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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

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


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


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How would you like the convergence? – Jonas Teuwen Feb 17 '11 at 21:13
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. – joriki Feb 17 '11 at 21:41
@joriki yes you are correct, I have corrected the error. – chutsu Feb 17 '11 at 22:51
@JonasT what do you mean by convergence? I think I just need the most general explanation for the derivation will do. – chutsu Feb 17 '11 at 22:52
@chutsu: You left the $n$ standing in the lower limit of the integral. – joriki Feb 17 '11 at 22:53

There's a mistake. Sin formula is $\sin(\omega nt)=\frac{e^{i \omega nt}-e^{-i\omega nt}}{2i}$

so the serie become:

$ 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} $

Well, you call $C_{n}=\left(\frac{a_n}{2}+\frac{b_n}{2i}\right)$ and $C_{-n}=\left(\frac{a_n}{2}-\frac{b_n}{2i}\right)$ and obtain:

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

that can be re-written in

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

because $e^{i\omega0t}=1$ and you can rename $\frac{a_0}{2}$ in ${c_0}e^{i\omega0t}$. Notice that if f(t) is real $a_n$ and $b_n$ are reals. so $c_n=\overline c_{-n}$.

For Fourier Transform the question is a little more complicated, because Fourier Transform is an operator that you define in $L^1$, you extend it on $L^2$ and in temperate distributions.

Fourier serie is connected with a function $L^1$on a interval (that become periodic on $\mathbb{R}$) or an hyperrectangle on $\mathbb{R}^n$. Imagine you can extend this to all $R^n$ obtain Fourier transform. The factor $\frac{1}{2\pi}$ is arbitrary, there are other choices, but all they must be coherent with the choice of the factor of the inverse Fourier Transform. $\frac{1}{\sqrt {2\pi}}$ is the most natural, because it's very similar to normalized complex exponencials that you use in Fourier Serie for the finite interval. And this choice makes Fourier Transform an Unitary Operator, because $F^{+}=F^{-1}$.

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