# A function as a sum of serie of modified FFT coeff. of another function - multiplied by sum number

I solve such a problem.

Lets have a function $Y=\sum_{k=-\infty}^\infty i\hat Y e^{ik\pi y}$ and then I have a function which is defined as $X=\sum_{k=-\infty}^\infty ik^2\hat Y e^{ik\pi y}$.

I know the $Y$. The $i$ is imaginary unit.

How can I compute the $X$? I think I do the FFT on $Y$ and obtain thus the $\hat Y$, right? And then I think I will do the backward FFT of function $f=ik^2\hat Y$. But what have I do with the summation index $k$ here in the $f$?

It is right that $FFT(ik^2\hat Y)=X$ ?

I'm not sure absolutely what to do with $k$ when the FFT sum is summated per $k$.

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It appears you are very close to your answer, so I am going to detail the process as a whole to clarify a couple small details. If this doesn't answer your question, please comment and I will modify my answer.

You know $Y$, so you evaluate it at a number of points over your interval and plug it into $FFT$. This produces your $\hat Y$, which is a vector. Let's denote each element in this vector as $\hat Y_k$.

You then have that $\hat X_k = i k^2 \hat Y_k$. $\hat X$ is a vector the same length as $\hat Y$. You then plug $\hat X_k$ back into your $FFT^{-1}$ to produce your X. It appears the only thing you were missing was representing $\hat Y$ as a vector. Note that the $i k^2$ can be any function of $k$ you want.

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ok, many thanks! but what plug I into f, only $\hat Y_k$? I have several functions derived from $Y$, they are different in the form, lets say $\hat X_k=ik^2\hat Y_k$, $\hat Z_k(j)=i(k^2+j^2)Y_k$ for some quantity $j$ and so one. Please what should I plug into FFT$^{-1}$? I mean how I say to the program that I need to use the summation number $k$ and how it should look like? Many thanks, I hope this is not too stupid question ;) – lovis Feb 11 '13 at 18:39
I updated the last paragraph slightly. Let me know if this helps or if you need more clarification. – Godric Seer Feb 11 '13 at 21:29
yes, of course! many thanks! – lovis Feb 12 '13 at 9:33

You may not realize this but you have, in fact, that $X(y)=-\frac1{\pi^2}Y''(y)$. If you know $Y(y)$ then it should be simple enough to compute $X(y)$.

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that I haven't realized - many thanks! – lovis Feb 12 '13 at 9:33