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Anybody know of a clever way of finding the Fourier transform of $$ F(x) = P(x)^n e^{-\pi x^2}$$ very $n$ is fairly large integer. It is to be used as part of a computation, so I would rather not multiply out the polynomial and then find the Fourier transform of each of the terms individually, hence I thought perhaps there is a way to use that my polynomial has few roots with high multiplicity to get a closed expression where $n$ features in an manageble way - perhaps something akin to the way factorials occur when $P(x)=x^n$. Any thoughts will be welcome.

EDIT: The polynomial $P$ can be assumed to be either linear or quadratic - in case that simplifies.

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What do we know about $P$? –  Davide Giraudo Feb 12 '12 at 21:16
    
Here's a thought: by the convolution theorem, you have $$\mathcal{F}\left\{P(x)^{2n}e^{-\pi x^2}\right\}=\mathcal{F}\left\{P(x)^{n}e^{-\frac{1}{2}\pi x^2}\right\}*\mathcal{F}\left\{P(x)^{n}e^{-\frac{1}{2}\pi x^2}\right\}$$ where * means convolution. If convoluting is more efficient (I'm not sure this is the case) then you can calculate once the expression $$\mathcal{F}\left\{P(x)e^{-\frac{1}{k}\pi x^2}\right\}$$ and then convolving it $\sim\log_2(n)$ times –  yohBS Feb 12 '12 at 21:53
    
Interesting idea about the convolutions - though it will probably still be too slow. I might try it out none the less though. –  testcase Feb 13 '12 at 11:04
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Warning: I am probably going to mess up factors of $\pm i$, $\pi$, etc. So take what I say modulo these constants.

We know that if $f(x)$ is sufficiently nice, then integration by parts yields $\mathcal{F}\{x^n f(x)\} = i^n \frac{d^n}{dk^n}\hat{f}(k)$. Since the Fourier transform is linear, if $p(x)$ is a polynomial, then we can express $\mathcal{F}(p(x)f(x))$ as a polynomial in the derivatives of $\hat{f}(k)$. In your case, the function $f(x)$ is a Gaussian, hence has Gaussian Fourier transform, so the question is reduced to:

If $p$ is a polynomial in $i d/dk$, is there a closed form expression for $p^n e^{-k^2}$? Let $h_n(k) = e^{k^2} p^n e^{-k^2}$ Note that $h_n(k)$ is a polynomial in $k$, and we can find an explicit recurrence relation via \begin{align} h_{n+1}(k) &= e^{k^2} p p^n e^{-k^2} \\ &= e^{k^2} p ( e^{-k^2} h_n(k) ) \end{align}

For example, in the linear case we have $p = id/dk + a$, and we obtain \begin{align} h_{n+1}(k) &= e^{k^2} (id/dk + a)(e^{-k^2} h_n(k)) \\ &= e^{k^2}(-2ki e^{-k^2} h_n(k) + i e^{-k^2} h_n'(k) + a e^{-k^2} h_n(k) \\ &= -2ki h_n(k) + i h_n'(k) + ah_n(k) \\ &= (-2ki + a) h_n(k) + i h_n'(k). \end{align} Note that since the $h_n(k)$ are polynomials, all of this can be computed symbolically. Also note that setting $a=0$ recovers (up to constants and conventions) the usual Hermite polynomials. Actually I expect that in this case it should be easy to work out an explicit relation to the Hermite polynomials (by shifting $k$), but I'll leave that to you.

The quadratic case will certainly be much harder as you'll get a more complicated recurrence relation. It might be possible to find a generating function (as in the case of Hermite polynomials) and to use it to find a closed-form expression.

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