_\ast\circ G_\ast$ and for any object $X$ of $A$ there exists a spectral sequence abutting $R^\ast (G\circ F) X$ with the $E_2$ sheet given by $E_2^{pq}=R^pG(R^q F(X))$. See e.g Gelfand-Manin, Methods of homological algebra, 3.7.
Applying this to the case when $A$, $B$ and $C$ are the categories of sheaves of $X$, $Y$ and the point respectively we get a spectral sequence $(E^{pq}_r,d_d)$ abutting to $H^*(X,F)$ with the $E_2$ term given by $$E_2^{pq}=H^p(Y,R^q f_\ast L).$$
If $X$ and $Y$ are sufficiently nice (say finite CW complexes), $F$ is constant and $f$ is a locally trivial fibration with fiber $F$, then we get (assuming for simplicity that $Y$ is simply-connected) $$E_2^{pq}=H^q(Y,H^q(F)).$$
Now, if we have an extraordinary cohomology theory $h^\ast$, we can construct (under the hypotheses of the previous paragraph) the Atiyah-Hirzebruch spectral sequence: the $E_2$ sheet is given by $$E_2^{pq}=H^q(Y,h^q(F))$$ and the spectral sequence abuts to $h^*(X)$. This looks pretty similar to the Leray spectral sequence, so it seems natural to ask whether it can be obtained in a way similar to the one described above.
Namely, given an extraordinary cohomology theory $h^\ast$ and a continuous map $f:X\to Y$ of topological spaces, is there an "extraordinary direct image" functor $f^{ex}_*$ from sheaves on $X$ to sheaves on $Y$ which would be "functorial in $f$" and which would give $h^{\ast}(X)$ after deriving when $Y$ is a point?
If not, is there still a way to view the Atiyah-Hirzebruch spectral sequence as (a version of) the spectral sequence of the composition of two derived functors? (It may happen that one has to consider something other than the categories of sheaves, but I have no idea what this could be.)
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The following theorem might be relevant: if we have a functor C from finite CW-pairs to chain complexes st. 0 -> C(A) -> C(X) -> C(X,A) -> 0 is short exact and the homology of C is a homology theory, then C represents ordinary homology. This is shown in a paper by Burdick, Conner and Floyd and illustrates that extraordinary homology theories do not fit very well in a classical chain complex context. – Lennart Meier Jul 17 2010 at 18:15
## 1 Answer
I'd like to propose that the answer is "No, but..". The viewpoint I'd like to suggest is that thinking of "derived functors" is probably insufficient here (because we're secretly interested in homotopical categories that are not derived categories of abelian categories), but that we are relying essentially on the observation that `$R\Gamma(X, -) = R\Gamma(Y, Rf_* -)$`. So what follows is a sketch-construction that gives a positive answer to a related question you could've been asking: "We can get the Serre spectral sequence from the Leray spectral sequence. Can we get the Atiyah-Hirzebruch spectral sequence in a similarly sheaf-theoretic way?"
We get the Leray spectral sequence by studying sheaves of complexes of $\mathbb{Z}$-modules. The "derived" category (of sheaves of complexes of $\mathbb{Z}$-modules) is the derived category of its heart (sheaves of $\mathbb{Z}$-modules) w.r.t. the usual $t$-structure, so it's reasonable that we get lots of mileage from looking at derived functors, composites of derived functors, etc.
Analogously, we can think of more general Atiyah-Hirzebruch type spectral sequences as arising from studying sheaves of spectra. This is not the derived category of its heart w.r.t. the usual $t$-structure (this heart is again sheaves of $\mathbb{Z}$-modules). But, this does give a way of thinking about Atiyah-Hirzebruch type spectral sequences:
We start with (say) the "constant sheaf of spectra" $\mathbf{E}$ on $X$, and we're interested in computing the (homotopy groups of the spectrum) $$R\Gamma(X, \mathbf{E}) = R\Gamma(Y, Rf_* \mathbf{E})$$ The spectral sequence of interests arises by filtering $Rf_* \mathbf{E}$ using the $t$-structure (Postnikov sections "on values"), $$\cdots \to \tau_{\geq k} Rf_* \mathbf{E} \to \tau_{\geq (k-1)} Rf_* \mathbf{E}\to \cdots \to \mathbf{E}$$ The $k$-th "associated graded piece" of this filtration is $\pi_k Rf_* \mathbf{E}$ (which, recall, is an object in the heart --- i.e., a sheaf of $\mathbb{Z}$-modules on $Y$). This gives rise to a spectral sequence (excuse the funny indexing!, and no comment on convergence) $$E^2_{p,q} = \pi_{-p} R\Gamma\left(Y, \pi_{-q} Rf_* \mathbf{E}\right) \Rightarrow \pi_{-p-q} R\Gamma\left(Y, Rf_*\mathbf{E}\right) = \pi_{-p-q} R\Gamma(X,\mathbf{E})$$ The funny indexing was picked so that I can rewrite it as $$H^p(Y, \pi_{-q} Rf_* \mathbf{E}) \Rightarrow E^{p+q}(X)$$
To recover the usual form of AH-SS we need the following observation (analogous to what we'd need to get the Serre spectral sequence via sheaf theory): If $f$ is nice (i.e., a fibration between reasonable spaces), then $\pi_{-q} Rf_* E$ will be the locally constant sheaf associated to $E^{q}(F)$ with its monodromy action.
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Thanks, Anatoly! This sounds very interesting but what exactly is the category of sheaves of spectra? – algori Jul 17 2010 at 3:20
The "short answer" is you start with a(n infinity-)category of presheaves of spectra, and impose descent with respect to covers. (What's a sheaf? For any cover, the map from $\mathcal{F}(U)$ to a certain equalizer is an isomorphism. The infinity-cat. version will be that the map from $\mathcal{F}(U)$ to the holim of a Cech-type cosimplicial diagram is an equivalence.) If enough infinity-categorical machinery is in place, this is as much of a definition as the usual definition of a sheaf (of sets/abelian groups). You can find as much detail as you want in ... (continued) ... – Anatoly Preygel Jul 17 2010 at 3:41
.. some combination of Lurie's "Higher Topos Theory" (with more good stuff in Appendix A to DAG VI), or for a more classical perspective something like May-Sigurdsson "Parameterized homotopy theory." What I call "sheaves of spectra" would go under "ex-spectra" in those terms. – Anatoly Preygel Jul 17 2010 at 3:43
Finally, I should mention that in this example we only ever needed to work with "locally constant" sheaves of spectra. Just as locally constant sheaves of e.g., vector spaces admit a simpler description on nice spaces (as "local systems"), so it is for spectra. Details are in above refs, but here's the intuitive sketch: At each point, you have a spectrum. For each path, you have an equivalence between the spectra over them; for any two composable paths, you have a way of "filling in the triangle" (i.e., a homotopy between the two resulting maps); etc. – Anatoly Preygel Jul 17 2010 at 3:49
Anatoly -- thanks, but could you please give more details or a more specific reference? I'm sure there is plenty of good stuff to be found in Lurie's Higher topos theory or in Appendix A to DAG VI but it is still not clear to me what exactly the category is that $E$ is an object of. Moreover, how one defines $Rf_{\ast}$ and $R\Gamma$ for sheaves of spectra? So far the beginning of your argument looks exactly like what one would do in the classical case, with abelian groups replaced with spectra. – algori Jul 19 2010 at 20:00
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": 69, "mathjax_display_tex": 7, "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": 0.9334957003593445, "perplexity_flag": "head"} |
http://mathhelpforum.com/geometry/124908-homework-question-pythag-th-i-think.html | # Thread:
1. ## homework question...pythag th i think
Assume the earth is a perfectly round sphere with a radius of 4000 miles and that 1 mile=5280 feet. How far does the horizon appear to be to a 5-ft person on a clear day? Explain.
i thought the answer would be just a small fraction over 4000 miles, so essentially the height of the man does not change how far the horizon appears to the man.
am i on the right track here?
2. Originally Posted by igottaquestion
Assume the earth is a perfectly round sphere with a radius of 4000 miles and that 1 mile=5280 feet. How far does the horizon appear to be to a 5-ft person on a clear day? Explain.
i thought the answer would be just a small fraction over 4000 miles, so essentially the height of the man does not change how far the horizon appears to the man.
am i on the right track here?
You are dealing with a right triangle. (See attachment)
The distance from the eye of the observer to the horizon is calculated according to Pythagorean theorem:
$d^2+4000^2 = \left(4000+\frac5{5280}\right)^2$
Solve for d
Spoiler:
I've got $d \approx 2.75\ mi$. If you want to increase the radius of the field of observation you should climb a hill or use a lighthouse.
Attached Thumbnails | {"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": 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": 0.9110321998596191, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/219500/dominated-convergence-with-continuous-variable | # Dominated Convergence with continuous variable?
Let $f(x,y)$ be a real valued function of two variables, defined for $a<y<b, c<x<d$. Assume that for each $x, f(x,.)$ is a Borel measurable function of $y$, and that there is a Borel measurable function $g: (a,b)\to$ $\Bbb R$ such that $|f(x,y)|<= g(y)$ for all $x,y$ and $\int_a^bg(y) dy <\infty.$ If $x_0\in (c,d)$ and $\lim_{x \to x_0}f(x,y)$ exist for all $y\in (a,b)$, show that $$\lim_{x \to x_0}\int_a^bf(x,y) dy = \int_a^b[\lim_{x \to x_0} f(x,y)]dy$$
(Can we use Dominated Convergence Theorem to prove this?)
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3
How about: $\lim_{x \to x_0} \phi(x) = L$ if and only if for every sequence $x_n$ with $x_n \to x_0$ we have $\lim_{n \to \infty} \phi(x_n) = L$. – GEdgar Oct 23 '12 at 16:59
2
Let me suggest you learn how to accept answers. – Did Oct 23 '12 at 18:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9190887212753296, "perplexity_flag": "head"} |
http://en.wikipedia.org/wiki/Trapezoidal_rule | Trapezoidal rule
This article is about the quadrature rule for approximating integrals. For the implicit trapezoidal rule for solving initial value problems, see Trapezoidal rule (differential equations). For the explicit trapezoidal rule for solving initial value problems, see Heun's method.
The function f(x) (in blue) is approximated by a linear function (in red).
In numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) is a technique for approximating the definite integral
$\int_{a}^{b} f(x)\,dx.$
The trapezoidal rule works by approximating the region under the graph of the function $f(x)$ as a trapezoid and calculating its area. It follows that
$\int_{a}^{b} f(x)\, dx \approx (b-a)\frac{f(a) + f(b)}{2}.$
Applicability and alternatives
The trapezoidal rule is one of a family of formulas for numerical integration called Newton–Cotes formulas, of which the midpoint rule is similar to the trapezoid rule. Simpson's rule is another member of the same family, and in general has faster convergence than the trapezoidal rule for functions which are twice continuously differentiable, though not in all specific cases. However for various classes of rougher functions (ones with weaker smoothness conditions), the trapezoidal rule has faster convergence in general than Simpson's rule.[1]
Moreover, the trapezoidal rule tends to become extremely accurate when periodic functions are integrated over their periods, which can be analyzed in various ways.[2][3]
For non-periodic functions, however, methods with unequally spaced points such as Gaussian quadrature and Clenshaw–Curtis quadrature are generally far more accurate; Clenshaw–Curtis quadrature can be viewed as a change of variables to express arbitrary integrals in terms of periodic integrals, at which point the trapezoidal rule can be applied accurately.
Numerical implementation
Uniform grid
For a domain discretized into N equally spaced panels, or N+1 grid points (1, 2, ..., N+1), where the grid spacing is h=(b-a)/N, the approximation to the integral becomes
$\int_{a}^{b} f(x)\, dx \approx \frac{h}{2} \sum_{k=1}^{N} \left( f(x_{k+1}) + f(x_{k}) \right)$ ${}= \frac{b-a}{2N}(f(x_1) + 2f(x_2) + 2f(x_3) + \ldots + 2f(x_N) + f(x_{N+1})).$
Non-uniform grid
When the grid spacing is non-uniform, one can use the formula
$\int_{a}^{b} f(x)\, dx \approx \frac{1}{2} \sum_{k=1}^{N} \left( x_{k+1} - x_{k} \right) \left( f(x_{k+1}) + f(x_{k}) \right).$
Error analysis
The error of the composite trapezoidal rule is the difference between the value of the integral and the numerical result:
$\text{error} = \int_a^b f(x)\,dx - \frac{b-a}{N} \left[ {f(a) + f(b) \over 2} + \sum_{k=1}^{N-1} f \left( a+k \frac{b-a}{N} \right) \right]$
There exists a number ξ between a and b, such that[4]
$\text{error} = -\frac{(b-a)^3}{12N^2} f''(\xi)$
It follows that if the integrand is concave up (and thus has a positive second derivative), then the error is negative and the trapezoidal rule overestimates the true value. This can also be seen from the geometric picture: the trapezoids include all of the area under the curve and extend over it. Similarly, a concave-down function yields an underestimate because area is unaccounted for under the curve, but none is counted above. If the interval of the integral being approximated includes an inflection point, the error is harder to identify.
In general, three techniques are used in the analysis of error:[5]
An asymptotic error estimate for N → ∞ is given by
$\text{error} = -\frac{(b-a)^2}{12N^2} \big[ f'(b)-f'(a) \big] + O(N^{-3}).$
Further terms in this error estimate are given by the Euler–Maclaurin summation formula.
It is argued that the speed of convergence of the trapezoidal rule reflects and can be used as a definition of classes of smoothness of the functions.[2]
Periodic functions
The trapezoidal rule often converges very quickly for periodic functions.[3] This can be explained intuitively as:
When the function is periodic and one integrates over one full period, there are about as many sections of the graph that are concave up as concave down, so the errors cancel.[5]
More detailed analysis can be found in.[2][3]
"Rough" functions
This section requires expansion. (January 2010)
For various classes of functions that are not twice-differentiable, the trapezoidal rule has sharper bounds than Simpson's rule.[1]
Sample implementations
Spreadsheets
The trapezoidal rule is easily implemented in a spreadsheet. As an example, we show the integral of $f(x) = x^2$.
Performing integration using the trapezoid rule on the function $f(x)=x^2$
Python
The (composite) trapezoidal rule can be implemented in Python as follows:
```#!/usr/bin/env python
from __future__ import division
def trapezoidal_rule(f, a, b, n):
"""Approximates the definite integral of f from a to b by
the composite trapezoidal rule, using n subintervals"""
h = (b - a) / n
s = f(a) + f(b)
for i in xrange(1, n):
s += 2 * f(a + i * h)
return s * h / 2
print trapezoidal_rule(lambda x:x**9, 0.0, 10.0, 100000)
# displays 1000000000.75
```
MATLAB and GNU Octave
The (composite) trapezoidal rule can be implemented in MATLAB as follows:
```function s=Traq(f,a,b,M)
%Input - f is the integrand input as a string 'f'
% - a and b are upper and lower limits of integration
% - M is the number of subintervals
%Output - s is the trapezoidal rule sum
h=(b-a)/M;
s=0;
for k=1:(M-1)
x=a+h*k;
s=s+feval(f,x);
end
s=h*(feval(f,a)+feval(f,b))/2+h*s;
```
An alternative implementation that uses the MATLAB's vectorization is
```function s=Trapezoid(f,a,b,M)
%Input - f is the integrand input as a string 'f', which must accept vector inputs
% - a and b are upper and lower limits of integration
% - M is the number of subintervals
%Output - s is the trapezoidal rule sum
h=(b-a)/M;
x=a:h:b;
fval=feval(f,x);
s=h*sum(fval(1:end-1)+fval(2:end))/2;
```
C++
In C++, one can implement the trapezoidal rule as follows.
```template <class ContainerA, class ContainerB>
double trapezoid_integrate(const ContainerA &x, const ContainerB &y) {
if (x.size() != y.size()) {
throw std::logic_error("x and y must be the same size");
}
double sum = 0.0;
for (int i = 1; i < x.size(); i++) {
sum += (x[i] - x[i-1]) * (y[i] + y[i-1]);
}
return sum * 0.5;
}
```
Here, `x` and `y` can be any object of a class implementing `operator[]` and `size()`.
1. ^ a b
2. ^ a b c
3. ^ a b c
4. ^ a b
References
The Wikibook A-level Mathematics has a page on the topic of: Trapezium Rule
• Atkinson, Kendall E. (1989), An Introduction to Numerical Analysis (2nd ed.), New York: John Wiley & Sons, ISBN 978-0-471-50023-0 .
• Rahman, Qazi I.; Schmeisser, Gerhard (December 1990), "Characterization of the speed of convergence of the trapezoidal rule", Numerische Mathematik 57 (1): 123–138, doi:10.1007/BF01386402, ISSN 0945-3245
• Burden, Richard L.; J. Douglas Faires (2000), Numerical Analysis (7th ed.), Brooks/Cole, ISBN 0-534-38216-9 .
• Weideman, J. A. C. (January 2002), "Numerical Integration of Periodic Functions: A Few Examples", 109 (1): 21–36, doi:10.2307/2695765, JSTOR 2695765
• Cruz-Uribe, D.; Neugebauer, C.J. (2002), "Sharp Error Bounds for the Trapezoidal Rule and Simpson's Rule", 3 (4) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 11, "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": 0.8648685812950134, "perplexity_flag": "middle"} |
http://www.cs.purdue.edu/homes/dgleich/nmcomp/homeworks/homework-1-soln.html | # Network & Matrix Computations
### CIVL 2123
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# Homework 1
## Problem 1: Norms
a) Show that $f(\vx) = \normof{\mA \vx}_p$ is a vector-norm, where $\mA$ is a non-singular matrix.
Solution Because $\mA$ is non-singular, $\mA \vx = 0$ implies that $\vx = 0$. Consequently, by the standard properties of a norm, we know that $f(\vx) \ge 0$, and $f(\vx) = 0$ if and only $\vx = 0$. The other two properties follow immediately from the properties of the vector norms and the properties of matrix multiplication.
b) Show that $f(\vx) = \normof{\mA \vx}_p$ is not a vector-norm if $\mA$ is singular.
Solution When $\mA$ is singular, there is a vector $\vx$ such that $\mA \vx = 0$. This vector violates the first property of being a norm.
These norms will arise in our study of spectral graph theorem. In those cases, the matrix $\mA$ is usually the diagonal matrix of degrees for each node – commonly written $\mD$.
## Problem 2
There are a tremendous number of matrix norms that arise. An interesting class are called the orthgonally invariant norms. Norms in this class satisfy:
for square orthogonal matrices $\mU$ and $\mV$. Recall that a square matrix is orthogonal when $\mU^T \mU = \mI$, i.e. $\mU^{-1} = \mU^T$.
a) Show that $\normof{ \mA }_F$ is orthogonally invariant. (Hint: use the relationship between $\normof{ \mA }_F$ and $\text{trace}( \mA^T \mA )$.)
Solution For the trace operator, $\trace(AB)=\trace(BA)$ so, we have
b) Show that $\normof{ \mA }_2$ is orthogonally invariant. (Hint: first show that $\normof{ \mU \vx }_2 = \normof{\vx}_2$ using the relationshp between $\normof{ \vx }$ and $\vx^T \vx$.)
Solution Note that $\normof{\vx}^2 = \sum_i {x_i}^2 = \vx^T \vx$. Consequently, $\normof{\mU \vx} = \sqrt{\vx^T \mU^T \mU \vx} = \normof{\vx}$.
Hence,
where the second to last expression follows because $\vy$ can be any vector because $\mV$ is a square orthogonal matrix.
## Problem 3
In this problem, we’ll work through the answer to the challenge question on the introductory survey.
Let $\mA$ be the adjacency matrix of a simple, undirected graph.
a) An upper bound on the largest eigenvalue
Show that $\lambda_{\max}(\mA)$ is at most, the maximum degree of the graph. Show that this bound is tight.
Solution $\lambda_{\max} \le \rho(\mA) \le \normof{\mA}$ where $\rho(\mA)$ is the spectral radius, the largest magnitude of any eigenvalue. The bound follows because the 1-norm of the $\mA$ is the largest degree.
Any constant degree graph, e.g. a clique, has this as the largest eigenvalue.
b) A lower bound on the largest eigenvalue Show that $\lambda_{\max}(\mA)$ is at least, the square-root of the maximum degree of the graph. Show that this bound is tight. (Hint: try and find a lower-bound on the Rayleigh-Ritz characterization $\lambda_{\max} = \max \vx^T \mA \vx / \vx^T \vx$.)
Solution Let $\mA_S$ be the adjacency matrix for a graph with fewer edges than $\mA$. Note that
for any vector $\vy$. Let $r$ be the vertex with maximum degree. Set $\mA_S$ to be the adjacency matrix only for the edges that constitute the maximum edgree, then $\mA_S$ is the matrix for a star-graph centered at $r$. Also set
Equivalently, $\vy = \ve_S - (1 - \sqrt{d_{\max}})\ve_r$ (where $\ve_S$ has 1s only on the set of vertices in the star.
Then $$\vy^T \mA_S \vy = \underbrace{\ve_S^T \mA_S \ve_S}_{=2 d_{\max}} - 2 (1 - \sqrt{d_{\max}}) \underbrace{\ve_r^T \mA_S \ve_S}_{=d_{\max}},$$and$$\vy^T \vy = 2d_{\max}$$
by a direct calculation.
Taking these ratios gives the lower-bound of $\sqrt{d_{\max}}$.
## Problem 4
In this question, we’ll show how to use these tools to solve a problem that arose when Amy Langville and I were studying ranking algorithms.
a) the quiz from class Let $\mA$ be an $n \times n$ matrix of all ones:
What are the eigenvalues of $\mA$? What are the eigenvectors for all non-zero eigenvalues? Given a vector $\vx$, how can you tell if it’s in the nullspace (i.e. it’s eigenvector with eigenvalue 0) without looking at the matrix?
Solution The eigenvalues are $n$ and $0$. A null-vector must have sum 0 because the eigenvalue $n$ is associated with the vector of all constants, and all other vectors must be orthogonal, e.g. $\ve^T \vx = 0$ for any vector in the nullspace.
b) my problem with Amy Amy and I were studying the $n \times n$ matrix:
that arose when we were looking at ranking problems like we saw in http://www.cs.purdue.edu/homes/dgleich/nmcomp/lectures/lecture-1-matlab.m What we noticed was that Krylov methods to solve
worked incredibly fast.
Usually this happens when $\mA$ only has a few unique eigenvalues. Show that this is indeed the case. What are the unique eigenvalues of $\mA$?
Note There was a typo in this question. It should have been an $n \times n$ matrix, which makes it non-singular. Anyway, we’ll solve the question as written.
Solution The eigenvalues of this matrix are just a shift away. We start with a single eigenvalue equal to $n+1$, and we shift all the eigenvalues in a positive direction by n+1, e.g. we write $\mA = (n+1) I - \mE$ where $\mE=\ve\ve^T$ is the matrix of all ones.
Hence, we’ll have $n+1$ eigenvalues equal to $n+1$.
c) solving the system Once we realized that there were only a few unique eigenvalues and vectors, we wanted to determine if there was a closed form solution of:
There is such a form. Find it. (By closed form, I mean, given $\vb$, there should be a simple expression for $\vx$.)
Solution If the sum of $\vb$ is non-zero, then there isn’t a solution. i.e. we need $\ve^T \vb = 0$ to have a solution. Now we just have to determine $\vx$ where
Let $\ve^T \vx = \gamma$, then
So we already know that $\vx$ is given by a rescaled $\vb$. Note that $\vx$ is a solution for any value of $\gamma$, so there is an infinite family of solutions. The simplest is just $\vb/(n+1)$.
## Problem 5
In this question, you’ll implement codes to convert between triplet form of a sparse matrix and compressed sparse row.
You may use any language you’d like.
a) Describe and implement a procedure to turn a set of triplet data this data into a one-index based set of arrays: `pointers, columns, and values` for the compressed sparse form of the matrix. Use as little additional memory as possible. (Hint: it’s doable using no extra memory.)
````function [pointers, columns, values] = sparse_compress(m, n, triplets)
% SPARSE_COMPRESS Convert from triplet form
%
% Given a m-by-n sparse matrix stored as triplets:
% triplets(nzi,:) = (i,j,value)
% Output the the compressed sparse row arrays for the sparse matrix.
% SOLUTION from https://github.com/dgleich/gaimc/blob/master/sparse_to_csr.m
pointers = zeros(m+1,1);
nz = size(triplets,1);
values = zeros(nz,1);
columns = values(nz,1);
% build pointers for the bucket-sort
for i=1:nz
pointers(triplets(i,1)+1)=pointers(triplets(i,1)+1)+1;
end
rp=cumsum(rp);
for i=1:nz
values(pointers(triplets(i,1))+1)=triplets(i,3);
columns(pointers(triplets(i,1))+1)=triplets(i,2);
pointers(triplets(i,1))=pointers(triplets(i,1))+1;
end
for i=n:-1:1
pointers(i+1)=pointers(i);
end
pointers(1)=0;
pointers=pointers+1;
````
b) Describe and implement a procedure to take in the one-indexed compressed sparse row form of a matrix: `pointers, columns, and values` and the dimensions `m, n` and output the compressed sparse row arrays for the transpose of the matrix:
````function [pointers_out, columns_out, values_out] = sparse_transpose(...
m, n, pointers, columns, values)
% SPARSE_TRANSPOSE Compute the CSR form of a matrix transpose.
%
%
triplets = zeros(pointers(end),3);
% SOLUTION
for row=1:m
for nzi=pointers(row):pointers(row+1)-1
triplets(nzi,1) = columns(nzi);
triplets(nzi,2) = row;
triplets(nzi,3) = values(nzi);
end
end
[pointers_out, columns_out, values_out] = sparse_compress(n, m, triplets);
````
## Problem 6: Make it run in Matlab/Octave/Scipy/etc.
In this problem, you’ll just have to run three problems on matlab. The first one will be to use the Jacobi method to solve a linear system. The second will be to use a Krylov method to solve a linear system. The third will be to use ARPACK to compute eigenvalues on Matlab.
For this problem, you’ll need to use the ‘minnesota’ road network.
It’s available on the website: http://www.cs.purdue.edu/homes/dgleich/nmcomp/matlab/minnesota.mat The file is in Matlab format. If you need another format, let me know.
a) Use the `gplot` function in Matlab to draw a picture of the Minnesota road network.
Solution
````load minnesota
gplot(A,xy)````
b) Check that the adjacency matrix A has only non-zero values of 1 and that it is symmetric. Fix any problems you encouter.
Solution
````all((nonzeros(A)) == 1)
A = spones(A);
all((nonzeros(A)) == 1)
nnz(A-A')````
c) We’ll do some work with this graph and the linear system described in class:
where $\mL$ is the combinatorial Laplacian matrix.
```` % In Matlab code
L = diag(sum(A)) - A;
S = speye(n) - gamma*L;````
For the right-hand side, label all the points above latitude line 47 with 1, and all points below latitude line 44 with -1.
```` % In Matlab code
b = zeros(n,1);
b(xy(:,2) > 47) = 1;
b(xy(:,2) < 44) = -1;````
Write a routine to solve the linear system using the Jacobi method on the compressed sparse row arrays. You should use your code from 5a to get these arrays by calling
```` [src,dst,val] = find(S);
T = [src,dst,val];
[pointers,columns,values] = sparse_compress(size(A,1), size(A,2), T);````
Show the convergence, in the relative residual metric:
when `gamma = 1/7` (Note that $\mA$ is the matrix in the linear system, not the adjacency matrix.)
Show what happens when `gamma=1/5`
Solution (No plots here)
````n = size(A,1);
L = diag(sum(A)) - A;
S = speye(n) - 1/7*L;
b = zeros(n,1);
b(xy(:,2) > 47) = 1;
b(xy(:,2) < 44) = -1;
[i j v] = find(S);
[pointers,columns,values] = sparse_compress(size(S,1), size(S,2),[i,j,v])
[x,resvec]=jacobi(pointers,columns,values,b);
semilogy(resvec);````
Jacobi sketch
````function [x,resvec] = jacobi(pointers,columns,values,b,tol,maxiter)
x = zeros(n,1);
for i=1:maxiter
y = zeros(n,1);
for row=1:length(b)
yi = b(row); di = 0;
for nzi=pointers(row):pointers(row+1)-1
if columns(nzi) ~= row, yi = yi - values(nzi)*x(columns(nzi));
else di=values(nzi);
end
end
y(row) = yi/di;
end
% compute the residual
r = zeros(n,1);
for row=1:length(b)
ri = b(row);
for nzi=pointers(row):pointers(row+1)-1
ri = ri - values(nzi)*y(columns(nzi));
end
end
resvec(i)=norm(ri);
if resvec(i) < tol, break; end
end
resvec = resvec(1:i);
if resvec(end) > tol, warning('did not converge'); end````
d) Try using Conjugate Gradient `pcg` and `minres` in Matlab on this same system with `gamma=1/7` and `gamma=1/5`. Show the convergence of the residuals.
Solution Both work for gamma=1/7, neither work for gamma=1/5.
````
S = speye(n) - 1/5*L;
b = zeros(n,1);
b(xy(:,2) > 47) = 1;
b(xy(:,2) < 44) = -1;
%%
[x,flag,relres,iter,resvec] = pcg(S,b);
semilogy(resvec);
%%
[x,flag,relres,iter,resvec] = minres(S,b,1e-8,500);
semilogy(resvec);
````
The `semilogy` was how to show the convergence.
e) Use the `eigs` routine to find the 18 smallest eigenvalues of the Laplacian matrix $\mL$.
````>> [V,D] = eigs(L,18,'SA'); diag(D)
ans =
-0.0000
0.0000
0.0008
0.0021
0.0023
0.0031
0.0051
0.0055
0.0068
0.0073
0.0100
0.0116
0.0123
0.0126
0.0134
0.0151
0.0165
0.0167```` | {"extraction_info": {"found_math": true, "script_math_tex": 83, "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": 0.8034659028053284, "perplexity_flag": "middle"} |
http://mathhelpforum.com/geometry/153647-area-quadrilateral.html | # Thread:
1. ## Area of the quadrilateral
2. The triangles BOC, OQC, on the same base BQ, have the same "height" (namely the perpendicular distance from C to BQ). Since the area of BOC is twice the area of OQC, it follows that $BO = 2OQ$. By a similar argument from the triangles COB, OBP, it follows that $CO = \frac54OP$.
Now let x be the area of triangle PAO, and let y be the area of triangle OAQ. Since $BO = 2OQ$, it follows (by the same reasoning as in the previous paragraph) that the area of triangle BAO is twice the area of triangle OAQ. In other words, $8+x = 2y$. A similar argument in triangles PAO and OAC shows that $5+y = \frac54x$.
That gives you two simultaneous equations for x and y. The answer to the problem is the sum x+y.
Attached Thumbnails
3. Thank you again Opalg, realy it is a logical solution.
The answer to the problem is the sum x+y as you say
x = 12 , y = 10
and x + y = 22
4. Originally Posted by razemsoft21
Alternatively, you can use ratios...
Using [BQ]
$5:10=y<img src=$x+8)" alt="5:10=yx+8)" />
Using [PC]
$(5+y):x=10:8$
This is because......
$h_1=perpendicular\ height\ of\ triangles\ OAB\ and\ OAQ$
$\displaystyle\huge\frac{1}{2}|OQ|h_1=y,\ \frac{1}{2}|OB|h_1=x+8\ \Rightarrow\frac{h_1}{2}=\frac{y}{|OQ|}=\frac{x+8} {|OB|}\Rightarrow\frac{|OB|}{|OQ|}=\frac{x+8}{y}$
$h_2=perpendicular\ height\ of\ triangles\ OQC\ and\ OBC$
$\displaystyle\huge\frac{1}{2}|OQ|h_2=5,\ \frac{1}{2}|OB|h_2=10\Rightarrow\frac{h_2}{2}=\fra c{5}{|OQ|}=\frac{10}{|OB|}\Rightarrow\frac{|OB|}{| OQ|}=\frac{10}{5}$
$h_3=perpendicular\ height\ of\ triangles\ OAP\ and\ OAC$
$\displaystyle\huge\frac{1}{2}|OP|h_3=x,\ \frac{1}{2}|OC|h_3=y+5\Rightarrow\frac{h_3}{2}=\fr ac{x}{|OP|}=\frac{y+5}{|OC|}\Rightarrow\frac{|OC}{ OP|}=\frac{y+5}{x}$
$h_4=perpendicular\ height\ of\ triangles\ OBP\ and\ OBC$
$\displaystyle\huge\frac{1}{2}|OP|h_4=8,\ \frac{1}{2}|OC|h_4=10\Rightarrow\frac{h_4}{2}=\fra c{8}{|OP|}=\frac{10}{|OC|}\Rightarrow\frac{|OC|}{| OP|}=\frac{10}{8}$ | {"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": 10, "end_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": 0.8809006810188293, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/49568-locate-discontinuities.html | # Thread:
1. ## Locate the discontinuities
Can someone tell me if my logic behind this question is correct
$<br /> y=ln(\tan^2(x))<br />$
the answer is
$<br /> n\pi/2 <br />$
where 'n' is any integer
logic: since arctangent is the inverse of tangent and
$<br /> -\pi/2 \le arctan \le \pi/2<br />$
multiplying these boundaries by tan would result in a discontinuity
2. Originally Posted by silencecloak
Can someone tell me if my logic behind this question is correct
$<br /> y=ln(\tan^2(x))<br />$
the answer is
$<br /> n\pi/2 <br />$
where 'n' is any integer
logic: since arctangent is the inverse of tangent and
$<br /> -\pi/2 \le arctan \le \pi/2<br />$
multiplying these boundaries by tan would result in a discontinuity
your answer is (partially) correct. your logic makes no sense to me though. what does arctan(x) have to do with anything?
3. Originally Posted by silencecloak
Can someone tell me if my logic behind this question is correct
$<br /> y=ln(\tan^2(x))<br />$
the answer is
$<br /> n\pi/2 <br />$
where 'n' is any integer
logic: since arctangent is the inverse of tangent and
$<br /> -\pi/2 \le arctan \le \pi/2<br />$
multiplying these boundaries by tan would result in a discontinuity
here is a hint:
the domain of the logarithm is the set of positive real numbers. if what is being logged is zero or negative, then the log is undefined.
tangent is discontinuous where cos(x) = 0 and it is zero where sin(x) = 0
4. Originally Posted by Jhevon
here is a hint:
the domain of the logarithm is the set of positive real numbers. if what is being logged is zero or negative, then the log is undefined.
tangent is discontinuous where cos(x) = 0 and it is zero where sin(x) = 0
so where does the $\pi/2$ for the answer come from
that's why i was tying arctan in :/
EDIT:
oh wait...we are playing with the unit circle now huh?
cos(90) = 0
but how do you know that the tan is discontinuous where cos(x) = 0 and is zero where sin(x) = 0
5. We know that because of the identity using $\tan{x}$. We know that:
$\tan{x} = \frac{\sin{x}}{\cos{x}}$
So, when $\cos{x} = 0$
$\tan{x} = \frac{\sin{x}}{0}$
Which is undefined and therefore discontinuous, and when $\sin{x} = 0$
$\tan{x} = \frac{0}{\cos{x}}$
Which is zero for all nonzero results of $\cos{x}$, so the logarithm is therefore discontinuous because it is not defined at zero. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "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": 0.9488961100578308, "perplexity_flag": "middle"} |
http://mathoverflow.net/revisions/81379/list | ## Return to Answer
3 added 226 characters in body
Oops, I was thinking of a Lie group. Edit: On a Lie group, it is pretty easy to test for integrability. Take a basis of complex linear left invariant 1-forms (i.e. left translate from a choice of such 1-forms on the Lie algebra). Then compute their exterior derivatives. You have an integrable almost complex structure if and only if the (0,2) parts of the exterior derivatives all vanish. To see this, you express the exterior derivatives in linear combinations of the wedge products of the original 1-forms and their complex conjugates. So examples are easy to check, and only require the differential familiar from Lie algebra cohomology, i.e. a pure Lie algebra calculation.
