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I'd like to know whether Fourier coefficients (and series) of $f(x):\mathbb{R} \to \mathbb{C}$ and $f(x+c)$ for $c \in \mathbb{R}$ are the same and if so, why?

I have this question where $f$ is a $ 2 \pi$ periodic and Continuously differential which it's Fourier coefficient is $3^{-n^2}$, I need to compute $g$'s Fourier coefficient, where g(x)=\pi f'(x+2011). eventually I got that $\hat{g(n)}=\pi i n \frac{1}{2 \pi}\int_{0}^{2\pi}e^{-inx}f(x+2011 )$ and I would like to say that $\hat{g(n)}=\pi i n\hat{f(n)}=\pi i n3^{-n^2}$, I'm just not sure if it is correct. Thanks a lot.

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Try to think: you probably know that if $f$ is a $C^1$, $2\pi$-periodic function, its Fourier series converges uniformly to $f$, hence if the coefficients were the same you'd have the same series and the same function at the end. So unless $f$ is 2012-periodic (and since $f$ is also $2\pi$ periodic it can only happen in one case: which one?), then no the coefficients are not the same.

As for computing the actual coefficients, try as AD. suggests.

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No not really.

Hints:

  1. Change the variable of integration in your expression for $\hat{g}(n)$.

  2. Use an exp-rule.

  3. What happens to the "area of integration" - perhaps you can you split it up and use periodicity?

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    @KannappanSampath Thank you Kannappan, I rewrote my answer. :)2012-02-05