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It must be proven that the solution of the integral equation $$f(x)=\int_{-\infty}^{+\infty} e^{-(x-t)^2} g(t)dt$$ is $$g(x)=\frac{1}{\sqrt{}\pi}\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{2^nn!} H_n(x)$$

where the $H_n(x)$ are the Hermite polynomials.

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What is $f$? $ $ – Jeff Dec 18 '11 at 16:51
It looks like a Fredholm equation of the first kind... the sort that's usually solved with a Fourier transform. – J. M. Dec 19 '11 at 3:12

You can start by plugging the series on the right side of the definition of $g(x)$, or rather $g(t)$, into the expression on the right side of $f(x)$. Interchange sum and integral, and you see what looks like the beginning of the Maclaurin series expansion for $f(x)$. (Your notation already assumes $f$ is infinitely differentiable.) It then suffices to show that $$ x^n=\frac{1}{2^n\sqrt{\pi}}\int_{-\infty}^\infty\exp(-(x-t)^2)H_n(t)\, dt, $$ as well as justifying the interchange of sum and integral.

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Could you please explain the above equation in more details? – neli Jan 1 '12 at 19:12
When I say 'It then suffices to show that...' I'm not claiming it's done, or even that there's a clear way to proceed. I'm saying this is the final step. This identity is true if the exercise is correct, and all known identities about Hermite polynomials will be found in reference books on special functions. – stopple Jan 3 '12 at 3:15

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