On a homogeneous space $G/H$ the computation is a little trickier. You take $\omega=g^{-1}dg$, the left invariant Maurer Cartan form on $G$, and then $\omega+\mathfrak{h}$ is semibasic for the quotient map $G \to G/H$ and splits into a complex linear part and a conjugate linear part . on each tangent space of $G/H$. Let $\eta$ be the complex linear part. Pick a splitting of $\mathfrak{g}$ into $\mathfrak{g}/\mathfrak{h}$ and some complement, identified with $\mathfrak{h}$, and let $\Omega$ be the projection of $\omega$ to complement. The equation $d \omega + (1/2)[\omega,\omega]=0$ gives an equation $d \eta = - \rho(\omega) Omega \wedge \eta + a \eta \wedge \eta + b \eta \wedge \bar{\eta} + c \bar{\eta} \wedge \bar{\eta}$. The Nijenhuis tensor is vanishes just when $c$. c=0\$. I am pretty sure that the generic invariant almost complex structure on a flag manifold (invariant under the compact form of the automorphism group) is not integrable, but I would have to check.
2 added 553 characters in body; added 46 characters in body; added 4 characters in body
It
Oops, I was thinking of a Lie group. Edit: On a Lie group, it is pretty easy to test for integrability. Take a basis of complex linear left invariant 1-forms (i.e. left translate from a choice of such 1-forms on the Lie algebra). Then compute their exterior derivatives. You have an integrable almost complex structure if and only if the (0,2) parts of the exterior derivatives all vanish. To see this, you express the exterior derivatives in linear combinations of the wedge products of the original 1-forms and their complex conjugates. So examples are easy to check, and only require the differential familiar from Lie algebra cohomology, i.e. a pure Lie algebra calculation.
On a homogeneous space $G/H$ the computation is a little trickier. You take $\omega=g^{-1}dg$, the left invariant Maurer Cartan form on $G$, and then $\omega+\mathfrak{h}$ is semibasic for the quotient map $G \to G/H$ and splits into a complex linear part and a conjugate linear part. Let $\eta$ be the complex linear part. The equation $d \omega + (1/2)[\omega,\omega]=0$ gives an equation $d \eta = - \rho(\omega) \wedge \eta + a \eta \wedge \eta + b \eta \wedge \bar{\eta} + c \bar{\eta} \wedge \bar{\eta}$. The Nijenhuis tensor is $c$. I am pretty sure that the generic invariant almost complex structure on a flag manifold (invariant under the compact form of the automorphism group) is not integrable, but I would have to check.
1
It is pretty easy to test for integrability. Take a basis of complex linear left invariant 1-forms (i.e. left translate from a choice of such 1-forms on the Lie algebra). Then compute their exterior derivatives. You have an integrable almost complex structure if and only if the (0,2) parts of the exterior derivatives all vanish. To see this, you express the exterior derivatives in linear combinations of the wedge products of the original 1-forms and their complex conjugates. So examples are easy to check, and only require the differential familiar from Lie algebra cohomology, i.e. a pure Lie algebra calculation. I am pretty sure that the generic invariant almost complex structure on a flag manifold (invariant under the compact form of the automorphism group) is not integrable, but I would have to check. | {"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": 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": 0.884257972240448, "perplexity_flag": "head"} |
http://mathhelpforum.com/advanced-applied-math/20844-magnitude-forces-print.html | # magnitude of the forces
Printable View
• October 18th 2007, 12:26 PM
rcmango
magnitude of the forces
A 32 kg crate rests on a horizontal floor, and a 69 kg person is standing on the crate.
Determine the magnitude of the normal force that the floor exerts on the crate.
N
Determine the magnitude of the normal force that the crate exerts on the person.
N
not sure why the answers would not be: 101 N and 69 N.
thanks for looking at this one.
• October 18th 2007, 02:15 PM
topsquark
Quote:
Originally Posted by rcmango
A 32 kg crate rests on a horizontal floor, and a 69 kg person is standing on the crate.
Determine the magnitude of the normal force that the floor exerts on the crate.
N
Determine the magnitude of the normal force that the crate exerts on the person.
N
not sure why the answers would not be: 101 N and 69 N.
thanks for looking at this one.
I'm going to use a specific notation for this one. It might seem a little confusing at first, but once you get used to it it saves a lot of time and trouble.
Define the symbol $F_{ab}$ to be the force on object a that is coming from object b.
So Free Body Diagram time. In each FBD I'm defining upward to be positive. We don't need a horizontal axis.
For the box, we have a normal force $N_{bf}$ (f represents the floor) pointing
up, a weight on the box $w_{bE}$ (E is the Earth) acting straight down, and a normal force $N_{bp}$ acting straight down.
So
$\sum F = N_{bf} - w_{bE} - N_{bp} = 0$ (since the box is not accelerating.)
We need $N_{bf}$, so
$N_{bf} = w_{bE} + N_{bp}$
So what is $N_{bp}$?
Do a FBD on the person:
We have a normal force $N_{pb}$ acting upward and a weight $w_{pE}$ acting downward.
Thus
$\sum F = N_{pb} - w_{pE} = 0$
So
$N_{pb} = w_{pE}$
So the magnitude of the normal force on the person is equal to the magnitude of the weight of the person.
Now, to finish the problem we need to find $N_{bp}$, not $N_{pb}$. The notation is meant to suggest a relationship between the two: they are Newton's 3rd Law pairs. So by Newton's 3rd these have equal magnitudes. Thus
$N_{bf} = w_{bE} + N_{bp} = w_{bE} + N_{pb} = w_{bE} + w_{pE}$
You can do the rest from here.
Edit: After going through all this it appears that the whole problem is that you gave your answer in terms of the mass, not the weight. Ah well, it was good for a demonstration. :)
-Dan
• October 19th 2007, 10:31 AM
rcmango
Okay, so in other words Fnet = mass * acceleration,
got it.
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http://mathoverflow.net/questions/88184?sort=newest | ## What is the significance of non-commutative geometry in mathematics?
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This is a question that has been winding around my head for a long time and I have not found a convincing answer. The title says everything, but I am going to enrich my question by little more explanations.
As a layman, I have started searching for expositories/more informal, rather intuitive, also original account of non-commutative geometry to get more sense of it, namely, I have looked through
Nevertheless, I am not satisfied with them at all. It seems to me, that even understanding a simple example, requires much more knowledge that is gained in grad school. Now for me, this field merely contains a lot of highly developed machineries which are more technical (somehow artificial) than that of other fields.
The following are my questions revolving around the significance of this field in Mathematics. Of course, they are absolutely related to my main question.
1. How can a grad student be motivated to specialize in this field? and
2. What is (are) the well-known result(s), found solely by non-commutative geometric techniques that could not be proven without them?
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I think Alain Valette's answer to your Q2 is a very good example. As for Q1, I have found that some combination of "there is some beautiful mathematics here" and "there might be a better chance of getting a grant or a postdoc if you do this, than if you do something idiotic like the cohomology of Banach algebras", quite forceful. – Yemon Choi Feb 11 2012 at 8:46
Fist of all, it's easy to understand simple example at grad level (at least a decent grad level) : it's \$C*-algebra theory for example, or even easier, it's algebraic geometry. Both are foundations and motivate NCG. Second, as for the contribution to existing maths of purely NCG theory, I'm not a specialist at all, but if I remember well what Connes said at a talk (but I didn't get much of it honestly), one achievement would be to prove Riemann's hypothesis in terms of the NCG analog model of the Weil proof of Riemann's hypothesis. – Amin Feb 11 2012 at 8:49
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Personally, I have come to believe more in good mathematics than in important mathematics, but one could perhaps attribute that to sour grapes... – Yemon Choi Feb 11 2012 at 8:58
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While the original question was quite good, I have to say that these comments are descending very rapidly into "subjective and argumentative" territory. MO is absolutely not about "ranking" subjects etc. Also, shouldn't this be community wiki... – Matthew Daws Feb 11 2012 at 14:59
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Dear @Yemon Choi: In fact, I don't intend to think like that. I wanted to know what the reactions are when I say the words that I was told some years ago about this field. This might be another subject for discussion. BTW, I've intended to work in a field which has common features of NCG, in some sense, regarding the grant, postdoc, etc. and now, is not very supported and funded, but it appeals me the most, is still growing and good works are being done. – Ehsan M. Kermani Feb 11 2012 at 18:50
show 9 more comments
## 6 Answers
I think I'm in a pretty good position to answer this question because I am a graduate student working in noncommutative geometry who entered the subject a little bit skeptical about its relevance to the rest of mathematics. To this day I sometimes find it hard to get excited about purely "noncommutative" results, but the subject has its tentacles in so many other areas that I never get bored.
Before saying anything further, I need to say a few words about the Atiyah-Singer index theorem. This theorem asserts that if $D$ is an elliptic differential operator on a manifold $M$ then its Fredholm index $dim(ker(D)) - dim(coker(D))$ can be computed by integrating certain characteristic classes of $M$. Non-trivial corollaries (obtained by "plugging in" well-chosen differential operators) include the generalized Gauss-Bonnet formula, the Hirzebruch signature theorem, and the Hirzebruch-Riemann-Roch formula. It was quickly realized (first by Atiyah, I think) that the proof of the theorem can be viewed as a statement about the Poincare duality pairing between topological K-theory and its associated homology theory (these days called K-homology).
I wasn't around, but I'm told that people were very excited about Atiyah and Singer's achievement (understandably so!) People quickly began trying to generalize and strengthen the theorem, and my claim is that noncommutative geometry is the area of mathematics that emerged from these attempts. Saying that marginalizes the other important reasons for developing the subject, but I think it was Connes' main motivation and in any event it is a convenient oversimplification for a MO answer. It also helps me answer your first question by playing to my personal biases: when I was choosing an area of research I told my adviser that I was interested in learning more about that Atiyah-Singer index theorem and I was led inexorably toward the tools of noncommutative geometry.
The origin of the relationship between NCG and Atiyah-Singer lies in equivariant index theory. Atiyah and Singer realized from the start that if $M$ admits an action by a compact Lie group $G$ and $D$ is invariant under the group action then it is better to think of the index of $D$ as a virtual representation of $G$ (i.e. an element of the $G$-equivariant K-theory of a point) rather than as an integer. If $G$ is not compact then this doesn't really work, but the noncommutative geometers realized that $D$ does have an index in the K-theory of the reduced group C$^\ast$-algebra $C_r^\ast(G)$. Indeed, to a noncommutative geometer equivariant index theory is all about a map $K_\ast(M) \to K_\ast(C_r^\ast(M)$ where $K_\ast(M)$ is the K-homology of $M$; in the case where $M$ is the universal classifying space of $G$, Baum and Connes conjectured that this map is an isomorphism. Proving this conjecture for more and more groups and understanding its consequences motivates a great deal of the development of NCG to this day.
The conjecture is interesting in its own right if you already care about index theory, but even if you don't injectivity of the Baum-Connes map implies the Novikov conjecture (see Alain Valette's answer) and surjectivity is related to the Borel conjecture. It has numerous other applications, for example to the theory of positive scalar curvature obstructions in Riemannian geometry or to the Kadison-Kaplansky conjecture in functional analysis (which would follow from surjectivity). Recently there has been a lot of interest in connections between the Baum-Connes conjecture and representation theory; the Baum-Aubert-Plyman conjecture in p-adic representation theory has its origins in these sorts of considerations.
Much of the rest of NCG can also be traced back to index theory. Kasparov's KK-theory arose as a way to understand maps and pairings between K-theory and K-homology, motivated in part by index theory. Connes' work on noncommutative measure theory arose from his work on index theory for measurable foliations (with applications to dynamical systems). Cyclic (co)homology was invented in part to gain access in a noncommutative setting to the Chern character map from K-theory to cohomology which translates the K-theoretic formulation of the index theorem into a cohomological formula. Connes' theory of spectral triples and noncommutative Riemannian geometry is based on the theory of Dirac operators which was invented by Atiyah and Singer to prove the index theorem. I guess my point with all of this is that all the esoteric machinery of NCG seems less artificial when viewed through the lens of index theory.
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Just to add up something to this nice answer I'd like to mention that (exactly through NC index theorems) NCgeometry contributed much in understanding foliations. Much work is by now devoted to the non commutative geometry of foliations. The idea is that this direction will lead to a better understanding also of singular foliations. – Nicola Ciccoli Feb 11 2012 at 16:48
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@Suvrit: I should have made a disclaimer at the outset of my answer that I am only informed about the analytic and topological aspects of NCG (a la Connes) and not NCAG. My understanding is that NCAG developed largely separately and for different reasons, though I think there are some like Jonathan Block and Ryszard Nest who straddle the line. That said, both areas use a lot of noncommutative algebra; in Connes' approach it enters via Hochschild and cyclic (co)homology. – Paul Siegel Feb 11 2012 at 20:08
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"Before saying anything further, I need to say a few words about the Atiyah-Singer index theorem." Paul, if I had a dollar for every time I've heard you preface something with that sentence, I'd be a very rich man. – Vaughn Climenhaga Feb 13 2012 at 1:31
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You're just jealous because you haven't thought of a better pick-up line. :) – Paul Siegel Feb 13 2012 at 9:36
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Just seen this conversation. Well, Paul, I think we know how you started or will start your marriage proposal... – Yemon Choi Feb 26 2012 at 12:08
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The spectral characterization of Riemannian manifolds is a good example in my opinion. Details are given in the following article: http://www.alainconnes.org/docs/ckm1.pdf .
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Connes has on his home page a very nice downloadable survey article "A view of mathematics" which also gives a good idea of the role of groupoids in this area, and links with many topics mentioned by others above.
I have also raised the question of the problem of applying these or related techniques to algebraically structured groupoids:
http://mathoverflow.net/questions/86617/convolution-algebras-for-double-groupoids
Since "motivating graduate students" was part of the original question, someone needs to point out at least one thing that has not been done. It can be useful to explain the "exterior" of a subject: unknowns, known unknowns, etc! One would also like the experts to point out anomalies in this area: I won't try to define this term, but they can lie around unrecognised.
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There are much better answers above than this one, but:
If you believe fiber bundles are important to classical mathematics, then you probably believe fibrations are, and maybe foliations are, as well. If you don't, note that a foliation of a smooth manifold is a decomposition of the manifold into integral submanifolds (roughly, solutions to differential equations). You can't get much more classical than this. In his book Noncommutative Geometry Connes tried to make it clear that to understand the leaf space of a foliation, more is needed than the classical quotient construction, groupoids and noncommutative geometry give more information about a patently classical "space". You probably say: So what? There are other ways. Connes tries then to show us that there is a connection between a fundamental von Neumann algebra invariant (the flow of weights) and one of the key invariants for a codimension 1 foliation (the Godbillon-Vey class), which appears in the first chapter on many introductory accounts of foliations. I find it hard to believe that this is coincidental. For me, this warrants closer investigation.
The index theorem for measured foliations discussed above perhaps grew from a seed like the above mentioned connection. (I wonder what we need to do to get Connes to weigh-in over here at MO?)
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And Kasparov, Moscovici, Baum, ... – Yemon Choi Feb 12 2012 at 19:45
Thanks, Yemon. I was thinking of Connes's book, and neglected to type that. – Jon Bannon Feb 12 2012 at 22:54
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We should at least be thankful that we have Alain Valette! – Paul Siegel Feb 13 2012 at 0:03
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I second that, Paul! – Jon Bannon Feb 13 2012 at 1:01
Alain's answer is the best application I heard about. I cannot add something comparable. Just two "motivations" which are somewhat nice to me. However they does not answer yours questions, sorry.
General claim - studying non-commutative objects is useful for understanding commutative ones.
Subclaim - non-commutative algebras can be equivalent (Morita, Koszul dual or whatever) to non-commutative ones, however non-commutative "models" can provide more easy way to study commutative things.
Examples 1. Consider commutative algebra A of functions on manifold M and group G. You may be interested in factor $M/G$ which is related to invariants $A^G$.
Claim. Under certain conditions COMMUTATIVE $A^G$ is Morita equivalent to NON-COMMUTATIVE $A\times C[G]$ - cross-product algebra of $A$ and group algebra of $G$. In some cases it is more easy to work with this cross-product sometimes it can be described more explicitly. You may see just the first sentences in Etingof Ginzburg famous paper: http://arxiv.org/abs/math/0011114
Example 2. Quantization. Our real world is actually quantum. So physicists are interested in this. Mathematical way to understand quantization is a procedure to construct the non-commutative algebras from commutative ones. The big mathematical challenge is to understand how to relate properties on non-commutative quantum algebras to properties of commutative ones. Probably the most striking and most simple formulated is the conjecture that automorphisms group of classical symplectic R^2n and quantum (i.e. just the algebra of differential operators in n-variables) are isomorphic. http://arxiv.org/abs/math/0512169 Automorphisms of the Weyl algebra Alexei Belov-Kanel, Maxim Kontsevich
It is somewhat related to the famous Jacobian conjecture. See http://arxiv.org/abs/math/0512171
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My favorite example concerns Novikov conjecture, on the homotopy invariance of higher signatures for closed manifolds with fundamental group $G$: see http://en.wikipedia.org/wiki/Novikov_conjecture (note that this Wikipedia entry rather stupidly says that it has been proved for finitely generated abelian groups: that's correct, but it was proved for MANY more groups, e.g. hyperbolic groups, countable subgroups of $GL_n(\mathbb{C})$, etc...). I think we agree that this is a conjecture in topology.
Now, look at this remarkable result by Guoliang Yu ( http://www.kryakin.com/files/Invent_mat_(2_8)/139/139_21.pdf )
"If the group $G$ admits a coarse embedding into Hilbert space, then it satisfies the Novikov conjecture".
A coarse embedding is a map $f:G\rightarrow L^2$ for which there exists control functions $\rho_{\pm}:\mathbb{R}^+\rightarrow\mathbb{R}$, with $\lim_{t\rightarrow\infty}\rho_\pm(t)=\infty$, which control'' $f$ in the sense hat, for every $x,y\in G$:
`$$\rho_-(|x^{-1}y|_S)\leq\|f(x)-f(y)\|_2\leq \rho_+(|x^{-1}y|_S),$$` where $|.|_S$ denotes word length with respect to some finite generating subset $S$ in $G$. The existence of a coarse embedding is a weak metric condition (actually we know of basically just one class of groups which do not admit such an embedding, the Gromov monsters''). And this weak metric condition, quite surprisingly, implies a strong consequence in topology.
Now my point is that the two known proofs of Yu's result (the original one, and the one by Skandalis-Tu-Yu, see http://www.math.univ-metz.fr/~tu/publi/coarse.pdf) both appeal in a fundamental way to the tools of non-commutative geometry: $C^*$-algebras, $K$-theory, groupoids, Kasparov's $KK$-theory (to be precise: equivariant $KK$-theory for groupoids'').
Now to answer your first question: how to motivate a graduate student? Well, the subject mixes classical geometry, algebraic topology, non-commutative algebra, functional analysis, so it is one of those subjects that give you a feeling of the unity of mathematics...
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Cool, I didn't know about that ! – Amin Feb 11 2012 at 8:52
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Just as an addendum, and with caveat that AV can explain this better than I can. There's an article by Connes, late 80s or early 90s, where he mentions some of the work on the Novikov conjecture done using the early incarnations of the machinery mentioned. Unfortunately my copy is buried somewhere in my office, but I vaguely recall it used index theory for $C(T^n)=C*(Z^n)$ to give a proof of Novikov's conjecture for abelian groups, as motivation for the need to do differentiyal geometry on noncommutative spaces, viz. K-theory of C*-algebras, cyclic cohomology & Chern character, etc – Yemon Choi Feb 11 2012 at 8:56
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I'd also put in a word for the Kadison-Kaplansky conjecture: in this case the problem originates in analysis rather than topology, but it seems that some of the most significant progress has either used NCG or used tools which received a lot of impetus from work in NCG – Yemon Choi Feb 11 2012 at 9:25
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"note that this Wikipedia entry rather stupidly says..." Why not edit it then? – David Corfield Feb 11 2012 at 14:46
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@Dear Alain: many thanks for your explanations. One of the reasons that I chose the Paul's answer is the sense that I got from the way he tried to relate important (classical) results that I've at least heard of many times to the known facts developed by NCG's method. As for my Q1, I'd say, unity of math sounds heroic and I don't really consider it as a motivation. – Ehsan M. Kermani Feb 14 2012 at 2:55
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http://math.stackexchange.com/questions/56063/minimum-degree-of-a-graph-and-existence-of-perfect-matching | # Minimum degree of a graph and existence of perfect matching
I was reading a result where the following proposition appears as a preliminary step (and left as exercise):
Claim: Suppose $G$ is a graph on $n$ vertices ($n$ even and $n \geqslant 3$) with minimum degree at least $n/2$. Show that $G$ contains a perfect matching.
Proof: By Dirac's theorem, $G$ has a Hamiltonian cycle $C$. Since $|C|=n$ is an even integer, the set of “odd” edges of $C$ gives a perfect matching for $G$. $\quad\Box$
I feel that using Dirac's theorem for this claim is an overkill. But after trying for a few days, couldn't come up with any other proofs. Can you give a more “direct” proof of the claim that avoids the machinery of Hamiltonian cycles?
Naturally, you might object to the vague requirement of “directness”. To clarify what I mean by it, I’ll give an example.
Claim 2. Suppose $G$ has a minimum degree at least $n/2$. Show that $G$ is connected.
Proof via Dirac's theorem. $G$ has a hamiltonian cycle as before, and hence is connected. $\Box$
A different proof. We'll show that any pair of vertices $u,v$ are connected by a path of length at most $2$. If $uv$ is an edge, we are done. Suppose not. Then $N(u) \cup N(v) \subseteq V \smallsetminus \{u,v\}$. Therefore $$|N(u) \cap N(v)| = |N(u)| + |N(v)| - |N(u) \cup N(v)| \geqslant \frac{n}{2}+\frac{n}{2}-(n-2) = 2 \gt 0.$$ In particular there exists vertex $w$ such that $uw$ and $wv$ are both edges, and we are done. $\quad \Box$
I find the proof via Dirac's theorem much less illuminating in this example. In fact, as Qioachu points out below, the Claim 2 might even appear as an intermediate steps of Dirac’s theorem, making the above proof cyclic. My intention here is only to point out that Dirac’s theorem can be used as a powerful black box in killing much easier results.
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The proof I know of Dirac's theorem needs a stronger version of Claim 2 as a lemma. – Qiaochu Yuan Aug 8 '11 at 22:17
@Qia Admittedly its first "proof" of Claim 2 is a bit silly and circular. I guess my point is this: Sometimes the easiest way to convince yourself that you are not trying to show some false statement is to check that it is a corollary to some big hammer that you are sure to be true. However it's not the right or best way. – Srivatsan Sep 13 '11 at 15:52
## 2 Answers
Here's an alternative proof using Hall's theorem -- I don't know whether that counts as a more "direct" proof, but at least it uses a theorem that directly deals with perfect matchings instead of a detour (pun intended) through Hamiltonian cycles.
Arbitrarily divide the vertices into two sets $A$ and $B$ of equal size. In each set, find a vertex with a minimal number of edges connecting it to the other set, and swap the two minimal vertices. Repeat this until the swap would no longer increase the total number of edges connecting the two sets. (Since there is a finite number of edges, this must happen at some point.) Denoting the two vertices whose swap would no longer increase the number of connections by $v_A$ and $v_B$ and the numbers of their edges within and between the sets by $v_{AA}$, $v_{AB}$, $v_{BA}$ and $v_{BB}$, and counting the number of connections between the sets before and after the swap, we have $v_{AB}+v_{BA}\ge v_{AA}+v_{BB}$. (There might be an edge between $v_A$ and $v_B$, but that works in favour of the inequality.) It follows that
$$v_{AB}+v_{BA}\ge \frac12(v_{AB}+v_{BA}+v_{AA}+v_{BB})=\frac12(\deg v_A+\deg v_B)\ge \frac12(n/2+n/2)=n/2\;.$$
Now consider a subset $X$ of $A$. If $|X|\le v_{AB}$, then since each element of $A$ has edges to at least $v_{AB}$ elements of $B$, $X$ has at least as many neighbours in $B$ as elements. If $|X|>v_{AB}$, then since each element of $B$ has edges to at least $v_{BA}$ elements of $A$ and $v_{AB}+v_{BA}\ge n/2=|A|$, at least one of these edges must lead to $X$, so $X$ has $n/2$ neighbours in $B$, and thus at least as many as it has elements. Thus the premise of Hall's theorem is fulfilled, and the bipartite graph induced on $A$ and $B$ must contain a perfect matching, which is also a perfect matching in $G$.
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Very nice. I find your idea of reducing the problem from general matching to bipartite matching refreshing. (Sorry about the late response! :-)) – Srivatsan Sep 13 '11 at 16:05
I believe it might be possible to provide a simpler proof by induction. Conisder the minimal situation where each vertex has degree $n/2$.
For the first case, $n=4$, this is equivalent to $C_4$ and there is an obvious perfect matching, and adding more edges between the vertices does not affect the existence of a matching.
Now consider the operation of adding 2 vertices ($u$ and $v$) and enough edges to a graph (which already contains a perfect matching) to fulfill the $n/2$ condition. If there is an edge between $u$ and $v$ add that edge to the matching and the matching is perefct.
If there is not an edge between the two new vertices then they must share at least 2 neigbors ($y$ and $z$) because there are $n$ original vertices and $u$ and $v$ must each dominate $\frac {n+2}{2}$ of these vertices. There are three further cases to consider.
Case 1 If the edge $yz$ is in the original matching then WLOG replace $yz$ with $uy$ and $vz$ and the matching is again perfect.
Case 2 If the edge $yz$ is not in the original matching and $u$ and $v$ share only 2 neighbors then create an augmenting path as follows: Select an alternating path from $y$ to $z$ taking the matched edge from $y$ as the first edge in the path. The vertex that is matched to $z$ ($p$) must be adjacent to either $u$ or $v$ because by only sharing 2 common neighbors $u$ and $v$ cover all of the vertices. WLOG assume the $v$ is adjacent $p$ and add $zp$ and $pv$ to the path. Add the edge $uy$ and the path becomes an augmenting path that when the edges are flipped creates a perfect matching.
Case 3 If $u$ and $v$ share more than 2 neighbors they must be adjacent to both ends of a matched edge. As in case 1 above you may remove that edge from the matching and replace with two edges of your choice attaching one of each $u$ and $v$ to an vertex from the previous matching.
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I do not understand how to get the conclusion "..they must be adjacent to both ends of a matched edge" in Case 3. (Also, a nitpick in the beginning of the proof: it is not always possible to make the graph $n/2$-regular by removing edges. For instance, what if the graph has $\rm 2$ vertices $a,b$ with degrees $> n/2$ and remaining with degree $n/2$, such that $ab$ is not an edge?) – Srivatsan Aug 30 '11 at 18:11
Thank you for your approach using induction. – Srivatsan Aug 30 '11 at 18:16
@Srivatsan Narayanan- Your'e welcome. In the beginning I intended to simply use your claim #2 to consider only the least possible edges $n/2$ per vertex. There may certainly be more in the graph, they just aren't relevant to the proof. I should have been more explicit with my wording. In case #3 if $u$ and $v$ have multiple common neighbors you can use the pigeonhole principle to show that they must be commonly adjacent to at least one matched previously matched edge. – user12998 Aug 31 '11 at 22:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 104, "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": 0.9563742280006409, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/26848?sort=votes | ## Looking for an interesting problem/riddle involving triple integrals.
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Does anyone know some good problem in real analysis, the solution of which involves triple integrals, and which is suitable for second semester Analysis students?
Thanks!
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Just since terminology differs in different countries: when you say "second semester analysis students", do you mean math concentrators at a university, taking a rigorous (proof-based) analysis sequence? Or do you mean students taking a multivariable calculus course (learning computational skills with Greens/Stokes theorem, etc.)? – Marty Jun 2 2010 at 18:16
I mean the latter (a multivariable calculus course for engineers). – Irene Adler Jun 2 2010 at 19:06
## 12 Answers
Ask the students what is the average distance from a point in the ball of radius R to the center of the ball. You will get the answer R/2. You then point out that the volume within R/2 of the center is much smaller than the remaining volume. Next some bright student proposes the radius that equates the volume inside and outside the shell. You explain why that does not give the correct answer either, which is more subtle. You commend the student who equated volumes for a good proposal, and ask if anyone has a better one. Then it is time to set up the very simple triple integral that gives the answer, and solve it by the method of spherical shells, or by spherical coordinates. And you compare the correct answer with the answer obtained by equating volumes, and see that they are quite close.
The emphasis in this example is more on setting up the right triple integral to answer the question, and not so much on evaluating a complicated integral.
You should let the students have a little time to ponder. I let them have about three minutes for each stage, because I don't want to lose the slower ones.
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Only a single integral is needed if you use spherical symmetry. – Richard Eager Oct 19 at 15:56
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I have seen students close to suicide over the following one: For $i=1,2,3$ denote by $C_i$ be the solid cylinder in $\mathbb R^3$ with axis given by the standard basis vector $x_i$ and radius 1. Compute the volume of $C_1\cap C_2 \cap C_3$.
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I remember seeing an elementary solution of this one in a certain installment of the famous Scientific American column by the late Martin Gardner. – J. H. S. Jun 2 2010 at 22:18
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If somebody knew in which number this appeared- I am interested. – Xandi Tuni Jun 3 2010 at 7:50
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@Xandi or @JHS: Since I don't feel like committing suicide, can one of you give me the best method for solving the problem? – André Henriques Aug 10 2010 at 20:02
@André: It actually works to integrate, as a function of radius, the surface area, whose smooth pieces come from cylinders cut by planes of symmetry and therefore unroll to sine curves. – Tracy Hall Aug 11 2010 at 3:36
I remember in high school trying (without success) to convince one of the engineers where I worked that specifying a quarter-round at the corner in each of the $X$, $Y$, and $Z$ directions of a blueprint was not technically the same (rather, only by convention) as putting a section of sphere there--that if you just cut along the projections, there would still be a distinct (if blunt) corner. – Tracy Hall Aug 11 2010 at 8:04
I do not know the exact location in his Collected Works but Dirichlet found the $n$-volume of
$$x_1, x_2, \ldots, x_n \geq 0$$ and $$x_1^{a_1} + x_2^{a_2} + \ldots + x_n^{a_n} \leq 1 .$$
For example with $n=3$ the volume is $$\frac{ \Gamma \left( 1 + \frac{1}{a_1} \right) \Gamma \left( 1 + \frac{1}{a_2} \right) \Gamma \left( 1 + \frac{1}{a_3} \right) }{ \Gamma \left(1 + \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right) }$$
Note that this has some attractive features. The limit as $a_n \rightarrow \infty$ is just the expression in dimension $n-1,$ exactly what we want. Also, we quickly get the volume of the positive "orthant" of the unit $n$-ball by setting all $a_j = 2,$ and this immediately gives the volume of the entire unit $n$-ball, abbreviated as $$\frac{\pi^{n/2}}{(n/2)!}$$
I think he also exactly evaluated the integral of any monomial $$x_1^{b_1} x_2^{b_2} \cdots x_n^{b_n}$$ on the same set.
So the question would be: given, say, positive integers $a,b,c,$ find the volume of $x,y,z \geq 0$ and $x^a + y^b + z^c \leq 1.$ If you like, fix the exponents, the triple $a=2, b=3, c=6$ comes up in a book by R.C.Vaughan called "The Hardy-Littlewood Method," page 146 in the second edition, where he assumes the reader knows this calculation.
This came back to mind because of a recent closed question on the area of $x^4 + y^4 \leq 1.$
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In Whittaker and Watson, end of the Gamma function chapter, by the way. – Charles Matthews Jun 2 2010 at 21:31
Thank you, Charles. I actually copied this as a few pages out of Dirichlet's collected works in translation. That was years ago and I can't find the pages. – Will Jagy Jun 2 2010 at 22:23
Coffee mug problem: the volume of the part of a cylinder cut off by a plane, representing the volume of the coffee you have left when you first see the bottom of the mug. (Plane is horizontal, cylinder tilted, plane touches the two circles bounding the cylinder). Much integration.
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The question "U(3) Sato-Tate measure" is a quite different example. – Charles Matthews Jun 2 2010 at 21:00
I always thought it was more interesting to ask the same problem for when you can see precisely half the bottom, since then it can't be solved in three seconds using symmetry. The again, it's a useful drill to complete a series of excruciating integrals, only to be shown the immediate answer for each one coming from symmetry. – Tracy Hall Aug 11 2010 at 3:45
The students can do $$\iiint_{[0,1]^3}\frac{x^n(1-x)^ny^n(1-y)^nz^n(1-z)^n}{(1-(1-xy)z)^{n+1}}\ dx\ dy\ dz,$$ at least for $n=0$ and $n=1$. These are the integrals producing Apery's approximations to $\zeta(3)$; the original paper is [F. Beukers, Bull. London Math. Soc. 11 (1979) 268--272].
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This is one of my all time favourites (quoted from Problems and Theorems in Analysis by Polya and Szego).
The 3D domain $\mathcal D$ is defined by the inequalities $$-1\leq x,y,z\leq 1,\quad -\sigma\leq x+y+z\leq \sigma.$$ Show that the volume of $\mathcal D$ is $$\iiint\limits_{\mathcal D}dx\, dy\, dz=\frac{8}{\pi}\int\limits_{-\infty}^{\infty} \left(\frac{\sin t}{t}\right)^3\frac{\sin \sigma t}{t} dt.$$
Hint. Show first that the number of integer lattice points that satisfy the conditions $$-n\leq x,y,z\leq n,\quad -s\leq x+y+z\leq s$$ for some $n$, $s\in\mathbb N$, is equal to $$\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}\left(\frac{\sin \frac{2n+1}{2}t}{sin\frac {t}{2}}\right)^3\frac{\sin \frac{2s+1}{2}t}{\sin \frac{t}{2}} dt.$$
Edit. Check out also the lecture notes by Oliver Knill for some classical examples of volume computation and fabulous illustrations.
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I guess you can find plenty of them in the Problem Department of the American Mathematical Monthly. For instance, here is a nice one by M. Hajja and P. Walker:
Evaluate $\displaystyle \int_{0}^{1}\int_{0}^{1} \int_{0}^{1} (1+u^{2}+v^{2}+w^{2})^{-2} du dv dw$.
Good thing about this one is that there may exist a CAS out there that does not give you the answer at once.
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Indeed, Mathematica 7 will handle the integral in $u$, and then in $v$, but can't handle the resulting integral in $w$. – Kevin O'Bryant Jul 6 2010 at 20:41
Not suitable for any student, but I can't resist giving it:
$\int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \frac{dxdydz}{3-\cos{x}-\cos{y}-\cos{z}} = \frac{\sqrt{6}}{96}\Gamma(\tfrac{1}{24})\Gamma(\tfrac{5}{24})\Gamma(\tfrac{7}{24})\Gamma(\tfrac{11}{24})$,
proven by G. N. Watson in 1939.
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Is the formula a special case of a general phenomenon? – Mariano Suárez-Alvarez Jun 3 2010 at 3:28
I don't believe so. – David Hansen Jun 3 2010 at 3:30
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It is! This integral appears in recent study of random walks, and there a more general integral is given. Don't ask me details; they are in recent papers of Jon Borwein on random walks, which are available from his site. – Wadim Zudilin Jun 3 2010 at 4:40
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Indeed, this integral gives (up to some constants) the probability that a 3-dimensional random walks ever returns to the starting point. The famous result of Polya is that this probability is 1 for dimensions 1,2 and strictly less than 1 for dimensions 3 and over. This fact is related to the fact that the values analogous integral for two variables ($\frac{1}{2-\cos(x)-\cos(y)}$) or one variable $\frac{1}{1-\cos(x)}$ are infinite. mathworld.wolfram.com/… – Guy Katriel Jun 3 2010 at 7:23
@Wadim: What I meant was, I don't know of any more general family of integrals this one fits into, which can be evaluated in terms of gamma functions. – David Hansen Jun 3 2010 at 16:28
A good example that my Honors Calc prof gave me is the following: $$\zeta(n) = \int_{[0,1]^n}\frac{d\boldsymbol x}{1-x_1\dotsm x_n}$$ The proof is an easy induction on $n$.
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I don't think you really need induction... after all, you have $\int_{[0,1]^n} \left( 1 - x_1 \cdots x_n \right)^{-1}dx_k = \int \sum_{i=0}^{\infty} (x_1 \cdots x_n)^i dx_k = \ldots = \sum_{i=1}^{\infty} 1/i^n$ – Eben Freeman Jul 6 2010 at 22:22
Well anywhere you see products or sums of arbitrarily many variables (let alone multiple integrals of arbitrary multiplicity) there's an implicit use of induction to define and manipulate them... – Noam D. Elkies Oct 19 at 14:29
This riddle is better phrased in term of double integral but can work for cubes instead of rectangles.
