3
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So from the fourier series, we can simplify it further and use trig identities to get the following:

$$ f(t) = \frac{a_0}{2} + \sum^{\infty}_{n=1} \left(\frac{a_n}{2}+\frac{b_n}{2i}\right)e^{i n \omega t} + \left(\frac{a_n}{2}-\frac{b_n}{2i}\right)e^{-i n \omega t} $$

So how do you go from the above line to

$$ f(t) = \sum^{\infty}_{n=-\infty} C_n e^{in\omega t} $$

and

$$ f(t) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} F(\omega) e^{-i \omega t} d\omega $$

Thanks

  • 0
    How would you like the convergence?2011-02-17
  • 1
    Is the last sum over $n$ supposed to be an integral instead? The function notation $F(\omega)$, the $\mathrm{d}\omega$ and the non-occurrence of $n$ seem to indicate so.2011-02-17
  • 0
    @joriki yes you are correct, I have corrected the error.2011-02-17
  • 0
    @JonasT what do you mean by convergence? I think I just need the most general explanation for the derivation will do.2011-02-17
  • 0
    @chutsu: You left the $n$ standing in the lower limit of the integral.2011-02-17
  • 0
    @chutsu: Well, these things can converge in say $L^2$ or pointwise.2011-02-17

2 Answers 2