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http://math.stackexchange.com/questions/189634/profinite-completion-of-special-linear-group/189641
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# Profinite completion of Special linear group
Is the profinite completion of $SL_2(\mathbb{Z})$ equal to $SL_2(\hat{\mathbb{Z}})$?
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## 1 Answer
No. The congruence kernel of the natural epimorphism $\widehat{\mathrm{SL}_2(\mathbb{Z})} \to \mathrm{SL}_2(\widehat{\mathbb{Z}})$ is a free profinite group of countably infinite rank. This is Theorem 8.8.1. of L. Ribes, P. Zalesskii, Profinite groups, Springer-Verlag, 2010.
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http://math.stackexchange.com/questions/191175/how-to-calculate-the-limit-of-n1-frac1n-as-n-to-infty?answertab=votes
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# How to calculate the limit of $(n+1)^{\frac{1}{n}}$ as $n\to\infty$
How to calculate the limit of $(n+1)^{\frac{1}{n}}$ as $n\to\infty$?
I know how to prove that $n^{\frac{1}{n}}\to 1$ and $n^{\frac{1}{n}}<(n+1)^{\frac{1}{n}}$. What is the other inequality that might solve the problem?
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Why not use the value of $\lim \bigl[(n+1)/n\bigr]^{1/n}$? – Lubin Sep 5 '12 at 0:26
## 3 Answers
With $$y=\lim_{n\to\infty} (n+1)^{1/n},$$ consider, using continuity of $\ln$, $$\ln y=\lim_{n\to\infty} \frac{1}{n}\ln(n+1)=0.$$ This tells you that your limit is $1$.
Alternately, $$n^{1/n}<n^{1/n}\left(1+\frac{1}{n}\right)^{1/n}<n^{1/n}\left(1+\frac{1}{n}\right),$$ where the middle guy is your expression.
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I would prefer a proof that not involves other concepts but convergence of know sequences. – Jön Sep 4 '12 at 22:34
For the other inequality, you could use $$(n+1)^{\frac1n}\leq (2n)^{\frac1n}=2^{\frac1n}\,n^{\frac1n}.$$
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What about $n^{1/n}\lt (n+1)^{1/n}\le (2n)^{1/n}=2^{1/n}n^{1/n}$, then squeezing.
Or else, for $n \ge 2$, $$n^{1/n}\lt (n+1)^{1/n}\lt (n^2)^{1/n}=(n^{1/n})(n^{1/n}).$$ Then we don't have to worry about $2^{1/n}$.
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http://mathhelpforum.com/pre-calculus/136934-2-polynomials-questions.html
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# Thread:
1. ## 2 polynomials questions
1. Consider the polynomial $P(x) = x^5 + ax^3 + bx - c$ where a,b and are real positive numbers
(i) show that P(x) has exactly one real root which is greater than zero
(ii) Hence show that P(x) has four complex roots with at least two of them with negative real parts
2. If the polynomial $x^3 + 3mx^2 + 3nx + r = 0$ has a double root. Show that:
(i) The double root is x = $\frac {mn-r}{2(n-m^2)}$
(ii) $(mn-r)^2 = 4(m^2 -n)(n^2 - mr)$
Thanks
2. Originally Posted by differentiate
1. Consider the polynomial $P(x) = x^5 + ax^3 + bx - c$ where a,b and are real positive numbers
(i) show that P(x) has exactly one real root which is greater than zero
$P(0)=0$ and $\lim_{x\to\infty}P(x)=\infty$. That guarantees the root. To see that it only has one note that $P'(x)=5x^4+3ax^2+b>0,\text{ }x>0$ so assuming there was another would violate Rolle's theorem.
3. Originally Posted by differentiate
1. Consider the polynomial $P(x) = x^5 + ax^3 + bx - c$ where a,b and are real positive numbers
(ii) Hence show that P(x) has four complex roots with at least two of them with negative real parts
Note that the same works for the negative numbers since $P'(x)$ is symmetric about the y-axis. Namely, $P'(x)=5x^4+3ax^2+b>0,\text{ }x<0$. Appealing to Rolle's theorem again guarantees that $P(x)\ne 0,\text{ }x<0$. It follows that $P^{-1}(\{0\})=\{0\}$. But, by the FTA we must have that $P:\mathbb{C}\to\mathbb{C}$ has five zeros counting multiplicity. But, by a very easy to prove theorem, we have that since the coefficients of $P$ are real that the zeros come in conjugate pairs, so that $P(x)=x(x-z_1)(x-\overline{z_1})(x-z_2)(x-\overline{z_2})$. The negative part buisness follows since $\text{Im}(z_k),\text{Im}(\overline{z_k})\ne 0,\text{ }k=1,2$ since they aren't real. Thus, either $\text{Im}(z_1)<0$ or $\text{Im}(\overline{z_1})<0$. A similar analysis shows that either $z_1,z_2$ has negative imaginary parts.
4. Originally Posted by Drexel28
$P(0)=0$
You mean P(0)= -c
and $\lim_{x\to\infty}P(x)=\infty$. That guarantees the root. To see that it only has one note that $P'(x)=5x^4+3ax^2+b>0,\text{ }x>0$ so assuming there was another would violate Rolle's theorem.
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http://math.stackexchange.com/questions/54688/bipartite-graph-cover-problem?answertab=active
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Bipartite graph cover problem
Let $G$ be bipartite with bipartition $A$, $B$. Suppose that $C$ and $C'$ are both covers of $G$. Prove that $C^{\wedge}$ = $(A \cap C \cap C') \cup (B \cap (C \cup C'))$ is also a cover of $G$.
Does anyone know which theorem is useful for proving this statement?
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1 Answer
This statement is fairly easy to prove without appealing to any special theorems. It might be useful to rewrite the statement
Every edge in $G$ has an endvertex in $C''=(A\cap C\cap C')\cup (B\cap(C\cup C')$,
which you are trying to prove, as the equivalent statement
If an edge $e\in E(G)$ has no endvertex in $B\cap (C\cup C')$, then it has an endvertex in $A\cap C\cap C'$.
Hint: every edge $e$ has an endvertex in $A$, an endvertex in $B$, and at least one endvertex in each of $C$ and $C'$.
Hope this helps!
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http://mathematica.stackexchange.com/questions/13253/how-do-i-reverse-the-axis-in-parametricplot
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# How do I reverse the axis in ParametricPlot?
In astronomy, right ascension is usually plotted with positive values that increase from right to left. I have seen discussions of successful and unsuccessful attempts to reverse the order of an axis in Mathematica, but I haven't seen anything that applies specifically to the `ParametricPlot[]` function, and perhaps I am not good enough at Mathematica to see how the other solutions using `ScalingFunctions` or `Transpose` might be applied here. I tried a few to no avail.
Plot 2 uses the default frame ticks, showing reversal of plotting order by reversing the signs of the $x$-coordinates of the plot objects. In the `Ticks` option, their specs seem to be ordered `{{left, right}, {bottom, top}}` with respect to the frame sides. I believe that replacing one of these terms, say left, with something like `{-1, 1}` would replace `-1` with `1` on the left side. But when I attempted to change the names of the ticks on the $x$-axis in plot 2 to positive numbers, the ticks and their names both disappeared, as in plot 3. I could replace the missing ticks with a cumbersome `Epilog` list, but I would prefer something more elegant. It strikes me as odd that the mathematicians who created Mathematica would arbitrarily limit their orientation, so there must be a native way of reversing order, no? The following three scripts produce these three plots in a row:
````Clear["Global`*"]
spiral[a_, t_, x_, y_] := {a*t*Cos[t] + x, a*t*Sin[t] + y} // N;
fs = 8; (* font size *)
objects = 5;
fl = {X, Rotate[Y, -Pi/2]}; (* frame label *)
unreversed =
ParametricPlot[
spiral[.002*#^(5/3), t, #, #] & /@ Range[objects], {t, 0, 10*Pi},
PlotRange -> {{0, objects + 1}, {0, objects + 1}},
PlotLabel -> Style["1. x axis not reversed", FontSize -> fs],
Frame -> True, FrameLabel -> fl, GridLines -> Automatic];
reversed1 =
ParametricPlot[
spiral[.002*#^(5/3), t, -#, #] & /@ Range[objects], {t, 0,
10*Pi}, PlotRange -> {{-objects - 1, 0}, {0, objects + 1}},
PlotLabel -> Style["2. x axis reversed", FontSize -> fs],
Frame -> True, FrameLabel -> fl, GridLines -> Automatic];
ticks = {{Automatic, None}, {{-#, #}, None}} & /@
Reverse[Range[objects]];
reversed2 =
ParametricPlot[
spiral[.002*#^(5/3), t, -#, #] & /@ Range[objects], {t, 0,
10*Pi}, PlotRange -> {{-objects - 1, 0}, {0, objects + 1}},
PlotLabel ->
Style["3. x axis reversed\nticks lost", FontSize -> fs],
Frame -> True, FrameLabel -> fl, GridLines -> Automatic,
FrameTicks -> ticks (* causes ticks to disappear *)];
GraphicsRow[{unreversed, reversed1, reversed2}]
````
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I notice in your examples that you are not reversing the spirals themselves. Is this deliberate? – Mr.Wizard♦ Oct 18 '12 at 12:20
@Mr.Wizard The orientation of the spirals was not a concern, but it could become meaningful in the future. – Gary Palmer Oct 18 '12 at 16:06
## 1 Answer
I cannot recall a built-in method to reverse an axis, at least for ParametricPlot, but maybe the right FrameTicks syntax will help:
````ticks = {{{-6, 6}, {-5, 5}, {-4, 4}, {-3, 3}, {-2, 2}, {-1, 1}, {0, 0}}, {{0,
0}, {1, 1}, {2, 2}, {3, 3}, {4, 4}, {5, 5}, {6, 6}}, {}, {}};
ParametricPlot[
spiral[.002*#^(5/3), t, -#, #] & /@ Range[objects], {t, 0, 10*\[Pi]},
PlotRange -> {{-objects - 1, 0}, {0, objects + 1}},
PlotLabel -> Style["3. x axis reversed\nticks lost", FontSize -> fs],
Frame -> True, FrameLabel -> fl, GridLines -> Automatic,
FrameTicks -> ticks]
````
It's possible that this version-8 function may work with ParametricPlot, though I can't test that, and `ParametricPlot` doesn't appear to be supported: ScalingFunctions
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This is my accepted answer. I see that I had the syntax all wrong. Why is it that the following doesn't work? It produces an output that looks equivalent to the above "ticks". xTicks = {-#, #} & /@ Reverse[Range[6]] yTicks = {#, #} & /@ Range[6] ticks = {{xTicks, yTicks}, {}, {}} – Gary Palmer Oct 18 '12 at 16:37
Wizard Never mind, I see it. It should be ticks = {xTicks, yTicks, {}, {}}. Thanks much. – Gary Palmer Oct 18 '12 at 16:52
lang-mma
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http://math.stackexchange.com/questions/tagged/general-topology?page=1&sort=unanswered&pagesize=15
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### A fiber bundle over Euclidean space is trivial.
What's the easiest way to see this? The only thing I could think to do was try to patch together trivializations. I couldn't find a way to make that work. Thank you! edit: For the record, here's why ...
0answers
110 views
### Question about proof of Tychonoff-Alaoglu
I'd like to check that I understand the proof in full detail. Can you tell me if the following is correct? Thanks for your help. Claim: The closed unit ball $B_{\|\cdot\|_{op}}(0,1)$ in $X^\ast$ is ...
0answers
71 views
### Continuous choice of basis for subspaces
Consider the flag variety (or flag manifold, depending on who you are) $V=\mathrm {Fl} (3,\mathbb C)$ of complete flags of subspaces of $\mathbb C^3$. That is, an element of M is a tuple (L , P) ...
0answers
227 views
### Generalizations for Tietze's extension theorem.
Tietze's extension theorem says: ''If $A$ is a closed subset of $X$ a normal space, and $f:A\to \mathbb{R}$ continuous, then we can extend $f$ to a continuous function $g:X\to \mathbb{R}$." I know ...
0answers
181 views
### Definition of Reshetikhin-Turaev TQFT
I am studying Reshetikhin-Turaev TQFT. In their paper or in the book " Quantum invariants of knots and 3-manifolds", they first define an invariant $\tau(M)$ for a closed orientable 3-manifold $M$ and ...
0answers
202 views
### Points in the plane at integer distances
Does there exist a set of $n$ points $p_1,p_2,...,p_n$ in the plane, all at mutual integer distances to each other, and an $e>0$, such that the following statement holds: For all $a,b$ with ...
0answers
138 views
### Connected subspaces question
Suppose that C, D are connected subsets of a topological space T such that $\bar{C} \cap \bar{D} \neq \emptyset$. Is it true that $C \cup D$ is necessarily connected? I think I have a counter example ...
0answers
183 views
### Are Arzelà–Ascoli theorems results of similar theorems on normed spaces, metric spaces or other spaces?
From Wikipedia, two generalizations of the Arzelà–Ascoli theorem are Let $X$ be a compact Hausdorff space. Then a subset $F$ of $C(X)$, the set of real-valued continuous functions on X, is ...
0answers
217 views
### Homotopy extension property vs. good pairs
I'm taking a course that uses the book "algebraic topology" by Allen Hatcher. I this book there are two different ways in which a pair (X,A) of a topological space X and a subspace A can be nice: They ...
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http://cstheory.stackexchange.com/questions/tagged/determinant
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# Tagged Questions
The determinant tag has no wiki summary.
1answer
119 views
### Determinants and Matrix Multiplication - Similarity and differences in algorithmic complexity and arithmetic circuit size
I am trying to understand the relation between algorithmic complexity and circuit complexity of Determinants and Matrix Multiplication. It is known that the determinant of an $n\times n$ matrix can ...
2answers
290 views
### Cancellation and determinant
Berkowitz algorithm provides a polynomial size circuit with logarithmic depth for determinant of a square matrix using matrix powers. The algorithm implicitly uses cancellation. Is cancellation ...
1answer
326 views
### Implications of approximating the determinant
It is known that one can compute exactly the determinant of an $n\times n$ matrix in determinstic $\log^2(n)$ space. What would be the complexity implications of approximating the determinant of a ...
0answers
127 views
### Grigoriev-Karpinski for the permanent
Grigoriev and Karpinski (ps.Z) showed that any depth-3 circuit over a fixed finite field computing $\mathrm{Det}_n$ requires $2^{\Omega(n)}$ size. I had the misconception(?) until recently that the ...
0answers
119 views
### Counting small terms in a determinant calculation over polynomials (counting spanning trees by weight)
I have a $n\times n$ matrix $A$. It's terms are $a_{ij}=-x^{w_{ij}}$ if $i\neq j$ and $a_{ii}=\sum_{j=0}^{n+1} x^{w_{ij}}$ on the diagonal. The matrix is symmetric as $w_{ij}=w_{ji}$. Numbers $w_{ij}$ ...
2answers
516 views
### Lower bound for determinant and permanent
In light of the recent chasm at depth-3 result (which among other things yields a $2^{\sqrt{n}\log{n}}$ depth-3 arithmetic circuit for the $n \times n$ determinant over $\mathbb{C}$), I have the ...
1answer
359 views
### Smallest known formula for the determinant
The smallest known formula for the determinant has size $n^{\mathcal O(\log n)}$ according to the folklore (or to Ran Raz in its paper Multi-Linear Formulas for Permanent and Determinant are of ...
2answers
488 views
### Gaussian Elimination in terms of Group Action
Gaussian elimination makes determinant of a matrix polynomial-time computable. The reduction of complexity in computing the determinant, which is otherwise sum of exponential terms, is due to ...
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http://unapologetic.wordpress.com/2007/05/23/arrow-categories/
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# The Unapologetic Mathematician
## Arrow Categories
One very useful example of a category is the category of arrows of a given category $\mathcal{C}$.
We start with any category $\mathcal{C}$ with objects ${\rm Ob}(\mathcal{C})$ and morphisms ${\rm Mor}(\mathcal{C})$. From this we build a new category called $\mathcal{C}^\mathbf{2}$, for reasons that I’ll explain later. The objects of $\mathcal{C}^\mathbf{2}$ are just the morphisms of $\mathcal{C}$. The morphisms of this new category are where things start getting interesting.
Let’s take two objects of $\mathcal{C}^\mathbf{2}$ — that is, two morphisms of $\mathcal{C}$ — and lay them side-by-side:
$\begin{matrix}A&&C\\\downarrow^f&&\downarrow^g\\B&&D\end{matrix}$
Now we want something that transforms one into the other. What we’ll do is connect each of the objects on the left to the corresponding object on the right by an arrow:
$\begin{matrix}A&\rightarrow^h&C\\\downarrow^f&&\downarrow^g\\B&\rightarrow^k&D\end{matrix}$
and require that the resulting square commute: $g\circ h=k\circ f$ as morphisms in $\mathcal{C}$. This is a morphism from $f$ to $g$. Sometimes we’ll write $(h,k):f\rightarrow g$, and sometimes we’ll name the square and write $\alpha:f\rightarrow g$.
If we have three morphisms $f$, $g$, and $h$ in $\mathcal{C}$, and commuting squares $(k_1,k_2):f\rightarrow g$ and $(k_3,k_4):g\rightarrow h$ then we can get a commuting square $(k_3\circ k_1,k_4\circ k_2):f\rightarrow h$. We check that this square commutes: $h\circ k_3\circ k_1=k_4\circ g\circ k_1=k_4\circ k_2\circ f$. This gives a composition of commuting squares. It’s easily checked that this is associative.
Given any morphism $f:A\rightarrow B$ in $\mathcal{C}$ we can just apply the identity arrows to each of $A$ and $B$ to get a commuting square $(1_A,1_B)$ between $f$ and itself. It is clear that this square serves as the identity arrow on the object $f$ in $\mathcal{C}^\mathbf{2}$, completing our proof that arrows and commuting squares in $\mathcal{C}$ do form a category.
### Like this:
Posted by John Armstrong | Category theory
## 3 Comments »
1. Thanks for this — nice crisp explanation. Would be even more helpful if it had a couple of simple, finite examples. Posets maybe?
Comment by Brian Rotman | January 28, 2010 | Reply
2. Well, these sorts of things show up in all sorts of other categorical constructions I’ve discussed, although I haven’t explicitly linked back here. Look at categories of natural transformations, or of representations, or (coming eventually) bundles.
Comment by | January 28, 2010 | Reply
3. How do you show that \$hom(f,g)\$ and \$hom(f’,g’)\$ are disjoint?
Comment by Tony | December 12, 2012 | Reply
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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.
• ## Feedback
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http://byorgey.wordpress.com/2008/09/16/the-poisson-distribution-and-sterling-numbers/
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The Poisson distribution and Stirling numbers
Posted on September 16, 2008 by
While working on an assignment for my machine learning class, I rediscovered the fact that if X is a random variable from a Poisson distribution with parameter $\lambda$, then
$\displaystyle E[X^n] = \sum_{k=1}^n S(n,k) \lambda^k,$
where $S(n,k)$ denotes a Stirling number of the second kind. (I actually prefer Knuth’s curly bracket notation, but I can’t seem to get it to work on this blog.) In particular, if $\lambda = 1$, then $E[X^n]$ is the nth Bell number $B_n$, the number of ways of partitioning a set of size n into subsets!
As it turned out, this didn’t help me at all with my assignment, I just thought it was nifty.
Share this:
This entry was posted in combinatorics, grad school, learning, math and tagged Bell numbers, moments, Poisson, Stirling numbers. Bookmark the permalink.
2 Responses to The Poisson distribution and Stirling numbers
1. keyholecontrol says:
I think \left\{k \atop b\right\} worked for me to for bracket notation.
I hope that shows up as code…
$\left\{k \atop b\right\}$
2. Brent says:
Aha, thanks!
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http://unapologetic.wordpress.com/2009/07/17/
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# The Unapologetic Mathematician
## Nondegenerate Forms II
Okay, we know what a nondegenerate form is, but what does this mean for the transformation that represents the form?
Remember that the form represented by the transformation $B$ is nondegenerate if for every nonzero ket vector $\lvert v\rangle$ there is some bra vector $\langle w\rvert$ so that $\langle w\rvert B\lvert v\rangle\neq0$. But before we go looking for such a bra vector, the transformation $B$ has turned the ket vector $\lvert v\rangle$ into a new ket vector $B\lvert v\rangle=\lvert B(v)\rangle$. If we find that $B(v)=0$, then there can be no suitable vector $w$ with which to pair it. So, at the very least, we must have $B(v)\neq0$ for every $v\neq0$. That is, the kernel of $B$ is trivial. Since $B$ is a transformation from the vector space $V$ to itself, the rank-nullity theorem tells us that the image of $B$ is all of $V$. That is, $B$ must be an invertible transformation.
On the other hand, if $B$ is invertible, then every nonzero ket vector $\lvert v\rangle$ becomes another nonzero ket vector $\lvert w\rangle=B\lvert v\rangle$. Then we find that
$\displaystyle\langle w\rvert B\lvert v\rangle=\langle w\vert w\rangle>0$
where this last inequality holds because the bra-ket pairing is an inner product, and is thus positive-definite. Indeed, a positive-definite (or negative-definite) form must be nondegenerate. Thus it is sufficient for $B$ to be invertible.
Incidentally, this approach gives us a good way of constructing a lot of positive-definite transformations. Given an invertible transformation $B$, we expand
$\displaystyle\langle w\vert w\rangle=\langle v\rvert B^*B\lvert v\rangle$
Since the form defined by the bra-ket pairing is invertible, so is the form defined by $B^*B$. And this is a sensible concept, since $B^*B$ is self-adjoint. Indeed, we take its adjoint to find
$\displaystyle\left(B^*B\right)^*=B^*\left(B^*\right)^*=B^*B$
This extends our analogy with the complex numbers. An invertible transformation composed with its adjoint is a self-adjoint, positive-definite transformation, just as a nonzero complex number multiplied by its conjugate is a real, positive number.
Posted by John Armstrong | Algebra, Linear Algebra | 2 Comments
## 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.
|
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http://mathoverflow.net/questions/94431?sort=votes
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## Technical question about cell complexes
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hello,
I have a technical question. My terminology:
I - set of standard inclusions $\partial I^n \to I^n$.
I-cell (Relative Cell Complexes) - transfinite compositions of pushouts of maps in $I$.
CW (CW complexes) - the usual definition (like I-cell but with "cells attached by order of dimension).
I-cof - retracts of maps in I-cell, the same as maps having Left Lifting Property w.r.t. maps having Right lifting property w.r.t. I (by Quillen small object argument).
My first question is to confirm that CW in I-cell in I-cof are all different
My second question is: In this page: http://ncatlab.org/nlab/show/model+structure+on+topological+spaces it is written under "Mixed model structure" (the model structure on $Top$ for which equivalences are weak equivalences and fibrations are Hurewicz fibrations) that cofibrant objects are spaces homotopy equivalent to CW complexes. Is this exactly true (or is it for example spaces homotopy equivalent to cell complexes?). I also read somewhere around nlab that Milnor advocated that spaces homotopy equivalent to CW complexes are nice, so I wanted to know if the cofibrant objects in this mixed model structure are exactly such, or a bit more general.
Thank you, Sasha
-
## 1 Answer
They are exactly such. One point is that one can use classical cell complexes, stop at $\omega$ with no transfinite nonsense, in setting up the Quillen model structure: it is a compactly generated (as well as a cofibrantly generated) model category. The distinction and full details are in May and Ponto, "More concise algebraic topology", published Feb. 1 this year. Another is that classical cell complexes are homotopy equivalent to CW complexes, as one can see by approximating attaching maps by maps that land in the n-skeleton. Formally, one has two filtrations on cell complexes, one given by the order of construction, the other given by dimensions of cells. The distinction is familiar and essential when one goes stable and works with spectra rather than spaces. Milnor wrote a classical paper "On spaces of the homotopy type of CW complexes" not just advocating but proving the niceness of the category of CW homotopy types. It is a result of Cole "Mixing model structures" that this category is exactly the cofibrant objects in his mixed model structure.
-
Thank you very much! A small question: "More concise algebraic topology" is not available online? – Sasha Apr 18 2012 at 19:35
"More Concise..." is a book that they published, not an article. So you should probably check the library rather than trying to find it online. – Josh Roberts Apr 19 2012 at 1:27
3
Everything else I've published is on my web page. There are already pirated versions on line (the first one I saw classified the book as science fiction!), but please do not download one. Reputable publishers will go out of business if they do not have enough time to at least recoup costs of production and distribution before their books go on line. Therefore I have not put it on line yet. – Peter May Apr 19 2012 at 2:43
OK, Thank you. Then I guess I will have to wait until it will appear in my library. – Sasha Apr 19 2012 at 6:58
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http://mathoverflow.net/questions/68674?sort=newest
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## Tensor products of permutation representations of symmetric groups.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am looking for a reference for the following fact which must be classical (which makes it harder, for me, to track a reference down). I am interested because there are similar (more complicated) statements about the cohomology of symmetric groups.
If $P$ is a partition, namely $p_{1} + \cdots + p_{k} = n$, we let $\rho_{P}$ denote the permutation representation of $S_{n}$, induced up from the trivial representation of $S_{P}$.
If $P$ and $Q$ are partitions of $n$ then consider any matrix $\hat{A}$ with nonnegative integer entries such that the entries of $i$th row of $A$ add up to $p_{i}$ and those of the $j$th column of $A$ add up to $q_{j}$. Then the entries of $\hat{A}$ form another partition of $n$, which we call $A$ and say that $A$ is a product-refinement of $P$ and $Q$. For example if $P = Q = 1 + 2$ then two possibilities for $\hat{A}$ are $\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 0 & 1 \ 1 & 1 \end{pmatrix}$.
Proposition: If $\rho_{P}$ and $\rho_{Q}$ are permutation representations of $S_n$ then $\rho_{P} \otimes \rho_{Q} \cong \bigoplus_{A} \rho_{A},$ where the sum is over $A$ which are product-refinements of $P$ and $Q$.
Questions: 1) what is the reference for this fact? and 2) what is standard terminology (for product-refinement in particular)?
-
The "Frobenius characteristic" is an isomorphism between the space of class functions of $S_n$ and the homogeneous symmetric functions of degree $n$. It sends the character of $\rho_P$ to the product of complete sums $h_P=h_{P_1} \cdot h_{P_2} \cdots$. By transport of structure it gives a new product on the symmetric functions of degree $n$. So you might find some references for this result by looking for "Kronecker products of complete sum symmetric functions". – Emmanuel Briand Jun 24 2011 at 0:09
I meant "It gives a new product on the symmetric functions of degree $n$, called the Kronecker product of symmetric functions." (or sometimes "internal product of symmetric functions"). – Emmanuel Briand Jun 24 2011 at 0:12
## 1 Answer
Hi Dev,
It looks to me like a proof of this fact is given in the answer to Exercise 7.84(b) of Richard Stanley's Enumerative Combinatorics, volume 2, along with a reference to Example I.7.23(e), page 131, of I. G. Macdonald's Symmetric Functions and Hall Polynomials (2nd edition).
-
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http://mathhelpforum.com/calculus/140913-integrals-print.html
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# integrals
Printable View
• April 23rd 2010, 06:59 AM
alexandrabel90
integrals
hi!
May i know how to solve this question?
and how do i solve it when the arrow on the left hand side of the diagram changes to the other direction?
http://i752.photobucket.com/albums/x...0/DSCF4517.jpg
• April 24th 2010, 11:24 AM
Failure
Quote:
Originally Posted by alexandrabel90
hi!
May i know how to solve this question?
How about splitting $\vec{F}(x,y)$, into a much simpler component $\vec{F}_1(x,y):=(-y,0)$ and a component $\vec{F}_2(x,y):= (y\sin(xy),x\sin(x,y))$ that happens to be a conservative field like this
$\vec{F}(x,y)=\vec{F}_1(x,y)+\vec{F}_2(x,y) := (-y,0)+\nabla \Phi(x,y)$, where $\Phi(x,y) := -\cos(xy)$.
Since the curve is closed, the conservative part can be dropped from the line integral: you only need to integrate $F_1(x,y)=(-y,0)$ along that curve.
Quote:
and how do i solve it when the arrow on the left hand side of the diagram changes to the other direction?
Are you asking what happens if you reverse the direction of the entire curve? - Well, in that case the line integral changes its sign.
• April 24th 2010, 02:02 PM
alexandrabel90
no. i meant just reversing the direction of the curve on the left hand side, while the direction on the right hand side remains the same.
• April 24th 2010, 10:53 PM
Failure
Quote:
Originally Posted by alexandrabel90
no. i meant just reversing the direction of the curve on the left hand side, while the direction on the right hand side remains the same.
In that case, things just might get a little more complicated (to put it politely): because in that case, the line integral for the remaining (non-conservative) component $\vec{F}_1(x,y)=(-y,0)$, may not (due to the symmetry of the curve) evaluate to 0 anymore...
• April 24th 2010, 11:16 PM
alexandrabel90
may i know how you got F1 and F2 from your reply above?
from how i see it, you split the curves up so that it will be a simple( non intersecting) curve?
• April 24th 2010, 11:18 PM
alexandrabel90
sorry im still very confused how to solve this current question.
and for the question where if the direction on the left hand side changes, i was thinking that since both the arrows will now oppose each other, they will cancel out and hence the line integral will be 0? but i guess i cant say it that way right?
• April 24th 2010, 11:28 PM
Failure
Quote:
Originally Posted by alexandrabel90
may i know how you got F1 and F2 from your reply above?
from how i see it, you split the curves up so that it will be a simple( non intersecting) curve?
No my separating $\vec{F}(x,y)$ has got nothing to do with the curve, it is simply a splitting up of the vector $\vec{F}(x,y)$ itself into a non-conservative and a conservative part. Integration along a closed curve allows you to just drop the conservative part and concentrate on the non-conservative part exclusively.
• April 24th 2010, 11:38 PM
Failure
Quote:
Originally Posted by alexandrabel90
sorry im still very confused how to solve this current question.
and for the question where if the direction on the left hand side changes, i was thinking that since both the arrows will now oppose each other, they will cancel out and hence the line integral will be 0? but i guess i cant say it that way right?
You can say it that way, alright, but you can't write it down like this: just take the intuition it provides to split the overall line integral into two line integrals (over parts of the curve) that cancel each other. To my eyes at least it seems that the line integral over the part of the curve in quadrants IV and I should cancel against the line integral over the part of the curve in quadrants II and III. If that hypothesis happens to be correct (I haven't checked), you just have to juggle the parametrizations of these two line integrals in such a way that their cancelling each other out is made obvious (without any need for actually calculating the values of the two line integrals themselves).
• April 25th 2010, 12:36 AM
alexandrabel90
thanks for the explanation.
by the way, in this case, we cant apply green's theorem because the curve is not simple right?
All times are GMT -8. The time now is 03:39 PM.
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http://amathew.wordpress.com/tag/holomorphic-functions/
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# Climbing Mount Bourbaki
Thoughts on mathematics
January 1, 2010
## A crash course in one complex variable
Posted by Akhil Mathew under analysis, complex analysis, math education | Tags: Cauchy formula, Cauchy theorem, holomorphic functions, Laurent series, Stokes theorem, Taylor series, z-bar derivative |
[5] Comments
A friend of mine is taking a course on analytic number theory in the spring and needs to learn basic complex analysis in a couple of weeks. I decided to do a post (self-contained, except for Stokes’ formula) on deducing the Cauchy theorems and their applications from Stokes’ theorem now instead of later–when I’ll talk about several complex variables. It might be objected that Stokes’ theorem is just Green’s theorem for $n=2$, commonly used in undergraduate treatments, but my goal was to take an expository challenge: write something rigorous on complex variables in as short a space as possible without sacrificing readability. So Stokes’ theorem for manifolds is preferable to Green’s theorem as stated in a vague way about “insides of a curve” (before, say, the Jordan curve theorem is proved) and the traditional proof of Green’s theorem via rectangular decompositions.
So, let’s consider an open set ${O \subset \mathbb{C}}$, and a ${C^2}$ function ${f: O \rightarrow \mathbb{C}}$. We can consider the differential
$\displaystyle df := f_x dx + f_y dy$
which is a complex-valued 1-form on ${O}$. It is also convenient to write the differential using the ${z}$ and ${\bar{z}}$-derivatives I talked about earlier, i.e.
$\displaystyle f_z := \frac{1}{2}\left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right) f, \quad f_{\bar{z}} := \frac{1}{2}\left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right) f.$
The reason these are important is that if ${w_0 \in O}$, we can choose ${A,B \in \mathbb{C}}$ with
$\displaystyle f(w_0+h) = f(w_0) + Ah + B \bar{h} + o(|h|), \ h \in \mathbb{C}$
by differentiability, and it is easy to check that ${A=f_z(w_0), B=f_{\bar{z}}(w_0)}$. So we can define a function ${f}$ to be holomorphic if it satisfies the differential equation
$\displaystyle f_{\bar{z}} = 0,$
which is equivalent to being able to write
$\displaystyle f(w_0 + h) =f(w_0) + Ah + o(|h|)$
for each ${w_0 \in O}$ and a suitable ${A \in \mathbb{C}}$. In particular, it is equivalent to a difference quotient definition. The derivative ${f_z}$ of a holomorphic function thus satisfies all the usual algebraic rules, under which holomorphic functions are closed. (more…)
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http://mathhelpforum.com/differential-equations/59784-second-order-differential-equation.html
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# Thread:
1. ## Second Order Differential Equation
Hi!
How do I determine the particular integral for second order differential equations with mixed f(x)s? Like, it's neither strictly polynomial, trigonometry or exponential.
Example of questions,
(i) $y'' - 4y' + 5y = (16x + 4)e^{3x}$
(ii) $y'' + 3y' = (10x + 6)sin x$
(iii) $y'' - 2y' + 4y = 541e^{2x}cos5x$
Thank you! (:
2. Hey, each of those have right members which are particular solutions to some homogeneous differential equation. For example, the first one has a right member which is a solution to the equation:
$(D-3)^2 y=0$
So the operator $(D-3)^2$ becomes an "annihilation" operator that we can apply to both side of the equation to convert it to a homogeneous equation:
$(D-3)^2 (D^2-4D+5)y=0$
This is the method of undetermined coefficients. Are you familiar with that method? Try it first on some simple ones. Any DE book should have a section on this subject.
3. Sorry but I do not understand what are you talking about at all.
4. Originally Posted by shawsend
Hey, each of those have right members which are particular solutions to some homogeneous differential equation. For example, the first one has a right member which is a solution to the equation:
$(D-3)^2 y=0$
So the operator $(D-3)^2$ becomes an "annihilation" operator that we can apply to both side of the equation to convert it to a homogeneous equation:
$(D-3)^2 (D^2-4D+5)y=0$
This is the method of undetermined coefficients. Are you familiar with that method? Try it first on some simple ones. Any DE book should have a section on this subject.
Aha!! I'm not the only one to know about the Annihilator approach
Originally Posted by panda*
Sorry but I do not understand what are you talking about at all.
Read post #6 and #7 here to see how to tackle equations like these.
--Chris
5. 1. First you must solve the homogeneous equations
(i) y''-4y-+5y=0
(ii) y''+3y'=0
(iii) y''-2y'+4y=0
and you will find the general solutions for them.
2. For non homogeneous equations you have to find the partial solutions according to what is on the right side of the equations
(i)
(ii)
(iii)
3. The final solution of your equations is the sum of the general solutions from 1st point and partial solutions from the 2nd point.
6. Thank you for referrals of posts and helps.
But I still don't really get it after I read the posts
Generally, I do know how to handle second order differential equations if f(x) was one of the normal terms like, purely exponential, or trig or polynomials. But when it is multiplied together like that, I can't immediately identify the pattern of the particular integral.
For example, the particular integral form for an exponential function of $f(x) = e^{kx}$ would be $y_p = pe^{2x}$ and we differentiate and sub it back in to compare coefficients in order to obtain p.
How do I deal when f(x) is not purely of those forms but multiplied together?
Thank you.
7. I'll work the first one using undetermined coefficients; differential equations open a unique window into the universe such that all her secrets are revealed:
$(D-3)^2(D^2-4D+5)y=0$
The general solution from the auxiliary equation is then:
$y=c_1 e^{3x}+c_2 e^{3x}+c_4 e^{2x}\cos(x)+c_5 e^{2x}\sin(x)$
with the desired solution (I'm taking this right out of Rainville almost word for word):
$y=y_c+y_p$
where $y_c=c_4 e^{2x}\cos(x)+c_5 e^{2x}\sin(x)$. Then there must be a particular solution of the original equation containing at most the remaining terms: $y_p=Ae^{3x}+Bxe^{3x}$. That's the undetermined coefficients which can be determined by substituting this $y_p$ into the original DE:
When I do that I get:
$2Ae^{3x}+2Bxe^{3x}+2Be^{3x}=16xe^{3x}+4e^{3x}$
Equating coefficients, I get $B=8$ and $A=-6$. Then the general solution of $y''-4y'+5y=(16x+4)e^{3x}$ is:
$y(x)=-6e^{3x}+8xe^{3x}+c_1e^{2x}\cos(x)+c_2e^{2x}\sin(x)$
8. Although I still do not understand the method you recommended, but I'm still thankful for your generous help and time! I figured out a different approach to do it already But still thanks anyway!
9. Originally Posted by panda*
I figured out a different approach to do it already
Close enough.
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http://mathoverflow.net/questions/60814/removable-singularities-for-elliptic-equations
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## Removable Singularities for Elliptic Equations
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The following fact is quite standard and does not have a very long proof:
$(*)$ If $u$ is harmonic on $B_1(0)\setminus {0}$ and uniformly bounded, then $u$ in fact extends to a harmonic function on the whole ball.
Some googling reveals that such statements are in fact true for large classes of elliptic operators with much more general singularity sets and growth conditions. There appears to be a vast literature on the subject.
I would like to know if the exact analogue of $(*)$ has a "simple" proof for linear elliptic operators. "Simple" is slightly ambiguous but is meant to mean tools present in standard elliptic theory textbooks, i.e. Gilbarg and Trudinger or Jost.
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Removable singularities are closely related to the maximum principle. It fails for elliptic systems (Serrin's counter-example). The right persons to ask for are Laurent Veron or Haïm Brézis. – Denis Serre Apr 6 2011 at 13:25
1
I believe one possible approach to is to use a cutoff function to show that the solution is a weak solution on the ball including the origin. Then standard elliptic regularity results show that it is a strong solution. – Deane Yang Apr 6 2011 at 15:15
@Deane I just checked the details and I believe that such an approach does indeed work. If you want to add this as an answer I will accept. – Yakov Shlapentokh-Rothman Apr 6 2011 at 15:45
## 3 Answers
The proof I had in mind was actually simpler and appears to work only in dimension greater than 2. Here is a sketch for a second order self-adjoint elliptic operator $Pu = \partial_i(a^{ij}\partial_ju)$. Suppose $u$ satisfies $Pu = 0$ on $B\backslash{0}$. We want to show that $u$ is a weak solution on $B$. It suffices to show that if $\phi$ is a smooth compactly supported function on $B$, then $$\int \phi Pu = 0.$$ Let $\chi$ be a smooth nonnegative compactly supported function that is equal to $1$ on a neighborhood of $0$ and, given $\epsilon > 0$, let $\chi_\epsilon(x) = \chi(x/\epsilon)$. Then $$\int \phi Pu = \int [\phi(1-\chi_\epsilon) + \phi\chi_\epsilon]Pu = \int \phi\chi_\epsilon Pu = \int P(\phi\chi_\epsilon)u.$$ If we now expand out $P(\phi\chi_\epsilon)$, we see that it is $O(\epsilon^{-2})$. Therefore, if $u \in L_\infty$, then $$\int P(\phi\chi_\epsilon)u = O(\epsilon^{n-2})$$ Therefore, taking the limit $\epsilon \rightarrow 0$ shows that $u$ is a weak solution on $B$. Elliptic regularity tells you that it is in fact a strong smooth solution.
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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.
Here are the details behind my original interpretation of Deane Yang's comment:
By Schauder estimates, $\vert\vert \nabla u\vert\vert_{L^{\infty}} \leq C\vert\vert u\vert\vert_{L^{\infty}}$, so we also know that the gradient of $u$ is uniformly bounded. Next, we claim that $u$ is a weak solution of $Lu = 0$ on $B_1(0)$. We have $Lu = \sum_{i,j=1}^n \partial_j(a^{ij}(x)\partial_i u) + \sum_{i=1}^nb^i(x)\partial_i u + c(x)u$. To show that $u$ is a weak solution on $B_1(0)$, we need to show that for every $\varphi \in C_c^{\infty}(B_1(0))$,
$\int_{B_1(0)} \sum_{i,j=1}^n a^{ij}(x)\partial_{i}u\partial_j\varphi + \sum_{i=1}^nb^i(x)(\partial_i u) \varphi+ c(x)u\varphi = 0$
Since all of the relevant terms are uniformly bounded, the left hand side is equal to
$\lim_{\epsilon\to 0}\int_{B_1(0)\setminus B_{\epsilon}(0)} \sum_{i,j=1}^n a^{ij}(x)\partial_{i}u\partial_j\varphi + \sum_{i=1}^nb^i(x)(\partial_i u) \varphi+ c(x)u\varphi$
After an integration by parts, this gives
$\lim_{\epsilon\to 0}\int_{B_1(0)\setminus B_{\epsilon}(0)}(Lu)\varphi + \int_{\partial B_{\epsilon}(0)}\sum_{i,j=1}^n a^{ij}(x)(\partial_{i}u)\varphi \nu_j =$
$\lim_{\epsilon\to 0}\int_{\partial B_{\epsilon}(0)}\sum_{i,j=1}^n a^{ij}(x)(\partial_{i}u)\varphi \nu_j$
Since everything involved is uniformly bounded, this goes to $0$. (EDIT: The dimension should be at least $2$ for this to work)
Now $L^2$ elliptic regularity implies that $u$ is smooth on the whole ball and we are done.
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I'm glad that worked out! – Deane Yang Apr 8 2011 at 2:42
Also, I'm glad you went ahead and posted the answer, because I did not work out the details and was just giving a semi-educated guess. – Deane Yang Apr 8 2011 at 2:44
Check out
Harvey, Reese; Polking, John Removable singularities of solutions of linear partial differential equations. Acta Math. 125 1970 39–56.
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http://en.wikisource.org/wiki/A_Treatise_on_Electricity_and_Magnetism/Part_I/Chapter_II
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# A Treatise on Electricity and Magnetism/Part I/Chapter II
From Wikisource
by James Clerk Maxwell Part I, Chapter II: Elementary Mathematical Theory of Electricity
A Treatise on Electricity and Magnetism — Part I, Chapter II: Elementary Mathematical Theory of Electricity James Clerk Maxwell
## CHAPTER II.
ELEMENTARY MATHEMATICAL THEORY OF STATICAL ELECTRICITY.
### Definition of Electricity as a Mathematical Quantity.
63.] We have seen that the actions of electrified bodies are such that the electrification of one body may be equal to that of another, or to the sum of the electrifications of two bodies, and that when two bodies are equally and oppositely electrified they have no electrical effect on external bodies when placed together within a closed insulated conducting vessel. We may express all these results in a concise and consistent manner by describing an electrified body as charged with a certain quantity of electricity, which we may denote by $e$. When the electrification is positive, that is, according to the usual convention, vitreous, $e$ will be a positive quantity. When the electrification is negative or resinous, $e$ will be negative, and the quantity $e$ may be interpreted either as a negative quantity of vitreous electricity or as a positive quantity of resinous electricity.
The effect of adding together two equal and opposite charges of electricity, $+e$ and $-e$, is to produce a state of no electrification expressed by zero. We may therefore regard a body not electrified as virtually charged with equal and opposite charges of indefinite magnitude, and an electrified body as virtually charged with unequal quantities of positive and negative electricity, the algebraic sum of these charges constituting the observed electrification. It is manifest, however, that this way of regarding an electrified body is entirely artificial, and may be compared to the conception of the velocity of a body as compounded of two or more different velocities, no one of which is the actual velocity of the body. When we speak therefore of a body being charged with a quantity $e$ of electricity we mean simply that the body is electrified, and that the electrification is vitreous or resinous according as $e$ is positive or negative.
## ON ELECTRIC DENSITY.
### Distribution in Three Dimensions.
64.] Definition. The electric volume-density at a given point in space is the limiting ratio of the quantity of electricity within a sphere whose centre is the given point to the volume of the sphere, when its radius is diminished without limit.
We shall denote this ratio by the symbol $\rho$, which may be positive or negative.
### Distribution on a Surface.
It is a result alike of theory and of experiment, that, in certain cases, the electrification of a body is entirely on the surface. The density at a point on the surface, if defined according to the method given above, would be infinite. We therefore adopt a different method for the measurement of surface-density.
Definition. The electric density at a given point on a surface is the limiting ratio of the quantity of electricity within a sphere whose centre is the given point to the area of the surface contained within the sphere, when its radius is diminished without limit.
We shall denote the surface-density by the symbol $\sigma$.
Those writers who supposed electricity to be a material fluid or a collection of particles, were obliged in this case to suppose the electricity distributed on the surface in the form of a stratum of a certain thickness $\theta$, its density being $\rho_0$ , or that value of $\rho$ which would result from the particles having the closest contact of which they are capable. It is manifest that on this theory
$\rho_0 \,\theta = \sigma\,\!$
When $\sigma$ is negative, according to this theory, a certain stratum of thickness $\theta$ is left entirely devoid of positive electricity, and filled entirely with negative electricity.
There is, however, no experimental evidence either of the electric stratum having any thickness, or of electricity being a fluid or a collection of particles. We therefore prefer to do without the symbol for the thickness of the stratum, and to use a special symbol for surface-density.
### Distribution along a Line.
It is sometimes convenient to suppose electricity distributed on a line, that is, a long narrow body of which we neglect the thickness. In this case we may define the line-density at any point to be the limiting ratio of the electricity on an element of the line to the length of that element when the element is diminished without limit.
If $\lambda$ denotes the line-density, then the whole quantity of electricity on a curve is $e= \int \lambda \, ds$, where dS is the element of the curve.
Similarly, if $\rho$ is the surface-density, the whole quantity of electricity on the surface is
$e = \iint \sigma\,dS$,
where dS is the element of surface.
If $\rho$ is the volume-density at any point of space, then the whole electricity within a certain volume is
$e= \iiint \rho \; dx \; dy\; dz$
where $dx dy dz$ is the element of volume. The limits of integration in each case are those of the curve, the surface, or the portion of space considered.
It is manifest that e, $\lambda$, $\sigma$ and $\rho$ are quantities differing in kind, each being one dimension in space lower than the preceding, so that if $a$ be a line, the quantities e, $a\lambda, a^2\sigma$ and $a^3 \rho$ will be all of the same kind, and if a be the unit of length, and $\lambda, \sigma, \rho$ each the unit of the different kinds of density, $a\lambda, a^2\sigma$ and $a^3 \rho$will each denote one unit of electricity.
### Definition of the Unit of Electricity.
65.] Let $A$ and $B$ be two points the distance between which is the unit of length. Let two bodies, whose dimensions are small compared with the distance $AB$, be charged with equal quantities of positive electricity and placed at $A$ and $B$ respectively, and let the charges be such that the force with which they repel each other is the unit of force, measured as in Art. 6. Then the charge of either body is said to be the unit of electricity. If the charge of the body at $B$ were a unit of negative electricity, then, since the action between the bodies would be reversed, we should have an attraction equal to the unit of force.
If the charge of $A$ were also negative, and equal to unity, the force would be repulsive, and equal to unity.
Since the action between any two portions of electricity is not affected by the presence of other portions, the repulsion between $e'$ units of electricity at $A$ and $e'$ units at $B$ is $ee'$, the distance $AB$ being unity. See Art. 39.
### Law of Force between Electrified Bodies.
66.] Coulomb shewed by experiment that the force between electrified bodies whose dimensions are small compared with the distance between them, varies inversely as the square of the distance. Hence the actual repulsion between two such bodies charged with quantities $e$ and $e'$ and placed at a distance $r$ is
$\frac {ee'}{r^2}$
We shall prove in Art. 74 that this law is the only one consistent with the observed fact that a conductor, placed in the inside of a closed hollow conductor and in contact with it, is deprived of all electrical charge. Our conviction of the accuracy of the law of the inverse square of the distance may be considered to rest on experiments of this kind, rather than on the direct measurements of Coulomb.
### Resultant Force between Two Bodies.
67.] In order to find the resultant force between two bodies we might divide each of them into its elements of volume, and consider the repulsion between the electricity in each of the elements of the first body and the electricity in each of the elements of the second body. We should thus get a system of forces equal in number to the product of the numbers of the elements into which we have divided each body, and we should have to combine the effects of these forces by the rules of Statics. Thus, to find the component in the direction of $x$ we should have to find the value of the sextuple integral
$\iiint \!\!\!\iiint \frac {\rho \rho' (x-x')\; dx \;dy \;dz \; dx' \;dy' \;dz' \;}{\{(x-x')^2+(y-y')^2+(z-z')^2\}^{3 \over 2}}$,
where $x, y, z$ are the coordinates of a point in the first body at which the electrical density is $\rho$, and $x' , y' , z'$, and $\rho'$ are the corresponding quantities for the second body, and the integration is extended first over the one body and then over the other.
### Resultant Force at a Point.
68.] In order to simplify the mathematical process, it is convenient to consider the action of an electrified body, not on another body of any form, but on an indefinitely small body, charged with an indefinitely small amount of electricity, and placed at any point of the space to which the electrical action extends. By making the charge of this body indefinitely small we render insensible its disturbing action on the charge of the first body.
Let $e$ be the charge of this body, and let the force acting on it when placed at the point $(x, y, z)$ be $Re$, and let the direction-cosines of the force be $l, m, n$, then we may call $R$ the resultant force at the point $(x, y, z)$.
In speaking of the resultant electrical force at a point, we do not necessarily imply that any force is actually exerted there, but only that if an electrified body were placed there it would be acted on by a force $R e$, where $e$ is the charge of the body.
Definition. The Resultant electrical force at any point is the force which would be exerted on a small body charged with the unit of positive electricity, if it were placed there without disturbing the actual distribution of electricity.
This force not only tends to move an electrified body, but to move the electricity within the body, so that the positive electricity tends to move in the direction of $R$ and the negative electricity in the opposite direction. Hence the force $R$ is also called the Electromotive Force at the point $(x,\, y,\,z)$.
When we wish to express the fact that the resultant force is a vector, we shall denote it by the German letter $\mathfrak {C}$. If the body is a dielectric, then, according to the theory adopted in this treatise, the electricity is displaced within it, so that the quantity of electricity which is forced in the direction of $\mathfrak {C}$ across unit of area fixed perpendicular to $\mathfrak {C}$ is
$\mathfrak {D}=\frac {1}{4\,\pi}\,K\mathfrak {C}$
where $\mathfrak {D}$ is the displacement, $\mathfrak {C}$ the resultant force, and $K$ the specific inductive capacity of the dielectric. For air, $K = 1$.
If the body is a conductor, the state of constraint is continually giving way, so that a current of conduction is produced and maintained as long as the force $\mathfrak {C}$ acts on the medium.
### Components of the Resultant Force.
If $X,\, Y,\, Z$ denote the components of R, then
$X=Rl,{\color{White}xxxx}Y=Rm,{\color{White}xxxx}Z=Rn;$
where $l , m, n$ are the direction-cosines of $R$.
### Line-Integral of Electric Force, or Electromotive Force along an Arc of a Curve.
69.] The Electromotive force along a given arc $AP$ of a curve is numerically measured by the work which would be done on a unit of positive electricity carried along the curve from the beginning, $A$, to $P$, the end of the arc.
If $s$ is the length of the arc, measured from $A$, and if the resultant force $R$ at any point of the curve makes an angle $c$ with the tangent drawn in the positive direction, then the work done on unit of electricity in moving along the element of the curve ${ds}$ will be
$R \cos\epsilon \,ds$
,
and the total electromotive force $V$ will be
$V= \int_{0}^{s} R\cos \epsilon \,ds$,
the integration being extended from the beginning to the end of the arc.
If we make use of the components of the force $R$, we find
$V \int_{0}^{s} (X \frac{dx}{dy}+Y\frac{dy}{ds} +Z\frac {dz}{ds})\,ds.$.
If $X, Y,$ and $Z$ are such that $X{dx}+Y{dy} + Z{dz}$ is a complete differential of a function of$x, y, z,$ then
$V=\int_{A}^{P} (X\,dx+Y\,dy+Z\,dz)=V_A-V_P$;
where the integration is performed in any way from the point $A$ to the point $P$, whether along the given curve or along any other line between $A$ and $P$.
In this case $V$ is a scalar function of the position of a point in space, that is, when we know the coordinates of the point, the value of $V$ is determinate, and this value is independent of the position and direction of the axes of reference. See Art. 16.
### On Functions of the Position of a Point.
In what follows, when we describe a quantity as a function of the position of a point, we mean that for every position of the point the function has a determinate value. We do not imply that this value can always be expressed by the same formula for all points of space, for it may be expressed by one formula on one side of a given surface and by another formula on the other side.
### On Potential Functions.
70.] The quantity $Xdx+Ydy+Zdz$ is an exact differential whenever the force arises from attractions or repulsions whose in tensity is a function of the distance only from any number of points. For if $r_1$ be the distance of one of the points from the point $(x, y, z)$, and if $R_1$ be the repulsion, then
$X_1 = R_1 l = R_1 \frac{dr_1}{dx}$,
with similar expressions for $Y_1$ and $Z_1$, so that
$X_1\,dx+Y_1\,dy+Z_1\,dz=R_1\,dr_1$;
and since $R_l$ is a function of $r_l$ only, $R_l dr_1$ is an exact differential of some function of $r_1$, say $V_1$.
Similarly for any other force $R_2$ , acting from a centre at distance $r_2$ ,
$X_2\,dx+Y_2\,dy+Z_2\,dz=R_2 \,dr_2=dV_2$.
But $X = X_1 + X_2 + \mbox{etc. and }Y$ and $Z$ are compounded in the same way, therefore
$X\,dx+Y\,dy+Z\,dz=dV_1+dV_2+\And \!\!c.=dV$.
$V$, the integral of this quantity, under the condition that $V =0$ at an infinite distance, is called the Potential Function.
The use of this function in the theory of attractions was introduced by Laplace in the calculation of the attraction of the earth. Green, in his essay 'On the Application of Mathematical Analysis to Electricity' gave it the name of the Potential Function. Gauss, working independently of Green, also used the word Potential. Clausius and others have applied the term Potential to the work which would be done if two bodies or systems were removed to an infinite distance from one another. We shall follow the use of the word in recent English works, and avoid ambiguity by adopting the following definition due to Sir W. Thomson.
Definition of Potential. The Potential at a Point is the work which would be done on a unit of positive electricity by the electric forces if it were placed at that point without disturbing the electric distribution, and carried from that point to an infinite distance.
71.] Expressions for the Resultant Force and its components in terms of the Potential.
Since the total electromotive force along any arc $AB$ is
$V_A - V_B \,\!$,
if we put $ds$ for the arc $AB$ we shall have for the force resolved in the direction of $ds$,
$R \, cos\epsilon\,=- \frac{dV}{ds}$;
whence, by assuming $ds$ parallel to each of the axes in succession, we get
$X=-\frac{dV}{dx},{\color{White}xxxx} Y=-\frac{dV}{dy},{\color{White}xxxx} Z=-\frac{dV}{dx}$;
$R={\left ({\left .\frac{\overline{dV}}{dx} \right |}^2+{\left .\frac{\overline{dV}}{dy} \right |}^2 + {\left .\frac{\overline{dV}}{dz} \right |}^2 \right )}^{1 \over 2}$
We shall denote the force itself, whose magnitude is $R$ and whose components are $X, Y, Z$, by the German letter $\mathfrak {C}$, as in Arts. 17 and 68.
### The Potential at all Points within a Conductor is the same.
72.] A conductor is a body which allows the electricity within it to move from one part of the body to any other when acted on by electromotive force. When the electricity is in equilibrium there can be no electromotive force acting within the conductor. Hence $R =0$ throughout the whole space occupied by the conductor. From this it follows that
$\frac {dV}{dx}=0,{\color{White}xxx}\frac {dV}{dy}=0,{\color{White}xxx}\frac {dV}{dz}=0$;
and therefore for every point of the conductor
$V\; = \; C$,
where $C$ is a constant quantity.
### Potential of a Conductor.
Since the potential at all points within the substance of the conductor is C, the quantity C is called the Potential of the conductor. C may be defined as the work which must be done by external agency in order to bring a unit of electricity from an infinite distance to the conductor, the distribution of electricity being supposed not to be disturbed by the presence of the unit.
If two conductors have equal potentials, and are connected by a wire so fine that the electricity on the wire itself may be neglected, the total electromotive force along the wire will be zero, and no electricity will pass from the one conductor to the other.
If the potentials of the conductors $A$ and $B$ be $V_A$ and $V_B$ then the electromotive force along any wire joining $A$ and $B$ will be
$V_A \; - \;V_B$
74: ELECTROSTATICS. [73.
in the direction $AB$, that is, positive electricity will tend to pass from the conductor of higher potential to the other.
Potential, in electrical science, has the same relation to Electricity that Pressure, in Hydrostatics, has to Fluid, or that Temperature, in Thermodynamics, has to Heat. Electricity, Fluids, and Heat all tend to pass from one place to another, if the Potential, Pressure, or Temperature is greater in the first place than in the second. A fluid is certainly a substance, heat is as certainly not a substance, so that though we may find assistance from analogies of this kind in forming clear ideas of formal electrical relations, we must be careful not to let the one or the other analogy suggest to us that electricity is either a substance like water, or a state of agitation like heat.
### Potential due to any Electrical System.
73.] Let there be a single electrified point charged with a quantity $e$ of electricity, and let $r$ be the distance of the point $x',y',z'$ from it, then
$V=\int_{r}^{\infty} R\,dr=\int_{r}^{\infty} \frac{e}{r^2}dr=\frac {e}{r}$.
Let there be any number of electrified points whose coordinates are $(x_1,y_1,z_1)$,$(x_2,y_2,z_2)$ &c. - and their charges $e_1, e_2$ &c., and let their distances from the point $(x',y',z')$ be $r_1$, $r_2$ , &c., then the potential of the system at $x',y',z'$' will be
$V=\sum{\left(\frac{e}{r} \right )}$.
Let the electric density at any point $(x, y, z)$ within an electrified body be p, then the potential due to the body is
$V=\iiint \frac {\rho}{r}dx\,dy\,dz$;
where $r= \{(x-x')^2 + (y-y')^2 +(z-z')^2 \}^{1 \over 2}$
the integration being extended throughout the body.
### On the Proof of the Law of the Inverse Square.
74.] The fact that the force between electrified bodies is inversely as the square of the distance may be considered to be established by direct experiments with the torsion-balance. The results, however, which we derive from such experiments must be regarded as affected by an error depending on the probable error of each experiment, and unless the skill of the operator be very great, 74.] PROOF OF THE LAW OF FORCE. 75
the probable error of an experiment with the torsion-balance is considerable. As an argument that the attraction is really, and not merely as a rough approximation, inversely as the square of the distance, Experiment VII (p. 34) is far more conclusive than any measurements of electrical forces can be.
In that experiment a conductor $B$, charged in any manner, was enclosed in a hollow conducting vessel $C$, which completely surrounded it. $C$ was also electrified in any manner.
$B$ was then placed in electric communication with $C$, and was then again insulated and removed from $C$ without touching it, and examined by means of an electroscope. In this way it was shewn that a conductor, if made to touch the inside of a conducting vessel which completely encloses it, becomes completely discharged, so that no trace of electrification can be discovered by the most delicate electrometer, however strongly the conductor or the vessel has been previously electrified.
The methods of detecting the electrification of a body are so delicate that a millionth part of the original electrification of $B$ could be observed if it existed. No experiments involving the direct measurement of forces can be brought to such a degree of accuracy.
It follows from this experiment that a non-electrified body in the inside of a hollow conductor is at the same potential as the hollow conductor, in whatever way that conductor is charged. For if it were not at the same potential, then, if it were put in electric connexion with the vessel, either by touching it or by means of a wire, electricity would pass from the one body to the other, and the conductor, when removed from the vessel, would be found to be electrified positively or negatively, which, as we have already stated, is not the case.
Hence the whole space inside a hollow conductor is at the same potential as the conductor if no electrified body is placed within it. If the law of the inverse square is true, this will be the case what ever be the form of the hollow conductor. Our object at present, however, is to ascertain from this fact the form of the law of attraction.
For this purpose let us suppose the hollow conductor to be a thin spherical shell. Since everything is symmetrical about its centre, the shell will be uniformly electrified at every point, and we have to enquire what must be the law of attraction of a uniform spherical shell, so as to fulfil the condition that the potential at every point within it shall be the same. Let the force at a distance $r$ from a point at which a quantity $e$ of electricity is concentrated be $R$, where $R$ is some function of $r$. All central forces which are functions of the distance admit of a potential, let us write $\tfrac {f(r)}{r}$ for the potential function due to a unit of electricity at a distance $r$.
Let the radius of the spherical shell be $a$, and let the surface-density be $\sigma$. Let $P$ be any point within the shell at a distance $p$ from the centre. Take the radius through $P$ as the axis of spherical coordinates, and let $r$ be the distance from $P$ to an element $dS$ of the shell. Then the potential at $P$ is
$V=\iint \sigma \, \frac{f(r)}{r} dS$,
$V=\int_{0}^{2 \pi}\int_{0}^{\pi}\sigma \, \frac{f(r)}{r}\,a^2\,sin\theta\,d\theta\,d\phi$.