Here is the riddle: Call a rectangle integral if the length of (at least) one of its edges is an integer.
Prove that a rectangle that can be tessellate by integral rectangles if also integral rectangle.
It is a huge hint to say that it has something to do with double integral so you may want to avoid it in the beginning.
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See mathdl.maa.org/images/upload_library/22/Ford/… for more proofs. – Yuval Filmus Aug 11 2010 at 6:27
mathoverflow.net/questions/14212/… – J. H. S. Nov 8 2010 at 1:26
Feynman's formulae for products. $$\iiint\limits_{[0,1]^3}\frac{x_1^2 x_2}{[(a_4-a_3)x_3 x_2x_1+(a_3-a_2)x_2 x_1+ (a_2-a_1)x_1+a_1]^4}dx_1dx_2dx_3=$$ $$=\int\limits_{0}^{1}dx_1\int\limits_{0}^{1-x_1}dx_2\int\limits_{0}^{1-x_1-x_2}dx_3 \frac{1}{[(a_3-a_4)x_3+(a_2-a_4)x_2+(a_1-a_n)x_1+a_4]^4}=\frac{1}{3!a_1a_2a_3a_4}.$$
There are several ways to compute the integrals, the simplest one is probably by induction in dimension (an obvious generalization of the formula holds true for any $n=\dim [0,1]^n$ with the numerator in the first integral being $x_1^{n-1}x_2^{n-2}\dots x_{n-1}$.)
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The volume of the real flag manifold $Fl_R^3 = O(3)/Z_2^3$ can be obtained on one hand by the explicit integration on the $O(N)$ invariant volume element on its big cell:
$\int_{-\infty}^{\infty}dx_1\int_{-\infty}^{\infty}dx_2\int_{-\infty}^{\infty}dx_3(1+x_1^2+(x_3-\frac{x_1x_2}{2})^2)^{-1} (1+x_2^2+(x_3+\frac{x_1x_2}{2})^2)^{-1}$
This integral is not hard to evaluate using elementary integration techniques. The result $2\pi^2$ can also be obtained from: $Vol(Fl_R^3)= Vol(RP^1) Vol(RP^2) = \frac{Vol(S^1)}{2}\frac{Vol(S^2)}{2}$
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http://mathoverflow.net/questions/6186?sort=newest | ## Counting solutions to x^{p+1}=y^4 in a finite field
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I need to compute the number of solutions to the equation $x^{p+1} = y^4$ in the field with $p^2$ elements (for p sufficiently large). The form of the equation suggests to me that the solution would depend on the congruence class of p mod 4, but I have reason to believe that the answer is a single polynomial in p.
I feel as if this should be easy, and I'm missing an obvious approach. Can anyone help me out?
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"Reason to believe the answer is a single polynomial in p": What's the conjectured polynomial, or is there not an explicit one? That might help, and certainly couldn't hurt. (Well, it could hurt. But it's more likely to help.) – Harrison Brown Nov 19 2009 at 23:06
This is a small piece of a larger calculation in which I'm trying to understand the action of a particular finite group on the intersection cohomology of a particular complex variety. The intersection cohomology is pure and concentrated in even degrees, so this computation is amenable to finite field techniques. The fact that the answer to my question (as Sivek points out) DOES depend on the residue of p mod 4 means either that this dependence will be magically cancelled out by another part of the calculation, or that I've made a mistake earlier on. – Nicholas Proudfoot Nov 20 2009 at 0:23
The [finite-fields] tag is appropriate here! – Sonia Balagopalan Nov 21 2009 at 13:16
## 1 Answer
Let g be a generator of the multiplicative group of the field; assuming x and y are nonzero, we can write x=ga and y=gb with 0 <= a,b < p2-1, and then xp+1=y4 becomes ga(p+1)=g4b, or equivalently a(p+1) = 4b (mod p2-1).
From this we see that p+1 | 4b is necessary, and if 4b=k(p+1) then (a,b) gives a solution iff a=k (mod p-1). Since a can range from 0 to p2-2, then, there are either 0 solutions or p+1 solutions for any fixed b. The total number of nonzero solutions is therefore (p+1)* #{b | p+1 divides 4b}, and then (x,y)=(0,0) is the remaining solution.
Now if p is 1 (mod 4) we have p+1 | 4b iff b is a multiple of (p+1)/2, and there are 2(p-1) such b up to p2-1, so there are 2(p-1)(p+1)+1 = 2p2-1 solutions.
On the other hand, if p is 3 (mod 4) then p+1 | 4b iff b is a multiple of (p+1)/4, so we have 4(p-1) such b and there are 4p2-3 solutions.
- | {"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": 0.9265767335891724, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/101851/generating-a-sigma-algebra | # Generating a $\sigma$-algebra [duplicate]
Possible Duplicate:
Preimage of generated $\sigma$-algebra
I wish to prove the following:
"Let $X$ be a set and $\mathcal{A}$ a family of subsets of $X$, and $\Sigma_{\mathcal{A}}$ the $\sigma$-algebra of subsets of $X$ generated by $\mathcal{A}$. Suppose that $Y$ is another set and $f : Y \rightarrow X$ a function. Then $\left\{f^{-1} \left[E \right] : E \in \Sigma_{\mathcal{A}} \right\}$ is the $\sigma$-algebra of subsets of $Y$ generated by $\left\{ f^{-1} \left[A\right] : A \in \mathcal{A} \right\}$."
I understand that the $\sigma$-algebra of subsets of $Y$ generated by $\left\{ f^{-1} \left[A\right] : A \in \mathcal{A} \right\}$ is defined to be $$\bigcap \left\{ \Sigma : \Sigma \text{ is a } \sigma \text{-algebra of subsets of }Y, \left\{ f^{-1} \left[A\right] : A \in \mathcal{A} \right\} \subseteq \Sigma\right\}.$$
I'm not sure how this leads to the desired result, though. Any help much appreciated.
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5
Don't you mean $f:Y\to X$, not $f:X\to Y$? – Zev Chonoles♦ Jan 24 '12 at 1:14
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I think the very last $X$ should be a $Y$ too. – cardinal Jan 24 '12 at 1:46
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Your set-up does not make sense as given, since $f^{-1}[E]$ for $E\in\Sigma_A$ does not make sense; because $E\subseteq X$, but $f\colon X\to Y$; that's what Zev is pointing at. – Arturo Magidin Jan 24 '12 at 4:39
Thank you. My apologies. – Harry Williams Jan 25 '12 at 2:02
2
– Byron Schmuland Feb 18 '12 at 23:54
show 4 more comments
## marked as duplicate by Nate Eldredge, t.b., Willie Wong♦Mar 26 '12 at 12:36
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
## 1 Answer
I think the best approach will be to just show inclusion of the two sigma algebras in both directions.
I am studying this kind of material as well at the moment, so my answer needs considered with this caveat in mind. I m sure better answers will be provided in due course !
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Something to consider: What do you mean by "$f$ is a measurable function"? (Indeed, in a sense, that's precisely what you're trying to create here.) :) – cardinal Jan 24 '12 at 2:23
I think what you are saying is that from the way we define the domain of $\quad f \quad$ it will necessarily be a measurable function right ? Bit new to this material, so just wondering whether I understand you correctly. (post amended accordingly...) – Beltrame Jan 24 '12 at 2:32
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Yes, essentially. – cardinal Jan 24 '12 at 2:44
I understand $f:D \rightarrow \mathbb{R}$ to be a measurable function if any, or equivalently all, of the following are true: (i) ${x : f(x) < a} \in \Sigma_D$ for every $a \in \mathbb{R}$; (ii) ${x : f(x) \leq a} \in \Sigma_D$ for every $a \in \mathbb{R}$; (iii) (i) ${x : f(x) > a} \in \Sigma_D$ for every $a \in \mathbb{R}$; (iv) (i) ${x : f(x) \geq a} \in \Sigma_D$ for every $a \in \mathbb{R}$, where $D \subseteq X$ and $\Sigma_D$ is the subspace $\sigma$-algebra of subsets of $D$. – Harry Williams Feb 6 '12 at 22:12
At least, this is the technical definition in my lecture notes; I can see there are some pre-images here, but they seem more specific than those in the problem here, so I'm not sure how it helps? – Harry Williams Feb 6 '12 at 22:17 | {"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": 0.9364172220230103, "perplexity_flag": "head"} |
http://mathhelpforum.com/pre-calculus/209649-need-help-exponential-equations.html | # Thread:
1. ## Need help with exponential equations
2x + 2x+1 + 2x+2 = 56
The solution is the following:
2x + 2 * 2x+ 4 * 2x = 56
Why is that? I don't understand how it came to this. Where did the 4 come from? How come there's so many 2s?
7 * 2x = 56
2x = 8
How did 56 turn into 8? Shouldn't it be 82 instead of just 8?
2x = 23
x = 3
I only understand the last two steps. Please help, I need to know this (and more) for a test. This isn't the only equation that confuses me, but it would help to know what lead to the final solution.
thanks.
2. ## Re: Need help with exponential equations
We are given:
$2^x+2^{x+1}+2^{x+2}=56$
Using the property of exponents $a^b\cdot a^c=a^{b+c}$ we may rewrite the equation as:
$2^x+2^1\cdot2^x+2^2\cdot2^x=56$
$2^x+2\cdot2^x+4\cdot2^x=56$
$2^x(1+2+4)=56$
$7\cdot2^x=56$
Divide through by 7:
$2^x=8=2^3$
Hence:
$x=3$
3. ## Re: Need help with exponential equations
Thanks, now I know how the conversion to multiplying occured.
However, I don't understand this part:
$2^x(1+2+4)=56$
Wouldn't
$2^x+2\cdot2^x+4\cdot2^x=56$
be 2x + 4x + 8x = 56?
How did you come up with the "1"?
Also, about dividing through by 7: 7 is not their common denominator - 2 is not in seven (at least not without a decimal). How does it stay unchanged? Is it because of the X?
Thanks
4. ## Re: Need help with exponential equations
Originally Posted by dirtyharry
...
I don't understand this part:
$2^x(1+2+4)=56$
Wouldn't
$2^x+2\cdot2^x+4\cdot2^x=56$
be 2x + 4x + 8x = 56?
How did you come up with the "1"?
Note that:
$2^x+2\cdot2^x+4\cdot2^x=56$
is not the same as:
$2^x+(2\cdot2)^x+(4\cdot2)^x=56$
To make things simpler to see, let $u=2^x$ then we have:
$u+2u+4u=56$
Now combine like terms on the left:
$7u=56$
Divide both sides by 7:
$u=8$
Back-substitute for $u$:
$2^x=8=2^3$
You see, when I factored in my first post, I used the fact that each term on the left has $2^x$ as a factor.
5. ## Re: Need help with exponential equations
Originally Posted by MarkFL2
Note that:
$2^x+2\cdot2^x+4\cdot2^x=56$
is not the same as:
$2^x+(2\cdot2)^x+(4\cdot2)^x=56$
To make things simpler to see, let $u=2^x$ then we have:
$u+2u+4u=56$
Now combine like terms on the left:
$7u=56$
Divide both sides by 7:
$u=8$
Back-substitute for $u$:
$2^x=8=2^3$
You see, when I factored in my first post, I used the fact that each term on the left has $2^x$ as a factor.
Thanks | {"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": 30, "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": 0.9443206787109375, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showthread.php?s=f0c0b1cffc26d4e3c9eedaf470b07057&p=4333562 | Physics Forums
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## problems with calclulating power
By the way seeing this, How could that be that the arm multiplier is less than 1, so the force we apply with an arm less long than 1m will be gone? Interesting world is the world of physics... What a dumbium.
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Quote by losbellos hej, well, őőőőő I dont what to say. again as you state, tangential speed is equal to ω x r this the velocity. - ω x r what is the power. Power is Force x velocity. true BUT HEJ THERE IS AN ARM THERE!!!! SO THE FORCE WILL BE... So power is P = Force x arm length < (BECAUSE YOU CANT TAKE THIS AWAY FROM IT) x ω x r . If it would be force only then you would disregard the arm... But we cant do that.... Because it is an arm and it makes the force much stronger on the smaller circle. Nobody doubts it I hope! So this is why i don't understand why we say p = T x ω ... You understand my problem with this? Force acting there call it torque or not it forces the cylinder to rotate... Force basically travels there and keeps the cylinder rotating with certain velocity.
You are wrong. There isn't much discussion about it. The force is the same whether or not there is an arm. The torque will be different depending on the length of the arm. If I apply a force of 1N to a lever, it is 1N no matter where I push. Depending on how far from the pivot I push, it will have a different torque, but the force is 1N. Torque is not force. We are not calling a force a torque for fun. We are calling torque a torque. Spend a little time thinking/reading about torque and then come back with questions.
As a side note, I feel like your tone is little too confident since nobody agrees with you and you don't have any sources. I think most people are more likely to discuss this if you ask questions without assuming that you are correct.
got no clue what to say, when I think that you guys acknowledge it that If I have an arm and use my muscles on the end and the force will be less on the other end of the arm because its is less than 1 meter.. then I am sorry, must be some sort of planetary influence taking over the physics forum.. I am sorry I find an other community. take care!!!
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Quote by losbellos got no clue what to say, when I think that you guys acknowledge it that If I have an arm and use my muscles on the end and the force will be less on the other end of the arm because its is less than 1 meter.. then I am sorry, must be some sort of planetary influence taking over the physics forum.. I am sorry I find an other community. take care!!!
its unfortunate that you wont accept the facts :(
you are not going to find any different answer on any other reputable physics site
as we all live in the same universe
Dave
Okay, let me try to make this clearer (please excuse typos, I've had a few beers) Consider a lever. There is a pivot at $x=0$ and the arm of the lever is 2m long. Assume that the lever extends from zero to $x=2$. Now, lets say we have a mechanism of some sort that can exert a force of 1N. If we apply that at the mid point (always at right angles to the arm in order to keep this as simple as possible), the torque will be $\tau=1N\cdot1m=1Nm$. If you measure the force at the end of the lever, the TORQUE will be the same and therefore $F=\frac{\tau}{r}=\frac12N$. So the measured force is less, but the torque, which is NOT the same as the force, is the same. Now we want to consider the power. To keep things simple, lets assume that applying the force of 1N allows this lever to move at a constant velocity $v=1m/s$ (maybe it is in some viscous liquid or whatever, the specifics aren't important). So we apply this force and we find that $P=Fv=1N\cdot 1m/s=1W$. Since the torque and angular velocity are constant and defined as $\tau=Fr,\ \omega=\frac{v}{r}$, this is the same as saying $P=\tau\omega=1Nm\cdot 1s^{-1}$. If somebody else attempts to extract energy at the far end of this lever, by my calculations the power will be $P=\frac12\cdot 2m/s=1W$ since the force is one half that of the midpoint and the velocity (linear) is twice that of the midpoint. This clearly shows that the work extracted from the end point in any given time interval is equal to the work put in. By your calculations, the power is $P=\tau r\omega$ which gives the same power as input but at the end of the lever $P=1Nm\cdot 2\cdot 1s^{-1}=2W$. So, by your calculation, every second this simple lever allows one to freely extract twice as much energy as one puts in. This would solve a lot of problems if it were true, but it is not.
The best way to see this (IMO) is this: If you accept (or look up a proof) that rotational K.E. is given by $0.5Iω^{2}$, and that $T = I dw/dt$. If you think about the change in rotational K.E. : $d(0.5Iω^{2})/dt = Iω dw/dt = Tω$
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http://en.wikipedia.org/wiki/Law_of_large_numbers | # Law of large numbers
An illustration of the law of large numbers using a particular run of rolls of a single die. As the number of rolls in this run increases, the average of the values of all the results approaches 3.5. While different runs would show a different shape over a small number of throws (at the left), over a large number of rolls (to the right) they would be extremely similar.
In probability theory, the law of large numbers (LLN) is a theorem that describes the result of performing the same experiment a large number of times. According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed.
The LLN is important because it "guarantees" stable long-term results for the averages of random events. For example, while a casino may lose money in a single spin of the roulette wheel, its earnings will tend towards a predictable percentage over a large number of spins. Any winning streak by a player will eventually be overcome by the parameters of the game. It is important to remember that the LLN only applies (as the name indicates) when a large number of observations are considered. There is no principle that a small number of observations will coincide with the expected value or that a streak of one value will immediately be "balanced" by the others. See the Gambler's fallacy.
## Examples
For example, a single roll of a six-sided die produces one of the numbers 1, 2, 3, 4, 5, or 6, each with equal probability. Therefore, the expected value of a single die roll is
$\tfrac{1+2+3+4+5+6}{6} = 3.5.$
According to the law of large numbers, if a large number of six-sided dice are rolled, the average of their values (sometimes called the sample mean) is likely to be close to 3.5, with the accuracy increasing as more dice are rolled.
It follows from the law of large numbers that the empirical probability of success in a series of Bernoulli trials will converge to the theoretical probability. For a Bernoulli random variable, the expected value is the theoretical probability of success, and the average of n such variables (assuming they are independent and identically distributed (i.i.d.)) is precisely the relative frequency.
For example, a fair coin toss is a Bernoulli trial. When a fair coin is flipped once, the theoretical probability that the outcome will be heads is equal to 1/2. Therefore, according to the law of large numbers, the proportion of heads in a "large" number of coin flips "should be" roughly 1/2. In particular, the proportion of heads after n flips will almost surely converge to 1/2 as n approaches infinity.
Though the proportion of heads (and tails) approaches 1/2, almost surely the absolute (nominal) difference in the number of heads and tails will become large as the number of flips becomes large. That is, the probability that the absolute difference is a small number, approaches zero as the number of flips becomes large. Also, almost surely the ratio of the absolute difference to the number of flips will approach zero. Intuitively, expected absolute difference grows, but at a slower rate than the number of flips, as the number of flips grows.
## History
Diffusion is an example of the law of large numbers, applied to chemistry. Initially, there are solute molecules on the left side of a barrier (purple line) and none on the right. The barrier is removed, and the solute diffuses to fill the whole container.
Top: With a single molecule, the motion appears to be quite random.
Middle: With more molecules, there is clearly a trend where the solute fills the container more and more uniformly, but there are also random fluctuations.
Bottom: With an enormous number of solute molecules (too many to see), the randomness is essentially gone: The solute appears to move smoothly and systematically from high-concentration areas to low-concentration areas. In realistic situations, chemists can describe diffusion as a deterministic macroscopic phenomenon (see Fick's laws), despite its underlying random nature.
The Italian mathematician Gerolamo Cardano (1501–1576) stated without proof that the accuracies of empirical statistics tend to improve with the number of trials.[1] This was then formalized as a law of large numbers. A special form of the LLN (for a binary random variable) was first proved by Jacob Bernoulli.[2] It took him over 20 years to develop a sufficiently rigorous mathematical proof which was published in his Ars Conjectandi (The Art of Conjecturing) in 1713. He named this his "Golden Theorem" but it became generally known as "Bernoulli's Theorem". This should not be confused with the principle in physics with the same name, named after Jacob Bernoulli's nephew Daniel Bernoulli. In 1837, S.D. Poisson further described it under the name "la loi des grands nombres" ("The law of large numbers").[3][4] Thereafter, it was known under both names, but the "Law of large numbers" is most frequently used.
After Bernoulli and Poisson published their efforts, other mathematicians also contributed to refinement of the law, including Chebyshev,[5] Markov, Borel, Cantelli and Kolmogorov and Khinchin, who finally provided a complete proof of the LLN for arbitrary random variables. These further studies have given rise to two prominent forms of the LLN. One is called the "weak" law and the other the "strong" law. These forms do not describe different laws but instead refer to different ways of describing the mode of convergence of the cumulative sample means to the expected value, and the strong form implies the weak.
## Forms
Two different versions of the Law of Large Numbers are described below; they are called the Strong Law of Large Numbers, and the Weak Law of Large Numbers. Both versions of the law state that – with virtual certainty – the sample average
$\overline{X}_n=\frac1n(X_1+\cdots+X_n)$
converges to the expected value
$\overline{X}_n \, \to \, \mu \qquad\textrm{for}\qquad n \to \infty$
where X1, X2, ... is an infinite sequence of i.i.d. integrable random variables with expected value E(X1) = E(X2) = ...= µ. Integrability means that E(|Xj|) < ∞ for j=1,2,....
An assumption of finite variance Var(X1) = Var(X2) = ... = σ2 < ∞ is not necessary. Large or infinite variance will make the convergence slower, but the LLN holds anyway. This assumption is often used because it makes the proofs easier and shorter.
The difference between the strong and the weak version is concerned with the mode of convergence being asserted. For interpretation of these modes, see Convergence of random variables.
### Weak law
Simulation illustrating the Law of Large Numbers. Each frame, you flip a coin that is red on one side and blue on the other, and put a dot in the corresponding column. A pie chart shows the proportion of red and blue so far. Notice that the proportion varies a lot at first, but gradually approaches 50%.
The weak law of large numbers states that the sample average converges in probability towards the expected value[6][proof]
$\overline{X}_n\ \xrightarrow{P}\ \mu \qquad\textrm{when}\ n \to \infty.$
That is to say that for any positive number ε,
$\lim_{n\to\infty}\Pr\!\left(\,|\overline{X}_n-\mu| > \varepsilon\,\right) = 0.$
Interpreting this result, the weak law essentially states that for any nonzero margin specified, no matter how small, with a sufficiently large sample there will be a very high probability that the average of the observations will be close to the expected value; that is, within the margin.
Convergence in probability is also called weak convergence of random variables. This version is called the weak law because random variables may converge weakly (in probability) as above without converging strongly (almost surely) as below.
### Strong law
The strong law of large numbers states that the sample average converges almost surely to the expected value[7]
$\overline{X}_n\ \xrightarrow{a.s.}\ \mu \qquad\textrm{when}\ n \to \infty.$
That is,
$\Pr\!\left( \lim_{n\to\infty}\overline{X}_n = \mu \right) = 1.$
The proof is more complex than that of the weak law.[8] This law justifies the intuitive interpretation of the expected value of a random variable as the "long-term average when sampling repeatedly".
Almost sure convergence is also called strong convergence of random variables. This version is called the strong law because random variables which converge strongly (almost surely) are guaranteed to converge weakly (in probability). The strong law implies the weak law.
The strong law of large numbers can itself be seen as a special case of the pointwise ergodic theorem.
Moreover, if the summands are independent but not identically distributed, then
$\overline{X}_n - \operatorname{E}\big[\overline{X}_n\big]\ \xrightarrow{a.s.}\ 0$
provided that each Xk has a finite second moment and
$\sum_{k=1}^{\infty} \frac{1}{k^2} \operatorname{Var}[X_k] < \infty.$
This statement is known as Kolmogorov's strong law, see e.g. Sen & Singer (1993, Theorem 2.3.10).
### Differences between the weak law and the strong law
The weak law states that for a specified large n, the average $\overline{X}_n$ is likely to be near μ. Thus, it leaves open the possibility that $|\overline{X}_n -\mu| > \varepsilon$ happens an infinite number of times, although at infrequent intervals.
The strong law shows that this almost surely will not occur. In particular, it implies that with probability 1, we have that for any ε > 0 the inequality $|\overline{X}_n -\mu| < \varepsilon$ holds for all large enough n.[9]
### Uniform law of large numbers
Suppose f(x,θ) is some function defined for θ ∈ Θ, and continuous in θ. Then for any fixed θ, the sequence {f(X1,θ), f(X2,θ), …} will be a sequence of independent and identically distributed random variables, such that the sample mean of this sequence converges in probability to E[f(X,θ)]. This is the pointwise (in θ) convergence.
The uniform law of large numbers states the conditions under which the convergence happens uniformly in θ. If[10][11]
1. Θ is compact,
2. f(x,θ) is continuous at each θ ∈ Θ for almost all x’s, and measurable function of x at each θ.
3. there exists a dominating function d(x) such that E[d(X)] < ∞, and
$\left\| f(x,\theta) \right\| \leq d(x) \quad\text{for all}\ \theta\in\Theta.$
Then E[f(X,θ)] is continuous in θ, and
$\sup_{\theta\in\Theta} \left\| \frac1n\sum_{i=1}^n f(X_i,\theta) - \operatorname{E}[f(X,\theta)] \right\| \xrightarrow{\mathrm{a.s.}} \ 0.$
### Borel's law of large numbers
Borel's law of large numbers, named after Émile Borel, states that if an experiment is repeated a large number of times, independently under identical conditions, then the proportion of times that any specified event occurs approximately equals the probability of the event's occurrence on any particular trial; the larger the number of repetitions, the better the approximation tends to be. More precisely, if E denotes the event in question, p its probability of occurrence, and Nn(E) the number of times E occurs in the first n trials, then with probability one,
$\frac{N_n(E)}{n}\to p\text{ as }n\to\infty.\,$
Chebyshev's Lemma. Let X be a random variable with finite expected value μ and finite non-zero variance σ2. Then for any real number k > 0,
$\Pr(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}.$
This theorem makes rigorous the intuitive notion of probability as the long-run relative frequency of an event's occurrence. It is a special case of any of several more general laws of large numbers in probability theory.
## Proof
Given X1, X2, ... an infinite sequence of i.i.d. random variables with finite expected value E(X1) = E(X2) = ... = µ < ∞, we are interested in the convergence of the sample average
$\overline{X}_n=\tfrac1n(X_1+\cdots+X_n).$
The weak law of large numbers states:
Theorem: $\overline{X}_n \, \xrightarrow{P} \, \mu \qquad\textrm{for}\qquad n \to \infty.$
### Proof using Chebyshev's inequality
This proof uses the assumption of finite variance $\operatorname{Var} (X_i)=\sigma^2$ (for all $i$). The independence of the random variables implies no correlation between them, and we have that
$\operatorname{Var}(\overline{X}_n) = \operatorname{Var}(\tfrac1n(X_1+\cdots+X_n)) = \frac{1}{n^2} \operatorname{Var}(X_1+\cdots+X_n) = \frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n}.$
The common mean μ of the sequence is the mean of the sample average:
$E(\overline{X}_n) = \mu.$
Using Chebyshev's inequality on $\overline{X}_n$ results in
$\operatorname{P}( \left| \overline{X}_n-\mu \right| \geq \varepsilon) \leq \frac{\sigma^2}{n\varepsilon^2}.$
This may be used to obtain the following:
$\operatorname{P}( \left| \overline{X}_n-\mu \right| < \varepsilon) = 1 - \operatorname{P}( \left| \overline{X}_n-\mu \right| \geq \varepsilon) \geq 1 - \frac{\sigma^2}{n \varepsilon^2 }.$
As n approaches infinity, the expression approaches 1. And by definition of convergence in probability, we have obtained
$\overline{X}_n \, \xrightarrow{P} \, \mu \qquad\textrm{for}\qquad n \to \infty.$
### Proof using convergence of characteristic functions
By Taylor's theorem for complex functions, the characteristic function of any random variable, X, with finite mean μ, can be written as
$\varphi_X(t) = 1 + it\mu + o(t), \quad t \rightarrow 0.$
All X1, X2, ... have the same characteristic function, so we will simply denote this φX.
Among the basic properties of characteristic functions there are
$\varphi_{\frac 1 n X}(t)= \varphi_X(\tfrac t n) \quad \text{and} \quad \varphi_{X+Y}(t)=\varphi_X(t) \varphi_Y(t) \quad$ if X and Y are independent.
These rules can be used to calculate the characteristic function of $\scriptstyle\overline{X}_n$ in terms of φX:
$\varphi_{\overline{X}_n}(t)= \left[\varphi_X\left({t \over n}\right)\right]^n = \left[1 + i\mu{t \over n} + o\left({t \over n}\right)\right]^n \, \rightarrow \, e^{it\mu}, \quad \text{as} \quad n \rightarrow \infty.$
The limit eitμ is the characteristic function of the constant random variable μ, and hence by the Lévy continuity theorem, $\scriptstyle\overline{X}_n$ converges in distribution to μ:
$\overline{X}_n \, \xrightarrow{\mathcal D} \, \mu \qquad\text{for}\qquad n \to \infty.$
μ is a constant, which implies that convergence in distribution to μ and convergence in probability to μ are equivalent (see Convergence of random variables.) Therefore,
$\overline{X}_n \, \xrightarrow{P} \, \mu \qquad\text{for}\qquad n \to \infty.$
This shows that the sample mean converges in probability to the derivative of the characteristic function at the origin, as long as the latter exists.
## Notes
1. Mlodinow, L. The Drunkard's Walk. New York: Random House, 2008. p. 50.
2. Jakob Bernoulli, Ars Conjectandi: Usum & Applicationem Praecedentis Doctrinae in Civilibus, Moralibus & Oeconomicis, 1713, Chapter 4, (Translated into English by Oscar Sheynin)
3. Poisson names the "law of large numbers" (la loi des grands nombres) in: S.D. Poisson, Probabilité des jugements en matière criminelle et en matière civile, précédées des règles générales du calcul des probabilitiés (Paris, France: Bachelier, 1837), page 7. He attempts a two-part proof of the law on pages 139-143 and pages 277 ff.
4. Hacking, Ian. (1983) "19th-century Cracks in the Concept of Determinism", Journal of the History of Ideas, 44 (3), 455-475 JSTOR 2709176
5. Tchebichef, P. (1846). "Démonstration élémentaire d'une proposition générale de la théorie des probabilités". Journal für die reine und angewandte Mathematik (Crelles Journal) 1846 (33): 259–267. doi:10.1515/crll.1846.33.259.
6. "The strong law of large numbers « What’s new". Terrytao.wordpress.com. Retrieved 2012-06-09.
7. Jennrich, Robert I. (1969). "Asymptotic Properties of Non-Linear Least Squares Estimators". The Annals of Mathematical Statistics 40 (2): 633–643. doi:10.1214/aoms/1177697731.
## References
• Grimmett, G. R. and Stirzaker, D. R. (1992). Probability and Random Processes, 2nd Edition. Clarendon Press, Oxford. ISBN 0-19-853665-8.
• Richard Durrett (1995). Probability: Theory and Examples, 2nd Edition. Duxbury Press.
• Martin Jacobsen (1992). Videregående Sandsynlighedsregning (Advanced Probability Theory) 3rd Edition. HCØ-tryk, Copenhagen. ISBN 87-91180-71-6.
• Loève, Michel (1977). Probability theory 1 (4th ed.). Springer Verlag.
• Newey, Whitney K.; McFadden, Daniel (1994). Large sample estimation and hypothesis testing. Handbook of econometrics, vol.IV, Ch.36. Elsevier Science. pp. 2111–2245.
• Ross, Sheldon (2009). A first course in probability (8th ed.). Prentice Hall press. ISBN 978-0-13-603313-4.
• Sen, P. K; Singer, J. M. (1993). Large sample methods in statistics. Chapman & Hall, Inc. | {"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": 33, "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": 0.8955804705619812, "perplexity_flag": "head"} |
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http://mathoverflow.net/questions/77209/eigenvalue-densities-of-sample-covariance-matrices-when-the-population-covariance/78843 | ## Eigenvalue densities of sample covariance matrices when the population covariance matrix is a perturbed identity matrix
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
TLDR: I'm looking for a random matrix theory reference for the eigenvalue densities of sample covariance matrices (both dimensions approaching infinity at the same rate) when the true (population) covariance matrix is a perturbed form of the identity matrix.
Consider an $N\times M$ matrix $X$, with elements in each column $X_{.\ j}\sim\mathcal{N}(\mathbf{0},\mathbf{I})$. Random matrix theory tells us that in the limit $N,M\to\infty,\ N/M\to\gamma$, the eigenvalue density of the sample covariance matrix
$$R=\frac{1}{M}XX^T$$
is given by the Marchenko-Pastur density.
Now consider an $N\times M$ matrix $Y$, with elements in each column $Y_{.\ j}\sim\mathcal{N}(\mathbf{0},\mathbf{C})$. Let the $N$ eigenvalues of $\mathbf{C}$ (assume diagonal) be $\lambda_i=1+v_i,\ i=\lbrace 1,\ldots,N\rbrace$ where $v_i$ is some zero-mean perturbation, i.e. $\sum_i v_i=0$. So $\mathbf{C}$ can be considered a perturbed identity matrix. As long as the perturbation is "not large", the eigenvalue density of the sample covariance matrix $$S=\frac{1}{M}YY^T$$ should be "loosely" described by the same Marchenko-Pastur distribution.
To illustrate, here's a quick MATLAB example for $N=1000$, $N/M=\gamma=1/4$:
````X=randn(1e3,4e3);
C=eye(1000)+0.1*diag(randn(1000,1));
A=sqrt(C);
S=(A*X)*(A*X)'/4e3;
[g,z]=hist(eig(S),linspace(0,3,100));
plot(z,g/trapz(z,g),'r--');hold on
mp=@(x)real(sqrt((x-0.25).*(2.25-x))./(0.5*pi*x));
x=linspace(0,3,100);
plot(x,mp(x),'k');hold off
````
The plot on the left below shows the eigenvalues of $\mathbf{C}$ (black solid line shows eigenvalues of $\mathbf{I}$) and the plot on the right shows the eigenvalue density of $S$ (dashed red) and the asymptotic density of $R$ (solid black).
So you can fiddle around and see that as long as the perturbation is "small" and $\gamma$ is not close to zero, the density will still be given by the Marchenko-Pastur density.
What are some canonical references that deal with this, so that I can properly reference it in my article?
One paper that kind of deals with a similar issue is
J. Baik, G. Ben Arous and S. Péché, "Phase transition of the largest eigenvalue for nonnull complex sample covariance matrices", Ann. Probab. vol:33, no:5, 2005
The authors show there that if the identity matrix is "spiked" (i.e., one eigenvalue larger than the others), then depending on the magnitude of spiking, the largest eigenvalue of the sample covariance matrix may or may not fall out of the Marchenko-Pastur bounds. I guess some heuristic argument can be applied to my case here by considering all the perturbations as small positive and negative spikes, but I'm looking for existing literature that deals with it.
-
## 2 Answers
You can rewrite $$S=\frac{1}{M}C^{1/2}XX^T(C^{1/2})^T$$ where $X_{.j}\sim \mathcal{N}(\boldsymbol 0, \boldsymbol I)$. It is then classical that the limiting eigenvalue distribution of $S$ (in the almost sure weak convergence) is given by $MP_\gamma\boxtimes\nu$, the free multiplicative convolution of the Marchenko-Pastur distribution $MP_\gamma$ of paramter $\gamma$ with the limiting eigenvalue distribution $\nu$ of $C$. Thus, you indeed obtain $MP_\gamma$ if and only if $\nu=\delta_1$, namely if and only if $$\frac{1}{N}\sum_{i=1}^N\delta_{v_i} \longrightarrow \delta_0\qquad \mbox{weakly as $N\rightarrow\infty$ }.$$ For a reference, you can look at the book of Anderson, Guionnet and Zeitouni, "Introduction to random matrices" (Chapter 5), but also any introduction to free probability I guess.
As you can see, free probability provides a powerful language to describe such perturbed random matrix models!
NB : As an example, a spiked model is the situation where only a fixed finite number of $v_i$'s are non zero, which fit with the above characterization.
-
- | {"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": 30, "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": 0.8298243880271912, "perplexity_flag": "head"} |
http://motls.blogspot.com.au/2012/07/mostly-good-news-on-fermi-130gev-line.html?m=1 | # The Reference Frame
Our stringy Universe from a conservative viewpoint
## Friday, July 20, 2012
### Mostly good news on the Fermi $$130\GeV$$ line
Dark matter may be starting to shine in front of our eyes
Someone at the Physics Stack Exchange asked about the recent status of the $$130\GeV$$ line located in the Fermi gamma-ray data by Christoph Weniger in April 2012 (a confirmation).
It's one of the most exciting and potentially emerging signals of new physics – signals of dark matter. I answered as follows.
Another very fresh paper presented at Dark Attack yesterday, one by Hektor et al.,
arXiv:1207.4466
also claims that the signal is there – not only in the center of the Milky Way but also in other galactic clusters, at the same 130 GeV energy. This 3+ sigma evidence from clusters is arguably very independent. All these hints and several additional papers of the sort look very intriguing.
