Exercises — Fourier series — motivation from periodic functions

What we look for: a period $T$ so that every piece repeats. $\cos (3x)$ repeats when $3x$ advances by $2\pi$, i.e. every $2\pi /3$. $\sin (5x)$ repeats every $2\pi /5$. The whole sum repeats at the smallest common length that is a whole-number multiple of both.

$\frac{2\pi }{3}$ and $\frac{2\pi }{5}$ both divide $2\pi$ (since $3$ and $5$ are integers). So $==T=2\pi ==$.

The slowest wave is $\cos (3x)$; its angular frequency is $\omega =3$ (radians per unit $x$). The slowest wave has the smallest frequency number.

Test: replace $x$ by $-x$. $h(-x)=(-x{)}^2\cos (-x)=x^2\cos x=h(x)$. Even times even is even, and cosine is even, so $h$ is even.

An even function has all $b_n=0$ — only cosines survive. Why: $b_n$ integrates $h(x)\sin \frac{n\pi x}{L}$, which is even × odd = odd, and an odd integrand over the symmetric interval $[-L,L]$ integrates to $0$.

Symmetry first (why: saves half the work). $f(-x)=-x=-f(x)$, so $f$ is odd ⟹ all $a_n=0$. Only sines.

Here $L=\pi$, so the machine is $b_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }x\sin (nx)dx$. Odd × odd = even integrand, so double the right half: $b_n=\frac{2}{\pi }{\int }_0^{\pi }x\sin (nx)dx.$

Integrate by parts (why: a polynomial times a wave — parts peels off the $x$). With $u=x,dv=\sin (nx)dx$: ${\int }_0^{\pi }x\sin (nx)dx=[-\frac{x\cos (nx)}{n}{]}_0^{\pi }+\frac{1}{n}{\int }_0^{\pi }\cos (nx)dx=-\frac{\pi \cos (n\pi )}{n}+\frac{1}{n}\cdot \underset{=0}{\underbrace{[\frac{\sin (nx)}{n}{]}_0^{\pi }}}.$ Since $\cos (n\pi )=(-1{)}^n$: $b_n=\frac{2}{\pi }\cdot (-\frac{\pi (-1{)}^n}{n})=-\frac{2(-1{)}^n}{n}=\frac{2(-1{)}^{n+1}}{n}.$

Result: $x=2(\sin x-\frac{\sin 2x}{2}+\frac{\sin 3x}{3}-\dots )=2{\sum }_{n=1}^{\infty }\frac{(-1{)}^{n+1}}{n}\sin (nx).$

$f$ is even, so $b_n=0$; we only chase $a_n$. $L=\pi$.

$a_0=\frac{1}{\pi }{\int }_{-\pi }^{\pi }{x}^{2}dx=\frac{1}{\pi }\cdot \frac{2{\pi }^3}{3}=\frac{2{\pi }^2}{3}.$ So the constant term $a_0/2={\pi }^2/3$ = the average of $x^2$ over $(-\pi ,\pi )$. Check: it is positive, as it must be for $x^2$.

$a_1=\frac{1}{\pi }{\int }_{-\pi }^{\pi }{x}^2\cos xdx=\frac{2}{\pi }{\int }_0^{\pi }{x}^2\cos xdx.$ Parts twice (why: bring the power $x^2$ down to a constant). The standard result ${\int }_0^{\pi }{x}^2\cos xdx=2\pi \cos \pi =-2\pi$, so $a_1=\frac{2}{\pi }\cdot (-2\pi )=-4.$

At $x=0$ the square wave jumps from $-1$ (just left) to $+1$ (just right). Every term $\sin (n\cdot 0)=0$, so the series literally sums to $0$.

The jump rule says a Fourier series converges at a discontinuity to the midpoint $\frac{1}{2}[f(x^{+})+f(x^{-})]=\frac{1}{2}[(+1)+(-1)]=0$. ✔ The two predictions agree: the series can only give the average of the two sides, never one or the other. See Gibbs Phenomenon for the overshoot near (not at) this jump.

Why $x=\pi$: $x^2$ is continuous there (period-$2\pi$ extension matches: ${\pi }^2$ from left, ${\pi }^2$ from right), so the series equals $f(\pi )={\pi }^2$ exactly.

At $x=\pi$: $\cos (n\pi )=(-1{)}^n$, so $(-1{)}^n\cos (n\pi )=(-1{)}^{2n}=1$. Thus ${\pi }^2=\frac{{\pi }^2}{3}+{\sum }_{n=1}^{\infty }\frac{4}{n^2}.$ Rearrange: $\sum \frac{4}{n^2}={\pi }^2-\frac{{\pi }^2}{3}=\frac{2{\pi }^2}{3}$, giving the Basel sum ${\sum }_{n=1}^{\infty }\frac{1}{n^2}=\frac{{\pi }^2}{6}.$

The idea (half-range expansion). On $(0,\pi )$ you are free to invent the other half. To kill all sines and keep cosines, extend $f$ to be even: reflect $x$ across the $y$-axis so $f(-x)=f(x)=|x|$ on $(-\pi ,\pi )$. Now the interval is $(-\pi ,\pi )$, so $L=\pi$, and the even extension guarantees $b_n=0$.

