It's not like I write anything here ever.
But, anyway, this is a test of LaTeXMathML. If I did it right, then $ax^2 + bx + c = 0$ iff $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Note: this works on my machine. Whether it works on yours depends on your browser, fonts, etc. In general, Internet Explorer sucks, although it does have some limited MathML support. Firefox (or Mozilla) is, of course, the preferred browser. You may need to install new fonts, as well.
08 December 2009
30 September 2009
29 September 2009
What the Hell is a Feynman Diagram?
As those of you who read my other blog know, I've been in Denmark this month, visiting the Center for the Topology and Quantization of Moduli Spaces, Department of Mathematical Sciences, Aarhus Universitet. I've gotten a fair amount of work done, attended a very fine Chern-Simons theory conference in Strasbourg, and met various people. But I miss Berkeley and its preponderence of interesting talks.
In any case, I'm in the final editing stages on my paper on Quantum Mechanics and Feynman Diagrams. Since by now I really know what those Diagrams mean, I gave an introductory talk in the PhD seminar here at Aarhus about them. My prepared notes are here; I said about half of them, and none of the interesting bits.
In any case, I'm in the final editing stages on my paper on Quantum Mechanics and Feynman Diagrams. Since by now I really know what those Diagrams mean, I gave an introductory talk in the PhD seminar here at Aarhus about them. My prepared notes are here; I said about half of them, and none of the interesting bits.
24 August 2009
Differential Equations Bleg. Or: In which we find out who uses an RSS reader.
I have a second-order differential equation on a manifold $N$, which might as well be $\R^n$. If it matters, my differential equation is the Euler-Lagrange equation for some Lagrangian $L$. I will call my differential equation $\D$, for want of a better name.
Let $T \in \R$ with $T > 0$ and $Q_1, Q_2 \in N$ be given. Assume that $\gamma: [0,T] \to N$ is a solution to the boundary value problem
(BVP) $\D$ and $\gamma(0) = Q_1$ and $\gamma(T) = Q_2$.
and moreover assume that $\gamma$ is an isolated point within the set of all solutions to (BVP).
Then I would like there to be an open neighborhood $U$ of $(Q_1,Q_2) \in N$ such that for each point $(q_1,q_2) \in U$, there is a solution $\gamma_{q_1,q_2}$ to the BVP
$\D$ and $\gamma_{q_1,q_2}(0) = q_1$ and $\gamma_{q_1,q_2}(T) = q_2$
such that $\gamma_{Q_1,Q_2} = \gamma$ and the function $U \times [0,T] \to N$ given by $(q_1,q_2,t) \mapsto \gamma_{q_1,q_2}(t)$ is smooth.
Is this true? Put succinctly, if I have an isolated solution to a boundary-value problem, do nearby boundary values have yield solutions that depend smoothly on the boundary values?
If it matters, as I said above, my differential equation is the Euler-Lagrange equations for some Lagrangian $L$. Recall that the Action of a path $\gamma$ is defined as the integral of $L$ over the path $\gamma$. Then the Euler-Lagrange equations are precisely the statement that the first derivative of the Action vanishes along any perturbation of the path with Derichlet boundary conditions.
For my particular problem, I know a bit more about the solution $\gamma$ than just that it is an isolated point among the set of solutions to the Euler-Lagrange equations. Just as setting the first derivative of the action to zero gives the Euler-Lagrange equations, the second derivative of the action is interesting as well. Like any second derivative, this one depends on the coordinates (it does not transform as a tensor), and for given coordinates is a symmetric bilinear form on the space of perturbations of $\gamma$. What I know is that this form has no kernel among the paths with Derichlet boundary conditions, in the sense that if $\zeta$ is a perturbation of $\gamma$ that does not perturb the endpoints, then there is some $\xi$ so that the second derivative of the action in the $\zeta\otimes \xi$ direction is non-zero. Put another way, our extra bit of knowledge about $\gamma$ is that a certain linear homogeneous second-order differential equation (with coefficients that depend on $\gamma$) along with the boundary requirements that the solution vanish at $0$ and at $T$ --- that this boundary value problem has only the trivial solution.