There are negative news, too. Fermi hasn't confirmed the "discovery status" of the line yet. Puzzles appear in detailed theoretical investigations, too. Cohen at al.
arXiv:1207.0800
claim that they have excluded neutralino – the most widely believed identity of a WIMP – as the source because the neutralino would lead to additional traces in the data because of processes involving other Standard Model particles and these traces seem to be absent. The WIMP could be a different particle than the supersymmetric neutralino, of course.
Another paper also disfavors neutralino because it is claimed to require much higher cross sections than predicted by SUSY models:
arXiv:1207.4434
But one must be careful and realize that the status of the "5 sigma discovery" here isn't analogous to the Higgs because in the case of the Higgs, the "canonical" null hypothesis without the Higgs is well-defined and well-tested. In this case, the $$130\GeV$$-line-free hypothesis is much more murky. There may still exist astrophysical processes that tend to produce rather sharp peaks around $$130\GeV$$ even though there are no particle species of this mass. I think and hope it is unlikely but it hasn't really been excluded.
Everyone who studies these things in detail may want to look at the list (or contents) of all papers referring to Weniger's original observation – it's currently 33 papers (update on Friday: 36):
Papers referring to Weniger
Addition for TRF readers...
Yesterday, I also wrote in a comment thread that I find it puzzling when there are many independent hints and each of them seems to be around 3 sigma. Given the a priori possible significance level or relative strength of the signal between $$10^{-10}$$ and $$10^{+10}$$, it seems surprising when all the significance levels of relevant observations manage to hit the narrow 3-5 sigma window.
However, when I am thinking about the same signal seen in the different galaxy clusters, I am more impressed by it than I was two days ago. The energy $$130\GeV$$ is very high. If it happened to be a peak of some astrophysical processes, they would probably have to be processes that depend on some typical star size, star density, and other things, and it seems likely that the peak which depends on these astrophysical parameters would be shifted to different energies in different galaxy clusters.
So it seems increasingly likely that some new physics is associated with $$130\GeV$$ photons. If this is where the evidence is going, we must determine what the particle is made of.
Do you remember that we used to call the satellite "GLAST" and then "Fermi, formerly GLAST"? The satellite has brought us so many wonderful insights that we can't even remember when we exactly switched to "Fermi" and internalized this new name. But it's here.
Meanwhile, in the dark matter direct search experiments, a new powerful explosion was detonated by the dark-matter-is-not-seen axis. XENON100 has published its new exclusion limits.
Click to zoom in.
All combinations of (mass, cross section with the nucleon) above the blue line at the bottom – the line inside the green-yellow Brazil band – are said to be forbidden by XENON100's data. This of course strengthens the conflict with the dark-matter-is-seen experiments such as DAMA, CoGeNT, CRESST, and perhaps others that claim that the dark matter apparently does show up and it lives at various islands in the XENON100-forbidden sea.
It's increasingly unclear which side is right. The conflict is getting sharper. The methodologies of all the experiments are inequivalent so it is in principle conceivable that the dark matter particle is behaving in very different ways in various experiments and all of them are compatible with each other. However, it seems more likely that some of the experiments are doing something wrong.
The tension exists not just in between the different dark matter direct search experiments. There is a tension with the 130 GeV Fermi line, too. While XENON100's exclusion curve is just starting to "bite" into the region predicted by various models of a neutralino, these models (and the XENON100-allowed cross sections) seem to predict way too low cross sections to explain the rather strong Fermi line located by Weniger.
You may decide that the right reaction is to dismiss XENON100: there are too many hints that dark matter exists. But that won't really reduce the discrepancies much because the direct dark-matter-is-seen experiments indicate the dark matter particle mass close to $$10\GeV$$ which is much lower than Weniger's $$130\GeV$$. So something could be completely wrong about our models or the experiments – and things could be much more complex or very different from what the experimenters think.
BTW I spent some time reading an impressive proof of the formula for all the tree-level amplitudes in $$\NNN=8$$ SUGRA. Equipped with this proof, it seems increasingly clear to me that this is just a version of the sphere-surface tree-level stringy scattering amplitude that may be extracted from type II string theory with some way to rewrite the degrees of freedom to the twistor variables, ignore some "compactified ones", and using some analytical continuations so that the path integrals get reduced to residues. I would love to see the reduction of the formula to tree-level string theory in some more explicit form but even without that, the status of the twistor formulae looks clearer to me – it's string theory with reparameterized world sheet variables.
#### 15 comments:
1. Dilaton
I already "liked" these good news at Physics SE, it is cool :-)
Rewriting the formula for these tree-level scattering amplitudes according to your ideas could probably lead to an awsome paper ... ;-) ?
2. David Nataf
Thanks for the post.
First, I want to say, that the signal can't be 20,000 sigma, I disagree :-) It is strictly between 0 and ~6-7 sigma for all sources, because if it were above ~6-7 sigma it would have been discovered already :-)
Second, something can be learned from the fact it isn't 0 sigma in the clusters. If this were due to dark matter decay rather than dark matter annihalation, the signal would be 0 toward the clusters. That is because the signal from dark matter decay scales as the density, whereas the signal from dark matter annihalation scales as the square of the density -- the latter benefits from substructure. I think decay of dark matter was already disfavoured from theory, but it is still useful to show that this line cannot be due to dark matter decay.
The substructure leads to a predicted parameter called "boosting". Boosting can be predicted from theory if you assume negligible self-interactions in the dark sector. They say that the boosting their result is consistent with is ~1800-3800, and that cosmological models of structure growth predict a boosting factor of around that quantity.
I am sure other astronomers will check this. I'm also cautious and optimistic.
As another note, Galaxy clusters are on average a few thousand times further away than the Galactic Center (2.g. 20 megaparsecs versus 8200 kiloparsecs). They are also a few thousand times more massive, so you'd expect the signal to be 1000x weaker. However, the "Galactic center", represents only a few percent of the signal from the Galaxy, they're using 6 clusters, and the signal to noise for these clusters is ~5x smaller than that for the Galactic center, so it might come close to working out.
Another potential issue is that this mass is very different than the 5 GeV dark matter reported by some of the direct detection experiments. I'm not sure who is more reliable at this point, but in any case I don't think it's certain dark matter has to be a single particle even though it's an unwritten assumption in most papers on the subject.
**
3. dopey_john
Does this have anything to do with the volume of polytopes talked about by Jacob Bourjaily on the Parks Taylor formula after 13:00 in this lecture?
http://tedxtalks.ted.com/video/TEDxUofM-Jacob-Bourjaily-Transf
4. scooby
So there is some good work being done after all the Perimeter Institute..
5. scooby
I meant at the Perimeter Institute. Is it possible to edit the Disqus comments?
6. There's a lot of good work at the Perimeter Institute these days.
I see "edit" before my comments, next to reply. Don't you?
7. gene
When I first became interested in the fundamental laws of physics, some six decades ago, the world was very different. Astronomy (cosmology) was totally empirical and could, in no way, be considered a science; it was observation accompanied only by wild speculation. Now, our observations of the cosmos are adding immensely to our understanding of the basics of our existence. I really think that this is the most exciting period that I have experienced vis-a-vis probing the depths of our knowledge concerning how things really are and how they came to be.
I just don't see how Sabine Hossenfelder and others can be discouraged and pessimistic about what is going on. I have never seen better times than these.
8. I agree, Gene, and I think that even during my childhood, we lived through conditions you suggested.
It would have been normal for kids like me to be really into astronomy. Let me admit, I have never been an astronomy geek. Astronomy looked like botany a little bit. People would be happy about looking at lots of different objects - usually just points - and their random properties that weren't related to anything else, and so on.
Of course, astronomy is a human activity of this kind even today.
On the other hand, there were far-reaching philosophical speculations about universe and God - those things were usually presented together with religious ideas etc. - and they were disconnected from science and a tighter network of ideas, too.
So already as a kid, I was more intrigued by elementary particles and the laws that hold everywhere and that one may study locally. But I learned lots of related things from astronomy-centered articles and TV programs, like from those by Dr Grygar but not only those, just to be sure.
Only in recent decade or so, those things got really connected with each other. Cosmology in particular became a precision science with data whose precision and statistical structure is analogous to the collider data and that are equally tightly and directly connected with some fundamental equations. It's of course exciting. The possible ability to find a complete new dominant kind of matter filling the galactic neighborhoods and possibly create it artificially on Earth, and we may really be 1-2 years from making those things clear, is something that even the famous decades half a century ago we may often be nostalgic about just couldn't have.
These are places where the humans have made some genuine progress and it is fascinating to watch and even more fascinating to be at least a remote part of this process.
9. scooby
I don't but I probably need to register with Disqus, will do that now.
10. scooby
I agree with you and Gene, but I used to think that cosmology became a science almost a century ago, when Friedmann and Lemaitre proposed their model of a hot, expanding universe and Hubble's observations confirmed it (around 1922 - 1930).
Certainly, there was a long wait after during which not much happened (as opposed to the situation in particle physics), until the discovery of the microwave background radiation in 1964.
11. Gene
The transition of cosmology from speculation to science has occurred (and is still occurring) as things get quantitative; i.e. when the numbers actually agree. I did not mean to disparage speculation, which is, necessarily, the start of everything. Besides, it's great fun!
12. David Nataf
When I first learned about the Kerr Metric, my first question was "how come this was only discovered in 1964?". And I got the same answer from multiple physicists -- a lot of people could have solved the problem before, but gravity, cosmology, black holes, etc were not attracting the best minds for a long time. It was an intellectual backwater.
13. OK, the existence of exact solutions like that is fun but I never considered them and I still don't consider them (and their form in particular) too important.
Einstein was satisfied with perturbation theory. He figured out how to derive the Newtonian limit of GR; and he figured out various intrinsic new GR corrections such as the bending of light and precession of Mercury perihelion.
If it were up to folks like Einstein, we would be treating GR in this way maybe even today and we wouldn't really lose that much. Maybe we would be heavily using computers and they would know the Kerr and similar solutions - no one would just realize that they have a simple algebraic form.
Schwarzschild was the guy who started this sport of analytic solutions but of course that it only became a mass sport once people decided it was relevant for something. So they had to really believe that the solutions were physical up to the horizon - and perhaps beyond - and then there was a reason to go through the work of finding the exact solutions.
Today, it's a mass sport to the extent that people are looking for exact solutions even if they know that they're inconsequential pretty much everywhere.
14. Ghandi
I think low energy susy does not exist. More likely it enters only at much higher energies, or only in some dual description of our universe. Which is probably more interesting and fundamental than the way phenomenologists describe it.
15. Cheap RS Gold
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## Who is Lumo?
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http://math.stackexchange.com/questions/258591/hessian-equivalence | Hessian equivalence
Let $F: R^n \longrightarrow R$ be twice differentiable and $x,y \in R^n$ with $F(x)=F(y)$. Further let $\phi [0,1] \rightarrow R^n$ be a nice curve with $\phi(0)=x$ and $\phi(1)=y$. If we know that $F(\phi(t))=F(x)$ for all $t\in [0,1]$. What does this imply for the Hessians at the points $\phi(t)$ where $t \in (0,1)$? Are the equivalent to the Hessian of F at x?
-
1 Answer
Certainly not. The conditions only determine $F$ along the curve, which might be e.g. a straight line. The function can do whatever it wants off the curve. For example, $F(x,y)=x^2y^2$ is constant on the $x$-axis, but $F_{yy}=2x^2$ isn't.
-
Hmm your example $F(x,y)$ is not constant on the x-aches, but on the curve $x\cdot y=1$?! But I guess I see the problem. Thanx! – Benedikt Dec 14 '12 at 10:11
@Benedikt: Assuming that you're referring to the $x$-axis: Sure it's constant there: $F(x,0)=x^2\cdot0^2=0$. There's no contradiction with also being constant on many other curves. – joriki Dec 14 '12 at 10:23
Yes, sorry. I realized this too now when I had my first morning coffee.. ;-) You never should post a question and a comment without coffee – Benedikt Dec 14 '12 at 10:52
But is there something that they still share? Like the Trace or one common eigenvalue? – Benedikt Dec 14 '12 at 10:58
But what happens if I know that the gradient is zero all along the curve? – Benedikt Dec 14 '12 at 19:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "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": 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": 0.9296900033950806, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/211370/linear-combination-of-vectors | # Linear Combination of vectors
I have a previous post here. There is a part b to that question and it asks:
Let $x=(1,1,1)^T$. Write x as a linear combination of $u_1, u_2, u_3$ using Parseval's formula to compute $||x||$.
I know how to compute $||x||$, it's simply the magnitude. However I am totally unsure how do to x as a linear combo, I've read through my book and tried looking online with no luck. Any ideas?
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## 1 Answer
Hint: Since you have orthonormal basis $\{u_1,u_2,u_3\}$, you can express any vector $x$ as a linear combination of the basis vectors as $$x=\alpha_1 u_1+\alpha_2 u_2+\alpha_3 u_3$$ where you have to determine the coefficients (scalars) $\alpha_i$'s.
Since, $u_i$'s are orthonormal, $<u_i,u_j>=u_i^t.u_j=\delta_{ij}$ (=1, only if $i=j$, else $0$). So,
$$x^t.u_i=<x,u_i>=(\sum\alpha_j u_j^t).u_i=\alpha_i$$.
-
you completely lost me after the "Since, $u_i$'s are". Can you please rephrase it a little differently, i'm not getting it – Charlie Yabben Oct 12 '12 at 0:48
Sorry... I assumed you know about inner product, otherwise how could the Perseval's identity come? The $<x,y>$ symbol stands for "inner product of x and y" which, in $\mathbb{R}^3$ can be taken as $x^T.y$. Is it clear now? – Tapu Oct 12 '12 at 0:56
inner product was mentioned in class but I not too familiar with it. But from what I'm reading from the book and online, isn't inner product just a dot product, just a more generalized version. And the book says perseval's formula but i have not idea what that is – Charlie Yabben Oct 12 '12 at 0:59
Inner product and dot product are different terminology of the same thing $<x,y>$. Perseval's formula says $\|x\|^2=\sum\alpha_i^2$ – Tapu Oct 12 '12 at 1:05
okay so the first step I is determine the coefficients $\alpha_i$'s which can be found by doing $<x,u_i>$. Am I right so far? – Charlie Yabben Oct 12 '12 at 1:09
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http://mathoverflow.net/questions/27361/do-actual-sudoku-puzzles-have-a-unique-rational-solution/27514 | ## Do actual Sudoku puzzles have a unique rational solution?
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Here is a question in the intersection of mathematics and sociology. There is a standard way to encode a Sudoku puzzle as an integer programming problem. The problem has a 0-1-valued variable $a_{i,j,k}$ for each triple $1 \le i,j,k \le 9$, expressing that the entry in position $(i,j)$ has value $k$. The Sudoku rules say that four types of 9-sets of the variables sum to 1, to express that each cell is filled with exactly one number, and that each number appears exactly once in each row, column, and $3 \times 3$ box. And in a Sudoku puzzle, some of these variables (traditionally 27 of them) are preset to 1.
It is known that generalized Sudoku, like general integer programming, is NP-hard. However, is that the right model for Sudoku in practice? I noticed that many human Sudokus can all be solved by certain standard tricks, many of which imply a unique rational solution to the integer programming problem. You can find rational solutions with linear programming, and if the rational solution is unique, that type of integer programming problem is not NP-hard, it's in P. Traditionally Sudoku puzzles have a unique solution. All that is meant is a unique integer solution, but maybe the Sudoku community has not explored reasons for uniqueness that would not also imply a unique rational solution.
Are there published human Sudoku puzzles with a unique solution, but more than one rational solution? Is there a practical way to find out? I guess one experiment would be to make such a Sudoku (although I don't know how difficult that is), and then see what happens when you give it to people.
-
3
I've certainly encountered published Sudoku problems that can't be solved using the "standard tricks" and that "require branching" (I deliberately leave the term "require branching" undefined because it's a notoriously slippery concept, but we all know it when we see it). Such Sudoku problems tend to be unpopular but they do exist. Many computer-generated Sudoku puzzles proceed by randomly eliminating entries from a filled grid until the original grid can no longer be uniquely reconstructed. It would be surprising to me if all such puzzles had unique rational solutions. – Timothy Chow Jun 7 2010 at 15:46
We have 729 variables $a_{i,j,k}$. Also we have $4\times81=324$ automatic linear constraints. Then we set a handful of variables to equal $1$ corresponding to the given entries. This is clearly not going to give us enough conditions for a unique solution. But suppose we are more generous: if we are given say $a_{1,1,1}=1$ suppose we are also given $a_{1,1,k}=0$ for $k>1$ and $a_{i,j,1}=0$ when $(i,j)\ne(1,1)$ but is in the same row, column or $3\times 3$ box as $(1,1)$. This gives us $28$ further conditions for each entry, which might be enough. – Robin Chapman Jun 7 2010 at 15:53
I thought of this question when I wrote a Sudoku solver in Python. (I only enjoyed Sudoku for a little while; at some point I decided to try meta-Sudoku.) What I learned was: (1) The solver generally didn't need to "branch" as much as you might think; it actually takes a lot of practice to spot every forced entry. (2) As you suggest, when I added more non-branching inferences, the amount of branching dropped sharply. I think that every inference in my fairly primitive program is of the unique rational type. – Greg Kuperberg Jun 7 2010 at 15:56
1
Vow - quantum sudoku! – Thomas Kragh Jun 7 2010 at 16:25
13
"The theory of Sudoku with $\textit{rational}$ number of rows is well known. Now..." – Victor Protsak Jun 7 2010 at 17:19
show 4 more comments
## 4 Answers
I wasn't planning on answering here but since someone mentioned my paper in the long comment thread above maybe I should anyway.
When I'm solving problems by hand one of the sets of patterns I frequently use involve uniqueness: something can't happen because it would lead to a puzzle with more than one solution, but a well-posed Sudoku puzzle has only one solution, so it's safe to assume that whatever it is doesn't happen. For instance, it's not possible to have four initially-empty cells in a rectangle of the same two rows, the same two columns, and the same two 3x3 squares, and also for these four cells to all contain one of the same two values, because then the two values they contain could be placed in two different ways and the rest of the puzzle wouldn't notice the change. So if there's only one way to prevent a rectangle like this from forming then the solution has to use that one way.
I have the sense (though I could be wrong) that this sort of inference does not imply a unique rational solution. But of course a puzzle could have a unique rational solution anyway — my computer solver knows these rules too, but uses them less frequently than I do by hand.
ETA: It does indeed seem to be true that these rules don't imply unique rational solutions. With them, my solver easily solves the following puzzle, which has a unique integer solution but does not have a unique fractional solution. Without them, my solver can still solve the same puzzle, but only by using rules that are (in my experience) much more difficult to apply by hand.
``` -----------------------------------
| 3 7 8 | 6 4 5 | 1 2 9 |
| | | |
| 6 9 . | . 7 . | 4 8 5 |
| | | |
| 4 . 5 | 9 . 8 | 3 7 6 |
|-----------------------------------|
| 7 . 9 | 5 3 . | 8 6 4 |
| | | |
| 8 4 . | 7 . . | 5 9 3 |
| | | |
| 5 3 6 | 8 9 4 | 2 1 7 |
|-----------------------------------|
| 1 6 3 | 2 5 7 | 9 4 8 |
| | | |
| 9 8 4 | . . . | 7 5 2 |
| | | |
| 2 5 7 | 4 8 9 | 6 3 1 |
-----------------------------------
```
Or if you want something that looks like the puzzles that get published, here's another one that uses the same mechanism:
``` -----------------------------------
| 1 . . | 8 . 7 | 5 . . |
| | | |
| . 5 . | 6 . . | 9 7 . |
| | | |
| . 6 . | . 4 . | . . . |
|-----------------------------------|
| . . 2 | . . . | . 6 . |
| | | |
| 6 . . | 4 . 3 | . . . |
| | | |
| . 4 . | . . . | 1 . . |
|-----------------------------------|
| . . . | . 1 . | . 5 . |
| | | |
| . 1 3 | . . 6 | . 9 . |
| | | |
| . . 7 | 5 . 4 | . . 1 |
-----------------------------------
```
-
So I gather that there is no particular reason to believe that this wouldn't occur in an ordinary newspaper Sudoku? If so, would such a puzzle be likely to be labeled "diabolical", or not necessarily all that likely? – Greg Kuperberg Jun 10 2010 at 4:35
I don't think this one is as hard as the newspaper "diabolical" puzzles. – David Eppstein Jun 10 2010 at 6:48
One could dig deeper into the sociology, but your discovery is already one answer to the question. Fractional Sudoku is a different game than integer Sudoku, not just in principle but at least occasionally in practice. – Greg Kuperberg Jun 10 2010 at 14:09
### 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.
Towards a weak proof of existence, I would go through Gordon Royle's database of Sudoku with a minimum number of givens, and (program a computer to) check for multiple rational solutions on any of the puzzles. If all of them had unique rational solutions, I would take that as strong evidence that all proper Sudoku have unique rational solutions. If one had multiple rational solutions, add some givens until you have a maximal desired puzzle, and publish it (or send it to Nick Baxter et. al. for the sociological answer).
-
Not an answer, just wanted to point to some references. David Eppstein has an arXiv piece on a clever "human-type" trick, I imagine he will post something for this question. He also produced random puzzles and counted the effect of his method on ability to solve, and there were still enough unsolvable when including his method. Eppstein's method made it into the puzzle book "Mensa Guide to solving Sudoku" by Peter Gordon and Frank Longo, one of the few books I found with any difficult puzzles at all. The method I am curious about, though, and I saved the issue, is in the April 2009 A.M.S. Notices, J. F. Crook, "A pencil-and-paper algorithm for solving sudoku puzzles," pages 460-468, pdf at http://www.ams.org/notices/200904/index.html
-
I'll look at Eppstein's article ( arxiv.org/abs/cs/0507053 ), and of course his attention her could be very interesting. As for the Notices article, Crook's Occupancy Theorem, Theorem 1, creates linear equalities that must hold for rational solutions. Moreover, the theorem is stated for three pairs of types of conditions, (cell,column), (cell,row), (cell,box); but it also applies to (row,column), and in a weaker form to (row,box) and (column,box). This phenomenon is exactly what led me to the posted question. – Greg Kuperberg Jun 7 2010 at 18:17
Dear Greg, the most interesting techniques I found came up first in British computer programmer forums. The best of these is the B.U.G. principle for puzzles with unique solutions. I've never been entirely sure I believed it, as the programmers were so very unmathematical in discussing it. However it worked every time, and usually reduced to situations where I could recover the choice it described by other means: sudocue.net/guide.php#BUG – Will Jagy Jun 7 2010 at 18:50
This "BUG" principle, the "Bivalue Universal Grave", can also be interpreted in terms of linear programming. The basic idea is that when a linear equality only has two surviving variables, then the variables are equivalent and you can remove one of them in the system of equations without sacrificing sparseness. If you diagram this as an edge between two variables, then in Sudoku the edges often make even-length cycles. So this by itself also won't create any difference between integer programming and linear programming. – Greg Kuperberg Jun 7 2010 at 19:01
I think you are onto something. For a similar problem that should be easier to decide, consider removing the constraints that involve the 3 by 3 boxes, back to the Latin square with some presets, which was historically the first version of the puzzles. The possible gap between rational and integral uniqueness may be wider this way, it may be narrower, but it should be easier to think about, and easier to program random problems that do have at least one integral solution. And one can start with 2 by 2 and 3 by 3 Latin squares first, see what happens on paper. – Will Jagy Jun 7 2010 at 19:49
Like Sudoku, the Latin square completion problem is known to be NP-complete. The proof of that almost certainly lets you construct non-trivial polytopes with only one integer point. My question is not strictly mathematics, it's sociological. – Greg Kuperberg Jun 7 2010 at 20:44
A good explanation of how humans solve sudoku is here -
http://onigame.livejournal.com/20626.html
-
Here as well, the "Sledgehammer" that this author describes also applies to rational solutions. – Greg Kuperberg Jun 8 2010 at 6:23 | {"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": 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": 0.9565435647964478, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/207311/convergence-in-function-space?answertab=active | # convergence in function space
Maybe is a silly question, but for some reason I am confused...
If $\mathcal{F}$ is a normed space of real functions and $\displaystyle{ f \in \mathcal{\bar F } }$ then there exists a sequence of functions $\displaystyle{ (f_n ) \subset \mathcal{F} }$ such that $\displaystyle{ f_n \to f \quad \text{as} \quad n \to \infty}$ which is equivelant to $\displaystyle{ || f_n -f|| \to 0 \quad \text{as} \quad n \to \infty}$
Here is my question: The convergence $f_n \to f$ is uniform or pointwise ?
Thank's in advance!
edit: $\displaystyle{ f, f_n : A \subset \mathbb R \to \mathbb R }$
Is now more clear my question?
Any ideas?
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1
What norm do you have? – enzotib Oct 4 '12 at 17:56
@enzotib: Some norm in general. It depends on the norm...??? – passenger Oct 4 '12 at 17:58
## 1 Answer
Convergence $f_n\to f$ here is defined by $\|f_f-f\|\to 0$ (whereas the latter is just convergence in $\mathbb{R}$). This notion can indeed lead to very different notions of convergence of functions.
A simple example is convergence with respect to the supremum norm $\|f\|_\infty=\sup|f(x)|$ and the 1-norm $\|f\|_1 = \int|f(x)|dx$. The former is just uniform convergence, while the latter is a notion of convergence which is usually not treated in calculus courses. In fact the latter allows unbounded sequences of functions which still converge to the zero function (and you should try to find such an example by yourself).
If I remember correctly, "pointwise almost everywhere convergence" is not induced by a norm...
-
Thank you for your reply! I missed to write that $f, f_n : \mathbb R \to \mathbb R$. So you say that the type of convergence ( uniform or pointwise) depends on the norm we use? – passenger Oct 4 '12 at 18:25
The important thing here is: There are more notions of convergence of sequences of functions than just uniform and pointwise. Uniform convergence is induced by the sup-norm while pointwise convergence is not characterized by any norm (if I remember correctly). But there are a lot more notions for convergence of sequences of functions. – Dirk Oct 5 '12 at 6:18 | {"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": 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": 0.9237145781517029, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/255130/how-to-do-lebesgue-decomposition-in-this-problem | How to do Lebesgue decomposition in this problem?
$\{x_n\}$ is a sequence in $\mathbb{R}$ and $\{p_n\}$ is a sequence of positive numbers. Define a $\sigma$-finite measure $\nu(E)=\sum_{x_n\in E}p_n$. Find the Lebesgue decomposition of $\nu$ with respect to Lebesgue measure on $\mathbb{R}$.
My solution:
Let $S=\{x_n\}_{n\in\mathbb{N}}$. Assume $\lambda(E)=0,\forall E\subseteq \mathbb{R}$.
Define $\nu_1(E)=\nu(E\backslash S)=0$ because $E\backslash S$ does not have corresponding $\{p_n\}\Rightarrow \nu_1\ll\lambda$.
Define $\nu_2(E)=\nu(E\cap S)=\sum_{x_n\in E}p_n>0\Rightarrow \nu_2\perp\lambda$.
Any comment is appreciated.
-
1
You cannot assume $\lambda(E) = 0 \quad \forall\, E \subseteq \mathbb{R}$. I suppose that is not what you wanted to write. – Thomas Dec 10 '12 at 9:15
Thank you, @Thomas. – Sam Dec 10 '12 at 15:20 | {"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": 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": 0.9066641926765442, "perplexity_flag": "head"} |
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http://math.stackexchange.com/questions/116899/if-f-is-a-non-constant-analytic-function-on-a-compact-domain-d-then-ref/116912 | # If $f$ is a non-constant analytic function on a compact domain $D$, then $Re(f)$ and $Im(f)$ assume their max and min on the boundary of $D$.
This is a homework problem I got, my attempted proof is:
Since $f$ is non constant and analytic, $f=u(x)+iv(y)$ where neither $u$ nor $v$ is constant(by Cauchy Riemann equations) and $u v$ are both analytic in $D$.
Therefore $u$ and $v$ both assume their max on the boundary of $D$ (by maximum modulus theorem).
Also, $u$ and $v$ have no minimums in the interior of $D$ unless they are $0$. I'm stuck here and don't know how to show that $u$ and $v$ are nonzero.
I looked at the answer at the back of the book. They used the Open Mapping Theorem(the image of an open set under a nonconstant analytic mapping is an open set):
According to the Open Mapping Theorem, the image under f of any open set D containing z0 in its interior is an open set containing $f (z_0)$ in its interior. Hence, both Re f and Im f assume larger and smaller values in D than the values $Re f (z_0)$ and $Im f (z_0)$.
I don't understant the proof given by the book, can someone explain a bit? Also, what do you think about my proof?
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Do you mean in your second paragraph that each of u,v is harmonic in D? – AQP Mar 6 '12 at 0:59
er...I don't know what that means. In the second paragraph I mean that by the Cauchy Riemann equations, if u is constant, then u_x=u_y=0, then so is v_x and v_y(and vice versa), so f is constant. – Scharfschütze Mar 6 '12 at 1:03
If you use polar coordinates for a circle, and Cauchy's formula may help you prove the result . Then the value at the center of a ball is the average of the values on the perimeter. – AQP Mar 6 '12 at 1:08
I don't see how that helps...can you explain more? – Scharfschütze Mar 6 '12 at 1:16
I see, that is correct if that's what you meant. I understood something else. – AQP Mar 6 '12 at 1:16
show 4 more comments
## 1 Answer
Here's the book's proof with a tad more detail: Suppose $z_0$ is not on the boundary. We will show neither the real nor the imaginary parts of $f$ are maximized at $z_0$. Since the real and complex parts are continuous, they obtain their maxima and minima somewhere on $D$ because $D$ is compact. Hence they must obtain their minima and maxima on the boundary.
Take a small neighborhood of $z_0$ inside your domain. The image of this open neighborhood is an open neighborhood of $f(z_0)$ by the open mapping theorem. Let $U$ be the neighborhood of $f(z_0)$ just described. For small values of $\varepsilon$, $U$ contains the points $f(z_0)+\varepsilon$ , $f(z_0)-\varepsilon$, $f(z_0)+i\varepsilon$, and $f(z_0)-i\varepsilon$, whose real and imaginary parts are more/less than those of $f(z_0)$.
-
For this your wrote:Since the real and complex parts are continuous, they obtain their maxima and minima somewhere on D because D is compact. Hence they must obtain their minima and maxima on the boundary. I don't understand why the minima must also be on the boundary, because the minimum modulus theorem says it's possible if f is 0 at some point inside D. – Scharfschütze Mar 6 '12 at 1:38
1
Max/min value of the function $\ne$ max/min modulus. – Did Mar 6 '12 at 6:16
The proof above was for the max/min values of the real and imaginary parts, but it is similar to the proof of the max/min modulus theorems: For small enough $\delta$, $(1+\delta)f(z_0)$ and $(1-\delta)f(z_0)$ are contained in $U$. Then the max/min modulus cannot be obtained at $z_0$ unless $f(z_0)$=0. So the max and min modulus are each either 0 or obtained on the boundary. The max modulus is only 0 if $f$ is 0 everywhere (in which case it is obtained on the boundary, among other places). – Brett Frankel Mar 6 '12 at 18:18 | {"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": 47, "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": 0.3, "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": 0.937400221824646, "perplexity_flag": "head"} |
http://crypto.stackexchange.com/questions/2758/decimal-to-binary-question/2760 | # Decimal to binary question [closed]
This is an easy question, but I can't explain it lol, how would you guys explain it?
There is a procedure for converting a decimal number to binary in the following way: repeatedly divide the number by 2, writing the quotient and remainder at each step. Stop when 0 is obtained as a quotient. The sequence of remainders, written backwards (from the way they were obtained) is the binary representation. Here is an example, finding the binary representation of 49:
|49 24 1 |
|24 12 0 |
|12 6 0 |
|6 3 0 |
|3 1 1 |
|1 0 1 |
So 49 = 110001 (base 2) - Explain why this method works.
-
Do you want to know why it works or the details of how it works? – Thomas Jun 1 '12 at 4:04
either way is good :) – hihello4 Jun 1 '12 at 4:16
Did my answer help? – Thomas Jun 1 '12 at 4:49
Sorry, I can't find how this relates to cryptography, the topic of this site. – Paŭlo Ebermann♦ Jun 1 '12 at 7:26
## closed as off topic by fgrieu, Paŭlo Ebermann♦Jun 1 '12 at 7:25
Questions on Cryptography Stack Exchange are expected to relate to cryptography within the scope defined in the FAQ. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about closed questions here.
## 1 Answer
In base 10 we write for example $133$ when we mean $$133 = 1 * 10^2 + 3*10^1 + 3*10^0.$$ If we want to write $49$ in base $2$ then note first that: $$49 = 1*2^5 + 1*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 1*2^0.$$ Because of this $49$ is $110001$.
Noe obviously, you "don't know this", but I wanted to write it down so that you can see what happens as you divide by two.
So you want to find the digits in base you, you could start by dividing $49$ by $2$. You get a remainder of $1$ because $49$ is odd. Hence you know that the coefficient in from of $2^0$ is one. You subtract that reminder and divide by $2$. Dividing by $2$ decreases all the exponents by one, so $$24 = 1*2^4 + 1*2^3 + 0*2^2 + 0*2^1 + 0*2^0.$$ Now you just continue like this dividing by $2$ to find the coefficient infront of $2^0$ (which then is the coefficient in front of $2^1$ in $49$).
Maybe this is all too many words...
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http://mathhelpforum.com/number-theory/22069-composite-number-proof-help.html | # Thread:
1. ## Composite number proof help
Hello, I am having trouble with following proof:
Prove that all integers
2^2^(2n+1) +3 2^2^(4n+1) +7 , n = 1, 2, 3 ...
are composite.
Any help is appreciated , Thanks - Kate
2. Originally Posted by katie07
Hello, I am having trouble with following proof:
Prove that all integers
2^2^(2n+1) +3 2^2^(4n+1) +7 , n = 1, 2, 3 ...
are composite.
Any help is appreciated , Thanks - Kate
I don't have time to do this in more detail, but my impression is that this might be doable by an induction proof.
-Dan
Edit: If it's going to be done by induction, it will take a better Mathematician than me!
3. this is supposed to be $2^{2^{2n + 1}} + 3 \cdot 2^{2^{4n + 1}} + 7$ right?
i'm having trouble proving this is composite for n = 1, but i'm not that good with divisibility on such a large scale, the number is huge. | {"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": 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": 0.9370684027671814, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/87735/is-the-integral-int-0-infty-frac-mathrmd-x1x21xa-equal-for-a | # Is the integral $\int_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$ equal for all $a \neq 0$?
Let $a$ be a non-zero real number.
Is it true that $\int_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$ is independent on $a$ ?
Any proof?
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2
– Did Dec 2 '11 at 22:36
## 4 Answers
Let $\mathcal{I}(a)$ denote the integral. Then $$\begin{eqnarray} \mathcal{I}(a) &=& \int_0^1 \frac{\mathrm{d} x}{(1+x^2)(1+x^a)} + \int_1^\infty \frac{\mathrm{d} y}{(1+y^2)(1+y^a)} \\ &\stackrel{y=1/x}{=}& \int_0^1 \frac{\mathrm{d} x}{(1+x^2)(1+x^a)} + \int_0^1 \frac{x^a \mathrm{d} x}{(1+x^2)(1+x^a)} \\ &=& \int_0^1 \frac{1+x^a}{(1+x^2)(1+x^a)} \mathrm{d} x = \int_0^1 \frac{1}{1+x^2} \mathrm{d} x = \frac{\pi}{4} \end{eqnarray}$$
Thus $\mathcal{I}(a) = \frac{\pi}{4}$ for all $a$. I do not see a need to require $a$ to be non-zero.