Now $r^2=a^2-2ap\, cos \theta+p^2$,
{{{2}}}
$V=2 \pi\sigma \frac {a}{p}\int_{a-p}^{a+p} f(r) dr$;
and V must be constant for all values of $p$ less than $a$.
Multiplying both sides by $p$ and differentiating with respect to $p$,
$V=2 \pi\sigma a{f(a+p)+f(a-p)}\,\!$.
Differentiating again with respect to $p$,
$0=f'(a+p)-f'(a-p)\,\!$
,
Since a and p are independent,
$f'(r) = C\,\!$, a constant.
Hence $f(r) = Cr+C'\,\!$
and the potential function is
$\frac {f(r)}{r}=C+\frac {C'}{r}$
The force at distance $r$ is got by differentiating this expression with respect to $r$, and changing the sign, so that
$R=\frac {C'}{r^2}$
or the force is inversely as the square of the distance, and this therefore is the only law of force which satisfies the condition that the potential within a uniform spherical shell is constant[1]. Now this condition is shewn to be fulfilled by the electric forces with the most perfect accuracy. Hence the law of electric force is verified to a corresponding degree of accuracy.
### Surface-Integral of Electric Induction, and Electric Displacement through a Surface.
75.] Let $R$ be the resultant force at any point of the surface, and $\epsilon$ the angle which R makes with the normal drawn towards the positive side of the surface, then $R cos \epsilon$ is the component of the force normal to the surface, and if $dS$ is the element of the surface, the electric displacement through $dS$ will be, by Art. 68,
$\frac {1}{4\pi}\,KR \,cos\epsilon\, dS$
Since we do not at present consider any dielectric except air, $K= 1$ .
We may, however, avoid introducing at this stage the theory of electric displacement, by calling $R cos \epsilon dS$ the Induction through the element $dS$. This quantity is well known in mathematical physics, but the name of induction is borrowed from Faraday. The surface-integral of induction is
$\iint R cos \epsilon dS$;
and it appears by Art. 21, that if $X, Y, Z$ are the components of $R$, and if these quantities are continuous within a region bounded by a closed surface $S$, the induction reckoned from within outwards is
$\iint R cos \epsilon dS=\iiint \left (\frac {}{}+\frac {}{}+\frac {}{} \right )dx\,dy\,dz$,
the integration being extended through the whole space within the surface.
### Induction through a Finite Closed Surface due to a Single Centre of Force.
76.] Let a quantity e of electricity be supposed to be placed at a point $0$, and let $r$ be the distance of any point $P$ from $0$, the force at that point is $R=\frac{e}{r^2}$ in the direction $OP$.
Let a line be drawn from $O$ in any direction to an infinite distance. If $O$ is without the closed surface this line will either not cut the surface at all, or it will issue from the surface as many times as it enters. If $O$ is within the surface the line must first issue from the surface, and then it may enter and issue any number of times alternately, ending- by issuing from it.
Let $\epsilon$ be the angle between $OP$ and the normal to the surface drawn outwards where $OP$ cuts it, then where the line issues from the surface $cos \epsilon$ will be positive, and where it enters $cos \epsilon$ will be negative.
Now let a sphere be described with centre $O$ and radius unity, and let the line $OP$ describe a conical surface of small angular aperture about $O$ as vertex.
This cone will cut off a small element $d \omega$ from the surface of the sphere, and small elements $dS_l$, $dS_2$ , &c. from the closed surface at the various places where the line $OP$ intersects it.
Then, since any one of these elements $dS$ intersects the cone at a distance $r$ from the vertex and at an obliquity $\epsilon$,
$dS=r^2\, \sec \epsilon\,d\omega$;
and, since $R = er^{-2}$ , we shall have
$R \,\cos \epsilon\, dS=\pm e\,d\omega$;
the positive sign being taken when $r$ issues from the surface, and the negative where it enters it.
If the point $O$ is without the closed surface, the positive values are equal in number to the negative ones, so that for any direction of $r$,
$\sum R\,\cos \epsilon\; dS=0 \,\!$,
and therefore $\iint R\, \cos \epsilon\, dS=0$,
the integration being extended over the whole closed surface.
If the point $O$ is within the closed surface the radius vector $OP$ first issues from the closed surface, giving a positive value of $e d\omega$, and then has an equal number of entrances and issues, so that in this case
$\sum R \cos \epsilon \,dS=e\, d\omega$
.
Extending the integration over the whole closed surface, we shall include the whole of the spherical surface, the area of which is $4\pi$, so that
$\iint R\, cos \epsilon\, dS=e \, \iint d\omega=4 \pi \,e$.
Hence we conclude that the total induction outwards through a closed surface due to a centre of force $e$ placed at a point is zero when is without the surface, and $4\pi e$ when $O$ is within the surface.
Since in air the displacement is equal to the induction divided by $4\pi$, the displacement through a closed surface, reckoned outwards, is equal to the electricity within the surface.
Corollary. It also follows that if the surface is not closed but is bounded by a given closed curve, the total induction through it is $\omega$, where $\omega$ is the solid angle subtended by the closed curve at $0$. This quantity, therefore, depends only on the closed curve, and not on the form of the surface of which it is the boundary.
### On the Equations of Laplace and Poisson.
77.] Since the value of the total induction of a single centre of force through a closed surface depends only on whether the centre is within the surface or not, and does not depend on its position in any other way, if there are a number of such centres $e_l$, $e_2$ , &c. within the surface, and $e_1'$, $e_2'$, &c. without the surface, we shall have
$\iint R \, \cos \epsilon\,dS=\,4 \pi e$;
where $e$ denotes the algebraical sum of the quantities of electricity at all the centres of force within the closed surface, that is, the total electricity within the surface, resinous electricity being reckoned negative.
If the electricity is so distributed within the surface that the density is nowhere infinite, we shall have by Art. 64,
$4\pi\,e=4\pi \iiint \rho\,dx\,dy\,dz$
and by Art. 75,
$\iint R \, \cos \epsilon\,dS=\iiint \left (\frac{dX}{dx}+\frac{dY}{dy}+\frac{dZ}{dz} \right )\,dx\,dy\,dz$.
If we take as the closed surface that of the element of volume $dx$ $dy$ $dz$, we shall have, by equating these expressions,
$\frac{dX}{dx}+\frac{dY}{dy}+\frac{dZ}{dz}=4\pi\,\rho$;
and if a potential $V$ exists, we find by Art. 71 ,
$\frac{d^2V}{dx^2}+\frac{d^2V}{dy^2}+\frac{d^2V}{dz^2}+4\pi\,\rho=0$
This equation, in the case in which the density is zero, is called Laplace's Equation. In its more general form it was first given by Poisson. It enables us, when we know the potential at every point, to determine the distribution of electricity. We shall denote, as at Art. 26, the quantity
$\frac{d^2V}{dx^2}+\frac{d^2V}{dy^2}+\frac{d^2V}{dz^2}$ by $-\nabla^2 V$,
and we may express Poisson's equation in words by saying that the electric density multiplied by $4\pi$ is the concentration of the potential. Where there is no electrification, the potential . has no concentration, and this is the interpretation of Laplace's equation.
If we suppose that in the superficial and linear distributions of electricity the volume-density $\rho$ remains finite, and that the electricity exists in the form of a thin stratum or narrow fibre, then, by increasing $\rho$ and diminishing the depth of the stratum or the section of the fibre, we may approach the limit of true superficial or linear distribution, and the equation being true throughout the process will remain true at the limit, if interpreted in accordance with the actual circumstances.
### On the Conditions to be fulfilled at an Electrified Surface.
78.] We shall consider the electrified surface as the limit to which an electrified stratum of density $\rho$ and thickness $v$ approaches when $\rho$ is increased and $v$ diminished without limit, the product $\rho\,v$ being always finite and equal to $\sigma$ the surface-density.
Let the stratum be that included between the surfaces
$F(x,\,y,\,z)=\,F\,=\,a$ (1)
and and $F=a\,+\, h$ (2)
If we put $R^2={\left .\frac{\overline{dV}}{dy} \right |}^2 + {\left .\frac{\overline{dV}}{dy} \right |}^2 + {\left .\frac{\overline{dV}}{dz}\right |}^2$, (3)
and if $l, m, n$ are the direction-cosines of the normal to the surface,
$Rl=\frac {dF}{dx}, {\color{White}xxxxx}Rm=\frac {dF}{dy}, {\color{White}xxxxx}Rn=\frac {dF}{dz}$ (4)
Now let $V_1$ be the value of the potential on the negative side of the surface $F = a, V'$ its value between the surfaces $F = a$ and $F = a + h$, and $V_2$ its value on the positive side of $F= a + h$.
Also, let $\rho_1,\rho'$, and $\rho_2$ be the values of the density in these three portions of space. Then, since the density is everywhere finite, the second derivatives of $V$ are everywhere finite, and the first derivatives, and also the function itself, are everywhere continuous and finite.
At any point of the surface $F = a$ let a normal be drawn of length $v$, till it meets the surface $F = a + h$, then the value of $F$ at the extremity of the normal is
$a+v \left (l \frac {dF}{dx}+m \frac {dF}{dy}+n\frac {dF}{dz} \right )$+&c., (5)
or $a+h=a+vR+\,\!$&c. (6)
The value of $V$ at the same point is
$V_2=V_1+v \left ( l\frac {dV'}{dx}+ m\frac {dV'}{dy}+n\frac {dV'}{dz} \right )+$ &c, (7)
or $V_2-V_1=\frac {h}{R} \frac {dV'}{dv}+$ &c. (8)
Since the first derivatives of $V$ continue always finite, the second side of the equation vanishes when $h$ is diminished without limit, and therefore if $V_2$ and $V_1$ denote the values of $V$ on the outside and inside of an electrified surface at the point $x, y, z,$
$V_1\,=\,V_2$ (9)
If $x +dx$, $y + dy$, $z + dz$ be the coordinates of another point on the electrified surface, $F=a$ and $F=a+h$ at this point also ; whence
$0=\frac{dF}{dx}dx+\frac{dF}{dy}dy+\frac{dF}{dz}dz+$ &c., (10)
$0=\left (\frac{dV_2}{dx}-\frac{dV_1}{dx} \right )dx+\left (\frac{dV_2}{dy}-\frac{dV_1}{dy} \right )dy+\left (\frac{dV_2}{dz}-\frac{dV_1}{dz} \right )dz+$ &c.; (11)
and when $dx$, $dy$, $dz$ vanish, we find the conditions
$\left .{\begin{matrix}\dfrac{dV_2}{dx}-\dfrac{dV_1}{dx}=Cl, \\ \\ \dfrac{dV_2}{dy}-\dfrac{dV_1}{dy}=Cm\\ \\ \dfrac{dV_2}{dz}- \dfrac{dV_1}{dz}=Cn \end{matrix}} \right \}$ (12)
where $C$ is a quantity to be determined.
Next, let us consider the variation of $F$ and $\frac{dV}{dx}$ along the ordinate parallel to $x$ between the surfaces $F= a$ and $F = a + h$.
We have $F=a+\frac{dF}{dx}dx+{1 \over 2}\frac{d^2F}{dx^2}(dx)^2+$ &c, (13)
and $\frac{dV}{dx}=\frac{dV'}{dx}+\frac{d^2V'}{dx^2}dx+\tfrac{1}{2}\frac{d^3V'}{dx^3}(dx)^2+$ &c., (14)
Hence, at the second surface, where $F=a + h$, and $V$ becomes $V_2$,
$\frac{dV_2}{dx}=\frac{dV_1}{dx}+\frac{d^2V'}{dx^2}dx+$ &c.; (15)
whence $\frac {d^2V'}{dx^2}dx +\mbox {etc.}=Cl$, (16)
by the first of equations (12).
Multiplying by $Rl$, and remembering that at the second surface
$Rl\,dx=h$ (17)
we find $\frac {d^2V'}{dx^2} h=CR\,l^2$. (18)
Similarly $\frac {d^2V'}{dy^2} h=CR\,m^2$ (19)
and $\frac {d^2V'}{dz^2} h=CR\,n^2$ (20)
Adding $\left ( \frac {d^2V'}{dx^2}+\frac {d^2V'}{dy^2}+\frac {d^2V'}{dz^2} \right )\,h=CR$; (21)
but $\frac {d^2V'}{dx^2}+\frac {d^2V'}{dy^2}+\frac {d^2V'}{dz^2}=-4 \pi\rho^'$ and $h=vR$; (22)
hence $C=-4 \pi \rho^'v\,\!=-4\pi\sigma$; (23)
where $\sigma$ is the surface-density; or, multiplying the equations (12) by $l, m, n$ respectively, and adding,
$l \left ( \frac {dV_2}{dx}- \frac {dV_1}{dx} \right ) + +m \left ( \frac {dV_2}{dy}- \frac {dV_1}{dy} \right ) +n \left (\frac {dV_2}{dz}- \frac {dV_1}{dz} \right ) + 4\pi\sigma = 0$. (24)
This equation is called the characteristic equation of $V$ at a surface. This equation may also be written
$\frac {dV_1}{dv_1}+ \frac {dV_2}{dv_2}+ 4\pi\sigma =0$; (25)
where $v_1, v_2$ are the normals to the surface drawn towards the first and the second medium respectively, and $V_1, V_2$ the potentials at points on these normals. We may also write it
$R_2 \cos \epsilon_2+R_1 \cos \epsilon_1 +4\pi\sigma =0\,\!$; (26)
where $R_1, R_2$ are the resultant forces, and $\epsilon_1, \epsilon_2$ the angles which they make with the normals drawn from the surface on either side.
79.] Let us next determine the total mechanical force acting on an element of the electrified surface.
The general expression for the force parallel to $x$ on an element whose volume is $dx\,dy\,dz$, and volume-density $\rho$, is
$dX=-\frac{dV}{dx}\rho\;dz\,dy\,dz$. (27)
In the present case we have for any point on the normal v
$\frac {dV}{dx}=\frac {dV_1}{dx}+v\frac {d^2v_1}{dx^2}+etc.$; (28)
also, if the element of surface is $dS$, that of the volume of the element of the stratum may be written $dS\,dv$ ; and if $X$ is the whole force on a stratum of thickness $v$,
$X=\iiint \left (\frac {}{}+v\frac {d^2v_1}{dx^2}+etc. \right )\rho'\,dS\,dv$. (29)
Integrating with respect to $v$, we find
$x=-\iint \rho'\,dS \left ( v \frac {dV_1}{dx}+\frac {v^2}{2}\frac {d^2V_1}{dx^2}+etc. \right )$ (30)
or, since $\frac {dV_2}{dx}=\frac {dV1}{dx}+v \frac {d^2V_1}{dx^2}+etc.$; (31)
$X=-\iint {1 \over 2}\,\rho'\,v\,dS \left (\frac {dV_1}{dx}+\frac {dV_2}{dx}\right )+ etc.$; (32)
When $v$ is diminished and $\rho'$ increased without limit, the product $\rho'v$ remaining always constant and equal to $\sigma$, the expression for the force in the direction of $x$ on the electricity $\rho\,dS$ on the element of surface $dS$ is
$X=-\sigma\,dS\,{1 \over 2}\left ( \frac {dV_1}{dx}+\frac {dV_2}{dx}\right )$; (33)
that is, the force acting on the electrified element $\rho\, dS$ in any given direction is the arithmetic mean of the forces acting on equal quantities of electricity placed one just inside the surface and the other just outside the surface close to the actual position of the element, and therefore the resultant mechanical force on the elec trified element is equal to the resultant of the forces which would act on two portions of electricity, each equal to half that on the element, and placed one on each side of the surface and infinitely near to it.
80.] When a conductor is in electrical equilibrium, the whole of the electricity is on the surface.
We have already shewn that throughout the substance of the conductor the potential $V$ is constant. Hence $\nabla^2V$ is zero, and therefore by Poisson s equation, $\rho$ is zero throughout the substance of the conductor, and there can be no electricity in the interior of the conductor.
Hence a superficial distribution of electricity is the only possible one in the case of conductors in equilibrium. A distribution throughout the mass can only exist in equilibrium when the body is a non-conductor. Since the resultant force within a conductor is zero, the resultant force just outside the conductor is along the normal and is equal to $4 \pi\sigma$, acting outwards from the conductor.
81.] If we now suppose an elongated body to be electrified, we may, by diminishing its lateral dimensions, arrive at the conception of an electrified line.
Let $ds$ be the length of a small portion of the elongated body, and let $c$ be its circumference, and $\sigma$ the superficial density of the electricity on its surface; then, if $\lambda$ is the electricity per unit of length, $\lambda=c\sigma$, and the resultant electrical force close to the surface will be
$4 \pi\sigma=4\pi\frac{\lambda}{c}$
If, while $\lambda$ remains finite,$c$ be diminished indefinitely, the force at the surface will be increased indefinitely. Now in every dielectric there is a limit beyond which the force cannot be increased without a disruptive discharge. Hence a distribution of electricity in which a finite quantity is placed on a finite portion of a line is inconsistent with the conditions existing in nature.
Even if an insulator could be found such that no discharge could be driven through it by an infinite force, it would be impossible to charge a linear conductor with a finite quantity of electricity, for an infinite electromotive force would be required to bring the electricity to the linear conductor.
In the same way it may be shewn that a point charged with a finite quantity of electricity cannot exist in nature. It is convenient, however, in certain cases, to speak of electrified lines and points, and we may suppose these represented by electrified wires, and by small bodies of which the dimensions are negligible com pared with the principal distances concerned.
Since the quantity of electricity on any given portion of a wire diminishes indefinitely when the diameter of the wire is indefinitely diminished, the distribution of electricity on bodies of considerable dimensions will not be sensibly affected by the introduction of very fine metallic wires into the field, so as to form electrical connexions between these bodies and the earth, an electrical machine, or an electrometer.
### On Lines of Force.
82.] If a line be drawn whose direction at every point of its course coincides with that of the resultant force at that point, the line is called a Line of Force. If lines of force be drawn from every point of a line they will form a surface such that the force at any point is parallel to the tangent plane at that point. The surface-integral of the force with respect to this surface or any part of it will therefore be zero.
If lines of force are drawn from every point of a closed curve $L_l$ they will form a tubular surface $S_0$. Let the surface $S_1$ bounded by the closed curve $L_1$ be a section of this tube, and let $S_2$ be any other section of the tube. Let $Q_0, Q_1, Q_2$ be the surface-integrals over $S_0, S_1, S_2$, then, since the three surfaces completely enclose a space in which there is no attracting matter, we have
$Q_0 \,+Q_1 \,+ Q_2 \,= \,0.$
But $Q_0= 0$, therefore $Q_2 =-Q_1$, or the surface-integral over the second section is equal and opposite to that over the first : but since the directions of the normal are opposite in the two cases, we may say that the surface-integrals of the two sections are equal, the direction of the line of force being supposed positive in both.
Such a tube is called a Solenoid[2], and such a distribution of force is called a Solenoidal distribution. The velocities of an in compressible fluid are distributed in this manner.
If we suppose any surface divided into elementary portions such that the surface-integral of each element is unity, and if solenoids are drawn through the field of force having these elements for their bases, then the surface-integral for any other surface will be re presented by the number of solenoids which it cuts. It is in this sense that Faraday uses his conception of lines of force to indicate not only the direction but the amount of the force at any place in the field.
We have used the phrase Lines of Force because it has been used by Faraday and others. In strictness, however, these lines should be called Lines of Electric Induction.
In the ordinary cases the lines of induction indicate the direction and magnitude of the resultant electromotive force at every point, because the force and the induction are in the same direction and in a constant ratio. There are other cases, however, in which it is important to remember that these lines indicate the induction, and that the force is indicated by the equipotential surfaces, being normal to these surfaces and inversely proportional to the distances of consecutive surfaces.
### On Specific Inductive Capacity.
83.] In the preceding investigation of surface-integrals I have adopted the ordinary conception of direct action at a distance, and have not taken into consideration any effects depending on the nature of the dielectric medium in which the forces are observed.
But Faraday has observed that the quantity of electricity induced by a given electromotive force on the surface of a conductor which bounds a dielectric is not the same for all dielectrics. The induced electricity is greater for most solid and liquid dielectrics than for air and gases. Hence these bodies are said to have a greater specific inductive capacity than air, which is the standard medium.
We may express the theory of Faraday in mathematical language by saying that in a dielectric medium the induction across any surface is the product of the normal electric force into the coefficient of specific inductive capacity of that medium. If we denote this coefficient by $K$, then in every part of the investigation of surface-integrals we must multiply $X$, $Y$, and $Z$ by $K$, so that the equation of Poisson will become
$\frac {d}{dx}\cdot\,K \frac{dV}{dx}+ \frac {d}{dy}\cdot\,K \frac{dV}{dy}+ \frac {d}{dz}\cdot\,K \frac{dV}{dz}+4 \pi\,\rho=0$.
At the surface of separation of two media whose inductive capacities are $K_1$ and $K_2$ , and in which the potentials are $V_1$ and $V_2$ the characteristic equation may be written
$K_2\frac{dV_2}{dv}-K_1\frac{dV_1}{dv}+4 \pi\,\rho=0$;
where $v$ is the normal drawn from the first medium to the second, and $\sigma$ is the true surface-density on the surface of separation; that is to say, the quantity of electricity which is actually on the surface in the form of a charge, and which can be altered only by conveying electricity to or from the spot. This true electrification must be distinguished from the apparent electrification $\rho'$, which is the electrification as deduced from the electrical forces in the neighbourhood of the surface, using the ordinary characteristic equation
$\frac{dV_2}{dv}-\frac{dV_1}{dv}+4 \pi\,\rho'=0$.
If a solid dielectric of any form is a perfect insulator, and if its surface receives no charge, then the true electrification remains zero, whatever be the electrical forces acting on it.
Hence $\frac{dV_2}{dv}=\frac{K_1}{K_2}\frac{dV_1}{dv}$, and $\frac{K_1-K_2}{K_2}\frac{dV_1}{dv}+4\pi\rho'=0$,
$\frac{dV_1}{dv}=\frac {4\pi\rho' K_2}{K_1-K_2}$, ${\color{White}xxxxx} \frac{dV_2}{dv}=\frac {4\pi\rho' K_1}{K_1-K_2}$.
The surface-density $\sigma'$ is that of the apparent electrification produced at the surface of the solid dielectric by induction. It disappears entirely when the inducing force is removed, but if during the action of the inducing force the apparent electrification of the surface is discharged by passing a flame over the surface, then, when the inducing force is taken away, there will appear an electrification opposite to $\sigma'$ [3].
In a heterogeneous dielectric in which $K$ varies continuously, if $\rho'$ be the apparent volume-density,
$\frac{d^2V}{dx^2}+\frac{d^2V}{dy^2}+\frac{d^2V}{dz^2} +4\pi\rho'=0$
Comparing this with the equation above, we find
$4\pi(\rho-K\rho')+\frac{dK}{dx}\frac{dV}{dx}+ \frac{dK}{dy}\frac{dV}{dy}+ \frac{dK}{dz}\frac{dV}{dz}=0$
The true electrification, indicated by $\rho$, in the dielectric whose variable inductive capacity is denoted by $K$, will produce the same potential at every point as the apparent electrification, indicated by $\rho'$, would produce in a dielectric whose inductive capacity is every where equal to unity.
## CHAPTER III.
SYSTEMS OF CONDUCTORS.
### On the Superposition of Electrical Systems.
84.] Let $E_l$ be a given electrified system of which the potential at a point $P$ is $V_1$, and let $E_2$ be another electrified system of which the potential at the same point would be $V_2$ if $E_l$ did not exist. Then, if $E_1$ and $E_2$ exist together, the potential of the combined system will be $V_1+V_2$.
Hence, if $V$ be the potential of an electrified system $E$, if the electrification of every part of $E$ be increased in the ratio of $n$ to 1 , the potential of the new system $nE$ will be $nV$.
### Energy of an Electrified System.
85.] Let the system be divided into parts, $A_1$, $A_2$ , &c. so small that the potential in each part may be considered constant through out its extent. Let $e_l$ ,$e_2$ , &c. be the quantities of electricity in each of these parts, and let $V_1$, $V_2$ &c. be their potentials.
If now $e_1$ is altered to $ne_1$, $e_2$ to $ne_2$, &c., then the potentials will become $nV_1$, $nV_2$, &c.
Let us consider the effect of changing $n$ into $n + dn$ in all these expressions. It will be equivalent to charging $A_1$ with a quantity of electricity $e_l dn$, $A_2$ with $e_2 dn$, &c. These charges must be supposed to be brought from a distance at which the electrical action of the system is insensible. The work done in bringing $e_1 dn$ of electricity to $A_1$, whose potential before the charge is $nV_1$, and after the charge $(n + dn)V_1$, lf must lie between
$nV_1e_1\,dn\,\!$ and $(n+dn)V_1e_1\,dn\,\!$.
In the limit we may neglect the square of $dn$, and write the expression
$V_1e_1n\,dn\,\!$
1. See Pratt s Mechanical Philosophy, p. 144.
2. From $\sigma\omega\lambda\eta\nu$, a tube. Faraday uses (3271) the term Sphondyloid in the same sense.
3. See Faraday s Kemarks on Static Induction, Proceedings of the Royal Institution, Feb. 12, 1858.
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http://mathhelpforum.com/geometry/158377-easy-ratio-question-need-help-please-print.html
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# easy ratio question, need help please.
Printable View
• October 4th 2010, 05:13 AM
Nguyen
easy ratio question, need help please.
Hey all, I have a easy question that I have found two different answers for (stupid me). Can anyone check my working out and tell me which one is correct. Thanks
The question is to find the value of side x in the rectangle below, the other side being 1. The rectangle is made up of two unit squares and a remaining rectangle that is a scaled down version of the original rectangle.
|-side x--------|
|-----|-----|---|
|-----|-----|---| 1
|-----|-----|---|
1st way: Combining the two unit squares we know that the side lengths of the small rectangle are (x-2) and 1.
|-side x-------|
|----------|---|
|----------|---| 1
|----------|---|
x-2 (meant to be side of small rectangle)
ratio: x/1 = 1/(x-2) we get x= 1+ sqrt(2)
but:
2nd way: Take one unit square away and x becomes (x-1), using the above diagram we get the ratio:
(x-1)/1 = 1/(x-2) and so x= (3 +sqrt(5))/2
I don't know what my error in thinking here is. Can anyone help?
Thanks heaps everyone, this forum is great!
• October 4th 2010, 07:45 AM
HallsofIvy
If I understand this correctly, you are saying that the rectangle with lenght x and width 1 can be partitioned into two unit squares and a "scaled down" rectangle similar to the orginal one. That is, the "scaled down" rectangle has sides of length $\alpha x$ and $\alpha$ for some $0< \alpha< 1$. The we must have $\alpha x= 1$, the width of the original rectangle and [tex]\alpha= x- 2[tex]. Putting that value of $\alpha$ into the first equation, $\alpha x= (x- 2)x= x^2- 2x= 1$. That also gives your ratio, $\frac{x}{1}= \frac{1}{x- 2}$ Completing the square by adding 1 to both sides, $x^2- 2x+ 1= (x- 1)^2= 2$, $x- 1= \pm\sqrt{2}$ so that $x= 1\pm\sqrt{1}$. Since x, as the lenght of a side of a rectangle, must be positive, $x= 1+ \sqrt{2}$, your first solution, is the only solution.
"Taking one square away" we still have $\alpha x= 1$ but now $x= 2+ \alpha$ or $\alpha= x- 2$ which reduces to the same thing.
Your "taking one square away" seems to be starting with a completly different problem in which the original rectangle is made up of only one square together with the "scaled down" rectangle.
All times are GMT -8. The time now is 01:48 AM.
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http://metric2011.wordpress.com/2012/12/06/notes-of-saars-hersonskys-lecture/
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site of the metric2011 program of Centre Emile Borel
## Notes of Sa’ar Hersonsky’s lecture
Posted on December 6, 2012 by
Combinatorial harmonic coordinates
Uniformizing combinatorial annuli.
1. Perspective
Can a combinatorial structure determine a rigid geometry ? Here are interesting cases where this works.
Theorem 1 (Thurston, Rodin-Sullivan, Schramm-He, Beardon-Stephenson, Colin de Verdière…) Cover a planar domain ${\Omega}$ with small equal circles. Apply Koebe’s theorem, get a piecewise affine map of the unit disk ${D}$ into ${\Omega}$ mapping centers of circles to centers of circles. As size of circles tends to zero, this map converges uniformly to a Riemann mapping. The ratio of radii of corresponding circles converges to the modulus of the derivative of the Riemann mapping.
Our work : We construct flat surfaces starting from combinatorial data. This can be viewed as a discrete uniformization, in the spirit of Schramm and Cannon-Floyd-Parry.
2. Boundary values on graphs
Let ${\Omega}$ be ${m}$-connected. Split its boundary into ${E_1\cap E_2}$ wher ${E_1}$ is the outermost component. Triangulate it. Let ${c:T^{(1)}\rightarrow{\mathbb R}_+}$ a symmetric conductance function. Then Laplacian makes sense,
$\displaystyle \begin{array}{rcl} \Delta u (x)=\sum_{x\sim y}c(x,y)(u(x)-u(y)). \end{array}$
Harmonic functions satisfy ${\Delta u=0}$ at inner vertices.
$\displaystyle \begin{array}{rcl} E(u)=\sum_{x\sim y}c(x,y)(u(x)-u(y))^2. \end{array}$
The discrete Dirichlet boundary value problem (D-BVP) consists in finding a harmonic function with prescribed boundary values ${g=k}$ constant at ${E_1}$ and ${g=0}$ at ${E_2}$.
The solution is used in the following theorem, in specifying the target space (replacement for the unit disk).
3. A warm up
Theorem 2 (Brooks-Smith-Stone-Tutte 1940) Let ${A}$ be an annulus, ${k}$ a positive number. Let ${S_A}$ be the straight Euclidean cylinder with height ${k}$ and circumference
$\displaystyle \begin{array}{rcl} C=\sum_{x\in E_1}\frac{\partial g}{\partial n}(x). \end{array}$
Then there exists a mapping ${f}$ which associates to each edge of ${A}$ a unique embedded Euclidean rectangle in ${S_A}$ in such a way that the collection of these rectangles form a tiling of ${S_A}$.
This map preserves energy. It seems that Dehn already had the idea of using Kirckhhoff’s laws in 1903, and pointed out difficulties which are still. This was clarified by Cannon-Floyd-Parry (1994) and Benjamini-Schramm (1996). I have an alternate proof.
The difficulty is that the mapping cannot be extended to a homeomorphism
We shall make a change of charts: We view the given triangulaton as a set of initial charts, and we shall improve on it.
4. A new theorem
Theorem 3 Let ${S_A}$ be the concentric Euclidean annulus with inner and outer radii ${r_1=1}$ and ${r_2=2\pi/}$ period of ${\theta}$ (see below).
Then there exist a cellular decomposition ${R}$ of ${A}$ and
1. a tiling ${T}$ of ${S_A}$ by annular shells,
2. a homeomorphism ${f:A\rightarrow S_A}$ mapping each quadrilateral in ${R^{(2)}}$ onto a single annular shell in ${S_A}$, and preserving area.
4.1. What goes into the proof
Fact: the level curves of ${g}$ foliate ${A}$. No critical points.
4.2. Construction of a combinatorial angle
We define a new function, ${\theta}$, on ${T^{(0)}}$, on the annulus minus a slit, the conjugate function of ${g}$. It is obtained by summing the normal derivatives of ${g}$ along a suitably chosen PL path, joigning the slit to a vertex.
Properties: Level curves of ${\theta}$ have no endpoints in the interior, and join ${E_1}$ to ${E_2}$. Any two are disjoint. The intersection number between level curves of ${\theta}$ and level curves of ${g}$ is 1.
4.3. Constructing a rectangular net
Consider the collection ${L=\{L(v_0),\ldots,L(v_k)\}}$ of level sets of ${g}$ containing all vertices of ${T}$. So ${L(v_0)=E_2}$ and ${L(v_k)=E_1}$. Do the same for ${\theta}$.
Definition 4 A rectangular combinatorial net on ${\Omega}$ i a cellular decomposition ${R}$ of ${\Omega}$ where each 2-cell is a simple quadrilateral, and a pair of functions ${\phi}$ and ${\psi}$ which satisfy
$\displaystyle \begin{array}{rcl} d\phi(e)d\psi(e)=0 \end{array}$
for all edges.
Theorem 5 There exists a choice of conductances such that ${g}$ and ${\theta}$ and their level sets form a rectangular combinatorial net.
5. Higher connectivity
Most of the discussion extends to higher connectivity domains.
Split ${\Omega}$ along singular level sets of ${g}$. Components need not be annuli (this makes it hard). These level sets have a naturally defined length, in terms of the period of a conjugate function ${\theta}$. This allows to pile up model annuli and get a model surface of high connectivity.
## About metric2011
metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See http://www.math.ens.fr/metric2011/
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http://mathhelpforum.com/number-theory/158985-congruence-classes.html
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# Thread:
1. ## Congruence Classes
Hello all!
I was looking over some examples of congruence classes and I ran into one I could not find.
1. Is there/Are there a congruence class(es) $(mod (3+\sqrt{3})/2)$ in $Q[\sqrt{-3}]$ ? If so, what are they? I'm not sure if they exist or not even.
2. If $\alpha$ is a quadratic integer in $Q[\sqrt{-d}]$, then a notion of congruence $(mod \alpha)$ can be defined. How can this be defined how can we further define +, -, and x for congruences classes?
Thank you everyone!
-Samson
2. What in the world. In one of your past posts, specifically this one:
http://www.mathhelpforum.com/math-he...rs-154495.html
You clearly state that you know the congruence classes are -1, 0, and 1. And in this post, you're asking what the congruence classes are. This doesn't make any sense, what's going on?
By the way, the congruence classes are -1, 0, and 1, which you seemed to know earlier but not now.
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http://math.stackexchange.com/questions/210681/if-a-n-to-ell-then-hat-a-n-to-ell
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# If $a_n\to \ell$ then $\hat a_n\to \ell$
I need some help to finish this proof:
THEOREM
Let $\{a_n\}$ be such that $\lim a_n=\ell$ and set
$$\hat a_n=\frac 1 n \sum_{k=1}^na_k$$
Then $\lim\hat a_n=\ell$
PROOF
Let $\epsilon >0$ be given. Since $\lim a_n=\ell$ , there exists an $N$ for which $$\left| {{a_n} - \ell } \right| < {\epsilon/2 }$$
whenever $n>N$. Now:
$$\begin{eqnarray*} \left| {\frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} - \ell } \right| &=& \left| {\frac{1}{n}\sum\limits_{k = 1}^n {\left( {{a_k} - \ell } \right)} } \right| \\ &=& \left| {\frac{1}{n}\sum\limits_{k = 1}^N {\left( {{a_k} - \ell } \right)} + \frac{1}{n}\sum\limits_{k = N + 1}^n {\left( {{a_k} - \ell } \right)} } \right| \\ &\leqslant& \frac{1}{n}\sum\limits_{k = 1}^N {\left| {{a_k} - \ell } \right|} + \frac{1}{n}\sum\limits_{k = N + 1}^n {\left| {{a_k} - \ell } \right|} \\ & <& \frac{N}{n}\zeta - \frac{{N }}{2n}\epsilon+\epsilon/2 \end{eqnarray*}$$
where $$\zeta=\mathop {\max }\limits_{1 \leqslant k \leqslant N} \left| {{a_k} - \ell } \right|$$
Now, let $n_0$ be such that if $n>n_0$,
$$\eqalign{ & \frac{{N\zeta }}{n} < {\epsilon} \cr & \frac{N}{n} < {1} \cr}$$ Then we get
$$\frac{N}{n}\zeta - \frac{N}{n}\frac{\epsilon }{2} + \frac{\epsilon }{2} < \epsilon - \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon$$
How far is this OK? Do you think there is an easier way to go about proving it? I now remember that by Stolz Cesàro:
$$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sum\limits_{k = 1}^{n + 1} {{a_k}} - \sum\limits_{k = 1}^n {{a_k}} }}{{n + 1 - n}} = \mathop {\lim }\limits_{n \to \infty } {a_{n + 1}} = \ell$$
-
Why $\left| {\frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} - \ell } \right| = \left| {\frac{1}{n}\sum\limits_{k = 1}^n {\left( {{a_k} - \ell } \right)} } \right|$ instead this: $\left| {\frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} - \ell } \right| = \left| {\frac{1}{n}\sum\limits_{k = 1}^n {\left( {{a_k} - n \ell } \right)} } \right|$? – M. Strochyk Oct 10 '12 at 20:57
1
@M.Strochyk That is wrong. The $1/n$ kills it off. – Peter Tamaroff Oct 10 '12 at 20:58
@M.Strochyk Because that gives too many $\ell$'s – AD. Oct 10 '12 at 20:59
Thanks, i was mistaken – M. Strochyk Oct 10 '12 at 21:13
1
– AD. Oct 11 '12 at 9:39
show 1 more comment
## 2 Answers
For the tail sum $$\frac1n\sum_{N+1}^n|a_k-\ell|\leq \frac{(n-N)}{n}\varepsilon<\varepsilon$$ for all large $n$, hence $$\limsup_n \frac1n\sum_{1}^n|a_k-\ell|\leq 0+\varepsilon=\varepsilon...$$
-
Ok, but use your $n_0$ then, it results in the same. – AD. Oct 10 '12 at 21:00
Your estimate of the tail is the problem, because the tail is positive and the upper bound of yours is negative. You should simply estimate the terms by $\varepsilon$ and count them. – AD. Oct 10 '12 at 21:04
1
I'm just getting $$\frac{1}{n}\sum\limits_{k = N + 1}^n {\left| {{a_k} - \ell } \right|} < \frac{{n - N}}{n}\frac{\epsilon }{2}$$ Why is that wrong? I think your $n-N-1$ should be a $n-(N+1)+1=n-N$ – Peter Tamaroff Oct 10 '12 at 21:07
OK, I can go with $\limsup$s now. – Peter Tamaroff Oct 11 '12 at 2:11
Right, yes I had an extra term which I just removed. The estimate of the comment is correct, but in the OP we have $$\frac{1}{n}\sum\limits_{k = N + 1}^n {\left| {{a_k} - \ell } \right|} \leq - \frac{{N }}{2n}\epsilon-\epsilon/2$$ which is certainly not the case. – AD. Oct 11 '12 at 5:40
show 2 more comments
The sequence $\{|a_n-\ell|\}$ converges to $0$ therefore it's bounded, so there exist some $M\gt 0$ such that $$|a_n-\ell|\leq M\quad\forall n\in \Bbb N.$$ This $M$ plays the role of the $\zeta$ in your proof, but notice it does not depends on $n$.
Let $\epsilon\gt 0$. There exist $n_1\in\Bbb N$ such that for all $n\in\Bbb N$, $n\geq n_1$ implies $$|a_n-\ell|\lt\frac{\epsilon}{2}.$$
Since $$\lim_{n\to\infty} \frac{n_1}{n}=0,$$ there exist $n_2\in\Bbb N$ such that, for all $n\in\Bbb N$, $n\geq n_2$ implies $$\frac{n_1}{n}\leq \frac{\epsilon}{2M}.$$
Let $N=\max\{n_1,n_2\}$, then for all $n\in\Bbb N$, $n\geq N\geq n_1$ implies $$\left(1-\frac{n_1}{n}\right)\leq 1$$ and then $$\begin{align*} \left| -\ell+\frac1{n}\sum_{j=1}^n a_j \right| &\leq \frac1{n}\sum_{j=1}^n \left|a_j-\ell\right|\\ &= \frac1{n}\sum_{j=1}^{n_1} \left|a_j-\ell\right| + \frac1{n}\sum_{j=n_1+1}^{n} \left|a_j-\ell\right|\\ &\lt \frac1{n} n_1M + \frac1{n}(n-n_1)\frac{\epsilon}{2}\\ &=\frac{n_1}{n}M + \left(1-\frac{n_1}{n}\right)\frac{\epsilon}{2}\\ &\lt \frac{\epsilon}{2M}M + 1\cdot \frac{\epsilon}{2}=\epsilon. \end{align*}$$
Therefore $$\hat a_n\to\ell.$$
-
Just rewriting your argument What for? – Did Apr 28 at 6:39
@Did to provide some justification on some of the steps of the OP original argument. Basically reordering ideas... – leo Apr 28 at 6:50
Really? Then you might want to make explicit which steps need more justification. // As an aside, the fact that your $M$ exists is not primarily a consequence of the convergence you cite. And using such a bound valid for every $n$ seems to only complicate things. – Did Apr 28 at 6:54
@Did the sequence at the beginning is convergent, so it's bounded. The $M$ is just one such bound. Don't see how its validity for all $n$ can complicate things. Is there something wrong in the above? – leo Apr 28 at 7:14
(1.) I do not know what "the sequence at the beginning is convergent" means. (2.) You did not explain what the problem with the other answer is. (3.) I did not use the word "wrong". – Did Apr 28 at 7:29
show 2 more comments
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http://mathhelpforum.com/math-topics/64858-electrical-resistance-print.html
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# electrical resistance
Printable View
• December 13th 2008, 07:09 PM
Stokoe10
electrical resistance
could anyone help me with the following question.
appreciate any ones help.
The electrical resistance R of a wire of length l and cross-sectional area A (measured in m2) is given by the equation
R=pl/A
where ρ is a constant (known as resistivity) whose value depends on the metal used to make the wire. Note that Equation 1 is not introduced in S104 and you do not need to worry about the meaning of the symbols.
(i) Rearrange Equation 1 to make
ρ the subject of the equation;
(ii) Use your answer to part (i) to find the units of
ρ. You should express your answer in terms of the SI unit of resistance, and any other units in the answer should be SI base units (using letters and symbols to represent all units). Ensure that you show your working.
• December 13th 2008, 09:22 PM
JD-Styles
(i) To isolate p, multiply both sides of the equation by A, thus pl=RA, then divide by l, so p=RA/l
(ii) Resistence is measured in ohms ( $\Omega$) and the area is measured in square meters ( $m^2$), and l is measured in meters (m) so the units of p are $\Omega m^2/m= \Omega m$
• December 13th 2008, 10:40 PM
Kiwi_Dave
Units Manipulation
Quote:
Originally Posted by Stokoe10
could anyone help me with the following question.
appreciate any ones help.
The electrical resistance R of a wire of length l and cross-sectional area A (measured in m2) is given by the equation
R=pl/A
where ρ is a constant (known as resistivity) whose value depends on the metal used to make the wire. Note that Equation 1 is not introduced in S104 and you do not need to worry about the meaning of the symbols.
(i) Rearrange Equation 1 to make ρ the subject of the equation;
(ii) Use your answer to part (i) to find the units of ρ. You should express your answer in terms of the SI unit of resistance, and any other units in the answer should be SI base units (using letters and symbols to represent all units). Ensure that you show your working.
I don't know your background so please excuse me if you are more advanced than I have assumed.
R=pl/A
Multiply both sides by A to get:
AR=pl
divide both sides by l to get:
$p=\frac{AR}l$
Now for part b, simply replace each symbol with the unit for that symbol to get:
unit for p = $\frac{m^2*Ohms}m$
Now simplify (cancel out a meters from top and bottom) to get your answer.
All times are GMT -8. The time now is 05:44 AM.
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http://math.stackexchange.com/questions/186683/can-a-transformation-matrix-be-expressed-in-terms-of-the-vector-to-be-transforme
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# Can a transformation matrix be expressed in terms of the vector to be transformed?
I'm currently learning linear algebra with my friend via an online course, and we have a disagreement that we would like settled.
Upon learning that vectors can be projected onto lines by a simple function, and that this function is a linear transformation, I recommended that we find a way to calculate the transformation matrix for the function.
This function was characterized by a vector $\vec{v}$ pointing along the line to be projected onto. The non-matrix form of the function we were given was this:
$$proj_{\vec{v}}(\vec{x}) = (\vec{x} \cdot \hat{u})\hat{u}$$
where $\hat{u}$ is the normalized form of $\vec{v}$ calculated by ($\frac{1}{||\vec{v}||}\vec{v}$).
When trying to construct this as a matrix, my friend came up with this (this is just for $\mathbb{R}^{3}$, but you get the idea):
$$proj_{\vec{v}}(\vec{x}) = \left[\begin{array}{ccc} \vec{x} \cdot \hat{u} & 0 & 0 \\ 0 & \vec{x} \cdot \hat{u} & 0 \\ 0 & 0 & \vec{x} \cdot \hat{u} \end{array}\right] \cdot \hat{u}$$
I see multiple problems with this. First of all, transformation matrices cannot be expressed in terms of the vector to be transformed, can they? To my understanding, they are supposed to contain constant values that are the same no matter what vector is being transformed. Second of all, linear transformations multiply $\vec{x}$ by the transformation matrix, not any other vector, right?
I'm not particularly interested in the actual transformation matrix for this problem - I plan to get some practice with what i've learned by calculating that :) - I'd just like confirmation as to whether or not the things my friend has done are valid. Thank you!
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## 1 Answer
The short answer is: You are right. The representing matrix has to be independent of $x$. Moreover a matrix $A$ represents a linear function $f$ if and only if $Ax=f(x)$ for all $x$, so $A$ has indeed to be multiplied with the vector in question.
Even though you didn't ask for it you can write:
$$(x\cdot u)u=u(x\cdot u)=u(u\cdot x)=u(u^tx)=(uu^t)x.$$
So $uu^t$ is the representing matrix. (I write the representing matrix, since the basis is understood to be fixed. In genereal there is more than one representing matrix.)
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http://unapologetic.wordpress.com/2009/11/19/
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# The Unapologetic Mathematician
## The Implicit Function Theorem I
Let’s consider the function $F(x,y)=x^2+y^2-1$. The collection of points $(x,y)$ so that $F(x,y)=0$ defines a curve in the plane: the unit circle. Unfortunately, this relation is not a function. Neither is $y$ defined as a function of $x$, nor is $x$ defined as a function of $y$ by this curve. However, if we consider a point $(a,b)$ on the curve (that is, with $F(a,b)=0$), then near this point we usually do have a graph of $x$ as a function of $y$ (except for a few isolated points). That is, as we move $y$ near the value $b$ then we have to adjust $x$ to maintain the relation $F(x,y)=0$. There is some function $f(y)$ defined “implicitly” in a neighborhood of $b$ satisfying the relation $F(f(y),y)=0$.
We want to generalize this situation. Given a system of $n$ functions of $n+m$ variables
$\displaystyle f^i(x;t)=f^i(x^1,\dots,x^n;t^1,\dots,t^m)$
we consider the collection of points $(x;t)$ in $n+m$-dimensional space satisfying $f(x;t)=0$.
If this were a linear system, the rank-nullity theorem would tell us that our solution space is (generically) $m$ dimensional. Indeed, we could use Gauss-Jordan elimination to put the system into reduced row echelon form, and (usually) find the resulting matrix starting with an $n\times n$ identity matrix, like
$\displaystyle\begin{pmatrix}1&0&0&2&1\\{0}&1&0&3&0\\{0}&0&1&-1&1\end{pmatrix}$
This makes finding solutions to the system easy. We put our $n+m$ variables into a column vector and write
$\displaystyle\begin{pmatrix}1&0&0&2&1\\{0}&1&0&3&0\\{0}&0&1&-1&1\end{pmatrix}\begin{pmatrix}x^1\\x^2\\x^3\\t^1\\t^2\end{pmatrix}=\begin{pmatrix}x^1+2t^1+t^2\\x^2+3t^1\\x^3-t^1+t^2\end{pmatrix}=\begin{pmatrix}0\\{0}\\{0}\end{pmatrix}$
and from this we find
$\displaystyle\begin{aligned}x^1&=-2t^1-t^2\\x^2&=-3t^1\\x^3&=t^1-t^2\end{aligned}$
Thus we can use the $m$ variables $t^j$ as parameters on the space of solutions, and define each of the $x^i$ as a function of the $t^j$.
But in general we don’t have a linear system. Still, we want to know some circumstances under which we can do something similar and write each of the $x^i$ as a function of the other variables $t^j$, at least near some known point $(a;b)$.
The key observation is that we can perform the Gauss-Jordan elimination above and get a matrix with rank $n$ if and only if the leading $n\times n$ matrix is invertible. And this is generalized to asking that some Jacobian determinant of our system of functions is nonzero.
Specifically, let’s assume that all of the $f^i$ are continuously differentiable on some region $S$ in $n+m$-dimensional space, and that $(a;b)$ is some point in $S$ where $f(a;b)=0$, and at which the determinant
$\displaystyle\det\left(\frac{\partial f^i}{\partial x^j}\bigg\vert_{(a;t)}\right)\neq0$
where both indices $i$ and $j$ run from $1$ to $n$ to make a square matrix. Then I assert that there is some $k$-dimensional neighborhood $T$ of $b$ and a uniquely defined, continuously differentiable, vector-valued function $g:T\rightarrow\mathbb{R}^n$ so that $g(b)=a$ and $f(g(t);t)=0$.
That is, near $(a;b)$ we can use the variables $t^j$ as parameters on the space of solutions to our system of equations. Near this point, the solution set looks like the graph of the function $x=g(t)$, which is implicitly defined by the need to stay on the solution set as we vary $t$. This is the implicit function theorem, and we will prove it next time.
Posted by John Armstrong | Analysis, Calculus | 4 Comments
## 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.
• ## Feedback
Got something to say? Anonymous questions, comments, and suggestions at Formspring.me!
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http://mathoverflow.net/questions/15153?sort=votes
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Periods and L-values
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
A famous theorem of Euler is that Zeta(2n) is a rational number times pi^(2n). Work of Kummer, Herbrand, Ribet and others shows that the rational multiplier has number theoretic significance.
For more general L-functions attached to motives, the philosophy has emerged (Deligne, Beilinson, Bloch, Kato, etc.) that (in vague terms) their values at certain integers are algebraic multiples of transcendental numbers and that the particular algebraic number that's a multiple of the transcendental number contains information about the motive that the L-function is attached to.
But a (nonzero) algebraic multiple of a transcendental number is again transcendental, so an arbitrary real number does not have a well defined decomposition as a product of an algebraic number and a transcendental number.
Still, because of the theorems of Kummer, et. al. one suspects that powers of pi are (at least close to) the "right" "transcendental parts" of the L-function values to be looking at. Maybe one should really be looking a powers of 2pi? But it seems clear that one should not be looking at powers of 691*pi because otherwise the statement of Kummer's criterion for the regularity of a prime would have an exceptional clause involving the prime 691.
Is there a conceptually motivated means of picking out the "right" "transcendental part" of a special value of an L-function?
Presumably the reason that Euler expressed his theorem in terms of pi is because pi was a commonly used symbol. (I've heard people argue that 2 pi is more conceptually primitive and that a label should have been made for the quantity 2 pi rather than for pi and am not sure what I think about this). In any case, there should be an a priori means to pin down the relevant transcendental number down "on the nose" (not up to a rational/algebraic multiple).
I have heard that Beilinson's conjecture give the transcendental number only up to a rational multiple and that the Bloch-Kato conjecture pins down the number. But I don't know enough to understand the statement of either conjecture and so am at present ill equipped to derive insight from reading the paper of Bloch and Kato. Are there more elementary considerations that give insight into how to pick out a particular transcendental number out of the set of all of its algebraic multiples?
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1 Answer
The ingredient in the Beilinson and Bloch--Kato conjectures is a motive (over ${\mathbb Q}$, say). If we take the integral cohomology of this motive (mod torsion, say) we get an integral lattice. If we take some kind of Neron model, and take the algebraic de Rham cohomology of this, we get a second integral lattice. Now computing the determinant of the pairing of one of these on the other, we get a transcendental number, well defined up to a unit in ${\mathbb Z}^{\times}$, i.e. a sign.
This should give you an idea of how one can attach a canonical period to a motive, and is the basic idea underlying the construction of periods for motives. (Since one doesn't have Neron models in general, this idea is just heuristic as it stands, but I think it gives the right idea. If you apply it to ${\mathbb G}_m$, you should recover the period $2\pi i$.)
EDIT: I should point out that the above really is just a heuristic, explaining how there are two ways of getting integral structures in cohomology: in singular cohomology, one just takes integral cycles (i.e. "true" cycles on the motive, with no funny coefficients), and in de Rham cohomology, one takes algebraic differential forms that are defined over the integers, like the Neron differential $dx/2y$ on an elliptic curve with minimal Weierstrass equation $y^2 = f(x)$.
To actually get the correct periods for a given $L$-function, one has to do a little more manipulation than I indicated; e.g. for an elliptic curve over ${\mathbb Q}$, one will integrate the Neron differential over the a basis for the real integral cycles (i.e. the cycles that are fixed by the action of complex conjugation on $E({\mathbb C})$; these are rank one subgroup of the cohomoloy of $E({\mathbb C})$). But hopefully what I wrote above gives some intuition for what is going on.
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Integral cohomology in which cohomology theory? – Jonah Sinick Feb 13 2010 at 2:44
Just Betti (i.e. singular) cohomology. The periods are obtained by integrating algebraic differential forms over integral cycles. Also, my heuristic is a bit rough and ready, and I've added an edit to this effect. – Emerton Feb 13 2010 at 3:19
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http://physics.stackexchange.com/questions/10565/does-a-magnetic-field-do-work-on-an-intrinsic-magnetic-dipole/28867
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# Does a magnetic field do work on an intrinsic magnetic dipole?
When you release a magnetic dipole in a nonuniform magnetic field, it will accelerate.
I understand that for current loops (and other such macroscopic objects) the magnetic moment comes from moving charges, and since magnetic fields do no work on charges (F perpendicular to v) it follows that the work done on the dipole (that caused its gain in kinetic energy) must have come from somewhere other than the magnetic forces (like electric forces in the material).
However, what about a pure magnetic moment? I'm thinking of a particle with intrinsic spin. Of course such a thing should be treated with quantum mechanics, but shouldn't classical electrodynamics be able to accommodate a pure magnetic dipole? If so, when I release the pure dipole in a nonuniform B-field and it speeds up, what force did the work? Is it correct to say that magnetic fields DO do work, but only on pure dipoles (not on charges)? Or should we stick with "magnetic forces never do work", and the work in this case is done by some other force (what?)?
Thanks to anyone who can alleviate my confusion!
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## 4 Answers
Yes, of course that if a field - magnetic field - is able to make a bar magnet rotate or move, it is doing work. The statement that magnetic fields don't do any work only applies to point-like pure electric charges.
Magnetic moments may be visualized as objects with a forced motion of charges (solenoids have the same magnetic field as bar magnets), and if something is moving, the magnetic force is becoming a force that does work.
In terms of formulae, the magnetic force on a charge is $q\vec v\times \vec B$ which is identically perpendicular to $\vec v$ and that's why it does no work. However, forces on magnetic dipoles and more general objects don't have the form $\vec v\times$ - they're not perpendicular to $\vec v$, so they do work in general.
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Thanks! That makes sense if it's true. But I thought that maybe since the formula for the force on a dipole $\vec F=\vec\nabla (\vec \mu \cdot \vec B)$ is usually derived by considering some limit of a charge/current distribution, it would somehow inherit the "no work" property of the Lorentz force law. How do you derive the forces on the "more general objects" you mentioned? Is there something more fundamental or different than the Lorentz force law? – Joss L May 30 '11 at 4:33
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@Lubos Motl: I disagree. You can construct a pure magetic dipole from a vanishingly small current loop. And because of the Lorentz law, no work can be done. Also, because no magnetic monopoles have been discovered, all magnetic sources in our world have to come from moving charges on which no work can be done. – Kasper Meerts May 30 '11 at 21:00
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Dear @kmm, what you write is silly, indeed. The objects known as "magnets" - such as iron, you know, they are ferromagnets, some children have already heard about it - have most of their magnetism coming from the spin of the electron that is not moving at all. Again, it is nonsense that one cannot do work with magnetic field. Magnetic moments in magnetic fields are clearly doing work. What matters is a nonzero $\vec j$, and it can be nonzero with arbitrarily small loops, including "vanishingly small" ones, as the spin example shows. – Luboš Motl Jun 1 '11 at 9:36
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Dear @Joss, the formula you wrote down is the fundamental one that applies to all situations, so there is nothing else to derive. A magnetic moment $\vec \mu$ in a magnetic field has the energy $-\vec\mu\cdot \vec B$. Any configuration of currents and magnets may be written as a volume density of magnetic moments, usually known as magnetization $\vec M$. See en.wikipedia.org/wiki/Magnetization Then the energy is $-\int \vec M \cdot \vec B$. The force is always the minus gradient of the energy in a magnetic field and the energy change is the work that has been done. – Luboš Motl Jun 1 '11 at 9:41
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@lubos doesn't the work come from the internal energy in the bar magnet? – John McVirgo Jun 3 '11 at 16:50
show 6 more comments
Is it correct to say that magnetic fields DO do work?
Yes! I show this quantitatively:
Each charged particle experiences action of magnetic force. This force is transmitted to a conductor in which the charges move. As a result, the magnetic field acts with a certain force on the current-carrying conductor. Let the volume charge density, (electrons in a metal, for example) is equal to $\rho$. Let distinguish a mental element of volume $dV$ of the conductor. There is a charge equal to $\rho dV$. Then the force acting on the element $dV$ of the conductor can be expressed by the Lorenz formula $\overrightarrow{F}=q(\overrightarrow{v}\times\overrightarrow{B})$ in the form:
$$\overrightarrow{dF}=\rho (\overrightarrow{v}\times\overrightarrow{B})dV$$
Since $\overrightarrow{j}=\rho\overrightarrow{v}$ where $\overrightarrow{j}$ is the current density vector we can write:$$\overrightarrow{dF}=(\overrightarrow{j}\times\overrightarrow{B})dV$$ If the current flows through a thin conductor, then the following holds:$$\overrightarrow{j}dV= \overrightarrow{dl}I$$ where I is a current in a thin conductor(wire) and $\overrightarrow{dl}$ is the vector of an element of the wire in direction of the current. Thus:
$$\overrightarrow{dF}=I(\overrightarrow{dl}\times\overrightarrow{B})$$ This is nothing more than Ampere's force. So the resulting Ampere's force acting on the contour of the current (current loop) in the magnetic field is determined as a line integral along the current loop:
$$\overrightarrow{F}=I\oint(\overrightarrow{dl}\times\overrightarrow{B})$$ If the magnetic field is nonuniform, then the integral is generally different from zero.
A conclusion:
It follows directly from the Lorenz law that magnetic fields do work on a current loop.
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@John McVirgo: The loop is made of a conductor. We are interested in forces applied to the conductor. Newton's third law applies. – Martin Gales Jun 1 '11 at 6:42
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@Joss L: In an electric motor, what other than the magnetic forces do work on current-carrying conductors? – Martin Gales Jun 1 '11 at 6:43
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@John McVirgo: Consider an extremely trivial example: A current(an electron) moves along a straight, long, thin, superconducting stationary wire with initial velocity $v_0$. Magnetic field is off. Now a static magnetic field, perpendicular to the direction of the wire is switched on for a while and then switched off. After this, the electron acquires a velocity $v$ along the wire and the wire acquires a velocity $u$, perpendicular to the direction of the wire. Because the magnetic field will not affect the energy of the electron, its speed(relative to the field) must remain the same, i.e. ... – Martin Gales Jun 6 '11 at 8:33
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... $v_0^2=v^2+u^2$. Electron's speed along a wire decreases or in other words: current in the wire decreases. Summary: The magnetic field draws energy from the current in the loop to do work. By analogy, you draw energy from the food to do a work. But you do not say that, after all, i never do a work, it is the food that does a work. – Martin Gales Jun 6 '11 at 8:33
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@John McVirgo: We can observe the process far from both, start and end points, to ignore transient effects. But what matters is an energy balance, conservation of energy. Work done by the field on the loop must equal to the change in the energy of the system(the loop + the static field). We consider a superconducting loop for simplicity. Its energy is $L\frac{i^2}{2}$ where $L$ is the self-inductance of the loop and $i$ is its current. This is the only energy source. So we can actually avoid considering fields themselves to analyze this problem. – Martin Gales Jun 7 '11 at 7:04
show 6 more comments
I'm going to take a risk and try to answer this, even though my answer is different to Lubos's and he does have a reputation that is overwhelming right compared to mine.
Static Magnetic fields don't don't do work, period (us), full stop (uk). So the work comes from the magnetic dipole itself whose internal energy is affected by the external force that positioned it in the static magnetic field in the first place.
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Your answer might be different, but that's good because you're right and he's wrong. – Kasper Meerts May 30 '11 at 20:52
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An electron has a magnetic dipole moment and, because of that, can be accelerated by a non-uniform static magnetic field. Then, if an initially static electron acquires a kinetic energy using this method, are you suggesting it comes from its internal energy (i.e. rest mass)? – mmc May 31 '11 at 3:30
@mmc, I was thinking more of a a loop of wire carrying a current where internal electric fields are created to maintain its orientation in a static magnetic field, from an applied force. It doesn't apply to an electron because it's not seen as a spinning charge and hence current loop in the classical sense. – Larry Harson May 31 '11 at 10:59
Someone actually told me that the rest mass of an electron will change when it's placed in a B field, and this is where the kinetic energy comes from. He was in a hurry and I'm going to talk to him later, but he's a professor and seems like he knows what he's talking about. – Joss L May 31 '11 at 14:46
Larry Harson's answer, although short, is right, to my point of view. How I explain it doesn't fit in a comment so I make a separate answer.