So, anyhoo, the condition in the previous paragraph guarantees that the original path $\gamma$ is an isolated point. I think that being an isolated point alone is enough to have the "depends smoothly on the parameters" thing that I want. But maybe this extra condition does.
Sorry if the post doesn't make any sense. It's late at night. If you might know this stuff, but want me to send you TeX with all the definitions, just drop me a line.
Let $T \in \R$ with $T > 0$ and $Q_1, Q_2 \in N$ be given. Assume that $\gamma: [0,T] \to N$ is a solution to the boundary value problem
(BVP) $\D$ and $\gamma(0) = Q_1$ and $\gamma(T) = Q_2$.
and moreover assume that $\gamma$ is an isolated point within the set of all solutions to (BVP).
Then I would like there to be an open neighborhood $U$ of $(Q_1,Q_2) \in N$ such that for each point $(q_1,q_2) \in U$, there is a solution $\gamma_{q_1,q_2}$ to the BVP
$\D$ and $\gamma_{q_1,q_2}(0) = q_1$ and $\gamma_{q_1,q_2}(T) = q_2$
such that $\gamma_{Q_1,Q_2} = \gamma$ and the function $U \times [0,T] \to N$ given by $(q_1,q_2,t) \mapsto \gamma_{q_1,q_2}(t)$ is smooth.
Is this true? Put succinctly, if I have an isolated solution to a boundary-value problem, do nearby boundary values have yield solutions that depend smoothly on the boundary values?
If it matters, as I said above, my differential equation is the Euler-Lagrange equations for some Lagrangian $L$. Recall that the Action of a path $\gamma$ is defined as the integral of $L$ over the path $\gamma$. Then the Euler-Lagrange equations are precisely the statement that the first derivative of the Action vanishes along any perturbation of the path with Derichlet boundary conditions.
For my particular problem, I know a bit more about the solution $\gamma$ than just that it is an isolated point among the set of solutions to the Euler-Lagrange equations. Just as setting the first derivative of the action to zero gives the Euler-Lagrange equations, the second derivative of the action is interesting as well. Like any second derivative, this one depends on the coordinates (it does not transform as a tensor), and for given coordinates is a symmetric bilinear form on the space of perturbations of $\gamma$. What I know is that this form has no kernel among the paths with Derichlet boundary conditions, in the sense that if $\zeta$ is a perturbation of $\gamma$ that does not perturb the endpoints, then there is some $\xi$ so that the second derivative of the action in the $\zeta\otimes \xi$ direction is non-zero. Put another way, our extra bit of knowledge about $\gamma$ is that a certain linear homogeneous second-order differential equation (with coefficients that depend on $\gamma$) along with the boundary requirements that the solution vanish at $0$ and at $T$ --- that this boundary value problem has only the trivial solution.
So, anyhoo, the condition in the previous paragraph guarantees that the original path $\gamma$ is an isolated point. I think that being an isolated point alone is enough to have the "depends smoothly on the parameters" thing that I want. But maybe this extra condition does.
Sorry if the post doesn't make any sense. It's late at night. If you might know this stuff, but want me to send you TeX with all the definitions, just drop me a line.
07 June 2009
In which we find out who uses RSS
This blog, as you probably have gathered, has been unused for some time. It will probably return to being unused soon. But for those of you who have it in your RSS feed, so that you are automatically alerted to new posts, I wanted to give a brief update.
I am happily in graduate school at UC Berkeley, studying quantum field theory. My quals are in less than a week, so I am reading and reviewing. This term I discovered that my favorite way to study is to write copious amounts. If you like to read, I invite you to check out the following:
I hope to say a few words here about some of this material. In particular, I'm thinking of writing about the following: the Weyl character formula; symplectic leaves of Poisson manifolds; what is quantization. Whether I actually write such entries is still up in the air, of course. Currently, I primarily post content directly to my website, but that doesn't have an RSS feature. If there are still people who watch this space, leave a comment, and I'll get into the habit of writing a blog entry any time I post something new to my website.
