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Consider the periodic function with period 2 given by

$$ f(x) = 2x, 0 \leq x \leq 1 $$ $$f(x) = 2x -4, 0 < x \leq 2$$

If c_k denote the k-th complex fourier coefficient, we know, using the derivative property, that

$$\bar{c_k} = (i k \omega) c_k$$

But how to use this, in this case, since the derivative is always the constant 2?

Thanks!

(P.S: $c_k = \frac{1}{T} \int_{0}^T f(x) exp(-i k \omega x) dx$)

@Edit

Using the delta function, I think that the correct value of the derivative should be

$$ 2 -4 \sum_{k=- \infty}^\infty \delta (x - 2k+1)$$

Giiovanna
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    $f$ is not continuous, it is not differentiable at $1$. If you consider only classical derivatives, you can't use the derivative property here. If you know how to deal with distributional derivatives, note that you get a multiple of the Dirac measure $\delta_1$ in the derivative. – Daniel Fischer May 07 '15 at 11:50
  • I've noted this after I posted. I think derivative should be a sequence of diracs plus a constant. But I dont know how to deal with it – Giiovanna May 07 '15 at 12:01
  • @DanielFischer if you could help me finding the derivative I would be glad – Giiovanna May 07 '15 at 12:05
  • What are the Fourier coefficients of a Dirac measure $\delta_p$? – Daniel Fischer May 07 '15 at 12:12
  • For a sequence of diracts with area A, it is A/T, where T is the fundamental period – Giiovanna May 07 '15 at 12:21
  • Not sure what "with area $A$" means there. Anyway, we get the $k$-th Fourier coefficient of a (sufficiently nice) function $g$ by integrating $\frac{1}{T} e^{-ik\omega x}$ against $g$ - that is, $$\hat{g}(k) = \int_0^T g(x) \biggl(\frac{1}{T} e^{-ik\omega x}\biggr),dx.$$ So generally, for $T$-periodic distributions, the natural way to define the Fourier coefficients is $$\hat{u}(k) = \frac{1}{T}u\bigl[e^{-ik\omega x}\bigr].$$ – Daniel Fischer May 07 '15 at 13:12

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