Consider a simple magnetic moment, familiar to those who study space plasmas: the electron, trapped in earth's magnetosphere, precisely in the Van Allen Radiation belt. Let B_0>0 be a static homogenous field oriented towards the z direction. If you put an electron with initial speed Vx>0, Vy and Vz being zero, the lorentz force will make the electron circle in the magnetic field, and if you look to the trajectory of the electron from above (z being the vertical, "up", direction, oriented towards the sky), it will rotate counter-clockwise. As the electron has negative charge, this will create a magnetic dipole, oriented against the external magnetic field B_z. Here you come to the result that free electrons in a static homogenous magnetic field are essentially "diamagnetic".
In Earth's magnetosphere, the electrons circle around magnetic flux lines, their speed along these lines being kept constant if the norm of the magnetic field is constant. They "follow" magnetic flux lines.
Now, if the electrons try to get too close to Earth, the magnetic field will intensify if you follow a magnetic flux line, because of the general shape of the magnetic flux lines of a dipole, that get closer to each other the closer you get to the source. To sketch this from the point of view of perturbation theory, imagine that, together with the static B_0 field defined later, you add a centripetal field, B_r<0. Imagine that you are approaching earth's south pole from outer space. Now, the electron will undergo two forces. If it has no V_z speed, then the B_r field will create a force oriented towards -Oz, that will superpose to the centripetal force of the B_0 field. "down", towards outer space. This force is what keeps the electrons from messing with our atmosphere too much: only the most energetic particles reach earth's atmosphere, those that came with a big enough V_z speed that the converging magnetic field could not stop. This is what creates Auroras at earth's poles.
Then, let us consider the process of this "brake" of electrons when approaching the south pole. Now, they have a non-negligible V_z speed. Look closely: because of the B_r component, it will speed their rotation up! Their V_z speed is only converted to V_x/V_y speed. The same process will perform in reverse, when the electron is finally reflected to outer space: they will lose rotational speed, and gain longitudinal speed. Of course, their kinetic energy is conserved.
If you look now to what happens for magnetic dipoles in a magnet, it will be a little bit different. In space, electrons had to be diamagnetic, as their motion was determined by the magnetic field. But if you look at, say the orbital momentum of an electron around an atom, then it is the electric force that makes it circle: it can circle clockwise or counter-clockwise, now. But the magnetic field will add a force, either centrifugal or centripetal, that will shift the energy levels. As we enter now the domain of quantum mechanics, I cannot explain formally without mistake what happens, but the remark about the B_r component of the field slowing down the "rotation" still seem to hold. If you consider a ferromagnetic atom as having its dipolar momentum oriented towards the magnetic field ("clockwise motion" in my model), then the magnetic force is centrifugal, and compensates a part of the electric field. For a given "radius", the equilibrium speed of the electron is thus lower (in the Rutherford model), which would tend to correspond to a "lower energy level". But what happens exactly in the shifts I do not know, one would need to look at these energy level shifts from the point of view of spectroscopy, to measure them and see indeed that, the higher the magnetic field when it is aligned with a dipole, the lower the intrinsic energy of the dipole.
So, finally, when you pull on a magnet, you are just compensating for all the atomic energy levels shifting up, the number of atoms being kept constant the overall energy of your magnet rises. Where to count the rise in the integrated energy of the magnetic field itself, I do not know for certain (as when two magnets are apart, the integral of B^2/(2*µ0) is usually higher than when they are "how they want to be", glued to one another).
One last thing about macroscopic loops of current: you need an electromotive force to make a motor move, the current tends to decrease by itself when a current loop moves according to Laplace's force, just as it increases by itself if it moves against this force (induction, NdFeB magnet falling in an almost adjusted copper tube, very funny experiment). If you see electrons as "bouncing on the walls of your copper wire", it is all about momentum conservation: if the electron gives momentum to the cristalline structure of the copper wire grain, it loses momentum itself. It will need to be reaccelerated by the electric field, to give this momentum again. So now, it is about the speed, gained from the electric field, that is converted to momentum of the macroscopic wire. The magnetic field is just "the man in the middle" that makes possible the transaction.
I hope what I said is clear...
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It's not true that you may represent magnets via orbital angular momentum. Both ferromagnetism and paramagnetism depend on the existence of the spin - the internal angular momentum that, according to quantum mechanics, is carried even by point-like particles - and the spin can't be explained by any charges' velocities. The spin would be exactly zero in classical physics. So there exists no accounting in which the motion of the magnet would come from the change of the internal energy: the state is exactly the same. After all, the spin is quantized and can't be "raised a bit". – Luboš Motl May 30 '12 at 10:42
Yes, about spin precisely this can't hold "as is". However, although the quantum of spin itself, and hence the quantum of magnetic field generated by one electron, do not change, maybe something in the internal energy of the electron itself changes. I mean: why would magnetic fields work on NOTHING but spins...just because we can't model it down enough, and see how we can fraction the energy of an electron...I agree, this spin problem is a big one, and unfortunately I don't know enough in quantum theory to find any answer to that. – MrBrody May 30 '12 at 19:26
Nothing can change about the "internal energy" of the electron. Every electron is always the same and fully described by its momentum $\vec p$ and the spin $\vec s$ and the energy of the electron only depends on the former, the momentum. The magnetic moment is $\vec\mu = C\vec s$ with a fixed constant $C$. This "preservation" of the full magnetic moment isn't necessarily a special quantum mechanical feature: a permanent magnet is made out of many spins like that and the total magnetic moment of a bar magnet is also preserved in those exercises, unless you work hard to damage or demagnetize it. – Luboš Motl Sep 19 '12 at 11:53
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http://mathoverflow.net/questions/72161/graphs-with-few-induced-subgraphs/72171
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## Graphs with few induced subgraphs
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Which graphs have the property that the number of $i$-vertex induced subgraphs is at most $i$ for some $i < n/2$ (where $n$ is the number of vertices)?
To avoid cases I am not interested in, I want the graph to be connected and non-bipartite and its complement to have the same properties.
Of course every graph has this property for $i=2$, and every triangle-free graph (or complement of one) for $i=3$. What about larger values of $i$?
Edit. More background information on the problem can be found here.
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Welcome to MathOverflow! – Douglas S. Stones Aug 5 2011 at 10:11
Should this be tagged open-problem? – Emil Aug 6 2011 at 11:42
## 3 Answers
Here's an example $G$ for all $i \geq 8$ and $n \geq 2i+1$ that contains only $8$ induced subgraphs (up to isomorphism).
If we choose $i$ vertices from the top row, we obtain $\overline{K_i}$.
If we choose $i-1$ vertices from the top row, we obtain $K_{1,i-1}$ or $\overline{K_i}$.
If we choose $i-2$ vertices from the top row, we obtain $\overline{K_i}$ or $\overline{K_{i-2}} \cup K_2$ or $K_{1,i-2} \cup K_1$ or $K_{1,i-1}$.
If we choose $i-3$ vertices from the top row, we obtain the subgraphs induced by $i-3 \text{ vertices from a,b,c,d,... } \cup {e,g,h}$ or $i-3 \text{ vertices from a,b,c,d,... } \cup {e,f,g}$ or $K_{1,i-2} \cup K_1$ or $\overline{K_{i-3}} \cup P_3$ (path with three vertices).
If we choose $i-4$ vertices from the top row, we obtain the graph induced by $i-4 \text{ vertices from a,b,c,d,... } \cup {e,f,g,h}$.
Since we must choose at least $i-4$ vertices from the top row, in total, that's 8 isomorphism classes of graphs. We can see that $G$ and its complement both contain a triangle (and are therefore not bipartite) and are connected (each vertex is not adjacent to either h or a).
Generalising this technique, we can replace {e,f,g,h} by any non-bipartite connected subgraph of diameter at least 2. In this case, there would exist some I, N such that for all i>=I and n>=max(N,2i+1) which would satisfy the required conditions.
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The vertex $e$ seems isolated in the complement... – MP Aug 5 2011 at 12:11
Ouch! Hmm... that's a problem! – Douglas S. Stones Aug 5 2011 at 12:12
Hmm... I think the idea can be salvaged for larger values of i; e.g. be attaching a leaf to g. – Douglas S. Stones Aug 5 2011 at 12:19
Right, I think it's fixed now. Wow, that was a bookkeeping nightmare. – Douglas S. Stones Aug 5 2011 at 12:48
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Thanks, Doug. But, of course, I am not satisfied with this answer! I would like the graph to be vertex-transitive, and edge-transitive. I already know that its diameter is at most 4 (in my application).
I will come a bit cleaner about the application. This arises in a question of Joao Araujo about semigroup theory, which reduces to the following: Which groups are $(i,i+1)$-homogeneous but not $i$-homogeneous? (The condition $(i,i+1)$-homogeneous means that given any $i$-set $I$ and $j$-set $J$, there is an element of the group mapping $I$ to a subset of $J$.
Assume that $2i+1\le n$. Such a group is easily seen to be primitive and have at most three orbits on $2$-sets. If it is $2$-homogeneous, then it must be $2$-transitive, and then one has the possibility of using the classification of such groups.
It is easy to see that such a group has at most $i$ orbits on $i$-sets; so if it is not $2$-homogeneous, then all the symmetrised orbital graphs have the property of my original question.
By the way, five examples of groups with this property are known: the cyclic and dihedral groups of degree 5, and the affine groups $\mathrm{AGL}(1,7)$, $\mathrm{ASL}(2,3)$ and $\mathrm{AGL}(2,3)$ (with $n=7,9,9$ for the last three).
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This is a pretty interesting question. Douglas S. Stones' answer provides an example that works (but has diameter 3). On the other hand, large diameter forces many induced subgraphs for all small values of $i$.
Lemma. Let $G$ be a graph with diameter $d \geq 8$. Then for all $4 \leq i \leq d/2$, the number of induced subgraphs of $G$ with $i$ vertices is more than $i$.
Proof. Let $P$ be an induced path of $G$ with $d$ vertices. Let $F_i$ be the set of all forests on $i$ vertices with maximum degree 2. Observe that $P$ contains all graphs in $F_i$ for all $i \leq d/2$ as induced subgraphs. Since $F_i > i$ for all $i \geq 4$, we are done.
The property seems harder to satisfy for larger values of $i$ which leads us to the following (updated) rash conjecture.
Rash Conjecture. Let $G$ be a connected, non-bipartite graph on $n$ vertices whose complement is also connected and non-bipartite. If $G$ has at most $i$ induced $i$-subgraphs for some $3 < i < n/2$, then $G$ has diameter at most 7.
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Note: 4 is not less than 5/2. – Douglas S. Stones Aug 5 2011 at 11:29
Indeed, edited accordingly. – Tony Huynh Aug 5 2011 at 11:36
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Only odd cycle of length 9 works. Otherwise you get (a) null graph, (b) P_2, (c) P_3, (d) P_4 and (e) P_2 \cup P_2. [where P_i = path with i vertices]. With a 9-cycle, you can't get an induced null graph. – Douglas S. Stones Aug 5 2011 at 11:49
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Ah, I missed $P_3$ for some reason. But I also don't think $C_9$ works either since it does contain the null graph as an induced subgraph. – Tony Huynh Aug 5 2011 at 11:57
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I have described the background to the problem in a little more detail <a href="cameroncounts.wordpress.com/2011/08/05/…;. – cameroncounts Aug 8 2011 at 9:04
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http://math.stackexchange.com/questions/264899/bounded-linear-maps-in-banach-spaces?answertab=oldest
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# Bounded linear maps in Banach spaces
Let $X$ be a Banach space and let $M: X \rightarrow X$ be a linear map. Prov that M is bounded iff there exists a set $S \subset X'$, dense in X', such that for each $\ell \in S$ the functional $m_l$ defined by $$m_\ell(x) = \ell M (x)$$ is continuous on X.
My try: If $M$ is bounded then $\ell M$ is bounded for all $\ell$, hence all $m_\ell$ are continuous, for all subsets $S$. So we need to find a dense one?
On the other hand: suppose all $m_\ell$ is continuous, since the weak limit is unique, $Mx_n \rightarrow y$ and $x_n \rightarrow x$ $\Longrightarrow$ $Mx = y$ and by the closed graph M is bounded/continuous.
It feels like I'm missing something with the denseness of $S$. Should I look at $\ell \in S^c$ also? and do some $\epsilon/2$ argument?
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Your argument "On the other hand: ..." has a gap: how do you argue that $Mx = y$? More precisely: What is that "weak limit" you are talking about? You only know about convergence for $\ell \in S$, not about all $\ell \in X'$. – Martin Dec 25 '12 at 10:10
take $s \not \in S$. $|m_s(x_n) - m_s(x)| = |m_s(x_n) - m_s(x) + \ell x_n - \ell x_n + \ell x - \ell x | \leq 3*\frac{1}{\epsilon}$ for some $\ell \in S$. Will this help me? Now we have weak convergens for $Mx_n$ ? – Johan Dec 25 '12 at 19:30
## 1 Answer
Assume that $\lim\limits_{n\to\infty} x_n=x$ and $\lim\limits_{n\to\infty} M(x_n)=y$. Take arbitrary $\ell\in S$, then $$\ell(y-M(x))= \ell(y)-\ell(M(x))= \ell(\lim\limits_{n\to\infty}M(x_n))-m_\ell(x)= \lim\limits_{n\to\infty}\ell(M(x_n))-m_\ell(x)= \lim\limits_{n\to\infty}m_\ell(x_n)-m_\ell(x)= m_\ell(\lim\limits_{n\to\infty}x_n)-m_\ell(x)= m_\ell(x)-m_\ell(x)=0$$ Since $S$ is dense in $X^*$ then for all $\ell\in X^*$ wee have $$\ell(M(x)-y)=0$$ By corollary of Hahn-Banach theorem it follows that $M(x)-y=0$. Now from closed graph theorem we get that $M$ is continuous.
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great, can you please state that corollary, I cannot find it! – Johan Dec 26 '12 at 12:38
Corollary of Hahn Banach theorem: Let $X$ be a normed space, and $x\in X$. If for all $\ell\in X^*$ we have $\ell(x)=0$, then $x=0$. – Norbert Dec 26 '12 at 12:50
thanks, I was thinking about that, why do we need hahn banach for that? Is it not true without axiom of choice? I was thinking it would be true for all projections... – Johan Dec 26 '12 at 12:54
this is another question – Norbert Dec 26 '12 at 12:55
That is true! thanks – Johan Dec 26 '12 at 13:03
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http://mathoverflow.net/questions/89345/example-of-a-manifold-which-is-not-a-homogeneous-space-of-any-lie-group/89352
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## Example of a manifold which is not a homogeneous space of any Lie group
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Every manifold that I ever met in a differential geometry class was a homogeneous space: spheres, tori, Grassmannians, flag manifolds, Stiefel manifolds, etc. What is an example of a connected smooth manifold which is not a homogeneous space of any Lie group?
The only candidates for examples I can come up with are two-dimensional compact surfaces of genus at least two. They don't seem to be obviously homogeneous, but I don't know how to prove that they are not. And if there are two-dimensional examples then there should be tons of higher-dimensional ones.
The question can be trivially rephrased by asking for a manifold which does not carry a transitive action of a Lie group. Of course, the diffeomorphism group of a connected manifold acts transitively, but this is an infinite-dimensional group and so doesn't count as a Lie group for my purposes.
Orientable examples would be nice, but nonorientable would be ok too.
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Well, a compact surface of genus at least two is the quotient of $\mathbb{H}$ by a group action, and $\mathbb{H}$ is a homogeneous space for $\text{SL}_2(\mathbb{R})$, so in some sense these examples still come from Lie groups. – Qiaochu Yuan Feb 24 2012 at 0:48
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Well a compact orientable surface of genus two or more has no transitive action of a compact Lie group because such an action would necessarily preserve a Riemann metric and therefore a conformal structure. The group of conformal automorphisms of such a surface is finite. – Tom Goodwillie Feb 24 2012 at 0:52
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I must say I am surprised that this question has garnered two votes to close as "not a real question." Would anybody care to explain? The answers below seem to imply (to me, at least) that the question isn't so trivial. – MTS Feb 24 2012 at 5:04
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@MTS: Sometimes people underestimate a question, oversimplifying it. I suspect in this case it's just a mistake on the part of the people who cast votes to close. But if they don't say anything it's hard to tell. I don't think you have to worry about this thread being closed. – Ryan Budney Feb 24 2012 at 5:11
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As you mention, locally symmetric spaces give a huge class. The example you describe with the genus at least two is actually $M = \Gamma \backslash PSL_2( \mathbb{R}) /PSO(2)$, where $\Gamma \cong \pi_1(M)$. – Marc Palm Feb 24 2012 at 13:03
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## 7 Answers
$\pi_2$ of a Lie group is trivial, so $\pi_2(G/H)$ is isomorphic to a subgroup of $\pi_1(H)$, which is finitely generated (isomorphic to $\pi_1$ of a maximal compact subgroup of the identity component of $H$). But $\pi_2$ of a closed manifold is often not finitely generated. For example, the connected sum of two copies of $S^1\times S^2$ has as a retract a punctured $S^1\times S^2$, which is homotopy equivalent to $S^1\vee S^2$ and so has universal cover homotopy equivalent to an infinite wedge of copies of $S^2$.
EDIT This ad hoc answer can be extended as follows: All I really used was that $\pi_2(G)$ and $\pi_1(G)$ are finitely generated. But $\pi_n(G)$ is finitely generated for all $n\ge 1$ (reduce to simply connected case and use homology), so $\pi_n(G/H)$ is finitely generated for $n\ge 2$. That leads to a lot more higher-dimensional non simply connected examples.
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Apart from already mentioned non simply connected examples most simply connected manifolds are also not homogeneous. One easy criterion is that simply connected homogeneous spaces are rationally elliptic, i.e. they have finite dimensional total rational homotopy. That is because any connected Lie group is rationally homotopy equivalent to a product of odd dimensional spheres. so a homogeneous space is elliptic by a long exact homotopy sequence.
Most simply connected manifolds are not rationally elliptic. For example the connected sum of more than two $CP^2$'s or $S^2\times S^2$'s. This is easily seen by looking at their minimal models. But even without computing minimal models it's known that an elliptic manifold $M^n$ has nonnegative Euler characteristic and has the total sum of its Betti numbers $\le 2^n$. So anything that violates either of these conditions such as the connected sum of several $S^3\times S^3$'s is definitely not rationally elliptic and hence can not be a homogeneous space or even a biquotient.
As for higher genus surfaces it should not be hard to show that they can not be homogeneous spaces $G/H$ even if you don't assume that $G$ acts by isometries. If $G/H=S^2_g$ and the $G$ action is effective then for any proper normal $K\unlhd G$ which does not act transitively on $S^2_g$ we must have $K/(K\cap H)=S^1$. But then $G/K$ is also 1-dimensional and hence also a circle which is obviously impossible. This reduces the situation to the case of $G$ being simple which can also be easily ruled out for topological reasons.
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In the last paragraph why does $K/(K\cap H)$ have to be compact? – Tom Goodwillie Feb 24 2012 at 13:01
sorry, I meant to say that that we should look at a closed proper normal subgroup $K$. – Vitali Kapovitch Feb 24 2012 at 13:21
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Even so, how do you rule out the possibility that $K/(K\cap H)$ is a noncompact $1$-manifold? – Tom Goodwillie Feb 24 2012 at 13:38
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hmm, I thought this was obvious but it does require some argument. how about this then. in such situation all orbits of $K$ must be 1-dimensional submanifolds of $S^2_g$ which gives a foliation of $S^2_g$ which is impossible since its Euler characteristic is not zero. – Vitali Kapovitch Feb 24 2012 at 14:23
Atiyah and Hirzebruch gave a rather dramatic answer to your question in their paper "Spin Manifolds and Group Actions": if $M$ is a compact smooth spin manifold of dimension $4k$ whose $\hat{A}$-genus is nonzero then no compact Lie group can act on $M$ nontrivially, let alone transitively! The proof uses Atiyah and Bott's Lefschetz fixed point theorem in a clever way.
Unfortunately I don't have a simple example of such a manifold lying around, though I know that there are plenty of examples among 4-manifolds. It's possible that some 4 dimensional lens spaces would do the job.
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I guess my answer doesn't rule out the possibility that $M$ is a homogeneous space for a non-compact Lie group, but perhaps it's still interesting. – Paul Siegel Feb 24 2012 at 3:03
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there are indeed many such 4-manifolds. connected sum of any number of $K3$-surfaces is the most standard example. so if anything like this is a homogenous space $G/H$ then the maximal compact subgroup of $G$ must be zero-dimensional which means that $G$ must be contractible. This is of course impossible which means that none of such manifolds can be homogeneous. – Vitali Kapovitch Feb 24 2012 at 3:13
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@Paul : You should perhaps say "no compact connected Lie group", as there might be finite order symmetries. This is equivalent to "no non trivial $S^1$ action". – BS Feb 24 2012 at 11:10
It is a result of Mostov that any compact homogeneous manifold must have nonnegative Euler characteristic:
http://www.ams.org/mathscinet-getitem?mr=2174096
That should provide plenty of counterexamples. :)
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Now I see this is also implied by Vitali's post... I wonder if Mostov knew this? – Dylan Wilson Feb 24 2012 at 6:59
This is also proven in R. Hermann "Compactification of homogeneous spaces. I." J. Math. Mech. 14 1965 655–678. Apparently Mostow did not know that paper. – Johannes Ebert Feb 24 2012 at 17:08
Are you sure they aren't assuming the group is compact? Mostow doesn't assume this. He also has some classification results. – Dylan Wilson Feb 24 2012 at 19:20
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@Dylan: I am pretty sure Hermann, as well as Mostow, does not assume compactness of $G$, only of $G/H$. If $G$ is compact, then in fact the result is much older and, nowadays, fairly easy. If $G$ is connected and compact and the rank of $H$ equals the rank of $G$, then there is a fibre bundle $H/T \to G/T \to G/H$, where $T$ is the maximal torus. By Hopf-Samelson, the Euler numbers of $G/T$ and $H/T$ are given by the order of the Weyl groups, hence both positive, so $\chi(G/H)>0$. – Johannes Ebert Feb 24 2012 at 20:54
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If $H$ has smaller rank, $S \subset H$ is a maximal torus of $H$, $S \subset T \subset G$ a maximal torus of $G$. The fibre bundle $T/S G/S \to G/T$ shows that $\chi(G/S)=0$. The fibre bundle $H/S \to G/S \to G/H$ shows $0=\chi(G/S) = \chi(H/S) \chi(G/H) = # W_H \chi(G/H)$ and so $\chi(G/H)\geq 0$. – Johannes Ebert Feb 24 2012 at 20:57
I would think that many examples from 3-manifold theory would work. Take any compact, oriented, irreducible 3-manifold $M$ whose torus decomposition is nontrivial and has at least one hyperbolic piece. Such examples at least have no locally homogeneous Riemannian metric, as a consequence of Thurston's analysis of the 8 geometries of 3-manifold theory. A specific example of this sort can be obtained from a hyperbolic knot complement in $S^3$ by deleting an open solid torus neighborhood of the knot and doubling across the resulting 2-torus boundary; the doubling torus produces a characteristic $Z^2$ subgroup of $\pi_1(M)$. These examples have universal cover homeomorphic to $R^3$, and so they have trivial $\pi_2$. By the homotopy exact sequence there would be a quotient group $\pi_1(G) / \pi_1(H)$ identified with a subgroup of $\pi_1(G/H)$ whose quotient set is $\pi_0(H)$. Perhaps, in order to get a proof, one can analyze this situation by considering the intersection of the $Z^2$ subgroup with the $\pi_1(G) / \pi_1(H)$ subgroup.
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This is a bit over my head, but a nice answer nonetheless. Thank you. – MTS Feb 24 2012 at 22:01
np. I'll throw in that this idea of analyzing the $Z^2$ subgroup intersected with $\pi_1(G) / \pi_1(H)$ is kind of what is going on in the proof of Thurston's 8 geometries theorem. It seems to me that your question, in the 3-manifold context, is a kind of generalization of the 8 geometries theorem. – Lee Mosher Feb 24 2012 at 23:28
Here is a proof that any closed surface $S$ of genus at least 2 cannot support a homogeneous Riemannian metric. In fact, let $g$ be any such metric. Being homogeneous, $g$ has constant curvature, and e.g. by Gauss-Bonnet such a curvature must be negative. Therefore, $(S,g)$ is homotetic to a hyperbolic surface, and it is well-known that the isometry group of any closed hyperbolic surface is finite (in fact, if $h$ is the genus of $S$, then $(S,g)$ admits at most $84(h-1)$ isometries).
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This is a nice argument, but I am not asking for a homogeneous Riemannian metric. This doesn't rule out the possibility of having a transitive action of a Lie group that is not an action by isometries for any Riemannian metric. – MTS Feb 24 2012 at 1:05
@MTS: if $M$ is a homogeneous space for a Lie group $G$ and $x \in M$ is a point with compact stabilizer, then you can choose a $G$-invariant inner product on $T_x(M)$ and homogeneity gives you a $G$-invariant Riemannian metric on $M$, doesn't it? I guess it is possible that there are no points with compact stabilizers, but in any case this shows at least that any compact surface of genus at least two is not a homogeneous space of a compact Lie group. – Qiaochu Yuan Feb 24 2012 at 1:36
Qiaochu, I think what you are saying is the following: if $M = G/H$, then the tangent bundle of $M$ is the homogeneous vector bundle induced by the adjoint representation of $H$ on $\mathfrak{g}/\mathfrak{h}$. If $H$ is compact then we can integrate with respect to the Haar measure of $H$ to get an invariant inner product on $\mathfrak{g}/\mathfrak{h}$. Then translating around with $G$ gives the invariant metric on $M$. Is that right? I don't think you can choose a $G$-invariant inner product on $T_x(M)$ since $G$ doesn't preserve the point $x$ in general. – MTS Feb 24 2012 at 4:45
@MTS: sorry, I meant a $\text{Stab}(x)$-invariant inner product. – Qiaochu Yuan Feb 24 2012 at 17:42
In the homotopy exact sequence of the fiber bundle $G\to G/H$ the group $\pi_i(G/H)$ sits between $\pi_i(G)$ and $\pi_{i-1}(H)$. For example, if $i=1$, then $\pi_1(G)$ is abelian, and $\pi_0(H)$ is finite (as $H$ is compact). Thus $\pi_1(G/H)$ has abelian subgroup of finite index. Surely there are lots of manifolds that do not have this property, e.g. any closed negatively curved manifold does not. Connected sum of several lens spaces is another example.
By the way, it is my opinion that the class of compact homogeneous spaces is very rich, and their finer topological properties are still poorly understood. For example, classifying homogeneous spaces up to diffeomorphism in a given homotopy type is quite challenging, and it is not easy to cook up a homogeneous spaces with prescribed topological features.
-
1
the OP is interested in a more general situation than homogeneous Riemannian manifolds (see his comments above). thus one can not assume that $H$ is compact. So things like quotients by uniform lattices in semisimple Lie groups or nilmanifolds are fair game. – Vitali Kapovitch Feb 24 2012 at 19:30
@Vitali: I was answering to OP's request "in finding an example of a compact manifold which is not a homogeneous space of any compact Lie group". If a compact Lie group acts transitively on a manifold, then the isotropy subgroup is closed, hence compact. – Igor Belegradek Feb 24 2012 at 22:55
@Igor sorry, this was not clear from your answer. the OP states the question 3 times and only mentions compactness once. He should modify the question IMO. It's also clear from this comments that he is interested in the general case and it creates confusion because some answers assume that G is compact and others do not. – Vitali Kapovitch Feb 25 2012 at 0:21
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http://catalog.flatworldknowledge.com/bookhub/reader/2273?e=ball-ch06_s03
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Introductory Chemistry, v. 1.0
by David W. Ball
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6.3 Gas Laws
Learning Objectives
1. Learn what is meant by the term gas laws.
2. Learn and apply Boyle’s law.
3. Learn and apply Charles’s law.
When seventeenth-century scientists began studying the physical properties of gases, they noticed some simple relationships between some of the measurable properties of the gas. Take pressure (P) and volume (V), for example. Scientists noted that for a given amount of a gas (usually expressed in units of moles [n]), if the temperature (T) of the gas was kept constant, pressure and volume were related: As one increases, the other decreases. As one decreases, the other increases. We say that pressure and volume are inversely related.
There is more to it, however: pressure and volume of a given amount of gas at constant temperature are numerically related. If you take the pressure value and multiply it by the volume value, the product is a constant for a given amount of gas at a constant temperature:
P × V = constant at constant n and T
If either volume or pressure changes while amount and temperature stay the same, then the other property must change so that the product of the two properties still equals that same constant. That is, if the original conditions are labeled P1 and V1 and the new conditions are labeled P2 and V2, we have
P1V1 = constant = P2V2
where the properties are assumed to be multiplied together. Leaving out the middle part, we have simply
P1V1 = P2V2 at constant n and T
This equation is an example of a gas law. A gas lawA simple mathematical formula that allows one to model, or predict, the behavior of a gas. is a simple mathematical formula that allows you to model, or predict, the behavior of a gas. This particular gas law is called Boyle’s lawA gas law that relates pressure and volume at constant temperature and amount., after the English scientist Robert Boyle, who first announced it in 1662. Figure 6.2 "Boyle’s Law" shows two representations of how Boyle’s law works.
Figure 6.2 Boyle’s Law
A piston having a certain pressure and volume (left piston) will have half the volume when its pressure is twice as much (right piston). One can also plot P versus V for a given amount of gas at a certain temperature; such a plot will look like the graph on the right.
Boyle’s law is an example of a second type of mathematical problem we see in chemistry—one based on a mathematical formula. Tactics for working with mathematical formulas are different from tactics for working with conversion factors. First, most of the questions you will have to answer using formulas are word-type questions, so the first step is to identify what quantities are known and assign them to variables. Second, in most formulas, some mathematical rearrangements (i.e., algebra) must be performed to solve for an unknown variable. The rule is that to find the value of the unknown variable, you must mathematically isolate the unknown variable by itself and in the numerator of one side of the equation. Finally, units must be consistent. For example, in Boyle’s law there are two pressure variables; they must have the same unit. There are also two volume variables; they also must have the same unit. In most cases, it won’t matter what the unit is, but the unit must be the same on both sides of the equation.
Example 3
A sample of gas has an initial pressure of 2.44 atm and an initial volume of 4.01 L. Its pressure changes to 1.93 atm. What is the new volume if temperature and amount are kept constant?
Solution
First, determine what quantities we are given. We are given an initial pressure and an initial volume, so let these values be P1 and V1:
P1 = 2.44 atm and V1 = 4.01 L
We are given another quantity, final pressure of 1.93 atm, but not a final volume. This final volume is the variable we will solve for.
P2 = 1.93 atm and V2 = ? L
Substituting these values into Boyle’s law, we get
(2.44 atm)(4.01 L) = (1.93 atm)V2
To solve for the unknown variable, we isolate it by dividing both sides of the equation by 1.93 atm—both the number and the unit:
Note that, on the left side of the equation, the unit atm is in the numerator and the denominator of the fraction. They cancel algebraically, just as a number would. On the right side, the unit atm and the number 1.93 are in the numerator and the denominator, so the entire quantity cancels:
What we have left is
Now we simply multiply and divide the numbers together and combine the answer with the L unit, which is a unit of volume. Doing so, we get
V2 = 5.07 L
Does this answer make sense? We know that pressure and volume are inversely related; as one decreases, the other increases. Pressure is decreasing (from 2.44 atm to 1.93 atm), so volume should be increasing to compensate, and it is (from 4.01 L to 5.07 L). So the answer makes sense based on Boyle’s law.
Test Yourself
If P1 = 334 torr, V1 = 37.8 mL, and P2 = 102 torr, what is V2?
Answer
124 mL
As mentioned, you can use any units for pressure or volume, but both pressures must be expressed in the same units, and both volumes must be expressed in the same units.
Example 4
A sample of gas has an initial pressure of 722 torr and an initial volume of 88.8 mL. Its volume changes to 0.663 L. What is the new pressure?
Solution
We can still use Boyle’s law to answer this, but now the two volume quantities have different units. It does not matter which unit we change, as long as we perform the conversion correctly. Let us change the 0.663 L to milliliters:
Now that both volume quantities have the same units, we can substitute into Boyle’s law:
The mL units cancel, and we multiply and divide the numbers to get
P2 = 96.7 torr
The volume is increasing, and the pressure is decreasing, which is as expected for Boyle’s law.
Test Yourself
If V1 = 456 mL, P1 = 308 torr, and P2 = 1.55 atm, what is V2?
Answer
119 mL
There are other measurable characteristics of a gas. One of them is temperature (T). Perhaps one can vary the temperature of a gas sample and note what effect it has on the other properties of the gas. Early scientists did just this, discovering that if the amount of a gas and its pressure are kept constant, then changing the temperature changes the volume (V). As temperature increases, volume increases; as temperature decreases, volume decreases. We say that these two characteristics are directly related.
A mathematical relationship between V and T should be possible except for one thought: what temperature scale should we use? We know from Chapter 2 "Measurements" that science uses several possible temperature scales. Experiments show that the volume of a gas is related to its absolute temperature in Kelvin, not its temperature in degrees Celsius. If the temperature of a gas is expressed in kelvins, then experiments show that the ratio of volume to temperature is a constant:
$VT=constant$
We can modify this equation as we modified Boyle’s law: the initial conditions V1 and T1 have a certain value, and the value must be the same when the conditions of the gas are changed to some new conditions V2 and T2, as long as pressure and the amount of the gas remain constant. Thus, we have another gas law:
This gas law is commonly referred to as Charles’s lawA gas law that relates volume and temperature at constant pressure and amount., after the French scientist Jacques Charles, who performed experiments on gases in the 1780s. The tactics for using this mathematical formula are similar to those for Boyle’s law. To determine an unknown quantity, use algebra to isolate the unknown variable by itself and in the numerator; the units of similar variables must be the same. But we add one more tactic: all temperatures must be expressed in the absolute temperature scale (Kelvin). As a reminder, we review the conversion between the absolute temperature scale and the Celsius temperature scale:
K = °C + 273
where K represents the temperature in kelvins, and °C represents the temperature in degrees Celsius.
Figure 6.3 "Charles’s Law" shows two representations of how Charles’s law works.
Figure 6.3 Charles’s Law
A piston having a certain volume and temperature (left piston) will have twice the volume when its temperature is twice as much (right piston). One can also plot V versus T for a given amount of gas at a certain pressure; such a plot will look like the graph on the right.
Example 5
A sample of gas has an initial volume of 34.8 mL and an initial temperature of 315 K. What is the new volume if the temperature is increased to 559 K? Assume constant pressure and amount for the gas.
Solution
First, we assign the given values to their variables. The initial volume is V1, so V1 = 34.8 mL, and the initial temperature is T1, so T1 = 315 K. The temperature is increased to 559 K, so the final temperature T2 = 559 K. We note that the temperatures are already given in kelvins, so we do not need to convert the temperatures. Substituting into the expression for Charles’s law yields
We solve for V2 by algebraically isolating the V2 variable on one side of the equation. We do this by multiplying both sides of the equation by 559 K (number and unit). When we do this, the temperature unit cancels on the left side, while the entire 559 K cancels on the right side:
The expression simplifies to
By multiplying and dividing the numbers, we see that the only remaining unit is mL, so our final answer is
V2 = 61.8 mL
Does this answer make sense? We know that as temperature increases, volume increases. Here, the temperature is increasing from 315 K to 559 K, so the volume should also increase, which it does.
Test Yourself
If V1 = 3.77 L and T1 = 255 K, what is V2 if T2 = 123 K?
Answer
1.82 L
It is more mathematically complicated if a final temperature must be calculated because the T variable is in the denominator of Charles’s law. There are several mathematical ways to work this, but perhaps the simplest way is to take the reciprocal of Charles’s law. That is, rather than write it as
$V1T1=V2T2$
write the equation as
$T1V1=T2V2$
It is still an equality and a correct form of Charles’s law, but now the temperature variable is in the numerator, and the algebra required to predict a final temperature is simpler.
Example 6
A sample of a gas has an initial volume of 34.8 L and an initial temperature of −67°C. What must be the temperature of the gas for its volume to be 25.0 L?
Solution
Here, we are looking for a final temperature, so we will use the reciprocal form of Charles’s law. However, the initial temperature is given in degrees Celsius, not kelvins. We must convert the initial temperature to kelvins:
−67°C + 273 = 206 K
In using the gas law, we must use T1 = 206 K as the temperature. Substituting into the reciprocal form of Charles’s law, we get
Bringing the 25.0 L quantity over to the other side of the equation, we get
The L units cancel, so our final answer is
T2 = 148 K
This is also equal to −125°C. As temperature decreases, volume decreases, which it does in this example.
Test Yourself
If V1 = 623 mL, T1 = 255°C, and V2 = 277 mL, what is T2?
Answer
235 K, or −38°C
Key Takeaways
• The behavior of gases can be modeled with gas laws.
• Boyle’s law relates a gas’s pressure and volume at constant temperature and amount.
• Charles’s law relates a gas’s volume and temperature at constant pressure and amount.
• In gas laws, temperatures must always be expressed in kelvins.
Exercises
1. Define gas law. What restrictions are there on the units that can be used for the physical properties?
2. What unit of temperature must be used for gas laws?
3. Boyle’s law relates the _____________ of a gas inversely with the ___________ of that gas.
4. Charles’s law relates the _____________ of a gas directly with the ___________ of that gas.
5. What properties must be held constant when applying Boyle’s law?
6. What properties must be held constant when applying Charles’s law?
7. A gas has an initial pressure of 1.445 atm and an initial volume of 1.009 L. What is its new pressure if volume is changed to 0.556 L? Assume temperature and amount are held constant.
8. A gas has an initial pressure of 633 torr and an initial volume of 87.3 mL. What is its new pressure if volume is changed to 45.0 mL? Assume temperature and amount are held constant.
9. A gas has an initial pressure of 4.33 atm and an initial volume of 5.88 L. What is its new volume if pressure is changed to 0.506 atm? Assume temperature and amount are held constant.
10. A gas has an initial pressure of 87.0 torr and an initial volume of 28.5 mL. What is its new volume if pressure is changed to 206 torr? Assume temperature and amount are held constant.
11. A gas has an initial volume of 638 mL and an initial pressure of 779 torr. What is its final volume in liters if its pressure is changed to 0.335 atm? Assume temperature and amount are held constant.
12. A gas has an initial volume of 0.966 L and an initial pressure of 3.07 atm. What is its final pressure in torr if its volume is changed to 3,450 mL? Assume temperature and amount are held constant.
13. A gas has an initial volume of 67.5 mL and an initial temperature of 315 K. What is its new volume if temperature is changed to 244 K? Assume pressure and amount are held constant.
14. A gas has an initial volume of 2.033 L and an initial temperature of 89.3 K. What is its volume if temperature is changed to 184 K? Assume pressure and amount are held constant.
15. A gas has an initial volume of 655 mL and an initial temperature of 295 K. What is its new temperature if volume is changed to 577 mL? Assume pressure and amount are held constant.
16. A gas has an initial volume of 14.98 L and an initial temperature of 238 K. What is its new temperature if volume is changed to 12.33 L? Assume pressure and amount are held constant.
17. A gas has an initial volume of 685 mL and an initial temperature of 29°C. What is its new temperature if volume is changed to 1.006 L? Assume pressure and amount are held constant.
18. A gas has an initial volume of 3.08 L and an initial temperature of −73°C. What is its new volume if temperature is changed to 104°C? Assume pressure and amount are held constant.
Answers
1. A gas law is a simple mathematical formula that allows one to predict the physical properties of a gas. The units of changing properties (volume, pressure, etc.) must be the same.
2. pressure; volume
3. amount of gas and temperature
4. 2.62 atm
5. 50.3 L
6. 1.95 L
7. 52.3 mL
8. 260 K
9. 444 K, or 171°C
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http://mathhelpforum.com/advanced-math-topics/132637-complex-analysis-maximum-principle-print.html
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# Complex Analysis-Maximum Principle
Printable View
• March 7th 2010, 11:06 PM
WannaBe
Complex Analysis-Maximum Principle
Let f be an analytic function at the open unit circle and continous at |z|=1.
Prove: If f(z)=1 on the upper half of the unit circle
( for $z=e^{i\gamma}$ where $0<= \gamma <=\pi$
then f is a constant at the unit circle...
Thanks !
• March 8th 2010, 06:38 AM
Opalg
Quote:
Originally Posted by WannaBe
Let f be an analytic function at the open unit circle and continous at |z|=1.
Prove: If f(z)=1 on the upper half of the unit circle
( for $z=e^{i\gamma}$ where $0<= \gamma <=\pi$
then f is a constant at the unit circle...
Using a conformal map from the unit disk to the upper half-plane that takes the upper part of the unit circle to the positive real axis, you can replace the problem by one in which g is an analytic function on the upper half-plane, continuous on the real axis and such that g(z)=1 on the positive real axis. Then apply the reflection principle, defining $g(\overline{z}) = \overline{g(z)}$, to extend g to an analytic function on the cut plane $\mathbb{C}\setminus(-\infty,0]$. This extended function is constant on the positive real axis and hence constant everywhere.
• March 8th 2010, 07:48 AM
WannaBe
Wow thanks a lot!
All times are GMT -8. The time now is 11:06 AM.
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Bayesian inference
Bayesian inference is statistical inference in which probabilities are interpreted not as frequencies or proportions or the like, but rather as degrees of belief. The name comes from the frequent use of the Bayes' theorem in this discipline.
Bayes' theorem is named after the Reverend Thomas Bayes. However, it is not clear that Bayes would endorse the very broad interpretation of probability now called "Bayesian". This topic is treated at greater length in the article Thomas Bayes.
Contents
Evidence and the scientific method
Bayesian statisticians claim that methods of Bayesian inference are a formalisation of the scientific method involving collecting evidence which points towards or away from a given hypothesis. There can never be certainty, but as evidence accumulates, the degree of belief in a hypothesis changes; with enough evidence it will often become very high (almost 1) or very low (near 0).
As an example, this reasoning might be
The sun has risen and set for billions of years. The sun has set tonight. With very high probability, the sun will rise tomorrow.
Bayesian statisticians believe that Bayesian inference is the most suitable logical basis for discriminating between conflicting hypotheses. It uses an estimate of the degree of belief in a hypothesis before the advent of some evidence to give a numerical value to the degree of belief in the hypothesis after the advent of the evidence. Because it relies on subjective degrees of belief, however, it is not able to provide a completely objective account of induction. See scientific method.
Bayes' theorem also provides a method for adjusting degrees of belief in the light of new information. Bayes' theorem is
$P(H_0|E) = \frac{P(E|H_0)\;P(H_0)}{P(E)}.$
For our purposes, H0 can be taken to be a hypothesis which may have been developed ab initio or induced from some preceding set of observations, but before the new observation or evidence E.
• The term P(H0) is called the prior probability of H0.
• The term P(E | H0) is the conditional probability of seeing the observation E given that the hypothesis H0 is true; as a function of H0 given E, it is called the likelihood function.
• The term P(E) is called the marginal probability of E; it is a normalizing constant and can be calculated as the sum of all mutually exclusive hypotheses $\sum P(E|H_i) P(H_i)$.
• The term P(H0 | E) is called the posterior probability of H0 given E.
The scaling factor P(E | H0) / P(E) gives a measure of the impact that the observation has on belief in the hypothesis. If it is unlikely that the observation will be made unless the particular hypothesis being considered is true, then this scaling factor will be large. Multiplying this scaling factor by the prior probability of the hypothesis being correct gives a measure of the posterior probability of the hypothesis being correct given the observation.
The keys to making the inference work is the assigning of the prior probabilities given to the hypothesis and possible alternatives, and the calculation of the conditional probabilities of the observation under different hypotheses.
Some Bayesian statisticians believe that if the prior probabilities can be given some objective value, then the theorem can be used to provide an objective measure of the probability of the hypothesis. But to others there is no clear way in which to assign objective probabilities. Indeed, doing so appears to require one to assign probabilities to all possible hypotheses.
Alternately, and more often, the probabilities can be taken as a measure of the subjective degree of belief on the part of the participant, and to restrict the potential hypotheses to a constrained set within a model. The theorem then provides a rational measure of the degree to which some observation should alter the subject's belief in the hypothesis. But in this case the resulting posterior probability remains subjective. So the theorem can be used to rationally justify belief in some hypothesis, but at the expense of rejecting objectivism.
It is unlikely that two individuals will start with the same subjective degree of belief. Supporters of Bayesian method argue that even with very different assignments of prior probabilities sufficient observations are likely to bring their posterior probabilities closer together. This assumes that they do not completely reject each other's initial hypotheses; and that they assign similar conditional probabilities. Thus Bayesian methods are useful only in situations in which there is already a high level of subjective agreement.
In many cases, the impact of observations as evidence can be summarised in a likelihood ratio, as expressed in the law of likelihood. This can be combined with the prior probability to reflect the original degree of belief and any earlier evidence already taken into account. For example, if we have the likelihood ratio
$\Lambda = \frac{L(H_0\mid E)}{L(\mbox{not } H_0|E)} = \frac{P(E \mid H_0)}{P(E \mid \mbox{not } H_0)}$
then we can rewrite Bayes' theorem as
$P(H_0|E) = \frac{\Lambda P(H_0)}{\Lambda P(H_0) + P(\mbox{not } H_0)} = \frac{P(H_0)}{P(H_0) +\left(1-P(H_0)\right)/\Lambda }.$
With two independent pieces of evidence E1 and E2, one possible approach is to move from the prior to the posterior probability on the first evidence and then use that posterior as a new prior and produce a second posterior with the second piece of evidence; an arithmetically equivalent alternative is to multiply the likelihood ratios. So
if $P(E_1, E_2 | H_0) = P(E_1 | H_0) \times P(E_2 | H_0)$
and $P(E_1, E_2 | \mbox{not }H_0) = P(E_1 | \mbox{not }H_0) \times P(E_2 | \mbox{not }H_0)$
then $P(H_0|E_1, E_2) = \frac{\Lambda_1 \Lambda_2 P(H_0)}{\Lambda_1 \Lambda_2 P(H_0) + P(\mbox{not } H_0)}$,
and this can be extended to more pieces of evidence.
Before a decision is made, the loss function also needs to be considered to reflect the consequences of making an erroneous decision.
Simple examples of Bayesian inference
From which bowl is the cookie?
To illustrate, suppose there are two bowls full of cookies. Bowl #1 has 10 chocolate chip and 30 plain cookies, while bowl #2 has 20 of each. Our friend Fred picks a bowl at random, and then picks a cookie at random. We may assume there is no reason to believe Fred treats one bowl differently from another, likewise for the cookies. The cookie turns out to be a plain one. How probable is it that Fred picked it out of bowl #1?
Intuitively, it seems clear that the answer should be more than a half, since there are more plain cookies in bowl #1. The precise answer is given by Bayes' theorem. Let H1 corresponds to bowl #1, and H2 to bowl #2. It is given that the bowls are identical from Fred's point of view, thus P(H1) = P(H2), and the two must add up to 1, so both are equal to 0.5. The datum D is the observation of a plain cookie. From the contents of the bowls, we know that P(D | H1) = 30/40 = 0.75 and P(D | H2) = 20/40 = 0.5. Bayes' formula then yields
$\begin{matrix} P(H_1 | D) &=& \frac{P(H_1) \cdot P(D | H_1)}{P(H_1) \cdot P(D | H_1) + P(H_2) \cdot P(D | H_2)} \\ \\ \ & =& \frac{0.5 \times 0.75}{0.5 \times 0.75 + 0.5 \times 0.5} \\ \\ \ & =& 0.6. \end{matrix}$
Before observing the cookie, the probability that Fred chose bowl #1 is the prior probability, P(H1), which is 0.5. After observing the cookie, we revise the probability to P(H1|D), which is 0.6.
False positives in a medical test
False positives are a problem in any kind of test: no test is perfect, and sometimes the test will incorrectly report a positive result. For example, if a test for a particular disease is performed on a patient, then there is a chance (usually small) that the test will return a positive result even if the patient does not have the disease. The problem lies, however, not just in the chance of a false positive prior to testing, but determining the chance that a positive result is in fact a false positive. As we will demonstrate, using Bayes' theorem, if a condition is rare, then the majority of positive results may be false positives, even if the test for that condition is (otherwise) reasonably accurate.
Suppose that a test for a particular disease has a very high success rate:
• if a tested patient has the disease, the test accurately reports this, a 'positive', 99% of the time (or, with probability 0.99), and
• if a tested patient does not have the disease, the test accurately reports that, a 'negative', 95% of the time (i.e. with probability 0.95).
Suppose also, however, that only 0.1% of the population have that disease (i.e. with probability 0.001). We now have all the information required to use Bayes' theorem to calculate the probability that, given the test was positive, that it is a false positive.
Let A be the event that the patient has the disease, and B be the event that the test returns a positive result. Then, using the second alternative form of Bayes' theorem (above), the probability of a true positive is
$\begin{matrix}P(A|B) &= &\frac{0.99 \times 0.001}{0.99\times 0.001 + 0.05\times 0.999}\, ,\\ ~\\ &\approx &0.019\, .\end{matrix}$
and hence the probability of a false positive is about (1 − 0.019) = 0.981.
Despite the apparent high accuracy of the test, the incidence of the disease is so low (one in a thousand) that the vast majority of patients who test positive (98 in a hundred) do not have the disease. (Nonetheless, the proportion of patients who tested positive who do have the disease is 20 times the proportion before we knew the outcome of the test! Thus the test is not useless, and re-testing may improve the reliability of the result.) In particular, a test must be very reliable in reporting a negative result when the patient does not have the disease, if it is to avoid the problem of false positives. In mathematical terms, this would ensure that the second term in the denominator of the above calculation is small, relative to the first term. For example, if the test reported a negative result in patients without the disease with probability 0.999, then using this value in the calculation yields a probability of a false positive of roughly 0.1(1-(0.99x0.001/(0.99x0.001+0.001x0.999))) = 0.050.
In this example, Bayes' theorem helps show that the accuracy of tests for rare conditions must be very high in order to produce reliable results from a single test, due to the possibility of false positives. (The probability of a 'false negative' could also be calculated using Bayes' theorem, to completely characterise the possible errors in the test results.)
In the courtroom
Bayesian inference can be used in a court setting by an individual juror to coherently accumulate the evidence for and against the guilt of the defendant, and to see whether, in totality, it meets their personal threshold for 'beyond a reasonable doubt'.
• Let G be the event that the defendant is guilty.
• Let E be the event that the defendant's DNA matches DNA found at the crime scene.
• Let p(E | G) be the probability of seeing event E assuming that the defendant is guilty. (Usually this would be taken to be unity.)
• Let p(G | E) be the probability that the defendant is guilty assuming the DNA match event E
• Let p(G) be the juror's personal estimate of the probability that the defendant is guilty, based on the evidence other than the DNA match. This could be based on his responses under questioning, or previously presented evidence.
Bayesian inference tells us that if we can assign a probability p(G) to the defendant's guilt before we take the DNA evidence into account, then we can revise this probability to the conditional probability p(G | E), since
p(G | E) = p(G) p(E | G) / p(E)
Suppose, on the basis of other evidence, a juror decides that there is a 30% chance that the defendant is guilty. Suppose also that the forensic evidence is that the probability that a person chosen at random would have DNA that matched that at the crime scene was 1 in a million, or 10-6.
The event E can occur in two ways. Either the defendant is guilty (with prior probability 0.3) and thus his DNA is present with probability 1, or he is innocent (with prior probability 0.7) and he is unlucky enough to be one of the 1 in a million matching people.
Thus the juror could coherently revise his opinion to take into account the DNA evidence as follows:
p(G | E) = (0.3 × 1.0) /(0.3 × 1.0 + 0.7 × 10-6) = 0.99999766667.
The benefit of adopting a Bayesian approach is that it gives the juror a formal mechanism for combining the evidence presented. The approach can be applied successively to all the pieces of evidence presented in court, with the posterior from one stage becoming the prior for the next.
The juror would still have to have a prior for the guilt probability before the first piece of evidence is considered. It has been suggested that this could be the guilt probability of a random person of the appropriate sex taken from the town where the crime occurred. Thus, for a crime committed by a adult male in a town containing 50,000 adult males the appropriate initial prior probability might be 1/50,000.
For the purpose of explaining Bayes' theorem to jurors, it will usually be appropriate to give it in the form of betting odds rather than probabilities, as these are more widely understood. In this form Bayes' theorem states that
Posterior odds = prior odds x Bayes factor
In the example above, the juror who has a prior probability of 0.3 for the defendant being guilty would now express that in the form of odds of 3:7 in favour of the defendant being guilty, the Bayes factor is one million, and the resulting posterior odds are 3 million to 7 or about 429,000 to one in favour of guilt.
In the United Kingdom, Bayes' theorem was explained to the jury in the odds form by a statistician expert witness in the rape case of Regina versus Denis John Adams. A conviction was secured but the case went to Appeal, as no means of accumulating evidence had been provided for those jurors who did not want to use Bayes' theorem. The Court of Appeal upheld the conviction and gave their opinion that "To introduce Bayes' Theorem, or any similar method, into a criminal trial plunges the Jury into inappropriate and unnecessary realms of theory and complexity, deflecting them from their proper task." No further appeal was allowed and the issue of Bayesian assessment of forensic DNA data remains controversial.
Gardner-Medwin argues that the criterion on which a verdict in a criminal trial should be based is not the probability of guilt, but rather the probability of the evidence, given that the defendant is innocent. He argues that if the posterior probability of guilt is to be computed by Bayes' theorem, the prior probability of guilt must be known. This will depend on the incidence of the of the crime and this is a odd piece of evidence to consider in a criminal trial. Consider the following 3 propositions:
A: The known facts and testimony could have arisen if the defendant is guilty,
B: The known facts and testimony could have arisen if the defendant is innocent,
C: The defendant is guilty.
Gardner-Medwin argues that the jury should believe both A and not-B in order to convict. A and not-B implies the truth of C, but the reverse is not true. It is possible that B and C are both true, but in this case he argues that a jury should acquit, even though they know that they are probably acquitting a guilty person.
Another court case in which probabilistic arguments played some role was the Howland Will forgery trial.
Search theory
In May 1968 the US nuclear submarine USS Scorpion (SSN-589) failed to arrive as expected at her home port of Norfolk, Virginia. The US Navy was convinced that the vessel had been lost off the Eastern seaboard but an extensive search failed to discover the wreck. The US Navy's deep water expert, John Craven, believed that it was elsewhere and he organised a search south west of the Azores based on a controversial approximate triangulation by hydrophones. He was allocated only a single ship, the USNS Mizar, and he took advice from a firm of consultant mathematicians in order to maximise his resources. A Bayesian search methodology was adopted. Experienced submarine commanders were interviewed to construct hypotheses about what could have caused the loss of the Scorpion. The sea area was divided up into grid squares and a probability assigned to each square, under each of the hypotheses, to give a number of probability grids, one for each hypothesis. These were then added together to produce an overall probability grid. The probability attached to each square was then the probability that the wreck was in that square. A second grid was constructed with probabilities that represented the probability of successfully finding the wreck if that square were to be searched and the wreck were to be actually there. This was a known function of water depth. The result of combining this grid with the previous grid is a grid which gives the probability of finding the wreck in each grid square of the sea if it were to be searched. This sea grid was systematically searched in a manner which started with the high probability regions first and worked down to the low probability regions last. Each time a grid square was searched and found to be empty its probability was reassessed using Bayes' theorem. This then forced the probabilities of all the other grid squares to be reassessed (upwards), also by Bayes' theorem. The use of this approach was a major computational challenge for the time but it was eventually successful and the Scorpion was found in October of that year. Suppose a grid square has a probability p of containing the wreck and that the probability of successfully detecting the wreck if it is there is q. If the square is searched and no wreck is found then, by Bayes, the revised probability of the wreck being in the square is given by
$p' = \frac{p(1-q)}{(1-p)+p(1-q)}.$
More mathematical examples
Naive Bayes classifier
See: naive Bayes classifier.
Posterior distribution of the binomial parameter
In this example we consider the computation of the posterior distribution for the binomial parameter. This is the same problem considered by Bayes in Proposition 9 of his essay.
We are given m observed successes and n observed failures in a binomial experiment. The experiment may be tossing a coin, drawing a ball from an urn, or asking someone their opinion, among many other possibilities. What we know about the parameter (let's call it a) is stated as the prior distribution, p(a).
For a given value of a, the probability of m successes in m+n trials is
$p(m,n|a) = \begin{pmatrix} n+m \\ m \end{pmatrix} a^m (1-a)^n.$
Since m and n are fixed, and a is unknown, this is a likelihood function for a. From the continuous form of the law of total probability we have
$p(a|m,n) = \frac{p(m,n|a)\,p(a)}{\int_0^1 p(m,n|a)\,p(a)\,da} = \frac{\begin{pmatrix} n+m \\ m \end{pmatrix} a^m (1-a)^n\,p(a)} {\int_0^1 \begin{pmatrix} n+m \\ m \end{pmatrix} a^m (1-a)^n\,p(a)\,da}.$
For some special choices of the prior distribution p(a), the integral can be solved and the posterior takes a convenient form. In particular, if p(a) is a beta distribution with parameters m0 and n0, then the posterior is also a beta distribution with parameters m+m0 and n+n0.
A conjugate prior is a prior distribution, such as the beta distribution in the above example, which has the property that the posterior is the same type of distribution.
What is "Bayesian" about Proposition 9 is that Bayes presented it as a probability for the parameter a. That is, not only can one compute probabilities for experimental outcomes, but also for the parameter which governs them, and the same algebra is used to make inferences of either kind. Interestingly, Bayes actually states his question in a way that might make the idea of assigning a probability distribution to a parameter palatable to a frequentist. He supposes that a billiard ball is thrown at random onto a billiard table, and that the probabilities p and q are the probabilities that subsequent billiard balls will fall above or below the first ball. By making the binomial parameter a depend on a random event, he cleverly escapes a philosophical quagmire that was an issue he most likely was not even aware of.
Computer applications
Bayesian inference has applications in artificial intelligence and expert systems. Bayesian inference techniques have been a fundamental part of computerized pattern recognition techniques since the late 1950s. There is also an ever growing connection between Bayesian methods and simulation Monte Carlo techniques since complex models cannot be processed in closed form by a Bayesian analysis, while the graphical model structure inherent to all statistical models, even the most complex ones, allows for efficient simulation algorithms like the Gibbs sampling and other Metropolis-Hastings algorithm schemes.
As a particular application of statistical classification, Bayesian inference has been used in recent years to develop algorithms for identifying unsolicited bulk e-mail spam. Applications which make use of Bayesian inference for spam filtering include Bogofilter, SpamAssassin and Mozilla. Spam classification is treated in more detail in the article on the naive Bayes classifier.
In some applications fuzzy logic is an alternative to Bayesian inference. Fuzzy logic and Bayesian inference, however, are mathematically and semantically not compatible: You cannot, in general, understand the degree of truth in fuzzy logic as probability and vice versa.
References
• On-line textbook: Information Theory, Inference, and Learning Algorithms, by David MacKay, has many chapters on Bayesian methods, including introductory examples; compelling arguments in favour of Bayesian methods (in the style of Edwin Jaynes); state-of-the-art Monte Carlo methods, message-passing methods , and variational methods; and examples illustrating the intimate connections between Bayesian inference and data compression.
• Jaynes, E.T. (1998) Probability Theory : The Logic of Science.
• Bretthorst, G. Larry, 1988, Bayesian Spectrum Analysis and Parameter Estimation in Lecture Notes in Statistics, 48, Springer-Verlag, New York, New York
• Berger, J.O. (1999) Statistical Decision Theory and Bayesian Statistics. Second Edition. Springer Verlag, New York. ISBN 0-387-96098-8 and also ISBN 3-540-96098-8.
• O'Hagan, A. and Forster, J. (2003) Kendall's Advanced Theory of Statistics, Volume 2B: Bayesian Inference. Arnold, New York. ISBN 0-340-52922-9.
• Robert, C.P. (2001) The Bayesian Choice. Springer Verlag, New York.
• Lee, Peter M. Bayesian Statistics: An Introduction. Second Edition. (1997). ISBN 0-340-67785-6.
• Dawid, A.P. and Mortera, J. Coherent analysis of forensic identification evidence. Journal of the Royal Statistical Society, Series B, 58,425-443.
• Foreman, L.A; Smith, A.F.M. and Evett, I.W. (1997). Bayesian analysis of deoxyribonucleic acid profiling data in forensic identification applications (with discussion). Journal of the Royal Statistical Society, Series A, 160, 429-469.
• Robertson, B. and Vignaux, G.A. (1995) Interpreting Evidence: Evaluating Forensic Science in the Courtroom. John Wiley and Sons. Chichester.
• Gardner-Medwin, A. What probability should the jury address?. Significance. Volume 2, Issue 1, March 2005
See also
Bolstad, William M. (2004) Introduction to Bayesian Statistics, John Wiley ISBN 0-471-27020-2
External links
• On-line textbook: Information Theory, Inference, and Learning Algorithms, by David MacKay, has many chapters on Bayesian methods, including introductory examples; compelling arguments in favour of Bayesian methods; state-of-the-art Monte Carlo methods, message-passing methods , and variational methods; and examples illustrating the intimate connections between Bayesian inference and data compression.