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I think there's a typo - you wanted to write $1+x^a$ instead of $1+x^2$ in the numerator. I hope you don't mind correcting. Very nice solution! – Martin Sleziak Dec 2 '11 at 14:13
can you explain the second step? thanks – tomerg Dec 2 '11 at 14:23
1
@tomerg $\mathrm{d}y = -\frac{1}{x^2}\mathrm{d}x$, $\frac{1}{1+y^2} = \frac{x^2}{1+x^2}$, and $\frac{1}{1+y^a}=\frac{x^a}{1+x^a}$. Also notice that $x \mapsto \frac{1}{x}$ reverts the orientation, i.e. changes boundaries $1$, $\infty$ into $1$, $0$ respectively. But the negative sign of Jacobian compensates that, thus overall +1 sign. – Sasha Dec 2 '11 at 15:35
+1 Nice argument. Is there a way to view this in terms of trigonometric identities involving arctangents? – Michael Hardy Dec 2 '11 at 16:44
@MichaelHardy Thank you. Do you mean to perform substitution $u=\tan(x)$, so that $\mathcal{I}(a) = \int_0^{\pi/2} \left( 1 + \tan^a(u) \right)^{-1} \mathrm{d} u$ ? The idea of the proof will then remain much the same, split integration into $(0,\pi/4)$ and $(\pi/4, \pi/2)$ and use $u \mapsto \pi/2-u$ substitution for the second integral. Then due to $\tan(\pi/2-u) = \cot(u)$ the result would follow. – Sasha Dec 2 '11 at 18:01
show 4 more comments
With a change of variable $$\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)(1+x^a)}\overset{x\to1/x}{=}\int_0^\infty\frac{x^a\mathrm{d}x}{(1+x^2)(1+x^a)}$$ Adding and dividing by two yields $$\begin{align} \int_0^\infty\frac{\mathrm{d}x}{(1+x^2)(1+x^a)} &=\frac{1}{2}\int_0^\infty\frac{\mathrm{d}x}{(1+x^2)}\\ &=\frac{\pi}{4} \end{align}$$
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I was just about to post this when I saw Sasha's answer appear so I deleted it. However, now that Sasha has probably capped for the day, I decided to undelete it. – robjohn♦ Dec 2 '11 at 19:08
3
This is the simplest answer IMO. I upvoted this. :-) – Srivatsan Dec 2 '11 at 19:09
How does the numerator after the substitution become $x^a$? I would have expected $x^{2+a}$. – Henning Makholm Dec 3 '11 at 12:39
1
@Henning: $$\begin{align}\int_\infty^0\frac{\mathrm{d}\frac{1}{x}}{\left(1+\left(\frac{1}{x}\right)^2\right)\left(1+\left(\frac{1}{x}\right)^a\right)}&=\int_\infty^0\frac{-\frac{1}{x^2}\;\mathrm{d}x}{(1+x^2)/x^2\;(1+x^a)/x^a}\\&=\int_0^\infty\frac{x^a\; \mathrm{d}x }{(1+x^2)(1+x^a)}\end{align}$$ I believe that is correct. – robjohn♦ Dec 3 '11 at 15:24
Oh, of course. – Henning Makholm Dec 3 '11 at 17:51
$\displaystyle I=\int_0^\infty \frac{dx}{(1+x^2)(1+x^a)}$
Substitution:
$\displaystyle x=\tan\theta$
$\displaystyle dx=\sec^2\theta d\theta$
$\displaystyle I=\int_0^{\pi/2}\frac{d\theta}{1+\tan^a \theta}$
$\displaystyle I=\int_0^{\pi/2}\frac{\cos^a \theta d\theta}{\sin^a \theta + \cos^a \theta}$
$\displaystyle I=\int_0^{\pi/2}\frac{\cos^a(\pi/2-\theta) d\theta}{\sin^{a}(\pi/2-\theta) + \cos^a (\pi/2-\theta)}$
$\displaystyle I=\int_0^{\pi/2}\frac{\sin^a\theta d\theta}{\sin^a\theta + \cos^a \theta}$
Therefore,
$\displaystyle 2I=\int_0^{\pi/2}d\theta$
$\displaystyle I=\pi/4$
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– robjohn♦ Dec 3 '11 at 15:13
Nice argument. +1 – Michael Hardy Dec 6 '11 at 4:47
$$\begin{align} I & = \int_0^{\infty} \frac{dx}{(1+x^2)(1+x^a)}\\ \frac{dI}{da} & = -\int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2} \end{align}$$
Let $\displaystyle J = \int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2}$ $$\begin{align} J & = \int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2}\\ & \stackrel{x=1/y}{=} \int_{0}^{\infty} \frac{1/y^a \log(1/y) d(1/y)}{(1+(1/y)^2)(1+(1/y)^a)^2}\\ & = \int_{\infty}^{0} \frac{y^a \log(y) dy}{(1+y^2)(1+y^a)^2}\\ & = -J \end{align}$$
Hence, $\frac{dI}{da}=0$. Hence, $I$ is independent of $a$.
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http://www.nag.com/numeric/cl/nagdoc_cl23/html/S/s17agc.html | # NAG Library Function Documentnag_airy_ai (s17agc)
## 1 Purpose
nag_airy_ai (s17agc) returns a value for the Airy function $\mathrm{Ai}\left(x\right)$.
## 2 Specification
#include <nag.h>
#include <nags.h>
double nag_airy_ai (double x, NagError *fail)
## 3 Description
nag_airy_ai (s17agc) evaluates an approximation to the Airy function, $\mathrm{Ai}\left(x\right)$. It is based on a number of Chebyshev expansions.
For large negative arguments, it is impossible to calculate the phase of the oscillatory function with any precision and so the function must fail. This occurs if $x<-{\left(3/2\epsilon \right)}^{2/3}$, where $\epsilon $ is the machine precision.
For large positive arguments, where $\mathrm{Ai}$ decays in an essentially exponential manner, there is a danger of underflow so the function must fail.
## 4 References
Abramowitz M and Stegun I A (1972) Handbook of Mathematical Functions (3rd Edition) Dover Publications
## 5 Arguments
1: x – doubleInput
On entry: the argument $x$ of the function.
2: fail – NagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_REAL_ARG_GT
On entry, ${\mathbf{x}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{x}}\le 〈\mathit{\text{value}}〉$.
x is too large and positive. The function returns zero.
NE_REAL_ARG_LT
On entry, x must not be less than $〈\mathit{\text{value}}〉$: ${\mathbf{x}}=〈\mathit{\text{value}}〉$.
x is too large and negative. The function returns zero.
## 7 Accuracy
For negative arguments the function is oscillatory and hence absolute error is the appropriate measure. In the positive region the function is essentially exponential-like and here relative error is appropriate. The absolute error, $E$, and the relative error, $\epsilon $, are related in principle to the relative error in the argument, $\delta $, by $E\simeq \left|x{\mathrm{Ai}}^{\prime }\left(x\right)\right|\delta $, $\epsilon \simeq \left|x{\mathrm{Ai}}^{\prime }\left(x\right)/\mathrm{Ai}\left(x\right)\right|\delta $.
In practice, approximate equality is the best that can be expected. When $\delta $, $\epsilon $ or $E$ is of the order of the machine precision, the errors in the result will be somewhat larger.
For small $x$, errors are strongly damped by the function and hence will be bounded by the machine precision.
For moderate negative $x$, the error behaviour is oscillatory but the amplitude of the error grows like $\text{amplitude}\left(E/\delta \right)\sim {\left|x\right|}^{5/4}/\sqrt{\pi }$. However the phase error will be growing roughly like $2\sqrt{{\left|x\right|}^{3}}/3$ and hence all accuracy will be lost for large negative arguments due to the difficulty in calculating sin and cos to any accuracy if $2\sqrt{{\left|x\right|}^{3}}/3>1/\delta $.
For large positive arguments, the relative error amplification is considerable, $\epsilon /\delta \sim \sqrt{{x}^{3}}$.
This means a loss of roughly two decimal places accuracy for arguments in the region of 20. However, very large arguments are not possible due to the danger of setting underflow, and so the errors are limited in practice.
None.
## 9 Example
The following program reads values of the argument $x$ from a file, evaluates the function at each value of $x$ and prints the results.
### 9.1 Program Text
Program Text (s17agce.c)
### 9.2 Program Data
Program Data (s17agce.d)
### 9.3 Program Results
Program Results (s17agce.r) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 26, "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": 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": 0.6962138414382935, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/126886/is-the-fermat-scheme-xpyp-zp-always-normal | # Is the Fermat scheme $x^p+y^p=z^p$ always normal
Let $K$ be a number field with ring of integers $O_K$.
Is the closed subscheme $X$ of $\mathbf{P}^2_{O_K}$ given by the homogeneous equation $x^p+y^p=z^p$ normal?
I know that this is true if $K=\mathbf{Q}$. (My method of proof is a bit awkward.)
I expect the answer to be no (unfortunately) in general. What is an equation for the normalization of $X$? In other words, what is an equation for the normalization of $\mathbf{P}^1_{O_K}$ in the function field of $\mathrm{Proj}K[x,y,z]/(x^p+y^p-z^p)$?
Note that the only difficulty arises modulo $p$. The fibre is then reduced.
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Maybe I'm missing something. If the only difficulty arises mod $p$, and you are only considering the case of being over a number field ... doesn't this mean there is no difficulty? – Matt Apr 1 '12 at 17:00
I'm considering the scheme over its ring of integers $O_K$. The generic fibre of this scheme is the curve $x^p+y^p=z^p$ over $K$. There are no problems in this case; this curve is smooth. But the scheme $X$ over $O_K$ has singularities. I'm asking if all them are normal. – Harry Apr 1 '12 at 17:26
Maybe the terminology modulo $p$ caused some confusion. What I meant is that if you consider the fibres of $X$ over $\mathrm{Spec} O_K$ you get a smooth curve unless you reduce modulo a prime lying over $p$. – Harry Apr 1 '12 at 17:27
Sorry, I see. I didn't carefully read the second to last sentence and just saw $Proj K[x,y,z]/(x^p+y^p-z^p)$ and assumed you were asking about the singularities of that. – Matt Apr 1 '12 at 18:26
1
I meant never normal when $K/\mathbb Q$ is ramified above $p$. – QiL'8 Apr 1 '12 at 21:32
show 1 more comment
## 1 Answer
This answer amplifies some of QiL's remarks in the comments above:
To investigate such a question, you can use Serre's criterion: normal $= R_1 + S_2$.
In this case your scheme is a projective hypersurface, thus Cohen--Macaulay, and so in particular $S_2$. So the only issue is $R_1$. Now the codimension one points either live in the generic fibre, which is smooth, or are generic points in one of the special fibres.
So you are reduced to computing the local ring around a generic point in positive characteristic. As you note, if this characteristic is different from $p$ then the fibre is smooth. In particular, it is generically smooth (equivalently, generically reduced, or again equivalently, $R_0$) and all is good. So you are reduced to the generic points in the char. $p$ fibres. Here you can hope to compute explicitly:
You can work in affine coords. where your curve is $x^p + y^p = 1$, and so your problem is to determine whether the ring $O_K[x,y]/(x^p + y^p -1)$ is regular after localizing at the prime ideals dividing $p$. Reducing mod $p$, you are asking if the minimal primes of the ring $(O_K/p)[x,y]/(x+y - 1)^p$ are prinicipal.
Let me assume that there is a unique prime in $O_K$ above $p$, just to ease my typing.
If $O_K/p$ is a field $k$ (i.e. $p$ is unramified), you'll be in good shape, since the minimal prime of $k[x,y]/(x+y-1)^p$ is generated by $x+y-1$. But if $p$ is ramified, of degree $e$, then you get problems: you need two generators: $(x+y - 1)$ and some nilpotent $\pi$ such that $\pi^e = 0$.
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http://physics.stackexchange.com/questions/53050/why-does-gravity-assist-transfer-twice-the-planets-velocity | # Why does gravity assist transfer twice the planet's velocity?
In orbital mechanics and aerospace engineering a gravitational slingshot (also known as gravity assist manoeuver or swing-by) is the use of the relative movement and gravity of a planet or other celestial body to alter the path and speed of a spacecraft, typically in order to save propellant, time, and expense. Gravity assistance can be used to accelerate (both positively and negatively) and/or re-direct the path of spacecraft.
Over-simplified example of gravitational slingshot: the spacecraft's velocity changes by up to twice the planet's velocity.
Where does the factor of $2$ come from in this relation?
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## 2 Answers
If you analyse the situation from the planetary frame of reference (which will essentially be the centre-of-mass frame), then the incoming probe has a velocity $U+v$ if it then does any sort of "elastic bounce" off the planet - i.e. if it loses no energy and goes back in the opposite direction - then the planet will not be affected (if it's massive enough) and the probe will go back in the opposite direction and with the same speed, $U+v$, as initially. If you add to this the planet's velocity $U$, then you get a speed of $2U+v$ in the original frame of reference.
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The emergence of the factor of 2 is easy to be seen from the celestial body's vantage point (in its reference frame).
Before the maneuver, the spaceship is moving by the velocity $\vec v-\vec u$ – the relative speed of the two objects – in this frame. The celestial body is largely unaffected so after the slingshot maneuver, we are still in the same frame.
Clearly, by the $Z_2$ symmetry of the situation, the final velocity of the spaceship is just minus the initial one, $\vec u-\vec v$. This velocity may be translated to the original reference frame by adding $\vec u$ that we subtracted at the beginning to get $\vec v -\vec u$ from $\vec v$. So we get $2\vec u-\vec v$ as the final velocity in the initial frame.
The signs are so that $\vec u$ has a negative $x$ component so both terms in $2\vec u-\vec v$ have a negative $x$ component and the picture represents the absolute value of the final $v_x$ which is therefore $2|u_x|+|v_x|$.
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this is ideal simplified example what is, its example complicate – Neo Feb 4 at 19:13 | {"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": 18, "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": 0.9094166159629822, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/11584/cases-of-any-known-fundamental-physical-constants-changing-within-our-locality/11585 | # Cases of any known fundamental physical constants changing within our locality?
Has there been any cases where the only explanation has been that at least one of the physical constants must have changed to explain an experiment/event/obervation? I am not intrested in large scale obsorvations such Gaolaxies holding together with insufficient obsorvable mass or stars seem to be moving at faster than speed of the light phenomenons. Just simple instances of experiments/events that the only possible logical explenation can be obtained if one had to assume a known fundamental physical constant must have varied.
I am trying to have this question to be only for non scale dependent situations.
Just to give you an idea, consider that in mathematics the value of constant $\pi$ changes with the geometry.
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5
– Qmechanic♦ Jun 26 '11 at 20:22
that depends on the definition of $\pi$ being used, what definition of $\pi$ makes it constant in all the places that $\pi$ gets used? – Arjang Jun 26 '11 at 23:13
@arjang All of them. – Holowitz Jun 27 '11 at 1:33
@JOHA : in that case that is in contradiction with waht Qmechanic said, as the sum of the angles of a triangle add up to $\pi$, and depending on geometry that value is changing. – Arjang Jun 27 '11 at 3:07
2
@Arjang: That is not the definition of $\pi$. If $\pi$ is a constant then its definition must yield a unique value that is the same in all contexts. The sum of angles in a triangle is $\pi$ in a Euclidean geometry, but not in other geometries. – Will Vousden Jun 27 '11 at 7:54
show 1 more comment
## 4 Answers
It's somewhat glib to put quantum mechanics this way, but one could say that Planck's constant was zero before Planck introduced it, insofar as the classical limit of QM is $\hbar\rightarrow 0$ (which is not so very far).
Something I've played with, to absolutely no good effect so far, is the idea that quantum fluctuations might be greater or less from place to place and from time to time, which fits a little better with your Question as you have put it. However, one would then describe quantum fluctuations at different space-time points relative to the unchanged constant $\hbar$, which would now be defined as "the amplitude of quantum fluctuations under such and such conditions". That's essentially Solomoan's Answer, however his apparent assumption that the idea is obvious is a little too fast for someone with almost any philosophical leanings. [The boiling point of water is a constant, 100°C, if one defines the conditions carefully enough. A very interesting account of the historical and philosophical process associated with the definition of temperature scales is given by Hasok Chang, in a book that won a major Philosophy of Science prize, "Inventing Temperature". If you read that —it's very accessible as these things go— you will ask a different Question.]
One way to approach this Question may be to suppose that new theories introduce new constants, in terms of which old constants can be expressed, together with the conditions in which they accurately describe the Physical world. Thus, in my speculative example as I put it above, one might define the amplitude of quantum fluctuations as $\hbar$ at the surface of the earth, and describe it's variation as a function of altitude —this would at least change our theory of gravity, and likely lots else besides, so the whole of Physics becomes part of the discussion. That is a thread in Philosophy of Science that is known as the Quine-Duhem thesis, or, as Wikipedia has it, the Duhem-Quine thesis, which can be loosely summarized as the idea that a theory stands or falls as a whole, a point that I take to be vitiated by the always present possibility of changing any theory in very small ad-hoc ways.
Your Question to some extent opens a philosophical Pandora's box, which you should be cautious about opening. My view is that anyone who wants to change Physics significantly must consider these and similar ideas at length, but for Physicists trying to do the everyday job of Physics it is perhaps as counterproductive to spend time on this as it is to spend time learning to play the violin.
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I guess from your use of the phrase "within our locality" that this is not what you mean, but various people have tried to measure astrophysically whether fundamental constants have evolved over time. A while back, there were some claims (e.g., this paper and others by the same people) that observations of spectral lines in very distant objects could best be explained by assuming that fundamental constants (specifically the fine-structure constant) had changed with time. There is, to say the least, no consensus that these observations and their interpretation are correct, though.
People certainly try to measure changes in physical constants locally, but as far as I know there's no evidence for such variation.
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There have been some attempts to "locally" (i.e. non-astronomically) measure the drift of the fine-structure constant $\alpha=e^2/\hbar c$, using experiments on trapped atoms and ions. If you want to read up on it I think this review article,
Lea, S. N. Limits to time variation of fundamental constants from comparisons of atomic frequency standards. Rep. Prog. Phys. 70 no. 9, p. 1473 (2007). doi:10.1088/0034-4885/70/9/R01.
is a good place to start.
I know of one example (of which I can't find a good up-to-date, readable reference, possibly because the experiment may still be ongoing), currently taking place at NPL. This uses an octupole transition in ytterbium: an S$\rightarrow$F transition which is forbidden by both dipole and quadrupole selection rules. This makes the excited state have a very long lifetime - six years - which makes the resonance very narrow in comparison to the ~470 nm wavelength. This makes the transition very valuable for precision spectroscopy.
Even better, this transition can be compared to a quadrupole transition in ytterbium ions, which is not as narrow but is still good enough for metrology purposes. The real catch is that the details of the frequencies of these two E2 and E3 transitions depend on $\alpha$ in opposite directions, so any change in it will make the frequency ratio change.
At the current state of the art in these experiments, the precision achievable is enough that any "astronomical-scale" changes in $\alpha$ (i.e. of the order of 1% per billion years) should be observable by measuring these kind of frequency ratios over a few years. This is an active area of research and experiments are currently running which will report in the near (i.e. ~few years) future, with precision matching or exceeding current astronomical bounds on $d\alpha/dt$.
-
About thirty years ago I read a paper giving a detailed answer to this question: Rozental’ I. L. “Physical laws and the numerical values of fundamental constants” Sov. Phys. Usp. vol. 23 pp.296–305 (1980). I recommend it very much. It is very good written and easy to understand. From this paper I learned that there is a whole science branch dealing with this question and other questions comparable to as well as more general than this one. Namely, it tries to address the questions like
• What are the limits that fundamental constants may vary, so that our world stays intact?
• What are decrements of variation of the fundamental constants in space and time that would be allowed, so that the world around us stays the same?
and comparable questions. I strongly recommend this paper just to get acquainted with this circle of problems. About 30 years have passed since that time, however. New important results may have appeared in this area that I am not aware of. Nevertheless, this paper would be a reasonable start.
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http://mathhelpforum.com/advanced-algebra/118176-t-projection-normal-then-t-orthogonal-projection.html | # Thread:
1. ## T is a projection and normal, then T is orthogonal projection
Question: Let $T$ be a normal operator on a finite-dimensional inner product space. Prove that if $T$ is a projection, then $T$ is also an orthogonal projection.
Attempt: Since $T$ is normal, $TT^* = T^*T$ and since $T$ is a projection, then $T=T^2$.
Since $T$ is an orthogonal projection if and only if $T^2 = T = T^*$, we already have the left equality by that fact that $T$ is a projection. We only need to show that $T=T^*$ and we'll be done.
Somehow, I figure that I can do this by showing that $<T(x),y>=<x,T(y)>$ but how?
2. Since $T$ is normal, it is unitarily equivalent to a diagonal matrix. Since $T=T^2$, its eigenvalues can be only $\pm 1$. Thus $T=UDU^*$ and $T^*=U^{**}D^*U^*=UDU^*=T$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "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": 0.9511876106262207, "perplexity_flag": "head"} |
http://en.m.wikipedia.org/wiki/Hodge_dual | # Hodge dual
In mathematics, the Hodge star operator or Hodge dual is a significant linear map introduced in general by W. V. D. Hodge. It is defined on the exterior algebra of a finite-dimensional oriented inner product space.
## Dimensionalities and algebra
Suppose that n is the dimensionality of the oriented inner product space and k is an integer such that 0 ≤ k ≤ n, then the Hodge star operator establishes a one-to-one mapping from the space of k-vectors and the space of (n−k)-vectors. The image of a k-vector under this mapping is called the Hodge dual of the k-vector. The former space, of k-vectors, has dimensionality
${n \choose k}$
while the latter has dimensionality
${n \choose n - k},$
and by the symmetry of the binomial coefficients, these two dimensionalities are in fact equal. Two vector spaces over the same field with the same dimensionality are always isomorphic; but not necessarily in a natural or canonical way. The Hodge duality, however, in this case exploits the inner product and orientation of the vector space. It singles out a unique isomorphism, that reflects therefore the pattern of the binomial coefficients in algebra. This in turn induces an inner product on the space of k-vectors. The 'natural' definition means that this duality relationship can play a geometrical role in theories.
The first interesting case is on three-dimensional Euclidean space V. In this context the relevant row of Pascal's triangle reads
1, 3, 3, 1
and the Hodge dual sets up an isomorphism between the two three-dimensional spaces, which are V itself and the space of wedge products of two vectors from V. See the Examples section for details. In this case the content is just that of the cross product of traditional vector calculus. While the properties of the cross product are special to three dimensions, the Hodge dual applies to all dimensionalities.
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## Extensions
Since the space of alternating linear forms in k arguments on a vector space is naturally isomorphic to the dual of the space of k-vectors over that vector space, the Hodge dual can be defined for these spaces as well. As with most constructions from linear algebra, the Hodge dual can then be extended to a vector bundle. Thus a context in which the Hodge dual is very often seen is the exterior algebra of the cotangent bundle (i.e. the space of differential forms on a manifold) where it can be used to construct the codifferential from the exterior derivative, and thus the Laplace-de Rham operator, which leads to the Hodge decomposition of differential forms in the case of compact Riemannian manifolds.
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## Formal definition of the Hodge star of k-vectors
The Hodge star operator on a vector space V with a nondegenerate symmetric bilinear form (herein aka inner product) is a linear operator on the exterior algebra of V, mapping k-vectors to (n−k)-vectors where n = dim V, for 0 ≤ k ≤ n. It has the following property, which defines it completely: given two k-vectors α, β
$\alpha \wedge (\star \beta) = \langle \alpha,\beta \rangle \omega$
where $\langle \cdot,\cdot \rangle$ denotes the inner product on k-vectors and ω is the preferred unit n-vector.
The inner product $\langle \cdot,\cdot \rangle$ on k-vectors is extended from that on V by requiring that $\langle \alpha,\beta \rangle = \det(\langle \alpha_i,\beta_j \rangle)$ for any decomposable k-vectors $\alpha = \alpha_1 \wedge \dots \wedge \alpha_k$ and $\beta = \beta_1 \wedge \dots \wedge \beta_k$.
The preferred unit n-vector ω is unique up to a sign. The choice of ω defines an orientation on V.
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## Explanation
Let W be a vector space, with an inner product $\langle\cdot, \cdot\rangle_W$. For every linear function $f \in W^*$ there exists a unique vector v in W such that $f(w) = \langle w, v \rangle_W$ for all w in W. The map $W^* \to W$ given by $f \mapsto v$ is an isomorphism. This holds for all vector spaces, and can be used to explain the Hodge dual.
Let V be an n-dimensional vector space with basis $\{e_1,\ldots,e_n\}$. For 0 ≤ k≤ n, consider the exterior power spaces $\bigwedge^k V$ and $\bigwedge^{n-k} V$. For each $\lambda \in \bigwedge^k V$ and $\theta \in \bigwedge^{n-k} V$, we have $\lambda \wedge \theta \in \bigwedge^n V$. There is, up to a scalar, only one n-vector, namely $e_1\wedge\ldots\wedge e_n$. In other words, $\lambda \wedge \theta$ must be a scalar multiple of $e_1\wedge\ldots\wedge e_n$ for all $\lambda \in \bigwedge^k V$ and $\theta \in \bigwedge^{n-k} V$.
Consider a fixed $\lambda \in \bigwedge^k V$. There exists a unique linear function $f_{\lambda} \in \left(\bigwedge^{n-k} V\right)^{\! *}$ such that $\lambda \wedge \theta = f_{\lambda}(\theta) \, (e_1\wedge\ldots\wedge e_n)$ for all $\theta \in \bigwedge^{n-k} V$. This $f_{\lambda}(\theta)$ is the scalar multiple mentioned in the previous paragraph. If $\langle\cdot, \cdot\rangle$ denotes the inner product on (n–k)-vectors, then there exists a unique (n–k)-vector, say $\star \lambda \in \bigwedge^{n-k} V$, such that $f_{\lambda}(\theta) = \langle \theta, \star \lambda\rangle$ for all $\theta \in \bigwedge^{n-k} V$. This (n–k)-vector $\star\lambda$ is the Hodge dual of λ, and is the image of the $f_{\lambda}$ under the canonical isomorphism between $\left(\bigwedge^{n-k} V\right)^{\! *}$ and $\bigwedge^{n-k} V$. Thus, $\star : \bigwedge^{k} V \to \bigwedge^{n-k} V$.
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## Computation of the Hodge star
Given an orthonormal basis $(e_1,e_2,\dots,e_n)$ ordered such that $\omega = e_1\wedge e_2\wedge \cdots \wedge e_n$, we see that
$\star (e_{i_1} \wedge e_{i_2}\wedge \cdots \wedge e_{i_k})= e_{i_{k+1}} \wedge e_{i_{k+2}} \wedge \cdots \wedge e_{i_n},$
where $\{i_1, \cdots i_k, i_{k+1} \cdots i_n\}$ is an even permutation of $\{1, 2, \cdots n\}$.
Of these $n! \over 2$ relations, only $n \choose k$ are independent. The first one in the usual lexicographical order reads
$\star (e_1\wedge e_2\wedge \cdots \wedge e_k)= e_{k+1}\wedge e_{k+2}\wedge \cdots \wedge e_n.$
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## Index notation for the star operator
Using index notation, the Hodge dual is obtained by contracting the indices of a k-form with the n-dimensional completely antisymmetric Levi-Civita tensor. This differs from the Levi-Civita symbol by a factor of |det g|½, where g is an inner product (the metric tensor). The absolute value of the determinant is necessary if g is not positive-definite, e.g. for tangent spaces to Lorentzian manifolds.
Thus one writes[1]
$(\star \eta)_{i_1,i_2,\ldots,i_{n-k}} = \frac{1}{k!} \eta^{j_1,\ldots,j_k}\,\sqrt {|\det g|} \,\epsilon_{j_1,\ldots,j_k,i_1,\ldots,i_{n-k}}$
where η is an arbitrary antisymmetric tensor in k indices. It is understood that indices are raised and lowered using the same inner product g as in the definition of the Levi-Civita tensor. Although one can take the star of any tensor, the result is antisymmetric, since the symmetric components of the tensor completely cancel out when contracted with the completely anti-symmetric Levi-Civita symbol.
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## Examples
A common example of the star operator is the case n = 3, when it can be taken as the correspondence between the vectors and the skew-symmetric matrices of that size. This is used implicitly in vector calculus, for example to create the cross product vector from the wedge product of two vectors. Specifically, for Euclidean R3, one easily finds that
$\star \mathrm{d}x=\mathrm{d}y\wedge \mathrm{d}z$
$\star \mathrm{d}y=\mathrm{d}z\wedge \mathrm{d}x$
$\star \mathrm{d}z=\mathrm{d}x\wedge \mathrm{d}y$
where dx, dy and dz are the standard orthonormal differential one-forms on R3. The Hodge dual in this case clearly relates the cross-product to the wedge product in three dimensions. A detailed presentation not restricted to differential geometry is provided next.
### Three-dimensional example
Applied to three dimensions, the Hodge dual provides an isomorphism between axial vectors and bivectors, so each axial vector a is associated with a bivector A and vice-versa, that is:[2]
$\mathbf{A} = \star \mathbf{a}\ ;\ \ \ \mathbf{a} = \star \mathbf{A} \ ,$
where $\star$ indicates the dual operation. These dual relations can be implemented using multiplication by the unit pseudoscalar in Cℓ3(R),[3]i = e1e2e3 (the vectors { eℓ } are an orthonormal basis in three dimensional Euclidean space) according to the relations:[4]
$\mathbf{A} = \mathbf{a}i\ ;\ \ \ \mathbf{a} = - \mathbf{A} i.$
The dual of a vector is obtained by multiplication by i, as established using the properties of the geometric product of the algebra as follows:
$\mathbf{a}i = \left(a_1 \mathbf{e_1} + a_2 \mathbf{e_2} +a_3 \mathbf {e_3}\right) \mathbf {e_1 e_2 e_3} \$
$= a_1 \mathbf{e_2 e_3} (\mathbf{e_1})^2 + a_2 \mathbf{e_3 e_1}(\mathbf{e_2})^2 +a_3 \mathbf{e_1 e_2}(\mathbf{e_3})^2 \$
$= a_1 \mathbf{e_2 e_3} +a_2 \mathbf{e_3 e_1} +a_3 \mathbf{e_1 e_2} = (\star \mathbf a )\ ;$
and also, in the dual space spanned by { eℓem }:
$\mathbf{A} i = \left(A_1 \mathbf{e_2e_3} + A_2 \mathbf{e_3e_1} +A_3 \mathbf {e_1e_2}\right) \mathbf {e_1 e_2 e_3} \$
$= A_1 \mathbf{e_1} (\mathbf{e_2 e_3})^2 +A_2 \mathbf{e_2} (\mathbf{e_3 e_1})^2 +A_3 \mathbf{e_3}(\mathbf{e_1 e_2})^2 \$
$=-\left( A_1 \mathbf{e_1} + A_2 \mathbf{e_2} + A_3 \mathbf{e_3} \right) = - (\star \mathbf A )\ .$
In establishing these results, the identities are used:
$(\mathbf{e_1e_2})^2 =\mathbf{e_1e_2e_1e_2}= -\mathbf{e_1e_2e_2e_1} = -1 \ ,$
and:
$\mathit{i}^2 =(\mathbf{e_1e_2e_3})^2 =\mathbf{e_1e_2e_3e_1e_2e_3}= \mathbf{e_1e_2e_3e_3e_1e_2} = \mathbf{e_1e_2e_1e_2} = -1 \ .$
These relations between the dual $\star$ and i apply to any vectors. Here they are applied to relate the axial vector created as the cross product a = u × v to the bivector-valued exterior product A = u ∧ v of two polar (that is, not axial) vectors u and v; the two products can be written as determinants expressed in the same way:
$\mathbf a = \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3\\u_1 & u_2 & u_3\\v_1 & v_2 & v_3 \end{vmatrix}\ ;\ \ \ \mathbf A = \mathbf{u} \wedge \mathbf{v} = \begin{vmatrix} \mathbf{e}_{23} & \mathbf{e}_{31} & \mathbf{e}_{12}\\u_1 & u_2 & u_3\\v_1 & v_2 & v_3 \end{vmatrix}\ ,$
using the notation eℓm = eℓem. These expressions show these two types of vector are Hodge duals:[2]
$\star (\mathbf u \wedge \mathbf v )=\mathbf {u \times v} \ ; \ \ \star (\mathbf u \times \mathbf v ) = \mathbf u \wedge \mathbf v \ ;$
as a result of the relations:
$\star \mathbf e_{\ell} = \mathbf e_{\ell} \mathit i =\mathbf e_{\ell} \mathbf{e_1e_2e_3} = \mathbf e_m \mathbf e_n \ ;$ with $\ell, m, n$ cyclic,
and:
$\star ( \mathbf e_{\ell} \mathbf e_m ) =-( \mathbf e_{\ell} \mathbf e_m )\mathit{i} =-\left( \mathbf e_{\ell} \mathbf e_m \right)\mathbf{e_1e_2e_3} =\mathbf e_{n}$ also with $\ell, m, n$ cyclic.
Using the implementation of $\star$ based upon i, the commonly used relations are:[5]
$\mathbf {u \times v} = -(\mathbf u \wedge \mathbf v ) i \ ; \ \ \mathbf u \wedge \mathbf v = (\mathbf {u \times v} ) i \ .$
### Four dimensions
In case n = 4, the Hodge dual acts as an endomorphism of the second exterior power, of dimension 6. It is an involution, so it splits it into self-dual and anti-self-dual subspaces, on which it acts respectively as +1 and −1.
Another useful example is n = 4 Minkowski spacetime with metric signature $(+, -, -, -)$ and coordinates $(t, x, y, z)$ where (using $\varepsilon_{0123} = 1$)
$\star \mathrm{d}t=\mathrm{d}x\wedge \mathrm{d}y \wedge\mathrm{d}z$
$\star \mathrm{d}x=\mathrm{d}t\wedge \mathrm{d}y \wedge\mathrm{d}z$
$\star \mathrm{d}y=\mathrm{d}t\wedge \mathrm{d}z \wedge\mathrm{d}x$
$\star \mathrm{d}z=\mathrm{d}t\wedge \mathrm{d}x \wedge\mathrm{d}y$
for one-forms while
$\star \mathrm{d}t \wedge\mathrm{d}x = - \mathrm{d}y\wedge \mathrm{d}z$
$\star \mathrm{d}t \wedge\mathrm{d}y = \mathrm{d}x\wedge \mathrm{d}z$
$\star \mathrm{d}t \wedge\mathrm{d}z = - \mathrm{d}x\wedge \mathrm{d}y$
$\star \mathrm{d}x \wedge\mathrm{d}y = \mathrm{d}t\wedge \mathrm{d}z$
$\star \mathrm{d}x \wedge\mathrm{d}z = - \mathrm{d}t\wedge \mathrm{d}y$
$\star \mathrm{d}y \wedge\mathrm{d}z = \mathrm{d}t\wedge \mathrm{d}x$
for two-forms.
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## Inner product of k-vectors
The Hodge dual induces an inner product on the space of k-vectors, that is, on the exterior algebra of V. Given two k-vectors $\eta$ and $\zeta$, one has
$\zeta\wedge \star \eta = \langle\zeta, \eta \rangle\;\omega$
where ω is the normalised n-form (i.e. ω ∧ ∗ω = ω). In the calculus of exterior differential forms on a pseudo-Riemannian manifold of dimension n, the normalised n-form is called the volume form and can be written as
$\omega=\sqrt{|\det [g_{ij}]|}\;dx^1\wedge\cdots\wedge dx^n$
where $[g_{ij}]$ is the matrix of components of the metric tensor on the manifold in the coordinate basis.
If an inner product is given on $\Lambda^k(V)$, then this equation can be regarded as an alternative definition of the Hodge dual.[6] The wedge products of elements of an orthonormal basis in V form an orthonormal basis of the exterior algebra of V.
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## Duality
The Hodge star defines a dual in that when it is applied twice, the result is an identity on the exterior algebra, up to sign. Given a k-vector $\eta \in \Lambda^k (V)$ in an n-dimensional space V, one has
$\star {\star \eta}=(-1)^{k(n-k)}s\eta$
where s is related to the signature of the inner product on V. Specifically, s is the sign of the determinant of the inner product tensor. Thus, for example, if n=4 and the signature of the inner product is either (+,−,−,−) or (−,+,+,+) then s=−1. For ordinary Euclidean spaces, the signature is always positive, and so s=+1. When the Hodge star is extended to pseudo-Riemannian manifolds, then the above inner product is understood to be the metric in diagonal form.
Note that the above identity implies that the inverse of $\star$ can be given as
$\star^{-1}:\Lambda^k\ni\eta \mapsto (-1)^{k(n-k)}s{\star \eta} \in\Lambda^{n-k}$
Note that if n is odd k(n−k) is even for any k whereas if n is even k(n−k) has the parity of k.
Therefore, if n is odd, it holds for any k that
$\star^{-1} = s\star$
whereas, if n is even, it holds that
$\star^{-1} = (-1)^k s\star$
where k is the degree of the forms operated on.
