Applying Plancheral's Theorem to Fourier Series With $e^{2\pi inx}$

Question

I'm working on a problem in Terence Tao's Analysis II book, where he has us show that the series $\frac{1}{3}+\sum_{n=1}^{\infty}\frac{4}{n^2\pi^2}\cos(2\pi nx)$ converges uniformly to $f(x)=(1-2x)^2$ on $[0,1)$. He then wants us to show that this implies $\sum_{n=1}^{\infty}\frac{1}{n^4}=\frac{\pi^4}{90}$, and hints that we should do this by expanding the cosine into exponentials and applying Plancheral's theorem.

I've done this and got the desired result, but in doing so I applied Plancheral's theorem to the series $\frac{1}{3}+\sum_{n=-\infty}^{\infty}\frac{2}{\pi^2n^2}e^{2\pi inx}$, i.e. set $\sum_{n=-\infty}^{\infty}(\frac{2}{n^2\pi^2})^2$, equal to $\int_0^1(1-2x)^4dx$, or $||f^2||_2^2$. But when Plancheral's theorem says that $\sum_{n=-\infty}^{\infty}|c_n|^2=||f^2||_2^2$, isn't this referring to the $c_n$ in the Fourier series $\sum_{n=-\infty}^{\infty}c_ne^{inx}$, not $\sum_{n=-\infty}^{\infty}c_ne^{2\pi inx}$?

I tried to define $y=2\pi x$ and then integrate $2\pi\int_0^1(1-\frac{y}{\pi})^4 dy$ instead (since $dy=2\pi dx$), but got a different answer.

Any explanation on where my reasoning is going wrong would be greatly appreciated!

Answer 1

If $y = 2\pi x$ then $dy = 2\pi dx \implies dx = \frac{dy}{2\pi}$ $\color{blue}{\text{You have multiplied, instead of dividing}}$

Also, the limits would be changed to $\color{blue}{0\to2\pi}$

Answer 2

If $f$ is an $L^{2}$ function on $[0,1]$ with period $1$ then $g(x)=f(\frac x {2\pi})$ defines a periodic function on $[0, 2\pi]$ with period $2\pi$. Plancherel Theorem on $[0, 2\pi]$ says $\frac 1 {2\pi}\int_0^{2\pi} |g(x)|^{2} dx=\sum\limits_{k=-\infty}^{\infty} |\hat {g} (n)|^{2}$ where $\hat {g} (n)=\frac 1 {2\pi}\int_0^{2\pi} g(x)e^{inx} dx$. Aplying the transformation $y= \frac x {2\pi}$ to both sides we get $\frac 1 {2\pi}\int_0^{2\pi} |f(x)|^{2} dx=\sum\limits_{k=-\infty}^{\infty} |\hat {f} (n)|^{2}$ where $\hat {f} (n)=\int_0^{1} f(x)e^{inx} dx$.