• Cause, chance and Bayesian statistics, to facilitate understanding Bayesian statistics. The statistical theory developed by Thomas Bayes enables analysis of conditional and marginal probabilities. Bayesian statistics enables logical inference
• Naive Bayesian learning paper
• A Tutorial on Learning With Bayesian Networks
• Paul Graham. "A Plan for Spam" (exposition of a popular approach for spam classification)
• Commentary on Regina versus Adams
03-10-2013 05:06:04
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http://mathoverflow.net/questions/47742/ellipse-naturally-associated-with-a-polygon
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## Ellipse naturally associated with a polygon
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My colleagues and I have stumbled onto a way to associate an ellipse, or equivalently a positive definite symmetric matrix, to a polygon that is different from other better known ways. We want to know if anyone has ever seen this before. Before describing this, I do want to note that there are other well-known ways of associating an ellipse to a polygon, including the matrix of second moments of the uniform distribution supported on the interior of the polygon and the so-called John ellipses, which are the ellipse of largest volume inscribed in the polygon and the one of smallest volume circumscribing the polygon.
Here is the ellipse we found: Given a polygon $P$ that contains the origin in its interior, let $\ell_1, \dots, \ell_n$ denote the lines that contain the sides of $P$. For each $i$, let $n_i$ denote the unit vector orthogonal to $\ell_i$, $h_i$ be the distance from the origin to the line $\ell_i$, and $s_i$ be the length of the side lying in $\ell_i$. For each $v \in R^2$, define
$q(v) = \sum_{i=1}^n (v\cdot n_i)^2\frac{s_i}{h_i}.$
Let $E_P = { q(v) \le 1}.$ The ellipse $E_P$ does not behave particularly nicely under translations of $P$. It however, behaves nicely under linear transformations. In particular, if $A$ is an invertible linear transformation with determinant $1$, then $E_{AP} = AE_P$. It is scale-invariant in that $E_{tP} = E_P$ for any $t > 0$.
The ellipse $E_P$ can in fact be defined without using the inner product and is a linear invariant of the polygon $P$. Its definition can be extended to any body $P$ containing the origin with a sufficiently regular boundary. It is an example of a matrix-valued valuation on the set of all convex bodies that contain the origin in their interior. Its volume is maximized if and only if the body $P$ is itself an ellipsoid centered at the origin.
Does anyone recognize this association of an ellipse to a polygon? I would be grateful for any information or references.
EDIT: Vladimir's comments below, especially the first sentence, indicate clearly that I omitted something crucial. The definition of the ellipsoid depends on the choice of the volume form in the ambient vector space (and the dual volume form in the dual vector space). Changing the volume form will rescale the ellipsoid by a factor. However, if you fix the volume form, then the ellipsoid is invariant under rescaling of the convex body.
This observation, however, makes the existence of a scale-invariant ellipsoid associated with the polygon much less surprising.
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## 4 Answers
Maybe it would have been more convincing if you had given the definition of $q_K$ in a unimodularly equivariant way from the beginning. That's not at all hard to do, and it's also easy to see how to create many more such (and how to create many more in all dimensions).
For example, suppose $K\subset V$ has vertices $v_0,v_1,\ldots,v_n=v_0$ (cyclically ordered counterclockwise, say). Let $\alpha_i\in V^\ast$ be the unique element that satisfies $\alpha_i(v_i)=\alpha_i(v_{i+1})=1$. Then $$q_K = \sum_{i=0}^{n-1} \Omega(v_i,v_{i+1})\ {\alpha_i}^2$$ (where $\Omega$ is the area form.) This is clearly unimodularly equivariant with respect to $K$ and, since, for $K' = tK$ (with $t>0$), one has $v'_i = tv_i$ and $\alpha_i' = (1/t)\alpha_i$, it follows that $q_{tK} = q_K$. Of course, anything like this would have worked. For example, you could have taken $$\tilde q_K = Area_\Omega(K)\left( \sum_{i=0}^{n-1} {\alpha_i}^2\right),$$ and this would also have had the same equivariance property.
In dimension $n$, I think that the right formula would be to define, for each face $F$ of $K$, the element $\alpha_F\in V^\ast$ to be the linear function that equals $1$ on $F$, let $\Omega(F)$ denote the volume of the cone with vertex $0\in V$ whose base is $F$, and then set $$q_K = \sum_{F\in\mathcal{F}(K)} \Omega(F)\ {\alpha_F}^2.$$ If you want it to be invariant under scaling, you should take $$q_K = \sum_{F\in\mathcal{F}(K)} \Omega(F)^{2/n}\ {\alpha_F}^2.$$ Perhaps, better, though, would be to take $$q_K = Vol_\Omega(K)^{(2-n)/n}\ \left(\sum_{F\in\mathcal{F}(K)} \Omega(F)\ {\alpha_F}^2\right),$$ since this is also invariant under subdivision of the faces of $K$.
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Robert, you're right. I don't normally work directly with polytopes, and it's a little trickier to write down a integral that is obviously linearly equivariant. I do explain how to do this in the following paper: deaneyang.com/papers/affine_survey.pdf – Deane Yang May 7 2012 at 18:53
Robert, am I happy you jumped in. I have never seen before the ellipsoids you define. If I understand correctly, most of them are defined only for polytopes and blow up if you take a limit to a continuous body. I knew of a complementary family that is defined for smooth convex bodies but vanishes for polytopes. Your family (defined by raising $\Omega(F)$ to powers less than 1) is well defined for polytopes but blow up (I believe) for polytopes. In particular, I don't see any way to define them using an integral (which is what I normally do). – Deane Yang May 7 2012 at 19:16
Correction: "blow up for smooth convex bodies" (not polytopes) – Deane Yang May 7 2012 at 19:36
The last formula, with $n = 2$, is exactly the one described in my question. – Deane Yang May 7 2012 at 23:00
@Deane: Actually, when $n=2$, it's one-half the one you gave (and also one-half the first unimodularly equivariant formula that I gave). Yes, the other formulae (besides the last one) don't have any good convergence properties if you take a sequence of polygons that converge to some convex curve. My point was just that there are lots of ways to define unimodularly equivariant ellipsoids associated to a polygon. In fact, there are lots of ways to define such ellipsoids that are equivariant under the full unimodular affine group (including translations). Have you asked Calabi your question? – Robert Bryant May 9 2012 at 12:20
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I am confused: is this not the construction in:
MR1781476 (2001j:52011) Lutwak, Erwin(1-PINY); Yang, Deane(1-PINY); Zhang, Gaoyong(1-PINY) A new ellipsoid associated with convex bodies. Duke Math. J. 104 (2000), no. 3, 375–390. 52A40 (52A39)
If so, why ask now :)
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1
Igor, you are absolutely right. We defined this construction in any dimension in the paper cited. But we only recently noticed that in the plane the ellipse is invariant under rescaling of the body. We were curious about whether in dimension 2 the ellipse was already known to someone. – Deane Yang Nov 30 2010 at 2:08
(Just a question, not an answer.) @Deane: Does the ellipse have a clear geometric relationship to the polygon? I believe it is always a circle for a centered regular polygon. For the equilateral triangle below, I compute $$3 \sqrt{3} (x^2 + y^2 ) = 1$$ and for the right triangle, centered as illustrated, $$6 x^2 + 4xy + 6y^2 = 1 \;.$$
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The geometric relationship is expressed by the property that $E_{AP} = AE_P$ for any polygon $P$ and determinant 1 linear transformation $A$. In particular, if $P$ is invariant under a rotation of an angle other than 180 degrees, this implies that the ellipse is also invariant by the same rotation and therefore must be a circle. – Deane Yang Nov 30 2010 at 16:54
One more comment: But among all shapes $P$ such that $E_P$ is a circle, the area enclosed by $E_P$ is maximized only if $P$ is itself a circle. But what's a little weird here is that the size of $E_P$, as well as its shape, is being determined by the shape but not the size of $P$. – Deane Yang Nov 30 2010 at 16:57
@Deane: Thank you! This is intriguing... – Joseph O'Rourke Nov 30 2010 at 16:58
Dear Dean, Dear All,
This is hopefully the last (4th) edition of my answer. I spend few hours in calculations which could be dismissed would I think a bit before doing them. I believe that now I understand what happens and what most of you probably knew before.
First of all, there exists no canonical construction $K\mapsto E_K$ of an ellipse by a convec body (I understand canonically as depends only on the linear structure and on $K$'') such that it satisfies $E_{tK} =E_K$. The reason is that this would imply the existence of a canonical euclidean structure in a 2-dimensional space equipped with a linear complex structure, which is wrong because the group of the linear transformations preserving a complex structure is bigger than the group of linear transformations preserving a euclidean structure.
Indeed, consider a two-dimensional linear complex structure on $\mathbb{R}^2$, i.e., fix a matrix $J$ such that $J^2= -1$. The group of linear transformations preserving the complex structure evidently contains the scalings $(x,y)\mapsto \textrm{const} \cdot (x,y)$ and therefore can not preserve any euclidean structure.
Suppose there exists a canonical construction of an ellipsoid $E_K$ by a convex body $K$ such that it behaves as follows w.r.t. to linear transformations: for every linear transformation $A$ with $det(A)= 1$ we have $A E_K = E_{AK}$; for every scaling $S:\mathbb{R}^2 \to \mathbb{R}^2$, $(x,y)\stackrel{S}\mapsto \textrm{const} \cdot (x,y)$ we have $E_{SK}= E_K$. Then, we can construct a canonical scalar product on our $\mathbb{R}^2$. Indeed, having $J$, we have a canonical notion of a circle centered at $\vec 0$. More precisely, in every real linear 2dim-space we have the notion of an ellipse. A circle around $\vec 0$ is an ellipse which is $J-$invariant. As you see, in order to define an ellipse we used the linear structure of $\mathbb{R}^2$ and the complex structure $J$ only, so any linear transformation that preserves $J$, in particular every scaling $(x,y)\stackrel{S}\mapsto \textrm{const} \cdot (x,y)$ takes a circle centered at $\vec 0$ to a circle centered at $\vec 0$. round
Now suppose your construction $K\mapsto E_K$ exists. As $K$, take the round ball centered at the origin. The freedom in choosing such a convex body is its scaling $K\mapsto \textrm{const}\cdot K$; if the construction do not depend on scaling we obtain that the freedom in choosing such a convex body does not affect the ellipse $E_K$ (which must also be a circle because of the symmetry) we obtain a canonical inner product on $(\mathbb{R}^2, J)$. There is no such product though, by the reasons I explained before: the group of complex transformations is bigger than the group of the orthogonal transformations.
Thus, there is no hope to construct a scaling-equivarinat and linear-transformations-invariant ellipse by a convex body in $\mathbb{R}^2$, so it must something be wrong with the announced properties of the construction from the question. I should confess that first I thought that the problem is with the invariance of the construction w.r.t. linear transformations, and bothered Deane and you all with attempts of counterexamples.
No, the problem is not with the behavior of the construction w.r.t. the linear transformations! The problem is that the construction is NOT CANONICAL. Indeed, it depends on the choice of the euclidean structure in the space: if you multiply the euclidean structure by a constant, the resulting ellipsoid will be the initial divided by the square of this constant!
Actually, since the construction is indeed invariant w.r.t. to linear transformations with determinant $1$, one can think that the information we need from the euclidean structure is its volume form only. This was pointed out by Deane to me in one of his comment, with a hint that normally should be sufficient for me to understand everything without doing the calculations. It is also stays inexplicit in the edited version of Deane's question.
By the way, the original'' version of the construction in the paper MR1781476 (2001j:52011) Lutwak, Erwin(1-PINY); Yang, Deane(1-PINY); Zhang, Gaoyong(1-PINY) A new ellipsoid associated with convex bodies. Duke Math. J. 104 (2000), no. 3, 375–390. 52A40 (52A39) does have additional term inside, which makes the construction to be volume-form-independent, but the resulting ellipse does not have the desired property $E_{tK}= E_K$ anymore.
Now, if you do not require the property $E_{tK}= E_K$, there exist tons of canonical constructions of an ellipse by a convex body. You even can do these construction the whole-affine-group-invariant, by first moving the body such that its barycenter is located in the origin of the coordinate system you are doing the construction starting from.
[Edit record --1st attempt: added the explanation why it could not worked at all and corrected the counterexample] [Edit record -- 2nd attempt: the previous counterexample did not work, now the new one that works is there] [Edit record -- 3 attempt:One more unsuccesful counterexample] [Edit record: -- 4th attempt: explanation that the construction is not invariant]
-
Vladimir, thanks for your response. I will take a closer look when I get a chance. It could be that my description above isn't exactly right. If you're interested, you can look at the paper mentioned by Igor Rivin. Except possibly for a dilation factor, the ellipsoid is definitely equivariant under linear transformations. If you normalize the ellipsoid by a scale factor equal to the right power of the volume of the body, then it is equivariant, period. – Deane Yang May 6 2012 at 20:16
Vladimir, when I do the calculation, I get $q(v) = 4[(v_1)^2 + (v_2)^2]$ for the square and $q(v) = 2[(v_1)^2 + 4(v_2)^2]$ for the image of the square under the map you give. This is consistent with my description of the ellipsoid. The transformation matrix can be factored into a scalar multiple of the diagonal and a matrix with determinant 1. The scalar multiple of the diagonal does not change the ellipsoid, but the matrix of determinant 1 does. Note that the volume of the new ellipsoid is the same as the volume of the original one. – Deane Yang May 7 2012 at 7:30
Dean, I edited my answer'' without seeing your comment. I will redo your calculations. But in the newer version of my answer there is an explanation why you construction can not exist at all. I also changes the transformation in my counterexample slightly. Could you please look at it? – Vladimir S Matveev May 7 2012 at 8:31
I redid the calculation for your new transformation. I get $q_{AK}(v_1, v_2) = 16(v_1)^2 + (v_2)^2.$ So $q_{AK}(Av) = 16(v_1/2)^2 + (2v_2)^2 = 4(v_1)^2 + 4(v_2)^2 = q_K(v_1, v_2)$. – Deane Yang May 7 2012 at 8:50
1
No, non of my current understanding contradicts your comments and would I read them more carefully and spend more time thinking about them I could probably save few hours of calculations. Would you say directly that the construction is not canonic it would be easier for me to understand the matter. Anyway it is a nice construction and it was a fun to think about it :-) – Vladimir S Matveev May 8 2012 at 16:42
show 19 more comments
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http://mathhelpforum.com/calculus/59515-volumes-revolution.html
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# Thread:
1. ## Volumes of Revolution
I have this question which is rather interesting and quite hard IMO.
I have to find the volumes of revolution generated by the regions A and B when they are each rotated about: the x axis and the y axis
If the equation was from a to b,then it means I need to find the volume of revolution for:
• x = a to b, rotated around the x axis, which would be:
• y = a to b, roated around the y axis, which would be:
are those right?
I also have to find this though and this is what I need help with:
• x = a to b, rotated around the y axis
• y = a to b, roated around the x axis
I dont know what to do for this though
2. Originally Posted by jhomie
I have this question which is rather interesting and quite hard IMO.
I have to find the volumes of revolution generated by the regions A and B when they are each rotated about: the x axis and the y axis
If the equation was from a to b,then it means I need to find the volume of revolution for:
• x = a to b, rotated around the x axis, which would be:
• y = a to b, roated around the y axis, which would be:
Those are correct. The other two involve volumes using cylindrical shells which have the formula $V=2\pi\int_a^b xf(x)dx$ or $2\pi\int_a^b yg(y)dy$ so:
$V_y=2\pi\int_a^b x^{n+1}dx$
$V_x=2\pi\int_a^b y^{1+1/n}dy$
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http://mathoverflow.net/questions/17031/modular-curves-of-genus-zero-and-normal-forms-for-elliptic-curves/57673
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## Modular curves of genus zero and normal forms for elliptic curves
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This is maybe the first question I actually need to know the answer to!
Let $N$ be a positive integer such that $\mathbb{H}/\Gamma(N)$ has genus zero. Then the function field of $\mathbb{H}/\Gamma(N)$ is generated by a single function. When $N = 2$, the cross-ratio $\lambda$ is such a function. A point of $\mathbb{H}/\Gamma(2 )$ at which $\lambda = \lambda_0$ is precisely an elliptic curve in Legendre normal form
$$y^2 = x(x - 1)(x - \lambda_0)$$
where the points $(0, 0), (1, 0)$ constitute a choice of basis for the $2$-torsion. When $N = 3$, there is a modular function $\gamma$ such that a point of $\mathbb{H}/\Gamma(3)$ at which $\gamma = \gamma_0$ is precisely an elliptic curve in Hesse normal form
$$x^3 + y^3 + 1 + \gamma_0 xy = 0$$
where (I think) the points $(\omega, 0), (\omega^3, 0), (\omega^5, 0)$ (where $\omega$ is a primitive sixth root of unity) constitute a choice of basis for the $3$-torsion.
Question: Does this picture generalize? That is, for every $N$ above does there exist a normal form for elliptic curves which can be written in terms of a generator of the function field of $\mathbb{H}/\Gamma(N)$ and which "automatically" equips the $N$-torsion points with a basis? (I don't even know if this is possible when $N = 1$, where the Hauptmodul is the $j$-invariant.) If not, what's special about the cases where it is possible?
-
2
This is a great question. I feel like I should know the answer off the top of my head, but apparently I don't. What I can contribute at the moment is the full list of N such that X(N) has genus 0: 1,2,3,4,5. So you are missing at most normal forms for N = 4 and N = 5. To try to work this out for myself, I would start by taking the curve in Kubert-Tate normal form and see what additional relations come from having full N-torsion. (Note that you will necessarily have to extend the ground field to Q(\zeta_N) to see full N-torsion.) – Pete L. Clark Mar 4 2010 at 0:35
5
By the way, in more sophisticated terms, since X(N) is a fine moduli space over Q(\zeta_N) for N > 2, what you are looking for is an equation for the universal elliptic curve over this rational (genus zero) curve. So such normal forms definitely do exist. Looking back at an old paper of mine, I found the one for N = 3 together with the remark that the one for N = 4 is "well known". Too bad I forgot to write it down! – Pete L. Clark Mar 4 2010 at 0:38
1
When N = 1, as you say the j-invariant is the Hauptmodul. So you just want to write down an elliptic curve with j-invariant some given quantity j. There is a standard recipe for this; see e.g. Silverman's book. However, when N = 1 and also when N = 2 the moduli space is not fine, so the family is not "universal" in the strict sense of moduli spaces, and also there will be multiple nonisomorphic elliptic curves over a non-algebraically closed ground field. For instance, as I believe came up here recently, full 2-torsion is not quite enough to put an elliptic curve in Legendre normal form. – Pete L. Clark Mar 4 2010 at 1:22
5
There is a paper of Daniel Kubert's from early 70s in which he writes out many normal forms (using 2 parameters) for elliptic curves with torsion points of various small orders. These could help you. Also, you can't get around the problem of not having a single universal family parameterized by j ; the curves with j=0 and 1728 have extra automorphisms, and there is no getting around it. – Emerton Mar 4 2010 at 3:03
1
Although I'm quite late, I want to remark that the N=4-case is treated in a paper of Shioda: projecteuclid.org/… – Lennart Meier Aug 22 at 8:42
show 3 more comments
## 4 Answers
I think the answer to your question is the content of Velu's thesis: Courbes elliptiques munies d'un sous-groupe $Z/NZ\times \mu_N$. In there, he explicitly writes down the universal elliptic curve over $X(p)$ for $p>3$.
-
### 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.
As I mentioned in connection with an answer to another question, it is not generally true for elliptic curves $f:E \rightarrow S$ over a base $S$ that there is a global embedding of $E$ into $\mathbf{P}^2_ S$. For example, if $S = {\rm{Spec}} ( A )$ for a Dedekind domain $A$ whose class group is nontrivial, it could fail (and sometimes does fail). The necessary and sufficient condition is that the line bundle $\omega_{E/S} = f_{\ast}(\Omega^1_{E/S})$ on $S$ is trivial.
Example: If $S$ is the complement of a non-empty finite set of rational points in the projective line over a field $k$ then it is the spectrum of a localization of $k[x]$ and hence has trivial Picard group. Thus, the obstructions vanish and a global embedding exists. This applies to the modular curve $Y(N)$ over $\mathbf{Q}$ (geometrically connected over $\mathbf{Q}(\zeta_ N)$ via the Weil $N$-torsion pairing) when $N = 3, 4, 5$. Of course, to then really find the normal form in explicit terms requires real work and not just this kind of "brain work".
In case the elliptic curve is the universal one over some (fine) moduli scheme $S$ and this line bundle obstruction vanishes, such as if we know the stronger fact that ${\rm{Pic}}(S) = 1$, then such a global embedding must exist and its determination can then be regarded as a "normal form".
On the other hand, consider universal elliptic curves $E \rightarrow Y$ over fine modular curves $Y$ whose "level structure" doesn't dominate one of the fine ones of genus 0, such as $Y(p)$ with a prime $p > 5$. To figure out if there is a Weierstrass form for $E$ over the entire affine base $Y$ (i.e., the projective plane doesn't need to be replaced with a projective space bundle, as is needed for general families of elliptic curves) one has to determine precisely if $\omega_{E/Y}$ is trivial. This amounts to the existence of modular functions which "transform" under the corresponding "congruence subgroup" (such as $\Gamma(p)$) like a weight-1 form and have no zeros or poles on the upper half-plane (if working over $\mathbf{C}$), and so can be analyzed concretely by thinking about Klein form in the case of full-level problems.
-
Thanks for the response! I don't exactly have much background here, but from what I can tell you're confirming and elaborating on what Pete mentioned in the comments. I guess I should mention that I would be perfectly happy with an embedding into a higher-dimensional projective space as long as its defining equations could be written explicitly in terms of a Hauptmodul. Does one exist for N = 1, 2? – Qiaochu Yuan Mar 4 2010 at 5:18
I'm only saying there is a way to define "normal form", a conceptual reason for its existence in some cases, and obstruction in general. (There are ways to work with elliptic curves other than Weierstrass equations. Useful!) The "Legendre elliptic curve" over Q(lambda) has no Legendre form over Q(lambda) relative to other ordered 2-torsion bases. Why? If you understand that, you'll better understand why N = 2 is subtle (N=1 is more so). Hint: what extra structure on an elliptic curve with 2-torsion basis corresponds to a compatible "Lengendre structure"? x = a/t^2 + ...., look at a. – BCnrd Mar 4 2010 at 6:17
"For example, if S=Spec(A) for a Dedekind domain A whose class group is nontrivial, it could fail (and sometimes does fail)." When A is the ring of integers in a number field K, a result of Silverman says that every ideal class c occurs as the obstruction for some quadratic twist of any fixed elliptic curve E|K. Cf. MR0804199 (86k:11030) Silverman, Joseph H., Weierstrass equations and the minimal discriminant of an elliptic curve. Mathematika 31 (1984), no. 2, 245--251 (1985). – Chandan Singh Dalawat Mar 4 2010 at 10:35
Hello,
I believe the following results that appear in papers of Rubin and Silverberg can be very useful here. Let $N=3,4,$ or $5$ and let $Y_N$ be the (non-compact) modular curve over $\mathbb{Q}$ which parametrizes $(E,P,C)$ where $E$ is an elliptic curve, $P$ is a point of order $N$ on $E$ and $C$ is a cyclic subgroup of order $N$ on $E$, and $C$ and $P$ generate $E[N]$. The curve $Y_N$ is isomorphic to one connected component of $Y(N)$, and $Y_N(\mathbb{C})$ is isomorphic to $\mathbb{H}/\Gamma(N)$. Let $X_N$ be the compactification of $Y_N$.
Rubin and Silverberg describe explicit isomorphisms $f_N:X_N \cong \mathbb{P}^1$, with $f(u) = (A_u,P_u,C_u)$ and give equations for $A_u$, here:
1) [Rubin and Silverberg] for $N=3$ and $5$ in Families of elliptic curves with constant mod p representations
and
2) [Silverberg] for $N=4$ in Explicit families of elliptic curves with prescribed mod $N$ representations'', in Modular forms and Fermat's last theorem, Cornell, Silverman, Stevens (Editors), Springer, p. 447 - 461.
I hope that helps,
Alvaro
-
The first thing you'd need in order to define a normal form is unirationality of the moduli space (otherwise you don't even have the correct number of parameters). In dimension 1, this means that you (at least) need the modular curve to be of genus 0, at which point we may look at the The On-Line Encyclopedia of Integer Sequences
Here is how you can do n-torsion assuming you know the m-torsion solution and m divides n (and of course, the moduli space is genus 0):
Let z be the moduli space parameter, and E(z) be the universal plane curve. Let $a_i(z),b_i(z)$ be n-torsion points on E(z) which span the set of n-torsion points; let $l_i(z)$ in the dual projective plane be the line connecting $a_i(z),b_i(z)$. Then the locus of $l_i(z)$ is a plane curve, which is -- by our assumption -- rational. Now use your favorite "Italian" method to find an explicit rationalization of a rational plane curve. The coordinates of the universal projective plane are determined by the four points $0, a_i(z), b_i(z), a_i(z)+b_i(z)$.
Note that from the original list of 2..10,12,13,16,18,25 you are now left with the task of finding solutions to 5,7,13.
-
Qiaochu specified that he was looking at the genus zero cases. (Unless that was edited in after you gave this answer?) – David Speyer Mar 4 2010 at 16:40
@David: sure, and the first step in solving the genus 0 cases is knowing which ones they are – David Lehavi Mar 5 2010 at 6:02
I think the confusion is that Qiaochu asked about $X(N)$, and David Lehavi seems to be thinking about $X_0(N)$. – Jamie Weigandt Oct 8 2010 at 1:10
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http://mathoverflow.net/questions/35556?sort=oldest
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Skewing the distribution of random values over a range
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The following code comment comes from PHP, a free and open source project. I have done my own research and I cannot find any evidence to support the argument made in this code comment. Thus the only tool to support such this statement is mathematics.
Prove or disprove that that taking the modulus of a random number `n` for any value of `M` such that `M<n` decreases entropy. Or to clarify the operation is: `n mod M` and the constraint is for any value where: `M<n` , `M>0`. The 2nd part of the proof is to show that `n' = a + n(b-a+1)/(M+1)` does not decrease entropy under the same constraint: Where n is any random value and M is any value smaller than n and M greater than zero.
Or to put it another way how is the value produced by `n' = a + n(b-a+1)/(M+1)` more random than the value produced by `n mod M` for the same values of n and M such that `M<n`.
Any helpful information on this topic will earn you a +1 by me. Proving that `n mod M` skews the distribution of random values over a range, would be a HUGE accomplishment in my humble opinion.
````/*
* A bit of tricky math here. We want to avoid using a modulus because
* that simply tosses the high-order bits and might skew the distribution
* of random values over the range. Instead we map the range directly.
*
*
* We need to map the range from 0...M evenly to the range a...b
* Let n = the random number and n' = the mapped random number
*
* Then we have: n' = a + n(b-a)/M
*
* We have a problem here in that only n==M will get mapped to b which
*
* means the chances of getting b is much much less than getting any of
* the other values in the range. We can fix this by increasing our range
* artifically and using:
*
* n' = a + n(b-a+1)/M
*
*
* Now we only have a problem if n==M which would cause us to produce a
* number of b+1 which would be bad. So we bump M up by one to make sure
* this will never happen, and the final algorithm looks like this:
*
*
* n' = a + n(b-a+1)/(M+1)
*
* -RL
*/
````
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There was a spelling error in your title, and an erroneous tag. I've taken the liberty of changing both. – Yemon Choi Aug 14 2010 at 4:37
I still think your phrase "decrease the distribution" is unfortunate. What do you mean by "the distribution", anyway? – Yemon Choi Aug 14 2010 at 4:39
And finally, your wikipedia link for "modulus" is wrong, as can be seen by comparing the topic of that link with the code that you've quoted. – Yemon Choi Aug 14 2010 at 4:40
1
@Yemon Choi I fixed the wiki link, and I really mean entropy not distribution. – George Aug 14 2010 at 4:44
@Yemon Choi also thank you for helping clarify my question and not just ripping me apart – George Aug 14 2010 at 4:52
show 2 more comments
3 Answers
Suppose you choose a number n randomly from {1,2,3} and then calculate n mod 2. Then you will get 1 twice more frequently than 0.
Another reason why code like this avoids taking the modulus is that for many old pseudorandom generator routines the lower-order bits were not as "random" as higher-order bits. This is no longer a problem with modern libraries.
Now I want my +1.
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+1 granted, thank you. – George Aug 14 2010 at 4:47
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.
If your PRNG algorithm can produce a "random" integer n truly uniformly distributed in the interval
````0,...,k−1
````
and then you take
````n(mod l)
````
which in the language C would be the operation
````n' = n % l
````
which gives a value in
````0,...,l−1
````
then n is still "random" if
````l|k
````
that is
````k==0(modl)
````
Otherwise no, as jmoy described. So it would appear you need the freedom to take
````l−1=b−a
````
and then k a multiple of l
Thanks to Will Jagy
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Yes, that's right, George. I do not know how this relates to practical issues in the original program or in your program, and I especially do not know what situation would allow one to define a value for my letter $l$ and then not allow my $k = l.$ But, assuming you are required to take $k > l,$ note that if $k$ is very much larger than $l$ the bias towards some numbers will be slight. Anyway, look, I make a standard offer, if you click on my name there are instructions for finding my email address. You may write to me if that might help, I have done enough programming. – Will Jagy Aug 14 2010 at 21:55
Let's clean this up. First, assume a = 0 (if not, just add it on at the end). Second, don't deal with closed intervals (0 <= n <= M, 0 <= n' <= b) but with half-open intervals (0 <= n < M, 0 <= n' < b). This second simplification is exactly the 'bit of tricky math' in the comment, which just replaces M by M+1 and b by b+1.
So now we have a random number 0 <= n < M, and we want to transform it into a random number 0 <= n' < b, where b <= M. This is an interesting question (if not necessarily the one that the OP asked). The solution in the OP's code fragment is to take n' = nb/M. Assuming that b doesn't divide M exactly (for then there is no problem), n' is not uniformly distributed: it takes a number of values (in fact, M mod b of them) with probability ⌈M/b⌉/M, and the remaining values with probability ⌊M/b⌋/M. Exactly the same is true if we take n'= n mod b; there is no statistical difference.
If M/b is big enough, then ⌈M/b⌉/M and ⌊M/b⌋/M are nearly equal; and this may be good enough for your application. If not, then the simplest solution is the following:
Loop:
Generate random 0 <= n < M
if (n < M - (M mod b)) return n mod b
goto Loop
The problem is that this algorithm is not guaranteed to halt! But you can get a closer and closer approximation to uniformity by setting a bigger and bigger bound on the number of times the loop is executed.
Note that for this algorithm, it is important to use n mod b instead of nb/M, because the check for uniformity (n < M - (M mod b)) is simple.
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http://mathhelpforum.com/calculus/171088-nth-derivative-proof-induction.html
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# Thread:
1. ## nth Derivative in Proof by Induction
Hey, I've been working through some proof by induction problems, and most of them are manageable, but then I saw one which involved differentiation which I have no idea what to do. Could someone please lead me in the right direction?
$\displaystyle \mbox{Prove that the nth derivative of } y=ln(x), x>0 \mbox{ is: }<br /> \frac{d^ny}{dx^n} = \frac{(-1)^{n+1}(n-1)!}{x^n}$
I have the basic steps done:
$===================$
Step 1: Prove for n = 1
$\dfrac{dy}{dx} [ln(x)] = \dfrac{1}{x}$
$--------$
$\dfrac{d^1y}{dx^1} = \dfrac{(-1)^2(1-1)!}{x^1}$
$\dfrac{d^1y}{dx^1} = \dfrac{1}{x} = \dfrac{dy}{dx} [ln(x)] \mbox{ \therefore True }$
$===================$
Step 2: Assume for n = k
$\dfrac{d^ky}{dx^k} = \dfrac{(-1)^{k+1}(k-1)!}{x^k}$
$===================$
Step 3: Prove for n = k+1
I know how to start this, but how do I continue?
Thanks in advance for help. BG
2. For
$<br /> n=k+1<br />$
$<br /> \dfrac{d}{dx} \; \dfrac{d^ky}{dx^k} <br />$
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http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=3312&bodyId=3622
|
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# Loci: Convergence
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# A Modern Vision of the Work of Cardano and Ferrari on Quartics
by Harald Helfgott (University of Bristol) and Michel Helfgott (East Tennessee State University)
## Different Strategies
Problem VI asks to "find a number which is equal to its square root plus twice its cube root." Denoting the positive number by x6, we have to solve the equation x6= x3 + 2x2. Dividing by x2 we get the equivalent equation x4= x + 2, a quartic indeed. Cardano subtracts 1 from each side and reaches x4 - 1= x + 1, thus (x4 - 1)/(x + 1)= 1. Consequently x3 - x2+ x - 1= 1, that is to say x3 + x = x2 + 2. He ascertains that $$x=\root 3\of{\sqrt{83\over 108}+{47\over 54}}-\root 3\of{\sqrt{83\over 108}-{47\over 54}}+{1\over 3}$$is a solution, without providing the details. Let us check whether this is a solution, using for that purpose the Cardano-Tartaglia method.
Defining y = x-1/3 we reach the depressed cubic y3 = -(2/3)y+ 47/27. Keeping in mind that (u - v)3 = -3uv(u - v) + (u3 - v3), a solution of the form y = u - v will be found provided that the algebraic system -3uv= -2/3, u3 - v3= 47/27 has a solution. Indeed u3- (2/9u)3= 47/27, that is to say u6 - (47/27)u3 - 8/729 = 0. One solution of this equation is $$u^3 = {47\over 54} + {1\over 2}\sqrt{\left({47\over 27}\right)^2 + {32\over 729}},$$ thus $$u = \root 3\of {{47\over 54}+\sqrt{83\over 108}}.$$ Replacing this value of u in u3 - v3= 47/27 we get $$v=\root 3\of{-{47\over 54}+\sqrt{83\over 108}}.$$ Then $$y=u-v=\root 3\of{{47\over 54}+\sqrt{83\over 108}} - \root 3\of{-{47\over 54}+\sqrt{83\over 108}},$$ which in turn leads to $$x=\root 3\of{\sqrt{83\over 108} + {47\over 54}}-\root 3\of{\sqrt{83\over 108}-{47\over 54}}+{1\over 3}.$$The other two solutions of the cubic are complex numbers, no wonder that Cardano ignores them.
A natural question, addressed in Ars Magna, is what happens if we follow Ferrari's procedure: We start with (x2)2 = x + 2 and introduce a new variable b. We observe that (x2 + b)2 = 2bx2 + x + b2 + 2. In order to have a perfect square we need to impose the condition 1/4 = 2b(b2 + 2), equivalent to demanding that the discriminant of 2bx2 + x + b2 + 2 is zero. Thus b3 + 2b= 1/8. Cardano ascertains that $$b=\root 3\of{{1\over 16}+\sqrt{2,075\over 6,912}}-\root 3\of{-{1\over 16}+\sqrt{2,075\over 6,912}}$$is a solution. We would have to replace this value of b, call it b1, in (x2 + b)2= 2bx2 + x + b2 + 2 knowing beforehand that the expression to the right is also a perfect square. In other words, it is necessary to solve (x2 + b1)2 = 2b1(x+ 1/4b1)2 or, equivalently, the pair of quadratic equations $$x^2+b_1=\sqrt{2b_1}\left(x+{1\over 4b_1}\right), \qquad x^2+b_1=-\sqrt{2b_1}\left(x+{1\over 4b_1}\right).$$The solutions of the first quadratic are $$x=\sqrt{b_1\over 2} \pm\sqrt{{1\over 2\sqrt{2b_1}}-{b_1\over 2}}.$$
Using the decimal approximation 0.062379 to b1 we get x = 1.35321 and x = -0.999998. The first is a decimal approximation to the solution found before, while the other root is a decimal approximation to -1. But we are not interested in the latter because we are seeking only positive solutions to the problem (from the very beginning we ruled out the trivial solutions of the equation x6 = x3 + 2x2, namely 0 and -1). It should be noted that the quadratic equation $$x^2 +b_1 = -\sqrt{2b_1} \left( x+\frac{1}{4b_1} \right)$$ has complex roots.
Problem VIII asks to "divide 6 into three proportional quantities the sum of the squares of the first and second of which is 4." Let x, y, z denote the quantities. Since x2 + y2= 4 we will have $$y = \sqrt{4 - x^2}$$. But x + y + z = 6, consequently $$z = 6 - x - \sqrt{4 - x^2}$$. Next Cardano ascertains that $$\left(6 - x - \sqrt{4 - x^2}\right) x = {4 - x^2}$$, which is correct since x, y, z are proportional quantities. Then $$6x - x^2 - x\sqrt{4 - x^2} = {4 - x^2}$$, that is to say $$6x - 4 = x\sqrt{4 - x^2}$$. Squaring both sides we get x4 + 32x2 + 16 = 48x, which is equivalent to (x2 + 16)2 = 48x + 240.
For any b we have (x2 + 16 + b)2 = 2bx2 + 48x + b2 + 32b + 240. We need to have the discriminant equal to zero in order to make sure that the right hand side is a perfect square, thus 0= 482 - 8b(b2 + 32b + 240). This expression is equivalent to b3 + 32b2 + 240b = 288. Then Cardano uses the transformation c= b + 32/3 and reaches the equation c3 = (101 1/3)c + 420 20/27, with a real solution c. Therefore b = c - 32/3 is a real solution of the resolvent cubic. Actually, Cardano does not introduce a new variable explicitly. Finally we have (x2 + 16+ b)2 = 2b(x + 48/4b)2, which in turn leads to (x2 + 16+ b)= ±√(2b)(x+ 12/b). Cardano only considers the equation with positive sign and writes it as x2 + 16 + b = √(2b)x+ √(b2 + 32b + 240), which is in agreement with x2 + 16 + b = √(2b)x+ √(2b)(12/b) because √(2b)(12/b) = √(288/b)= √[(b3 + 32b2 + 240b)/b] = √(b2 + 32b + 240). He leaves the task of finding b, and solving the above-mentioned quadratic equation, to the readers of Ars Magna.
An alternative path is to start with x4 = -32x2 + 48x - 16 and introduce a new variable z so that the expression on the right of (x2+ z)2 = (2z - 32)x2 + 48x + z2 - 16 is a perfect square. This can be accomplished by solving 0 = 482 - 4(2z - 32)(z2 -16), i.e. z3 - 16z2 - 16z - 32= 0. The usual approach to cubics leads to $$z \approx 17.0486$$, therefore $$(x^2+17.0486)^2 = 2.0972x^2+48x+274.655 = 2.0972\left(x+{48\over 2\cdot 2.0972}\right)^2, \quad {\rm or}\quad (x^2+17.0486)^2 = \left(\sqrt{2.0972}\left(x+{48\over 4.1944}\right)\right)^2.$$ Hence $$x^2+17.0486 = \sqrt{2.0972}\left(x+{48\over 4.1944}\right) \quad {\rm or} \quad x^2+ 17.0486 = -\sqrt{2.0972}\left(x+{48\over 4.1944}\right).$$ The first quadratic has the real solutions 0.943916 and 0.504855 while the second quadratic has complex solutions. Obviously, both real solutions are approximations.
Next page >> Using a New Transformation
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Helfgott, Harald and Michel Helfgott, "A Modern Vision of the Work of Cardano and Ferrari on Quartics," Loci (June 2009), DOI: 10.4169/loci003312
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http://math.stackexchange.com/questions/9872/0-times-n-matrices
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# $0\times n$ matrices
I think it is desirable to have that $M_{m\times n}\left(\mathbb{K}\right)\not=M_{m'\times n'}\left(\mathbb{K}\right)$ if $m\not=m'$ or $n\not=n'$. In other words, the set of all $m\times n$ matrices on $\mathbb{K}$ should be different from the set of all $m'\times n'$ matrices on the same field when $m\not=m'$ or $n\not=n'$. But if I define $M_{m\times n}\left(\mathbb{K}\right)$ as $\left(\mathbb{K}^n\right)^m$ then $M_{0\times n}\left(\mathbb{K}\right)=M_{0\times n'}\left(\mathbb{K}\right)$ for every $n$ and $n'$. That is, two matrices with zero rows are equal, no matter how many columns they have, because $X^0$ is the set containing the empty tuple, for every set $X$. I could have defined $M_{m\times n}\left(\mathbb{K}\right)$ as $\left(\mathbb{K}^m\right)^n$ instead, but then the problem is with $M_{m\times 0}\left(\mathbb{K}\right)$ and $M_{m'\times 0}\left(\mathbb{K}\right)$.
Two questions:
1. Am I defining matrices correctly?
2. Are $m\times 0$ or $0\times n$ matrices that important?
Thanks.
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5
2: No, they are not important at all, so why bother? – Hans Lundmark Nov 11 '10 at 18:57
2
@Hans: I disagree. Talking about 0xn and nx0 matrices may seem like nonsense but it is a concrete way of talking about an important fact: that the category of vector spaces has a terminal object which is also an initial object. – Qiaochu Yuan Nov 12 '10 at 22:31
1
@Qiaochu: Matrices are useful for doing computations with concrete linear transformations. The mappings to and from the trivial vector space are easy enough to think of abstractly, so what is there to gain by trying to represent them by matrices? Anyway, the last paragraph in Arturo's answer sums it up pretty well, I think. – Hans Lundmark Nov 12 '10 at 23:15
## 2 Answers
As Laurent S points out, the sensible way to define $\mathbb{K}^0$ is as the one element set (either $\{0\}$ or $\{\emptyset\}$).
You could likewise argue that since you want $M_{m\times n}(\mathbb{K})$ to correspond to linear transformations from $\mathbb{K}^n$ to $\mathbb{K}^m$, and since $\mathbb{K}^0$ is a perfectly good vector space, then we should define the matrices $M_{0\times n}(\mathbb{K})$ and $M_{m\times 0}(\mathbb{K})$. Then you are certainly forced to define both as the trivial set, since only the trivial linear transformation exists in either case (the zero map for $m=0$, and the trivial inclusion for $n=0$). In that respect, they would fit with the general theory and not require exceptions when describing vector spaces (where you would need to explicitly exclude the zero-dimensional vector space, for instance).
Why, then, are $M_{0\times n}(\mathbb{K})$ and $M_{0\times n'}(\mathbb{K})$ both equal even when $n\neq n'$, if the first "is" a linear transformation from $\mathbb{K}^n$ to $\{0\}$ and the second "is" a linear transformation from $\mathbb{K}^{n'}$ to $\{0\}$, and hence different? The key is the `"`. They are both identified with those maps, but as a vector space itself they are in fact isomorphic, so there is no horrible universe-shattering error that crept in by having the "matrix" rings be equal. Of course, now your theorem that $M_{m\times n}(\mathbb{K})$ is uniquely determined (in some way) by the pair $(m,n)$ is no longer true: you need to say that they are equal if and only if either $(m,n)=(m',n')$ or $mn=m'n'=0$.
Such is life: there is often a special case that needs to be dealt with separately, and your choice is whether you exclude it by design or you exclude it by mention. Here, your choice is between specifying "nonzero" in a lot of theorems about vector spaces, or specifying "or $mn=m'n'=0$" in the theorem about matrices. The latter is usually less disruptive. (In set theory, one often has to deal with the emptyset in a special manner; that corresponds to $0$ here).
As to their interest: they have very little structure to them, so they are not generally interesting in and of themselves. They can show up as part of a general scheme (as above, fitting into the role of representing linear transformations between the standard finite dimensional vector spaces) so that exceptions don't need to be made, but as such they play more the role of "exception-avoiders" than "interesting in and of themselves."
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1. No,I guess the only sensible way of defining $\mathbb{K}^0$ when $\mathbb{K}$ is a field is by $\mathbb{K}^0=0$, since it should be a 0-dimensional vector space.
2. No, I have never seen $m\times 0$-matrices used anywhere, other than here.
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http://math.stackexchange.com/questions/153537/exclusive-prime-factors
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# Exclusive prime factors
Let $S$ be an finite or infinite subset of the primes. Let $f(x)=1$ if $x$ has no factors in $S$. If not, $f(x)=0$. Is there a way to calculate the limit $\displaystyle\sum_{n=1}^{x} f(n)/x$, as $x$ grows big?
Examples;
S contains all primes of form $2778t+79$
S={2,5,17}
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You should clarify "$2778t + 79$." What are the restrictions on $t$? Furthermore, can you show there are an infinity of primes of the form $2778t + 79$? – ThisIsNotAnId Jun 4 '12 at 0:01
– Norbert Jun 4 '12 at 0:04
For finite set i think its easy, because f(n) is periodic, right? – Sigma23 Jun 4 '12 at 0:06
## 1 Answer
Yes, for a finite set it's easy. For your example, $S=\{{2,5,17\}}$, $x$ has no factors in $S$ if and only if $x$ is relatively prime to $170=2\times5\times17$. The number of integers less than and relatively prime to 170 is $\phi(170)=1\times4\times16=64$, and then it's periodic beyond 170, so your limit is $64/170$, also known as $32/85$.
For infinite sets of primes, life is more difficult. I suspect that for the infinite set of primes in an arithmetic progression the limit is zero, that is, almost all big numbers have a factor in every arithmetic progression, but I can't provide a reference for this.
EDIT: A well-known example. The numbers that can be written as a sum of two squares, that's known to be a set of density zero (see, e.g., http://www.math.niu.edu/~rusin/known-math/93_back/2squares). Now those include all the numbers that have no prime factors of the form $4t+3$, so it follows that the set of numbers with no prime factor $4t+3$ has density zero.
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http://psychology.wikia.com/wiki/Prior_probabilities
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# Prior probability
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## Redirected from Prior probabilities
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A prior probability is a marginal probability, interpreted as a description of what is known about a variable in the absence of some evidence. The posterior probability is then the conditional probability of the variable taking the evidence into account. The posterior probability is computed from the prior and the likelihood function via Bayes' theorem.
As prior and posterior are not terms used in frequentist analyses, this article uses the vocabulary of Bayesian probability and Bayesian inference.
Throughout this article, for the sake of brevity the term variable encompasses observable variables, latent (unobserved) variables, parameters, and hypotheses.
## Prior probability distribution
In Bayesian statistical inference, a prior probability distribution, often called simply the prior, of an uncertain quantity p (for example, suppose p is the proportion of voters who will vote for the politician named Smith in a future election) is the probability distribution that would express one's uncertainty about p before the "data" (for example, an opinion poll) are taken into account. It is meant to attribute uncertainty rather than randomness to the uncertain quantity.
One applies Bayes' theorem, multiplying the prior by the likelihood function and then normalizing, to get the posterior probability distribution, which is the conditional distribution of the uncertain quantity given the data.
A prior is often the purely subjective assessment of an experienced expert. Some will choose a conjugate prior when they can, to make calculation of the posterior distribution easier.
## Informative priors
An informative prior expresses specific, definite information about a variable. An example is a prior distribution for the temperature at noon tomorrow. A reasonable approach is to make the prior a normal distribution with expected value equal to today's noontime temperature, with variance equal to the day-to-day variance of atmospheric temperature.
This example has a property in common with many priors, namely, that the posterior from one problem (today's temperature) becomes the prior for another problem (tomorrow's temperature); pre-existing evidence which has already been taken into account is part of the prior and as more evidence accumulates the prior is determined largely by the evidence rather than any original assumption, provided that the original assumption admitted the possibility of what the evidence is suggesting. The terms "prior" and "posterior" are generally relative to a specific datum or observation.
## Uninformative priors
An uninformative prior expresses vague or general information about a variable. The term "uninformative prior" is a misnomer; such a prior might be called a not very informative prior. Uninformative priors can express information such as "the variable is positive" or "the variable is less than some limit". Some authorities prefer the term objective prior.
In parameter estimation problems, the use of an uninformative prior typically yields results which are not too different from conventional statistical analysis, as the likelihood function often yields more information than the uninformative prior.
Some attempts have been made at finding probability distributions in some sense logically required by the nature of one's state of uncertainty; these are a subject of philosophical controversy. For example, Edwin T. Jaynes has published an argument (Jaynes 1968) based on Lie groups that suggests that the prior for the proportion $p$ of voters voting for a candidate, given no other information, should be $p^{-1}(1-p)^{-1}$. If one is so uncertain about the value of the aforementioned proportion $p$ that one knows only that at least one voter will vote for Smith and at least one will not, then the conditional probability distribution of $p$ given this information alone is the uniform distribution on the interval [0, 1], which is obtained by applying Bayes' Theorem to the data set consisting of one vote for Smith and one vote against, using the above prior.
Priors can be constructed which are proportional to the Haar measure if the parameter space $X$ carries a natural group structure. For example, in physics we might expect that an experiment will give the same results regardless of our choice of the origin of a coordinate system. This induces the group structure of the translation group on $X$, and the resulting prior is a constant improper prior. Similarly, some measurements are naturally invariant to the choice of an arbitrary scale (i.e., it doesn't matter if we use centimeters or inches, we should get results that are physically the same). In such a case, the scale group is the natural group structure, and the corresponding prior on $X$ is proportional to $1/x$. It sometimes matters whether we use the left-invariant or right-invariant Haar measure. For example, the left and right invariant Haar measures on the affine group are not equal. Berger (1985, p. 413) argues that the right-invariant Haar measure is the correct choice.
Another idea, championed by Edwin T. Jaynes, is to use the principle of maximum entropy. The motivation is that the Shannon entropy of a probability distribution measures the amount of information contained the distribution. The larger the entropy, the less information is provided by the distribution. Thus, by maximizing the entropy over a suitable set of probability distributions on $X$, one finds that distribution that is least informative in the sense that it contains the least amount of information consistent with the constraints that define the set. For example, the maximum entropy prior on a discrete space, given only that the probability is normalized to 1, is the prior that assigns equal probability to each state. And in the continuous case, the maximum entropy prior given that the density is normalized with mean zero and variance unity is the standard normal distribution.
A related idea, reference priors, was introduced by Jose M. Bernardo. Here, the idea is to maximize the expected Kullback-Leibler divergence of the posterior distribution relative to the prior. This maximizes the expected posterior information about $x$ when the prior density is $p(x)$. The reference prior is defined in the asymptotic limit, i.e., one considers the limit of the priors so obtained as the number of data points goes to infinity. Reference priors are often the objective prior of choice in multivariate problems, since other rules (e.g., Jeffreys' rule) may result in priors with problematic behavior.
Philosophical problems associated with uninformative priors are associated with the choice of an appropriate metric, or measurement scale. Suppose we want a prior for the running speed of a runner who is unknown to us. We could specify, say, a normal distribution as the prior for his speed, but alternatively we could specify a normal prior for the time he takes to complete 100 metres, which is proportional to the reciprocal of the first prior. These are very different priors, but it is not clear which is to be preferred. Similarly, if asked to estimate an unknown proportion between 0 and 1, we might say that all proportions are equally likely and use a uniform prior. Alternatively, we might say that all orders of magnitude for the proportion are equally likely, which gives a prior proportional to the logarithm. The Jeffreys prior attempts to solve this problem by computing a prior which expresses the same belief no matter which metric is used. The Jeffreys prior for an unknown proportion $p$ is $p^{1/2}(1-p)^{1/2}$, which differs from Jaynes' recommendation.
Practical problems associated with uninformative priors include the requirement that the posterior distribution be proper. The usual uninformative priors on continuous, unbounded variables are improper. This need not be a problem if the posterior distribution is proper. Another issue of importance is that if an uninformative prior is to be used routinely, i.e., with many different data sets, it should have good frequentist properties. Normally a Bayesian would not be concerned with such issues, but it can be important in this situation. For example, one would want any decision rule based on the posterior distribution to be admissible under the adopted loss function. Unfortunately, admissibility is often difficult to check, although some results are known (e.g., Berger and Strawderman 1996). The issue is particularly acute with hierarchical Bayes models; the usual priors (e.g., Jeffreys' prior) may give badly inadmissible decision rules if employed at the higher levels of the hierarchy.
## Improper priors
If Bayes' theorem is written as
$P(A_i|B) = \frac{P(B | A_i) P(A_i)}{\sum_j P(B|A_j)P(A_j)}\, ,$
then it is clear that it would remain true if all the prior probabilities P(Ai) and P(Aj) were multiplied by a given constant; the same would be true for a continuous random variable. The posterior probabilities will still sum (or integrate) to 1 even if the prior values do not, and so the priors only need be specified in the correct proportion.
Taking this idea further, in many cases the sum or integral of the prior values may not even need to be finite to get sensible answers for the posterior probabilities. When this is the case, the prior is called an improper prior. Some statisticians use improper priors as uninformative priors. For example, if they need a prior distribution for the mean and variance of a random variable, they may assume p(m, v) ~ 1/v (for v > 0) which would suggest that any value for the mean is equally likely and that a value for the positive variance becomes less likely in inverse proportion to its value. Since
$\int_{-\infty}^{\infty} dm\, = \int_{0}^{\infty} \frac{1}{v} \,dv = \infty,$
this would be an improper prior both for the mean and for the variance.
## References
• Andrew Gelman, John B. Carlin, Hal S. Stern, and Donald B. Rubin. Bayesian Data Analysis, 2nd edition. CRC Press, 2003. ISBN 1-58488-388-X
• James O. Berger, Statistical Decision Theory and Bayesian Analysis, Second Edition. Springer-Verlag, 1985. ISBN 0-387-96098-8
• Edwin T. Jaynes, "Prior Probabilities," IEEE Transactions of Systems Science and Cybernetics, SSC-4, 227-241, Sept. 1968. Reprinted in Roger D. Rosenkrantz, Compiler, E. T. Jaynes: Papers on Probability, Statistics and Statistical Physics. Dordrecht, Holland: Reidel Publishing Company, pp. 116-130, 1983. ISBN 90-277-1448-7
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http://mathoverflow.net/revisions/54699/list
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## Return to Answer
2 added 172 characters in body
This only answers part of your question 0 unfortunatly. The construction is certainly functorial, but the two notions of symmetric/alternating power do not always agree. Let's write $\operatorname{Sym}^n (V)$ for the symmetric tensors, and $\operatorname{Alt} ^n (V)$ for the alternating tensors. I wish this were established notation, but it probably isn't. Let $p$ be the characteristic of the field. Note $V^{\otimes n}$ is a $kGL(V) - \Sigma_n$ bimodule ($\Sigma_n$ is the symmetric group). Then if $r$ is less than $p$, or if $p=0$, $\operatorname{Sym}^r (V) \cong S^r(V)$ and $\operatorname{Alt}^r (V) \cong \Lambda ^r(V)$ as $GL(V)$-modules (this is proved by writing down maps explicitly).
If $r \geq p$ then $S^r$ and $\operatorname{Sym}^r$ are the contravariant (i.e. transpose) duals of one another as $GL$ modules. I imagine the same is true of the alternating power/antisymmetric tensors.
1
Let's write $\operatorname{Sym}^n (V)$ for the symmetric tensors, and $\operatorname{Alt} ^n (V)$ for the alternating tensors. I wish this were established notation, but it probably isn't. Let $p$ be the characteristic of the field. Note $V^{\otimes n}$ is a $kGL(V) - \Sigma_n$ bimodule ($\Sigma_n$ is the symmetric group). Then if $r$ is less than $p$, or if $p=0$, $\operatorname{Sym}^r (V) \cong S^r(V)$ and $\operatorname{Alt}^r (V) \cong \Lambda ^r(V)$ as $GL(V)$-modules (this is proved by writing down maps explicitly).
If $r \geq p$ then $S^r$ and $\operatorname{Sym}^r$ are the contravariant (i.e. transpose) duals of one another as $GL$ modules. I imagine the same is true of the alternating power/antisymmetric tensors.
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http://mathematica.stackexchange.com/questions/tagged/variable-definitions?page=1&sort=active&pagesize=50
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# Tagged Questions
Questions on assigning and unassigning definitions to names that represent them, including issues raised by context and localization. Use tag [assignment] for issues relating to Set versus SetDelayed.
learn more… | top users | synonyms
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### Create an adaptive amount of local variables for error propagation
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### How to clear all variables except one?
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### Elegant manipulation of the variables list
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### defining recursively a function with multiple if conditions
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### how can I generate a sequence of assignments?
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260 views
### How to create a Table of Tables with indexed variables
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### Can I redefine a list as a set of variables?
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### Evaluating a function on permutations of its arguments
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### Reloading package after accidental Remove[] of symbol
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### How to create functions of arbitrary number of variables?
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### Using external variable as a variable inside WhenEvent & NDSolve
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### What's the best practice for nested local constants? [duplicate]
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### local variable naming & symbolic argument
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### Better solution than returning a list of 3 values?
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### Context “Unique to This Notebook” makes variables black even if they're not defined?
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0answers
58 views
### What's reliable way of finding declared but undefined functions/symbols in several interrelated packages?
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5answers
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### How do I clear all user defined symbols?
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143 views
### Is it possible to Clear all variables (but not functions)?
I have written a Mathematica script in which I define functions and variables. Here is a vastly simplified example: ...
3answers
666 views
### Why modules with no variables?
I was reading some code, in particular, recipe 4.13 on unification pattern-matching in Sal Mangano's Mathematica Cookbook, and there were many instances of Modules with no variables in them, such as ...
3answers
313 views
### Variable variable names
I have two variables, $u \in \{0,40\}$ and $\gamma \in \{0,1\}$. I take $10$ values of $u$ and $5$ values of $\gamma$ and using their combinations as a pair of input parameters, I perform some ...
1answer
182 views
### Can I create a dynamic number of rows in TabView through an iteration?
I have written code, with the help of stackoverflow of course, and I want to make it user friendly so that other people in my lab can use it. I'm playing with DialogCreate and similar functions. I ...
1answer
458 views
### Constructing variable names from a string
I have groups of descriptive variable names and dont want to write explicit code for each group, e.g. (and this is a contrived simplification) ...
0answers
270 views
### Using constraints in Solve[]
I'm not sure if the title fits. I'm basically just looking for a way to define a value and constrain it to $[a,b]$ when I use Solve later. More specifically, I ...
4answers
247 views
### How to use pattern matching to assign values to Subscript[f,x_]?
I want to define two subscripted functions Subscript[f,1] and Subscript[f,2]. To keep the assignments local, I would like to ...
1answer
152 views
### How do I use a matrix that has a variable?
I seem to be getting an error whenever I try running this piece of code. Matrix12[n_] := {{1, 0}, {0, {\[Kappa]1[n]}/{\[Kappa]2[n]}}} I've already defined the ...
1answer
178 views
### How to substitute variables in interpolated function?
I'm looking for the way to substitute variables in interpolated function again to original one. (I want to know how to get P(r, theta) from Psol(v, w).) The original variables : ...
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http://rjlipton.wordpress.com/2013/02/24/who-invented-boolean-functions/
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## a personal view of the theory of computation
by
Hint: I am not sure
Calixto Badesa is a faculty member at the University of Barcelona. His department is called the Department of Logic, History, and Philosophy of Science. It is neat to see the idea of the “logic of science” added to the familiar ‘HPS’ acronym. Well, Badesa himself specializes in Logic—that is, the history of Logic as a discipline.
Today I want to discuss why propositional calculus as we now know it took so long to create.
I am currently reading Badesa’s book entitled: The Birth Of Model Theory. His main focus is on the fundamental Löwenheim-Skolem Theorem. This refers to several theorems. The first was proved by Leopold Löwenheim, in a paper in 1915. It was later re-proved by Thoralf Skolem.
For Badesa the theorem is pivotal because it leverages the distinction between syntax and semantics. Model Theory is about this distinction, and understanding it makes Logic a mature field. But that’s almost saying Logic is less than 100 years old, while logic has been studied for 3,000 years. This makes Ken and me wonder how old our modern understanding of basic concepts really is.
## Löwenheim-Skolem
The main result is that any first-order theory with only a countable number of symbols that has an infinite model must also have a countable model. This sounds plausible and is correct, but on second thought it is paradoxical. Consider ZF, or any other reasonable first order theory of set theory. This theory can prove that there are “uncountable sets,” yet by the Löwenheim-Skolem Theorem it follows that the theory has a countable model. The trick is that outside the theory we can “see” that the model is countable, but inside the theory we cannot. Very neat.
The story of the Löwenheim-Skolem Theorem, like the theorem itself, is somewhat paradoxical. The initial proof of Löwenheim’s is unclear at several key places. Badesa spends a majority of his 220-page book discussing what Löwenheim actually did prove and what he did not. Also what Skolem proved and what he did not. Perhaps we can discuss this another time. Read his book.
Badesa’s first chapter is on the work of George Boole, William Jevons, Charles Peirce, and Ernst Schröder. All of them worked on the early days of what became the theory of propositional calculus. Let’s discuss what they did and what they did not do.
## A Joke
Who invented Boolean Functions? This sounds like an old joke of Groucho Marx’s. His radio/TV show You Bet Your Life was a game show where contestants, usually a couple, tried to win money by answering general-knowledge questions.
The best part of the show was not the game, but Groucho’s patter with the contestants. Contestants who won \$25 or less were asked a trivial question by Groucho at the end of the show—he wanted them to have something more to take home. One “trivial” question he often used was:
Who is buried in Grant’s Tomb?
Ken grew up thinking the answer was, “his horse.” Actually the correct answer is “no one,” since Ulysses Grant and his wife are in an an above-ground vault, rather than buried. Recall “buried” means in the ground. All answered “Grant” and were given a small prize, except one man who actually got the question right and shocked Groucho. The man pointed out that Grant’s Tomb is an above-ground mausoleum. Oh well.
So who actually did invent Boolean functions? Indeed.
## Boole’s Work
Back to Badesa’s book. In his first chapter he goes into quite a bit of detail on what Boole did, and also Jevons, Peirce, and Schröder. Boole used symbols such as ${x}$ and ${y}$ to represent “the class of individuals to which a particular name or description is applicable.” He used ${xy}$ to denote the class of things represented by ${x}$ and ${y}$ where both are simultaneously applicable. Thus if ${x}$ stands for “white” and ${y}$ for “horse,” then ${xy}$ stands for “white horses.” He used ${x+y}$ for “or” but stated that the objects in ${x}$ and ${y}$ should be disjoint. Although he did sometimes diverge and allows the classes ${x}$ and ${y}$ to overlap, it was really Jevons who later observed that the disjointness restriction is unnecessary.
