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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.