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## Hodge star on manifolds
One can repeat the construction above for each cotangent space of an n-dimensional oriented Riemannian or pseudo-Riemannian manifold, and get the Hodge dual (n−k)-form, of a k-form. The Hodge star then induces an L2-norm inner product on the differential forms on the manifold. One writes
$(\eta,\zeta)=\int_M \eta\wedge \star \zeta = \int_M \langle \eta, \zeta \rangle \; d \mathrm{Vol}$
for the inner product of sections $\eta$ and $\zeta$ of $\Lambda^k(M)$. (The set of sections is frequently denoted as $\Omega^k(M)=\Gamma(\Lambda^k(M))$. Elements of $\Omega^k(M)$ are called exterior k-forms).
More generally, in the non-oriented case, one can define the hodge star of a k-form as a (n−k)-pseudo differential form; that is, a differential forms with values in the canonical line bundle.
### The codifferential
The most important application of the Hodge dual on manifolds to is to define the codifferential δ. Let
$\delta = (-1)^{nk + n + 1}s\, {\star d\star} = (-1)^k\,{\star^{-1}d\star}$
where d is the exterior derivative or differential, and s=+1 for Riemannian manifolds.
$d:\Omega^k(M)\rightarrow \Omega^{k+1}(M)$
while
$\delta:\Omega^k(M)\rightarrow \Omega^{k-1}(M).$
The codifferential is not an antiderivation on the exterior algebra, in contrast to the exterior derivative.
The codifferential is the adjoint of the exterior derivative, in that
$\langle \eta,\delta \zeta\rangle = \langle d\eta,\zeta\rangle.$
where ζ is a (k+1)-form and η a k-form. This identity follows from Stokes' theorem for smooth forms, when
$\int_M d(\eta \wedge \star \zeta)=0 =\int_M (d\eta \wedge \star \zeta - \eta\wedge \star (-1)^{k+1}\,{\star^{-1}d{\star \zeta}})=\langle d\eta,\zeta\rangle -\langle\eta,\delta\zeta\rangle$
i.e. when $M$ has empty boundary or when $\eta$ or $\star\zeta$ has zero boundary values (of course, true adjointness follows after continuous continuation to the appropriate topological vector spaces as closures of the spaces of smooth forms).
Notice that since the differential satisfies $d^2=0$, the codifferential has the corresponding property
$\! \delta^2 = s^2{\star d{\star {\star d{\star}}}} = (-1)^{k(n-k)} s^3{\star d^2\star} = 0$
The Laplace-deRham operator is given by
$\! \Delta=(\delta+d)^2 = \delta d + d\delta$
and lies at the heart of Hodge theory. It is symmetric:
$\langle\Delta \zeta,\eta\rangle = \langle\zeta,\Delta \eta\rangle$
and non-negative:
$\langle\Delta\eta,\eta\rangle \ge 0.$
The Hodge dual sends harmonic forms to harmonic forms. As a consequence of the Hodge theory, the de Rham cohomology is naturally isomorphic to the space of harmonic k-forms, and so the Hodge star induces an isomorphism of cohomology groups
$\star : H^k_\Delta(M)\to H^{n-k}_\Delta(M),$
which in turn gives canonical identifications via Poincaré duality of Hk(M) with its dual space.
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## Derivatives in three dimensions
The combination of the $\star$ operator and the exterior derivative d generates the classical operators grad, curl, and div, in three dimensions. This works out as follows: d can take a 0-form (function) to a 1-form, a 1-form to a 2-form, and a 2-form to a 3-form (applied to a 3-form it just gives zero). For a 0-form, $\omega=f(x,y,z)$, the first case written out in components is identifiable as the grad operator:
$d\omega=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz.$
The second case followed by $\star$ is an operator on 1-forms ($\eta=Adx+Bdy+Cdz$) that in components is the $curl$ operator:
$d\eta=\left({\partial C \over \partial y} - {\partial B \over \partial z}\right)dy\wedge dz + \left({\partial C \over \partial x} - {\partial A \over \partial z}\right)dx\wedge dz+\left({\partial B \over \partial x} - {\partial A \over \partial y}\right)dx\wedge dy.$
Applying the Hodge star gives:
$\star d\eta=\left({\partial C \over \partial y} - {\partial B \over \partial z}\right)dx - \left({\partial C \over \partial x} - {\partial A \over \partial z}\right)dy+\left({\partial B \over \partial x} - {\partial A \over \partial y}\right)dz.$
The final case prefaced and followed by $\star$, takes a 1-form ($\eta=Adx+Bdy+Cdz$) to a 0-form (function); written out in components it is the divergence operator:
$\star\eta=Ady\wedge dz-Bdx\wedge dz+Cdx\wedge dy$
$d{\star\eta}=\left(\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}\right)dx\wedge dy\wedge dz$
$\star d{\star\eta}=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}.$
One advantage of this expression is that the identity $d^2=0$, which is true in all cases, sums up two others, namely that $\operatorname{curl}(\operatorname{grad}(f))=0$ and $\operatorname{div}(\operatorname{curl}(\mathbf{F}))=0$. In particular, Maxwell's equations take on a particularly simple and elegant form, when expressed in terms of the exterior derivative and the Hodge star.
One can also obtain the Laplacian. Using the information above and the fact that $\Delta f = \operatorname{div}\; \operatorname{grad} f$ then for a 0-form, $\omega=f(x,y,z)$:
$\Delta \omega =\star d{\star d\omega}= \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}$
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## Notes
1. ^ a b Pertti Lounesto (2001). "§3.6 The Hodge dual". Clifford Algebras and Spinors, Volume 286 of London Mathematical Society Lecture Note Series (2nd ed.). Cambridge University Press. p. 39. ISBN 0-521-00551-5.
2. Venzo De Sabbata, Bidyut Kumar Datta (2007). "The pseudoscalar and imaginary unit". Geometric algebra and applications to physics. CRC Press. p. 53 ff. ISBN 1-58488-772-9.
3. William E Baylis (2004). "Chapter 4: Applications of Clifford algebras in physics". In Rafal Ablamowicz, Garret Sobczyk. Lectures on Clifford (geometric) algebras and applications. Birkhäuser. p. 100 ff. ISBN 0-8176-3257-3.
4. David Hestenes (1999). "The vector cross product". New foundations for classical mechanics: Fundamental Theories of Physics (2nd ed.). Springer. p. 60. ISBN 0-7923-5302-1.
5. Darling, R. W. R. (1994). Differential forms and connections. Cambridge University Press.
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## References
• David Bleecker (1981) Gauge Theory and Variational Principles. Addison-Wesley Publishing. ISBN 0-201-10096-7. Chpt. 0 contains a condensed review of non-Riemannian differential geometry.
• Jurgen Jost (2002) Riemannian Geometry and Geometric Analysis. Springer-Verlag. ISBN 3-540-42627-2. A detailed exposition starting from basic principles; does not treat the pseudo-Riemannian case.
• Charles W. Misner, Kip S. Thorne, John Archibald Wheeler (1970) Gravitation. W.H. Freeman. ISBN 0-7167-0344-0. A basic review of differential geometry in the special case of four-dimensional spacetime.
• Steven Rosenberg (1997) The Laplacian on a Riemannian manifold. Cambridge University Press. ISBN 0-521-46831-0. An introduction to the heat equation and the Atiyah-Singer theorem.
• Tevian Dray (1999) The Hodge Dual Operator. A thorough overview of the definition and properties of the Hodge dual operator.
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http://physics.stackexchange.com/questions/45053/the-cleverest-way-to-calculate-left-hatam-hata-dagger-n-right-wi | # The cleverest way to calculate $\left[\hat{a}^{M},\hat{a}^{\dagger N}\right]$ with $\left[\hat{a},\hat{a}^{\dagger}\right]=1$
Who can provide me some elegant solution for
$$\left[\hat{a}^{M},\hat{a}^{\dagger N}\right]\qquad\text{with} \qquad\left[\hat{a},\hat{a}^{\dagger}\right]~=~1$$
other than brute force calculation?
================================================
O lala!!! Thanks for @Prathyush and @Qmechanic !!! I got the same result with Qmechanic... I think Prathyush's suggestion should be equivalent to the my suggestion of the correspondence up to a canonical transformation. Here is my calculation (I was not confident to post it...)
$\begin{array}{c} \mbox{representation of }\left(\hat{a},\hat{a}^{\dagger}\right)\mbox{ on polynomial space }span\left\{ \frac{x^{n}}{\sqrt{n!}}\right\} _{n\ge0}\\ \hat{a}\left[f\left(x\right)\right]=\frac{d}{dx}f\left(x\right)\;;\;\hat{a}^{\dagger}\left[f\left(x\right)\right]=xf\left(x\right)\;;\;\left[\hat{a},\hat{a}^{\dagger}\right]\left[f\left(x\right)\right]=id\left[f\left(x\right)\right]\\ \left|0\right\rangle \sim 1\;;\;\left|n\right\rangle \sim x^{n}/\sqrt{n!} \end{array}$
$\begin{array}{c} \mbox{calculate the normal ordering }\left[\hat{a}^{M},\hat{a}^{\dagger}{}^{N}\right]\mbox{:}\\ \sim\left[\frac{d^{M}}{dx^{M}},x^{N}\right]=\frac{d^{M}}{dx^{M}}\left(x^{N}\star\right)-x^{N}\frac{d^{M}}{dx^{M}}\left(\star\right)\\ \sim\left\{ \overset{min\left\{ M,N\right\} }{\underset{k=0}{\sum}}\frac{N!}{\left(N-k\right)!}C_{M}^{k}\left(\hat{a}^{\dagger}\right)^{N-k}\left(\hat{a}\right)^{M-k}\right\} -\left(\hat{a}^{\dagger}\right)^{N}\left(\hat{a}\right)^{M}\\ \end{array}$
====================== One comment on 02-12-2012: The representation I was using is actually related to Bergmann representation with the inner product for Hilbert space (polynomials) being:
$\left\langle f\left(x\right),g\left(x\right)\right\rangle :=\int dxe^{-x^{2}}\overline{f\left(x\right)}g\left(x\right)\,,x\in\mathbb{R}\,,\, f,g\in\mathbb{C}\left[x\right]$
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I tried to map this commutator to the problem $\left[\frac{d}{dx},x\right]\circ f\left(x\right)=1\circ f\left(x\right),\,\hat{a}\sim\frac{d}{dx}\,,\,\hat{a}^{\dagger}\sim x$ but I don't know how to make this correspondence mathematically rigorous, i.e. to proof the existence of such a correspondence. – Yunlong Lian Nov 25 '12 at 9:56
the correspondence you write is not correct. its something like x+ip and x-ip look up wiki – Prathyush Nov 25 '12 at 10:12
@Prathyush If so, the correspondence should be $\hat{a}\sim x+\frac{d}{dx}\,,\,\hat{a}^{\dagger}\sim x-\frac{d}{dx}$, right? – Yunlong Lian Nov 25 '12 at 12:24
yes with an extra $\sqrt{\frac{1}{2}}$ factor – Prathyush Nov 25 '12 at 14:55
I wonder what the canonical transformation is that converts x+ip to x? – Prathyush Nov 25 '12 at 15:03
show 1 more comment
## 1 Answer
The standard way is to use generating functions (in this case a la coherent states). Usually one would like the resulting formula to be normal ordered.
1. Recall the following version $$\tag{1} e^Ae^B~=~e^{[A,B]}e^Be^A$$ of the Baker-Campbell-Hausdorff formula. It holds if the commutator $[A,B]$ commutes with both the operators $A$ and $B$.
2. Put $A=\alpha a$ and $B=\beta a^{\dagger}$, where $\alpha,\beta\in\mathbb{C}$.
3. Let $[a, a^{\dagger}]=\hbar {\bf 1}$, so that the commutator $[A,B]=\alpha\beta\hbar {\bf 1}$ is a $c$-number.
4. Now Taylor-expand the exponential factors in (1).
5. For fixed orders $n,m\in \mathbb{N}_0$, consider terms in (1) proportional to $\alpha^n\beta^m$.
6. Deduce that the normal ordered commutator is $$\tag{2} [a^n,(a^{\dagger})^m]~=~\sum_{k=1}^{\min(n,m)} \frac{n!m!\hbar^k}{(n-k)!(m-k)! k!}(a^{\dagger})^{m-k}a^{n-k}.$$
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.8620842695236206, "perplexity_flag": "middle"} |
http://mathematica.stackexchange.com/questions/tagged/intervals?sort=votes&pagesize=15 | # Tagged Questions
Mathematica's Interval Arithmetic is capable of handling the uncertainty in a rational manner
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### Can Mathematica Handle Open Intervals? Interval complements?
Open Intervals Following up on this question, I was wondering whether Mma can handle open intervals. For example, the union of the intervals, $$1<x<5$$ and $$5<x<8$$ should not ...
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### How to plot function over interval returned by Reduce[]
Let $f(x) = (x^2 + 1) / x$. I need to plot the function for the values of $x$ in which the square of the function is larger than 4, and its cube is smaller than 64. So I used ...
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### Union and Intersection of intervals
I have two sets $A = \{1 \leq x \leq 5\}$ and $B = \{5 \leq x \leq 8\}$. Now I want to find the Union and Intersection of $A$ ... | {"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": 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": 0.9061611294746399, "perplexity_flag": "head"} |
http://www.nag.com/numeric/CL/nagdoc_cl23/html/S/s21ccc.html | # NAG Library Function Documentnag_jacobian_theta (s21ccc)
## 1 Purpose
nag_jacobian_theta (s21ccc) returns the value of one of the Jacobian theta functions ${\theta }_{0}\left(x,q\right)$, ${\theta }_{1}\left(x,q\right)$, ${\theta }_{2}\left(x,q\right)$, ${\theta }_{3}\left(x,q\right)$ or ${\theta }_{4}\left(x,q\right)$ for a real argument $x$ and non-negative $q\le 1$.
## 2 Specification
#include <nag.h>
#include <nags.h>
double nag_jacobian_theta (Integer k, double x, double q, NagError *fail)
## 3 Description
nag_jacobian_theta (s21ccc) evaluates an approximation to the Jacobian theta functions ${\theta }_{0}\left(x,q\right)$, ${\theta }_{1}\left(x,q\right)$, ${\theta }_{2}\left(x,q\right)$, ${\theta }_{3}\left(x,q\right)$ and ${\theta }_{4}\left(x,q\right)$ given by
$θ 0 x,q = 1 + 2 ∑ n=1 ∞ -1 n q n 2 cos 2 n π x , θ 1 x,q = 2 ∑ n=0 ∞ -1 n q n + 1 2 2 sin 2 n + 1 π x , θ 2 x,q = 2 ∑ n=0 ∞ q n + 1 2 2 cos 2 n + 1 π x , θ 3 x,q = 1 + 2 ∑ n=1 ∞ q n 2 cos 2 n π x , θ 4 x,q = θ 0 x,q ,$
where $x$ and $q$ (the nome) are real with $0\le q\le 1$. Note that ${\theta }_{1}\left(x-\frac{1}{2},1\right)$ is undefined if $\left(x-\frac{1}{2}\right)$ is an integer, as is ${\theta }_{2}\left(x,1\right)$ if $x$ is an integer; otherwise, ${\theta }_{\mathit{i}}\left(x,1\right)=0$, for $\mathit{i}=0,1,\dots ,4$.
These functions are important in practice because every one of the Jacobian elliptic functions (see nag_jacobian_elliptic (s21cbc)) can be expressed as the ratio of two Jacobian theta functions (see Whittaker and Watson (1990)). There is also a bewildering variety of notations used in the literature to define them. Some authors (e.g., Abramowitz and Stegun (1972), 16.27) define the argument in the trigonometric terms to be $x$ instead of $\pi x$. This can often lead to confusion, so great care must therefore be exercised when consulting the literature. Further details (including various relations and identities) can be found in the references.
nag_jacobian_theta (s21ccc) is based on a truncated series approach. If $t$ differs from $x$ or $-x$ by an integer when $0\le t\le \frac{1}{2}$, it follows from the periodicity and symmetry properties of the functions that ${\theta }_{1}\left(x,q\right)=±{\theta }_{1}\left(t,q\right)$ and ${\theta }_{3}\left(x,q\right)=±{\theta }_{3}\left(t,q\right)$. In a region for which the approximation is sufficiently accurate, ${\theta }_{1}$ is set equal to the first term $\left(n=0\right)$ of the transformed series
$θ 1 t,q = 2 λ π e - λ t 2 ∑ n=0 ∞ -1 n e - λ n + 1 2 2 sinh 2 n + 1 λ t$
and ${\theta }_{3}$ is set equal to the first two terms (i.e., $n\le 1$) of
$θ 3 t,q = λ π e - λ t 2 1 + 2 ∑ n=1 ∞ e - λ n 2 cosh 2 n λ t ,$
where $\lambda ={\pi }^{2}/\left|{\mathrm{log}}_{\mathrm{e}}q\right|$. Otherwise, the trigonometric series for ${\theta }_{1}\left(t,q\right)$ and ${\theta }_{3}\left(t,q\right)$ are used. For all values of $x$, ${\theta }_{0}$ and ${\theta }_{2}$ are computed from the relations ${\theta }_{0}\left(x,q\right)={\theta }_{3}\left(\frac{1}{2}-\left|x\right|,q\right)$ and ${\theta }_{2}\left(x,q\right)={\theta }_{1}\left(\frac{1}{2}-\left|x\right|,q\right)$.
## 4 References
Abramowitz M and Stegun I A (1972) Handbook of Mathematical Functions (3rd Edition) Dover Publications
Byrd P F and Friedman M D (1971) Handbook of Elliptic Integrals for Engineers and Scientists pp. 315–320 (2nd Edition) Springer–Verlag
Magnus W, Oberhettinger F and Soni R P (1966) Formulas and Theorems for the Special Functions of Mathematical Physics 371–377 Springer–Verlag
Tølke F (1966) Praktische Funktionenlehre (Bd. II) 1–38 Springer–Verlag
Whittaker E T and Watson G N (1990) A Course in Modern Analysis (4th Edition) Cambridge University Press
## 5 Arguments
1: k – IntegerInput
On entry: the function ${\theta }_{\mathrm{K}}\left(x,q\right)$ to be evaluated. Note that ${\mathbf{k}}=4$ is equivalent to ${\mathbf{k}}=0$.
Constraint: $0\le {\mathbf{k}}\le 4$.
2: x – doubleInput
On entry: the argument $x$ of the function.
Constraints:
• x must not be an integer when ${\mathbf{q}}=1.0$ and ${\mathbf{k}}=2$ ;
• $\left({\mathbf{x}}-0.5\right)$ must not be an integer when ${\mathbf{q}}=1.0$ and ${\mathbf{k}}=1$. .
3: q – doubleInput
On entry: the argument $q$ of the function.
Constraint: $0.0\le {\mathbf{q}}\le 1.0$.
4: fail – NagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_INFINITE
The evaluation has been abandoned because the function value is infinite.
NE_INT
On entry, ${\mathbf{k}}=〈\mathit{\text{value}}〉$.
Constraint: $0\le {\mathbf{k}}\le 4$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
NE_REAL
On entry, ${\mathbf{q}}=〈\mathit{\text{value}}〉$.
Constraint: $0.0\le {\mathbf{q}}\le 1.0$.
On entry, ${\mathbf{x}}=〈\mathit{\text{value}}〉$.
Constraint: $\left({\mathbf{x}}-0.5\right)$ must not be an integer when ${\mathbf{q}}=1.0$ and ${\mathbf{k}}=1$.
On entry, ${\mathbf{x}}=〈\mathit{\text{value}}〉$.
Constraint: x must not be an integer when ${\mathbf{q}}=1.0$ and ${\mathbf{k}}=2$.
## 7 Accuracy
In principle nag_jacobian_theta (s21ccc) is capable of achieving full relative precision in the computed values. However, the accuracy obtainable in practice depends on the accuracy of the C standard library elementary functions such as sin and cos.
None.
## 9 Example
The example program evaluates ${\theta }_{2}\left(x,q\right)$ at $x=0.7$ when $q=0.4$, and prints the results.
### 9.1 Program Text
Program Text (s21ccce.c)
### 9.2 Program Data
Program Data (s21ccce.d)
### 9.3 Program Results
Program Results (s21ccce.r) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 70, "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": 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": 0.6319827437400818, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/110129/smooth-group-scheme-genrically-abelien | ## smooth group scheme genrically abelien
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let J a smooth group scheme over a smooth connected base S.
I assume, that over an open subset U of S, J is a torus, do I have that J is abelian?
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1
If $S$ is any scheme and $j:U \rightarrow S$ is a quasi-compact open immersion (so $j_{\ast}$ applied to quasi-coherent sheaves commutes with flat base change) such that $j$ is schematically dense (i.e., $O_S \rightarrow j_{\ast}(O_U)$ is injective) then $S$-maps $f, g:X \rightrightarrows Y$ are equal if $X$ is $S$-flat, $Y$ is $S$-separated, and $f_U = g_U$. This is a good exercise. It follows that a flat separated $S$-group $G$ is commutative if $G_U$ is commutative (via Sawin's argument with the commutator morphism, upgrading "set" to "scheme" suitably). – xbnv Oct 20 at 6:36
## 1 Answer
Yes. Consider the map: $J \times_S J \to J$ that sends $(a,b)$ to $aba^{-1}b^{-1}$. The pullback of the zero section along this map is a closed set. It is exactly the set of commuting pairs. It includes the inverse image of $U$, which is nonempty open.
Since $S$ is smooth and connected it is irreducible, so since $J$ is a torus over $U$ it has only one irreducible component over $U$, and any other irreducible component of $J$ would have to live entirely over $U^c$ which is impossible since it's flat, so $J$ is irreducible, so the inverse image of $U$ is dense, so the set of commuting pairs is the whole set, so it's commutative.
Edit: Per xbnv's comment, yes only if it's separated.
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This argument assumes via the consideration of the identity section as a closed immersion that $J$ is $S$-separated (and that $U$ is non-empty, which the OP clearly meant to assume). Without separatedness it is false over $S = {\rm{Spec}}(R)$ for any discrete valuation ring $R$. Indeed, choose a non-commutative finite group $G$ and let $J$ be the gluing along the open generic point $U$ of copies of $S$ indexed by $G$, equipped with the obvious $S$-group structure. This is quasi-finite etale over $S$ with $J_U$ trivial, but the special fiber is the non-commutative finite constant group $G$. – xbnv Oct 20 at 6:02
Dear @Will, I'm not sure what $S$ is supposed to be smooth over in the question, but in any case, why is a scheme smooth over some base which is also connected necessarily irreducible? A quasi-compact, connected, regular scheme is definitely irreducible, but I believe there are smooth morphisms with the source (and necessarily the target) not regular (I could be wrong). And without some quasi-compactness hypothesis, having all local rings domains and being connected does not imply irreducibility. – – Keenan Kidwell Oct 21 at 4:08
My wording was weird so I should clarify. I mean to ask: why is a connected scheme which is smooth over some base necessarily irreduicble? – Keenan Kidwell Oct 21 at 4:13
I assumed that $S$ was smooth over the spec of a field, since a base wasn't mentioned. In that case the implication is standard - if there are two different irreducible components, since it's connected they must meet at a point, and that point must have zero divisors in its local ring and thus me singular. If there is a base, and it's reducible, clearly the argument doesn't work. If it's irreducible I think the implication that $S$ is reducible still goes through. – Will Sawin Oct 21 at 5:20
1
@Will: Did you mean to say "irreducible" at the end of your preceding comment? A smooth connected scheme over an irreducible base need not be irreducible. For example, the projective planar nodal cubic $y^2 = x^2(x-1)$ over an alg. closed field is irreducible but it admits reducible connected finite etale covers of every degree $n > 1$ (namely, the so-called $n$-gon of projective lines). Basically, the problem is that whereas normality is local for the etale topology, irreducibility is not (in the absence of normality). – xbnv Oct 21 at 6:18
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": 47, "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": 0.9393302798271179, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/27324/what-are-some-naturally-occurring-high-degree-polynomials/27333 | ## What Are Some Naturally-Occurring High-Degree Polynomials?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
To construct J. H. Conway's look-and-say sequence, begin by putting down a 1 as the first entry. The other entries are found by saying the previous entry aloud, and writing what you hear.
````1
11
21
1211
111221
312211 (previous entry was three 1's, two 2's and one 1)
...
````
Conway provides his usual fantastical analysis in The Weird and Wonderful Chemistry of Audioactive Decay [Eureka 46, 5-18], where he demonstrates several otherworldly properties of this sequence. One was this: the ratio of the lengths of consecutive entries has a limit, $\lambda$. Furthermore, $\lambda$ is the root of a polynomial of degree 71.
Now, when I was in high school we were taught the quadratic formula and told there is a cubic formula, but you don't have to learn it. Why? "You won't be needing it." And mostly I've found that to be true. Am I wrong, or do high-degree polynomials rarely occur (in uncontrived settings)?
What are some other examples of useful roots of polynomials of high degree? Power series and the like can obviously produce useful polynomials of arbitrarily large degree, but I'm looking for surprises such as the degree-71 polynomial at the heart of the look-and-say sequence above.
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I would classify the question as wiki. Higher degree polynomials occur quite often; for example, see the comment of Jonas Meyer to mathoverflow.net/questions/21003. – Wadim Zudilin Jun 7 2010 at 7:36
OK, agreed. I converted to wiki. – I. J. Kennedy Jun 7 2010 at 7:56
What happens in other bases? (I guess Conway did a general analysis and that one always gets polynomials with degree depending on the basis.) One could also look at more exotic number systems (eg. the "Fibonacci base" or the "factorial base"). – Roland Bacher Jun 7 2010 at 7:58
1
Well, the title and your question don't fit quite well: do you like polynomials or the roots? In my response, I more worry about the polynomial, since the roots might be helpful but nobody will solve the algebraic relation. – Wadim Zudilin Jun 7 2010 at 7:59
3
@Roland, Wadim's point is that no number greater than 3 can occur in the example sequence. So the base is irrelevant. – TonyK Jun 7 2010 at 11:18
show 6 more comments
## 12 Answers
From Fernando Rodriguez-Villegas' preprint:
"Chebyshev in his work on the distribution of prime numbers used the following fact $$u_n:=\frac{(30n)!n!}{(15n)!(10n)!(6n)!}\in\mathbb Z, \qquad n = 0, 1, 2, \dots$$ (see also my question -- WZ). This is not immediately obvious (for example, this ratio of factorials is not a product of multinomial coefficients) but it is not hard to prove. The only proof I know proceeds by checking that the valuations $v_p(u_n)$ are non-negative for every prime $p$; an interpretation of $u_n$ as counting natural objects or being dimensions of natural vector spaces is far from clear.
As it turns out, the generating function $$u(\lambda):=\sum_{n=0}^\infty u_n\lambda^n$$ is algebraic over $\mathbb Q(\lambda)$; i.e. there is a polynomial $F\in\mathbb Q[x,y]$ such that $$F(\lambda,u(\lambda))=0.$$ However, we are not likely to see this polynomial explicitly any time soon as its degree is $483,840$ (!)"
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I should add that the roots of this monstrous polynomial will shed light on the integrality of $u_n$ without $p$-order study. – Wadim Zudilin Jun 7 2010 at 7:53
2
Many interesting polynomials are related to minimizing Mahler's measure (in connection with Lehmer's conjecture). But they are less astronomic, so I won't do a separate post. – Wadim Zudilin Jun 7 2010 at 8:03
### 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.
Some high-degree polynomials have been discovered using integer-relation algorithms applied to the powers of interesting constants. Here is a quotation from Section 3 of a paper of David Bailey, which can be found here: http://www.nersc.gov/homes/dhbailey/dhbpapers/pslq-cse.pdf
"One of the first results of this sort was the identification of the constant B3 = 3.54409035955 · · · [1]. B3 is the third bifurcation point of the logistic map $x_{k+1} = rx_k(1 − x_k )$, which exhibits period doubling shortly before the onset of chaos. To be precise, B3 is the smallest value of the parameter r such that successive iterates $x_k$ exhibit eight-way periodicity instead of four-way periodicity. Computations using a predecessor algorithm to PSLQ found that B3 is a root of the polynomial
$0 = 4913 + 2108t^2 − 604t^3 − 977t^4 + 8t^5 + 44t^6 + 392t^7 − 193t^8 − 40t^9 + 48t^{10} − 12t^{11} + t^{12}.$
Recently, B4 = 3.564407268705 · · ·, the fourth bifurcation point of the logistic map, was identified using PSLQ by British physicist David Broadhurst [5]. Some conjectural reasoning had suggested that B4 might satisfy a 240-degree polynomial, and some further analysis had suggested that the constant α = −B4 (B4 − 2) might satisfy a 120-degree polynomial. In order to test this hypothesis, Broadhurst applied a PSLQ program to the 121-long vector $(1, α, α^2 , · · · , α^{120})$. Indeed, a relation was found, although 10,000 digit arithmetic was required. The recovered integer coefficients descend monotonically from $257^{30} ≈ 1.986 × 10^{72}$ to one."
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I can only comment that David Broadhurst is not only a physicist but also a very strong mathematician! – Wadim Zudilin Jun 7 2010 at 8:19
The high-degree polynomial $x^{65537}-1$ is interesting because its nontrivial roots can be expressed in terms of square roots, and thus (in principle) the regular 65537-gon is constructible by ruler and compass. It is the largest known constructible $n$-gon with a prime number of sides. The roots, however, occupy several megabytes when written out in full.
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>(in principle) Actually, there's a book here in Goettingen which describes the construction. It's very heavy. Some poor soul got a PhD for this, I think. – Łukasz Grabowski Sep 30 2010 at 21:25
Yes, there is a book, by Johann Gustav Hermes, in the mathematics library at Göttingen. However, the experts I've spoken to doubt that Hermes really solved the equation. – John Stillwell Sep 30 2010 at 21:51
@JohnStilwell I know this is an old comment, but could you please expand on it? I find this very interesting, and would be nice to mention something about it next time it comes up in lecture. (I know of no works criticizing Hermes's construction) – Andres Caicedo May 11 at 17:17
1
@Andres: I was in Göttingen in 2002 (so my memory is fading a bit) and I was shown the book of Hermes' calculations by the number theorist S. J. Patterson. It is a very meticulous hand written book, but Patterson's opinion was that Hermes does not reach a clear solution of the equation. I wrote a little more about this, see tech.groups.yahoo.com/group/Hyacinthos/message/… – John Stillwell May 12 at 6:35
Thank you, John. (It would be nice if they digitized the book so others can see it.) – Andres Caicedo May 14 at 22:22
If you accept polynomials in more than one variable...
There is a polynomial inequality in 26 variables that describes the set of primes (Wikipedia - from where I also sourced the rest of this answer). By a result of Matiyasevich there is another polynomial inequality in 10 variables that also describes the primes, but whose order in on the order of $10^{45}$, which dwarfs Wadim's answer. In the other direction there is an order 4 polynomial inequality that will do the job, but in 58 variables.
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Some high degree polynomials appears in Delsarte scheme for estimating kissing numbers in $R^n$. Here is the excerpt from the Florian Pfender and Gunter M. Ziegler. Kissing numbers, sphere packings, and some unexpected proofs:
Theorem 3 (Delsarte, Goethals and Seidel [11]). If $$f(t)=\sum_{k=0}^d c_k G_k^{(n)}(t)$$ is a nonnegative combination of Gegenbauer polynomials, with $c_0 > 0$ and $c_k ≥ 0$ otherwise, and if $f (t) ≤ 0$ holds for all $t \in [−1, 1/2 ]$ , then the kissing number for $R^n$ is bounded by $$\kappa(n)\leq \frac{f(1)}{c_0}$$
For example for $n=24$ polynomial $f_{24}(t)=(t-\frac{1}{2})(t-\frac{1}{4})^2 t^2 (t+\frac{1}{4})^2 (t+\frac{1}{2})^2(t+1)$ gives the precise number for kissing number in $R^{24}$ - 196 560.
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Cyclotomic polynomials (http://en.wikipedia.org/wiki/Cyclotomic_polynomial), which when n is composite are rather more complicated than you might guess.
I think the point here is that polynomials play an implicit part all over algebra; and in algebraic geometry. The case of a quadratic equation is somewhat misleading, if useful on its own ground. It is possible to reason about polynomials without knowing exactly what the solutions look like. It is possible to solve numerically for roots of polynomial equations, without having a formula; and indeed the formulae we know for degree 3 and 4 are not necessarily useful for numerical work.
Fairly concrete examples are seen in the theory of Laplace transforms, where differential operators become, thanks to the transform, operators of multiplication by polynomials. Here engineering applications that have on the face of it no connection to polynomials may be studied by means of them, and the complexity of the polynomial reflects that of the problem you start with.
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All these examples do not really answer in the sense that cyclic polynomials, orthogonal polynomials etc form sequences and high dimensional polynomials in engineering applications (perhaps you think of formulae for numerical integrations) occur also only because we want fast convergencey or high precision and they have also analogues of low degree. Conway's example falls in some sense "out of the sky" and known "extraterrestrial" examples of such high degree are perhaps quite rare. – Roland Bacher Jun 7 2010 at 7:52
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Not, I think, a view explicit in the question; but of course the only way to find another John Conway would be cloning technology. High-degree polynomials do occur outside "contrived settings". For some problems such as class field towers (Golod-Shafarevich) they are there, and the question would be why anyone would write them down. – Charles Matthews Jun 7 2010 at 8:10
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Thanks for that comment, Roland. It captures the spirit of my question, which I failed to convey adequately. – I. J. Kennedy Dec 27 2010 at 4:43
This answer is not so bad. The cyclotomic polynomial of degree 105 is the first one with a coefficient not in {-1,0,1}. – drvitek Feb 6 2011 at 16:00
For practical matters, smooth functions (exp, sin, atan...) are often computed numerically using polynomials of high order.
Calculators compute the exponential function in the range [0,1] by evaluating a 12 degree polynomial whose coefficients are chosen so as to get 8 or 10 digits correct. These approximating polynomials can be found in formulas handbooks (e.g. "Methods and Programs for Mathematical Functions". Also the EDM lists a few in its appendices). They do a better job, at the given precision 10^(-8), than the standard Taylor polynomials.
Also have a look at your favorite language mathematical library, to see how standard functions (sin, exp, Atan...) are implemented (someone may suggest a link ?).
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I was recently amazed by an answer to this question.
I encountered another curious fact while working with hypergeometric functions. The following absolute value of a complex-valued $_4F_3$ function: $$\left|_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\sqrt{\phi }\right)\right|$$ where $\phi$ is the golden ratio, is actually an algebraic number with the minimal polynomial of degree 80 and a coefficient exceeding $10^{55}$ (see its explicit form here). I'm not sure if the root is expressible in radicals.
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My favorite is Theorem 9.2 on page 180 of "Primes of the form $x^2 + n y^2$ " by David A. Cox. In short, there is a polynomial $f_n(z)$ in one variable, integer coefficient and monic irreducible of degree $h(-4 n)$, such that if an odd prime $p$ does not divide $n$ or the discriminant of $f_n(z),$ then $p$ is represented by the positive binary quadratic form $x^2 + n y^2$ if and only if both $( -n | p) = 1$ and $f_n(z)$ has a root $\pmod p.$
Note that this does not duplicate the effect of calculating genera, that process divides up the primes by very simple congruences. The theorem comes into play with primes that are represented by some form in the principal genus, therefore automatically if there is only one genus. A prime is represented by at most one class $a x^2 + b x y + c y^2$ (and its opposite $a x^2 - b x y + c y^2$ if distinct) among forms of a given discriminant. There are similar theorems for odd discriminants and $x^2 + x y + k y^2,$ in case this $k > 0$ is even it is the same theorem, for utterly trivial reasons.
I found a cute way to incorporate this into a problem about polynomials in three variables, http://mathoverflow.net/questions/12486/integers-not-represented-by-2-x2-x-y-3-y2-z3-z
It is this problem I had with me when arriving at MO, in the same sense that Grace Slick arrived at Jefferson Airplane with "Somebody to Love" and "White Rabbit." I found all such problems for class number 3, and I'm guessing Kevin Buzzard's methods could finish all of them. I would like to know whether the same problem produces interesting phenomena in class number 5.
Meanwhile, Kaltofen and Yui made a serious effort to find such polynomials with small coefficients in "Explicit Construction of the Hilbert Class Fields of Imaginary Quadratic Fields by Integer Lattice Reduction," with table B5 giving three pages of polynomials.
