network profile
| bio | website | math.utoronto.ca/jmfisher |
|---|---|---|
| location | Toronto | |
| age | ||
| visits | member for | 1 year, 11 months |
| seen | Mar 12 at 18:34 | |
| stats | profile views | 104 |
I'm a PhD candidate interested in mathematics inspired by theoretical physics (QFT, strings, integrable systems, etc.).
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Dec 15 |
awarded | Revival |
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Aug 27 |
awarded | Enlightened |
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Aug 27 |
awarded | Nice Answer |
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Aug 24 |
comment |
Formula for trace of compact operators on $L^2(\mathbb{R})$ given by integral kernels? If $K$ is nice, then for each fixed $z$, $K(z,x)$ is a well-behaved function of $x$, hence by the 3rd last equation can express $K(z,x)$ as the sum of $n$ of the integral of $K(z,y) e_n(y)$ with respect to $y$. Now plug in $z = x$ to obtain the result. |
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Aug 23 |
revised |
Help solving differential equation added 222 characters in body |
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Aug 23 |
answered | Help solving differential equation |
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Aug 23 |
revised |
Formula for trace of compact operators on $L^2(\mathbb{R})$ given by integral kernels? added 356 characters in body |
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Aug 23 |
comment |
Formula for trace of compact operators on $L^2(\mathbb{R})$ given by integral kernels? Yes, I am well aware and absolutely agree. I answered as above, assuming $K(x,y)$ to be as nice as necessary, since it didn't seem that Mike was asking for the weakest possible hypotheses under which this holds, but rather whether it can be made rigorous at all. |
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Aug 22 |
answered | Formula for trace of compact operators on $L^2(\mathbb{R})$ given by integral kernels? |
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Aug 21 |
answered | relation between integral and summation |
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Aug 21 |
answered | Method of characteristics with constant PDE |
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Aug 13 |
answered | Definition of the rank of the isometry group of a manifold. |
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Aug 10 |
comment |
Are there n-th roots of differential operators? This technique depends sensitively on both the given differential operator, as well as the space of functions on which it acts. In the case of the Dirac operator, it is not the Laplacian on functions that admits an algebraic square root, but rather the Laplacian on spinors. This also works for vector-valued functions, since we can decompose vectors into products of spinors. These kinds of $n$th roots are completely different than what I described in my answer, but both can be useful depending on the situation. |
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Aug 10 |
comment |
Are there n-th roots of differential operators? Yes, if you write $\sqrt[1/2]{D_y} \sqrt[1/2]{D_y} g(y)$ as a multiple integral, there will be a term like $\int e^{2\pi i y(k-k')} dy$ which is $\delta(k-k')$ and everything takes care of itself. This is what makes the pseudodifferential calculus work. |
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Aug 10 |
answered | Are there n-th roots of differential operators? |
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Aug 10 |
comment |
Computation of determinant of a matrix with elements from an arbitrary commutative ring I don't know the answer to this particular question, but there is a general phenomenon in algebraic geometry that computational complexity of certain algebraic problems is (inversely) related to the "niceness" of an associated geometric object. The presence of zero divisors definitely complicates the geometry. So I wouldn't find it completely surprising if cofactor expansion really is the most efficient method for an arbitrary commutative ring, without any additional assumptions. |
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Aug 3 |
comment |
The boundedness of an integral After integrating by parts and playing around a little bit, you can easily get an estimate like this provided you make an assumption like $|a| > a_0$, $|b| > b_0$. If you want an estimate that makes no such assumption, then I'm not sure there is any reason to expect such an estimate to exist. |
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Aug 3 |
revised |
Asymptotic behavior of a sequence based on a subsequence. deleted 177 characters in body |
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Aug 3 |
comment |
Asymptotic behavior of a sequence based on a subsequence. I didn't realize that $m$ is supposed to be fixed. I'll update my answer when I get a chance. |
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Aug 3 |
answered | Asymptotic behavior of a sequence based on a subsequence. |