The point that I think is the most important is what Boole did not say. He did not say:
There are two elements in the world: ${0}$ and ${1}$. Here are the operations that I use on these elements. Let now turn and study their properties. Consider …
Schröder studied what Boole did and built a set of axioms that attempt to capture the general structure. He used the symbol ${\stackrel{\subset}{=}}$ in his theory (I actually cannot get his symbol exactly). One interesting point that Badesa discusses is the possibility that Schröder may have been the first ever to try to prove that a result is independent from a set of axioms. The issue was whether or not a particular axiom was needed. Peirce thought the axiom in question could be proved from the others, but lost his argument. Schröder in 1883 purported to show that the axiom was unprovable by finding a model that violates it, but satisfies the other axioms. This would show that it did not follow from the other axioms.
What is interesting about this is not the particular result, but that he may have been the first to prove an independence result via model theory.
## Why So Long?
The real numbers are the basis of all mathematics of the Greeks, since geometry by definition is all about the real plane. Later the whole thrust of western mathematics was on functions from real numbers to real numbers. This of course gave rise to the notions of continuous functions, derivatives, integrals, and more. These are beautiful, important, and actually quite delicate concepts. Part of the difficulty is that all touch in one way or another on the infinite. Real numbers demand a good understanding of the infinite, as do all the other related concepts.
Any definition of a real number involves, in some way, the infinite. Whether you define them via infinite decimals, Dedekind cuts, Cauchy sequences of rationals, or otherwise, there is no way to avoid notions based on infinity. At least there are none that I know.
The same issues of infinity arise in the trouble that early mathematics had in defining precisely the notion of function. Again a proper definition requires that we have infinite sets. The number of real functions is not just “infinity,” but is a larger cardinality in the precise sense of Cantor.
Yet the notion of a boolean space ${\{0,1\}}$ of just two elements is all too concrete. At first glance it seems almost too easy. There are no fundamental issues in defining operations on this space. You cannot understand what a real function really is just by giving the values in a table. Yet for a Boolean function, all is said by a finite table—simple like an accountant’s view of the world.
So why was the creation of a theory of Boolean functions so hard? My theory is that it was too simple. I argue it is a huge surprise historically that practical notions based on infinity were developed long before Cantor, while it lay hidden until the 20th Century that the theory of functions mapping ${0,1}$'s to ${0,1}$'s is so rich and so important. How can functions that you can see because they are finite be so important?
I would argue that this is a huge surprise that only recently has become clear. How could early mathematicians know that discrete simple functions were as important as continuous ones?
## Logic in the Looking-Glass
I—Ken writing this section—think that we can perceive the answer by examining the work of another Victorian-era logician, Charles Dodgson. That is, Lewis Carroll. Carroll’s work in logic mainly involved both formalizing and visualizing arguments based on syllogisms. Syllogisms involve quantifiers, and their terms refer directly to categories and attributes.
Today we might say that the various syllogistic theories of his time are fragments of the full predicate calculus in which quantifier forms are limited in complexity. Bounded-quantifier logic and arithmetic are subtle in computational theory even today. Thus the academics were following the long shadow of Aristotle into meaningfully intricate work, and did not see need to re-base it.
Carroll was influenced by John Venn and his diagrams for visualizing set-theoretic relations. He is known to have contacted Charles Babbage, but perhaps only as far as Babbage’s devices could be used for numerical calculation.
Even the universality of the binary NAND and NOR functions was not published until 1913 by Henry Sheffer—again, just a hundred years ago. Peirce had observed it thirty years earlier, but didn’t publish it. Even here the thinking seems to have been elsewhere from simply functions of ${0}$ and ${1}$. NAND and NOR were viewed as “the same” since they are dual functions in a Boolean algebra—alike in a looking-glass. Even the Sheffer stroke originally stood for NOR before its current usage as NAND.
All of these conceptualizations tended away from our modern ideas of logic as based on ${0}$ and ${1}$, of computation being based on this logic, and even of “it from bit.”
## Open Problems
So who really did invent Boolean functions? Why were they not invented a thousand years ago? And what about our current understanding may seem “immature” a century from now?
[fixed Pierce->Peirce]
### Like this:
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34 Comments leave one →
1. February 24, 2013 10:38 pm
The spelling is “Charles Sanders Peirce”, the last name pronounced like “purse”, the original family name being “Pers”.
• February 25, 2013 12:31 am
Ah, thanks! I did note the “ampheck” usage while editing but not the spelling of his name…
2. February 24, 2013 10:50 pm
Here is a useful reference:
Geraldine Brady • From Peirce to Skolem : A Neglected Chapter in the History of Logic
3. TCNE
February 24, 2013 11:06 pm
>>> Why were they not invented a thousand years ago?
It is really meant to be an OPEN question? In what sense?
Mankind had a collective mind to have a choice a thousand years ago?
If it had had, would every individual in that collection have contributed? So if taken our current version of BF versed in the language of that time, you think things could have been different? Just wondering.
BTW, in the previous one, it obviously should be M0 -> M3 (not M1 ->M3). Sorry for the error.
4. February 24, 2013 11:30 pm
See the PlanetMath entry, “Ampheck”, for one of Peirce’s latter discussions of the NAND/NNOR operators.
5. Anonymous
February 25, 2013 12:51 am
Maybe they were not invented because people didn’t have enough interest in them. What could they be used for? Logic was more philosophy than mathematics. On the other hand, this is not true about real numbers and functions over them.
The general concepts of real number and function as we know are rather new. As you know there were arguments among mathematicians if there can be a no where differentiable continuous function. The naive notion of real function was rather simple: lines on plane that we can draw. They are far from set theoretic notion. Similarly functions were not set theoretic but simple rules to obtain output from input. These notions were not formal and rigorous but intuitive ones.
6. February 25, 2013 6:26 am
I think that Boolean logic was not invented because the European and Arab intellectual worlds were so in thrall to Aristotle. Schiller’s book on Formal Logic, published a century ago, has no reference to either Leibniz or Boole, the two possible claimants, but regards formal logic as Aristotelian syllogistic logic.
My comments on this are on my blog under Formal Logic.
• February 26, 2013 12:01 am
Hi, Peter—indeed this is exactly what I agree with. I was unable to find a reference I read last summer or so that I recall as really drawing a connection between syllogistic theories and bounded-quantifier arguments.
I also read as a child someone’s assertion that had Archimedes been adopted rather than killed by the Romans, we might have had calculus some decades later and spaceflight by 500 A.D. ¿Hay por Google también? Actually I think “hay” is important—yet another reference I remember is we needed hay to be invented to get anything off the ground, and that didn’t happen until closer to 900 AD.
• February 26, 2013 12:03 am
Whoops: campus network outage made me edit a post from my home Mac for the first time ever, and I forgot to sign out as Pip and back in as me.
7. February 25, 2013 9:46 am
The question recalls recent discussions of discovery and invention in the mathematical field, bringing back to mind questions I’ve wondered about for as long as I can remember.
Speaking as an unreconstructed Platonic realist, I am tempted to say that Boolean functions are mathematical objects that cannot be invented, only discovered. But speaking as a semiotic constructivist I would have to concede that we do indeed invent all sorts of syntactic systems for talking and thinking about these mathematical objects. And some calculi can even be better than others for the purpose of calculation, a fact that repays us to consider the alternatives as they work out in practice.
On the third hand, I have more lately been thinking that the concepts of discovery and invention, being human constructs like the proverbial concepts of particles and waves, may not be adequate in the final analysis to articulate the reality of the process at hand. It may well be that all of these questions are more like the question, Who discovered Orion in the night sky?
• John Sidles
February 27, 2013 12:34 am
Regarding discovery-versus-invention, Domokos Szasz attributes the following all-purpose quotation to Alfréd Rényi: “Other mathematicians prove what they can, von Neumann what he wants.”
8. Alex
February 25, 2013 10:34 am
Until the 18th century, I think that many mathematicians and natural philosophers approached problems geometrically, since Euclid was the one source everyone knew. Look at the development of calculus: the two main problems that people wanted solved were the problems of tangents and rectification of various curves. Even Newton wrote the Principia using the language of geometry, since nobody would have understood the algebraic approach.
9. Oliver Heaviside
February 25, 2013 9:26 pm
C. Shannon’s master’s thesis applied boolean algebra to the design of switching circuits (sort of a precursor to digital logic design).
http://en.wikipedia.org/wiki/A_Symbolic_Analysis_of_Relay_and_Switching_Circuits
• February 26, 2013 12:16 am
Cf. Peirce to Allan Marquand (30 Dec 1886)
10. February 26, 2013 4:48 am
The same issues of infinity arise in the trouble that early mathematics had in defining precisely the notion of function. Again a proper definition requires that we have infinite sets.
No, this is not true. The proper definition of function is in terms of cartesian closure, and this definition works perfectly well even with finite models. (For example, even if you drop the axiom of infinity from set theory, you can still construct the set of functions between any two sets.)
The difficulties 19th century mathematicians had with functions had more to do with the fact that they had not yet developed the formalism for handling free variables (a problem no one made progress on until Frege), which made giving an algebraic formulation of function very difficult. Church solved this problem with the invention of the lambda-calculus in the 1930s, and Lambek’s recognition in the early 70s that the lambda-calculus gives a syntax for cartesian closed categories was a major impetus for the import of categorical methods into theoretical computer science.
11. Serge
February 27, 2013 1:21 pm
Our brain is composed of finitely many neurones. Nonetheless, our mind can sometimes perceive the infinite. I don’t know if this fact is connected in any way with the Löwenheim-Skolem theorem, but the illusion of infinity is known to have misled the great Dedekind himself. In his essay on “The Nature and Meaning of Numbers”, he appeals to “[his] own realm of thoughts” to “prove” that “the totality of things which can be objects of [his] thought is infinite”…
I regard Dedekind as the true inventor of the Peano axioms, for he was the first to base a definition of the natural numbers on the successor operation. But to him, infinity was such an obviously granted concept that it didn’t require any axiom. In the present days that situation has been reversed and most people now find it hard to “believe” in infinity… witness the recurring debate on the validity of Cantor’s proof on this blog.
12. March 1, 2013 3:45 pm
Computing, in its way, and science, and its broader way,
both involve the relation between what appears limited
and what appears not. Whether you believe in divinity
or not, and whether you believe that humanity contains
a spark of divinity or not, we have to acknowledge that
our powers as oracles are limited and, even if they were
not, problems of relative computability would still arise
in the striving of oracles to communicate with one another.
• March 2, 2013 10:36 am
Edit —
Computing, in its way, and science, in its broader way …
13. Robert Furber
March 3, 2013 6:35 pm
What is interesting about this is not the particular result, but that he may have been the first to prove an independence result via model theory.
I think Beltrami’s construction of a model of the hyperbolic plane was a little earlier. This shows that the parallel postulate is not implied by the other axioms of Euclidean geometry.
I first thought Bolyai, Lobachevski or Gauss or somebody must have provided a model, but I cannot find any reference to this being so.
14. March 4, 2013 8:14 am
I just ran across some notes on model theory that I wrote out a decade ago when a number of online working groups were tackling the subject for the sake of applications to web ontologies and semantic webs. I focused especially on how the sibling subjects of formal semantics and representation theory grew out of their native ground in the natural semantics and pragmatics of mathematics.
• Model Theory Unplugged
15. March 4, 2013 2:22 pm
Peirce’s published works are available as downloadable PDFs from the Institute for Studies in Pragmaticism at Texas Tech —
• Published Works of Charles Sanders Peirce
16. March 5, 2013 3:11 am
This is sociological issue. People are not ready to accept new ideas if they do not see their application. Like with Cantor uncountability argument it was useful at a time. As a side note, I make public crackpot notes that are ignored. In the private e-mails I get responses. Look, it took 10 min (including my explanation) of my friend to point out to determinantial rings in my monomial example. Roughly speaking society support “my” writing to 100 top researchers either each spending their time, explaining “me” what is going on, or ignoring “me”. In the first case, you spend much more resources than needed, in the second case, you possibly through away baby. I remember in depth analysis of a particular countability arguments to convince author he is wrong, whereas my simple intuitive argument were ignored (I’m not pushing it, I do not care, but others too). I know we are speaking different language, mine of cause, extraterrestrial alien. It is not truth that is searched by this society, but sparseness, leading to domination. No building starts with rigor, it starts with draft drawing. It is only when people like the draft they push the process.
Now how can I believe you, mathematicians, telling me you are searching for truth, whereas even in the Cantor diagonal argument is not that simple as you are trying tell, and it is more teaching convenience that the truth.
Draft (stop reading here, until I sent it to you by e-mail).
We start with random $inf$ x $inf$ table of “0″ and “1″ each drawn from Bernoulli process. Rows represent numbers. Starting from top to down we are going to write the Cantor diagonal number. First we write digit that is different from the first column in the first row. So we have the number that is certainly different from the first number, than by induction if we have a number that is different from top k numbers in first k columns we are write k+1 digit in Cantor diagonal number being different from k+1 digit in k+1 row. Therefore, by the induction the Cantor diagonal number different from all numbers in the table. Next we take the Cantor diagonal number to show that it is in the table, moreover countably many times. We start from the first column, and select those numbers that have the same first digit as Cantor diagonal number – there are countably many of them. induction step. From the selected countably infinite set, where each number coincides with Cantor diagonal number up to digit k we select countably infinite subset of rows which coincide in k+1 digit with k+1 digit of Cantor diagonal number. Therefore, by induction argument there are countably infinite numbers in the table that coincide with Cantor diagonal number. QEDraft.
If we assign weights to the columns ($2^{-k}$) than we immediately see (constructing Cauchy sequences from sequence of selected sets, they are by the way once drawn are well ordered by the row index) there are infinitely many numbers in the table distance 0 from Cantor diagonal number.
Finally, thanks to MrX, the main issue here is that if numbers are represented as a path of binary tree, this tree is self similar, i.e. every sub-tree is isomorphic to the whole tree. That probably is the reason for incompleteness theorem, although it is interesting how algebraic operations stratify the tree.
Sincerely yours, never looking truth, only the fame, crackpot trisector.
• March 5, 2013 10:04 am
The interplay of mathematical inquiry and social interaction, especially when taken over historical time, is certainly an interesting subject. But this looks like the big middle of a discussion whose beginning I must have missed. Just by way of identifying the ballpark, is your concern more about uncountability or more about indirect proofs?
It’s my sense of things that people who are tea-totalers about constructive mathematics or intuitionistic logic have voluntarily chosen to work under a non-classical aesthetic and here as in all matters of taste there is little chance of arguing to a settled conclusion unless the decision of nature or reality or some other umpire is final.
• March 5, 2013 11:02 am
I have no concerns. the internet is the place where information should be sorted, and presented for advances. The ideal case is the social network of mental workers, providing algorithm for mechanical devices, producing goods for mental workers to at least support their life, and the resource distribution mechanism not connected to human beings. as the first step no ads payed by the producer.
1) наши дни к веселью мало оборудованы;
2) наши дни под радость мало оборудованы;
3) наши дни под счастье мало оборудованы;
4) наша жизнь к веселью мало оборудована;
5) наша жизнь под радость мало оборудована;
6) наша жизнь под счастье мало оборудована;
7) для веселий планета наша мало оборудована;
8) для веселостей планета наша мало оборудована;
9) не особенно планета наша для веселий оборудована;
10) не особенно планета наша для веселья оборудована;
11) планетишка наша к удовольствиям не очень оборудована;
и, наконец, последняя, 12-я –
12) для веселия планета наша мало оборудована.
• March 5, 2013 11:12 am
In English all functions begin with fun.
• March 5, 2013 11:27 am
17. March 5, 2013 2:23 pm
My statistics teacher liked to complain that he thought it unfair how such a simple idea could get one’s name attached to it for all time. On the other hand, he would say, I can’t remember which brother it was, so maybe there is compensation. On a related note, my favorite quote about the discrete and continuous properties of mathematical study…
“The sole natural object of mathematical thought is the whole number. It is the external world that has imposed the continuous upon us…
“…we have devoted almost all our time and all our strength to the study of the continuous. Who will regret it; who will think that this time and this strength have been wasted? Analysis unfolds before us infinite perspectives that arithmetic never suspects; it shows us at a glance a majestic assemblage whose array is simple and symmetric; on the contrary, in the theory of numbers, where reigns the unforseen, the view is, so to speak, blocked at every step.” Poincare’ at the first Int’l Congress of Mathematics, Zurich, 1897.
18. March 13, 2013 2:00 pm
A comment that I meant to make here is turning into a work in progress, so here’s the first installment —
• Indicator Functions : 1
19. March 15, 2013 9:50 am
(Note. I’m trying to see if I can copy an image to a comment box. If it doesn’t work then try the link in my previous comment.)
One of the things that it helps to understand about 19th Century mathematicians, and those who built the bridge to the 20th, is that they were capable of high abstraction — in Peirce’s case a cut above what is common today — and yet they remained close enough to the point where abstract forms were teased away from the concrete materials of mathematical inquiry that they were able to maintain a sense of connection between the two. There are few better places to see this connection than in the medium of Venn diagrams. But Venn diagrams are such familiar pictures that it’s easy to overlook their subtleties, so it may be useful to spend some time developing the finer points of what they picture.
There are actually several types of Boolean functions that are represented in a typical Venn diagram. They all have the Boolean domain $\mathbb{B} = \{ 0, 1 \}$ or one of its powers $\mathbb{B}^k$ as their functional codomains but their functional domains may vary, not all being limited to finite cardinalities. As an aid to sorting out their variety, consider the array of functional arrows in the following figure.
Suppose $X$ is a universe of discourse represented by the rectangular area of a Venn diagram. Note that the set $X$ itself may have any cardinality. The most general type of Boolean function is a map $f : X \to \mathbb{B}.$ This is known as a Boolean-valued function since only its functional values need be in $\mathbb{B}.$
A function of the type $f : X \to \mathbb{B}$ is called a characteristic function in set theory or an indicator function in probability and statistics since it can be taken to characterize or indicate a particular subset $S$ of $X,$ namely, the fiber or inverse image of the value $1,$ for which we have the notation and definition $f^{-1}(1) = \{ x \in X : f(x) = 1 \}.$
The notation $f_S$ is often used for the characteristic function of a subset $S$ of $X.$ All together then, we have $f_S^{-1}(1) = S \subseteq X.$
To be continued …
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http://mathoverflow.net/questions/114893?sort=newest
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## A “holomorphic” Peano curve?
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A Peano curve is a continuous map $[0,1]\to [0,1]^2$ whose image is the whole square.
I would like to know if on can obtain "holomorphic" Peano curves. Namely, is it possible to find a continuous map $\phi$ from the unit disk $|z|\le 1$ to $\mathbb C^1$ such that $\phi$ is holomorphic for $|z|<1$ and the image of the boundary $|z|=1$ has non-empty interior in $\mathbb C^1$ under the map $\phi$.
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## 2 Answers
Here it is: MR0015154 Salem, R.; Zygmund, A. Lacunary power series and Peano curves. Duke Math. J. 12, (1945). 569–578.
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2
And here is the link: projecteuclid.org/… – Brian Rushton Nov 30 at 4:06
### 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.
Define $$\phi(z):=\frac{1}{2\pi i}\int_{S^1}(\zeta-z)^{-1}\cdot \varphi(\zeta)d\zeta$$
for $|z|<1$, where $\varphi: S^1\to\mathbb{C}$ is a Peano curve (i.e. its image has nonempty interior), and $\phi(z):=\varphi(z)$ for $z\in S^1$. [Edit: this construction doesn't work because $\phi$, as I defined it, may not be continuous up to the boundary - see the comments]
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In general there is no reason for this function $\phi$ to be contunuous on $\overline{\mathbb{D}}$! For example, if negative terms are present in the Fourier expansion of $g$, then certainly a continuous extension of the Cauchy Transform of $g$ to the circle will not be equal to $g$. – Malik Younsi Nov 29 at 18:16
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I see. You are right! I will not delete the answer because it might be an "instructive mistake" (I'm not an analyst, btw) - I will edit adding a disclaimer. – Qfwfq Nov 29 at 18:25
@Qfwfq : Yes, it is instructive. Thank you for the edit! – Malik Younsi Nov 30 at 12:48
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http://mathoverflow.net/questions/85693/from-sato-grassmannian-to-spectral-curve/85695
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## From Sato grassmannian to spectral curve
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Assume a tau-function of the KP integrable hierarchy is fixed by the point of the Sato grassmannian (that is by a semi-infinite set of Laurant series $\varphi_k(z)=\sum_{m>0}a_{km}z^{-k+m}$). Can one restore a spectral curve that corresponds to this solution?
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## 1 Answer
This is explained in Segal-Wilson. Essentially, realize your point in the Grassmannian as as a space W of functions w(z), then look for all functions g(z) such that g(z)W is included in W. These functions form a commutative algebra, and Spec of this algebra is your spectral curve. Of course for most points of the Grassmannian the commutative algebra is just C.
But this is all beautifully explained in Segal-Wilson.
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Maarten, thank you for your ansver. I think I have to re-read Segal-Wilson. Just to be sure: assume I know the coefficients $a_{km}$ explicitely. Is there a step-by-step procedure to construct a spectral curve? – Sasha Jan 14 2012 at 22:38
I am not sure what your notation means. Is \phi_k a basis for the point of the Grassmannian? In any case, for a generic point of the Grassmanian the spectral curve will be trivial. – Maarten Bergvelt Jan 14 2012 at 22:44
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Well, look for all g such that $g\phi_k$ is a linear combination of the basis elements. You get a bunch of equations for the coefficients of g. Yes, it is an interesting exercise to see if the Kontsevich-Witten point has a spctral curve. I do not know the answer, but you would expect that the curve would be trivial, no? It would be very strange if the genus of this curve would be any finite number. I am not sure if you can get infinite genus curves in this way. – Maarten Bergvelt Jan 15 2012 at 0:04
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@Sasha, you and David may have a different definition of what it means for a solution to be "described" by a spectral curve. Krichever's definition (explained e.g. in Segal-Wilson) is that the stabilizer of $W$ (the algebra of all $g$) is sufficiently big, for example $W$ is a rank 1 module over it. All string solutions to KdV have the minimal possible stabilizer $\mathbf{C}[z^2]$ (and so $W$ has rank 2). – Pavel Safronov Jan 15 2012 at 16:04
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The Kontsevich-Witten tau funtion is in fact a solution of the Kortewg-de Vries hierarchy, not just of KP. This means that the corresponding point of the Grassmannian comes from an element Of the loopgroup of Sl_2, not from an element of the general linear group. Such points always have z^2 as stabilizer, so for trivial reasons we get a P^1 as spectral curve. The "honest" spectral curves for KdV are 2-fold covers of this base curve. See, as always, Segal-Wilson. – Maarten Bergvelt Jan 16 2012 at 2:47
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http://mathoverflow.net/questions/56563/why-does-homotopy-behave-well-with-respect-to-fibrations-and-homology-with-respec
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Why does homotopy behave well with respect to fibrations and homology with respect to cofibrations?
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(I apologize that this is a vague question).
I seems to me somehow that homotopy groups behave well with respect to (Serre)-fibrations. For example you get a long exact sequence of homotopy groups from it. On the other hand cofibrations and homotopy groups seems to be no good friends at all (e.g. $S^1\to D^2\to S^2$).
But then again, the situation in homology seems to be the other way round. They behave well with respect to cofibrations (you get a long exact sequence) and fibrations are harder to investigate (Serre spectral sequence etc.).
My question is: What is the intuition behind this difference? (in particular with respect to the fact that homology is just homotopy of another space.)
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2 Answers
I agree that the long exact sequence in homotopy groups of a fibration follows from the fact that fibrations are defined using a mapping property in which the fibration is the target.
One way to understand why homology behaves well with respect to cofibrations is to spell out your remark that "homology is just homotopy of another space". This is true, but not obvious. There are a number of constructions of ordinary homology which take the following form.
One finds a functor $F$ from (pointed) spaces to (pointed) spaces which takes cofibrations to quasifibrations. (A quasifibration is something for which you have a long exact sequence of homotopy groups, for example a Serre fibration). And then $H_* (X) \cong \pi_* F(X).$ If $X\to Y \to Z$ is a cofibration (maybe a cofibration of CW complexes), then $\dots \to \pi_* F(X) \to \pi_* F(Y) \to \pi_* F(Z) \to \ldots$ is the long exact sequence in homology associated to the cofibration.
Here are several contexts in which one can describe such a functor $F$.
First, a formal approach. Let $\mathbf{S}$ denote the category of spectra: it is connected to the category $\mathbf{T}$ of spaces by adjoint functors $\Sigma^\infty: \mathbf{T} \to \mathbf{S}$ and $\Omega^\infty: \mathbf{S} \to \mathbf{T}.$ There a spectrum called the "Eilenberg-MacLane" spectrum, denoted $H\mathbb{Z}$: its job is to represent singular cohomology, and one can take $F(X) = \Omega^\infty ((\Sigma^\infty X) \wedge H\mathbb{Z})$.
Why does $F$ have the cofibration-to-quasifibration property? Well, the way that this is set up, $\Sigma^\infty$ preserves cofibrations of CW complexes, and $\Omega^\infty$ preserves fibrations of fibrant objects, and in the category of spectra every cofibration is equivalent to a fibration.
To be more explicit about $H\mathbb{Z}$, you can define $F(X) = \lim (\ldots \Omega^k(X \wedge K(\mathbb{Z},k)) \to \Omega^{k+1} (X\wedge K(\mathbb{Z},k+1))a \ldots),$ where the limit is a colimit and the maps defining the system arise from the maps $\Omega K(\mathbb{Z},k) \simeq \Omega K(\mathbb{Z},k+1)$.
Second, the Dold-Thom theorem says that one can take $F(X) = Sp^\infty (X).$ Here $Sp^n(X) = X^n/\Sigma_n$, and $Sp^\infty(X) = \lim \ldots Sp^n(X) \to Sp^{n+1}(X) \ldots$, again the limit is a colimit.
Third, if you're willing to allow $X$ to be a simplicial set, then one can take then one can take $F(X) = \mathbb{Z}X$, the simplicial set whose $n$ simplices are the free abelian group on the $n$-simplices of $X$. (This approach is due to Dan Kan; see the proceedings of the Hurewicz conference)
All of this is to focus attention on functors which take cofibrations to quasifibrations. In fact all $-1$-connected generalized homology theories (at least the ones associated to cohomology theories: are there homology theories which are not? I don't know) are of the form $E_* X = \pi_* G(X)$, where $G$ is a functor which takes cofibrations to quasifibrations. Indeed one takes $G(X) = \Omega^\infty (\Sigma^\infty X \wedge R)$, where $R$ is the spectrum representing the cohomology theory. This approach goes back to G. W. Whitehead.
One of the more compact discussions of such a functor, which I like, is in an article by G. Segal in Springer LNM 575; he gives a construction of connective real $K$-homology there. Really he's showing how to generalize the work of Dold and Thom: Segal's argument applies just as well to $Sp^\infty(X)$.
I apologize that throughout I have done a poor job of saying how to handle basepoints.
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Fibrations are defined as target-type concepts -- they have good formal homotopy properties when you map into them, for example, if you apply the functor $\pi_n(-) = [S^n,-]$.
Now I'll subject you to my point of view on homology.
Dually, cofibrations are a domain-type concept, so they behave well when you map out of them, say if you apply (represented) cohomology: $\widetilde H^n(-;G) = [-, K(G,n)]$.
Homology is a bit of a monster: it is a covariant functor that works well with domain-type input. It is a small miracle that such functors exist at all; this is why they are hard to construct. But the answer for homology has to be: homology works well with cofibrations because we built it to work well with cofibrations, and a more informative answer to your question would depend on the construction you use.
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@Jeff: I don't get this: "Homology is a bit of a monster: it is a covariant functor that works well with domain-type input. It is a small miracle that such functors exist at all; this is why they are hard to construct." – John Klein Feb 24 2011 at 21:35
@John: As far as I know, the exactness of homology (or its invariance with respect to suspension) depends on a substantial theorem (such as Blakers-Massey, or Freudenthal, or something else nontrivial). There is no way that I know of to construct a functor that is covariant and exact on cofibrations without invoking a real theorem. – Jeff Strom Feb 24 2011 at 22:47
What about the ordinary definition of singular homology? – Eric Wofsey Feb 25 2011 at 0:13
@Eric: Explain how you prove exactness, and I'll endeavor to point out the major theorem hiding in your proof. – Jeff Strom Feb 25 2011 at 0:57
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Eric, having spent an hour yesterday putting the proof of homotopy invariance on the blackboard in class, I'm going to agree with Jeff: it's a major pain! Yes, there's a nice geometric idea about prisms, and Hatcher's rather loose notation makes it look a little simpler than it really is, but the details are messy and not at all formal. For instance, how did Hatcher decide to break up the sums the way he did in his proof? One has to get one's hands pretty dirty before coming away with a clean argument. There's hard work in there! Same goes for barycentric subdivision of chains. – Dan Ramras Feb 26 2011 at 2:10
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http://mathhelpforum.com/latex-help/135979-part-code-involved-matrices.html
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# Thread:
1. ## part of code involved in matrices
what does in \begin{array}{ccccc} what does the cccccc do and how is the number of characters determined? for example, do I need 5 c's for a matrix that has 5 columns or something?
2. Originally Posted by superdude
what does in \begin{array}{ccccc} what does the cccccc do and how is the number of characters determined? for example, do I need 5 c's for a matrix that has 5 columns or something?
Yes, the number of letters in the second set of brackets indicates the number of columns. You can list either a "c", "r", or "l" to signify the justification of the numbers in the matrix. So, for example, in the command you listed, you have five columns with the numbers centered in each column.
3. You can have more c's than required:
$\left[\begin{array}{ccc}1&2\\3&4\\5&6\end{array}\right]$
which allows you to leave blank spaces:
$\left[\begin{array}{ccc}1&2&3\\&4\\5&6\end{array}\right]$
- Hollywood
« Test | test »
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http://math.stackexchange.com/questions/190022/how-to-find-expected-time-to-reach-a-state-in-a-ctmc?answertab=oldest
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# How to find expected time to reach a state in a CTMC?
Given a simple CTMC with three states 0,1,2. There are three transitions $0 \rightarrow 1$ (with rate $2u$), $1 \rightarrow 2$ (with rate $u$), $1 \rightarrow 0$ (rate $v$).So $2$ is an absorbing state. Can someone please help me understand how to find the expected time to reach state-2 from state-0? Using the understanding I get, I want to do the analysis on some complex markov chains.
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Did you try the standard method? That is, introduce the expected time t(x) to reach state 2 starting from state x, find a linear system which the t(x) solve, deduce every t(x), and in particular t(0), which is the expected time you are after. – Did Sep 2 '12 at 13:52
## 2 Answers
You do it in a similar way as you would for finding the expected first passage time for a discrete chain. If $Q = q_{ij}$ is the transition rate matrix, $P = p_{ij}$ is the transition matrix of the embedded chain, \operatorname{Boole} is the indicator function and $T[i, Q]$ is the first passage time starting at state $i$, then
$$\begin{align*} \operatorname{\operatorname{Mean}}[FPT[i, Q]] &= \operatorname{Expectation}[FPT \mid X[0] = i]\\ &= \operatorname{Expectation}[\operatorname{SojournTime} \mid X[0] = i] + \sum _{j=1}^{\infty} p_{ij} \operatorname{Expectation}[FPT \mid X[0] = j]\\ &=-\frac1{q_{ii}} + \sum _{j=1}^{\infty} \frac{-q_{ij}}{q_{ii}} \operatorname{Boole}[j\neq i] \operatorname{Mean}[FPT[j, Q]] \end{align*}$$
where the sojourn/holding time is, of course, exponentially distributed with parameter $-q_{ii}$.
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Since the OP stays silent, here is an answer. As explained in a comment, the standard method is to introduce the expected time $t_x$ to reach state $2$ starting from each state $x$, to find a linear system which the vector $(t_x)_x$ solve, to deduce $(t_x)_x$, and in particular $t_0$ which is the expected time asked for.
Here, $t_0=\frac1{2u}+t_1$ and $t_1=\frac{v}{u+v}(\frac1v+t_0)+\frac{u}{u+v}\frac1u$, hence $t_0=\frac1{2u}+\frac2{u+v}+\frac{v}{u+v}t_0$, which yields finally $t_0=\frac{5u+v}{2u^2}$.
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http://math.stackexchange.com/questions/96270/what-is-zero-of-multiplicity/96272
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# What is 'zero of multiplicity'?
What is a 'zero of multiplicity'?
I looked up this definition on Mathwords.com:
How many times a particular number is a zero for a given polynomial. For example, in the polynomial function $f(x) = (x – 3)^4(x – 5)(x – 8)^2$, the zero 3 has multiplicity 4, 5 has multiplicity 1, and 8 has multiplicity 2. Although this polynomial has only three zeros, we say that it has seven zeros counting multiplicity.
What is a 'zero of multiplicity' and where do the "seven zeros" come from (I know the "7" comes from adding the three exponents, "4", the hidden "1", and "2" in the equation, but why call them zeroes?)?
As in, I don't understand where the zero part of the term comes in...
(bold emphasis mine).
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## 6 Answers
A "zero" of a polynomial is a value of $x$ at which the polynomial, when evaluated, is equal to zero.
The compound phrase "of multiplicity $k$" (where $k$ is a positive integer) modifies "zero" (e.g., "zero of multiplicity 3", "multiplicity 3" and not just 'multiplicity' is modifying "zero"); what it means is that the zero is actually a solution "multiple times".
It is a theorem (called the Factor Theorem) that if $a$ is a zero of the polynomial $p(x)$, then you can write the polynomial $p(x)$ as $p(x)=(x-a)q(x)$; that is, a product. Any zero of $q$ is also a zero of $p(x)$.
We say that $a$ is a zero "of multiplicity $k$" of $p(x)$ if you can write $p(x)$ as $p(x)=(x-a)^kq(x)$, but not as $p(x)=(x-a)^{k+1}q(x)$.
For example, take $p(x)=x^2-2x+1$. Then $x=1$ is a zero of $p(x)$; in fact, since $p(x) = (x-1)^2$, $1$ is a zero "of multiplicity $2$".
Similarly, $p(x)=x^4 - 9x^3 + 30x^2 - 44x + 24$ has $x=3$ and $x=2$ as zeros (plug them in, you get zero: $p(3) = 81 - 243 + 270 - 132 + 24 = 0$, $p(2) = 16 - 72 + 120 - 88 + 24 = 0$). In fact, $p(x) = (x-2)^3(x-3)$, so $3$ is a zero "once" and $2$ is a zero "three times", so $2$ is a zero "of multiplicity three" and 3 is a zero "of multiplicity one".
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this helps, I'm getting there :) – studiohack Jan 4 '12 at 4:57
Recall that a number $r$ is a root or zero of a polynomial $P(x)$ if $P(r) = 0$: i.e., when you plug in $r$, you get zero.
Every root $r$ of $P(x)$ occurs with a certain multiplicity which is the number of times we can factor out $(x-r)$ from $P(x)$. In your example
$f(x) = (x-3)^4 (x-5)(x-8)^2$
the polynomial is conveniently written as a product of distinct linear factors raised to certain powers. These powers are then the multiplicity of the roots of the polynomial, so $3$ occurs with multiplicity $4$, $5$ occurs with multiplicity $1$ and $8$ occurs with multiplicity $2$.
Based on several years of teaching freshman calculus, I can say that this concept is one of the things that university-level teachers think is covered in precalculus mathematics but seems not to be, at least not in a way that makes precalculus students remember / understand it by the time they get to university calculus.
Casting about for a decent, elementary explanation of this material on the web, I found this page. (By contrast, wikipedia does a rather poor job...)
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Wow! Sir , you are back again ? , its been a long time to hear from you, I hope you are fine and will be fine, In order to add something to Prof.Pete's answer, this concept of zero with multiplicity has many good applications like even in Birch Swinnerton dyer conjecture , one connects the rank of underlying Mordell Weil group to the multiplicity of zero of L-function at $s=1$ @Pete.L.Clark – Iyengar Jan 4 '12 at 4:50
In your example, 3 is a zero of $f$, because $f(3) = 0$. That's all it means to call something a zero of a polynomial. If $n$ is some number, and $g(x)$ is a polynomial, we say $n$ is a zero of $g$ if $g(n)=0$.
Now, if we have two polynomials with different zeros, we know the polynomials are different, but how can we tell if $$f(x) = (x-2)(x-3)$$ and $$g(x) = (x-2)(x-3)^2$$ are different? They both have 2 and 3 as their roots, so that won't be quite enough to distinguish them. That's where the multiplicity of a root comes in. We say 3 is a zero of multiplicity 1 for $f$, whereas 3 is a zero of multiplicity 2 for $g$.
Another way to look at this, is if we look at $f(x)$ without the factor of $x-3$, namely, the polynomial $$f'(x) = \frac{f(x)}{x-3} = x-2,$$ we can see $f'(3) = 3-2 = 1 \neq 0$. However, if we look at $$g'(x) = \frac{g(x)}{x-3} = (x-2)(x-3),$$ we have $g'(3) = 0$. So in some sense, the multiplicity can be thought of as "how many times can we remove the zero from the polynomial, until it's no longer a zero?"
Caution: When I write the divisions above, do not try to plug the zeros in to them. I.e., you cannot say $$f'(3) = \frac{f(3)}{3-3} = \frac{0}{0},$$ which I claimed was 1 earlier, as you cannot divide by 0.
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A polynomial $\rm\:f(x)\:$ has a root ("zero") $\rm\:r\:$ of multiplicity $\rm\:n\:$ if $\rm\ f(x)\ =\ (x-r)^n\ g(x)\$ where $\rm\:g(r)\ne 0\:.\:$ Recall by the Factor Theorem that $\rm\:f(r) = 0\ \iff\ x-r\$ divides $\rm\ f(x)\:.\:$ The multiplicity simply counts how many factors of $\rm\ x-r\$ occur (the "degree" or "order" of the root $\rm\:r\:$).
Your example $\rm\ (x-3)^4\:(x-5)\:(x-8)^2\$ has $\ 4 + 1 + 2\ =\ 7\$ roots (zeros) counting multiplicities since the roots $\ 3\:,\:5\:,\:8\$ have multiplicity $\rm\ 4\:,\:1\:,\:2\$ respectively. Note that if we view the roots as a multiset $\rm\ \{3,3,3,3,5,8,8\}\$ then the multiplicity of a root is simplicity its multiplicity in this multiset, i.e. the number of times that it occurs.
Abhyankar, a master of algebraic geometry, remarks in his charming exposition [1] that
much of algebraic geometry ultimately gets reduced to the following Fundamental Principle (plain or supplemented).
Fundamental Principle. $\$ The number of roots (or irreducible factors) of a polynomial $\rm\:f(x)\:$ in one variable, counted with their multiplicities (resp. degrees and multiplicities), equals the degree of $\rm\:f(x)\:.\:$
It is certainly the algebraical key to the various 'counting properly.'
I highly recommend reading Abhyankar's paper. It explains simply and beautifully much more than how to algebraically count properly. Indeed, it won various prestigious awards for expository excellence (AMS Lester R. Ford, MAA Chauvenet Prize).
[1] Abhyankar. Historical Ramblings in Algebraic Geometry and Related Algebra
Amer. Math. Monthly 83 (1976), 409-448
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"zero" comes from the fact that those values (3,5,8 in the example) are the values where the polynomial is equal to zero.
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"the values where the polynomial is equal to zero"? What do you mean? this is exactly my hangup, can you explain? – studiohack Jan 4 '12 at 4:45
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If you substitute x=3, x=5, or x=8 into the polynomial, you get 0. And if you substitute any other number, you don't get 0. So 3,5,8 are the zeroes of the polynomial. – Ted Jan 4 '12 at 5:09
okay, that explains it... now what about the multiplicity part? – studiohack Jan 4 '12 at 5:12
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Polynomials have the property that if $r$ is a zero of the polynomial $p(x)$, then $p(x)$ is divisible by $x-r$. The multiplicity of a zero is how many times you can divide out $x-r$ from the polynomial before the remaining polynomial no longer has $r$ as a zero. For example, take the zero 3 of your polynomial. You can see that it's divisible by $(x-3)^4$, so you can divide out $x-3$ four times from the polynomial (after dividing by $(x-3)^4$, the polynomial that remains no longer has 3 has a zero), so the zero 3 has multiplicity 4. – Ted Jan 4 '12 at 5:34
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Another point I just realized from the title of your question: A phrase like "zero of multiplicity 4" should be parsed as "zero (of multiplicity 4)", not as "(zero of multiplicity) 4". In other words, there's no such thing as a "zero of multiplicity". Instead, a polynomial has a zero, and that zero has a particular number associated with it, which is called its "multiplicity". – Ted Jan 4 '12 at 5:42
With polinomials, you can always write it like:
$(x-x_{1})(x-x_{2})(x-x_{3})...$
where $x_{1}$, $x_{2}$, $x_{3}$ etc, are the solutions of the polinomial equation (aka roots aka zeros).
E.g. you can always write $x^2-5x+6$ as $(x-2)(x-3)$ because $2$ and $3$ are the solution of the equation $x^2-5x+6=0$.
Now if the polinomial is $x^2-6x+9$, we have that $x_{1}=x_{2}=3$, so we write that
$x^2-6x+9 = (x-3)(x-3) = (x-3)^2$
and we have that $3$ is a "double" solution to the equation, aka, 3 is a zero with a multiplicity of two.
Your original equation can we written as $f(x) = (x-3)(x-3)(x-3)(x-3)(x-5)(x-8)(x-8)$
which has seven terms, and the whole thing has a value of $0$ if and only if some of them are zero. So "zero 3 has multiplicity 4" is another way of saying that "if we set $x$ to $3$, four terms will be $0$"
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http://physics.stackexchange.com/questions/38348/is-the-principle-of-least-action-a-boundary-value-or-initial-condition-problem/38393
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# Is the principle of least action a boundary value or initial condition problem?
Here is a question that's been bothering me since I was a sophomore in university, and should have probably asked before graduating: In analytic (Lagrangian) mechanics, the derivation of the Euler-Lagrange equations from the principle of least action assumes that the start and end coordinates at the initial and final times are known. As a consequence, any variation on the physical path must vanish at its boundaries. This conveniently cancels out the contributions of the boundary terms after integration by parts, and setting the requirement for minimal action, we obtain the E.L. equations.
This is all nice and dandy, but our intention is finding the location of a particle at a time in the future, which we do not know a priori; after we derive any equations of motion for a system, we solve them by applying initial values instead of boundary conditions.
How are these two approaches consistent?
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Most treatments contain a footnote or remark that the method can be extended to include variation of the endpoints and do not elaborate (e.g. both Goldstein and Marion & Thorten). I'm afraid that I have never pursued the matter. – dmckee♦ Sep 26 '12 at 3:06
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Hi Benji, and welcome to Physics Stack Exchange! This is (and was) actually an excellent question :-) The only thing I'd suggest you remember for the future is not to put in a signature or greeting (including the "hope I'm not doing anything wrong" type); generally, we like to have questions be as concise as possible. Also, it often helps to phrase the title as a question, but it's not a requirement. – David Zaslavsky♦ Sep 26 '12 at 3:32
– Qmechanic♦ Sep 26 '12 at 5:38
## 4 Answers
I) Initial value problems and boundary value problems are two different classes of questions that we can ask about Nature.
Example: To be concrete:
1. an initial value problem could be to ask about the classical trajectory of a particle if the initial position $q_i$ and the initial velocity $v_i$ are given,
2. while a boundary value problem could be to ask about the classical trajectory of a particle if the initial position $q_i$ and the final position $q_f$ are given (i.e. Dirichlet boundary conditions).
II) For boundary value problems, there are no teleology, because we are not deriving a (100 percent certain deterministic) prediction about the final state, but instead we are merely stating that if the final state is such and such, then we can derive such and such.
III) First let us discuss the classical case. Typically the evolution equations (also known as the equations of motion(eom), e.g. Newton's 2nd law) are known, and in particular they do not depend on whether we want to pose an initial value question or a boundary value question.
Let us assume that the eom can be derived from an action principle. (So if we happen to have forgotten the eom, we could always rederive them by doing the following side-calculation: Vary the action with fixed (but arbitrary) boundary values to determine the eom. The specific fixed values at the boundary doesn't matter because we only want to be reminded about the eom; not to determine an actual solution, e.g. a trajectory.)
IV) Next let us consider either an initial value problem or a boundary value problem, that we would like to solve.
Firstly, if we have an initial value problem, we can solve the eom directly with the given initial conditions. (It seems that this is where OP might want to set up a boundary value problem, but that would precisely be the side-calculation mentioned in section III, and it has nothing to do with the initial value problem at hand.)
Secondly, if we have a boundary value problem, there are two possibilities:
1. We could solve the eom directly with the given boundary conditions.
2. We could set up a variational problem using the given boundary conditions.
V) Finally, let us briefly mention the quantum case. If we would try to formulate the path integral
$$\int Dq ~e^{\frac{i}{\hbar}S[q]}$$
as an initial value problem, we would face various problems:
1. The concept of a classical path would be ill-defined. This is related to that the concept of the functional derivative $$\frac{\delta S[q]}{\delta q(t)}$$ would be ill-defined, basically because we cannot apply the usual integration-by-part trick when the (final) boundary terms do not vanish.
2. To specify both the initial position $q_i$ and the initial velocity $v_i$ would violate the Heisenberg uncertainty principle.
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Thanks Qmechanic! I posed my question in a strict classical context, so while I agree that the limitation of uncertainty is indeed problematic, it is technically not existent in the Lagrangian formalism. – Benji Remez Sep 27 '12 at 6:07
So should your second point be interpreted as follows: I know that a particle is at a given point at an initial time, and will be in some, yet unknown, location at a later time, which I will be able to measure to arbitrary accuracy (because I deny uncertainty). Demanding minimal action, I find the differential equation that governs the physical path between the two locations - but now, technically, I can solve the equation with initial values instead, to "find" which final position would have yielded this initial momentum had the problem been solved under boundary conditions. Am I close? – Benji Remez Sep 27 '12 at 6:18
I updated the answer. – Qmechanic♦ Sep 27 '12 at 20:47
The usual derivation of the Euler-Lagrange equations forces us to assume that both the initial and final conditions are fixed. However, when one actually derives the equations, he sees that there are differential equations in time so from the knowledge of the initial state, including the velocities (or whatever derivatives are needed to specify the initial point of the phase space), one may derive the values at an infinitesimally later moment.
That's why the "teleological", acausal character of the principle of least action is just an illusion. The trajectory at time $t$ doesn't really depend on any "assumed" values of the fields at later times. This fact may not be obvious immediately, when the principle is formulated, but it is nevertheless true and easy to see via the mathematical derivation of what the principle implies.
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As I understand you, you mean that just because the equations are differential in time, this necessitates that the problem is an initial value problem? I doubt that's really what you meant. Can you refer me to a derivation where this distinction is manifest? – Benji Remez Sep 26 '12 at 5:01
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@Lubosh Motl: You wrote: "The trajectory at time $t$ doesn't really depend on any "assumed" values ... at later times." But yes, it does: if only $q(t_1)$ is known, then the differential equation determines a whole family of possible $q(t_2)$ and only final known data help to fix it and find the intermediary trajectory. – Vladimir Kalitvianski Sep 26 '12 at 14:51
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Benji, I don't understand your concern. The fact that the equations are differential means that one may define and solve initial value problems. That means that both initial coordinates and velocities (or momenta) must be given. Alternatively, one may specify initial and final conditions but only coordinates: momenta/velocities must be left free. Trying to impose both initial and final conditions for both coordinates and momenta would clearly mean an overdetermined problem that generically has no solutions. – Luboš Motl Sep 26 '12 at 16:39
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I fully understand that imposing both constrains creates an overdetermined problem. Clearly, only one set of conditions is needed - what is unclear to me is which one is used in the context of least action. Why is it derived under one assumption and solved under the other? – Benji Remez Sep 27 '12 at 6:02
The best way to understand this teleological property is the path integral--- the final boundary condition is required because the action is the amplitude phase for a path, and the stationary condition is saying that you are taking the path of stationary phase, so that the contributions add coherently.
Then the relation between this and the differential formulation is manifest, as Lubos Motl explains. The condition for an extremal path is enforced by a local differential equation. This is not mysterious, because the sum over all paths is not teleological at all, it becomes teleological when you consider that you seem to have taken a stationary path, but this is a consequence of the cancellation away from the stationary path.
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Mathematically it is a boundary value problem by definition, but physically it is practiced as an initial condition problem. Remember the Newton equation: $$p(t+dt)=p(t)+F(t)dt$$ This means the next moment value $p(t+dt)$ is determined with the local values of $p(t)$ and $F(t)$, and no future data is involved. This equation was first discovered as a differential one with the initial (known) conditions, and only later an "integral derivation" under condition of known initial and final coordinates was discovered. Mathematically the boundary value problem is correct and possible, but not physically.
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http://math.stackexchange.com/questions/30598/formal-schemes-mittag-leffler?answertab=active
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Formal Schemes Mittag-Leffler
Here is a question that is similar to my last one. I've been trying to learn about Grothendieck's Existence Theorem, but it seems that there aren't very many places that talk about formal schemes and even less that come up with examples.
Suppose $(\mathfrak{X}, \mathcal{O}_\mathfrak{X})$ is a Noetherian formal scheme and let $\mathcal{I}$ be an ideal of definition. Then we have a system of schemes $X_n=(|\mathfrak{X}|, \mathcal{O}_\mathfrak{X}/\mathcal{I}^n)$.
If the inverse system $\Gamma(X_n, \mathcal{O}_{X_n})\to \Gamma(X_{n-1}, \mathcal{O}_{X_{n-1}})$ satisfies the Mittag-Leffler condition (the images eventually stabilize), then we get some particularly nice properties such as $Pic(\mathfrak{X})=\lim Pic(X_n)$.
More generally, we don't have to be worried about converting between thinking about coherent sheaves on the formal scheme and thinking about them as compatible systems of coherent sheaves on actual schemes.
My question is, is there a known example of a formal scheme for which that system of global sections does not satisfy the Mittag-Leffler condition? One thing to note is that it can't be affine (the maps are all surjective) or projective (finite dimensionality forces the images to stabilize).
A subquestion is whether or not there is a general reason to believe such an example exists. People I talk to usually say things along the lines of: you definitely have to be careful here because in principle this could happen. But no one seems to have ever thought up an example.
Lastly (still related...I think), is there a known example where you can't think of coherent (or maybe just invertible) sheaves as systems because the two aren't the same?
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In principle. ${}$ – Mariano Suárez-Alvarez♦ Apr 3 '11 at 7:38
Ah, thanks! Embarrassed – Matt Apr 3 '11 at 17:38
1 Answer
I don't have enough reputation to comment, so I must submit an answer which I do not really have. There is one class of varieties you must consider when searching for a counterexample: quasiprojective varieties. That is, open subsets of projective varieties which are not affine, e.g a surface minus a point. You also want the maps in question to fail to be surjective, so you should look for an ideal $\mathcal I$ which satisfies $H^1(X, \mathcal I^r / \mathcal I^{r+1}) \neq 0$ for all $r>>0$. This sounds like it's harder to do, but not impossible if you know examples.
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http://mathoverflow.net/revisions/89993/list
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## Return to Question
2 added suggested tag; made clear that x is a function of t
I asked this question a week ago over on math.stackexchange and got no reply, so I am asking here with slightly different wording. I am trying to prove that there exists a solution to a problem. I can't solve this problem in general, but I can always continuously vary the parameters of the general problem to turn it into a specific case which is easy to solve. I believe that the solution of the easy problem can then be varied continuously to be a solution to the hard problem.
It seems to me that this would be a widely used technique, but I haven't seen it before, and I don't know how to prove that it works. To be specific, suppose that $g:\mathbb{R}^n \times \mathbb{R} \to \mathbb{R}^n$. I have a solution $x_0$ to $g(x_0, t=0)=0$. I desire to show the existence of a solution to $g(x_1, t=1)=0$. My strategy is to take the known solution $x_0$ and vary it from $t=0$ to $t=1$. What I need is a theorem of the following form:
Suppose $g(x_0, t=0)=0$, and suppose that $\exists f$ such that \begin{equation} \frac{dx}{dt}=f(x,t) \implies \frac{dg(x,t)}{dt}=0. frac{dg(x,t)}{dt}=0, \end{equation} with $x$ a function of $t$. If $f$ and $g$ are "well behaved enough" in the vicinity $|g(x,t)| < \epsilon$ for some $\epsilon$, then $\exists x$ such that $g(x,1)=0$.
$f$ and $g$ in my case are "well behaved enough" that I can probably prove just about any sort of continuity condition that is needed. What properties of $f$ and $g$ are needed, and what theorem will help me here? It looks like Picard-Lindelof may help, but it seems to only give the existence of a unique solution to the differential equation, and I need to show that that solution satisfies $g(x_1, t=1)=0$. Furthermore, $f$ is not well behaved when $g$ is far from zero, and so it seems I cannot use Picard-Lindelof without prior assumption that $g(x,t)$ stays small (which is kind of assumes the fact that I am trying to prove).
1
# Continuous variation from solution of easy problem to solution of hard problem
I asked this question a week ago over on math.stackexchange and got no reply, so I am asking here with slightly different wording. I am trying to prove that there exists a solution to a problem. I can't solve this problem in general, but I can always continuously vary the parameters of the general problem to turn it into a specific case which is easy to solve. I believe that the solution of the easy problem can then be varied continuously to be a solution to the hard problem.
It seems to me that this would be a widely used technique, but I haven't seen it before, and I don't know how to prove that it works. To be specific, suppose that $g:\mathbb{R}^n \times \mathbb{R} \to \mathbb{R}^n$. I have a solution $x_0$ to $g(x_0, t=0)=0$. I desire to show the existence of a solution to $g(x_1, t=1)=0$. My strategy is to take the known solution $x_0$ and vary it from $t=0$ to $t=1$. What I need is a theorem of the following form:
Suppose $g(x_0, t=0)=0$, and suppose that $\exists f$ such that \begin{equation} \frac{dx}{dt}=f(x,t) \implies \frac{dg(x,t)}{dt}=0. \end{equation} If $f$ and $g$ are "well behaved enough" in the vicinity $|g(x,t)| < \epsilon$ for some $\epsilon$, then $\exists x$ such that $g(x,1)=0$.
$f$ and $g$ in my case are "well behaved enough" that I can probably prove just about any sort of continuity condition that is needed. What properties of $f$ and $g$ are needed, and what theorem will help me here? It looks like Picard-Lindelof may help, but it seems to only give the existence of a unique solution to the differential equation, and I need to show that that solution satisfies $g(x_1, t=1)=0$. Furthermore, $f$ is not well behaved when $g$ is far from zero, and so it seems I cannot use Picard-Lindelof without prior assumption that $g(x,t)$ stays small (which is kind of assumes the fact that I am trying to prove).
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http://mathoverflow.net/questions/119429/eigenfunctions-restricted-on-closed-geodesics/119437
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## Eigenfunctions restricted on closed geodesics
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Consider the flat torus $T^2=\frac{\mathbb{R}^2}{l_1\mathbb{Z}\oplus l_2\mathbb{Z}}$. It is easy to see that the eigenvalues of the Laplacian on torus, $-\frac{\partial^2}{\partial x^2}-\frac{\partial^2}{\partial y^2}$, are $\lambda_{m_1,m_2}=(2\pi)^2(\frac{m_1^2}{l_1^2}+\frac{m_2^2}{l_2^2})$ with the associated eigenfunction $$f_{(m_1,m_2)}(x,y)=e^{2\pi i(\frac{m_1}{l_1}x+\frac{m_2}{l_2}y)}.$$ where $m_1,m_2\in \mathbb{Z}$. Furthermore, The closed geodesics of $T^2$ parametrized by the arc length, are $$\gamma_{(n_1,n_2)}(t)=\frac{1}{l}(n_1l_1t,n_2l_2t)$$ where $n_1,n_2\in \mathbb{Z}$ and $l=\sqrt{n_1^2l_1^2+n_2^2l_2^2}$. A simple computation shows that an eigenfunction, say $f_{(m_1,m_2)}$, restricted on a closed geodesic, $\gamma_{(n_1,n_2)}$, gives
$$f_{(m_1,m_2)}\circ \gamma_{(n_1,n_2)}(t)=e^{2\pi i(\frac{m_1n_1+m_2n_2}{l})t}$$ Which is an eigenfunction on the circle $\mathbb{R}/l\mathbb{Z}$ with the eigenvalue $\tilde{\lambda}=\left( \frac{2\pi}{l}(m_1n_1+m_2n_2)\right)^2$.
Now my question is: Is this true in the general cases? More precisely;
Let $\gamma:[0,l]\to M$ be a closed geodesics on the Riemannian manifold $(M,g)$ which is parametrized by the arc length. If $f\in C^\infty(M)$ is an eigenfunction for the Laplacian on $M$, i.e. $$\Delta(f)=\lambda f$$ Then
Question 1) Is $f\circ \gamma$ an eigenfunction on the circle $S^1=\mathbb{R}/l\mathbb{Z}$? Or, Is it in the form of $$f\circ \gamma(t)=c e^{2\pi i \tilde{\lambda}t}.$$
Question 2) If so, how does $\tilde{\lambda}$ depend on $\gamma$ and $\lambda$?
Thanks.
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This is just a complement to Robert's answer: restrictions of eigenfunctions to geodesics (with a view to $L_p$ estimates) are treated here: Restrictions of the Laplace-Beltrami eigenfunctions to submanifolds N. Burq, P. Gérard, and N. Tzvetkov Source: Duke Math. J. Volume 138, Number 3 (2007), 445-486. – alvarezpaiva Jan 21 at 8:42
## 1 Answer
The answer is 'no', as you can see by taking the case of $M$ being the unit $2$-sphere in $\mathbb{R}^3$ and the geodesic $\gamma$ being a great circle, say, the horizontal great circle given by $z=0$. If you consider the harmonic polynomials of degree $2$ in $x,y,z$ restricted to the $2$-sphere, these are eigenfunctions of the Laplacian on the $2$-sphere, but their restrictions to the horizontal great circle aren't usually eigenfunctions of the Laplacian on the circle.
More generally, you take the $k$-th eigenspace of the Laplacian on the $2$-sphere for $k>1$, you'll find that the restriction of these functions to each great circle projects into a sum of a finite number of eigenspaces of the Laplacian on the great circle (I think it's about $\tfrac12(k{+}2)$ of them), but not into a single one of these eigenspaces.
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Thanks. You are right. – Asghar Ghorbanpour Jan 21 at 15:26
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http://mathhelpforum.com/pre-calculus/142951-polynominal-root-solution-problem.html
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# Thread:
1. ## Polynominal root and solution problem
Hi i am trying to find the roots of the following equation,
2x4 -8x3 +13x2 -13x + 6 = 0
I know two of the roots shold be x=1 and x=2 and that there should be 2 more roots that are complex but I dont know how to calculate them.
Regards Maria
2. Originally Posted by maria88
Hi i am trying to find the roots of the following equation,
2x4 -8x3 +13x2 -13x + 6 = 0
I know two of the roots shold be x=1 and x=2 and that there should be 2 more roots that are complex but I dont know how to calculate them.
Regards Maria
Since you have found the solutions x=1 and x=2 you can factor out (x-1)(x-2) on the left side of your equation (do a polynomial division of the left side by $(x-1)(x-2)$, i.e. be $x^2-3x+2$). The remaining solutions are zeros of the remaining quadratic factor $2x^2-2x+3$ (and you know a formula to determine those).
3. Originally Posted by maria88
Hi i am trying to find the roots of the following equation,
2x4 -8x3 +13x2 -13x + 6 = 0
I know two of the roots shold be x=1 and x=2 and that there should be 2 more roots that are complex but I dont know how to calculate them.
Regards Maria
Consider the function p(x) = 2x^4 - 8x^3 + 13x^2 - 13x + 6.
(x - 1)(x - 2) = x^2 - 3x + 2 is a factor of p(x). Divide it into p(x) to get the other quadratic factor. And you should be able to get the complex factors of this quadratic factor.
4. I would do it slightly differently. Rather than multiplying (x- 1)(x- 2) I would use "Synthetic Division" to divide by x- 1 and than again use synthetic division to divide the result by x- 2. That leaves a quadratic equation which can be solved by the quadratic formula.
5. Let the four roots be 1, 2, a+ib and a-ib.
Sum of the roots = 8/2 = 4 = 3 + 2a.
So a = 1/2.
The product of roots = 2(a^2 + b^2) = 6/2 = 3
a^2 + b^2 = 3/2
b^2 = 3/2-1/4 = 5/4
So b = sqrt(5)/2
So the roots are 1, 2, 1/2 + i*sqrt(5)/2 and 1/2 - i*sqrt(5)/2
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http://mathhelpforum.com/calculus/135817-show-sum-convergent-divergent-series-divergent.html
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# Thread:
1. ## Show that the sum of a convergent and divergent series is divergent
I am completely stumped on this one, so I could really use a hand with it.
Show that if the series $\sum a_k$ converges and the series $\sum b_k$ diverges, then the series $\sum (a_k+b_k)$ diverges.
A rigorous proof is expected, and we are to prove this by contradiction.
I don't know how I could answer this. This is all I really know that could possibly help me answer it.
