Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
What is $f$? $ $ –  Jeff Dec 18 '11 at 16:51
3  
Please don't use the [homework] tag on its own, add some information about what is topic of the question. –  Asaf Karagila Dec 18 '11 at 16:54
    
$$f$$is f(x)function. –  Tera Dec 18 '11 at 17:52
1  
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

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.