Timeline for Approximating high-dimensional integrals by low-dimensional ones
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 14, 2011 at 22:25 | vote | accept | algori | ||
| Sep 14, 2011 at 21:32 | answer | added | Noam D. Elkies | timeline score: 5 | |
| Sep 14, 2011 at 19:21 | comment | added | algori | Noam -- thanks again! Could you give an idea or reference as to why the resulting product is equal $\int_{A_t}e^{sx}d\mu_t$ (with $\mu_t$ the $\dim A_t$-dimensional Hausdorff measure)? | |
| Sep 14, 2011 at 17:41 | comment | added | Noam D. Elkies | @algori: For the generalized Cantor set where at the $n$-th stage you keep only $2^n$ intervals of length $r^n$ (for some positive $r \leq 1/2$), the $n$-th factor in the product is $\left(\exp((r^{n-1}-r^n)s)+1\right)/2$. [Cantor is $r=1/3$; the full interval is $r=1/2$ — check that the product then gives $(e^s-1)/s$ as it should.] It looks like your $t$ is $2r$, so you can take $r=t/2$ in the resulting product. | |
| Sep 14, 2011 at 17:11 | history | edited | algori | CC BY-SA 3.0 |
clarified the question
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| Sep 14, 2011 at 16:38 | comment | added | algori | Noam -- thanks! I would be perfectly happy with answers of that form, or with integrating exponentials instead of polynomials but I was wondering if you could elaborate a little. In particular, how does the integral depend on $t$? | |
| Sep 14, 2011 at 15:48 | comment | added | Noam D. Elkies | The uniform measure on $A_t$ is a countable convolution of distributions with 2-point support, so it's easy to integrate exponentials if you're willing to accept answers like $\prod_{n=1}^\infty \bigl((\exp(2s/3^n)+1)/2 \bigr)$ for the integral of $\exp sx$ on the Cantor set. In particular you can get such product formulas for Fourier components. You can also integrate $x^k$ by differentiating $k$ times and taking $s=0$, though the resulting formulas do get complicated as $k$ grows. | |
| Sep 14, 2011 at 15:13 | history | edited | algori | CC BY-SA 3.0 |
added 178 characters in body; edited tags; edited title
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| Sep 13, 2011 at 22:37 | comment | added | algori | Peter -- that's exactly the problem: I don't know how to compute the averages of polynomials (or any other functions) along $A_t$'s in the first place. | |
| Sep 13, 2011 at 22:08 | comment | added | Peter Luthy | If X is very nice, perhaps you can utilize Gaussian quadrature? For example, in a fixed interval of $\mathbb{R}$, by sampling n special points one can compute exactly the integrals of all polynomials of degree <2n. It seems like this might be generalizable to, say, rectangular regions in higher dimensions; it is computationally infeasible to do this in high dimensions, though, as it would require an absurd number of samplings. That is ultimately why Monte Carlo methods are used to do integrations in $\mathbb{R}^100$ | |
| Sep 13, 2011 at 20:23 | comment | added | algori | jc -- thanks! This is quite an interesting article (although it does not seem to directly address the question of the posting). | |
| Sep 13, 2011 at 19:19 | comment | added | j.c. | Perhaps the following AMS Notices article will be helpful? ams.org/notices/200511/fea-sloan.pdf | |
| Sep 13, 2011 at 18:21 | comment | added | algori | Steve -- the way I understand it, Monte Carlo integration computes the integral straight away, as the limit of a sequence of expectations. I was interested to know how one can express, in a simple example, a 1-dimensional integral in terms of lower dimensional integrals. | |
| Sep 13, 2011 at 17:27 | comment | added | Steve Huntsman | As you undoubtedly know, Monte Carlo/Las Vegas integration speaks to the randomized version of this problem. | |
| Sep 13, 2011 at 16:50 | history | asked | algori | CC BY-SA 3.0 |