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Suppose I have a function $f(x)$ which has fourier coefficients $$f_m=\frac 1 d \int^d_0f(x)e^{im\frac {2\pi} dx}dx$$ Which in discrete form (fft) is: $$f_m=\frac 1 N\sum ^Nf_ie^{im2\pi x}$$ I then have a relation to the derivative: $$f'_m(x)=-i m 2\pi f_m$$

Now suppose I have a second function $C(x)$ where $$C(x)=\frac {f'(x)} {1+f'(x)^2}$$ I would like to know if I can find the fourier coefficients $C_m$ by plugging in the fourier coefficients of $f'(x)$ into $C(x)$

yankeefan11
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    I think you need to check your discrete fft, if the $i$ in the exponential is the imaginary unit then the summation index is missing and you need to divide the argument of the exponential by $N$. – Graham Hesketh Jul 23 '13 at 16:13
  • The coefficient is not a function of $x$, $x$ is integrated over. – Graham Hesketh Jul 23 '13 at 16:18

1 Answers1

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Let $g=f'$ to simplify the setup. The question is about the relation of the Fourier coefficients of $g$ and $C=g/(1+g^2)$. Since the transform from $g$ to $C$ is nonlinear, there is no clear relation between the coefficients. For example, if $g(x)=\cos x$, then the only nonzero (cosine) coefficient of $g$ is $a_1=1$ while the expansion of $C$ is $$ (2-\sqrt{2})\cos x +(7\sqrt{2}-10)\cos 3x+ (58-41\sqrt{2}) \cos 5x +(239\sqrt{2}-338)\cos 7x +\dots $$ Clearly, you can't get these coefficients by manipulating with $a_1=1$. Other than reconstructing $g$ from its Fourier coefficients, and expanding $g/(1+g^2)$ as usual.

But one can give some qualitative statements, such as: if the coefficients of $g$ are $O(n^{-k-\epsilon})$ for some integer $k$ and $\epsilon>0$, then the coefficients of $C$ are $O(n^{1-k})$. Indeed, $\hat g_k=O(n^{-k-\epsilon})$ implies $g$ is $(k-1)$ times continuously differentiable, consequently, so is $C$, and the claim follows.

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