Fourier Series: Orthogonality & Parseval's Theorem
Fourier Series: Orthogonality & Parseval's Theorem
The Fourier series decomposes a periodic function into sinusoids $\{e^{inx}\}$ that form a complete orthonormal basis of $L^2[-\pi,\pi]$. The Fourier coefficients $\hat{f}(n)=\frac{1}{2\pi}\int f(x)e^{-inx}\,dx$ are projections onto these basis elements, and Parseval's theorem $\|f\|_2^2=\sum|\hat{f}(n)|^2$ expresses energy conservation — the total power equals the sum of power in each frequency component.
Fourier Coefficients & Series
For $f\in L^2[-\pi,\pi]$, the complex Fourier coefficients are $\hat{f}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}\,dx$ for $n\in\mathbb{Z}$. The Fourier series is $f(x)\sim\sum_{n=-\infty}^{\infty}\hat{f}(n)e^{inx}$. Real form: $f(x)\sim\frac{a_0}{2}+\sum_{n=1}^\infty[a_n\cos(nx)+b_n\sin(nx)]$ with $a_n=\frac{1}{\pi}\int f\cos(nx)$, $b_n=\frac{1}{\pi}\int f\sin(nx)$. Symmetry: $f$ even $\Rightarrow b_n=0$ (cosine series); $f$ odd $\Rightarrow a_n=0$ (sine series).
Parseval's Theorem
For $f\in L^2[-\pi,\pi]$: $$\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^2\,dx=\sum_{n=-\infty}^{\infty}|\hat{f}(n)|^2.$$ This is the Plancherel/Parseval identity: the Fourier map $f\mapsto(\hat{f}(n))$ is an isometric isomorphism from $L^2[-\pi,\pi]$ to $\ell^2(\mathbb{Z})$. Physically: total energy equals sum of spectral energies. The completeness of $\{e^{inx}\}$ follows: $f\perp e^{inx}$ for all $n$ implies $\hat{f}(n)=0$ for all $n$, hence $\|f\|=0$.
Example 1
Find the Fourier series of $f(x)=x$ on $(-\pi,\pi)$ and use Parseval to sum $\sum 1/n^2$.
Solution: $f$ is odd, so $a_n=0$. $b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}x\sin(nx)\,dx=\frac{2(-1)^{n+1}}{n}$ (integration by parts). So $x\sim 2\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\sin(nx)$. Parseval: $\|f\|^2=\frac{1}{2\pi}\int_{-\pi}^{\pi}x^2\,dx=\frac{\pi^2}{3}$. Right side: $\sum b_n^2/2=\sum\frac{2}{n^2}$. So $\frac{\pi^2}{3}=2\sum\frac{1}{n^2}$, giving $\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$ (Basel problem).
Example 2
Compute the Fourier series of the square wave $f(x)=\text{sgn}(x)$ on $(-\pi,\pi)$.
Solution: $f$ is odd, $a_n=0$. $b_n=\frac{2}{\pi}\int_0^{\pi}\sin(nx)\,dx=\frac{2(1-(-1)^n)}{n\pi}$. So $b_n=4/(n\pi)$ for odd $n$, 0 for even $n$. $f(x)\sim\frac{4}{\pi}\sum_{k=0}^\infty\frac{\sin((2k+1)x)}{2k+1}$. At $x=\pi/2$: $1=\frac{4}{\pi}(1-1/3+1/5-\cdots)$, giving Leibniz formula $\pi/4=1-1/3+1/5-\cdots$.
Practice
- Compute the Fourier series of $f(x)=x^2$ on $(-\pi,\pi)$ and evaluate $\sum_{n=1}^\infty 1/n^4$.
- Prove the Riemann-Lebesgue Lemma: $\hat{f}(n)\to 0$ as $|n|\to\infty$ for any $f\in L^1[-\pi,\pi]$.
- Show that $\{1,\cos x,\sin x,\cos 2x,\sin 2x,\ldots\}$ is orthogonal on $[-\pi,\pi]$ and find all norms.
- Prove that $|\hat{f}(n)|=O(|n|^{-k})$ if $f\in C^k$ with $f(-\pi)=f(\pi)$ (and similarly for derivatives).
Show Answer Key
1. $f(x)=x^2=\frac{\pi^2}{3}+\sum_{n=1}^\infty\frac{4(-1)^n}{n^2}\cos nx$. Setting $x=\pi$: $\pi^2=\frac{\pi^2}{3}+4\sum\frac{1}{n^2}$, so $\sum 1/n^2=\pi^2/6$. Using Parseval: $\frac{1}{\pi}\int_{-\pi}^\pi x^4dx=\frac{2\pi^4}{5}=\frac{\pi^4}{9}\cdot2+\sum\frac{16}{n^4}\cdot2$, giving $\sum 1/n^4=\pi^4/90$.
2. For $f\in L^1$: $|\hat{f}(n)|=|\frac{1}{2\pi}\int f(x)e^{-inx}dx|$. Integration by parts or density of smooth functions in $L^1$ shows this $\to0$ as $|n|\to\infty$.
3. Orthogonality: $\int_{-\pi}^\pi\cos mx\cos nx\,dx=\pi\delta_{mn}$ (for $m,n\ge1$), $\int\sin mx\sin nx\,dx=\pi\delta_{mn}$, $\int\cos mx\sin nx\,dx=0$, $\int 1\cdot 1\,dx=2\pi$. Norms: $\|1\|=\sqrt{2\pi}$, $\|\cos nx\|=\|\sin nx\|=\sqrt{\pi}$.
4. If $f\in C^k$ and periodic, integrate by parts $k$ times: $\hat{f}(n)=\frac{1}{(in)^k}\widehat{f^{(k)}}(n)$. Since $f^{(k)}\in L^2$, $\widehat{f^{(k)}}(n)$ is bounded, so $|\hat{f}(n)|=O(|n|^{-k})$.