If $\sum_{k=0}^{\infty}a_k$ converges and $\sum_{k=0}^{\infty}b_k$ converges, then $\sum_{k=0}^{\infty}(a_k+b_k)$
Moreover, if $\sum_{k=0}^{\infty}a_k=L$ and $\sum_{k=0}^{\infty}b_k=M$, then $\sum_{k=0}^{\infty}(a_k+b_k)=L+M$
I could really use some help with this one. I wasn't able to see my Calculus professor concerning the matter due to an unexplained absence on his part.
2. Hint: Argue by contradiction.
Start by assuming that the series $\sum (a_k+b_k)$ converges.
3. Originally Posted by General
Hint: Argue by contradiction.
Start by assuming that the series $\sum (a_k+b_k)$ converges.
Alright, I've tried your advice out and here's the work I currently have.
Assume $\sum (a_k+b_k)$ converges.
Since $\sum (-1)a_k=-\sum a_k$ converges,
$-\sum a_k + \sum (a_k+b_k) = \sum b_k$
This implies that $\sum b_k$ converges. This is a contradiction, since $\sum b_k$ diverges.
Is this proof correct?
4. Very good!
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http://mathhelpforum.com/advanced-algebra/2052-algebra-units.html
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# Thread:
1. ## Algebra.....units
Find all the units in M2(Z2)
Thanks guys!!!
2. Originally Posted by suedenation
Find all the units in M2(Z2)
Thanks guys!!!
You need to find all the matrices, of form
$\left( \begin{array}{cc}a&b\\c&d\end{array} \right)$ where $a,b,c,d\in\mathbb{Z}_2$ such as it has an inverse.
There are 16 possible such matrices, check each one if it is invertible.
If I am not making a mistake, then I think that,
$\left( \begin{array}{cc}0&1\\1&0\end{array} \right)$
and,
$\left( \begin{array}{cc}1&0\\0&1\end{array} \right)$
and,
$\left( \begin{array}{cc}0&1\\1&1\end{array} \right)$
and,
$\left( \begin{array}{cc}1&1\\1&0\end{array} \right)$
and,
$\left( \begin{array}{cc}1&1\\0&1\end{array} \right)$
and,
$\left( \begin{array}{cc}1&0\\1&1\end{array} \right)$
Are the only units in this ring.
3. I was thinking of a more elegant way of solving this. There is a theorem in Linear Algebra which states that a square matrix is invertible (a unit) if and only if its determinant is non-zero.
The problem is I do not know if this is going to work here, possibly because $\mathbb{Z}_2$ is not a field.
But let us assume, thus we need that,
$\left| \begin{array}{cc}a&b\\c&d\end{array} \right|\not =0$
Since there are only two elements in the set by elimination we have that,
$\left| \begin{array}{cc}a&b\\c&d\end{array} \right|=1$
Thus, by evaluating the determinant,
$ad-bc=1$
But remember in this ring since it has only two elements, that $-x=x$ for all $x$.
Thus,
$ad+bc=1$
Now, just observe the possibilities for $a,b,c,d$ which make this equation true. Since there are only 16 of them it is easy.
1)If $a=0$ then, $b,c\not=0$ thus, $b,c=1$ and $d$ can be anything (either 1 or 0) this gives us two matrices,
$\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$ and $\left(\begin{array}{cc}0&1\\1&1\end{array}\right)$.
2)If $a=1$ then , $d=1$ would mean that either $b$ or $c$ must be zero. Also, if $d=0$ would mean that both $a$ and $b$ cannot be zero (thus they are both one). This gives us four matrices,
$\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$ and, $\left(\begin{array}{cc}1&1\\0&1\end{array}\right)$ and,
$\left(\begin{array}{cc}1&0\\1&1\end{array}\right)$ and, $\left(\begin{array}{cc}1&1\\1&0\end{array}\right)$
Now there is no need to observe what happens with $b,c,d$ because we know what matrix would it be depending on the value of $a$.
4. Originally Posted by ThePerfectHacker
There is a theorem in Linear Algebra which states that a square matrix is invertible (a unit) if and only if its determinant is non-zero.
The problem is I do not know if this is going to work here, possibly because $\mathbb{Z}_2$ is not a field.
It works fine, because $\mathbb{Z}_2$ actually is a field.
Here's another way which generalises to any finite field $GF(q)$ and any dimension $n$.
A square matrix is invertible if and only if its rows are linearly independent. Take each row in turn. The first row must be non-zero so there are $q^n-1$ possibilities. The second row must not the in the subspace generated by the first row, which has dimension one, so there are $q^n-q$ possiblities. The $(r+1)$-th row must not be in the subspace generated by the previous $r$ rows, linearly independent by induction, so of dimension $r$ giving $q^n-q^r$ possibilities. The overal number of possiblities is thus $(q^n-1)(q^n-q)\cdots(q^n-q^r)\cdots(q^n-q^{n-1})$
In the case $q=2, n=2$ this is $(2^2-1)(2^2-2^1) = 6$.
5. Originally Posted by rgep
It works fine, because $\mathbb{Z}_2$ actually is a field.
Here's another way which generalises to any finite field $GF(q)$ and any dimension $n$.
A square matrix is invertible if and only if its rows are linearly independent. Take each row in turn. The first row must be non-zero so there are $q^n-1$ possibilities. The second row must not the in the subspace generated by the first row, which has dimension one, so there are $q^n-q$ possiblities. The $(r+1)$-th row must not be in the subspace generated by the previous $r$ rows, linearly independent by induction, so of dimension $r$ giving $q^n-q^r$ possibilities. The overal number of possiblities is thus $(q^n-1)(q^n-q)\cdots(q^n-q^r)\cdots(q^n-q^{n-1})$
In the case $q=2, n=2$ this is $(2^2-1)(2^2-2^1) = 6$.
Thanks for the tip!
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http://mathoverflow.net/questions/88433/reference-for-estimation-gaussian-of-the-heat-kernel
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## Reference for estimation gaussian of the heat kernel
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $(M,g^{TM})$ a Riemannian manifold of dimension $n$ and $\Delta$ the Laplace–Beltrami operator. I would like to find a reference (analytic or probabilistic) for the following classic result.
If $p_t(x,y)$ is the kernel of the semigroup $e^{t\Delta}$, then there exist $C,c>0$, such that $$p_{t}(x,y)\leq \frac{C}{t^{n/2}}e^{-cd(x,y)^2/t}.$$
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1
I don't have a copy so I don't know if it's in there, but one place I'd look is "Aspects of Sobolev-Type Inequalities" by Saloff-Coste. – Mark Meckes Feb 14 2012 at 15:05
## 3 Answers
A probabilistic proof is given in the book: "Stochastic Analysis on Manifolds" by E. Hsu. See Theorem 5.3.4, which also gives the lower bound.
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It's worth noting that this proof is for compact Riemannian manifolds. – Nate Eldredge Feb 15 2012 at 22:18
### 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'm pretty sure this can be found in Davies, Heat kernels and spectral theory. I'll check when I get to my office in an hour or so.
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Thank you!I find it also in Davies's book. – shu Feb 15 2012 at 7:48
You may also be interested in the following paper of Grigoryan, Gaussian upper bounds for the heat kernel on arbitrary manifolds, which establishes the desired Gaussian bounds whenever one can show that there exists $C>0$ such that for all $x\in M$ and $t>0$, $p_t(x,x) \leq Ct^{-n/2}$. The latter estimate may be obtained via a Sobolev inequality or a Nash inequality or through other means (and I'm sure it's discussed in the Davies and Saloff-Coste books already mentioned). This paper also establishes Gaussian upper bounds even when the function appearing in the 'on-diagonal bound' is not of the form $t^{-n/2}$, or if one only has control of $p_t(x,x)$ at two points $x_1$ and $x_2$.
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Thank you, mfolz. It is a very interesting paper. – shu Feb 16 2012 at 10:17
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http://mathhelpforum.com/differential-geometry/121239-convergence-measure.html
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# Thread:
1. ## convergence in measure
Let $(X,F, \mu)$ be a measure space with $\mu (X) < \infty$.
Prove that $f_n \rightarrow f$ in measure $\Longleftrightarrow$ $\int_X \frac{|f_n-f|}{1+|f_n-f|} d \mu$ $\rightarrow 0$.
2. Hello,
See here : http://www.mathhelpforum.com/math-he...-1-yn-0-a.html
(the integral is like the expectation, and measure = probability)
3. i understand the first half. but i have a question on the second half.
by letting $Y_n=|f_n-f|$, $Z_n = \frac{Y_n}{1+Y_n}$and mimicking the steps. i have
$\int Z_n d\mu = \int Z_n 1_{\{Z_n \leq \epsilon \} } d\mu + \int Z_n 1_{ \{Z_n > \epsilon \} } d\mu$
$\int Z_n 1_{\{Z_n \leq \epsilon \} } d\mu \leq \int \epsilon 1_{\{Z_n \leq \epsilon \} } d\mu = \epsilon \int 1_{\{Z_n \leq \epsilon \} } d\mu$
but how do i show that $\epsilon \int 1_{\{Z_n \leq \epsilon \} } d\mu \leq \epsilon$ without knowing that $\mu$ is a probability measure? is there anyway i can estimate integral $\epsilon \int 1_{\{Z_n \epsilon \} }$? please help me.
4. It's not that difficult, think about an hypothesis you've not been using
Spoiler:
$\{Z_n<\epsilon\}\subset X \Rightarrow \mu(Z_n<\epsilon)=\int_X \bold{1}_{\{Z_n<\epsilon\}} ~d\mu \leq \mu(X) <\infty$
So you can say that it's $<\epsilon'$ where $\epsilon'=\epsilon/\mu(X)$
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http://mathoverflow.net/questions/54393/analytical-solutions-of-a-differential-equation-from-archimedes-spiral/54428
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## Analytical solutions of a differential equation (from Archimedes' Spiral)
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
There is a differential equation in polar coordinates: $r'^2+r^2=(kt)^2$, $r(t=0)=0$, k- Const.
I've found that a) if $\phi \in (0,t)$, t is quite small, then $r(\phi) \approx k/2 *\phi^2$ b) if $\phi \in (c, + \infty)$, c is quite large positive number, then $r(\phi)=k\phi+o(\phi)$, as $\phi \to \infty$.
I am looking for approximation between a & b. How the Archimedes' Spiral transforms into other curve. Any ideas are highly welcomed.
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## 3 Answers
By a change of time scale we can assume that $k=1$. Rewrite the equation as $$r'=\sqrt{t^2-r^2},\quad r(0)=0.$$ The solution must be increasing, hence positive, and bounded by $t$, which by the way is a supersolution. It is also clear that $0\le r(t)\le t^2/2$. From this, it follows that $$r'\ge t \sqrt{1-t^2/4}\ ,\quad 0\le t\le 2.$$ Integrating we obtain $4/3\le r(2)<2$. The function $t-4/(3t)$ is a subsolution on $[2,\infty)$ and its value at $t=2$ is $4/3$. It follows that $$t-\frac{4}{3t}\le r(t)\le t\quad\forall t\ge2.$$
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@Julian: the final displayed equation you probably want $$t - \frac{4}{3t}$$ on the LHS. – Willie Wong Feb 5 2011 at 19:09
@Willie This is becoming a routine :-) – Julián Aguirre Feb 6 2011 at 1:12
Thank you, the spiral curve which is the solution of the equation is of great intrest for me. The length is $d^2S/dt^2=C^2$, C is a const. The similar movement but along a line will be: $x(t)=ch(t)-1$, $y(t)=sh(2t)/4-t/2$, $t \in (0, +\infnty)$ ( the natural equation: $k=1/(2(s+b))*\sqrt{b/s}$, b - const). Do you find it intresting? – Mikhail Gaichenkov Feb 6 2011 at 7:41
### 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.
If you go to the Archimedean Spiral page of the Virtual Math Museum:
http://virtualmathmuseum.org/Curves/archimedean_spiral/archimedean_spiral.html
you will find a lot of material about Archimede's Spiral and the more general class of Archimedean spirals to which it belongs, and these may be of use to you. In particular, try experimenting with the 3D-XplorMath-j applet that is there, and download the file Archimedean_Spiral.pdf.
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Thank you, very intresting! – Mikhail Gaichenkov Feb 6 2011 at 7:48
[Updated 3/11/2011]
Your approximate solutions are the leading order terms of the asymptotic expansions of $r(t)$ for $t\ll1$ and $t\gg1$. More terms can be computed using perturbation techniques: $$r(t)/k = \frac{1}{2}t^2-\frac{1}{32}t^4+\frac{1}{768}t^6+O(t^8),\quad t\ll1,\quad (1)$$ and $$r(t)/k = t-\frac{1}{2}t^{-1}-\frac{5}{8}t^{-3}+O(t^{-5}),\quad t\gg1.\quad (2)$$ The expansion for small $t$ reproduces the first few terms of the Maclaurin series for $r(t)$. The full series can be obtained by substituting the formal power series expansion into your equation and matching the terms at equal powers of $t$. Coefficients of the resulting power series $$r(t)=\sum_{n=1}^{\infty} R_{2n}t^{2n}\quad (3)$$ can be computed using the recurrent formulae $$R_2=k/2,\quad R_4=-k/32,$$ $$R_{2n}=-\frac{R_{2n-2}}{8n} -\frac{1}{4kn}\sum_{i=1}^{n-2} \left( R_{2i}+4(i+1)(n-i)R_{2i+2} \right)R_{2n-2i} ,\quad n=3,4,\dots$$
You said that you were interested in an approximate solution valid for $t=O(1)$; neither expansion (1) nor expansion (2) is valid there. It would not be possible to match (1) and (2) without having to derive an intermediate asymptotic, because their present regions of validity do not overlap. This means that series (3) is likely to be your best bet at the direct analytic computation of $r(t)$. I was unable to compute the radius of its convergence analytically. At the same time, coefficients $R_{2n}$ appear to decay in a rapid fashion, so the radius is not likely to be small. If unsure, you can always estimate the value of the radius numerically using the Domb-Sykes plot.
In situations like yours people sometimes construct two-point Pade approximants $$r(t)/k \approx \sum_{k=2}^Na_kt^k/\left(1+\sum_{k=1}^{N-1}b_kt^k\right),$$ i.e. ratios of two polynomials that, when expanded into a power series at $t=0$ or at the $t=\infty$, match the first few terms of the appropriate power series expansions, see G.A. Baker & P. Graves-Morris "Pade Approximants. Part II: Extensions and Applications", Addison-Wesley (1981) for more details. For example, for $N=2$, $$r(t)/k \approx t^2/(2+t)$$ matches the leading order behaviour of both expansions given above. It is actually a very poor approximation, because it has an $O(1)$ absolute error for large $t$. Higher-order approximants get progressively better (and uglier); e.g. for $N=5$ one obtains $$r(t)/k \approx \frac{2t^2(772+432t+184t^2+65t^3)}{(3088+1728t+929t^2+368t^3+130t^4)}.$$ This expression recovers the first couple of terms in expansions (1) and (2) and its relative error in the intermediate region is below $5\%$.
It is worth noting that two-point Pade approximants are not always well-behaved in the intermediate region. In this particular case they seem to be converging to the exact solution reasonably well.
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http://math.stackexchange.com/questions/120/is-1-a-prime-number/5735
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# Is 1 a prime number?
Is 1 classified as a prime number?
And if so, why? If not, why not?
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5
Given how often "let p be an odd prime" shows up in theorems, sometimes I wonder if we'd be better off defining 2 as non-prime too ;) – user7530 Oct 4 '11 at 20:12
So, is there something about why two is prime? – Gustavo Bandeira Apr 3 at 16:17
## 8 Answers
One of the whole "points" of defining primes is to be able to uniquely and finitely prime factorize every natural number.
If 1 was prime, then this would be more or less impossible.
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2
In the same way, it's often convenient for a variety of reasons not to consider hte unit ideal as prime (e.g. for unique factorization in Dedekind domains). – Akhil Mathew Jul 20 '10 at 21:15
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That's a good way of looking at it, but I wouldn't call it a proof. We simply define 1 as a prime, there is no proof of that fact; it's a definition. – Edan Maor Jul 20 '10 at 22:48
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Akhil's reason is the real reason... – user126 Jul 21 '10 at 8:51
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@Stacked let's say I wanted to prime factorize 6. It is 2*3, but it's also 2*3*1. It is also 2*3*1*1, and 2*3*1*1*1; that is, 6 no longer has a unique prime factorization – Justin L. Aug 10 '10 at 23:06
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@Stacked: “Isn’t it now impossible to factorize 1?” Good question… it’s not impossible; it takes lateral thinking. 1 is the product of the empty list of factors (just as 0 is the sum of an empty list). See the answers to math.stackexchange.com/questions/6832/… for discussion of this! – Peter LeFanu Lumsdaine Nov 21 '10 at 8:57
show 8 more comments
The main point of talking about prime numbers is Euclid's theorem that every positive integer can be written uniquely as a product of primes. As Justin remarks, this would break horribly if $1$ were considered prime, for example we could factor $2$ as $2\times1\times1\times1\times1\times1$. Instead we say that $1$ is not a prime, but it is the product of zero primes (see Why is $x^0 = 1$ except when $x = 0$? to understand why any prime multiplied by itself $0$ times is $1$) so Euclid's theorem works out nicely!
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Euclid's theorem could easily be modified to work with 1 as a prime. The reason why 1 isn't a prime is either because it makes a whole lot of different results nicer, or an accident of history, not because of one specific theorem – Casebash Jul 20 '10 at 23:25
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– Doug Spoonwood Aug 22 '11 at 21:16
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Hence the sentence "Euclid's theorem that every positive integer can be written uniquely as a product of primes" and then referring back to it later. – Noah Snyder Aug 23 '11 at 0:33
It's important to understand that this is not something that can be proved: it's a definition. We choose not to regard 1 as a prime number, simply because it makes writing lots of theorems much easier.
Noah gives the best example in his answer: Euclid's theorem that every positive integer can be written uniquely as a product of primes. If 1 is defined to be a prime number, then we'd have to change that theorem to: "every positive integer can be written uniquely as a product of primes, except for infinite multiplications by 1". So we choose to go with the easier path of defining 1 to not be a prime.
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actually 1 was considered a prime number until the beginning of 20th century. Unique factorization was a driving force beneath its changing of status, since it's formulation is quickier if 1 is not considered a prime; but I think that group theory was the other force. Indeed I prefer to describe numbers as primes, composites and unities, that is numbers whose inverse exists (so if we take the set of integer numbers Z, we have that 1 and -1 are unities and we still have unique factorization up to unities).
We can always amend the defition of a prime number and say it is a number with exactly two divisors: in this way 1 is not a prime by definition :-)
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– Charles Oct 4 '11 at 14:44
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In the classical Greek system, 1 was not even called a "number", and "prime" was a subclassification of "odd", so both 1 and 2 were excluded then. – GEdgar Oct 4 '11 at 15:05
If you're talking about the integers, you're missing one. Every integer is prime, composite, a unit, or zero. – Tanner Swett Mar 7 '12 at 0:33
It's worth emphasizing that, in addition to unique factorization, there are also somewhat deeper structural reasons underlying the convention that $\:1\:$ is neither prime nor a prime ideal, but $\:0\:$ is. Below I discuss the motivations for these differing conventions.
One important motivation for including the zero ideal as prime is that this facilitates powerful reductions. For example, in many ring theoretic problems involving an ideal $\rm\; I\;$, one can reduce to the case $\rm\;I = P\;$ prime, and then reduce to $\rm\;R/P\;$, therefore reducing to the case when the ring is a domain. In this case one simply says that one can factor out $\rm\; P\;$, so wlog assume $\rm\; P = 0\;$ is prime, hence the ring is a domain. For example, at the end of this post is an excerpt from Kaplansky's classic textbook "Commutative Rings", section 1-3: G-Ideals, Hilbert Rings, and the Nullstellensatz, where I've explicitly highlighted a few prototypical examples of such reductions - cf. reduce to...
Thus we have solid evidence for the utility of the convention that the zero ideal is prime. So why don't we adopt the same convention for the unit ideal $1$ or, equivalently, why don't we permit the zero ring as a domain? There are a number of reasons. First, in domains and fields it often proves very convenient to assume that one has a nonzero element available. This permits proofs by contradiction to conclude by deducing $1 = 0\:$. More importantly, it implies that the unit group is nonempty, so unit groups always exist. It'd be very inconvenient to have to always add the proviso (except if $\;\rm R = 0)$ to the ubiquitous arguments involving units and unit groups. More generally it's worth emphasizing that the usual rules for equational logic are not complete for empty structures. That is why groups and other algebraic structures are always axiomatized to prevent nonempty structures (see this thread for further details).
Below is the mentioned Kaplansky excerpt on reduction to domains by factoring out prime ideals.
Let $\rm\; I\;$ be any ideal in a ring $\rm\; R\;$. We write $\rm\: R^{*}\:$ for the quotient ring $\rm\; R/I\;$. In the polynomial ring $\rm\; R[x]\;$ there is a smallest extension $\rm\; IR[x]\;$ of $\rm\; I\;$. The quotient ring $\rm\; R[x]/IR[x]\;$ is in a natural way isomorphic to $\rm\; R^*[x].\;$ In treating many problems, we can in this way reduce to the case $\rm\; I = 0\;$, and we shall often do so.
THEOREM 28. $\;$ Let $\rm\; M\;$ be a maximal ideal in $\rm\; R[x]\;$ and suppose that the contraction $\rm\; M \cap R\ =\ N\;$ is maximal in $\rm\; R\;.\$ Then $\rm\; M\;$ can be generated by $\rm\; N\;$ and one more element $\rm\; f\:.\$ We can select $\rm\; f\;$ to be a monic polynomial which maps mod $\rm\; N\;$ into an irreducible polynomial over the field $\rm\; R/N\;$.
Proof. $\;$ We can reduce to the case $\rm\; N = 0,\;$ i. e., $\rm\; R\;$ a field, and then the statement is immediate.
THEOREM 31. $\;$ A commutative ring $\rm\; R\;\;$ is a Hilbert ring if and only if the polynomial ring $\rm\; R[x] \;\;$ is a Hilbert ring.
Proof. $\;$ If $\rm\; \;\rm R[x]\;$ is a Hilbert ring, so is its homomorphic image $\rm\; R\;$. Conversely, assume that $\rm\; R\;$ is a Hilbert ring. Take a G-ideal $\rm\: Q\:$ in $\rm\; R[x]\;$; we must prove that $\rm\: Q\:$ is maximal. Let $\rm\: P = Q \cap R\:$; we can reduce the problem to the case $\rm\ P = 0\:,\$ which, incidentally, makes $\rm\ R\$ a domain. Let $\rm\; u\;$ be the image of $\rm\; x\;$ in the natural homomorphism $\rm\ R[x] \ \to\ R[x]/Q\ .\ \$ Then $\rm\; R[u]\;$ is a G-domain. $\$ By Theorem 23, $\rm\ u\$ is algebraic over $\rm\ R\$ and $\rm\ R\$ is a G-domain. Since $\rm\:R\:$ is both a G-domain and a Hilbert ring, $\rm\ R\$ is a field. $\$ But this makes $\rm\ R[u] = R[x]/Q\$ a field, $\$ proving $\rm\ Q\$ to be maximal.
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Prime numbers are the multiplicative building blocks of the natural numbers in the sense that every natural number is either a prime or a product of primes (the empty product gives 1). Multiplicatively 1 does not contribute anything and so it is not a building block.
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As this question has been just bumped up, might as well mention this great article by Chris Caldwell (who maintains The Prime Pages) and Yeng Xiong:
• "What is the smallest prime?", J. Int. Seq., Vol. 15 (2012), Article 12.9.7,
arXiv:1209.2007 [math.HO]
The abstract:
What is the first prime? It seems that the number two should be the obvious answer, and today it is, but it was not always so. There were times when and mathematicians for whom the numbers one and three were acceptable answers. To find the first prime, we must also know what the first positive integer is. Surprisingly, with the definitions used at various times throughout history, one was often not the first positive integer (some started with two, and a few with three). In this article, we survey the history of the primality of one, from the ancient Greeks to modern times. We will discuss some of the reasons definitions changed, and provide several examples. We will also discuss the last significant mathematicians to list the number one as prime.
It shows that "There does not appear to be any period of time during which most mathematicians deemed one to be a prime" (to a large extent because "For much of history it did not even make sense to ask if the number one was a prime", as 1 was not considered a number). But just as a teaser, here are a couple more quotes (I've removed the references to make this readable):
Others who listed one as prime in this period are F. Wallis (1685), J. Prestet (1689), C. Goldbach (1742), J. H. Lambert (1770), A. Felkel (1776) and E. Waring (1782). [...]
Even after Gauss' pivotal text, many continued to write that unity was prime. Among these are: A. M. Legendre (1830), E. Hinkley (1853), M. Glaisher (1876), K. Weierstrass (1876), R. Frick & F. Klein (1897), A. Cayley (1890), L. Kronecker (1901), G. Chrystal (1904) and D. N. Lehmer (1914). [...]
G. H. Hardy listed one as a prime in several editions of his text A Course of Pure Mathematics.
Read the article; it's really interesting! They also have a "The history of the primality of one—a selection of sources".
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I don't think anyone's offered this: We know that $\sqrt{p}$ is irrational if $p \in \mathbb{Z}^+$ is prime. But 1 is its own square root, and is rational, thus 1 is not a prime.
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This is a consequence of our definition of what it means for $p$ to be prime - you might as well have said "1 is not prime, thus 1 is not prime"! – Clive Newstead Jul 28 '12 at 13:29
## protected by Nate EldredgeJul 28 '12 at 13:10
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http://mathoverflow.net/questions/84420/spectral-properties-of-the-ldlt-matrix-factorization
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## Spectral properties of the LDL^T matrix factorization
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Assume that a square, symmetric matrix $A$ can be factored into $A=LDL^T$ where $L$ is unit lower triangular and $D$ is diagonal. For indefinite $A$, $D$ may have $2x2$ blocks on the diagonal. How much information about the spectrum of $A$ can we obtain from $D$?
For example, it is known by Sylvester's law of matrix inertia that the inertia of $D$ is the same as that of $A$ (they have the same number of positive and negative eigenvalues). This is interesting, but I am wondering what other information is hidden in $D$.
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I was always wondering the same :) Great question. – Alexander Chervov Dec 28 2011 at 8:18
But why you write about 2x2 blocks in indefinite case ? Usually just elements of D can be taken negative. – Alexander Chervov Dec 28 2011 at 8:21
## 2 Answers
Well, for the closely related Cholesky factorization, there is the following:
Fast Accurate Eigenvalue Computations Using the Cholesky Factorization (1997) (by Roy Matthias), which says that the eigenvalues are very close to the squares of the diagonal elements of the Cholesky factor. (the paper is available on CiteSeer).
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I've heard similar, but there were some conditions, may be small eigs, or like that... But "squares" seems to me misprint - take A=D then they are equal. Any way... What can be the reason for relation in case "L" is not very small ? Is there some intuitive explanation ? – Alexander Chervov Dec 28 2011 at 8:23
In Cholesky, $D=I$ and the diagonal scaling is "included" in the $L$ factor, which need not have ones on its main diagonals. So that's where you get those square roots from. – Federico Poloni Dec 28 2011 at 8:57
@Federico, You mean - probably the paper mentioned above deals with LL^t ? Then it is Okay- we need squares. Just in the question LDL^t was mentioned which is sometimes called Cholesky or Cholesky without square roots also. In this case you do not need square roots. Any way puzzle seems to be resolved :) What about my questions ? What the reason can be to have eigs = D ? – Alexander Chervov Dec 28 2011 at 10:05
### 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.
Not an answer, but just thinking loudly (can I say like this in English?)
Consider A=
$a ~~ b$
$b ~~ a$
Then eigenvalues are equal to a-b, a+b (it is easy to check since trace and determinant are correct).
Then L =
$1~~~~~~ 0$
$b/a~~1$
D=
$a ~~~~ 0$
$0 ~~~~ a - b^2/a$
(It is easy in 2x2 case since determinant(A) = $d_1d_2$ ).
So we see: eigenvalues are:
$a \pm b$ and elements of D are $a$ and $a-b^2/a$
Well, do they look similar ?
Stupid case when they are similar is b=0.
Another case is more interesting - take b=a - very degenerate but will be positve under small perturbation a=b+small. So in this case eigs are : $2a, 0$, and elements of D are $a$ and $0$ . So we see that the smallest number (i.e. $0$) is the same.
Probably that is the phenomena which I heard about i.e. something similar holds true for NxN matrices.
Probably paper mentioned in Igor's answer is something related.
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http://mathhelpforum.com/differential-geometry/87678-properties-continuous-functions.html
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Thread:
1. Properties of Continuous Functions
Show that any polynomial of odd degree has at least one real root.
2. intermediate value theorem.
3. Algebra
Another good method is to note the fact that in $\mathbb{C}$ a degree n polynomial has n roots (counting with multiplicity) as $\mathbb{C}$ is algebraicly closed . But for every complex number, it's conjugate is also a root, you can check this easily enough. But since there are an odd number of roots that means one of the conjugates must be itself, but this means one of the roots is purely real.
This is a more algebraic argument of course, but it works all the same.
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http://mathhelpforum.com/advanced-applied-math/38659-theoretical-yield.html
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# Thread:
1. ## Theoretical yield
I think that the theoretical yield is the max amount of the products that can be formed. But I don't see how to find it.
The problem that is stumping me is this:
Given the equation $Cu_2 + O_2 \to 2Cu + SO_2$, what is the theoretical yield of $SO_2$ if $8.20 g$ of $0_2$ are used ?
Will someone simply and clearly explain how to go about solving this?
2. You have 8.20 g of $O_{2}$ so we must "convert" it into terms of $SO_{2}$. However, remember that a chemical reaction is representative of each species in moles. So that is where we will start off with:
$\text{8.20 g } O_{2} \times \frac{\text{1 mol } O_{2} }{\text{32.00 g }O_{2} } = \text{0.25625 mol } O_{2}$
Now, we know that for every mole of $SO_{2}$ produced, 1 mole of $O_{2}$ was used up. So we have a 1-1 ratio.
$\text{0.25625 mol } O_{2} \times \frac{\text{1 mol } SO_{2} }{\text{1 mol } O_{2} } = \text{0.25625 mol } SO_{2} \text{ produced}$
Note how I set up the ratio so that the units "mol $O_{2}$" cancels out. That is pretty much the key in solving stoichiometry questions.
So you have the moles of $SO_{2}$ produced. Now all that is left is converting it into terms of grams and that should be your theoretical yield.
3. that makes complete sense! thank you.
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http://mathhelpforum.com/pre-calculus/213044-epsilon-print.html
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# Epsilon
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• February 13th 2013, 04:46 AM
Petrus
2 Attachment(s)
Epsilon
Hello i got problem with nr 62. I got some progress but i dont understand
• February 13th 2013, 06:50 AM
emakarov
Re: Epsilon
Maybe you can start by determining N from the graph instead of algebraically. Post your answers here.
• February 13th 2013, 07:12 AM
Petrus
Re: Epsilon
Hi im currently on with phone but it looks around when n=5 kinda hard to se the graf
• February 13th 2013, 07:20 AM
emakarov
Re: Epsilon
Quote:
Originally Posted by Petrus
Hi im currently on with phone but it looks around when n=5 kinda hard to se the graf
Yes, for ε = 0.5, N = 5 and even N = 3 works.
So, you need to determine for which x > 0 it is the case that $\frac{\sqrt{4x^2+1}}{x+1}>2-\varepsilon=2-0.5=1.5$. Multiply both sides by (two times) the denominator, take square of both sides and solve (approximately) the resulting quadratic equation.
• February 13th 2013, 09:42 AM
Petrus
Re: Epsilon
Quote:
Originally Posted by emakarov
Yes, for ε = 0.5, N = 5 and even N = 3 works.
So, you need to determine for which x > 0 it is the case that $\frac{\sqrt{4x^2+1}}{x+1}>2-\varepsilon=2-0.5=1.5$. Multiply both sides by (two times) the denominator, take square of both sides and solve (approximately) the resulting quadratic equation.
so i get x_1= -7.3 and x_2=2.79
• February 13th 2013, 09:55 AM
emakarov
Re: Epsilon
Quote:
Originally Posted by Petrus
so i get x_1= -7.3 and x_2=2.79
That's not what I get. The inequality is $2\sqrt{4x^2+1}>3(x+1)$, or $7x^2-18x-5>0$.
• February 13th 2013, 10:04 AM
Petrus
Re: Epsilon
i will try again, try se if i can se what i did wrong
• February 13th 2013, 10:27 AM
Petrus
2 Attachment(s)
Re: Epsilon
Here is one progress i made.. Im kinda confused right now[ATTACH=CONFIG]26998[/ATTACHAttachment 26999
• February 13th 2013, 10:38 AM
emakarov
Re: Epsilon
You start with the quadratic inequality, which skips a lot of steps. I don't agree with the inequality, but I can't point out the step with an error because these steps are skipped. I wrote a couple of intermediate results in post #6. I also recommended multiplying both sides by 2 (or 4) to get rid of fractions like 1.5.
Here is a sequence of steps I recommend. Start with $\frac{\sqrt{4x^2+1}}{x+1}>1.5$.
1. Multiply both sides by (x + 1). (The direction of the inequality does not change because we are looking for x > -1, where x + 1 > 0.)
2. Multiply both sides by 2 to get rid of 1.5.
3. Take square of both sides. (This could lead to appearance of spurious solutions, but we ignore this for now.)
4. Move everything to the left-hand side and add like terms.
5. Solve the quadratic equation obtained when > is replaced with =. Let's denote the solutions by $x_1$ and $x_2$ where $x_1 < 0$ and $x_2 > 0$.
6. Since the leading coefficient of the quadratic polynomial f(x) is positive and the inequality has the form f(x) > 0, the solutions to the inequality are $x < x_1$ and $x > x_2$. We are interested in $x_2$.
Edit: Sorry, I missed that there are two attached images and not one. I'm looking at the second one...
• February 13th 2013, 10:47 AM
Petrus
Re: Epsilon
idk im following my book how it tell me :P that is when f(x)>0 and now ima solve for f(x)<0
• February 13th 2013, 10:55 AM
emakarov
Re: Epsilon
OK, I see that you started from $\left|\frac{\sqrt{4x^2+1}}{x+1}-2\right|<0.5$, which you converted into $\frac{\sqrt{4x^2+1}}{x+1}-2<0.5$. In fact, if we denote $\frac{\sqrt{4x^2+1}}{x+1}$ with z, $|z-2|<0.5$ is equivalent to $-0.5<z-2<0.5$. Adding 2 to all sides, we get $1.5<z<2.5$. So you are right that $\frac{\sqrt{4x^2+1}}{x+1}<2.5$ is a part of the original inequality. However, if you saw the graph from post #2, you must have seen that as x tends to infinity, the graph approaches 2 from below. This means that $\frac{\sqrt{4x^2+1}}{x+1}<2$ for positive x, so the inequality $\frac{\sqrt{4x^2+1}}{x+1}<2.5$ is automatically true for positive x. What we are interested in is the other part: $1.5<\frac{\sqrt{4x^2+1}}{x+1}$. It becomes true only starting from some N > 0. I assumed you saw all this after post #2 and therefore I recommended solving
$\frac{\sqrt{4x^2+1}}{x+1}>2-\varepsilon=2-0.5=1.5$ (*)
in post #4. Later, in post #6, I wrote a couple of inequalities you obtain while solving (*). You ignored all this and started solving a different inequality.
• February 13th 2013, 11:02 AM
Petrus
Re: Epsilon
Quote:
Originally Posted by emakarov
OK, I see that you started from $\left|\frac{\sqrt{4x^2+1}}{x+1}-2\right|<0.5$, which you converted into $\frac{\sqrt{4x^2+1}}{x+1}-2<0.5$. In fact, if we denote $\frac{\sqrt{4x^2+1}}{x+1}$ with z, $|z-2|<0.5$ is equivalent to $-0.5<z-2<0.5$. Adding 2 to all sides, we get $1.5<z<2.5$. So you are right that $\frac{\sqrt{4x^2+1}}{x+1}<2.5$ is a part of the original inequality. However, if you saw the graph from post #2, you must have seen that as x tends to infinity, the graph approaches 2 from below. This means that $\frac{\sqrt{4x^2+1}}{x+1}<2$ for positive x, so the inequality $\frac{\sqrt{4x^2+1}}{x+1}<2.5$ is automatically true for positive x. What we are interested in is the other part: $1.5<\frac{\sqrt{4x^2+1}}{x+1}$. It becomes true only starting from some N > 0. I assumed you saw all this after post #2 and therefore I recommended solving
$\frac{\sqrt{4x^2+1}}{x+1}>2-\varepsilon=2-0.5=1.5$ (*)
in post #4. Later, in post #6, I wrote a couple of inequalities you obtain while solving (*). You ignored all this and started solving a different inequality.
You are actually right... I am stick to a exemple on my book.... ima resolve this... tbh im really geting confused. I need to read the book again...
• February 13th 2013, 11:05 AM
emakarov
Re: Epsilon
I also recommend clicking the "Reply With Quote" button under my posts to see how formulas are typed using LaTeX. Using LaTeX is not difficult, and you won't need to attach pictures. Also check out the LaTeX Help subforum on this site.
• February 13th 2013, 11:07 AM
Petrus
1 Attachment(s)
Re: Epsilon
Defination 7 says me what im doing idk, but the problem says i shall use defination 7. Attachment 27001
• February 13th 2013, 11:09 AM
Petrus
Re: Epsilon
Quote:
Originally Posted by emakarov
I also recommend clicking the "Reply With Quote" button under my posts to see how formulas are typed using LaTeX. Using LaTeX is not difficult, and you won't need to attach pictures. Also check out the LaTeX Help subforum on this site.
indeed that but there is a problem.. im kinda never on pc anymore.. i use my smartphone if i get stuck :S (im mostly in school without any pc)
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http://mathhelpforum.com/pre-calculus/8404-equations-geometry.html
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# Thread:
1. ## Equations with Geometry
The square with vertices (-1,-1)(1,-1)(-1,1)(1,1) is ciut by the line
y=x/2+1
Inro a triangle and a pentagon. What is the number of square units in the area of the pentagon? express your answer as a decimal to th nearest hundreth
2. Originally Posted by Rimas
The square with vertices (-1,-1)(1,-1)(-1,1)(1,1) is ciut by the line
y=x/2+1
Inro a triangle and a pentagon. What is the number of square units in the area of the pentagon? express your answer as a decimal to th nearest hundreth
Look at the figure below.
The trick is that the area of the pentagon is the area of the square minus the area of the triangle.
So the question is, can you find the area of the triangle? or do you need help...
Attached Thumbnails
3. Originally Posted by Quick
Look at the figure below.
The trick is that the area of the pentagon is the area of the square minus the area of the triangle.
So the question is, can you find the area of the triangle? or do you need help...
I need help with the area of the triangle to
4. Originally Posted by Rimas
I need help with the area of the triangle to
Alright, notice that it's a right triangle, so its area is 1/2 base times height.
That's good, but what's the base length?
(by the way, for this problem, I'm calling the base the vertical side)
To do that, you need to figure out for what value of y does x equal -1?
So: $y=\frac{x}{2}+1$
Substitute: $y=\frac{-1}{2}+1$
Then: $y=\frac{1}{2}$
Now how long is the base? Well it's $1-\frac{1}{2}=\frac{1}{2}$ because that is the distance from the vertex to the point of intersection.
Can you find height?
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http://simple.wikipedia.org/wiki/Riemann_hypothesis
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# Riemann hypothesis
The Riemann zeta function, in the complex plane. The real part $\textrm{Re}(s)$ of the number is drawn horizontally, the imaginary part $\textrm{Im}(s)$ vertically. White dots show the zeros where $\textrm{Re}(s)=$½. Click to get a full view.
The Riemann hypothesis is a mathematical conjecture. Many people think that finding a proof of the hypothesis is one of the hardest and most important unsolved problems of pure mathematics.[1]
The hypothesis is named after Bernhard Riemann. It is about a special function, the Riemann zeta function. This function inputs and outputs complex number values. The inputs that give the output zero are called zeros of the zeta function. Many zeros have been found. The "obvious" ones to find are the negative even integers. This follows from Riemann's functional equation. More have been computed and have real part 1/2. The hypothesis states all the undiscovered zeros must have real part 1/2.
The functional equation also says all zeros (except the "obvious" ones) must be in the critical strip: real part is between 0 and 1. The Riemann hypothesis says more: they are on the line given, in the image on the right (the white dots). If the hypothesis is false, this would mean that there are white dots which are not on the line given.
If proven correct, this would allow mathematicians to better describe how the prime numbers are placed among whole numbers. The Riemann hypothesis is so important, and so difficult to prove, that the Clay Mathematics Institute has offered \$1,000,000 to the first person to prove it.[2]
## References
1. Bombieri, Enrico. "The Riemann Hypothesis - official problem description". Clay Mathematics Institute. Retrieved 2008-10-25.
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http://mathhelpforum.com/discrete-math/202605-help-quantifiers-print.html
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# Help with Quantifiers
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• August 27th 2012, 09:55 AM
JoannaEris
1 Attachment(s)
Help with Quantifiers
Can someone help me? I'm stuck with this:
Prove or give a counterexample to the following statement: For each positive integer a, there exists a positive integer b such thatAttachment 24612
I'd appreciate any help.
• August 27th 2012, 10:33 AM
emakarov
Re: Help with Quantifiers
$\frac{1}{2b^2+b}<\frac{1}{ab^2}$ iff $2b^2+b>ab^2$. In the latter inequality, both sides are quadratic polynomials in b. I understand that this problem is formally about quantifiers, but essentially it is about comparing two quadratic functions. For this reason, if you need help with this, it would make sense to post the problem to pre-university Algebra or Pre-calculus subforums.
• August 27th 2012, 11:16 AM
Plato
Re: Help with Quantifiers
Quote:
Originally Posted by JoannaEris
Prove or give a counterexample to the following statement: For each positive integer a, there exists a positive integer b such thatAttachment 24612
Is there a solution if $a=3~?$
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http://math.stackexchange.com/questions/161116/simple-property-of-gcd-and-modular-arithmetic/163092
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# Simple Property of GCD and Modular Arithmetic
I'm stuck on proving a rather elementary property, as I'm not really sure how to start off the approach. Suppose $g^a\equiv 1$ mod $m$ and $g^b\equiv 1$ mod $m$. Does this imply that $g^{\gcd(a,b)}\equiv 1$ mod $m$?
Here's my attempt: By definition, we know that $m\mid g^a-1$ and $m|g^b-1$, so there exists some $x,y\in\mathbb{Z}$ such that $mx=g^a-1$ and $my=g^b-1$. Then \begin{align} m(x+y)&=g^{\gcd(a,b)}g^{\frac{a}{\gcd(a,b)}}+g^{\gcd(a,b)}g^{\frac{b}{\gcd(a,b)}}-2\\ &=g^{\gcd(a,b)}\Big(g^{\frac{a}{\gcd(a,b)}}+g^{\frac{b}{\gcd(a,b)}}\Big)-2. \end{align} However, I feel like this approach is only making the problem more complicated than it actually is, as the terms become harder and harder to manipulate to get our desired result $mz=g^{\gcd(a,b)}-1$ for some $z\in\mathbb{Z}.$
Any help would be appreciated!
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5
By Bezout's, there exists $u,v$ such that $au+bv=\gcd(a,b)$. Thus mod $m$: $g^{\gcd(a,b)}\equiv g^{au+bv}\equiv\cdots$ – anon Jun 21 '12 at 6:48
Thanks. As described, I just wasn't sure which of the many properties to try out, but this one proves it quite easily. – Riem Jun 21 '12 at 7:14
When you have several things to try, you should actually try them rather than become paralyzed with indecision. You really shouldn't be stuck on a problem until you're at the point where you can't think of anything reasonable to try. – Hurkyl Jun 21 '12 at 9:01
Also, if it's choosing among many options that's giving you trouble, then you should say so! Without prompting, very few people will even think about offering advice on that topic. In this case, I suspect the answer most people would give is "there was one or more obvious things to try, so I tried them and it was quickly clear that <method of choice> would work". Of course, what's obvious to them might not have been obvious to you -- so such advice may not be helpful if you don't share what alternatives you saw and what you thought about them. – Hurkyl Jun 21 '12 at 9:06
Sure, I understand, but this comes at the cost of being long-winded, making it harder for others to contribute to the problem at hand without having understood (or at least glimpsed through) the majority of what I would have written down. I feel the current content was accurate enough to explain my overally situation without becoming verbose. – Riem Jun 21 '12 at 9:35
## 3 Answers
We have by the extended Euclidean algorithm that there exists $x, y \in \Bbb{Z}$ such that $ax + by = gcd(a,b)$. So $$g^{gcd(a,b)} = g^{ax+by} = g^{ax} \cdot g^{by} = (g^a)^x \cdot (g^b)^y \equiv 1^x\cdot 1^y = 1 \pmod{m}.$$
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Ah, thank you. There's many different properties on $\gcd$ and I wasn't necessarily sure which one would provide the simplest proof. This makes perfect sense though, since we can easily split them in the manner you wrote out. – Riem Jun 21 '12 at 7:13
We do not even need Bézout's Theorem. All we need is to know that the original Euclidean Algorithm (with subtraction) terminates with the $\gcd$. Let $a\gt b$, and let $c=a-b$. Note that $\gcd(b,c)=\gcd(a,b)$.
We have $g^{c}g^{b}=g^{a}$, and therefore if $g^a\equiv 1\pmod{m}$ and $g^b\equiv 1 \pmod{m}$ then $g^c\equiv \pmod{m}$.
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It is easy: $\rm\ \ g^A,\,g^B\equiv 1\:\Rightarrow\,order(g)\, |\, A,B\:\Rightarrow\: order(g)\, |\, (A,B)\:\Rightarrow\: g^{\,(A,B)}\equiv 1\quad$ QED
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See this post for more. For an application see here. – Gone Jun 26 '12 at 0:50
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http://mathhelpforum.com/algebra/20230-little-help-these-if-possible-please.html
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# Thread:
1. ## Little help on these if possible please?
I have some Pythagorean Triples proofs that I kind of have a hard time getting started on. I am just looking for a little insight to help me get started. Any help at all would be greatly appreciated. Here goes.
A pythagorean triple is a triple of positive integers (a,b,c) such that a^2 +b^2=c^2. In othere words, a set of three numbers is a pythagorean tiple if there exists a right triangle having these three thtegers as its side lengths. The triples one most commonly encounters when one first studies right triangles are (3,4,5) and (5,12,13).
(a) Are there othere pythagorean triples besides (3,4,5) which consist of three consecutive positive integers? Either give an exam;le of another such triple, or else prove n o other such triple can exist.
(b) Prove that there are infinitely many pythagorean triples (a,b,c) for which c -b=1. (Suggestion: direct proof by constructing them explicitly. Start by finding some small ones and look for a pattern. )
(c) Prove that there does not exist a pythagorean triple (a,b,c) such that a is odd, b is odd, and c sis even. ( Suggestion: Assume (a,b,c) is such a triple and derive a contradiction.)
On (a) it is pretty obvious that there is not any more triples that consist of three consecutive positive integers as a^2 + b^2 gets bigger much quicker than c^2. I think that this is proof by contradiction but not completely sure. Like I said any help would be greatly apprecitated.
2. Originally Posted by padsinseven
I have some Pythagorean Triples proofs that I kind of have a hard time getting started on. I am just looking for a little insight to help me get started. Any help at all would be greatly appreciated. Here goes.
A pythagorean triple is a triple of positive integers (a,b,c) such that a^2 +b^2=c^2. In othere words, a set of three numbers is a pythagorean tiple if there exists a right triangle having these three thtegers as its side lengths. The triples one most commonly encounters when one first studies right triangles are (3,4,5) and (5,12,13).
(a) Are there othere pythagorean triples besides (3,4,5) which consist of three consecutive positive integers? Either give an exam;le of another such triple, or else prove n o other such triple can exist.
Let n be the smallest of the triple. then (n + 1) is the next in the triple and (n + 2) is the last. if there are consecutive positive Pythagorean triples, then they must satisfy:
$n^2 + (n + 1)^2 = (n + 2)^2$
solving this, we find that $n = 3$ or $n = -1$
$n = -1$ makes no sense, so only $n = 3$ works. thus the only Pythagorean triple that fulfills the conditions is 3,4,5
QED
3. Originally Posted by padsinseven
(c) Prove that there does not exist a pythagorean triple (a,b,c) such that a is odd, b is odd, and c sis even. ( Suggestion: Assume (a,b,c) is such a triple and derive a contradiction.)
Assume, for the sake of contradiction, that such a triple exists.
Since $a$ is odd, $b$ is odd and $c$ is even, we can write: $a = 2n + 1$, $b = 2m + 1$ and $c = 2k$ for $m,n,k \in \mathbb {Z}$. Since they are a Pythagorean triple, we have that:
$(2n + 1)^2 + (2m + 1)^2 = (2k)^2$
$\Rightarrow 4n^2 + 4m^2 + 4n + 4m + 2 = 2 \left( 2 k^2\right)$
$\Rightarrow 2 \left( 2n^2 + 2m^2 + 2n + 2m + 1 \right) = 2 \left( 2k^2 \right)$
$\Rightarrow 2 \left[ 2 \left( n^2 + m^2 + n + m \right) + 1 \right] = 2 \left( 2k^2 \right)$
Thus we must have that: $2 \left( n^2 + m^2 + n + m \right) + 1 = 2k^2$
But $2 \left( n^2 + m^2 + n + m \right) + 1$ is odd, and $2k^2$ is even, so that cannot be. Thus we arrive at a contradiction.
Therefore, no such Pythagorean triple exists
QED
4. Originally Posted by padsinseven
(b) Prove that there are infinitely many pythagorean triples (a,b,c) for which c -b=1. (Suggestion: direct proof by constructing them explicitly. Start by finding some small ones and look for a pattern. )
here are the first few. apparently i'm too tired to spot any helpful pattern, maybe you'll have more luck with it
(3,4,5), (7,24,25), (9, 40, 41), (11, 60, 61), (13, 84, 85), ...
note that all the first numbers are odd numbers greater than 1. you have a triple for all such odd numbers.
also, the numbers have this relationship.
a is odd, b is even and c is odd. if $a = 2n + 1$, $b = 2m \implies c = 2m + 1$ for $m,n \in \mathbb {Z}$, then:
$(2n + 1)^2 + (2m)^2 = (2m + 1)^2$ ................(1)
i guess you can search for a way to relate n and m. the problem will probably be solved if you accomplish that
EDIT: well, $n = \frac {\sqrt{4m + 1}}2 - \frac 12$ ................(2)
...does that help. i'm just feeling in the dark here, hoping to stumble on something
EDIT 2: well, i think i have at least the outline. first we prove that the triples have to obey equation (1). then we prove that for special integers m greater than 2, we can find an integer n that satisfies equation (2), so we have to find all integers m that cause n to be an integer. this of course happens when 1 + 4m is an odd perfect square
EDIT 3: ...i don't think this is going anywhere. i'm tired, wait for someone else, i got two out of three, that's good enough
5. Thank you Jhevon. You were a lot of help on these.
6. Originally Posted by padsinseven
I have some Pythagorean Triples proofs that I kind of have a hard time getting started on. I am just looking for a little insight to help me get started. Any help at all would be greatly appreciated. Here goes....
(b) Prove that there are infinitely many pythagorean triples (a,b,c) for which c -b=1. (Suggestion: direct proof by constructing them explicitly. Start by finding some small ones and look for a pattern. )
...
Hi,
there are infintely many Pythagorean triples:
a) if (a, b, c) is a Pythegorean triple then $(ta, tb, tc), t\ \in \ \mathbb{N}$ is a Pythagorean triple too;
b) let $\boxed{\begin{array}{l}a=u^2-v^2 \\b=2uv\\c=u^2+v^2\end{array}}$ and $u>v~\wedge~u-v\text{ is odd}$
then you have a Pythagorean triple.
Examples
Code:
```u v | (a, b, c)
-------------------------
2 1 | (3, 4, 5)
3 2 | (5, 12, 13)
4 1 | (15, 8, 17)
.. ... | ........```
7. Originally Posted by earboth
Hi,
there are infintely many Pythagorean triples:
a) if (a, b, c) is a Pythegorean triple then $(ta, tb, tc), t\ \in \ \mathbb{N}$ is a Pythagorean triple too;
b) let $\boxed{\begin{array}{l}a=u^2-v^2 \\b=2uv\\c=u^2+v^2\end{array}}$ and $u>v~\wedge~u-v\text{ is odd}$
then you have a Pythagorean triple.
Examples
Code:
```u v | (a, b, c)
-------------------------
2 1 | (3, 4, 5)
3 2 | (5, 12, 13)
4 1 | (15, 8, 17)
.. ... | ........```
recall that c - b = 1. your last example does not follow that
8. Originally Posted by Jhevon
recall that c - b = 1. your last example does not follow that
Hi,
of course you are right. My post was meant to give the general pattern of constructing Pythagorean triples.
In this case you have to take:
$c-b=u^2-2uv+v^2=(u-v)^2=1$ that means u = v+1.
Thus all pairs (u, v) with this condition will give a Pythagorean triple like;
u = 4, v = 3 ==> (7, 24, 25)
u = 5, v = 4 ==> (9, 40, 41)
and so on...
9. Originally Posted by earboth
Hi,
of course you are right. My post was meant to give the general pattern of constructing Pythagorean triples.
In this case you have to take:
$c-b=u^2-2uv+v^2=(u-v)^2=1$ that means u = v+1.
Thus all pairs (u, v) with this condition will give a Pythagorean triple like;
u = 4, v = 3 ==> (7, 24, 25)
u = 5, v = 4 ==> (9, 40, 41)
and so on...
oh ok
10. It is shown in a standard number theory course that what Earboth posted in (b) is the complete list of Pythagoren triples.
11. Hello, padsinseven!
We can use a bit of algebra on #3 ... rather than making a long list.
(b) Prove that there are infinitely many PTs $(a,b,c)$ for which $c -b\:=\:1$
Let $b \,=\,m$
Then $c \,=\,m+1$
The Pythagorean Triple would be: . $a^2 + m^2 \:=\:(m+1)^2$
. . which simplifies to: . $a^2\:=\:2m+1\quad\Rightarrow\quad a \:=\:\sqrt{2m+1}$
We want $2m$ to be one less than a square.
. . Then: . $m \:=\:4,\,12,\,24,\,40,\,\cdots$
We find that $m$ is of the form: . $2n(n+1)$ for any natural number $n.$
Therefore, there is an infinite number of such PTs.
(c) Prove that there does not exist a PT (a,b,c) such that
$a$ is odd, $b$ is odd, and $c$ is even.
Let $a \,= \,2p + 1,\;b \,= \,2q + 1,\;c \,= \,2r$ for natural numbers $p,\,q,\,r$
. . Assume that such a PT exists.
Then: . $(2p+1)^2 + (2q+1)^2\:=\:(2r)^2\quad\Rightarrow\quad 4p^2 + 4p + 1 + 4q^2 + 4q + 1 \:=\:4r^2$
. . $\Rightarrow\quad4(p^2+p+q^2+q) + 2 \:=\:4r^2\quad\Rightarrow\quad 4r^2 - 4(p^2+p+q^2+q) \:=\:2$
. . $\Rightarrow\quad4(r^2 - p^2-p-q^2-q) \:=\:2$
The left side is a multiple of 4; the right side is not.
We have reached a contradiction . . . Q.E.D.
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http://mathhelpforum.com/pre-calculus/49645-hyperbola.html
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# Thread:
1. ## Hyperbola
Hyperbola is at the center of coordinate system. Its foci (?) are on abscissa. There are two points on the hyperbola; these are $T_1(\frac{3\sqrt5}{2},2)$ and $T_2(4,\frac{4\sqrt7}{9})$. What is hyperbola's equation?
(I'm having trouble with solving two equations that follow.)
2. Originally Posted by courteous
Hyperbola is at the center of coordinate system. Its foci (?) are on abscissa. There are two points on the hyperbola; these are $T_1(\frac{3\sqrt5}{2},2)$ and $T_2(4,\frac{4\sqrt7}{9})$. What is hyperbola's equation?
(I'm having trouble with solving two equations that follow.)
Since its foci are on the abscissa [x-axis] and the hyperbola is centered at the origin, we know that the hyperbola has the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, where $a>b$
Now plug the two points into this equation, and you will generate two new equations:
At $\left(4,\frac{4\sqrt{7}}{9}\right)$, we get $\frac{16}{a^2}-\frac{112}{81b^2}=1$
At $\left(\frac{3\sqrt{5}}{2},2\right)$, we get $\frac{45}{4a^2}-\frac{4}{b^2}=1$
After solving the system of equations, I got $a=\sqrt{\frac{981}{53}}$
But I'm getting $b^2=-\frac{1744}{171}\implies b\text{ is complex.}$
Are you sure that you wrote down the coordinates correctly?
--Chris
3. Originally Posted by Chris L T521
Are you sure that you wrote down the coordinates correctly?
Sorry!!! No!!! Sorry, Chris! The points are $T_1(\frac{3\sqrt5}{2},2)$ and $T_2(4,\frac{4\sqrt7}{3})$. The $T_2$ y-coordinate's denominator is 3 (not 9).
I've done it so many times that I've automatically restarted with partially already calculated number.
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http://mathoverflow.net/questions/32442/how-to-generate-a-net-on-a-8-dimensional-sphere/33661
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## How to generate a net on a 8-dimensional sphere
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Using Matlab, how to generate a net of 3^10 points that are evenly located (or distributed) on the 8-dimensional unit sphere?
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How are you measuring evenness? In any case, you could just pick them uniformly at random... – Mariano Suárez-Alvarez Jul 19 2010 at 5:12
Mariano, Probably use the euclidean distance to measure the evenness. But how to pick the points uniformly at random in matlab? – unknown (yahoo) Jul 19 2010 at 6:14
2
Google for "sample points uniformly on a sphere" and choose the link to Mathoverflow offered by google :P – Mariano Suárez-Alvarez Jul 19 2010 at 6:26
6
MathOverflow is not a place where it is appropriate to ask for MATLAB code, but we welcome interesting questions in applied mathematics. I don't know the precise meaning of "evenly located" as you use it, but if you search for "mesh-generation algorithms" you will find a lot of information. In particular, I think Ruppert's algorithm can be adapted to your purposes, and you can set lower bounds on the dihedral angles you get. – S. Carnahan♦ Jul 19 2010 at 12:57
## 7 Answers
If it's really important for the points to be evenly distributed, and you don't mind doing a lot of calculation to get them that way, you can start with a randomly distributed set and then iterate over the entire set repeatedly, allowing each point in turn to make whatever small adjustment improves your chosen definition of uniformity, and repeat this until the set of points converges. If you're even pickier than that, and not satisfied by just a locally optimal arrangement, the canonical next thing to try is simulated annealing.
For picking points at random, I agree with Peter Shor that taking the time to implement a one-to-one volume-preserving map from a product of intervals to a high-dimensional sphere would be much more wasteful (of time; you would learn a lot) than throwing away 98% of your random numbers. It's an interesting question, though, whether systematically chosen points in a product of intervals can be well-distributed under one of these volume-preserving (but distance-destroying) maps. The first interesting case of such a map is the axial projection from the curved surface of a cylinder of height 2 and radius 1 to the surface of the unit sphere it contains: projecting straight to the axis, one direction gets stretched out in exact counterbalance to the compression of the other direction. Call the coordinates of the cylinder surface z ∈ [$-1$, $1$] and θ ∈ [$0$, $2\pi$]. Choosing an ordinary regular rectangular grid in z and θ does terrible things to the projection. On the other hand, for any $N$, setting zi = $(-N+2 i - 1)$/$N$ and θi = $2\pi (\phi i$ mod $1$), where $\phi$ is the golden mean, actually gives a very nice distribution of points after projection. It's possible that in any dimension there is such a lattice in the cube that projects nicely, for any N, to the sphere.
-
### 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.
If you're looking for points on the 8-dimensional sphere, another thing you could do is go to Neil Sloane's table of spherical codes, scroll down until you get to dimension 8, and obtain a sphere covering which has 2160 points fairly evenly distributed (obtained from second shell of vectors in the E8 lattice). Now, if you apply $3^{10}/2160 \approx 27$ random orthogonal matrices to this set, you'll get a set of points distributed on the 8-dimensional sphere which is presumably somewhat more uniform than a set of random points. Since I don't know why you want these vectors, I don't know how much an improvement this is over random points, and whether it's worth all the extra work.
You can get random orthogonal transformations by starting with an $8 \times 8$ matrix with random Gaussian entries. First, normalize the top row to make it a length 1 vector. Next, subtract a multiple of the top row from the second row to make it perpendicular to the top row, and then normalize to make the second row a length 1 vector, and so on.
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All of these operations should be straightforward using MATLAB. – Peter Shor Jul 19 2010 at 17:34
In case the goal is to draw points inside the sphere, the discussion http://mathoverflow.net/questions/33129/intuitive-proof-that-the-first-n-2-coordinates-on-a-sphere-are-uniform-in-a-bal seems relevant.
In other words, one simply draws random points on the 10-dimensional sphere (by drawing a normal vector and normalizing it) and discards the last two coordinates.
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Forgetting Matlab, the 'best' way to...hold on, do you mean -in- the sphere or -on- the sphere?
For -in- the sphere, create 8-tuples where each element is from the uniform distribution from 0 to 1. Ignore those tuples whose Euclidean norm $\sqrt{\sum x_i^2}$ is greater than 1. Do this until you have $3^{10}$ points. This is a uniform distribution over the sphere.
