# Theoretical Understanding of Fourier Series

I understand mathematically how it is derived, and how it works (ie how it is applied and the significance of Fourier Series in solving PDEs). What I want to know is why it works, theoretically and mathematically. I am unable to prove to myself exactly why a Fourier Series converges on an interval as opposed to a Taylor Series, which is only at a point. Can someone explain theoretically why a Fourier Series does what it does (or via proof)?

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Taylor series can converge on an interval, too. You may be thinking of the fact that any Taylor series always converges at its expansion point. –  BobaFret Oct 8 '12 at 0:48
I asked the similar question math.stackexchange.com/questions/183459, however it still have no response on why Fourier Series should only be found by intervals. –  doraemonpaul Oct 8 '12 at 1:30
In fact finding Fourier Series coefficients is only a type of summation kernel inversion. Some PDEs can be solved by Fourier Series only because their calculations meet the relative kernel, and does not mean that all linear PDEs using separation of variables to solve them must face Fourier Series. –  doraemonpaul Oct 8 '12 at 1:43
If you set $z=e^{2 \pi i t}$ then a Fourier series becomes a Laurent series on the unit circle. This might help to understand its convergence behaviour. –  WimC Oct 8 '12 at 2:05
I am not sure what you mean by "it works." Are you asking a question more along the lines of "why should we have picked these functions to do stuff with in the first place?" or "what are the analytic details behind everything converging properly?" They're very different questions. –  Qiaochu Yuan Oct 8 '12 at 2:26

I don't know the exact reason about Fourier Series coefficients should only be found piecewisely by finite intervals, but I know the principle of finding formulae of Fourier Series coefficients are only as simple as follows:

For $f(x)=\int_a^bF(u)K(u,x)~du$ or $f(x)=\sum\limits_{u=a}^bF(u)K(u,x)$ ,

Consider $f(x)K(v,x)=\int_a^bF(u)K(u,x)K(v,x)~du$ or $f(x)K(v,x)=\sum\limits_{u=a}^bF(u)K(u,x)K(v,x)$

$\int_p^qf(x)K(v,x)~dx=\int_p^q\int_a^bF(u)K(u,x)K(v,x)~du~dx$ or $\int_p^qf(x)K(v,x)~dx=\int_p^q\sum\limits_{u=a}^bF(u)K(u,x)K(v,x)~dx$ or $\sum\limits_{x=p}^qf(x)K(v,x)=\sum\limits_{x=p}^q\int_a^bF(u)K(u,x)K(v,x)~du$ or $\sum\limits_{x=p}^qf(x)K(v,x)=\sum\limits_{x=p}^q\sum\limits_{u=a}^bF(u)K(u,x)K(v,x)$

$\int_p^qf(x)K(v,x)~dx=\int_a^bF(u)\int_p^qK(u,x)K(v,x)~dx~du$ or $\int_p^qf(x)K(v,x)~dx=\sum\limits_{u=a}^bF(u)\int_p^qK(u,x)K(v,x)~dx$ or $\sum\limits_{x=p}^qf(x)K(v,x)=\int_a^bF(u)\sum\limits_{x=p}^qK(u,x)K(v,x)~du$ or $\sum\limits_{x=p}^qf(x)K(v,x)=\sum\limits_{u=a}^bF(u)\sum\limits_{x=p}^qK(u,x)K(v,x)$

If we can find $p$ and $q$ so that either $\int_p^qK(u,x)K(v,x)~dx=C$ or $\sum\limits_{x=p}^qK(u,x)K(v,x)=C$ of which $C\neq0$ and independent of $u$ and $v$ when $u=v$ under the integral range of $a$ and $b$ or the summation range of $a$ and $b$ ,

Then we can say that either $F(v)C=\int_p^qf(x)K(v,x)~dx$ or $F(v)C=\sum\limits_{x=p}^qf(x)K(v,x)$ holds for at least $x\in(p,q)$

i.e. either $F(v)=\dfrac{1}{C}\int_p^qf(x)K(v,x)~dx$ or $F(v)=\dfrac{1}{C}\sum\limits_{x=p}^qf(x)K(v,x)$ holds for at least $x\in(p,q)$

i.e. either $F(u)=\dfrac{1}{C}\int_p^qf(x)K(u,x)~dx$ or $F(u)=\dfrac{1}{C}\sum\limits_{x=p}^qf(x)K(u,x)$ holds for at least $x\in(p,q)$

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