I am happily in graduate school at UC Berkeley, studying quantum field theory. My quals are in less than a week, so I am reading and reviewing. This term I discovered that my favorite way to study is to write copious amounts. If you like to read, I invite you to check out the following:
- Lie Groups and Lie Algebras (pdf). These are edited lecture notes from a one-semester class last fall. They should be fairly complete and accurate, but please let me know about any errors: typos are easy to fix, and mathematical errors should be corrected for morality sake.
- Poisson Lie linear algebra in the graphical language (pdf). This article outlines all the definitions from the theory of Lie bialgebras, but does it using only Penrose's ``birdtrack'' notation, championed by among others Cvitanovic.
- Some notes on Quantum Groups (pdf). These are not particularly edited, and are rather ideosyncratic, and include the graphical language article as a chapter.
I hope to say a few words here about some of this material. In particular, I'm thinking of writing about the following: the Weyl character formula; symplectic leaves of Poisson manifolds; what is quantization. Whether I actually write such entries is still up in the air, of course. Currently, I primarily post content directly to my website, but that doesn't have an RSS feature. If there are still people who watch this space, leave a comment, and I'll get into the habit of writing a blog entry any time I post something new to my website.
03 March 2008
From products of infinite sums to subtraction and division
In some of the research I'm doing, I have occasion to write expressions like
A + ABA + ABABA + ABABABA + \dots = (A^{-1} - B)^{-1}
with no consideration of convergence.
Although in my setting I have (multiplicative) inverses and (additive) negatives, it nonetheless becomes interesting to ask what sorts of operations we can do with just plus and times, provided that we can do such things infinitely.
For example, with just positive integers and some questionable summations, we can get all negative fractions:
-1 = 1 + 2 + 4 + 8 + 16 + \dots
-m/n = m + (n+1)\times m + (n+1)^2\times m + \dots
and some irrationals:
\sqrt{2\pi} = 1 \times 2 \times 3 \times 4 \times 5 \times \dot
The first two sums are from the fact that 1+r+r^2+\dots=1/(1-r). Indeed, this comes from multiplying and shifting. If, e.g., we add 1 to the first equation, and then telescope the sum, we have
1+1 + 2 + 4 + 8 + 16 + \dots = 2+2 + 4 + 8 + 16 + \dots = 4+4 + 8 + 16 + \dots = 8+8 + 16 + \dots = etc.
The product is from \zeta-function regularization: we recognize the right hand side as the exponential of d\zeta(s)/ds at s=0.
Since we can multiply power series, we can get all fractions; for any given fraction, there are lots of ways of writing it as a series. For example,
1 = (-1)^2 = (1+2+4+8+\dots)^2 = 1 + (2+2) + (4+4+4) + (8+8+8+8) + \dots
= \sum (n+1)2^n.
Since \sum 2^n = -1, we clearly have \sum n2^{n-1} = 1 as well, which can be checked directly (it is d[\sum 2^n]/d2 = d[1/(1-2)]/d2 = 1/(1-2)^2.).
How about inverses of variables? Especially if you're not sure of the 2-adic sum, perhaps you'd rather use
-1/X = 1 + (1+X) + (1+X)^2 + (1+X)^3 + \dots
to get your negative and inverse Xs. This formula is hard to verify, since it still requires the same shifting tricks:
1 + ( X + X(1+X) + X(1+X)^2 + \dots ) = 1 + X + X(1+X) + X(1+X)^2 + \dots
= 1+X + X(1+X) + X(1+X)^2 + \dots
= (1+X)(1+X) + X(1+X)^2 + \dots
= (1+X)(1+X)^2 + \dots
= etc.
See, infinite sums are not associative --- the distance between terms matters --- but these manipulations are justified since infinite sums are finitely associative.
We get, either by squaring or differentiating,
1/X^2 = 1 + 2(1+X) + 3(1+X)^2 + 4(1+X)^3 + \dots
and hence all powers of X. Well, we get \pm 1/X^n; for, say, -X, the shortest is to use X^2(-1/X) = X^2 + X^2(1+X) + X^2(1+X)^2 + X^2(1+X)^3 + \dots, which telescopes in the same way.