For -on- the sphere, create 8-tuples as before, but then divide each point by the norm (of course throw out $\langle 0,0,0,0,0,0,0,0\rangle$ ). This will place a point on the surface. Do this $3^{10}$ times. This is not an exact uniform distribution but is a very good approximation to one, and is very easy to do.
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1
Because the volume of an $n$-sphere is small when $n$ is large, I believe there will be significant inefficiencies your first method: more than 97% of the volume of the cube is exterior to the enclosed 8-ball. – Joseph O'Rourke Jul 19 2010 at 17:27
Actually, 98%, as Peter's calculation shows! – Joseph O'Rourke Jul 19 2010 at 17:29
4
If I did the calculation right, only 1.6% of the points given by this method will be inside the sphere. However, in this case the overhead of choosing 60 times as many points as you need is pretty much negligible compared to the cost of programming a better method. – Peter Shor Jul 19 2010 at 17:30
I believe you meant to say draw 8-tuples from the uniform distribution from $-1$ to $1$, otherwise you are picking from just 1/256th of the the volume of the 8-sphere by limiting it to the region of space with all positive coordinates. Alternately, you could pick from the uniform distribution from 0 to 1 but change your selection criteria to accepting only those points where $$\sqrt{\sum (x_i-0.5)^2} \le 0.5$$. – sleepless in beantown Aug 29 2010 at 14:40
In general, for generating extra-regular but not-too-regular distributions of points (in a technical sense, "low discrepancy", meaning that the variance in the length of gaps between points is smaller than a uniform distribution), you can use a class of methods called quasi monte carlo methods. There are libraries in MATLAB.
http://en.wikipedia.org/wiki/Quasi-Monte_Carlo_method
http://www.mathworks.com/matlabcentral/fileexchange/17457-quasi-montecarlo-halton-sequence-generator
Though if you want a totally uniform set of points, these won't help you.
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Talking out of my depth so I am not sure this work, but here is a start of an idea... Say you were finding points uniformly on $S^3$, couldn't you use Hopf fibration? The idea is you select Hopf fiber uniformly from $S^2$ then it is easy to populate points uniformly on the Hopf fiber, which is a circle in this case. Unfortunately, there is no such relation for $S^8$.
Instead one could potentially use the following
$S^1\hookrightarrow S^3 \rightarrow S^2$
and then
$S^3\hookrightarrow S^7 \rightarrow S^4$
but then one needs to construct uniformly spaced points on $S^4$ from uniformly spaced points in $S^3$ and same for $S^8$ and $S^7$, which doesn't seem that hard.
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Shells of evenly spaced lattice points:
To generate evenly spaced sets of non-random points on an n-sphere, start with the permutations of { 0 1 1 ... 2 2 ... }, then make 2^n flips of that. For example, in 4-space start with the 12 permutations of { 0 1 1 2 }. Each point is √6 from the origin, and each has 4 neighbours √2 away (+1 here, -1 there):
````0 1 1 2
0 1 2 1
0 2 1 1
1 0 1 2
1 1 0 2
````
Make 2^4 sign-flipped copies of this, i.e. multiply by { 1 1 1 1 } .. { -1 -1 -1 -1 } except where there's a 0. This gives a shell of 96 points, 0 1 1 2 .. 0 -1 -1 -2. Each is √6 from the origin, and each now has 6 neighbours √2 away.
For the 8-sphere, start with the 280 permutations of { 0 1 1 1 1 2 2 2 }. Each has of course the same distance from the origin, and each has 12 neighbours √2 away — a nice, regular graph. The shell of 280 * 2^7 = 35840 sign-flipped points is not quite 3^10, but.
(I'd appreciate links to papers or programs on such graphs.)
-
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http://unapologetic.wordpress.com/2011/07/06/
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# The Unapologetic Mathematician
## Tensor Bundles
We have a number of other constructions similar to the tangent bundle that will come in handy. These are all sort of analogues of certain constructions we already know about on vector spaces. Let’s review these first.
Taking the tensor product of vector spaces is old hat by now, as is using the dual space $V^*$. We’ll put them together by defining the space of “tensors of type $(r,s)$” as
$\displaystyle T^r_s(V)=V\otimes\dots\otimes V\otimes V^*\otimes\dots\otimes V^*$
where we have $r$ copies of the vector space $V$ and $s$ copies of the dual space $V^*$. Vectors in $V$, then, are tensors of type $(1,0)$, while vectors in the dual space are tensors of type $(0,1)$.
We also know about the space of antisymmetric tensors of rank $k$ over a vector space. In particular, we’re most interested in carrying this construction out over the dual space: $\Lambda^*_k(V)=\Lambda_k(V^*)$. And of course we can take the direct sum of these spaces over all $k$ to get the exterior algebra $\Lambda^*(V)$.
Now, we will take these constructions and apply them to the tangent spaces to a manifold. We define the bundle of tensors of type $(r,s)$ over $M$:
$\displaystyle T^r_s(M)=\bigcup\limits_{p\in M}T^r_s(\mathcal{T}_pM)$
the “exterior $k$-bundle” over $M$:
$\displaystyle \Lambda^*_k(M)=\bigcup\limits_{p\in M}\Lambda_k(\mathcal{T}^*_pM)$
and the exterior algebra bundle over $M$:
$\displaystyle \Lambda^*(M)=\bigcup\limits_{p\in M}\Lambda(\mathcal{T}^*_pM)$
The trick here is that for each of these constructions, if we have a basis of $V$ we automatically get a basis of each space $T^r_s(V)$, $\Lambda^*_k(V)$, and $\Lambda(V)$. If we start with a coordinate patch $(U,x)$ on $M$, we get a basis $\frac{\partial}{\partial x^i}$ of $\mathcal{T}_pM$ for each $p\in U$. Then, just as we did with the tangent bundle and the cotangent bundle we can come up with a coordinate patch “induced by $(U,x)$” on each of our new bundles. In this way, we can turn them from disjoint unions of vector spaces into manifolds of their own right, each with a smooth projection down to $M$.
Now we can define a “tensor field of type $(r,s)$” on an open region $U\subseteq M$ as a section of the projection $\pi:T^r_s(U)\to U$. That is, it’s a smooth map $t:U\to T^r_s(U)$ such that $\pi(t(p))=p$. Similarly, we define a “differential $k$-form” over $U$ to be a section of the projection $\pi:\Lambda^*_k(U)\to U$.
## 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.
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http://math.stackexchange.com/questions/217758/minimal-set-of-trig-identities-to-define-all-the-trig-functions
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# Minimal set of trig identities to define all the trig functions
What are a minimal set of trig identities that can uniquely define the trig functions? I know that you can define, for example, $\sin(x)$ as the unique solution to the differential equation $f''(x) = -f(x), f(0)=0, f'(0)=1$, but I'd like to avoid analytical definitions to as much of a degree as possible (though this definition is interesting in that it nowhere explicitly mentions $\pi$).
Obviously, if we define one trig function, then we can define the rest in terms of it (like $\cos(x) = \sin(\frac{\pi}{2} - x)$). I am just curious if we can give some clever set of trig identities (like half angle formulas or the Pythagorean identity), which completely define the trig functions. It would then follow that all other trig identities can be derived from these.
And is it possible to do this without using any analysis? My intuition says no, at least for a finite set of identities, because any finite set of identities would only define a countable set of points, so you'd at least have to assert that the functions are continuous. So what are either a good infinite set of identities to define one of the trig functions, or a good finite set with the addition of the condition that the function is continuous (which to me is the least possible analysis we can get away with in this case)? Or if you have another clever way that doesn't fit what I said above, I'd like to hear about it too.
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## 1 Answer
$f(x) = \cos x, g(x) = \sin x$ is the unique continuous (but even measurable would suffice) pair of functions satisfying
$$f(x)^2 + g(x)^2 = 1$$ $$f(x + y) = f(x) f(y) - g(x) g(y)$$ $$g(x + y) = f(x) g(y) + g(x) f(y)$$
with initial conditions $f(0) = 1, g(0) = 0$ and period $2 \pi$. If you don't want to explicitly mention $\pi$ then I am reasonably sure you will need to mention a derivative. You can't get $\pi$ with absolutely no analytic input whatsoever.
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1
Do you have a reference for this? – Neal Oct 21 '12 at 5:05
1
@Neal: nope. I would prefer to leave it as an exercise. Hint: consider the function $f(x) + i g(x)$. – Qiaochu Yuan Oct 21 '12 at 5:06
True, you can't get period $\pi$. But if one is willing to live with the angles in units other than radians, it seems one could say "with period P" for any $P$ and get the trig functions in these alternate units. – coffeemath Oct 21 '12 at 6:32
Oh sure, I knew you would have to mention $\pi$. When I said not analytic, I meant more about requirements on the functions rather than analytically defined constants. – asmeurer Oct 21 '12 at 7:09
2
@asmeurer: it's known that pathological solutions to the Cauchy functional equation ( en.wikipedia.org/wiki/Cauchy's_functional_equation) are not measurable (imi.kyushu-u.ac.jp/~ssaito/eng/maths/Cauchy.pdf). The proof should apply to this case as well. – Qiaochu Yuan Oct 21 '12 at 7:25
show 6 more comments
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http://mathhelpforum.com/advanced-algebra/198188-how-understand-set-exponential.html
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# Thread:
1. ## How to understand set as exponential?
I came across a paper, where the notation is:
$2^\Theta$
where $\Theta$ is a set,
Normally we see exponential such as $a^n$, where n is a real number.
How should I interpret 2^Theta, is this a matrix where in each entry the base is 2?
Thanks for help.
2. ## Re: How to understand set as exponential?
It's the set of all functions from theta to {0,1}.
3. ## Re: How to understand set as exponential?
Originally Posted by colruyt
I came across a paper, where the notation is:
$2^\Theta$
where $\Theta$ is a set,
Normally we see exponential such as $a^n$, where n is a real number.
How should I interpret 2^Theta, is this a matrix where in each entry the base is 2?
There are two principal meaning the symbol $2^A$ where $A$ is set.
It stands for the set of all functions from $A$ to $\{0,1\}.$
Some authors use it to be short hand for the power set, the set of all subsets of $A$.
4. ## Re: How to understand set as exponential?
Thanks a lot.
I don't understhand the first meaning:
It stands for the set of all functions from A to {0,1}
Do you mean 2^A is a function, but why to {0,1}??
Actually I think it might be the case for power set, since this comes with a definition for a function:
$\mu: C\cup U \rightarrow 2^\Theta \cup (C\cup \Theta)$
5. ## Re: How to understand set as exponential?
Originally Posted by colruyt
Thanks a lot.
I don't understhand the first meaning:
Do you mean 2^A is a function, but why to {0,1}??
Actually I think it might be the case for power set, since this comes with a definition for a function:
$\mu: C\cup U \rightarrow 2^\Theta \cup (C\cup \Theta)$
This is confusion caused by your not posting the complete problem.
It seems that you are working of something else of which this is a sub-part.
6. ## Re: How to understand set as exponential?
Originally Posted by colruyt
Thanks a lot.
I don't understhand the first meaning:
Do you mean 2^A is a function, but why to {0,1}??
No, he means that 2^A is a set of functions- the set of all functions from A to {0, 1}. And it is to {0, 1} because of the "2". 3^A would be a set of all functions from A to {0, 1, 2}, or more generally, from A to any set containing 3 objects.
Actually I think it might be the case for power set, since this comes with a definition for a function:
$\mu: C\cup U \rightarrow 2^\Theta \cup (C\cup \Theta)$
7. ## Re: How to understand set as exponential?
Originally Posted by Plato
This is confusion caused by your not posting the complete problem.
It seems that you are working of something else of which this is a sub-part.
Sorry my mistake: basically it's a function that matches two sets into two sets, think of match students into schools. where $\Theta$ is the preference of students, and a typo in my previous function, the correct form is:
$\mu: C \cup \Theta \rightarrow 2^\Theta \cup (C \cup \Theta)$
My original guess was power set too, but I am just no sure about the notation...
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http://math.stackexchange.com/questions/243484/vector-fields-generating-a-transformation
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# Vector fields generating a transformation
It would be great if someone can explain to me what the following means:
Vector fields $V_i, i=1,2,3$ generate 3 single-parameter groups of transformations in $\mathbb R$ -- $$\tilde x =x\exp(\alpha);$$$$\tilde x = x+\beta;$$$$\tilde x = {x\over 1-\gamma x}$$ respectively. These 3 vector fields together generate a 3-parameter group of transformations: $\tilde x = {ax+b\over cx+d}$ where $ad-bc=1$
I think I understand the first bit, and I think the $V_i$'s are $x\partial_x; \partial_x;x^2\partial _x$ respectively. However, I have no clue what the second bit is talking about -- how do you combine them and why is that so?
Qiaochu has helpfully pointed out that this has to do with Lie algebra. Could anyone further explain how they combine these vector fields/transformations to give the 3-parameter transformation group?
Thank you.
Anyone?
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The second bit asserts that three vector fields span a Lie algebra (en.wikipedia.org/wiki/Lie_algebra) which generates a Lie group (en.wikipedia.org/wiki/Lie_group). – Qiaochu Yuan Nov 24 '12 at 2:18
Thank you, @QiaochuYuan , how though did they combine them to get that resulting transformation? – Henry Nov 24 '12 at 2:26
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http://mathhelpforum.com/trigonometry/128868-double-angle-question.html
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# Thread:
1. ## Double angle question
The question is
If tanA=4 and tabB=3/5, and A and B are acute angles, prove that A-B=pi/4 (pi in radians, so really it's 45degrees).
My answer is:
$tan(x-y)=\frac {tan4-tan \frac {3}{5}}{1 + tan4 * tan \frac {3}{5}}<br /> =1<br /> \therefore tan(x-y)=1=\frac {\pi}{4}$
I have a feeling there is more to add. My lecturer said something about 135 being a possible answer, and I was wondering why the question included "acute"; as in first quadrant?
2. Hello RAz
Originally Posted by RAz
The question is
If tanA=4 and tabB=3/5, and A and B are acute angles, prove that A-B=pi/4 (pi in radians, so really it's 45degrees).
My answer is:
$\color{red}tan(x-y)=\frac {tan4-tan \frac {3}{5}}{1 + tan4 * tan \frac {3}{5}}\color{black}<br /> =1<br /> \therefore tan(x-y)=1\color{red}=\frac {\pi}{4}$
I have a feeling there is more to add. My lecturer said something about 135 being a possible answer, and I was wondering why the question included "acute"; as in first quadrant?
Your general method is OK, but your use of notation is very sloppy. Say what you mean! What you mean, of course, is:
$\tan(A-B)=\frac {\tan A-\tan B}{1 + \tan A \tan B} = \frac{4-\tfrac35}{1+4\cdot\tfrac35}= 1$
$\Rightarrow A-B = \frac{\pi}{4}$
I can't see where the possible angle of $135^o$ might come from. $\tan135^o = -1$, not $+1$. And if $A$ and $B$ are acute with $\tan A > \tan B$, then $A>B$. So $A-B$ is also acute. Hence, $\tan(A-B) = 1\Rightarrow A-B = 45^o$.
Grandad
3. If your teacher meant $225^\circ$, I can see the possibility but definitely not $135^\circ$
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http://mathoverflow.net/revisions/85625/list
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## Return to Answer
4 clarified some statements
The search term you want to look for is "Klimyk's Formula." This formula boils down to the following:
Fix $G$ a compact complex semisimple Lie group. Suppose $V(\lambda)$ and $V(\mu)$ are irreducible representations with highest weights $\lambda$ and $\mu$ respectively. Let $W_\lambda = {\lambda_1,\lambda_2,\ldots \lambda_d}$ be the multiset of weights of $V(\lambda)$ with $d = dim(V(\lambda))$. Then the irreducible components of $V(\lambda)\otimes V(\mu)$ are given by ${V(\mu+\lambda_i)}_{i=1}^d$.
To apply this in practice, you need to be comfortable with the concept of defining $V(\lambda)$ when $\lambda$ is not a dominant weight (which sometimes causes modules to cancel when they appear with both positive and negative signs in the sum), but it applies to lots of groups (even beyond the scope of compact complex semisimple in some cases if im not mistaken), and Littlewood-Richardson is just the special case of this formula in type $A$.
An example for $G_2$ (since that is also my favorite compact semisimple Lie group) is to let $\lambda = [1,0]$ be the highest weight of the 7-dimensional representation and $\mu = [0,1]$ the highest weight of the 14-dimensional adjoint representation.
The seven weights of $V(\lambda)$ are $[1,0]$, $[-1,1]$, $[2,-1]$, $[0,0]$, $[-2,1]$, $[1,-1]$, and $[-1,0]$ so Klimyk tells us the 98-dimensional tensor product decomposes as:
$V([1,1]) \oplus V([-1,2]) \oplus V([2,0]) \oplus V([0,1]) \oplus V([-2,2]) \oplus V([1,0]) \oplus V([-1,1])$
This is where familiarity with interpreting modules with non-dominant highest weights comes in; $V[-1,2]$ and $V[-1,1]$ turn out to be 0-dimensional modules, while $V([-2,2]) \cong -V([0,1])$*. Thus the terms which do not disappear are $V([1,1])$ which is a 64-dimensional module, $V([2,0])$ which is a 27-dimensional module, and $V([1,0])$ which is the 7-dimensional defining representation, a total of 98 dimensions.
If you had instead chosen to switch $\lambda$ and $\mu$ and add the 14 weights of $V([0,1])$ to [1,0], you would have obtained 14 modules, but as before, some would have been zero and others would have cancelled in pairs ultimately leading to the same three modules as above being the only things left over. In my opinion, this reflexivity always holding is the coolest thing about Klimyk's formula.
One neat corollary to Klimyk's formula is that a tensor product of two irreducible modules cannot decompose into a sum of more than $d$ irreducibles where $d$ is the minimum of the dimensions of the two modules.
*EDIT: After posting, I decided to add a bit more about modules with non-dominant highest weight. Basically, the weights of a group $G$ are permuted via the Weyl group action on the weights. Weights are determined by integer $r$-tuples where $r$ is the rank of $G$; tuples containing a -1 lie in the walls of the Weyl chambers and so the modules with these highest weights end up being 0. All There are a few other subspaces which also correspond to walls; weights $\mu$ not lying in the walls satisfy $w(\mu) = \lambda$ for some dominant weight $\lambda$ (all coordinates nonnegative) and a unique $w$ in the Weyl group (i.e. only one $w$ in the Weyl group will take $\mu$ to a dominant weight, so $\lambda$ is also uniquely determined). Then $V(\mu)$ is defined by the following relationship:
$V(\mu) = (-1)^w\cdot V(\lambda)$
Here $(-1)^w$ is the sign representation which appears with all Weyl groups; in the $A$-series whose Weyl groups are the $S_n$'s this is the ordinary sign representation.
3 added 944 characters in body
This is where familiarity with interpreting modules with non-dominant highest weights comes in; $V[-1,2]$ and $V[-1,1]$ turn out to be 0-dimensional modules, while $V([-2,2]) \cong -V([0,1])$. V([0,1])$*. Thus the terms which do not disappear are V([1,1])$V([1,1])$which is a 64-dimensional module, V([2,0])$V([2,0])$which is a 27-dimensional module, and V([1,0])$V([1,0])\$ which is the 7-dimensional defining representation, a total of 98 dimensions.
One neat corollary to Klimyk's formula is that a tensor product of two irreducible modules cannot decompose into a sum of more than $d$ irreducibles where $d$ is the minimum of the dimensions of the two modules.
*EDIT: After posting, I decided to add a bit more about modules with non-dominant highest weight. Basically, the weights of a group $G$ are permuted via the Weyl group action on the weights. Weights are determined by integer $r$-tuples where $r$ is the rank of $G$; tuples containing a -1 lie in the walls of the Weyl chambers and so the modules with these highest weights end up being 0. All other weights $\mu$ satisfy $w(\mu) = \lambda$ for some dominant weight $\lambda$ (all coordinates nonnegative) and a unique $w$ in the Weyl group (i.e. only one $w$ in the Weyl group will take $\mu$ to a dominant weight, so $\lambda$ is also uniquely determined). Then $V(\mu)$ is defined by the following relationship:
$V(\mu) = (-1)^w\cdot V(\lambda)$
Here $(-1)^w$ is the sign representation which appears with all Weyl groups; in the $A$-series whose Weyl groups are the $S_n$'s this is the ordinary sign representation.
2 added 45 characters in body
The search term you want to look for is "Klimyk's Formula." This formula boils down to the following:
Fix $G$ a compact complex semisimple Lie group. Suppose $V(\lambda)$ and $V(\mu)$ are irreducible representations with highest weights $\lambda$ and $\mu$ respectively. Let $W_\lambda = {\lambda_1,\lambda_2,\ldots \lambda_d}$ be the multiset of weights of $V(\lambda)$ with $d = dim(V(\lambda))$. Then the irreducible components of $V(\lambda)\otimes V(\mu)$ are given by ${V(\mu+\lambda_i)}_{i=1}^d$.
To apply this in practice, you need to be comfortable with the concept of defining $V(\lambda)$ when $\lambda$ is not a dominant weight (which sometimes causes modules to cancel when they appear with both positive and negative signs in the sum), but it applies to lots of groups (even beyond the scope of compact complex semisimple in some cases if im not mistaken), and Littlewood-Richardson is just the special case of this formula in type $A$.
An example for $G_2$ (since that is also my favorite compact semisimple Lie group) is to let $\lambda = [1,0]$ be the highest weight of the 7-dimensional representation and $\mu = [0,1]$ the highest weight of the 14-dimensional adjoint representation.
The seven weights of $V(\lambda)$ are $[1,0]$, $[-1,1]$, $[2,-1]$, $[0,0]$, $[-2,1]$, $[1,-1]$, and $[-1,0]$ so Klimyk tells us the 98-dimensional tensor product decomposes as:
$V([1,1]) \oplus V([-1,2]) \oplus V([2,0]) \oplus V([0,1]) \oplus V([-2,2]) \oplus V([1,0]) \oplus V([-1,1])$
This is where familiarity with interpreting modules with non-dominant highest weights comes in; V[-1,2] $V[-1,2]$ and V[-1,1] $V[-1,1]$ turn out to be 0-dimensional modules, while $V([-2,2]) = \cong -V([0,1])$. Thus the terms which do not disappear are V([1,1]) which is a 64-dimensional module, V([2,0]) which is a 27-dimensional module, and V([1,0]) which is the 7-dimensional defining representation, a total of 98 dimensions.
If you had instead chosen to switch $\lambda$ and $\mu$ and add the 14 weights of $V([0,1])$ to [1,0], you would have obtained 14 modules, but as before, some would have been zero and others would have cancelled in pairs ultimately leading to the same three modules as above being the only things left over. In my opinion, this reflexivity always holding is the coolest thing about Klimyk's formula.
One neat corollary to Klimyk's formula is that a tensor product of two irreducible modules cannot decompose into a sum of more than $d$ irreducibles where $d$ is the minimum of the dimensions of the two modules.
1
The search term you want to look for is "Klimyk's Formula." This formula boils down to the following:
Fix $G$ a compact complex semisimple Lie group. Suppose $V(\lambda)$ and $V(\mu)$ are irreducible representations with highest weights $\lambda$ and $\mu$ respectively. Let $W_\lambda = {\lambda_1,\lambda_2,\ldots \lambda_d}$ be the multiset of weights of $V(\lambda)$ with $d = dim(V(\lambda))$. Then the irreducible components of $V(\lambda)\otimes V(\mu)$ are given by ${V(\mu+\lambda_i)}_{i=1}^d$.
To apply this in practice, you need to be comfortable with the concept of defining $V(\lambda)$ when $\lambda$ is not a dominant weight (which sometimes causes modules to cancel when they appear with both positive and negative signs in the sum), but it applies to lots of groups (even beyond the scope of compact complex semisimple in some cases if im not mistaken), and Littlewood-Richardson is just the special case of this formula in type $A$.
An example for $G_2$ (since that is also my favorite compact semisimple Lie group) is to let $\lambda = [1,0]$ be the highest weight of the 7-dimensional representation and $\mu = [0,1]$ the highest weight of the 14-dimensional adjoint representation.
The seven weights of $V(\lambda)$ are $[1,0]$, $[-1,1]$, $[2,-1]$, $[0,0]$, $[-2,1]$, $[1,-1]$, and $[-1,0]$ so Klimyk tells us the 98-dimensional tensor product decomposes as:
$V([1,1]) \oplus V([-1,2]) \oplus V([2,0]) \oplus V([0,1]) \oplus V([-2,2]) \oplus V([1,0]) \oplus V([-1,1])$
This is where familiarity with non-dominant highest weights comes in; V[-1,2] and V[-1,1] turn out to be 0-dimensional modules, while $V([-2,2]) = -V([0,1])$. Thus the terms which do not disappear are V([1,1]) which is a 64-dimensional module, V([2,0]) which is a 27-dimensional module, and V([1,0]) which is the 7-dimensional defining representation, a total of 98 dimensions.
If you had instead chosen to switch $\lambda$ and $\mu$ and add the 14 weights of $V([0,1])$ to [1,0], you would have obtained 14 modules, but some would have been zero and others would have cancelled in pairs ultimately leading to the same three modules as above being the only things left over. In my opinion, this reflexivity always holding is the coolest thing about Klimyk's formula.
One neat corollary to Klimyk's formula is that a tensor product of two irreducible modules cannot decompose into a sum of more than $d$ irreducibles where $d$ is the minimum of the dimensions of the two modules.
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http://mathhelpforum.com/differential-equations/176399-mixture-print.html
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# Mixture
Printable View
• March 30th 2011, 10:20 PM
Naples
Mixture
A tank holds 300 gallons of a brine solution, the rate at which brine is pumped into the tank is 3 gal/min with a concentration of salt of 2 lb/gal and the well stirred solution is pumped out at a rate of 2 gal/in. It stands to reason that since brine is accumulating in the tank at the rate of 1 gal/in, any finite tank must eventually overflow. Suppose that the tank has an open top and has a total capacity of 400 gallons and A(0)=50.
(a) When will the tank overflow?
(b) What will be the number of pounds in the tank at the instant it overflows?
(c) Assume that although the tank is overflowing, brine solution continues to be pumped in at a rate of 3 gal/min and the solution continues to be pumped out at a rate of 2 gal/min. Devise a method for determining the number of pounds of salt in the tank at t = 150 minutes.
(d) Determine the number of pounds of salt in the tank as t approaches infinity. Does your answer agree with your intuition?
(e) Use a graphing utility to plot the graph of A(t) on the interval [0,500)
First I set up a DE using the information provided
$DA/dt = 6 - (2A/(300 + t))$
$DA/dt + (2A/(300 + t)) = 6$
Used an integrating factor of $(300 + t)^2$
Multiplied by IF and integrated and cleaned it up to get:
$A= 600 + 2t + C/(300+t)^2$ used initial value A(0)=0 to solve for C
$A= 600 + 2t - (4.95x10^7)/(300+t)^2$
(a) tank overflows when 300+t = 400 so when t=100 minutes
(b) A(100)=490 lbs
(c) This is where I get iffy, I'm pretty sure you can't just use A(150) because you need to take into account that the tank is overflowing, I just don't know how to.
(d) I did the limit as t approaches infinity of A(t) so:
$600 + 2(inf) - 0 = infinity$
obviously since the container has a finite capacity, it's impossible for the amount of salt to be infinity, the problem is that the equation doesn't account for the container's limited capacity
(e) I plugged my equation A(t) into my calculator and basically just got a straight line going diagonally across the screen
Thanks in advance
• March 31st 2011, 01:31 AM
Ackbeet
In my copy of Zill, 4th edition, this is problem 3.1.26, exactly. If that's also true for you, I think maybe your answer to (b) is incorrect. Zill did an example in the text (it's example 5 in my copy), where he gives you the solution to the problem before it starts overflowing. The solution he gives is
$A(t)=600+2t-(4.95\times 10^{7})(300+t)^{-2}.$
Double-check what you get when you plug in $t=100$ into this equation.
As for (c), you always have this:
$\dot{A}(t)=\text{rate of salt entering}-\text{rate of salt leaving}.$
How much salt is entering? How much is leaving? I don't think it's wise to continue from here, until you have the correct DE and have solved it correctly.
• March 31st 2011, 04:24 AM
Naples
Quote:
Originally Posted by Ackbeet
In my copy of Zill, 4th edition, this is problem 3.1.26, exactly. If that's also true for you, I think maybe your answer to (b) is incorrect. Zill did an example in the text (it's example 5 in my copy), where he gives you the solution to the problem before it starts overflowing. The solution he gives is
$A(t)=600+2t-(4.95\times 10^{7})(300+t)^{-2}.$
Double-check what you get when you plug in $t=100$ into this equation.
As for (c), you always have this:
$\dot{A}(t)=\text{rate of salt entering}-\text{rate of salt leaving}.$
How much salt is entering? How much is leaving? I don't think it's wise to continue from here, until you have the correct DE and have solved it correctly.
The equation you gave me for A(t) is the same one that I derived in my original post.
For (c) can I just solve for A(150) then using that equation for A(t) or does the overflowing of the container change something?
• March 31st 2011, 04:39 AM
Ackbeet
Looks like I made a mistake in my working. Your figure for (b) is correct. You can't, however, just plug in 150 into that formula, because the DE changes at t = 100. The overflowing of the container definitely changes the DE. So, I go back to my previous question: how much salt is leaving, and how much is entering?
• March 31st 2011, 07:57 AM
Naples
Rate in is still 6 lb/gal
Rate out would be 2a/400 then cause after it overflows the amount in the container remands at 400
So dA/dt + 2A/400 = 6
• March 31st 2011, 08:00 AM
Ackbeet
The rate out would be 3A/400, because the tank is leaking. Otherwise, your new DE is correct. What's the initial condition?
• March 31st 2011, 08:03 AM
Naples
Initial condition would just be the amount when tank first started to overflow so A(0)=491
• March 31st 2011, 08:07 AM
Ackbeet
You could do it that way, I suppose. You're "resetting your clock" by doing that, which is probably not what the original problem had in mind. I would probably do y(100) = 491. That way, the 150 in the expression y(150) will turn out right. Doing y(100) = 491 will also make it easier to graph.
All times are GMT -8. The time now is 05:46 AM.
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http://math.stackexchange.com/questions/174821/selfadjoint-compact-operator-with-finite-trace?answertab=oldest
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# Selfadjoint compact operator with finite trace
I have a compact selfadjoint operator $T$ on a separable Hilbert space. For some fixed orthonormal basis, the operator's diagonal is in $\ell^1(\mathbb{N})$.
Can we conclude that $T$ is trace class?
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By diagonal, do you mean $\langle Te_n,e_n\rangle$, where $\{e_n\}$ is the fixed orthonormal basis? – Davide Giraudo Jul 24 '12 at 21:55
May be I'm mistaken somwhere, but what is wrong with the following proof. Since $T$ is self adjoint and compact we can say that $T(x)=\sum_n\lambda_n\langle x,e_n\rangle e_n$. So $|\mathrm{Tr}(T)|=|\sum_n\langle T e_n,e_n\rangle|=|\sum_n\lambda_n|\leq\Vert\lambda\Vert_1<\infty$ – Norbert Jul 24 '12 at 21:57
1
@Davide: yes, including the sum you didn't type. – Martin Argerami Jul 24 '12 at 22:01
1
@Norbert: you are assumming that $T$ is diagonal in the given basis, which is not the case. – Martin Argerami Jul 24 '12 at 22:02
## 1 Answer
No, we cannot conclude that the operator is trace class.
For example, let a Hilbert space have orthonormal basis $e_1,f_2,e_2,f_2,e_3,f_3,\ldots$, and $T$ interchanges $e_i,f_i$, while multiplying both by a positive real $\lambda_i$. That is, in these coordinates, the matrix of $T$ is a list of diagonal blocks, with the $i$-th diagonal block being anti-diagonal $\lambda_i,\lambda_i$.
For $\lambda_i\rightarrow 0$, the operator is compact, almost from the definition.
All the diagonal entries are $0$.
The operator is self-adjoint because the matrix is symmetric real.
However, the operator is not trace class unless $\sum_i |\lambda_i|<\infty$, which easily fails for many sequences of positive reals $\lambda_i\rightarrow 0$.
Edit: It is noteworthy that the analogous characterization (I pointedly don't say "definition") of "Hilbert-Schmidt" does not depend on choice of basis. Thus, "defining" trace-class as composition of two Hilbert-Schmidt operators is sometimes usefully more intrinsic, less basis/coordinate-dependent.
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Nice example. Thanks! – Martin Argerami Jul 25 '12 at 1:45
@Martin: You probably know this, but just in case: If your operator happens to be positive (not only self-adjoint), then your desired conclusion does hold, i.e. the trace is summable independently of the orthonormal basis, see Corollary 3.4.4 on page 117 of Pedersen's Analysis Now. – t.b. Jul 25 '12 at 2:19
1
Yes, I have wished so hard for my operator to be positive... – Martin Argerami Jul 25 '12 at 2:21
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http://math.stackexchange.com/questions/285189/finding-solutions-using-trigonometric-identities
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Finding solutions using trigonometric identities
I have an exam tomorrow and it is highly likely that there will be a trig identity on it. To practice I tried this identity:
$$2 \sin 5x\cos 4x-\sin x = \sin9x$$
We solved the identity but we had to move terms from one side to another.
My question is: what are the things that you can and cannot do with trig identities?
And what things are must not when doing trig identities?
Thank you
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This is not an identity (false for x=0. It is an equation whose roots are to be found. – marty cohen Jan 23 at 18:33
Sorry about that. I accidentally added a "+" between $sin5x$ and $cos4x$ – gekkostate Jan 23 at 18:42
4 Answers
Those are very general questions: what "can you do" and what "can't you do"...in terms of using trig identities:
What you can do is use an identity to replace one expression with its strict equivalent, as determined by the identity in question. It's simply grounded in "substituting one expression for its equivalent", which is legitimate beyond just its handiness with utilizing trig identities to transform one expression to an equivalent expression.
You can algebraically manipulate equations involving trigonometric expressions in any way that is "legal" for manipulating any equation.
So for a simple example, ${1-\sin^2\theta} = \cos^2 \theta$ because we know the identity:
$$\sin^2\theta + \cos^2\theta = 1.$$ We do not change this equivalence when we manipulate it to get: $$\sin^2\theta + \cos^2\theta = 1 \iff \cos^2\theta = 1 - \sin^2\theta$$
In the identity you solved, you can use:
$$\sin c+\sin d=2\sin\frac{c+d}2\cos\frac{c-d}2$$
Applied here: $$\sin9x+\sin x=2\sin\frac{9x+x}2\cos\frac{9x-x}2=2\sin5x\cos4x$$
$$2\sin a\cos b=\sin(a+b)+\sin(a-b)$$
Applied here: $$2\sin5x\cos4x=\sin(5x+4x)+\sin(5x-4x)=\sin9x+\sin x$$
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I can move terms from one side to another as long as both side are equal right? – gekkostate Jan 23 at 18:16
Yes, indeed! Just as I took the identity above, subtracted $\sin^2 \theta$ from both sides of the identity $\sin^2\theta + \cos^2\theta = 1$ – amWhy Jan 23 at 18:23
Are you asking specifically about the identity in question (that you solved)? Or about trig identities in general? I took your question to refer to "in general". If you don't find my answer helpful, I can delete it. – amWhy Jan 23 at 19:02
Yes, my question was general and yes your answer is helpful. – gekkostate Jan 23 at 19:03
Hint: We have $\sin 5x\cos 4x+\cos 5x\sin 4x=\sin 9x$. Replace one of the $\sin 5x\cos 4x$ by $\sin 9x-\cos 5x\sin 4x$. Leave the other one alone. Something nice will happen.
You will end up having used the basic addition and subtraction laws $$\sin(a\pm b)=\sin a\cos b\pm \cos a\sin b.$$
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In fact, the "subtraction" laws are just a matter of substituting $-b$ into the "addition" laws, which is why I don't consider them worth mentioning. – pre-kidney Jan 23 at 19:14
@pre-kidney: True, but in this context it serves to extend the hint. – André Nicolas Jan 23 at 20:08
The intended techniques all follow from using the sine and cosine addition formulas and normalization, which you should have seen before.
However, I wanted to point out that a more unified approach to the general problem of proving trig. identities is to work in the complex plane. For instance $e^{ix}=\cos(x)+i\sin(x)$ allows you to derive identities such as $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ and $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$. Let $w=e^{ix}$. Then you can turn your identity into an equivalent "factorization problem" for a sparse polynomial in $w$, of degree $18$.
Try it out!
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1
Working in the complex plane is not an option because we haven't learned that yet. – gekkostate Jan 23 at 18:15
Then as I said, all your techniques will follow from addition formulas and normalization. Also, I checked your "identity", and it does not hold. Is there a typo somewhere? In case you meant "equation", not "identity", I checked for roots and saw that it didn't have any computable ones, either. – pre-kidney Jan 23 at 18:26
$$\sin C+\sin D=2\sin\frac{C+D}2\cos\frac{C-D}2$$
So, $$\sin9x+\sin x=2\sin\frac{9x+x}2\cos\frac{9x-x}2=2\sin5x\cos4x$$
$$2\sin A\cos B=\sin(A+B)+\sin(A-B)$$
So, $$2\sin5x\cos4x=\sin(5x+4x)+\sin(5x-4x)=\sin9x+\sin x$$
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Observe that the two formulae are related and one can be easily derived from the other. – lab bhattacharjee Jan 23 at 19:09
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http://mathhelpforum.com/pre-calculus/26438-2-optimization-problems.html
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# Thread:
1. ## 2 Optimization Problems
An open topped rectangular box with a square base is to be constructed with a volume of $32m^{3}$. Find the Dimension that require the lease amount of surface material.
and...
In microcomputers, most of the components are squeezed into a single box-shaped block. If the block has a length equal to twice the width and if the total surface area of the block must be $200cm^{2}$ in order to dissipate the heat produced, find the dimensions for the maximum volume of the block.
I couldn't figure out how to get equations out of these two problems.
Thanks.
2. Originally Posted by johntuan
...
In microcomputers, most of the components are squeezed into a single box-shaped block. If the block has a length equal to twice the width and if the total surface area of the block must be $200cm^{2}$ in order to dissipate the heat produced, find the dimensions for the maximum volume of the block.
...
Hello,
surface area of a rectangular block: $a = 2\cdot l\cdot w + 2\cdot l\cdot h + 2\cdot w\cdot h$
Since l = 2w and a = 200 you get:
$200 = 2\cdot 2w \cdot w + 2\cdot 2w \cdot h + 2\cdot w\cdot h~\implies~200 = 4w^2 + 6wh~\implies~h = \frac{200}{6w} - \frac{4w^2}{6w}$
The volume of a rectangular block is $V=l \cdot w \cdot h$
Plug in the values you already know:
$V(h)=2w^2 \cdot \left( \frac{200}{6w} - \frac{4w^2}{6w} \right)$
Expand the bracket and calculate the minimum using the first derivative ....and so on.
3. How would you expand the bracket?
4. Originally Posted by johntuan
How would you expand the bracket?
Hello,
multiply and cancel equal factors:
$<br /> V(h)=2w^2 \cdot \left( \frac{200}{6w} - \frac{4w^2}{6w} \right) = \frac{200}{3} w - \frac43 w^3$......... Therefore
$V'(h) = \frac{200}{3}-4w^2$........ To calculate the minimum V'(h) = 0:
$w = \sqrt{\frac{50}{3}}\approx 4.08...$ ........ Now plug in this value to calculate l and h.
$\left(l=\sqrt{\frac{200}{3}},~h=\frac{20}{9} \cdot \sqrt{6} \right)$
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http://mathoverflow.net/questions/123183/what-can-one-say-about-differentiable-topological-structure-of-cy3s
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## What can one say about (differentiable) topological structure of CY3s?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
It is known that there is unique differantial topological structure on the elliptic curves or K3 surfaces over $\mathbb{C}$. Since we know tons of Hodge diamonds for Calabi-Yau threefolds, we cannot really expect an easy classification of (differantial) topological structure.
What is known about (differantial) topological structures of Calabi-Yau 3-folds? Are they really different from those of Kahler 3-folds? Are there Calabi-Yau 3-folds that are (diffeomorphic or) homeomorphic but not (complex) deformation equivalent?
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## 1 Answer
There is at least one pair of examples of diffeomorphic but not deformation equivalent three-dimensional Calabi-Yau manifold (Ruan-Gross). The example is explained on pages 47-48 of this paper: http://arxiv.org/pdf/math/9806111.pdf
Otherwise it is of course natural to try to distinguish Calabi Yau three folds by their diffo type. Note that in dimension six two smooth compact manifolds that are homeomorphic are necessarily diffeomorphic, so the classifications up to homeo and diffeo are the same. Classification of simply connected 6-manifolds with torsion free homology according to diffeo is given by a theorem of Wall (the essential bit here is the cubic intersection form on $H^2(M^6,\mathbb Z)$). I am not aware of (current) classification work in this direction for Calabi-Yau 3-fold. But I think someone who would like to do this should use computer (the majority of examples of CY 3-folds are an outcome of a certain computer program). And it seems to me that it should be possible in principle to improve the existing algorithm so that it computes not only betti numbers, but also multiplication on $H^*$ and so the type as well.
Concerning the topology of CY 3-folds, on can say at least that the fundamental group of CY manifolds is finite (or, depending on definition of what you call CY manifold, virtually Abelian). At the same time general Kahler manifolds can have very sophisticated fundamental groups.
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Thank you for the nice answer, Dmitri. I was not aware of the Ruan-Gorss example and it is very interesting. I wonder what makes fundamental group of CY3s finite with your definition. Isn't there Oguis-Sakurai's paper about CY3 with infinite fundamental groups? – K Kim Feb 28 at 20:50
Dear Kim, by Bogomolov-Beuaville theorem every CY manifold has a finite cover that is a product of Tori, hyperkahler manifolds and manifolds $M^n$ such that $H^k(M^n,O)=0$ for $k\ne 0,n$. So for some people "proper" CY manifolds are only those that satisfy the last condition: $H^k(M^n,O)=0$ for $k\ne 0,n$. Such manifolds also have the property that the holonomy group of CY metrics on them coincide with $SU(n)$ (and not smaller than this). Such manifolds do have finite fundamental groups. Maybe for Oguis-Sakurai a Kahler manifold is $CY$ iff it has a holomorphic volume form... – Dmitri Feb 28 at 21:36
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http://mathhelpforum.com/geometry/99230-finding-radius-earth.html
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# Thread:
1. ## Finding the radius of the earth
OK---i have a problem.
you lie on a beach and watch a sunset. then you stand and watch it set again--you are higher and can see that supposedly.
my book says that you then use these values:
$h=1.70 m$ (your height)
$t=11.1 s$ (the time between sunsets)
then using the picture and the Pythagorean theorem, with $d$ as the distance between your eyes and the tangent to the earth from the second sunset, you get
$d^2 + r^2 = (r + h)^2 = r^2 + 2rh + h^2$
or
$d^2 = 2rh + h^2$
but you remove $h^2$ because your height is insignificant compared to the radius of the earth and the term $2rh$
then you use the number of hours in a day, the rotation of the earth in a day, and $t$ to find $\theta$:
$\frac \theta {360^o}$
which solves to $0.04625^o$
then you can get $d = r\tan \theta$ from the picture and, substituting for $d$ in the first equation, $r^2\tan^2\theta = 2rh$
which when you plug in the values (like $\theta = 0.04625$) makes $5.22 * 10^6 m$
The problem is in the fact that we replaced the value for the $\theta$ from the angle between the sunsets with the $\theta$ from the angle from the center of the earth. i cant figure out how that works.
Here is the picture:
2. see:
Attachment 12590
Angle $\text{uvA}=\theta$ so $\text{vuA}$ is $90-\theta$ and so $\text{uOB}=\theta$
CB
3. Originally Posted by CaptainBlack
see:
Attachment 12590
Angle $\text{uvA}=\theta$ so $\text{vuA}$ is $90-\theta$ and so $\text{uOB}=\theta$
CB
wow ok thanks. i dont know why the book didnt explain that.
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http://openwetware.org/index.php?title=User:Pranav_Rathi/Notebook/OT/2011/03/01/Laser_Shutter_.2&diff=657480&oldid=528907
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User:Pranav Rathi/Notebook/OT/2011/03/01/Laser Shutter .2
From OpenWetWare
(Difference between revisions)
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| - | The full detail is coming soon. | + | https://lh4.googleusercontent.com/-4t9t2lciTB4/UKfY1P8IfeI/AAAAAAAAB4g/BCTJwMN6JNA/s640/Laser%2520enclosure%2520AOM2.jpg |
| | | + | |
| | ==Motivation== | | ==Motivation== |
| - | Motivation behind designing the shutter is speed, accuracy and variability (time). In our optical tweezers we need to center the trap over the tethered bead, to get the geometry right(because this affects the force measurement)and make our feed-back program run. To center the tether we need to turn the laser intensity on/off quickly for variable time intervals. To do this we used to use AOM, because it’s extremely quick (nsec) and variable. So I needed some-thing which can replace this AOM function. | + | Motivation behind designing this shutter is speed, accuracy and variability (activation-time). In our optical tweezers we need to center the trap over the tethered bead, to get the geometry right (this affects the force measurement) and make our feedback program run. To center the tether we need to turn the laser intensity (trap) on/off quickly for various time intervals. To do this we used to use AOM (acoustooptic module), because it’s extremely quick (nano second on/off time). But AOM has some inherent oscillation problem which makes it use as a switch, unsuccessful. So I needed something else, (a shutter) which can replace this AOM function. |
| - | This shutter does it exact. It is fast with opening time of 4 and closing time 2 μsec. Since the shutter runs on/with the active voltage (controlled by the toggle foot switch), it remains active with the applied voltage, with the freedom to chose any active time. The speed of the shutter can also be controlled. | + | |
| - | In design the laser passes through an aperture, and the only moving part is the cylinder. No gears, and no electronics in the actual shutter, makes this design very stable and accurate, even under the heat produced by the laser beam. | + | This shutter does it exact and it is fast (activation-time) with an opening time of 4 and closing time of 2 μsec. The shutter runs on a +5 volts voltage. It moves to an on-position with an active voltage (controlled by the toggle foot switch) working against a restore-spring and remains on when the voltage is applied and turns to off when the voltage is removed. This gives a freedom of choice for the duration shutter is active with very simple electronics used to control the speed (activation time.) |
| - | This shutter needs no special power supply, it can be run through a cell phone charge with an output of roughly 300mA/5V. | + | |
| - | Cost and construction time is also important. With this design, a shutter can be prepared under \$40 with 10 hours of construction time (10 x 25 (hourly wage of a technician)=\$250+40=\$290). Still better than many commercially available shutter systems with same performance. | + | In the design the laser passes through an aperture on the only moving part a cylinder. No gears and no electronics in the design make it very simple, stable and accurate, even under the heat produced by a laser beam. The shutter needs no special power supply and can be run on a cell phone charge with an output of roughly 300mA/5V. |
| | | + | |
| | | + | Cost and construction time is also important. With this design, a shutter can be prepared under \$40 with 10 hours of construction time (10 x 25 (hourly wage of a technician) =\$250+40=\$290), still better than many commercially available shutter systems with same performance. |
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| | == Design & Construction== | | == Design & Construction== |
| Line 39: | | Line 42: | |
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| | === Construction=== | | === Construction=== |
| - | The buildup is divided is the following categories. In the start make sure that you have a good set of screw driver, players, wire cutter & stripper and solder. | + | The buildup is divided is the following categories. In the start make sure that you have a good set of screw driver, pliers, wire cutter, stripper and solder. |
| | ==== Shutter==== | | ==== Shutter==== |
| - | I will start with the choice of the motor. A 12V DC motor is a good choice because it can provide a wide rang of torques. The motor and the spring works against each other so it important to have the right set. So choose a 12V DC motor with a shaft length of at least 15mm. Mount the motor on the cage plate as shown in the picture, you will need 4/40 screws to tight it. once this is done, unscrew and take the motor out. | + | I will start with the choice of the motor. A 12V DC motor is a good choice because it can provide a wide range of torques. The motor and the spring works against each other so it important to have the right set. So choose a 12V DC motor with a shaft length of at least 15mm. Mount the motor on the cage plate or platform as shown in the picture, you will need 4/40 screws to tight it (in present setup). Once this is done, unscrew and take the motor out. |
| | | + | |
| | | + | Now next task is to choose the right spring. The spring should be half the length of the shaft and the shaft should easily fit through the spring. I want to keep the spring-choice procedure really simple and easy. To choose the spring first need to know the torque of the motor. This is really simple; it can be mathematically calculated if the voltage and the current are known. But I like the experimental way; the setup is shown in the slide2. In the slide I have a DC motor with a leverage pivoted at the shaft and a free weight hanging with a string on the other side. In this, all I am doing is balancing the weigh at that point by giving the motor just enough power. |
| | | + | So the torque is equal to the weight at the distance from the shaft: |
| | | | |
| - | Now next task is to choose the right spring. The spring should be half the length of the shaft and the shaft should easily fit through the spring. I want to keep all this really simple, so there is a very easy way to choose the right spring. To choose the spring first need to know the torque of the motor. This is really simple, it can be mathematically calculate if the voltage and the current is known. But I like the experimental way; the setup is shown in the slide2. In the slide I have a DC motor with a leverage pivoted at the shaft and a free weight hanging with a string on the other side. In this, all I am doing is balancing the weigh at that point by giving the motor just enough power. | | |
| - | so the torque is equal to the weight at the distance from the shaft: | | |
| | *:: <math> \mathbf{\tau}=r \times F = r \times m .g </math> | | *:: <math> \mathbf{\tau}=r \times F = r \times m .g </math> |
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| - | Once this is know we can choose the spring with less torque than this. The restoring torque applied by the spring can also be measure in the same fashion (as motor) as shown in slide1, with hanging known weight. The spring stiffness and the torque can be related through: | + | Once this is know we can choose the spring with less torque. The restoring torque applied by the spring can also be measure in the same fashion (as motor) as shown in slide1, with hanging known weight. The spring stiffness and the torque can be related through: |
| | | + | |
| | *:: <math> \mathbf{\tau}=2{\pi}n.K.r^2 </math> | | *:: <math> \mathbf{\tau}=2{\pi}n.K.r^2 </math> |
| | | | |
| - | Where K is the stiffness, r distance from the shaft center and n number of times spring twisted one complete 360 degrees circle before it was hacked. Since the torque is n dependent, it is a matter of great convenience, because now it is really easy to pick the right spring. All you have to do is twist the required number of times before hacked to get the right restoring torque. | + | Where K is the stiffness, r distance from the shaft center and n number of times spring twisted one complete 360 degrees circle before it was hacked. Since the torque is n dependent, it is a matter of great convenience, because now it is really easy to pick the right spring. All you have to do is twist the required number of times before hacked to get the right restoring torque. |
| | | | |
| - | Now we are ready to put the the shutter together. | + | There is even easier way to deal with mambo-jumbo, get a motor get a spring and put them together. |
| | | + | |
| | | + | Now we are ready to put the shutter together. Start with the motor; most of the motors have little holes for the screws to hold slide3. Choose a screw which can fit into that but still sticking out few millimeters slide4. Next is the rotation stage which joins the cylinder to the motor and holds the spring assembly. I choose the wood for this function, because it’s easy to machine and a good thermal-insulator slide5. I drilled two hole on both sides; to fit the shaft and 1/4 screw on either sides. To drill for 1/4 I used 15/64 size drill-bit slide6, 7, 8 & 9. Now the motor can be attached to one side of the stage and cylinder to other slide10 & 11. |
| | | + | |
| | | + | |
| | | + | Now we can machine the spring assembly. Slide 12 shows the parts of spring assembly. We need two metal strips of .75 and 1 cm long with a hole in the end for the screws. These strips with the screw on the motor will define the boundaries of the movement. Cylinder will rotate end to end between these strips. To screw the strips directly on the stage we need to drill two holes for the screws. The holes are 135 degree apart over the face of the stage as shown in the slide13. We also drill a small hole for the spring attachment near the edge. |
| | | + | |
| | | + | Now all parts are ready to get together slide 14. This shutter is little different of what is shown in the video/pictures but follows the same concept. Put the shorter end of the spring in the hole on the stage like slide 15. Now the motor shaft will go through the spring into the stage. The other side of the spring is hold against the screw. You can twist the required number of turns before it holds to get the right restoring torque. Now the spring wants to move the stage clockwise but the strip against the screw does not let it (this is position 1). As the voltage is applied the motor turns counterclockwise working against the spring until the second strip stops it. When the voltage is applied the stage remains in the second position. As soon as the voltage is turned down the spring restores the stage into the previous position. These two positions can be chosen for the shutter to be on/off slide 16, 17 & 18. |
| | | + | |
| | | + | A fully assembled shutter is ready. Now the next task is to prepare a control box. |
| | | | |
| | ==== Control Box==== | | ==== Control Box==== |
| | | + | I used a 2X3 inch aluminium box to house the electronics slide20. The electronics is really simple. The circuit is shown in the slide21. The foot switch is used to connect the motor ground to the power-supply hence activates the shutter. Since the shutter works on active voltage, the voltage is always on, when the foot paddle is active. So it is very important to give the right voltage to the motor to avoid damage to the motor over the long duration of active voltage. To do this i use a 100 ohm POT. POT let me choose the right voltage and current. This can be done once the shutter is ready. |
| | | + | |
| | | + | Now connect the shutter to the shutter input and foot-paddle to its input. |
| | | + | |
| | ==== Power Supply==== | | ==== Power Supply==== |
| | | + | The good thing about the system is that it can use any power supply which can give enough power to operate the motor. The POT inside the control box let choose the appropriate voltage and current and hence makes easy to choose a power-supply. |
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| | == Comments== | | == Comments== |
| - | | + | <div align="center"><html><div style="width:600px" id="__ss_8859065"><strong style="display:block;margin:12px 0 4px"><a href="http://www.slideshare.net/pranavrathi/motor-shutter-buildup" title="Motor shutter buildup">Motor shutter buildup</a></strong><object id="__sse8859065" width="600" height="300"><param name="movie" value="http://static.slidesharecdn.com/swf/ssplayer2.swf?doc=motorshutterbuildup-110815183358-phpapp02&stripped_title=motor-shutter-buildup&userName=pranavrathi" /><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed name="__sse8859065" src="http://static.slidesharecdn.com/swf/ssplayer2.swf?doc=motorshutterbuildup-110815183358-phpapp02&stripped_title=motor-shutter-buildup&userName=pranavrathi" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="padding:5px 0 12px">View more <a href="http://www.slideshare.net/">presentations</a> from <a href="http://www.slideshare.net/pranavrathi">pranavrathi</a>.</div></div></html></div> |
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| | | + | [[Image:Shutter2.JPG|left|thumb|400x280px|Full Shutter System]] |
| | | + | [[Image:Shutter1.JPG|right|thumb|400x280px|Shutter]] |
| | | + | [[Image:Shutter4.JPG|left|thumb|400x280px|Control box]] |
| | | + | [[Image:Shutter3.JPG|right|thumb|400x280px|Shutter different view]] |
| | | + | [[Image:Shutter5.JPG|center|thumb|400x280px|Control box different view]] |
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| - | [[Image:Shutter2.JPG|right|thumb|500px|Full Shutter System]] | + | |
| - | [[Image:Shutter1.JPG|right|thumb|500px|Shutter]] | + | |
| - | [[Image:Shutter4.JPG|right|thumb|500px|Control box]] | + | |
| - | [[Image:Shutter3.JPG|right|thumb|500px|Shutter different view]] | + | |
| - | [[Image:Shutter5.JPG|right|thumb|500px|Control box different view]] | + | |
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| | [[Category:Construction of Optical Tweezers and devices]] | | [[Category:Construction of Optical Tweezers and devices]] |
| - | [[Category:Optical Tweezer]] | | |
Motivation
Motivation behind designing this shutter is speed, accuracy and variability (activation-time). In our optical tweezers we need to center the trap over the tethered bead, to get the geometry right (this affects the force measurement) and make our feedback program run. To center the tether we need to turn the laser intensity (trap) on/off quickly for various time intervals. To do this we used to use AOM (acoustooptic module), because it’s extremely quick (nano second on/off time). But AOM has some inherent oscillation problem which makes it use as a switch, unsuccessful. So I needed something else, (a shutter) which can replace this AOM function.
This shutter does it exact and it is fast (activation-time) with an opening time of 4 and closing time of 2 μsec. The shutter runs on a +5 volts voltage. It moves to an on-position with an active voltage (controlled by the toggle foot switch) working against a restore-spring and remains on when the voltage is applied and turns to off when the voltage is removed. This gives a freedom of choice for the duration shutter is active with very simple electronics used to control the speed (activation time.)
In the design the laser passes through an aperture on the only moving part a cylinder. No gears and no electronics in the design make it very simple, stable and accurate, even under the heat produced by a laser beam. The shutter needs no special power supply and can be run on a cell phone charge with an output of roughly 300mA/5V.
Cost and construction time is also important. With this design, a shutter can be prepared under \$40 with 10 hours of construction time (10 x 25 (hourly wage of a technician) =\$250+40=\$290), still better than many commercially available shutter systems with same performance.
Design & Construction
Components
There are three major parts of the system.
• Shutter.
• Control box.
• Power supply.
The components used:
Shutter
• 12V DC motor
• Wood rotation stage
• Spring with torque of .1 N m
• Rubber padding
• Pillar Post Extension, Length=1" from Thorlabs (shutter cylinder)
• 30mm Cage Plate Optic Mount from Thorlabs
• Post-holder, base-plate ext...
Control Box
• 1 power jack M&F
• 1 1/4" mono Panel-Mount Audio Jack M&F
• 1 Foot Paddle
• 1 100Ω pot with, 1 220Ω resistor
• 1 on/off toggle switch
• 1 LED
• 1 box enclosure
• some connection wires, solder gun and solder wire
Power Supply
Any power-supply which can provide 300mA at 5V and above. The motor torque is power dependent and the shutter speed is resorting spring's stiffness (torque) dependent. So choose the spring carefully before decide on the power supply. I would recommend a variable power-supply which can be bought easily from any where.
Construction
The buildup is divided is the following categories. In the start make sure that you have a good set of screw driver, pliers, wire cutter, stripper and solder.
Shutter
I will start with the choice of the motor. A 12V DC motor is a good choice because it can provide a wide range of torques. The motor and the spring works against each other so it important to have the right set. So choose a 12V DC motor with a shaft length of at least 15mm. Mount the motor on the cage plate or platform as shown in the picture, you will need 4/40 screws to tight it (in present setup). Once this is done, unscrew and take the motor out.
Now next task is to choose the right spring. The spring should be half the length of the shaft and the shaft should easily fit through the spring. I want to keep the spring-choice procedure really simple and easy. To choose the spring first need to know the torque of the motor. This is really simple; it can be mathematically calculated if the voltage and the current are known. But I like the experimental way; the setup is shown in the slide2. In the slide I have a DC motor with a leverage pivoted at the shaft and a free weight hanging with a string on the other side. In this, all I am doing is balancing the weigh at that point by giving the motor just enough power. So the torque is equal to the weight at the distance from the shaft:
• $\mathbf{\tau}=r \times F = r \times m .g$
Once this is know we can choose the spring with less torque. The restoring torque applied by the spring can also be measure in the same fashion (as motor) as shown in slide1, with hanging known weight. The spring stiffness and the torque can be related through:
• $\mathbf{\tau}=2{\pi}n.K.r^2$
Where K is the stiffness, r distance from the shaft center and n number of times spring twisted one complete 360 degrees circle before it was hacked. Since the torque is n dependent, it is a matter of great convenience, because now it is really easy to pick the right spring. All you have to do is twist the required number of times before hacked to get the right restoring torque.
There is even easier way to deal with mambo-jumbo, get a motor get a spring and put them together.
Now we are ready to put the shutter together. Start with the motor; most of the motors have little holes for the screws to hold slide3. Choose a screw which can fit into that but still sticking out few millimeters slide4. Next is the rotation stage which joins the cylinder to the motor and holds the spring assembly. I choose the wood for this function, because it’s easy to machine and a good thermal-insulator slide5. I drilled two hole on both sides; to fit the shaft and 1/4 screw on either sides. To drill for 1/4 I used 15/64 size drill-bit slide6, 7, 8 & 9. Now the motor can be attached to one side of the stage and cylinder to other slide10 & 11.
Now we can machine the spring assembly. Slide 12 shows the parts of spring assembly. We need two metal strips of .75 and 1 cm long with a hole in the end for the screws. These strips with the screw on the motor will define the boundaries of the movement. Cylinder will rotate end to end between these strips. To screw the strips directly on the stage we need to drill two holes for the screws. The holes are 135 degree apart over the face of the stage as shown in the slide13. We also drill a small hole for the spring attachment near the edge.
Now all parts are ready to get together slide 14. This shutter is little different of what is shown in the video/pictures but follows the same concept. Put the shorter end of the spring in the hole on the stage like slide 15. Now the motor shaft will go through the spring into the stage. The other side of the spring is hold against the screw. You can twist the required number of turns before it holds to get the right restoring torque. Now the spring wants to move the stage clockwise but the strip against the screw does not let it (this is position 1). As the voltage is applied the motor turns counterclockwise working against the spring until the second strip stops it. When the voltage is applied the stage remains in the second position. As soon as the voltage is turned down the spring restores the stage into the previous position. These two positions can be chosen for the shutter to be on/off slide 16, 17 & 18.
A fully assembled shutter is ready. Now the next task is to prepare a control box.