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A very nice example, Will! Your reference reminded me about mathoverflow.net/questions/22001/… and about other monstrous polynomials, called modular polynomials (or modular equations). I expect that some of MOthematicians could argue about how natural are those monsters... – Wadim Zudilin Jun 9 2010 at 2:39
Right. On page 298 of Cox he gives $H_{-71}(X)$ with giant coefficients. There are some nice examples in Kaltofen and Yui, but nothing I could use for my problem with class number 5. Blair Spearman sent me one appropriate polynomial for h=5, but the numbers still got too big to see any patterns. – Will Jagy Jun 9 2010 at 4:20
The problem of packing $n$ equal circles into a unit square, or equivalently, finding the largest possible minimal distance between $n$ points in a unit square, produces some high-degree polynomials (the minimal polynomials of the radii of circles). For example, for $n=11$ the minimal polynomial has degree 18, and for $n=13$ it has degree 40 (cf. http://www.inf.u-szeged.hu/~pszabo/Pub/45survey.pdf, page 17).
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High-degree algebraic numbers also come up in lots of related optimization problems. For example, if you put eight identical charged particles on the 2-sphere and try to minimize Coulomb energy between them (i.e., the usual electrostatic potential), the conjectured minimum is an algebraic number of degree 48. – Henry Cohn Jan 7 2012 at 14:55
There is a rational function that arises naturally as a weight function in G. Chinta's "Mean values of biquadratic zeta functions" Invent. Math. 160 (2005), no.1, 145-164.
He writes out the numerator and denominator explicitly at the end of the paper (and it takes 3 pages),
http://www.sci.ccny.cuny.edu/~chinta/publ/biquad.pdf
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Don't people compute eigenvalues of very large matrices all the time? Those are the roots of the characteristic polynomial...
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It's hard to think of a worse way of computing eigenvalues numerically than calculating and then estimating the zeros of the characteristic polynomial – Alex R. Apr 27 at 4:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "mathjax_display_tex": 6, "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": 0.9331172704696655, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/519/how-do-you-prove-that-a-group-specified-by-a-presentation-is-infinite/606 | # How do you prove that a group specified by a presentation is infinite?
The group:
$$G = \left\langle x, y \; \left| \; x^2 = y^3 = (xy)^7 = 1\right. \right\rangle$$
is infinite, or so I've been told. How would I go about proving this? (To prove finiteness of a finitely presented group, I could do a coset enumeration, but I don't see how this helps if I want to prove that it's infinite.)
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It feels like the element $xy^2$ should be of infinite order here, but I'm not very sure of how to prove this. – Akhil Mathew Jul 22 '10 at 22:02
Is the operation simply multiplication here? – Noldorin Jul 23 '10 at 8:02
@Noldorin: This is an abstract group specified by a presentation, and, the operation is the group operation. This is not supposed to be a subset of C. – Simon Nickerson Jul 23 '10 at 12:04
@Simon: I misunderstood clearly. However, surely by the definition x can only have at most 2 values and y at most 3 values, even in the complex plane? It can't be infinite that way. – Noldorin Jul 23 '10 at 12:09
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@Noldorin: The group is the "freest" group that is generated by the elements x and y such that x^2 = y^3 = (xy)^7 = 1. As another example <x : x^n = 1> would be the cyclic group of order n. – Eric O. Korman Jul 23 '10 at 13:15
show 5 more comments
## 6 Answers
<x,y|x^2=y^3=1>≅PSL_2(Z) and this isomorphism identifies G with $PSL_2/T^7=1$ (where $T:z\mapsto z+1$). Result is the symmetry group of the tiling of the hyperbolic plane. From this description one can see that G is infinite (e.g. because there are infinitely many triangles in the tiling and G acts on them transitively).
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What does PSL_2 mean? – Casebash Jul 23 '10 at 23:38
SL_2 is 2x2 matrices with determinant 1. PSL_2 is the quotient of this group module its center, namely the scalar matrices. – Noah Snyder Jul 24 '10 at 0:14
Grigory has already answered your particular question. However, I wanted to point out that your question "How do you prove that a group specified by a presentation is infinite?" has no good answer in general. Indeed, in general the question of whether a group presentation defines the trivial group is undecidable.
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And an increadibly nice where one can read about this is John Stillwell's Classical topology and combinatorial group theory – Mariano Suárez-Alvarez♦ Jul 30 '10 at 16:23
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It's certainly true that there is no universal algorithm for solving such problems. But quite universal method of showing that a group is infinite (or non-trivial) is constructing transitive action on infinite (resp. non-trivial) set. – Grigory M Jul 31 '10 at 8:12
One general way to do this, (which is not guaranteed to work, as Noah points out), is to exhibit infinitely many different homomorphic images of the group you start with . In this case, any group generated by an element of order $2$ and an element of order $3$ whose product has order $7$ is a homomorphic image of the group $G$. Such a group is a Hurwitz group, and these are well studied, for example in the work of M. Conder and of G. Higman, among others. Infinitely many finite simple groups are known to be Hurwitz groups, I believe.
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Another approach, which can't work in general (see Noah's answer), but will surely work in this case, is to find a normal form for each element of the group, and then see whether there are finitely or infinitely many.
In practice, that means imagining a word in x and y, and then applying the relations as much as possible to simplify it, and then trying to figure out (and prove!) what the possible different "irreducible words" (i.e. words that can no longer be simplified) are.
In your case, the first thing one would note is that x can only appear to the 1st power (since any higher power can be simplified using x^2 = 1), while y can only appear to the powers +1 or -1 (for the same reason). Also, we can't have too many expressions of the form xy or yx in a row, because of the third relation.
One can keep going like this. I didn't, but what I imagine is that one can have expressions of the form x y x y^{-1} x y x y^{-1} ... that are arbitrarily long, and inequivalent, explaining the infinite order of the group.
It shouldn't be so hard to settle the question from this point of view by sitting down with pencil and paper and just playing around with different words to get a feel for what kinds of reductions can take place. (In geometric arguments like Grigory's, one uses a geometric context, such as an action on a hyperbolic tiling, as a more conceptual way of understanding the relations and distinguishing inequivalent words. But in this case I'm sure it won't be hard to see everything directly from the presentation.)
Added after rereading the question: what I am suggesting is precisely that even when the answer might be infinite, you can still hope to find a coset enumeration, at least in a case like this with relatively simple relations.
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Usually it's easy to show that some relations hold but quite hard to show that some relations doesn't (i.e. that constructed "canonical form" is indeed canonical)... – Grigory M Jul 31 '10 at 8:14
Of course that is true in general, since the word problem is not solvable in general, but not in this case. – Matt E Jul 31 '10 at 12:25
The group:
$$G = \left\langle x, y \mid x^l = y^m = (xy)^n = 1 \right\rangle$$ is triangular group so, if $\frac{1}{l} + \frac{1}{m} + \frac{1}{n} \lt 1$, then this group is infinite...
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– Andres Caicedo Apr 7 '12 at 14:47
Your group is the alternating subgroup of a Coxeter group
$$\langle x, y, z | x^2 = y^2 = z^2 = (xy)^2 = (yz)^3 = (zx)^7 = 1 \rangle$$
which is not on the (well-understood) list of finite Coxeter groups, and the alternating subgroup always has index $2$. (The connection to the hyperbolic plane is that Coxeter groups of rank $3$ always act as symmetries of a tiling of either the sphere, the Euclidean plane, or the hyperbolic plane by triangles.)
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http://en.wikipedia.org/wiki/Area | # Area
The combined area of these three shapes is 15.56 squares.
Area is a quantity that expresses the extent of a two-dimensional surface or shape, or planar lamina, in the plane. Area can be understood as the amount of material with a given thickness that would be necessary to fashion a model of the shape, or the amount of paint necessary to cover the surface with a single coat.[1] It is the two-dimensional analog of the length of a curve (a one-dimensional concept) or the volume of a solid (a three-dimensional concept).
The area of a shape can be measured by comparing the shape to squares of a fixed size.[2] In the International System of Units (SI), the standard unit of area is the square metre (written as m2), which is the area of a square whose sides are one metre long.[3] A shape with an area of three square metres would have the same area as three such squares. In mathematics, the unit square is defined to have area one, and the area of any other shape or surface is a dimensionless real number.
There are several well-known formulas for the areas of simple shapes such as triangles, rectangles, and circles. Using these formulas, the area of any polygon can be found by dividing the polygon into triangles.[4] For shapes with curved boundary, calculus is usually required to compute the area. Indeed, the problem of determining the area of plane figures was a major motivation for the historical development of calculus.[5]
For a solid shape such as a sphere, cone, or cylinder, the area of its boundary surface is called the surface area.[1][6] Formulas for the surface areas of simple shapes were computed by the ancient Greeks, but computing the surface area of a more complicated shape usually requires multivariable calculus.
Area plays an important role in modern mathematics. In addition to its obvious importance in geometry and calculus, area is related to the definition of determinants in linear algebra, and is a basic property of surfaces in differential geometry.[7] In analysis, the area of a subset of the plane is defined using Lebesgue measure,[8] though not every subset is measurable.[9] In general, area in higher mathematics is seen as a special case of volume for two-dimensional regions.[1]
Area can be defined through the use of axioms, defining it as a function of a collection of certain plane figures to the set of real numbers. It can be proved that such a function exists.
## Formal definition
See also: Jordan measure
An approach to defining what is meant by "area" is through axioms. "Area" can be defined as a function from a collection M of special kind of plane figures (termed measurable sets) to the set of real numbers which satisfies the following properties:
• For all S in M, a(S) ≥ 0.
• If S and T are in M then so are S ∪ T and S ∩ T, and also a(S∪T) = a(S) + a(T) − a(S∩T).
• If S and T are in M with S ⊆ T then T − S is in M and a(T−S) = a(T) − a(S).
• If a set S is in M and S is congruent to T then T is also in M and a(S) = a(T).
• Every rectangle R is in M. If the rectangle has length h and breadth k then a(R) = hk.
• Let Q be a set enclosed between two step regions S and T. A step region is formed from a finite union of adjacent rectangles resting on a common base, i.e. S ⊆ Q ⊆ T. If there is a unique number c such that a(S) ≤ c ≤ a(T) for all such step regions S and T, then a(Q) = c.
It can be proved that such an area function actually exists.[10]
## Units
A square metre quadrat made of PVC pipe.
Every unit of length has a corresponding unit of area, namely the area of a square with the given side length. Thus areas can be measured in square metres (m2), square centimetres (cm2), square millimetres (mm2), square kilometres (km2), square feet (ft2), square yards (yd2), square miles (mi2), and so forth.[11] Algebraically, these units can be thought of as the squares of the corresponding length units.
The SI unit of area is the square metre, which is considered an SI derived unit.[3]
### Conversions
Although there are 10 mm in 1 cm, there are 100 mm2 in 1 cm2.
The conversion between two square units is the square of the conversion between the corresponding length units. For example, since
1 foot = 12 inches,
the relationship between square feet and square inches is
1 square foot = 144 square inches,
where 144 = 122 = 12 × 12. Similarly:
• 1 square kilometer = 1,000,000 square meters
• 1 square meter = 10,000 square centimetres = 1,000,000 square millimetres
• 1 square centimetre = 100 square millimetres
• 1 square yard = 9 square feet
• 1 square mile = 3,097,600 square yards = 27,878,400 square feet
In addition,
• 1 square inch = 6.4516 square centimetres
• 1 square foot = 0.09290304 square metres
• 1 square yard = 0.83612736 square metres
• 1 square mile = 2.589988110336 square kilometres
### Other units
See also: Category:Units of area
There are several other common units for area. The "Are" was the original unit of area in the metric system, with;
• 1 are = 100 square metres
Though the are has fallen out of use, the hectare is still commonly used to measure land:[11]
• 1 hectare = 100 ares = 10,000 square metres = 0.01 square kilometres
Other uncommon metric units of area include the tetrad, the hectad, and the myriad.
The acre is also commonly used to measure land areas, where
• 1 acre = 4,840 square yards = 43,560 square feet.
An acre is approximately 40% of a hectare.
On the atomic scale, area is measured in units of barns, such that:[11]
• 1 barn = 10−28 square meters.
The barn is commonly used in describing the cross sectional area of interaction in nuclear physics.[11]
In India,
• 20 Dhurki = 1 Dhur
• 20 Dhur = 1 Khatha
• 20 Khata = 1 Bigha
• 32 Khata = 1 Acre
## Area formulae
### Polygon Formulae
#### Rectangles
The area of this rectangle is lw.
The most basic area formula is the formula for the area of a rectangle. Given a rectangle with length l and w, the formula for the area is:[2]
A = lw (rectangle)
That is, the area of the rectangle is the length multiplied by the width. As a special case, as $l = w$ in the case of a square, the area of a square with side length s is given by the formula:[1][2]
A = s2 (square)
The formula for the area of a rectangle follows directly from the basic properties of area, and is sometimes taken as a definition or axiom. On the other hand, if geometry is developed before arithmetic, this formula can be used to define multiplication of real numbers.
Equal area figures.
#### Dissection formulae
Most other simple formulae for area follow from the method of dissection. This involves cutting a shape into pieces, whose areas must sum to the area of the original shape.
For an example, any parallelogram can be subdivided into a trapezoid and a right triangle, as shown in figure to the left. If the triangle is moved to the other side of the trapezoid, then the resulting figure is a rectangle. It follows that the area of the parallelogram is the same as the area of the rectangle:[2]
A = bh (parallelogram).
Two equal triangles.
However, the same parallelogram can also be cut along a diagonal into two congruent triangles, as shown in the figure to the right. It follows that the area of each triangle is half the area of the parallelogram:[2]
$A = \frac{1}{2}bh$ (triangle).
Similar arguments can be used to find area formulae for the trapezoid and the rhombus, as well as more complicated polygons.[citation needed]
### Area of curved shapes
#### Circles
A circle can be divided into sectors which rearrange to form an approximate parallelogram.
Main article: Area of a circle
The formula for the area of a circle (more properly called area of a disk) is based on a similar method. Given a circle of radius r, it is possible to partition the circle into sectors, as shown in the figure to the right. Each sector is approximately triangular in shape, and the sectors can be rearranged to form and approximate parallelogram. The height of this parallelogram is r, and the width is half the circumference of the circle, or πr. Thus, the total area of the circle is r × πr, or πr2:[2]
A = πr2 (circle).
Though the dissection used in this formula is only approximate, the error becomes smaller and smaller as the circle is partitioned into more and more sectors. The limit of the areas of the approximate parallelograms is exactly πr2, which is the area of the circle.[12]
This argument is actually a simple application of the ideas of calculus. In ancient times, the method of exhaustion was used in a similar way to find the area of the circle, and this method is now recognized as a precursor to integral calculus. Using modern methods, the area of a circle can be computed using a definite integral:
$A \;=\; \int_{-r}^r 2\sqrt{r^2 - x^2}\,dx \;=\; \pi r^2$
#### Ellipses
Main article: Ellipse#Area
The formula for the area of an ellipse is related to the formula of a circle; for an ellipse with semi-major and semi-minor axes x and y the formula is:[2]
$A = \pi xy \,\!$
#### Surface area
Main article: Surface area
Archimedes showed that the surface area and volume of a sphere is exactly 2/3 of the area and volume of the surrounding cylindrical surface.
Most basic formulae for surface area can be obtained by cutting surfaces and flattening them out. For example, if the side surface of a cylinder (or any prism) is cut lengthwise, the surface can be flattened out into a rectangle. Similarly, if a cut is made along the side of a cone, the side surface can be flattened out into a sector of a circle, and the resulting area computed.
The formula for the surface area of a sphere is more difficult to derive: because the surface of a sphere has nonzero Gaussian curvature, it cannot be flattened out. The formula for the surface area of a sphere was first obtained by Archimedes in his work On the Sphere and Cylinder. The formula is:[6]
A = 4πr2 (sphere).
where r is the radius of the sphere. As with the formula for the area of a circle, any derivation of this formula inherently uses methods similar to calculus.
### General formulae
#### Areas of 2-dimensional figures
• A triangle: $\tfrac12Bh$ (where B is any side, and h is the distance from the line on which B lies to the other vertex of the triangle). This formula can be used if the height h is known. If the lengths of the three sides are known then Heron's formula can be used: $\sqrt{s(s-a)(s-b)(s-c)}$ where a, b, c are the sides of the triangle, and $s = \tfrac12(a + b + c)$ is half of its perimeter.[2] If an angle and its two included sides are given, the area is $\tfrac12 a b \sin(C)$ where C is the given angle and a and b are its included sides.[2] If the triangle is graphed on a coordinate plane, a matrix can be used and is simplified to the absolute value of $\tfrac12(x_1 y_2 + x_2 y_3 + x_3 y_1 - x_2 y_1 - x_3 y_2 - x_1 y_3)$. This formula is also known as the shoelace formula and is an easy way to solve for the area of a coordinate triangle by substituting the 3 points (x1,y1), (x2,y2), and (x3,y3). The shoelace formula can also be used to find the areas of other polygons when their vertices are known. Another approach for a coordinate triangle is to use Infinitesimal calculus to find the area.
• A simple polygon constructed on a grid of equal-distanced points (i.e., points with integer coordinates) such that all the polygon's vertices are grid points: $i + \frac{b}{2} - 1$, where i is the number of grid points inside the polygon and b is the number of boundary points.[13] This result is known as Pick's theorem.[13]
#### Area in calculus
Integration can be thought of as measuring the area under a curve, defined by f(x), between two points (here a and b).
The area between two graphs can be evaluated by calculating the difference between the integrals of the two functions
• The area between a positive-valued curve and the horizontal axis, measured between two values a and b (b is defined as the larger of the two values) on the horizontal axis, is given by the integral from a to b of the function that represents the curve:[1]
$A = \int_a^{b} f(x) \, dx$
• The area between the graphs of two functions is equal to the integral of one function, f(x), minus the integral of the other function, g(x):
$A = \int_a^{b} ( f(x) - g(x) ) \, dx$ where $f(x)$ is the curve with the greater y-value.
• An area bounded by a function r = r(θ) expressed in polar coordinates is:[1]
$A = {1 \over 2} \int r^2 \, d\theta$
• The area enclosed by a parametric curve $\vec u(t) = (x(t), y(t))$ with endpoints $\vec u(t_0) = \vec u(t_1)$ is given by the line integrals:
$\oint_{t_0}^{t_1} x \dot y \, dt = - \oint_{t_0}^{t_1} y \dot x \, dt = {1 \over 2} \oint_{t_0}^{t_1} (x \dot y - y \dot x) \, dt$
(see Green's theorem) or the z-component of
${1 \over 2} \oint_{t_0}^{t_1} \vec u \times \dot{\vec u} \, dt.$
#### Surface area of 3-dimensional figures
• cone:[14] $\pi r\left(r + \sqrt{r^2 + h^2}\right)$, where r is the radius of the circular base, and h is the height. That can also be rewritten as $\pi r^2 + \pi r l$[14] or $\pi r (r + l) \,\!$ where r is the radius and l is the slant height of the cone. $\pi r^2$ is the base area while $\pi r l$ is the lateral surface area of the cone.[14]
• cube: $6s^2$, where s is the length of an edge.[6]
• cylinder: $2\pi r(r + h)$, where r is the radius of a base and h is the height. The 2$\pi$r can also be rewritten as $\pi$ d, where d is the diameter.
• prism: 2B + Ph, where B is the area of a base, P is the perimeter of a base, and h is the height of the prism.
• pyramid: $B + \frac{PL}{2}$, where B is the area of the base, P is the perimeter of the base, and L is the length of the slant.
• rectangular prism: $2 (\ell w + \ell h + w h)$, where $\ell$ is the length, w is the width, and h is the height.
#### General formula
The general formula for the surface area of the graph of a continuously differentiable function $z=f(x,y),$ where $(x,y)\in D\subset\mathbb{R}^2$ and $D$ is a region in the xy-plane with the smooth boundary:
$A=\iint_D\sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2+1}\,dx\,dy.$
Even more general formula for the area of the graph of a parametric surface in the vector form $\mathbf{r}=\mathbf{r}(u,v),$ where $\mathbf{r}$ is a continuously differentiable vector function of $(u,v)\in D\subset\mathbb{R}^2$:[7]
$A=\iint_D \left|\frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\right|\,du\,dv.$
### List of formulae
There are formulae for many different regular and irregular polygons, and those additional to the ones above are listed here.
Additional common formulae for area:
Shape Formula Variables
Regular triangle (equilateral triangle) $\frac{1}{4} \sqrt{3}s^2\,\!$ $s$ is the length of one side of the triangle.
Triangle[1] $\sqrt{s(s-a)(s-b)(s-c)}\,\!$ $s$ is half the perimeter, $a$, $b$ and $c$ are the length of each side.
Triangle[2] $\tfrac12 a b \sin(C)\,\!$ $a$ and $b$ are any two sides, and $C$ is the angle between them.
Triangle[1] $\tfrac12bh \,\!$ $b$ and $h$ are the base and altitude (measured perpendicular to the base), respectively.
Rhombus $\tfrac12ab$ $a$ and $b$ are the lengths of the two diagonals of the rhombus.
Parallelogram $bh\,\!$ $b$ is the length of the base and $h$ is the perpendicular height.
Trapezoid $\tfrac12(a+b)h \,\!$ $a$ and $b$ are the parallel sides and $h$ the distance (height) between the parallels.
Regular hexagon $\frac{3}{2} \sqrt{3}s^2\,\!$ $s$ is the length of one side of the hexagon.
Regular octagon $2(1+\sqrt{2})s^2\,\!$ $s$ is the length of one side of the octagon.
Regular polygon $\frac{1}{4}nl^2\cdot \cot(\pi/n)\,\!$ $l$ is the side length and $n$ is the number of sides.
Regular polygon $\frac{1}{4n}p^2\cdot \cot(\pi/n)\,\!$ $p$ is the perimeter and $n$ is the number of sides.
Regular polygon $\frac{1}{2}nR^2\cdot \sin(2\pi/n) = nr^2 \tan(\pi/n)\,\!$ $R$ is the radius of a circumscribed circle, $r$ is the radius of an inscribed circle, and $n$ is the number of sides.
Regular polygon $\tfrac12a p \,\!$ $a$ is the apothem, or the radius of an inscribed circle in the polygon, and $p$ is the perimeter of the polygon.
Circle $\pi r^2\ \text{or}\ \frac{\pi d^2}{4} \,\!$ $r$ is the radius and $d$ the diameter.
Circular sector $\frac{\theta}{2}r^2\ \text{or}\ \frac{L \cdot r}{2}\,\!$ $r$ and $\theta$ are the radius and angle (in radians), respectively and $L$ is the length of the perimeter.
Ellipse[2] $\pi ab \,\!$ $a$ and $b$ are the semi-major and semi-minor axes, respectively.
Total surface area of a cylinder $2\pi r (r + h)\,\!$ $r$ and $h$ are the radius and height, respectively.
Lateral surface area of a cylinder $2 \pi r h \,\!$ $r$ and $h$ are the radius and height, respectively.
Total surface area of a sphere[6] $4\pi r^2\ \text{or}\ \pi d^2\,\!$ $r$ and $d$ are the radius and diameter, respectively.
Total surface area of a pyramid[6] $B+\frac{P L}{2}\,\!$ $B$ is the base area, $P$ is the base perimeter and $L$ is the slant height.
Total surface area of a pyramid frustum[6] $B+\frac{P L}{2}\,\!$ $B$ is the base area, $P$ is the base perimeter and $L$ is the slant height.
Square to circular area conversion $\frac{4}{\pi} A\,\!$ $A$ is the area of the square in square units.
Circular to square area conversion $\frac{1}{4} C\pi\,\!$ $C$ is the area of the circle in circular units.
The above calculations show how to find the area of many common shapes.
The areas of irregular polygons can be calculated using the "Surveyor's formula".[12]
## Optimization
Given a wire contour, the surface of least area spanning ("filling") it is a minimal surface. Familiar examples include soap bubbles.
The question of the filling area of the Riemannian circle remains open.[citation needed]
## See also
• Equi-areal mapping
• Integral
• Orders of magnitude (area)—A list of areas by size.
• Perimeter
• Planimeter, an instrument for measuring small areas, e.g. on maps.
• Volume
## References
1. Eric W. Weisstein. "Area". Wolfram MathWorld. Retrieved 3 July 2012.
2. "Area Formulas". Math.com. Retrieved 2 July 2012.
3. ^ a b Bureau International des Poids et Mesures Resolution 12 of the 11th meeting of the CGPM (1960), retrieved 15 July 2012
4. Mark de Berg; Marc van Kreveld; Mark Overmars; Otfried Schwarzkopf (2000). "Chapter 3: Polygon Triangulation". Computational Geometry (2nd revised ed.). Springer-Verlag. pp. 45–61. ISBN 3-540-65620-0
5. Boyer, Carl B. (1959). A History of the Calculus and Its Conceptual Development. Dover. ISBN 0-486-60509-4.
6. Eric W. Weisstein. "Surface Area". Wolfram MathWorld. Retrieved 3 July 2012.
7. ^ a b
8. Walter Rudin, Real and Complex Analysis, McGraw-Hill, 1966, ISBN 0-07-100276-6.
9. Moise, Edwin (1963). Elementary Geometry from an Advanced Standpoint. Addison-Wesley Pub. Co. Retrieved 15 July 2012.
10. ^ a b c d Bureau international des poids et mesures (2006). The International System of Units (SI). 8th ed. Retrieved 2008-02-13. Chapter 5.
11. ^ a b Braden, Bart (September 1986). "The Surveyor's Area Formula". The College Mathematics Journal 17 (4): 326–337. doi:10.2307/2686282. Retrieved 15 July 2012.
12. ^ a b Trainin, J. (2007-11). "An elementary proof of Pick's theorem". 91 (522): 536–540.
13. ^ a b c Eric W. Weisstein. "Cone". Wolfram MathWorld. Retrieved 6 July 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": 110, "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": 0.9001768231391907, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/55471/list | ## Return to Answer
4 Added: Complement
Dear Hugo, the wonderful formalism of schemes allows us to have a Riemann-Roch theorem for a projective curve $X$ over an arbitrary field $k$, even without any assumption of smoothness. It says, like in the good old times of Riemann surfaces, that for a Cartier divisor $D$ on $X$ we have $$\chi (\mathcal O_X(D))= deg(D)+ \chi (\mathcal O_X)$$
There is a dualizing sheaf $\omega$ and Serre duality yields the formula $$h^0(X,\mathcal O_X(D))-h^0(X,\omega \otimes\mathcal O_X(-D))=1-p_a(X)+deg D$$ where $p_a(X)=1-\chi(\mathcal O_x)$ is the so called arithmetic genus of the curve.
Everything is in our friend Qing Liu's fantastic book Algebraic Geometry and Arithmetic Curves but I bet he's too modest to give you this obvious answer!
Edit The first displayed formula is actually valid in even greater generality: it holds for any projective curve $X$ (smooth or not) over an arbitrary artinian ring $k$. The proof is on page 164 of
Altman, A.; Kleiman, S., Introduction to Grothendieck duality theory. Lecture Notes in Mathematics No. 146, Springer-Verlag, Berlin-New York, 1970
Complement As an answer to Hugo's question in his comment below, let me add that indeed, in the case of a smooth projective curve over a field $k$, the dualizing sheaf $\omega$ is nothing else than the canonical sheaf . More generally for a smooth projective variety of dimension $r$ over $k$, the dualizing sheaf is just the canonical sheaf $\omega=\Omega^r_{X/k}$. This is (a special case of) Theorem I.4.6, page 14 in Altman-Kleiman's monograph.
3 suppressed redundant word "book" that I had used twice
Dear Hugo, the wonderful formalism of schemes allows us to have a Riemann-Roch theorem for a projective curve $X$ over an arbitrary field $k$, even without any assumption of smoothness. It says, like in the good old times of Riemann surfaces, that for a Cartier divisor $D$ on $X$ we have $$\chi (\mathcal O_X(D))= deg(D)+ \chi (\mathcal O_X)$$
There is a dualizing sheaf $\omega$ and Serre duality yields the formula $$h^0(X,\mathcal O_X(D))-h^0(X,\omega \otimes\mathcal O_X(-D))=1-p_a(X)+deg D$$ where $p_a(X)=1-\chi(\mathcal O_x)$ is the so called arithmetic genus of the curve.
Everything is in our friend Qing Liu's book fantastic book Algebraic Geometry and Arithmetic Curves but I bet he is he's too modest to give you this obvious answer!
Edit The first displayed formula is actually valid in even greater generality: it holds for any projective curve $X$ (smooth or not) over an arbitrary artinian ring $k$. The proof is on page 164 of
Altman, A.; Kleiman, S., Introduction to Grothendieck duality theory. Lecture Notes in Mathematics No. 146, Springer-Verlag, Berlin-New York, 1970
2 added reference for statement over artin ring
Dear Hugo, the wonderful formalism of schemes allows us to have a Riemann-Roch theorem for a projective curve $X$ over an arbitrary field $k$, even without any assumption of smoothness. It says, like in the good old times of Riemann surfaces, that for a Cartier divisor $D$ on $X$ we have $$\chi (\mathcal O_X(D))= deg(D)+ \chi (\mathcal O_X)$$
There is a dualizing sheaf $\omega$ and Serre duality yields the formula $$h^0(X,\mathcal O_X(D))-h^0(X,\omega \otimes\mathcal O_X(-D))=1-p_a(X)+deg D$$ where $p_a(X)=1-\chi(\mathcal O_x)$ is the so called arithmetic genus of the curve.
Everything is in our friend Qing Liu's book fantastic book Algebraic Geometry and Arithmetic Curves but I bet he is too modest to give you this obvious answer!
Edit The first displayed formula is actually valid in even greater generality: it holds for any projective curve $X$ (smooth or not) over an arbitrary artinian ring $k$. The proof is on page 164 of
Altman, A.; Kleiman, S., Introduction to Grothendieck duality theory. Lecture Notes in Mathematics No. 146, Springer-Verlag, Berlin-New York, 1970
1
Dear Hugo, the wonderful formalism of schemes allows us to have a Riemann-Roch theorem for a projective curve $X$ over an arbitrary field $k$, even without any assumption of smoothness. It says, like in the good old times of Riemann surfaces, that for a Cartier divisor $D$ on $X$ we have $$\chi (\mathcal O_X(D))= deg(D)+ \chi (\mathcal O_X)$$
There is a dualizing sheaf $\omega$ and Serre duality yields the formula $$h^0(X,\mathcal O_X(D))-h^0(X,\omega \otimes\mathcal O_X(-D))=1-p_a(X)+deg D$$ where $p_a(X)=1-\chi(\mathcal O_x)$ is the so called arithmetic genus of the curve.
Everything is in our friend Qing Liu's book fantastic book Algebraic Geometry and Arithmetic Curves but I bet he is too modest to give you this obvious answer! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "mathjax_display_tex": 8, "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": 0.905025839805603, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/55524/rule-with-independent-random-variables-and-conditional-expectations | # Rule with independent random variables and conditional expectations
I want to use a rule for conditional expectation I found in (German) wikipedia, not in my script/textbook of probability theory, I guess it should be simple and follow more or less straight from the general definition (I want have a proof to be sure that I don't build up on a wikipedia mistake)
Let X be independent of Z and of Y (XY integrable and X,Y,Z random variables) $$E(XY|Z) = E(X) E(Y|Z)$$
My idea: Showing that the rhs meets the conditions of the general definition of E(XY|Z), that is (i) it should be $\sigma(Z)$-measureable, check. (ii) $E\left( E(X) E(Y|Z) 1_A \right) \stackrel{!}{=} E(XY 1_A) \forall A \in \sigma(Z)$ Now the lhs$=E(X) E(E(Y|Z)1_A) = E(X)E(Y 1_A)$ (according to (ii) of the definition of $E(Y|Z)$, for all $A\in \sigma(Z)$)
$= E(XY1_A)$ as wished (X, Y are independent).
But, in this proof I did not use that X,Z are independent, so it would follow as well $E(XY|Z)=E(Y)E(X|Z)=E(Y)E(X)=E(XY)$ which shouldn't be this way.
Maybe it's all much simpler and I have just the wrong point of view on it. Q: Does anybody see the flaw in my proof? Can anybody hint me to a proof or a reference to a proof?
(@Didier, i try to get Williams book, I have to see if my library can get it for me).
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The faulty step is when you assert that $E(X)E(Y1_A)=E(XY1_A)$. Here you must not only assume that $X$ is independent of $Y$ (which is not enough to conclude) but that $X$ is independent of $(Y,Z)$. (+1 for showing the steps you tried.) – Did Aug 4 '11 at 8:56
The usual counterexample works: take X and Y i.i.d. centered Bernoulli random variables and Z=XY. Then (X,Y) is independent (by definition), as are (Y,Z) and (Z,X) (easy to check), but (X,Y,Z) is not (for example P(X=Y=1,Z=-1)=0 instead of $\frac18$). And E(XY|Z)=Z although E(X)E(Y|Z)=0. – Did Aug 4 '11 at 12:05
– Johannes L Aug 4 '11 at 12:34
Yes, [X is independent of (Y,Z)] means that the sigma-algebras sigma(X) and sigma(Y,Z) are independent. And sigma(Y,Z) coincides with sigma(sigma(Y),sigma(Z)). – Did Aug 4 '11 at 13:37
So I would have to have (i) X independent from (Y,Z).. I wonder if that follows if I assume Y and Z independent (which would be more natural to my application than to have to introduce the assumption (i)) - i make a new question out of this. – Johannes L Aug 5 '11 at 11:47
## 1 Answer
The faulty step is when you assert that $E(X)E(Y1_A)=E(XY1_A)$. Here one must not only assume that $X$ is independent on $Y$ but that $X$ is independent on $(Y,Z)$.
To see that there is a difference, the usual example works here: take $X$ and $Y$ i.i.d. centered Bernoulli random variables and $Z=XY$. Then $X$ and $Y$ are independent (by definition), as are $Y$ and $Z$, as are $Z$ and $X$ (easy to check), but $X$, $Y$ and $Z$ are not*. And $E(XY|Z)=Z$ although $E(X)E(Y|Z)=0$.
*For example, $P(X=1,Y=1,Z=-1)=0$ but $P(X=1)P(Y=1)P(Z=-1)=1/8$.
- | {"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": 34, "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": 0.9517104625701904, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/22353?sort=votes | Do you know of any asymmetric, nonparametric measure of dependence?
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
A measure of dependence is a way to assign a number (usually normalized between 0 and 1) to a couple of random variable, such that $\delta(X,Y)=0$ if and only of $X$ and $Y$ are independent, and $\delta(X,Y)=1$ as soon as there is a complete dependence between $X$ and $Y$. Rényi introduced in the paper On measures of dependence (Acta Math. Acad. Sci. Hungar. 10 1959 441--451) a set of axioms that a measure of dependence should fulfill. Among them are nonparametricity ($\delta(f(X),g(Y))=\delta(X,Y)$ as soon as $f,g$ are inversible bimeasurable functions, so that $\delta$ dos not rely on a metric, or affine structure unlike the correlation) and symmetry ($\delta(X,Y)=\delta(Y,X)$), and the "complete dependence" is defined by the existence of a relation $X=f(Y)$ of $Y=f(X)$.
I feel that the symmetry assumption is unnatural. In view of, for example, Bell inequalities, one would want to have some sort of triangle inequality, so that if $\delta(X,Y)$ and $\delta(Y,Z)$ are both very high, $\delta(X,Z)$ should be quite high too. This rules out symmetry since one can easily construct random variables such that $X=f(Y)$, $Z=g(Y)$ but $X$ and $Z$ are independent (in this example, one would want to say that $X$ depends heavily on $Y$, but that $Y$ does not depend that much on $X$).
Question: do you know a measure of dependence, or a similar tool, that is nonparametric and nonsymmetric? Does it satisfy a triangle inequality? Any reference would be useful.
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1
According to the work of Schweitzer and Wolff, it was not the symmetry condition which is unnatural, rather the "existence of a relation". His original conditions were satisfied only by the maximal correlation coefficient. projecteuclid.org/… – Gjergji Zaimi Apr 23 2010 at 21:03
1
@Gjergji Zaimi: this reference is interesting, but I would not call a measure of dependency nonparameteric when it relies so heavily on the order structure (although it is certainly very weakly parametric, since the order is quite flexible compared to an affine structure). Imagine for example that you have a sociological study and you want to examine the correlation between the gender and some other characteristic, you do not have a structure, not even an order, on the set of gender. – Benoît Kloeckner Apr 24 2010 at 6:32
1 Answer
Conditional entropy might be close to what you are looking for. You will have to transform it in an obvious way to fit Rényi's framework. See http://en.wikipedia.org/wiki/Conditional_entropy and especially the section titled "Intuition".