These manipulations require that there be a number 1 which multiplies with X to X and which we can add to X. Which is true is X is, say, a matrix, or a CW complex, but not if X is a two-zero-tensor. If X is such a tensor, then X^{-1} is the tensor which convolves on both sides to the identity matrix. This is the setting of my original line (A is two-zero, and B is zero-two). In this situation, I don't have a good construction of X^{-1}.
A + ABA + ABABA + ABABABA + \dots = (A^{-1} - B)^{-1}
with no consideration of convergence.
Although in my setting I have (multiplicative) inverses and (additive) negatives, it nonetheless becomes interesting to ask what sorts of operations we can do with just plus and times, provided that we can do such things infinitely.
For example, with just positive integers and some questionable summations, we can get all negative fractions:
-1 = 1 + 2 + 4 + 8 + 16 + \dots
-m/n = m + (n+1)\times m + (n+1)^2\times m + \dots
and some irrationals:
\sqrt{2\pi} = 1 \times 2 \times 3 \times 4 \times 5 \times \dot
The first two sums are from the fact that 1+r+r^2+\dots=1/(1-r). Indeed, this comes from multiplying and shifting. If, e.g., we add 1 to the first equation, and then telescope the sum, we have
1+1 + 2 + 4 + 8 + 16 + \dots = 2+2 + 4 + 8 + 16 + \dots = 4+4 + 8 + 16 + \dots = 8+8 + 16 + \dots = etc.
The product is from \zeta-function regularization: we recognize the right hand side as the exponential of d\zeta(s)/ds at s=0.
Since we can multiply power series, we can get all fractions; for any given fraction, there are lots of ways of writing it as a series. For example,
1 = (-1)^2 = (1+2+4+8+\dots)^2 = 1 + (2+2) + (4+4+4) + (8+8+8+8) + \dots
= \sum (n+1)2^n.
Since \sum 2^n = -1, we clearly have \sum n2^{n-1} = 1 as well, which can be checked directly (it is d[\sum 2^n]/d2 = d[1/(1-2)]/d2 = 1/(1-2)^2.).
How about inverses of variables? Especially if you're not sure of the 2-adic sum, perhaps you'd rather use
-1/X = 1 + (1+X) + (1+X)^2 + (1+X)^3 + \dots
to get your negative and inverse Xs. This formula is hard to verify, since it still requires the same shifting tricks:
1 + ( X + X(1+X) + X(1+X)^2 + \dots ) = 1 + X + X(1+X) + X(1+X)^2 + \dots
= 1+X + X(1+X) + X(1+X)^2 + \dots
= (1+X)(1+X) + X(1+X)^2 + \dots
= (1+X)(1+X)^2 + \dots
= etc.
See, infinite sums are not associative --- the distance between terms matters --- but these manipulations are justified since infinite sums are finitely associative.
We get, either by squaring or differentiating,
1/X^2 = 1 + 2(1+X) + 3(1+X)^2 + 4(1+X)^3 + \dots
and hence all powers of X. Well, we get \pm 1/X^n; for, say, -X, the shortest is to use X^2(-1/X) = X^2 + X^2(1+X) + X^2(1+X)^2 + X^2(1+X)^3 + \dots, which telescopes in the same way.
These manipulations require that there be a number 1 which multiplies with X to X and which we can add to X. Which is true is X is, say, a matrix, or a CW complex, but not if X is a two-zero-tensor. If X is such a tensor, then X^{-1} is the tensor which convolves on both sides to the identity matrix. This is the setting of my original line (A is two-zero, and B is zero-two). In this situation, I don't have a good construction of X^{-1}.
12 February 2008
New Blog
If you do not subscribe to the RSS feed for this blog, then you have by now stopped checking it for updates. I intend to post occasionally, but school is absorbing my mathematical thoughts, rather than writing.
Since I have no extra time, I have started a new blog, sister to this one. Whereas Orange Juice Files is primarily for mathematical (and nonmathematical) essays, Local Seasoning will be a repository of recipes and meals from my kitchen. Enjoy.
Since I have no extra time, I have started a new blog, sister to this one. Whereas Orange Juice Files is primarily for mathematical (and nonmathematical) essays, Local Seasoning will be a repository of recipes and meals from my kitchen. Enjoy.
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