Control Box
I used a 2X3 inch aluminium box to house the electronics slide20. The electronics is really simple. The circuit is shown in the slide21. The foot switch is used to connect the motor ground to the power-supply hence activates the shutter. Since the shutter works on active voltage, the voltage is always on, when the foot paddle is active. So it is very important to give the right voltage to the motor to avoid damage to the motor over the long duration of active voltage. To do this i use a 100 ohm POT. POT let me choose the right voltage and current. This can be done once the shutter is ready.
Now connect the shutter to the shutter input and foot-paddle to its input.
Power Supply
The good thing about the system is that it can use any power supply which can give enough power to operate the motor. The POT inside the control box let choose the appropriate voltage and current and hence makes easy to choose a power-supply.
Comments
Motor shutter buildup
View more presentations from pranavrathi.
Full Shutter System
Shutter
Control box
Shutter different view
Control box different view
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http://math.stackexchange.com/questions/41257/find-remaining-vertices-of-a-square-given-2
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# Find remaining vertices of a square, given 2
I need a hint for this problem.
Let the vertices of a square ABCD represent on the Argand diagram the complex numbers a,b,c, and d respectively. A,B,C,D are taken anti-clockwise in the order named.
If $$a = 3 + i, b = 4 - 2i$$, find c and d.
For a different problem, where square was at the origin, I used the idea that $$i(z_1)$$ is an anti-clockwise rotation and since it's a square etc. But here it's not at the orgin. Any help is much appreciated. Thanks.
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## 4 Answers
You can make a translation so that one of the vertices goes to the origin, then make the inverse translation.
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Got it, I did a translation of (3+i) and (4-2i) to get 2 points and then moved them back after rotation to get, C = (7-i) and D = 6 + 2i, which checks out. Thanks! – mathguy80 May 25 '11 at 13:43
Hint: $b$ to $c$ is a rotation through $\pi/2$ of $a$ to $b$.
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Hint: You can find the side of the square from $|b-a|$. If you think of walking around the square anti-clockwise, at each corner you make a left turn of $\frac{\pi}{2}$ and go the same distance.
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Haven't studied polar form of complex numbers yet, this and @Ben Bosisel idea seems more elegant one I have studied that! Until then that translation solution worked fine. Thanks. – mathguy80 May 25 '11 at 13:45
You don't need polar form, all you need to know is that when you multiply a complex number by $i$ you rotate it by 90 degrees, counterclockwise. – Gerry Myerson May 26 '11 at 0:29
A rotation in the complex plane around point $b$ with angle $\theta$ has the expression $z \mapsto b+(z-b)\cdot e^{i\theta}$, where $e^{i\theta}=\cos \theta+i\sin \theta$.
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Thanks, need to study the trignometry form of complex numbers first! – mathguy80 May 25 '11 at 13:50
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http://physics.stackexchange.com/questions/tagged/entropy?page=1&sort=active&pagesize=15
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# Tagged Questions
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### black hole no-hair theorems vs. entropy and surface area
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### Why is the equation for Entropy of an ideal gas that undergoes reversible change in T at constant Pressure like this?
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### How to solve state parameters using these givens for an ideal gas?
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### What is the physical or mathematical meaning of the Gibbs-Duhem equation?
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4answers
945 views
### What is information?
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1answer
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### Do gasses always mix because of their Gibbs free energy?
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### Entropy change relation to the number of lost bits
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### Entropy calculation, erasing bits? [closed]
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### How can dQ/T be interpreted as a system's level of disorder?
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6answers
223 views
### Entropy and Crystal Growth
I was reading about growing single crystals and I'm a little confused about this - In most crystal growing processes, a "seed crystal" is used, and the rest of the material crystallizes on the seed ...
1answer
48 views
### What are the units of the Bekenstein bound?
Working with the Wikipedia definition of the Bekenstein bound: $S \leq \frac{2 \pi R k_bE}{\hbar c}$ $2\pi R \$ is $m^2$ $k_b$ is $\frac{J}{K}$ $E$ is $J$ $\hbar$ is $J*s$ $c$ is $\frac{m}{s}$ ...
2answers
142 views
### How to compute configurations (entropy) of a system?
If we have a system $X$ consisting of subsystems $X_1$ and $X_2$. We also know that $X_1$ and $X_2$ have eigenstates $H_1 = 1 \times 10^{20}$ and $H_2 = 1 \times 10 ^{22}$. Can we calculate the ...
1answer
116 views
### relation between first law of thermodynamics and statistical mechanics definition of entropy
From the definition of entropy as $S= - Tr (\rho\, ln \rho)$ one obtains that $S = \frac{\langle E \rangle}{T} + \log Z.$ The first law of thermodynamics has $dS = {dE \over T}$. Why is there no ...
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http://math.stackexchange.com/questions/176592/how-are-the-full-semantics-of-sol-and-hol-specified/176735
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# How are the full semantics of SOL and HOL specified?
In relation to this question about the "fundamental" character of possible logical systems, I realized that I just had an intuitive (and so, inadequate) understanding of the way logics higher than FOL can unambiguously specify the kind of semantics which make up the intended interpretation of their formalisms, inside the formalisms themselves. This is quite a meaningful question, since the ability of an automated system of reasoning at the object-language level can only recognize what is coded in the formalism itself, at that level; and so, those approaches which start from the construction of a model at the meta-level, are a priori ruled out in the sense I'm describing here.
What I'm thinking about are computer systems reasoning with the logic, such as the HOL-based proof assistants, as Isabelle.
So, how are the intended semantics of SOL and HOL specified in a computer system?
P.S. : I have realized that this topic isn't actually new in this site, and has been brought up in other questions like this one.
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– Makarius Mar 8 at 14:06
## 3 Answers
From the point of view of derivability and syntax, there is no distinction between full higher order semantics and first-order (Henkin) semantics. This is, in one sense, the reason that there is no completeness theorem for full semantics - because the completeness theorem matches derivability with Henkin semantics, and so any genuinely different semantics will not match up with derivability. Syntactic things like proof assistants, which only care about derivability, are somewhat indifferent to semantic issues.
I believe that the main benefit of using higher order logic in proof assistants is that it makes it easier to formalize theorems that have been proven in ordinary mathematics. Even if these theorems could be formalized in, say, Peano arithmetic, by creating entirely new proofs, it is often easier to modify the existing proof to work in higher order logic.
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I think you caught what I'm asking. In essence, the whole concept of "full semantics" is what troubles me. How can it be taken as a "well-defined" term, if higher order logics which fulfill categoricity need a FO meta-theory of set-theoretical flavour to prove the categoricity of what "they say" with their "more expressive power"? Where does the "expressivity" come from, if you've used a non-categorical set theory to prove it? If I get it right, it seems to be nearly the same problem as when considering the "standard universe" of set theory. It's just intuition? What's "formal" about that? – Mono Jul 30 '12 at 14:57
Right, higher-order logic with full semantics is not "formal" in that sense, and the issue is very related to the issue of the "standard model" of set theory. – Carl Mummert Jul 30 '12 at 15:13
Thank you very much. This question was related to another which I cite in the first paragraph; the sense of the word "fundamental" I used there, is more or less the same we've discussed here under the adjective "formal", except for the added meaning of being able to "simulate" other logics from a mathematical point of view (e.g. fuzzy logics can be formalized in a FO mathematical theory which supports the real numbers, and define the "fuzzy membership" as an ordered pair). – Mono Jul 30 '12 at 15:51
– Mono Jul 30 '12 at 15:52
As far as I'm aware, computer proof systems do not care about semantics at all. Their task is to construct and check formal proofs that follow the syntacical rules for what constitutes a valid derivation, and it is then up to the human user to convince himself that the formal system corresponds to his intuition about semantics.
(Well, some proof assistants do care about semantics some of the time, such as if they contain calculation-based decision procedures for specific theories such as ordered fields or Presburger arithmetic. But then -- depending on the level of paranoia used in the development -- the role of these semantic subparts parts is usually just to suggest a syntactic proof that can be verified independently).
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What confuses me is that, even if your argument is absoltely sound (computers only follow syntactical rules and don't care about semantics), then it seems to follow that every logic of a higher order than FOL is necessarily computed like if it was a many-sorted FOL (i.e. doesn't recognizing, for example, that second-order variables range over subsets of the same set which first-order variables range over). I suspect I'm missing something here, though. – Mono Jul 29 '12 at 20:41
@Mono: It is true that the syntactic character of proof checking for HOL is the same as for a many-sorted FOL. What characterizes HOL is that there are usually specific logical axioms and/or rules of inference dealing with higher-order variables and ensuring, for example, that everything that is definable an an explicit formula is also considered when we quantify a higher-order variable of an appropriate signature. ...(contd) – Henning Makholm Jul 29 '12 at 20:49
... We're free to consider these special rules to be part of the theory proper instead of part of the logic, and the result would be a first-order theory that could prove the same thing -- essentially the original higher-order theory would be embedded in a (possibly weak) version of first-order set theory. The position that all higher-order logic is just an abbreviation for set theory was championed by Quine, and largely carried the day until the advent of practical computer proof systems made clear how cumbersome it is to reduce everything explicitly to first order all the time. – Henning Makholm Jul 29 '12 at 20:53
Note also that the fact the the logical axioms for higher-order variables are always the same (in contrast to if they were just incidental parts of the subject theory) makes it more practical for developers of proof assistants to provide special-cased general strategies for using them which allow users to specify common reasoning patterns easily. – Henning Makholm Jul 29 '12 at 20:55
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@Mono: It's unclear to me why you're suddenly speaking about categoricity and models -- those are semantic properties which proof checkers/assistants don't care about. All they care about is the existence of a valid proof in some formal system; whether the formal system corresponds to any semantic notion of model is not their problem. – Henning Makholm Jul 30 '12 at 11:17
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I want to clear something up: HOL is not merely multi-sorted FOL
The key difference is in the two systems expressivity. FOL cannot express transitive closure. Here is a nice note explaining why. On the other hand, HOL can express transitive closure. Here's the source code from Isabelle/HOL's implementation if you are interested.
EDIT 1: Note the caveat in the comments: FOL extended with ZF or machinery from arithmetic can express transitive closure. That no extended calculus can do this is not the claim I'm making here, however.
EDIT 2: Likewise, it's inappropriate blindly make use of intuition from Henkin's semantics for his higher order logic, as its application to computer-based HOLs is not straight forward. For one thing, proof assistants are based off of Church's HOL, which predates Henkin's work and has its own peculiarities. Semantics for Church's HOL may given using applicative structures, as per Harvey Friedman (1975) and subsequent papers.
[H]ow are the intended semantics of SOL and HOL specified in a computer system?
You can't really specify the semantics, as others have noted, but there are different ways that $x \in A$ gets parsed into base syntax.
In Isabelle, you can load either Isabelle/ZF or Isabelle/HOL. Depending on which system you load, $x \in A$ gets interpreted differently.
In Isabelle/ZF, it's the meaning you learned in set theory class: $\in$ is a binary relation in FOL and it obeys the various axioms in set theory.
In Isabelle/HOL, `S :: 'x set` (ie, "`S` is a set of objects of type `'x`") is really just a wrapper for an object of type `f :: 'x -> bool` (ie, "`f` is an indicator function that take `'x` to True/False"). Set comprehension and membership are effectively defined by the equivalence $a \in \{x \ |\ P(x) \} \iff P(a)$. You can read about it in Isabelle/HOL's source if you're into that sort of thing.
In both Isabelle/ZF and Isabelle/HOL, the familiar syntax $\{x \in S\ |\ \phi(x)\}$ is also interpreted as syntactic sugar. In both cases, it's the parser's responsibility for compiling the extended syntax into the base syntax; and how it is done differs based on foundation.
Finally, while computer proof assistants are not generally used to reason about their own semantics, there is an exception. John Harrison developed two relative consistency proofs of HOL-Light within HOL-Light here.
He first demonstrates how to construct a full model of HOL-Light without the axiom of infinity in HOL-Light. He then shows how to construct a full model of all of HOL-Light in HOL-Light extended with a strongly inaccessible cardinal.
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The claim "FOL cannot express transitive closure" is somewhat vague. ZFC can certainly define the transitive closure of a relation, and is a first order theory. Anything that can be expressed in HOL can be expressed, by exactly the same sentence, in FOL. The only difference is in semantics - but HOL has several semantics, the weakest of which has the same strength as FOL. Everything Isabelle does is compatible with these semantics; Isabelle has no way to tell what semantics I use to interpret its output. If I use Henkin semantics, Isabelle will have the same semantic limitations as FOL. – Carl Mummert Jul 30 '12 at 12:12
@Carl Mummert: I suppose I should make a distinction. While in ZF you can express the transitive closure of a relation between sets, it can't express the closure of logical relations such at $\in$; FOL can't do that. On the other hand, Isabelle/HOL doesn't have this problem. Likewise, Isabelle/HOL does not just encode Henkin's higher order calculus; it also includes extensions such as the axiom of choice. Isabelle/HOL can be embedded to ZFC, but that's not the same being embeddable in FOL simpliciter (like Henkin's higher order calculus can). – Matt W-D Jul 30 '12 at 12:30
Henkin's method works with arbitrary choice and comprehension axioms added in; indeed these are typically used in second-order logic, even with Henkin semantics. Furthermore, it is certainly possible to express, within ZFC, the transitive closure of an arbitrary set or class $A$ under the $\in$ relation: a set $b$ is in the transitive closure of $A$ if and only if there is a finite chain $b \in a_n+1 \in a_n \in \cdots \in a_1$ where $a_1 \in A$. That can be written as a single sentence of ZFC. – Carl Mummert Jul 30 '12 at 12:36
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(That's the transitive closure in the set theoretic sense. For the transitive closure in the order theoretic sense, we would put $b \in^* a$ if there is a finite chain $b \in a_{n+1} \in \cdots \in a_1 = a$ where $a_{n+1}, a_n, \ldots, a_1$ are all in $A$. This is again a single sentence of ZFC. In general, if we can define the transitive closure of any relation $R$ then we can define the transitive closure of $\in$ by just replacing $R$ with $\in$ in the definition.) – Carl Mummert Jul 30 '12 at 12:46
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I think the issue is that FOL is not a single logic. There are many "first order" logics with slightly different syntax. They are all first order in the sense that they have effective deductive systems, semantics using ordinary first-order structures, and completeness theorems. The FOL that people usually see first doesn't have types, $\lambda$ terms, or set quantifiers, but these are easy to add while preserving completeness. Usually $\lambda$ terms are only seen in proof theory or computational settings, for example in higher-order arithmetic, as in Kohlenbach's Applied Proof Theory. – Carl Mummert Jul 30 '12 at 14:06
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http://mathhelpforum.com/geometry/93010-proving-two-bisectors-equal.html
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# Thread:
1. ## Proving two bisectors are equal
In a triangle ABC, angle B is 60 degrees. Angles A and C are bisected to points D and E respectively, on the edge of the triangle (see diagram). They cross at point O. Prove that OD = OE
I've gone through it a little and got this.
2x + 2y + 60 = 180
2x + 2y = 120
x + y = 60
I don't know if this helps at all.
Remember, angles A and C are bisected, so the two adjacent angles are equal to each other (x, x and y, y)
Thanks for any help, BG
2. ## Geometry Proof
Hello BG5965
Originally Posted by BG5965
In a triangle ABC, angle B is 60 degrees. Angles A and C are bisected to points D and E respectively, on the edge of the triangle (see diagram). They cross at point O. Prove that OD = OE
I've gone through it a little and got this.
2x + 2y + 60 = 180
2x + 2y = 120
x + y = 60
I don't know if this helps at all.
Remember, angles A and C are bisected, so the two adjacent angles are equal to each other (x, x and y, y)
Thanks for any help, BG
Join $OB$, and note that it is the bisector of $\angle B$, since the angle bisectors of a triangle are concurrent.
$\Rightarrow \angle OBD = \angle OBE = 30^o$
Also $\angle EOA = x + y$ (exterior angle of $\triangle AOC$)
$\Rightarrow EOA = 60^o = \angle EBD$
$\Rightarrow EODB$ is a cyclic quadrilateral (exterior angle = interior opposite angle)
Now join $ED$, and use angles in the same segment:
$\angle OBD = \angle OED = 30^o$ and $\angle OBE = \angle ODE = 30^o$
$\Rightarrow \angle ODE = \angle OED$
$\Rightarrow EO = OD$ (isosceles $\triangle OED$)
Grandad
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http://mathoverflow.net/revisions/57293/list
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## Return to Answer
2 added 155 characters in body
It is not an answer to your question, but I hope it will help:
In general arithmetic complexity of convolution in non-anelian groups "equivalent" to the complexity of matrix multiplication. Here is the reason why:
The way of doing Fourier Transform in abelian group $A$ can be described in the is the following way: Let $f,g \in F[A]$ We know that $F[A]$ is isomorphic to the space $F^A$ with pointwise multiplication. Let $T$(which is acctually Fourier Transform) be this isomorphism. If we want calculate $f*g$ then calculate $T^{-1}(T(f)\cdot T(g))$. In case of non abelian group like $S_n$ It holds that $F[G]$ is isomorphic to the direct sum of matrix algebras that is $F[G]\simeq\oplus M_{n_i}$. Thus using the same formula you can calculate convolution in $S_n$, but now you will need to multiply matrixes.
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It is not an answer to your question, but I hope it will help:
The way of doing Fourier Transform in abelian group $A$ can be described in the is the following way: Let $f,g \in F[A]$ We know that $F[A]$ is isomorphic to the space $F^A$ with pointwise multiplication. Let $T$(which is acctually Fourier Transform) be this isomorphism. If we want calculate $f*g$ then calculate $T^{-1}(T(f)\cdot T(g))$. In case of non abelian group like $S_n$ It holds that $F[G]$ is isomorphic to the direct sum of matrix algebras that is $F[G]\simeq\oplus M_{n_i}$. Thus using the same formula you can calculate convolution in $S_n$, but now you will need to multiply matrixes.
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http://mathoverflow.net/revisions/43155/list
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## Return to Answer
2 added 25 characters in body
I think I understand what Kevin is saying in the Theorem. Let $\mathfrak o$ be the ring of integers of $K$ and $\mathfrak p$ the maximal ideal. Define the following subgroups of $G=GL(2, K)$:
$I= \begin{pmatrix}\mathfrak o^{\times} & \mathfrak o // \mathfrak p & \mathfrak o \end{pmatrix}$
$I_n=\begin{pmatrix} \mathfrak o^{\times} & \mathfrak o // \mathfrak p^n & 1+\mathfrak p^n\end{pmatrix}$
$Z_n= \begin{pmatrix}1+\mathfrak p^n & 0 // 0 & 1+\mathfrak p^n\end{pmatrix}$
Lemma. Let $\pi$ be a smooth representation of $I$, I$with a central character, such that$Z_n$acts trivially,$Z_{n-1}$does not act trivially and the space of$I_n$-invariants is non-zero. Then the restriction of$\pi$to$I_n Z_0$contains a one dimensional subrepresentation of the form$\chi: \begin{pmatrix} a & b // c & d\end{pmatrix} \mapsto \chi_1(d)$, where$\chi_1: \mathfrak o^{\times}\rightarrow \mathbb C^{\times}$is a smooth character of conductor$\mathfrak p^n\$.
Proof. Look at the action of the abelian group $Z_0$ on $\pi^{I_n}$.
The pair $(I_nZ_0, \chi)$ is a type for the Bernstein component, which contains the principal series representations that Kevin describes. In other words, if $\pi$ is an irreducible smooth representation of $G$, then $Hom_{I_n Z_0}(\chi, \pi)\neq 0$ if and only if $\pi$ is a principal series rep with one character unramified, the other of conductor $\mathfrak p^n$. For this you could look either in the appendix by Heniart to:
http://www.math.u-psud.fr/~breuil/PUBLICATIONS/multiplicite.pdf
or in the book of Bushnell and Henniart. The main point being that the representation $\chi$ (as a representation of $\begin{pmatrix} \mathfrak o^{\times} & 0 // 0 & \mathfrak o^{\times}\end{pmatrix}$) shows up in the $U$-coinvariants of $\pi$, where $U$ is unipotent upper (lower?) triangular matrices.
1
I think I understand what Kevin is saying in the Theorem. Let $\mathfrak o$ be the ring of integers of $K$ and $\mathfrak p$ the maximal ideal. Define the following subgroups of $G=GL(2, K)$:
$I= \begin{pmatrix}\mathfrak o^{\times} & \mathfrak o // \mathfrak p & \mathfrak o \end{pmatrix}$
$I_n=\begin{pmatrix} \mathfrak o^{\times} & \mathfrak o // \mathfrak p^n & 1+\mathfrak p^n\end{pmatrix}$
$Z_n= \begin{pmatrix}1+\mathfrak p^n & 0 // 0 & 1+\mathfrak p^n\end{pmatrix}$
Lemma. Let $\pi$ be a smooth representation of $I$, such that $Z_n$ acts trivially, $Z_{n-1}$ does not act trivially and the space of $I_n$-invariants is non-zero. Then the restriction of $\pi$ to $I_n Z_0$ contains a one dimensional subrepresentation of the form $\chi: \begin{pmatrix} a & b // c & d\end{pmatrix} \mapsto \chi_1(d)$, where $\chi_1: \mathfrak o^{\times}\rightarrow \mathbb C^{\times}$ is a smooth character of conductor $\mathfrak p^n$.
Proof. Look at the action of the abelian group $Z_0$ on $\pi^{I_n}$.
The pair $(I_nZ_0, \chi)$ is a type for the Bernstein component, which contains the principal series representations that Kevin describes. In other words, if $\pi$ is an irreducible smooth representation of $G$, then $Hom_{I_n Z_0}(\chi, \pi)\neq 0$ if and only if $\pi$ is a principal series rep with one character unramified, the other of conductor $\mathfrak p^n$. For this you could look either in the appendix by Heniart to:
http://www.math.u-psud.fr/~breuil/PUBLICATIONS/multiplicite.pdf
or in the book of Bushnell and Henniart. The main point being that the representation $\chi$ (as a representation of $\begin{pmatrix} \mathfrak o^{\times} & 0 // 0 & \mathfrak o^{\times}\end{pmatrix}$) shows up in the $U$-coinvariants of $\pi$, where $U$ is unipotent upper (lower?) triangular matrices.
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http://mathoverflow.net/questions/123713/era-pra-pa-transfinite-induction-and-equivalences
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ERA, PRA, PA, transfinite induction and equivalences
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I'm quite sure I don't understand very well the links between proof theoretical ordinals of theories, the axioms of transfinite induction and the objects a theory can prove to exist.
For instance I'm considering Peano Axioms ($\mathbf{PA}$), of proof theoretic ordinal $\epsilon_0$, Primitive Recursive Arithmetic ($\mathbf{PRA}$) of proof theoretic ordinal $\omega^\omega$ and Elementary Recursive Arithmetic ($\mathbf{ERA}$), which is a fragment of $\mathbf{PRA}$.
I was wondering if $\mathbf{PRA}+TI\{\alpha\in\epsilon_0\}$ (where $TI$ stands for transfinite induction) was equivalent in some sense to $\mathbf{PA}+TI\{\alpha\in\epsilon_0\}$ or/and to $\mathbf{ERA}+TI\{\alpha\in\epsilon_0\}$ ?
And more generally, if it was true that for any set $A$ of (countable) ordinals such that $\epsilon_0 \subset A$, $\mathbf{ERA}+TI\{\alpha\in A\} = \mathbf{PRA}+TI\{\alpha\in A\} = \mathbf{PA}+TI\{\alpha\in A\}$ ?
Any enlightenment would be most welcome :) Thanks in advance.
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Sorry for the bad TI notation, it was 1st order transfinite induction up to $epsilon_0$. Thanks for your answer, all is now much clearer to mo. – Primitive Recursive Fab Mar 6 at 13:58
1 Answer
This entirely depends on what exactly you mean by TI, as there are several options (I actually do not understand what the $\{\alpha\in\epsilon_0\}$ part of the notation is supposed to mean either, but I will assume it just means transfinite induction up to $\epsilon_0$):
1. $TI\{\alpha\in\epsilon_0\}$ is the schema $$\forall x\,(\forall y\prec x\,\phi(y)\to\phi(x))\to\forall x\,\phi(x),$$ where $\phi$ is an arbitrary formula, and $\prec$ the standard ordering of type $\epsilon_0$. It is easy to see that transfinite induction implies ordinary induction over a weak base theory (say, $I\Delta_0$), hence in this case, $I\Delta_0+TI\{\alpha\in\epsilon_0\}=\mathrm{PA}+TI\{\alpha\in\epsilon_0\}$ (and the same holds for any base theory in between).
2. $TI\{\alpha\in\epsilon_0\}$ is the same schema restricted to formulas of bounded complexity $\Gamma$. Typically used choices for $\Gamma$ include $\Pi^0_2$, $\Pi^0_1$, or open formulas in the language of PRA or EA (also called ERA or EFA). In all these cases, $\mathrm{PRA}+TI\{\alpha\in\epsilon_0\}$ is strictly weaker than $\mathrm{PA}+TI\{\alpha\in\epsilon_0\}$, since the former theory can be axiomatized by formulas of bounded complexity, and no consistent set of formulas of bounded complexity can imply full ordinary induction (which is equivalent to the full uniform reflection schema). In the case where $\Gamma$ are open EA-formulas, $\mathrm{EA}+TI\{\alpha\in\epsilon_0\}$ is likewise strictly weaker than $\mathrm{PRA}+TI\{\alpha\in\epsilon_0\}$. On the other hand, if $\Gamma\supseteq\Pi^0_1$, then $TI\{\alpha\in\epsilon_0\}$ implies $I\Sigma_1\supseteq\mathrm{PRA}$ over a weak base theory.
3. $TI\{\alpha\in\epsilon_0\}$ is the second-order induction axiom $$\forall X\,\forall x\,(\forall y\prec x\,y\in X\to x\in X)\to\forall x\,x\in X.$$ Then one needs to include some comprehension schema in the base theory to make any sense, and its strength determines the strength of the $TI\{\alpha\in\epsilon_0\}$. In particular, if we take at least $\Sigma^0_1$-comprehension, we are in the same situation as in 1. If we take recursive comprehension, it is the same as 2 with $\Gamma=\Delta^0_1$.
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Emil: can you recommend some references for the results in #2 in connection with fragments of arithmetic? – Ali Enayat Mar 6 at 19:40
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@Ali: It seems that I have misremembered the complexities of the restricted transfinite induction schemata used, they seem to be rather lower. As one can find in e.g. Girard’s “Proof theory and logical complexity”, already PRA with the quantifier-free transfinite induction rule up to $\epsilon_0$ proves Con(PA). TI up to $\epsilon_0$ for formulas of restricted complexity proves uniform reflection schemata for PA of restricted complexity. Girard explicitly mentions only the full reflection schema (which needs full TI), but a similar argument gives the restricted versions. ... – Emil Jeřábek Mar 11 at 19:02
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... If I get it right this time, $\Sigma^0_1$-reflection (i.e., 1-consistency) is provable over PRA using quantifier-free TI axiom, and more generally $\Sigma^0_{n+1}$-reflection follows from TI up to $\epsilon_0$ for $\Pi^0_n$-formulas. The claim I made above that EA with quantifier-free TI is strictly weaker than the same plus PRA is probably bogus: PRA (or rather the equivalent theory in the language of PA) is a $\Pi^0_2$-axiomatized fragment of PA, hence it is implied by uniform $\Sigma^0_1$-reflection for PA. I’m not sure whether the cut-elimination argument involved in the proof ... – Emil Jeřábek Mar 11 at 19:11
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... of $\Sigma^0_1$-reflection by open TI works (or can be worked around) in EA, but it should work over $\mathrm{EA}^+=I\Delta_0+\mathrm{SUPEXP}$. As for other things I mentioned, the fact that PA proves the full uniform reflection for its finite fragments (which implies it is not included in any theory of bounded complexity) can be found in various places, e.g. in Girard again, but a comprehensive source for related results is igitur-archive.library.uu.nl/lg/2008-0326-201008/… . – Emil Jeřábek Mar 11 at 19:17
Emil: thanks for your detailed explanations (I only now saw them). – Ali Enayat Apr 9 at 3:51
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http://physics.stackexchange.com/questions/13931/what-does-the-rayleigh-phase-function-tell-us?answertab=oldest
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# What does the Rayleigh Phase Function tell us
I am working on some radiative transfer equations, and struggling as I'm fairly new to this field. I have read about the Rayleigh Phase Function which is:
$P(\theta) = \frac{3}{4}(1 + cos^2 \theta)$
I can plot this function, and generate values from this function in my computer program (which is what I need to do) - but what is it actually telling me? If I know I have a ray of incident light at a certain angle, will this function tell me what angle it will scatter at? Or is it more of a statistical generalisation where the areas where the function is higher are the angles at which the scattering is more likely to happen?
Furthermore, when I see the function plotted it is always from 0-180, rather than 0-360 as I would expect for the scattering. This makes me think that the function may be giving the difference between the incoming angle and the outgoing angle rather than the absolute angle it is scattered at. Is that correct?
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The function is symmetrical, which would explain why it is typically only plotted from 0-180 degrees, it'd just be mirrored over 0 degrees. – PearsonArtPhoto Aug 25 '11 at 1:21
Thanks - that makes sense. – robintw Aug 25 '11 at 8:23
## 1 Answer
Not sure why the word "phase" is in the name of this function. The intensity of unpolarized light scattered by Rayleigh particles as a function of detector angle $\theta$ is proportional to this function. Colloidal scientists call this measurement "Static Light Scattering," but atmospheric scientists and others may have different names for it.
You can get some physical intuition for this function by picturing an EM wave interacting with a point scatterer. Imagine sweeping a detector in a circle of a radius $R$ centered on the scatterer. Then place an analyzer (polarizer) in front of the detector so you are detecting only either vertically or horizontally polarized light. If the incident light is vertically polarized, i.e. polarization normal to the plane of the circle, the amplitude (and intensity) of scattered light is isotropic. If the light is horizontally polarized, i.e. (polarization in the plane of the circle), the intensity detected depends on detection angle.
Convention is to call $\theta=0$ degrees the angle of forward scattered (also unscattered) light. That way, $\theta=180$ is the angle where we detect backscattered light, and $\theta=90$ and $\theta=270$ are each right in between those two "poles". Imagine horizontally polarized radiation interacting with our point scatterer. This polarization leads to maximum scattered intensity at $\theta=0$ and $\theta=180$, and minimum of zero scattered intensity at $\theta=90$ (and $\theta=270$). You can use the right-hand rule and dot products to convince yourself of this.
Let your fingers point in the direction of the polarization, and your thumb point in $\hat{k}$, the direction in which the scattered photon is propagating. You can let your thumb point in any direction around the circle; imagine that your thumb points toward the detector, so it determines which $\theta$ you are analyzing. If your fingers are in the plane of the circle you are looking at horizontally polarized light. Try to put your fingers in the plane and point your thumb at $\theta=90$. You should find that your fingers are pointing in a direction that is orthogonal to the horizontal radiation of the incident light, which means none of such light will scatter in this direction. The $cos^2(\theta)$ term characterizes this angular dependence of horizontal polarization, and the $1$ term states the isotropy of the vertically polarized radiation. We have to add these for unpolarized incident light.
This general dependence is true for dilute solutions of Rayleigh scatterers, or gases.
The intensity measurement comes from counting a large number of photons at each angle. The function comes from classical optics/electromagnetism. I suspect it could also be derived from a quantum mechanical (QM) treatment of photons. QM may be better suited for determining the "angles at which the scattering is more likely to happen" as you put it, since we think of the classical picture of as purely deterministic.
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http://www.openwetware.org/index.php?title=Physics307L:People/Long/Formal_Report&diff=prev&oldid=374637
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# Physics307L:People/Long/Formal Report
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## Revision as of 02:24, 13 December 2009
Author: Ryan Long
Experimentalists: Ryan Long & Tom Mahony
The University of New Mexico
Department of Physics & Astronomy
email: rlong1@unm.edu
## Abstract
The speed of light is a very large value, nevertheless, the speed of light can be measured using relatively simple time of flight methods. Experiments of this sort have been carried out since Isaac Beeckman and Galileo Galilei first tried in the early 1600s. [1] In Junior Lab at University of New Mexico, we measure the speed of light by measuring flight time of LED pulses over the course of a short distance. A major obstacle to overcome in this experiment is the occurence of "time walk", this can cause major systematic error, if not addressed properly. We obtain a value of 29.448 +/- .1424 cm/ns, which is inconsistent with the accepted value of 29.979 cm/ns, indicating some source of systematic error. We discuss possibilities for removing this systematic error in future work.
## Introduction
Perhaps one of the most well known and frequently used constants of physics both classical and modern, is the speed of light, denoted by a lower case “c”. The speed of light constant is used for many purposes from calculating the distance to astronomical events, to understanding quantum mechanics. Today the value for the speed of light is defined as 29.9792 cm/ns, this exact value comes from the National Institute of Standards and Technology. Measuring the speed of light can be achieved numerous ways with modern technology, some include radio interferometry, or methane stabilized lasers [2]. We measured the speed of light using a relatively simple method which involves measuring time delay of an LED pulse using a photomultiplier tube and a Time amplitude converter or simply (TAC). The photomultiplier is a device sensitive enough to measure individual photons. When the cathode of the PMT receives incident photons, photoelectrons are ejected from an anode inside the PMT, [3] the resultant charge pulse intervals from the photoelectrons are then converted into amplitudes by the TAC and displayed on an oscilloscope. The voltage amplitudes can then be converted to time and divided by the distance to obtain the speed of light. This simple, yet effective experiment yielded decent results for my partner, Tom and I.
## Methods, Materials and Procedure
Fig. 1: A sketch of our equipment setup.
Fig. 2: Sketch of the time-walk effect, varying amplitudes are shown, (a result of moving the LED to and from the PMT.
Fig. 3: a photo of the LED mounted on a meter stick at the opening of the cardboard tube.
In order to measure the speed of light, we set up a long, opaque tube made of cardboard with a pulsating LED light source in one end and the photomultiplier tube on the other end of the tube. We connected the photomultiplier tube with BNC cables first to a Canberra 2058 delay module. We then connected the delay module to a Tektronix TDS 1002 digital oscilloscope in parallel with the “Stop” input on the Ortec 567 TAC/SCA Module. The Light source is also connected to the TAC in the “start” input. The delay module was set to 9 ns, to make certain that the stop signal from the PMT arrived after the start signal from the LED. This procedure is necessary due to a substantially longer BNC cable connecting the LED and the TAC. See Figure 1 for an illustration of our setup. Note: the LED was not labeled, we were unsure of the origin of the LED, it is pictured in figure 3.
The light source is mounted on a meter stick so that we can measure various distances of light travel time, which ideally would lower our uncertainty. However this introduces a possible source of systematic error. As the LED is moved closer to the PMT, the PMT amplitude rises due to heightened intensity of photon bombardment. This issue is known as “time walk”. In order to reduce this time walk or fluctuation in voltage, a polarizer is mounted to the front of the PMT. As the LED is pushed down the tube toward the PMT, we rotate the PMT to keep the intensity as continuous as possible. Figure 2 shows the differences in amplitude as a result of "time walk".
### Data Collection
We used six trials of recording data for our experiment. Each trial, we moved the LED closer to the PMT in 10 cm increments and recorded the amplitude for each increment. The first trial was only measured with nine increments of 10 cm because we started at a different mark on the meter stick. For trials 1-5, one experimenter (TM) positioned the PMT while the other experimenter (RL) rotated the PMT in order to maintain maximum continuous voltage, and also recorded values from the oscilloscope. For the last trial we switched roles to test for any systematic error in our procedural roles, however there was no significant change in data collection. Our raw data can be found in the google docs spreadsheet below.
## Results and Analysis
Figure 4: Comparison of results from trials 1-6, calculated weighted mean, and the accepted value.
For my analysis, I performed a linear regression analysis of my data using the "LINEST" function of Excel (Microsoft Office 2008). To compute my final value for the speed of light, I computed a weighted average and corresponding weighted uncertainty. My excel sheet can be downloaded here.
My calculated value is: $29.4480\pm .1424\frac{cm}{ns}$
The accepted value from NIST is: $29.9792\frac{cm}{ns}$
Figure 4 shows each measurement from trials one through six, my calculated weighted mean, and the accepted value. The error bars on the figure also demonstrate that my calculated value is inconsistent with the exact value from NIST. I speculate that this inconsistency is attributed to some systematic error in our experiment. Figure 5 shows raw data for each trial, our measurements for trials 2-6 were very consistent.
Figure 5: A Plot of our raw measurements during trials 1-6, illustrating our consistency in measurements for 2-6.
## Conclusions
Our data collection was very consistent, in the sense that we did not have wild fluctuations in measurement but there was obvious systematic error in our experiment. Time walk was most likely responsible for a large portion of it, perhaps the photomultiplier could be modified so that it is stationary with only a rotating polarizer, instead of rotating the entire PMT. Another possible source of error is from the positioning of the LED with respect to the PMT, it is nearly impossible to make perfect increments of ten centimeters. This error would probably be reduced by simply taking more measurements. Lastly measuring the voltages on the oscilloscope is not a excellently efficient process, so in our final day of collecting data, we attempted to DAQ hardware to measure our voltages, but due to time constraints and a lack of a proper DAQ card, we did not have any success with this venture.
## Acknowledgements
My special thanks to my partner Tom Mahony for his essential contributions to data acquisition, analysis, and overall exceptional work ethic. I also would like to thank Dr. Koch, and Pranav Rathi, for their assistance in setting up the experiment.
## References
[1] Boyer, CB (1941). "Early Estimates of the Velocity of Light". Isis 33 (1): 24. doi:10.1086/358523. (from wikipedia)
[2] K.M. Evenson et al “Speed of Light from Direct Frequency and Wavelength Measurements of the Methane-Stabilized Laser” Physical Review Letters, Vol. 29, no. 19, (1972)
[3] R. Foord, R. Jones, C.J. Oliver, E.R. Pike “The Use of Photomultiplier Tubes for Photon Counting” Applied Optics, Vol. 8, No. 10, pg. 1 (1969)
[4] National Institute of Standards and Technology. (Constants, Units and Uncertainty) http://www.nist.gov/index.html
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http://unapologetic.wordpress.com/2011/03/02/standard-differentiable-structures/?like=1&source=post_flair&_wpnonce=966197e62c
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# The Unapologetic Mathematician
## Standard Differentiable Structures
It’s high time we introduced the “standard” smooth structures on real vector spaces, which are (of course) our models for all other smooth manifolds.
The easiest one to discuss is $\mathbb{R}$. The standard smooth structure is given by starting with an extremely simple atlas: the single coordinate patch $U$ contains all of $\mathbb{R}$, and the coordinate map $\phi_U:U\to\mathbb{R}$ is just the identity! But we don’t just have this patch, of course. We also have all coordinate patches which are compatible with it.
Since the inverse $\phi_U^{-1}$ is again the identity, it’s easy to pick these out. A coordinate patch is a subset $V\subseteq\mathbb{R}$ and a real-valued function $\phi_V$ on $V$. The two transition functions between $U$ and $V$ are $\phi_V\circ\phi_U^{-1}=\phi_V$ and $\phi_U\circ\phi_V^{-1}=\phi_V^{-1}$. Both of these functions must be differentiable for the patch to fit into the standard differentiable structure. They must both be smooth to fit into the standard smooth structure.
Now, consider a finite-dimensional real vector space $V$. Again, the standard smooth structure starts with using all of $V$ as a coordinate patch. For the coordinate map, we can choose any linear isomorphism $T:V\to\mathbb{R}^n$. Of course, we know that finding such a $T$ is equivalent to picking a basis — given a basis of $V$ we can just send it to the standard basis of $\mathbb{R}^n$, and given a linear isomorphism we can use the preimages of the standard basis to get a basis of $V$.
So there’s a choice to be made: which linear isomorphism to start with. Does it matter? no! If $T_1$ and $T_2$ are two such linear isomorphisms, then $T_2\circ T_1^{-1}$ is a linear automorphism on $\mathbb{R}^n$. And clearly this transition function is smooth. Thus all the possible choices are compatible with each other and generate the same smooth — and thus the same differentiable — structure.
We say “standard” structures here. This is because — and I know this might sound sort of hard to believe — there actually do exist “nonstandard” or “wild” differentiable structures as well. The proofs establishing these examples are tremendously complicated and I’m not about to go into them now. But the fact remains: there do exist manifolds which are homeomorphic to $\mathbb{R}^4$ — they are equivalent to $\mathbb{R}^4$ as topological spaces — and yet they are not equivalent as differentiable manifolds. Any homeomorphism from one topological space to the other will not be smooth.
### Like this:
Posted by John Armstrong | Differential Topology, Topology
## 5 Comments »
1. “Of course, we know that finding such a T is equivalent to picking a basis — given a basis of V we can just send it to the standard basis of \mathbb{R}^n, and given a linear isomorphism we can use the preimages of the standard basis to get a basis of V.”
This is where there seems to be a chicken/egg problem. It seems that the existence of the isomorphism T depends on already having a coordinate system in V, so what is the purpose of getting yet another coordinate system in V by using the preimage?
Comment by | March 2, 2011 | Reply
2. This converse pair is just a quick rundown of the equivalence that we’ve already pointed out before. It’s not that we’re trying to get new coordinate systems on $V$; we’re showing that the coordinate systems we’ve been using in linear algebra work as coordinate systems in differential topology and geometry. Further, we’re showing that they’re all compatible with each other as smooth coordinate maps.
Comment by | March 2, 2011 | Reply
3. [...] a smooth manifold is nothing but a smooth map , where is some interval in the real line with its standard differentiable structure. The interval can, in principle, be half-infinite or infinite, but commonly we just consider [...]
Pingback by | April 8, 2011 | Reply
4. [...] bundle to a Euclidean space. That is, we let be a finite-dimensional real vector space with its standard differentiable structure and see what these constructions look [...]
Pingback by | April 11, 2011 | Reply
5. [...] we can conclude that is an open submanifold of , which comes equipped with the standard differentiable structure on . Matrix multiplication is clearly smooth, since we can write each component of a product matrix [...]
Pingback by | June 9, 2011 | Reply
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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.
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http://math.stackexchange.com/questions/240983/overview-and-introduction-to-strong-logics/243881
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# Overview and introduction to strong logics
Is there a nice and simple paper which summarizes the definitions and properties of strong logics? When I say strong logics I mean something like stationary logic, or Magidor-Malitz quantifier, $\cal L_{\kappa,\lambda}$, etc.
What I am looking for is a paper without many proofs (although preferably with some proofs, just to get the idea) which gives out the definitions of the various extensions of first-order logic, and outlines their properties (compactness, completeness) and differences.
Of course a combination of several short papers is also welcomed, but I really wish to avoid (for now) the long and technical expositions on each logic.
I want to say out that at this moment I am particularly interested in the three logics mentioned above, the stationary logic, MM-quantifier, and $\cal L_{\kappa,\lambda}$ logics. Other types of strong logics are very welcomed, but those three are currently the main points of interest.
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Compactness is generally sacrificed when passing to infinitary logics: it's a fun exercise to find an (infinite) list of propositions (in, say, infinitary propositional logic) whose finite fragments are consistent while the list as a whole is inconsistent. (Hint: pigeonhole principle.) – Zhen Lin Nov 20 '12 at 0:33
Zhen, I know that. However one may still retain partial compactness, or weak-compactness. Especially if one does not fear large cardinals. :-) – Asaf Karagila Nov 20 '12 at 0:49
## 2 Answers
Peter Koellner's Strong Logics of First and Second Order seems to be the sort of article you might be interested in reading. Here's a link to an abstract to the article: Bull. Symbolic Logic Volume 16, Issue 1 (2010), 1-36..
If I recall, there are abundant references cited in the article; if this article doesn't suit your needs, one of the cited references may be more appropriate.
EDIT - the logics in question are all mentioned on the first page of the following reference:
See Part B and/or Chapter IV: The Quantifier "There Exist Uncountably Many" and Some of Its Relatives, (edited?) by M. Kaufmann], openly accessible @ ProjectEuclid. Alternatively, consider looking into the source of Chapter IV: J. Barwise, S. Feferman, eds., Model-Theoretic Logics (New York: Springer-Verlag, 1985), 123-176.
(Here is a permanent link to the article/chapter @ProjectEuclid.)
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Wonderful! I will take a look tomorrow and let you know. Thank you! – Asaf Karagila Nov 20 '12 at 0:27
1
@Asaf: Oh, my gosh! I didn't see that it was your question. I just saw the question. Let me know what you think. Perhaps Peter Smith can chime in too. – amWhy Nov 20 '12 at 0:29
I went over the paper, it is very interesting and I will definitely read it soon. I am particularly interested in stationary logic and the Magidor-Malitz quantifier, neither of which is mentioned in this paper. – Asaf Karagila Nov 20 '12 at 7:36
Asaf: I included an additional reference that seems to speak directly to the logics you specify. – amWhy Nov 20 '12 at 14:41
I haven't read the chapter yet, but I have copied the entire book to my computer. I am actually interested in studying a draft by my advisor and two people. One of those people wrote a very relevant chapter. Either way, I think this should give me most of what I want to know (even if I have to slightly distill it from details). Thank you very much! – Asaf Karagila Nov 21 '12 at 23:26
show 1 more comment
This paper of Vaananen should be quite useful: http://www.math.ucla.edu/~asl/bsl/1001/1001-004.ps
Other texts that you may find useful are Shelah's paper "Generalized quantifiers and compact logic" (which introduces stationary logic) and Barwise's recent book on AEC, which has some material on infinitary logic and generalized quantifiers as well.
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Thanks. I hope you are coming to my lecture on Wednesday. – Asaf Karagila Nov 24 '12 at 18:47
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http://mathoverflow.net/questions/6346?sort=oldest
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## Simplicial set notation and vocabulary question.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Notation question:
What does $(\Delta^1)^{ \{1, \ldots,n-1 \}}$ denote? UPDATE: I (David Speyer) tried to fix the LaTeX. Please see if I got it right.
Vocabulary question:
Suppose $z:\Delta^{n+1} \rightarrow S$ is a morphism of simplicial sets. What does the following translate to in algebraic terms: $z|\Delta^{ \{0,\ldots,n \} }$ is a constant simplex at a vertex $x$.
So mainly, I just don't know what that is supposed to mean, "is a constant simplex at the vertex x". Everything else makes fine sense.
I've searched through a number of books on homotopy theory, algebraic topology, etc. and I've been unable to find these precise usages.
I ask these questions only because I'm reading HTT by Lurie, and these usages come up and they're quite confusing.
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Check for yourselves on Google, there are a total of 4 unique results for "simplex at the vertex" and 7 unique results for "simplex at a vertex". – Harry Gindi Nov 21 2009 at 3:49
Where do you see it in HTT? – S. Carnahan♦ Nov 21 2009 at 3:58
I've removed the offending remark. Anyway, Scott Carnahan, see page 24 for the question about notation and see page 27 for "constant simplex at the vertex x". Also, David Speyer has fixed the LaTeX issue, so I will remove the no-longer relevant remarks. – Harry Gindi Nov 21 2009 at 4:06
## 1 Answer
Okay, I've found the relevant notation in Higher Topos Theory. The first is at the end of 1.1.5.10, and is the simplicial set of maps from an n-1 element set to the 1-simplex (i.e., an n-1-cube). The second is at the end of warning 1.2.2.2, and describes a constant map of simplicial sets whose image is x. The curly braces give a reference to the specific ordered set that defines the the simplex.
A simplex at a vertex is a degenerate simplex. You take the simplicial subset defined by x, and demand that the map from your simplex to the target factor through the inclusion of x.
Edit in response to comment: You can think of vertices in (at least) two ways. One way is as an element of S0, i.e., a zero-simplex of the simplicial set. Another way is as a simplicial subset X of S, such that X0 is the chosen element of S0, and all Xi have a single element, namely the image of X0 under the unique degeneracy map. The statement is that the map Z takes a particular nondegenerate n-dimensional face of $\Delta^{n+1}$ to the unique element of Xn.
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For the n-1-cube part, are we assuming that the n-1 element set is ordered or no? – Harry Gindi Nov 21 2009 at 4:20
It is ordered, but it doesn't matter. We're looking at maps from an ordered set to a simplex, and there is no condition on orientation. – S. Carnahan♦ Nov 21 2009 at 4:33
Two questions: What do you mean by "simplicial subset defined by x"? Whatever that means, let's let this "simplicial subset defined by x" be called X. Then did you mean that a "simplex at x" is a map that factors through the inclusion of x or the inclusion of X? – Harry Gindi Nov 21 2009 at 18:34
Much thanks for the help. – Harry Gindi Nov 21 2009 at 20:28
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http://math.stackexchange.com/questions/2883/show-that-3p2-q2-implies-3p-and-3q/2941
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# Show that $3p^2=q^2$ implies $3|p$ and $3|q$
This is a problem from "Introduction to Mathematics - Algebra and Number Systems" (specifically, exercise set 2 #9), which is one of my math texts. Please note that this isn't homework, but I would still appreciate hints rather than a complete answer.
The problem reads as follows:
If 3p2 = q2, where $p,q \in \mathbb{Z}$, show that 3 is a common divisor of p and q.
I am able to show that 3 divides q, simply by rearranging for p2 and showing that
$$p^2 \in \mathbb{Z} \Rightarrow q^2/3 \in \mathbb{Z} \Rightarrow 3|q$$
However, I'm not sure how to show that 3 divides p.
Edit:
Moron left a comment below in which I was prompted to apply the solution to this question as a proof of $\sqrt{3}$'s irrationality. Here's what I came up with...
[incorrect solution...]
...is this correct?
Edit:
The correct solution is provided in the comments below by Bill Dubuque.
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– Gone Aug 21 '10 at 2:19
@Bill: Can you explain why that inference isn't correct? – Cam Aug 21 '10 at 2:21
How did you go from the 2nd last equation to the last? – Gone Aug 21 '10 at 2:33
@Bill: By making silly mistake :) - thanks for offering the correct solution. – Cam Aug 21 '10 at 2:41
Alternatively you could assume $p$ minimal at the start, then canceling 3 from $p$ and $q$ yields $\sqrt{3} = \frac{q/3}{p/3}$ contra minimality of $p$. There are many variations - see the references in my post linked above. – Gone Aug 21 '10 at 3:14
## 5 Answers
Write $q$ as $3r$ and see what happens.
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Then it's the same problem as before, but with the variables switched - thanks :) – Cam Aug 20 '10 at 21:20
6
@Cam: Exactly. Do you see how that helps you prove that $\sqrt{3}$ is irrational? – Aryabhata Aug 20 '10 at 21:24
I think so. I've edited my question in response (the character limit for comments is too high). Is my solution correct? – Cam Aug 21 '10 at 1:25
Below is a conceptual proof of the irrationality of square-roots. It shows that this result follows immediately from unique fractionization -- the uniqueness of the denominator of any reduced fraction -- i.e. the least denominator divides every denominator. This in turn follows from the key fact that the set of all possible denominators of a fraction is closed under subtraction so comprises an ideal of $\,\mathbb Z,\,$ necessarily principal, since $\,\mathbb Z\,$ is a $\rm PID$. But we can easily eliminate this highbrow language to obtain the following conceptual high-school level proof:
Theorem $\$ Let $\;\rm n\in\mathbb N.\;$ Then $\;\rm r = \sqrt{n}\;$ is integral if rational.
Proof $\$ Consider the set $\rm D$ of all the possible denominators $\rm d$ for $\rm r, \;$ i.e. $\;\rm D = \{\, d\in\mathbb Z \;:\: dr \in \mathbb Z\,\}$. Note $\rm D$ is closed under subtraction: $\rm\, d,e \in D\, \Rightarrow\, dr,\,er\in\mathbb Z \,\Rightarrow\, (d-e)\:r = dr - er \in\mathbb Z.\;$ Further $\rm d\in D \,\Rightarrow\, dr\in D\,$ since $\rm\, (dr)r = dn\in\mathbb Z \;$ by $\;\rm r^2 = n\in\mathbb Z.\;$ Therefore, invoking the Lemma below, with $\rm d$ the least positive element in $\rm D,$ we infer that $\;\rm d\,|\,dr \;$ in $\mathbb Z,\;$ i.e. $\rm\ r = (dr)/d \in\mathbb Z.\quad$ QED
Lemma $\$ Suppose $\;\rm D\subset\mathbb Z \;$ is closed under subtraction and that $\rm D$ contains a nonzero element.
Then $\rm D \:$ has a positive element and the least positive element of $\rm D$ divides every element of $\rm D\:$.
Proof $\rm\,\ \ 0 \ne d\in D \,\Rightarrow\, d-d = 0\in D\,\Rightarrow\, 0-d = -d\in D.\,$ Hence $\rm D$ contains a positive element. Let $\rm d$ be the least positive element in $\rm D$. Since $\rm\: d\,|\,n \!\iff\! d\,|\,{-}n,\,$ if $\rm\, c\in D$ is not divisible by $\rm d$ then we may assume that $\rm c$ is positive, and the least such element. But $\rm\, c-d\,$ is a positive element of $\rm D$ not divisible by $\rm d$ and smaller than $\rm c$, contra leastness of $\rm c$. So $\rm d$ divides every element of $\rm D.\$ QED
The theorem's proof exploits the fact that the denominator ideal $\rm D$ has the special property that it is closed under multiplication by $\rm\: r\:.\$ The fundamental role that this property plays becomes clearer when one learns about Dedekind's notion of a conductor ideal. Employing such yields a trivial one-line proof of the generalization that a Dedekind domain is integrally closed since conductor ideals are invertible so cancellable. This viewpoint serves to generalize and unify all of the ad-hoc proofs of this class of results - esp. those proofs that proceed essentially by descent on denominators. This conductor-based structural viewpoint is not as well known as it should be - e.g. even some famous number theorists have overlooked this. See my post here for further details.
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Moron's answer certainly covers your question, but as someone who's not your instructor I'd like to see a few more details in your 'proof' of the first half - can you be more specific about how $q^2/3 \in \mathbb{Z} \Rightarrow 3|q$? While that's easy, it's not necessarily trivial, and you've elided some details there...
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Assume 3 does not divide $q$. Then 3 does not divide $q^2$, so $\frac{q^2}{3}$ is not an integer. But this is a contradiction, because by arranging for $p^2$ we see that $p^2=\frac{q^2}{3}$ and $p^2 \in \mathbb{Z}$ so it follows that $\frac{q^2}{3} \in \mathbb{Z}$. Therefore $3|q$. – Cam Aug 21 '10 at 2:54
@Cam: That doesn't answer Steven's query. You need to prove that $3|q^2 \Rightarrow 3|q$. For one simple way see my comment to Katie's post here. – Gone Aug 21 '10 at 3:21
@Bill: What about: if $3|q^2$ then 3 is a prime factor of $q^2$, so 3 must be a prime factor of $q$, and therefore $3|q$? Otherwise put, it would be impossible for 3 to be a factor of $q^2$ unless it was a factor of $q$ as well, because 3 is prime. – Cam Aug 21 '10 at 15:04
That works but it implictly assumes a very powerful result - that integers have unique factorization. For a simpler way see my comment to Katie's post. See also my proof here of the general result based upon unique fractionization. – Gone Aug 21 '10 at 15:40
Think about how many times each prime factor must appear on each side of the equation, if you were to break p and q into their prime factorizations. The left side has a 3 in it, how many must the right side have, at least?
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2
But using unique factorization is a bit of a sledgehammer. Instead one need only note that $x^2 = 0 \Rightarrow x=0 \pmod 3$ since $x\ne 0 \Rightarrow x = \pm 1 \Rightarrow x^2 = 1 \pmod 3$. – Gone Aug 20 '10 at 21:56
Here we go. $3p^2=q^2$ implies that $3$ divides $q$, since $3$ is prime and if a prime divides a product, it divides one of the factors. But then, if $3$ divides $q$, then we also have that $3^2$ divides $q^2$. Hence, by factoring out the 9 on the rhs, we can cancle the 3 on the left hand side and still be left with a three. i.e $3\alpha=p^2$. But then, $3$ divides p, as required.
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http://mathhelpforum.com/differential-geometry/186622-prove-certain-function-polynomial.html
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# Thread:
1. ## prove a certain function is a polynomial
I have seen the following problem and I didn't manage to solve it, so I will appreciate any help.
Suppose f is an analytic function defined everywhere in $\mathbb{C}$ and such that for each $z_0\in\mathbb{C}$ at least one coefficient in the expansion f(z)=\sum_{n=0}^\infty c_n(z-z_0)^n is equal to 0. Prove that f is a polynomial.
It is easy to prove that for each z_0 there exists n such that f^{(n)}(z_0)=0, which means that each z_0 is root for a derivative of f, but how do I conclude that f s a polynomial?
2. ## Re: prove a certain function is a polynomial
If f(*) is analytic everywhere in $\mathbb{C}$ , then its Taylor expansion in any point $z=z_{0}$ is...
$f(z)= \sum_{n=0}^{\infty} a_{n}\ (z-z_{0})^{n}$ (1)
... where...
$a_{n}= \frac{1}{n!}\ \frac{d^{n}}{d z^{n}} f(z)_{z=z_{0}}$ (2)
Now if $\vorall z_{0}$ is $a_{n}=0$ then the derivative of order n of f(*) vanishes everywhere and that means that f(*) is a polynomial of order k<n...
Kind regards
$\chi$ $\sigma$
3. ## Re: prove a certain function is a polynomial
you can solve it using 2 hints:
-whats the union of sets {f^(k)(z)=0} where k runs through natural numbers?
-what are the properties of the zero set of a holomorphic function?
4. ## Re: prove a certain function is a polynomial
I was thinking something like this: the union of sets {f^(k)(z)=0} where k runs through natural numbers is uncountable from the hypothesis, that means that there must be a number N which satisfies f^(N)=0 for all z, which demonstrates that f is a polynomial. But I am not sure if there are some missing arguments.
About the set of the zeros of a holomorphic function, I suppose it must be finite? If a holomorphic function has countable many zeros, that means that it is the function constant=0?
5. ## Re: prove a certain function is a polynomial
Put $F_n:=\{z\in\mathbb C, f^{(n)}(z)=0\}$. $F_n$ is a closed subset of $\mathbb C$ and $\bigcup_{n\in\mathbb N}F_n=\mathbb C$. Using Baire category's theorem, we can find $n_0$ such that $F_{n_0}$ has a nonempty interior. We can find a ball on which $f^{(n_0)}$ is identically $0$ on this ball. Now you can conclude, using the fact that the zeros of an holomorphic function are isolated.
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http://math.stackexchange.com/questions/90214/does-proving-second-countable-rightarrow-lindelof-require-the-axiom-of-ch/90220
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# Does proving (second countable) $\Rightarrow$ (Lindelöf) require the axiom of choice?
Background:
This question came up in my homework (but was not a homework problem). The problem was proving one direction of the Heine-Borel theorem. As with all proofs of compactness, one begins with, "Suppose $A$ is closed and bounded, and $\mathcal{U}$ is an open cover ..." My proof, which I believe is typical, constructed a convergent sequence of points and derived a contradiction. The problem: to construct a sequence, one needs the cover to be countable.
So I invoked that second-countability implies Lindelöf. This is not controversial, of course, but I included a little argument along the lines of the answers in this question: Every open cover has a countable subcover (Lindelöf's lemma).
In particular, for each $x\in A$ there is some $U$ in $\mathcal{U}$ with $x\in U$; second-countability implies for each $x\in A$ there is some basis element $Q$ with $x\in Q\subset U$; the set of all such $Q$ covers $A$ and each $Q\subset U$ for at least one $U\in\mathcal{U}$, so picking one $U\supset Q$ for each $Q$ gives a countable cover of $A$.
I noticed that the answers in that thread included similar phrases: "Now for each $B\in\mathcal{B}U$ choose some $U(B)\in U$ such that $B\subseteq U$", "for each element $O$ of $\Sigma$ choose an element $U$ of $\Omega$ containing it"
My argument and the arguments suggested in the thread linked both seem to assume the Axiom of Choice. I'm not sure that I understand the axiom of choice, though. So, two questions:
1) Are we actually using the axiom of choice in these arguments?
2) Does (second countable) $\Rightarrow$ (Lindelöf) require the axiom of choice?
Thanks!
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## 2 Answers
The results requires some choice. It is consistent that $\mathbb R$ is not Lindelöf but still second countable (without the axiom of choice).
The axiom of choice for countable families of subsets of $\mathbb R$ is in fact equivalent that second countable implies Lindelöf, and to a few other interesting assertions:
H. Herrlich and G. E. Strecker, When is $\mathbb{N}$ Lindelöf?, Comment. Math. Univ. Carolinae 38,3 (1997), 553-556.
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Beautiful! Thank you! – Neal Dec 10 '11 at 17:57
@Neal: No problem, I would have answered sooner, but my system crashed ;-) – Asaf Karagila Dec 10 '11 at 18:05
I think it's worth pointing out that, although second-countability implies Lindelöf requires some choice, the Heine-Borel theorem itself doesn't. To prove that a closed and bounded subset of $\mathbb{R}$ is compact, consider first the case $[0,1]$.
Given an open cover of $[0,1]$, consider the set $S = \{x \in [0,1]\mid\text{there is a finite subcover of }[0,x]\}$. Then $0 \in S$, $S$ is open, and the least upper bound of $S$ is an element of $S$ (check each of these facts). It follows that $1 \in S$ is the least upper bound, so there is a finite subcover.
To prove that any closed and bounded set is compact, realize that it is a closed subset of $[-a,a] \cong [0,1]$ for sufficiently large $a$.
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