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I just started Fourier Series and I have three questions on Fourier transforms. I am incredibly lost on the subject and I feel like i'm missing something more fundamental. This is going to be a long one.

I tried to derive the formula(s) for the Fourier transform by using this page. The first problem I have is deriving (and understanding) the Rayleigh identity:

$ 2L\displaystyle\sum_{n=-\infty}^{\infty} |c_n|^2 = \displaystyle\int_{-L}^L |f(x)|^2 dx$

where $2L$ is a period, $c_n$ is the Fourier coefficient from $f(x)=\displaystyle\sum_{n=-\infty}^{\infty}c_ne^{i\dfrac{n\pi}{L}}$. I found this derivation but I have trouble understanding it. I get that the Fourier coefficient is the projection or inner product of the function and the exponential but I don't understand the inner product between the two summations. Shouldn't there be an integral of the product of those two sums? Plus, what is that little $\delta_{nm} $ thing? What's going on? Does it have to do with the fact that the conjugates 'cancel' out to one? But then why does all the $m$'s disappear and transform into $n$'s?

O.K so I just accepted the identity as true and I carried on the proof. I managed to reach the Fourier transformations but the page I linked before mentioned that sometimes the transformation are written with a $\dfrac{1}{\sqrt{2\pi}}$ coefficient instead of a $\dfrac{1}{2\pi}$. How is this possible? What allows this? Does it matter?

Lastly, what's the point of Fourier transformations? I'm a Physics major in a Quantum Mechanics I class and I'm told it's a 'continuous analogue' of the discrete Fourier series. Is this true? If so, this doesn't seem to justify the 'transform' part in the name but I have seen Fourier transformations applied to signals and that sort of stuff.

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  • $\begingroup$ If you know what is an orthonormal / unitary matrix, you should really look at the discrete Fourier transform first. The Parseval theorem then becomes $\sum_{k=0}^{N-1} |X_k|^2 = \sum_{k=0}^{N-1} |x_n|^2$ with $X_k = \frac{1}{\sqrt{N}} \sum_{n=0}^{N-1} x_n e^{-2 i \pi n k / N}$. After that, you should think to the Fourier series as the limiting case when $N \to \infty$ (all the proofs being very similar) $\endgroup$ – reuns Jun 7 '16 at 2:23
  • $\begingroup$ (And leave the Fourier transform out for now, you will see it later) Just for making everything clear : discrete Fourier transform : unitary operator $\mathbb{C}^N, \|.\|_2 \to \mathbb{C}^N, \|.\|_2$. Fourier series : unitary operator $L^2([-L/2,L/2]) \to l^2(\mathbb{Z}) \approx \lim_{N \to \infty} \mathbb{C}^N, \|.\|_2$. Fourier transform : unitary operator $L^2([-\infty,+\infty]) \to L^2([-\infty,+\infty])$. and there is also the Fourier transform of distributions, unifying all this. $\endgroup$ – reuns Jun 7 '16 at 2:31
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The inner product is defined as the integral over the product, so the integral is where you expected it to be, just hidden in the inner product notation.

$\delta_{nm}$ is the Kronecker delta, which is $1$ exactly if its indices are equal. That also answers your other question: The $m$'s disappear and transform into $n$'s because the Kronecker delta forces $m$ and $n$ to be equal, so performing the sum over $m$ leaves only the terms with $m=n$.

The different normalisations are a matter of convention. The product of the factors in front of the transform and the inverse transform has to be $\frac1{2\pi}$, and different conventions put this factor either all in the transform or all in the inverse transform or split it up symmetrically as $\frac1{\sqrt{2\pi}}$ in both transforms.

Yes, you can view the Fourier transform as a continuous analogue of the discrete Fourier series (with some important differences). The term "transform" here refers to an integral transform.

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