I've taken differential equations, but it's going to be a while until I'm going to be able to take a course in PDEs. However, they sound very interesting and I've been trying to study them in the meantime. The extended version of my ODEs textbook (Boyce 11e) actually does a bit of Fourier Series so I've been looking at that and messing around with some of the formulas. I think I found a "shortcut", but I'm not very sure what it means.
The formulas for the coefficients of a Fourier Series are given as follows $$a_n=\frac{1}{L}\int_{-L}^Lf(x)\cos\left(\frac{m\pi x}{L}\right)dx,\quad b_n=\frac{1}{L}\int_{-L}^Lf(x)\sin\left(\frac{m\pi x}{L}\right)dx$$ I noticed that you can use Euler's Formula to simplify this into one integral. $$\implies a_n+ib_n=\frac{1}{L}\int_{-L}^Lf(x)e^{i\frac{m\pi x}{L}}dx$$ You can additionally do a change of variables to simplify things further ($s=\frac{x}{L}$) $$a_n+ib_n=\int_{-1}^1f(Ls)e^{im\pi s}ds$$ Thus, you can just do this integral and the real part will be the coefficients for the cosine term and the imaginary part will be the coefficient for the sine term.
But then I noticed you can get the whole expression $a_m\cos\left(\frac{m\pi x}{L}\right)+b_m\sin\left(\frac{m\pi x}{L}\right)$ with $$a_m\cos\left(\frac{m\pi x}{L}\right)+b_m\sin\left(\frac{m\pi x}{L}\right)=\Re\left[\left(e^{-i\frac{m\pi x}{L}}\right)\left(a_m+ib_m\right)\right]$$ Making the whole Fourier series $$f(x)=\Re\left[\sum_{m=0}^\infty\left(e^{-i\frac{m\pi x}{L}}\int_{-1}^1f(Ls)e^{im\pi s}ds\right)\right]$$ or alternatively $$f(x)=\Re\left[\sum_{m=0}^\infty\left(\int_{-1}^1f(Ls)e^{im\pi\left(s-\frac{x}{L}\right)}ds\right)\right]$$ I don't know a lot about Fourier transforms, but the integral is looking kind of similar to that? Am I just tapping into something very obvious and standard? Is this how Fourier Series are usually presented and calculated? Is there something bigger going on here? And how can this be used to solve PDEs?
I know that's a lot of questions... but any help is appreciated!