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This is effectively close to what I am looking for; my question could mostly be recast into: « has (possibly renormalized) conditional entropy be used has a measure of dependency, and can it be nonparametrically extended to non-discrete random variable? » – Benoît Kloeckner Apr 26 2010 at 8:21
Conditional entropy is often used as a measure of dependency: one specific example is in the realm of databases, where it's used to capture weak functional dependencies (Dalkilic/Robertson, Principles Of Database Systems, 2000). There, it's discrete. In general, you can define it via an integration, but like with differential entropy, you have to be careful – Suresh Venkat Apr 27 2010 at 0:59 | {"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": 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": 0.9383468627929688, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/149610/exact-value-of-an-integral?answertab=votes | # Exact value of an integral
I have a problem with getting the exact value of this integral : $$\int_{x}^{+\infty}\frac{e^{-t}}{t}dt$$
Any help would be much appreciated.
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2
– Norbert May 25 '12 at 12:09
7
I downvoted because "I need it as soon as possible" is not what I want to see on this site. – Asaf Karagila May 25 '12 at 12:09
2
Bassem, welcome to Math.SE. Whenever you ask a question please provide as much info as possible on what you have done, where you got the problem from etc. Also, try to be polite - people are here to help you and are often willing to do so. – AD. May 25 '12 at 12:20
4
@Asaf: Maybe you regard OP's "I need it as soon as possible" as a claim that we have to answer his question asap. I regard it as the need OP have and don't assume he thinks I have to do anything. All in all, I think that before downvoting the question, such comment as AD should be better appreciated. – Ilya May 25 '12 at 12:23
2
@Asaf: I certainly agree to disagree and I do respect your point of view. Just wanted to warn you about cascades of downvoters caused by a 1st downvote to questions of new users, asked not in the lines of MSE. Challenge to outrace is accepted, but I won't take it serious, so most likely you win :) anyway, I respect your point of view and don't want to convince you I'm right. – Ilya May 25 '12 at 12:48
show 5 more comments
## 1 Answer
Did you try WolframAlpha?
````integrate e^(-t)/t, t=x..infinity
````
yields $\log (x)+\Gamma (0,x)$ for $x>0$. When a result is specified in terms of a special function like this, it's probably not exactly computable. Nonetheless, you can generate numerical approximations easily enough.
````log(x) + gamma(0,x) at x=3
````
or look at a plot
````plot log(x) + gamma(0,x)
````
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What's $\Gamma(0,x)$? Is it the same as $\Gamma(x)$? – Stefan Smith May 26 '12 at 1:23
@user20520 If you enter the code I suggested, you'll see that $\Gamma(a,x)$ refers to the incomplete Gamma function and you'll see pointers to more information. – Mark McClure May 26 '12 at 1:46
2
Thank you for you remarks. I am really sorry for let you feel that i am not polite. But, it s juste my first time to post on. – Bassem May 30 '12 at 20:12
@Bassem No, no, not at all! I just think that it is generally useful to point out to someone how they might find things out for themselves in the future - and I'm certainly not one of the downvoters of the question, either. – Mark McClure May 30 '12 at 20:24 | {"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": 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": 0.3, "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": 0.9622546434402466, "perplexity_flag": "middle"} |
http://mathhelpforum.com/advanced-algebra/85182-left-right-cosets-print.html | # Left and Right Cosets
Printable View
• April 22nd 2009, 09:59 PM
mpryal
Left and Right Cosets
Let G = S3, the symmetric group of degree 3 and let H = {i,f} where f(x1) = x2, f(x2) = x1, f(x3) = x3
a) find all the left cosets of H in G
b) find all the right cosets of H in G
c) Is every left coset of H a right coset of H?
Please show explicit steps I'm very confused on how to prove these coset problems. Thanks so much!
• April 22nd 2009, 11:06 PM
Gamma
cycle notation
I find it a lot easier to use cycle notation when working with the symmetric group. example $(1,2,4, 7)\in S_7$ is the permutation that takes 1 to 2, 2 to 4, 4 to 7 and 7 to 1. The unmentioned numbers are left the same.
Your coset $H=\{(1), (1,2)\}$ it has size two and $|S_n|=n! \Rightarrow |S_3|=3!=6$. So you should expect 3 cosets each of size 2. Basically you just gotta multiply them out and see what happens.
Here are the left cosets of H
$(1)H=\{(1), (1,2)\}$
$(1,3)H=\{(1,3), (1,2,3)\}$
$(2,3)H=\{(2,3), (1,3,2)\}$
Right cosets of H found similarly
$H(1)=\{(1), (1,2)\}$
$H(1,3)=\{(1,3), (1,3,2)\}$
$H(2,3)=\{(2,3), (1,2,3)\}$
Compare these cosets and see that the last two do not match up, so these are not the same. In particular this tells you that H is infact not a normal subgroup of $S_3$ because $(1,3)H \not = H(1,3)$
All times are GMT -8. The time now is 03:45 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": 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": 10, "end_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": 0.88345867395401, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/8417/conditional-expectation-of-convolution-product-equals | ## Conditional expectation of convolution product equals..
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Let $X, Y$ be two $L^1$ random variables on the probablity space $(\Omega, \mathcal{F}, P)$. Let $\mathcal{G} \subset \mathcal{F}$ be a sub-$\sigma$-algebra. Consider the conditional expectation operator $E(\cdot|\mathcal{G}) \colon L^1(\mathcal{F}) \to L^1(\mathcal{G})$. When is $E(X \ast Y|\mathcal{G}) = E(X|\mathcal{G}) \ast E(Y|\mathcal{G})$? Here $\ast$ is the convolution product on $L^1$ (which makes $(L^1, \ast)$ a Banach algebra, so I'm asking when is $E(\cdot|\mathcal{G})$ an algebra homomorphism?).
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I'm not quite sure how you're defining convolution - normally one can only define convolution of (equivalence classes of) L^1-functions which are defined on some semigroup that carries a suitable measure. Could you please write out your definition? – Yemon Choi Dec 10 2009 at 3:13
@davidk01 I think that this is indeed the usual notion of conditional expectation wrt a sub $\sigma$-algebra; so I don't think that bit needs changing - or at least, it's not so important – Yemon Choi Dec 10 2009 at 6:33
@Yemon Choi: You are right. I completely ignored the question of what kind of structure Omega has. – david karapetyan Dec 10 2009 at 8:12
1
-1 since it is still (at time of writing) not clear what semigroup structure the author is assuming on the measure space, and hence I still don't follows what the "convolution product is supposed to be... – Yemon Choi Dec 11 2009 at 5:34
## 3 Answers
As long as $\mathcal{G}$ is invariant under whatever operation you use in the convolution ("+", say), $\mathcal{G}$-measurable functions will convolve to $\mathcal G$-measurable functions and the equalities of integrals that define the conditional expectation will be automatic: for any $H \in \mathcal G$ we have $$\int_{\Omega} E(X|\mathcal{G}) * E(Y|\mathcal{G}) (t) \chi_H(t) dP(t) =$$ $$= \int_{\Omega\times\Omega} E(X|\mathcal{G})(x)E(Y|\mathcal{G})(y) \chi_{H}(x+y) dP(x)dP(y) =$$ $$= \int_{\Omega} E(X|\mathcal{G})(x) \int_{\Omega} E(Y|\mathcal{G})(y) \chi_{H}(x+y) dP(y)dP(x) =$$ $$= \int_{\Omega} E(X|\mathcal{G})(x) \int_{\Omega} Y(y) \chi_{H}(x+y) dP(y)dP(x) =$$ $$= \int_{\Omega} Y(y) \int_{\Omega} E(X|\mathcal{G})(x) \chi_{H}(x+y) dP(x)dP(y) =$$ $$= \int_{\Omega} Y(y) \int_{\Omega} X(x) \chi_{H}(x+y) dP(x)dP(y) =$$ $$= \int_{\Omega} X*Y(t) \chi_H(t) dP(t) = \int_{\Omega} E(X*Y|\mathcal{G})(t) \chi_H(t) dP(t)$$ If $\mathcal{G}$ is not invariant under the operation, though, I see no reason for the convolution to be $\mathcal{G}$-measurable. Is an example for that what you're requesting?
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Assuming $\Omega$ has the structure for defining convolutions I don't think it is ever an algebra homomorphism. Take $X$ to be supported on $\mathcal{G}^c$, i.e. take some set of non-zero measure in $\mathcal{G}^c$ and let $X$ be a function whose support lies in that set, then $E(X\ast Y|\mathcal{G})\neq 0$ but $E(X|\mathcal{G})=0$ so $E(X|\mathcal{G})\ast E(Y|\mathcal{G})=0$.
Edit: Scratch what I said. I was confusing sub-$\sigma$-algebra with sub-algebra of random variables and even in the finite case my statement is completely incorrect. In almost every instance $E(X|\mathcal{G})$ will not be zero as Jonas points out in the comments.
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What do you mean when you say that X is supported on the complement of a sigma algebra of subsets of \Omega? – Jonas Meyer Dec 10 2009 at 10:48
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I don't understand. There may be an element of F\G on which X is supported, but that doesn't imply that E(X|G) = 0. Of course the support of X lies in $\Omega\in\mathcal{G}$. – Jonas Meyer Dec 11 2009 at 2:26
As I mentioned, G will always contain one particular set which contains the support of X, namely $\Omega$, and G will typically contain many more sets on which X is not identically zero, even if the support itself is an element of F\G. Consider for example Omega = the unit circle, F = the set of Lebesgue measurable sets, G = the set of sets that are countable or have countable complement, X = a function that is identically 1 on a semicircle and 0 otherwise. Then the support of X is not an element of G, but E(X|G) is not zero because its integral over sets with countable complement is > 0. – Jonas Meyer Dec 11 2009 at 3:11
I'm glad that's cleared up, but I think it can be confusing when you delete your comments that appeared as part of an exchange. – Jonas Meyer Dec 11 2009 at 17:34
Ya, you have a point but hopefully the edit makes it clear. – david karapetyan Dec 12 2009 at 0:51
I hadn't really thought through the question of which structure I want $\Omega$ to have specifically. I guess what is needed is some convolution semigroup structure which accurately resembles the convolution product (for $\Omega = \mathbb{R}^n$ and the Lebesgue measure) and $P$ being some kind of probability Haar measure which makes $(L^1(\Omega), \ast)$ a Banach algebra. But I consider Thorny's post a sufficient answer of my question.. Thanks for your help.
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http://medlibrary.org/medwiki/Gallager_code | # Gallager code
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In information theory, a low-density parity-check (LDPC) code is a linear error correcting code, a method of transmitting a message over a noisy transmission channel,[1][2] and is constructed using a sparse bipartite graph.[3] LDPC codes are capacity-approaching codes, which means that practical constructions exist that allow the noise threshold to be set very close (or even arbitrarily close on the BEC) to the theoretical maximum (the Shannon limit) for a symmetric memory-less channel. The noise threshold defines an upper bound for the channel noise, up to which the probability of lost information can be made as small as desired. Using iterative belief propagation techniques, LDPC codes can be decoded in time linear to their block length.
LDPC codes are finding increasing use in applications requiring reliable and highly efficient information transfer over bandwidth or return channel–constrained links in the presence of data-corrupting noise. Although implementation of LDPC codes has lagged behind that of other codes, notably turbo codes, the absence of encumbering software patents has made LDPC attractive to some.[4]
LDPC codes are also known as Gallager codes, in honor of Robert G. Gallager, who developed the LDPC concept in his doctoral dissertation at MIT in 1960.[5]
## History
Impractical to implement when first developed by Gallager in 1963,[6] Gallager's LDPC codes were forgotten until Gallager's work was discovered in 1996.[7] Turbo codes, another class of capacity-approaching codes discovered in 1993, became the coding scheme of choice in the late 1990s, used for applications such as deep space satellite communications. However, in the last few years, the advances in low-density parity-check codes have seen them surpass turbo codes in terms of error floor and performance in the higher code rate range, leaving turbo codes better suited for the lower code rates only.[8]
## Applications
In 2003, an LDPC code beat six turbo codes to become the error correcting code in the new DVB-S2 standard for the satellite transmission of digital television.[9] In 2008, LDPC beat convolutional turbo codes as the FEC scheme for the ITU-T G.hn standard.[10] G.hn chose LDPC over turbo codes because of its lower decoding complexity (especially when operating at data rates close to 1 Gbit/s) and because the proposed turbo codes exhibited a significant error floor at the desired range of operation.[11] LDPC is also used for 10GBase-T Ethernet, which sends data at 10 gigabits per second over twisted-pair cables. As of 2009, LDPC codes are also part of the Wi-Fi 802.11 standard as an optional part of 802.11n and 802.11ac, in the High Throughput (HT) PHY specification.[12]
Some OFDM systems add an additional outer error correction that fixes the occasional errors (the "error floor") that get past the LDPC correction inner code even at low bit-error rates. For example: The Reed-Solomon code with LDPC Coded Modulation (RS-LCM) uses a Reed-Solomon outer code.[13] The DVB-T2 standard and the DVB-C2 standard use a BCH code outer code to mop up residual errors after LDPC decoding.[14]
## Function
LDPC codes are defined by a sparse parity-check matrix. This sparse matrix is often randomly generated, subject to the sparsity constraints—LDPC code construction is discussed later. These codes were first designed by Gallager in 1962.
Below is a graph fragment of an example LDPC code using Forney's factor graph notation. In this graph, n variable nodes in the top of the graph are connected to (n−k) constraint nodes in the bottom of the graph. This is a popular way of graphically representing an (n, k) LDPC code. The bits of a valid message, when placed on the T's at the top of the graph, satisfy the graphical constraints. Specifically, all lines connecting to a variable node (box with an '=' sign) have the same value, and all values connecting to a factor node (box with a '+' sign) must sum, modulo two, to zero (in other words, they must sum to an even number).
Ignoring any lines going out of the picture, there are 8 possible 6-bit strings corresponding to valid codewords: (i.e., 000000, 011001, 110010, 101011, 111100, 100101, 001110, 010111). This LDPC code fragment represents a 3-bit message encoded as six bits. Redundancy is used, here, to increase the chance of recovering from channel errors. This is a (6, 3) linear code, with n = 6 and k = 3.
Once again ignoring lines going out of the picture, the parity-check matrix representing this graph fragment is
$\mathbf{H} = \begin{pmatrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 1 & 0 \\ \end{pmatrix}.$
In this matrix, each row represents one of the three parity-check constraints, while each column represents one of the six bits in the received codeword.
In this example, the eight codewords can be obtained by putting the parity-check matrix H into this form $\begin{bmatrix} -P^T | I_{n-k} \end{bmatrix}$ through basic row operations:
$\mathbf{H} = \begin{pmatrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 1 & 0 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 & 1 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 \\ \end{pmatrix}.$
From this, the generator matrix G can be obtained as $\begin{bmatrix} I_k | P \end{bmatrix}$ (noting that in the special case of this being a binary code $P = -P$), or specifically:
$\mathbf{G} = \begin{pmatrix} 1 & 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 0 \\ \end{pmatrix}.$
Finally, by multiplying all eight possible 3-bit strings by G, all eight valid codewords are obtained. For example, the codeword for the bit-string '101' is obtained by:
$\begin{pmatrix} 1 & 0 & 1 \\ \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 0 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 & 1 & 0 & 1 & 1 \\ \end{pmatrix}.$
## Decoding
As with other codes, optimally decoding an LDPC code on the binary symmetric channel is an NP-complete problem, although techniques based on iterative belief propagation used in practice lead to good approximations. In contrast, belief propagation on the binary erasure channel is particularly simple where it consists of iterative constraint satisfaction.
For example, consider that the valid codeword, 101011, from the example above, is transmitted across a binary erasure channel and received with the first and fourth bit erased to yield ?01?11. Since the transmitted message must have satisfied the code constraints, the message can be represented by writing the received message on the top of the factor graph.
In this example, the first bit cannot yet be recovered, because all of the constraints connected to it have more than one unknown bit. In order to proceed with decoding the message, constraints connecting to only one of the erased bits must be identified. In this example, either the second or third constraint suffices. Examining the second constraint, the fourth bit must have been 0, since only a 0 in that position would satisfy the constraint.
This procedure is then iterated. The new value for the fourth bit can now be used in conjunction with the first constraint to recover the first bit as seen below. This means that the first bit must be a 1 to satisfy the leftmost constraint.
Thus, the message can be decoded iteratively. For other channel models, the messages passed between the variable nodes and check nodes are real numbers, which express probabilities and likelihoods of belief.
This result can be validated by multiplying the corrected codeword r by the parity-check matrix H:
$\mathbf{z} = \mathbf{Hr} = \begin{pmatrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 1 & 0 \\ \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 1 \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}.$
Because the outcome z (the syndrome) of this operation is the 3 × 1 zero vector, the resulting codeword r is successfully validated.
### Updating node information
In recent years, there has also been a great deal of work spent studying the effects of alternative schedules for variable- and constraint-node update. The original technique that was used for decoding LDPC codes was known as flooding. This type of update required that, before updating a variable node, all constraint nodes needed to be updated and vice versa. In later work by Vila Casado et al.,[15] alternative update techniques were studied, in which variable nodes are updated with the newest available check-node information.
The intuition behind these algorithms is that variable nodes whose values vary the most are the ones that need to be updated first. Highly reliable nodes, whose log-likelihood ratio (LLR) magnitude is large and does not change significantly from one update to the next, do not require updates with the same frequency as other nodes, whose sign and magnitude fluctuate more widely. These scheduling algorithms show greater speed of convergence and lower error floors than those that use flooding. These lower error floors are achieved by the ability of the Informed Dynamic Scheduling (IDS)[15] algorithm to overcome trapping sets of near codewords.[16]
When non-flooding scheduling algorithms are used, an alternative definition of iteration is used. For an (n, k) LDPC code of rate k/n, a full iteration occurs when n variable and n − k constraint nodes have been updated, no matter the order in which they were updated.
### Lookup table decoding
It is possible to decode LDPC codes on a relatively low-powered microprocessor by the use of lookup tables.
Whilst codes such as LDPC are generally implemented on high-powered processors, with long block lengths, there are also applications which use lower-powered processors and short block lengths (1024).
It is possible therefore to pre-calculate the output bit based upon pre-determined input bits. A table is generated which contains n entries (for a block length of 1024 bits, this would be 1024 bits long), and contains all possible entries for different input states (both errored and non-errored).
As a bit is input, it is then added to a FIFO register, and the value of the FIFO register is then used to look up in the table the relevant output from the pre-calculated values.
By this method, very high iterations can be used, with little processor overhead, the only cost being that of the memory for the lookup table, such that LDPC decoding is possible even on a 4 MHz PIC chip.
## Code construction
For large block sizes, LDPC codes are commonly constructed by first studying the behaviour of decoders. As the block size tends to infinity, LDPC decoders can be shown to have a noise threshold below which decoding is reliably achieved, and above which decoding is not achieved.[17] This threshold can be optimised by finding the best proportion of arcs from check nodes and arcs from variable nodes. An approximate graphical approach to visualising this threshold is an EXIT chart.
The construction of a specific LDPC code after this optimisation falls into two main types of techniques:
• Pseudo-random approaches
• Combinatorial approaches
Construction by a pseudo-random approach builds on theoretical results that, for large block size, a random construction gives good decoding performance.[7] In general, pseudo-random codes have complex encoders, however pseudo-random codes with the best decoders can have simple encoders.[18] Various constraints are often applied to help ensure that the desired properties expected at the theoretical limit of infinite block size occur at a finite block size.
Combinatorial approaches can be used to optimise properties of small block-size LDPC codes or to create codes with simple encoders.
Some LDPC codes are based on Reed-Solomon codes, such as the RS-LDPC code used in the 10-gigabit Ethernet standard.[19] Compared to randomly generated LDPC codes, structured LDPC codes -- such as the LDPC code used in the DVB-S2 standard -- can have simpler and therefore lower-cost hardware -- in particular, codes constructed such that the H matrix is a circulant matrix.[20]
Yet another way of constructing LDPC codes is to use finite geometries. This method was proposed by Y. Kou et al. in 2001.[21]
## See also
### Applications
• G.hn/G.9960 (ITU-T Standard for networking over power lines, phone lines and coaxial cable)
• 802.3an (10 Giga-bit/s Ethernet over Twisted pair)
• CMMB(China Multimedia Mobile Broadcasting)
• DVB-S2 / DVB-T2 / DVB-C2 (Digital video broadcasting, 2nd Generation)
• DMB-T/H (Digital video broadcasting)[22]
• WiMAX (IEEE 802.16e standard for microwave communications)
• IEEE 802.11n-2009 (Wi-Fi standard)
### Other capacity-approaching codes
• Turbo codes
• Online codes
• Fountain codes
• LT codes
• Raptor codes
• Repeat-accumulate codes (a class of simple turbo codes)
• Tornado codes (LDPC codes designed for erasure decoding)
• Polar codes
## References
1. David J.C. MacKay (2003) Information theory, Inference and Learning Algorithms, CUP, ISBN 0-521-64298-1 [Amazon-US | Amazon-UK], (also available online)
2. Todd K. Moon (2005) Error Correction Coding, Mathematical Methods and Algorithms. Wiley, ISBN 0-471-64800-0 [Amazon-US | Amazon-UK] (Includes code)
3. NewScientist, Communication speed nears terminal velocity, by Dana Mackenzie, 9 July 2005
4. Larry Hardesty (21 January 2010), "Explained: Gallager codes", MIT News, retrieved 2010-08-18
5. ^ a b
6. ^ a b
7. T. Richardson, “Error floors of LDPC codes,” in Proc. 41st Allerton Conf. Comm., Control, and Comput., Monticello, IL, 2003.
8. Thomas J. Richardson and M. Amin Shokrollahi and Rüdiger L. Urbanke, "Design of Capacity-Approaching Irregular Low-Density Parity-Check Codes," IEEE Transactions in Information Theory, 47(2), February 2001
9. Thomas J. Richardson and Rüdiger L. Urbanke, "Efficient Encoding of Low-Density Parity-Check Codes," IEEE Transactions in Information Theory, 47(2), February 2001
10. Y. Kou, S. Lin and M. Fossorier, "Low-Density Parity-Check Codes Based on Finite Geometries: A Rediscovery and New Results," IEEE Transactions on Information Theory, vol. 47, no. 7, November 2001, pp. 2711- 2736.
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http://mathoverflow.net/revisions/106265/list | ## Return to Question
2 added 22 characters in body
In Fonctions L p-adiques des corps quadratiques imaginaires et de leurs extensions abéliennes, J. Reine Angew. Math. 358 (1985), 76–91, Roland Gillard shows the following result (I mainly follow the MR review here):
Let $K$ be an imaginary quadratic field and let $p > 3$ be a prime number which splits in K into $(p)=\mathfrak{p}\mathfrak{p'}$. Let $K_{\infty}$ be the unique `$\mathbb{Z}_{p}$`-extension of $K$ unramified outside $p$ \mathfrak{p}$(thus noncyclotomic). Let$F$be a finite abelian extension of$K$and let$M$be the maximal abelian$p$-extension of$F$unramified outside$p$. \mathfrak{p}$. Then Theorem 3.4 states that $\mathrm{Gal}(M/FK_{\infty})$ is $\mathbb{Z}_{p}$-torsion-free; in particular its $\mu$-invariant is 0.
Question: does anyone know if the vanishing of this $\mu$-invariant is also proven somewhere when $p=3$ (even special cases would be of interest)?
Note: with some work, you can get a PDF of the article in question without a subscription by following the link from here: http://www.ams.org/dmr/JournalListJ.html
1
# Vanishing of certain $\mu$-invariants attached to abelian extensions of imaginary quadratic fields
In Fonctions L p-adiques des corps quadratiques imaginaires et de leurs extensions abéliennes, J. Reine Angew. Math. 358 (1985), 76–91, Roland Gillard shows the following result (I mainly follow the MR review here):
Let $K$ be an imaginary quadratic field and let $p > 3$ be a prime number which splits in K into $(p)=\mathfrak{p}\mathfrak{p'}$. Let $K_{\infty}$ be the unique `$\mathbb{Z}_{p}$`-extension of $K$ unramified outside $p$ (thus noncyclotomic). Let $F$ be a finite abelian extension of $K$ and let $M$ be the maximal abelian $p$-extension of $F$ unramified outside $p$. Then Theorem 3.4 states that $\mathrm{Gal}(M/FK_{\infty})$ is $\mathbb{Z}_{p}$-torsion-free; in particular its $\mu$-invariant is 0.
Question: does anyone know if the vanishing of this $\mu$-invariant is also proven somewhere when $p=3$ (even special cases would be of interest)?
Note: with some work, you can get a PDF of the article in question without a subscription by following the link from here: http://www.ams.org/dmr/JournalListJ.html | {"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": 36, "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": 0.8358002305030823, "perplexity_flag": "head"} |
http://medlibrary.org/medwiki/Soft_body_dynamics | # Soft body dynamics
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Soft body dynamics is a field of computer graphics that focuses on visually realistic physical simulations of the motion and properties of deformable objects (or soft bodies).[1] The applications are mostly in video games and film. Unlike in simulation of rigid bodies, the shape of soft bodies can change, meaning that the relative distance of two points on the object is not fixed. While the relative distances of points are not fixed, the body is expected to retain its shape to some degree (unlike a fluid). The scope of soft body dynamics is quite broad, including simulation of soft organic materials such as muscle, fat, hair and vegetation, as well as other deformable materials such as clothing and fabric. Generally, these methods only provide visually plausible emulations rather than accurate scientific/engineering simulations, though there is some crossover with scientific methods, particularly in the case of finite element simulations. Several physics engines currently provide software for soft-body simulation.[2][3][4][5][6][7]
Softbody objects react to forces and are able to collide with other 3D objects. This example has been created with Blender.
## Deformable solids
The simulation of volumetric solid soft bodies can be realised by using a variety of approaches.
### Spring/mass models
Two nodes as mass points connected by a parallel circuit of a spring and a damper.
In this approach, the body is modeled as a set of point masses (nodes) connected by ideal weightless elastic springs obeying some variant of Hooke's law. The nodes may either derive from the edges of a two-dimensional polygonal mesh representation of the surface of the object, or from a three-dimensional network of nodes and edges modeling the internal structure of the object (or even a one-dimensional system of links, if for example a rope or hair strand is being simulated). Additional springs between nodes can be added, or the force law of the springs modified, to achieve desired effects. Applying Newton's second law to the point masses including the forces applied by the springs and any external forces (due to contact, gravity, air resistance, wind, and so on) gives a system of differential equations for the motion of the nodes, which is solved by standard numerical schemes for solving ODEs.[8] Rendering of a three-dimensional mass-spring lattice is often done using free-form deformation,[9] in which the rendered mesh is embedded in the lattice and distorted to conform to the shape of the lattice as it evolves. Assuming all point masses equal to zero one can obtain the Stretched grid method aimed at several engineering problems solution relative to the elastic grid behavior.
### Finite element simulation
This is a more physically accurate approach, which uses the widely used finite element method to solve the partial differential equations which govern the dynamics of an elastic material. The body is modeled as a three-dimensional elastic continuum by breaking it into a large number of solid elements which fit together, and solving for the stresses and strains in each element using a model of the material.[10] The elements are typically tetrahedral, the nodes being the vertices of the tetrahedra (relatively simple methods exist[11][12] to tetrahedralize a three dimensional region bounded by a polygon mesh into tetrahedra, similarly to how a two-dimensional polygon may be triangulated into triangles). The strain (which measures the local deformation of the points of the material from their rest state) is quantified by the strain tensor $\boldsymbol{\epsilon}$. The stress (which measures the local forces per-unit area in all directions acting on the material) is quantified by the Cauchy stress tensor $\boldsymbol{\sigma}$. Given the current local strain, the local stress can be computed via the generalized form of Hooke's law: $\boldsymbol{\sigma} = \mathsf{C} \boldsymbol{\varepsilon} \ ,$ where $\mathsf{C}$ is the "elasticity tensor" which encodes the material properties (parametrized in linear elasticity for an isotropic material by the Poisson ratio and Young's modulus).
The equation of motion of the element nodes is obtained by integrating the stress field over each element and relating this, via Newton's second law, to the node accelerations.
Pixelux (developers of the Digital Molecular Matter system) use a finite-element-based approach for their soft bodies, using a tetrahedral mesh and converting the stress tensor directly into node forces.[13] Rendering is done via a form of free-form deformation.[9]
### Energy minimization methods
This approach is motivated by variational principles and the physics of surfaces, which dictate that a constrained surface will assume the shape which minimizes the total energy of deformation (analogous to a soap bubble). Expressing the energy of a surface in terms of its local deformation (the energy is due to a combination of stretching and bending), the local force on the surface is given by differentiating the energy with respect to position, yielding an equation of motion which can be solved in the standard ways.[14][15]
### Shape matching
In this scheme, penalty forces or constraints are applied to the model to drive it towards its original shape[16] (i.e. the material behaves as if it has shape memory). To conserve momentum the rotation of the body must be estimated properly, for example via polar decomposition. To approximate finite element simulation, shape matching can be applied to three dimensional lattices and multiple shape matching constraints blended.[17]
### Rigid-body based deformation
Deformation can also be handled by a traditional rigid-body physics engine, modeling the soft-body motion using a network of multiple rigid bodies connected by constraints, and using (for example) matrix-palette skinning to generate a surface mesh for rendering. This is the approach used for deformable objects in Havok Destruction.[18]
## Cloth simulation
In the context of computer graphics, cloth simulation refers to the simulation of soft bodies in the form of two dimensional continuum elastic membranes, that is, for this purpose, the actual structure of real cloth on the yarn level can be ignored (though modeling cloth on the yarn level has been tried).[19] Via rendering effects, this can produce a visually plausible emulation of textiles and clothing, used in a variety of contexts in video games, animation, and film. It can also be used to simulate two dimensional sheets of materials other than textiles, such as deformable metal panels or vegetation. In video games it is often used to enhance the realism of clothed characters, which otherwise would be entirely animated.
Cloth simulators are generally based on mass-spring models, but a distinction must be made between force-based and position-based solvers.
### Force-based cloth
The mass-spring model (obtained from a polygonal mesh representation of the cloth) determines the internal spring forces acting on the nodes at each timestep (in combination with gravity and applied forces). Newton's second law gives equations of motion which can be solved via standard ODE solvers. To create high resolution cloth with a realistic stiffness is not possible however with simple explicit solvers (such as forward Euler integration), unless the timestep is made too small for interactive applications (since as is well known, explicit integrators are numerically unstable for sufficiently stiff systems). Therefore implicit solvers must be used,[20] requiring solution of a large sparse matrix system (via e.g. the conjugate gradient method), which itself may also be difficult to achieve at interactive frame rates. An alternative[21][22] is to use an explicit method with low stiffness, with ad hoc methods to avoid instability and excessive stretching (e.g. strain limiting corrections).
### Position-based dynamics
To avoid needing to do an expensive implicit solution of a system of ODEs, many real-time cloth simulators (notably PhysX, Havok Cloth, and Maya nCloth) use position based dynamics (PBD),[23] an approach based on constraint relaxation. The mass-spring model is converted into a system of constraints, which demands that the distance between the connected nodes be equal to the initial distance. This system is solved sequentially and iteratively, by directly moving nodes to satisfy each constraint, until sufficiently stiff cloth is obtained. This is similar to a Gauss-Seidel solution of the implicit matrix system for the mass-spring model. Care must be taken though to solve the constraints in the same sequence each timestep, to avoid spurious oscillations, and to make sure that the constraints do not violate linear and angular momentum conservation. Additional position constraints can be applied, for example to keep the nodes within desired regions of space (sufficiently close to an animated model for example), or to maintain the body's overall shape via shape matching.
## Collision detection for deformable objects
Main article: Collision detection
Realistic interaction of simulated soft objects with their environment may be important for obtaining visually realistic results. Cloth self-intersection is important in some applications for acceptably realistic simulated garments. This is challenging to achieve at interactive frame rates, particularly in the case of detecting and resolving self collisions and mutual collisions between two or more deformable objects.
Collision detection may be discrete/a posteriori (meaning objects are advanced in time through a pre-determined interval, and then any penetrations detected and resolved), or continuous/a priori (objects are advanced only until a collision occurs, and the collision is handled before proceeding). The former is easier to implement and faster, but leads to failure to detect collisions (or detection of spurious collisions) if objects move fast enough. Real-time systems generally have to use discrete collision detection, with other ad hoc ways to avoid failing to detect collisions.
Detection of collisions between cloth and environmental objects with a well defined "inside" is straightforward since the system can detect unambiguously whether the cloth mesh vertices and faces are intersecting the body and resolve them accordingly. If a well defined "inside" does not exist (e.g. in the case of collision with a mesh which does not form a closed boundary), an "inside" may be constructed via extrusion. Mutual- or self-collisions of soft bodies defined by tetrahedra is straightforward, since it reduces to detection of collisions between solid tetrahedra.
However, detection of collisions between two polygonal cloths (or collision of a cloth with itself) via discrete collision detection is much more difficult, since there is no unambiguous way to locally detect after a timestep whether a cloth node which has penetrated is on the "wrong" side or not. Solutions involve either using the history of the cloth motion to determine if an intersection event has occurred, or doing a global analysis of the cloth state to detect and resolve self-intersections. Pixar has presented a method which uses a global topological analysis of mesh intersections in configuration space to detect and resolve self-interpenetration of cloth.[24] Currently, this is generally too computationally expensive for real-time cloth systems.
To do collision detection efficiently, primitives which are certainly not colliding must be identified as soon as possible and discarded from consideration to avoid wasting time. To do this, some form of spatial subdivision scheme is essential, to avoid a brute force test of $O[n^2]$ primitive collisions. Approaches used include:
• Bounding volume hierarchies (AABB trees,[25] OBB trees, sphere trees)
• Grids, either uniform[26] (using hashing for memory efficiency) or hierarchical (e.g. Octree, kd-tree)
• Coherence-exploiting schemes, such as sweep and prune with insertion sort, or tree-tree collisions with front tracking.
• Hybrid methods involving a combination of various of these schemes, e.g. a coarse AABB tree plus sweep-and-prune with coherence between colliding leaves.
## Other applications
Other effects which may be simulated via the methods of soft-body dynamics are:
• Destructible materials: fracture of brittle solids, cutting[27] of soft bodies, and tearing of cloth. The finite element method is especially suited to modelling fracture[13] as it includes a realistic model of the distribution of internal stresses in the material, which physically is what determines when fracture occurs, according to fracture mechanics.
• Plasticity[16] (permanent deformation) and melting[28]
• Simulated hair,[29] fur, and feathers
• Simulated organs for biomedical applications[30]
Simulating fluids in the context of computer graphics would not normally be considered soft-body dynamics, which is usually restricted to mean simulation of materials which have a tendency to retain their shape and form. In contrast, a fluid assumes the shape of whatever vessel contains it, as the particles are bound together by relatively weak forces.
## Engines supporting soft body physics
• Bullet 2.69
• Carbon, by Numerion Software[31]
• CryEngine 3
• Digital Molecular Matter
• Havok Cloth
• Maya nCloth
• OpenTissue
• OpenCloth - (http://code.google.com/p/opencloth) - A collection of source codes implementing cloth simulation algorithms as well as soft body dynamics in OpenGL.
• Physics Abstraction Layer (PAL) - Uniform API, supports multiple physics engines
• PhysX
• Phyz (Dax Phyz)
• SOFA (Simulation Open Framework Architecture)
• Step
• Syflex (Cloth simulator)
• Unreal Engine 3
• BeamNG - (http://beamng.com)
## References
1. ^ a b
2. ^ a b
3. ^ a b
4. "Numerion Software Ltd". Retrieved 15 July 2011.
Content in this section is authored by an open community of volunteers and is not produced by, reviewed by, or in any way affiliated with MedLibrary.org. Licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License, using material from the Wikipedia article on "Soft body dynamics", available in its original form here:
http://en.wikipedia.org/w/index.php?title=Soft_body_dynamics
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