$a_0=\frac{1}{\pi }{\int }_{-\pi }^{\pi }|x|dx=\frac{2}{\pi }{\int }_0^{\pi }xdx=\frac{2}{\pi }\cdot \frac{{\pi }^2}{2}=\pi .$ The constant term $a_0/2=\pi /2$ = average of $|x|$ on $(-\pi ,\pi )$. ✔ The figure shows the even reflection.

$a_n=\frac{2}{\pi }{\int }_0^{\pi }x\cos (nx)dx.$ Parts ($u=x,dv=\cos nxdx$): ${\int }_0^{\pi }x\cos nxdx=[\frac{x\sin nx}{n}{]}_0^{\pi }-\frac{1}{n}{\int }_0^{\pi }\sin nxdx=0-\frac{1}{n}[-\frac{\cos nx}{n}{]}_0^{\pi }=\frac{\cos n\pi -1}{n^2}=\frac{(-1{)}^n-1}{n^2}.$ So $a_n=\frac{2}{\pi }\cdot \frac{(-1{)}^n-1}{n^2}$, which is $0$ for even $n$ and $-\frac{4}{\pi n^2}$ for odd $n$. Final cosine series: $x=\frac{\pi }{2}-\frac{4}{\pi }(\cos x+\frac{\cos 3x}{9}+\frac{\cos 5x}{25}+\dots ),0<x<\pi .$

Tool choice: the product-to-sum identity $\sin A\cos B=\frac{1}{2}[\sin (A+B)+\sin (A-B)]$. Why this tool: it turns a product (hard to integrate) into a sum of single sines (each trivially integrable).

With $A=\frac{n\pi x}{L},B=\frac{m\pi x}{L}$: ${\int }_{-L}^L\sin \frac{n\pi x}{L}\cos \frac{m\pi x}{L}dx=\frac{1}{2}{\int }_{-L}^L[\sin \frac{(n+m)\pi x}{L}+\sin \frac{(n-m)\pi x}{L}]dx.$ Reason it dies: every $\sin (k\pi x/L)$ is an odd function of $x$, and the interval $[-L,L]$ is symmetric, so each integrates to $0$ — regardless of whether $n=m$, $n\ne m$, or either is zero. (Even the case $n=m$ gives $\sin (2n\pi x/L)+\sin 0$, both integrate to $0$.) Hence the mixed integral is always $0$. This is the pillar that lets sines and cosines never interfere with each other — the machinery of Orthogonality of Functions.

Square wave: $f(x)=\frac{4}{\pi }(\sin x+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}+\dots )$, and $f(\pi /2)=+1$ (continuous there — $\pi /2$ sits inside the "$+1$" block, not on a jump, so the series equals $f$ exactly).

Evaluate each sine at $x=\pi /2$: $\sin \frac{\pi }{2}=1,\sin \frac{3\pi }{2}=-1,\sin \frac{5\pi }{2}=1,\sin \frac{7\pi }{2}=-1,\dots$ So the odd terms alternate $+1,-1,+1,\dots$ with denominators $1,3,5,7,\dots$: $1=\frac{4}{\pi }(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots )⟹1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots =\frac{\pi }{4}.$ This is the Leibniz series for $\pi$, falling out of a square wave.

Why Parseval: it says total "energy" $\int f^2$ equals the sum of the squared amplitudes — the same orthogonality idea, now for the norm. Left side: $f^2=1$, so $\frac{1}{\pi }{\int }_{-\pi }^{\pi }1dx=\frac{1}{\pi }\cdot 2\pi =2$.

Right side: $b_n=\frac{4}{\pi n}$ for odd $n$, else $0$, so $b_n^2=\frac{16}{{\pi }^2{n}^2}$ (odd $n$). Thus $2={\sum }_{\text{odd }n}\frac{16}{{\pi }^2{n}^2}=\frac{16}{{\pi }^2}{\sum }_{\text{odd }n}\frac{1}{n^2}.$ Solve: ${\sum }_{\text{odd }n}\frac{1}{n^2}=\frac{2{\pi }^2}{16}=\frac{{\pi }^2}{8}.$ ✔ (And since ${\sum }_{\text{all}}\frac{1}{n^2}-{\sum }_{\text{even}}\frac{1}{n^2}=\frac{{\pi }^2}{6}-\frac{1}{4}\cdot \frac{{\pi }^2}{6}=\frac{{\pi }^2}{8}$, it is consistent with Basel from L3.2.)

These exercises rehearsed all four moves: (1) read symmetry to delete half the coefficients, (2) turn the crank with integration by parts, (3) evaluate at smart points (continuous vs jump) to harvest number identities, (4) reuse orthogonality both to isolate coefficients and to prove Parseval. The same modes reappear when you solve the Heat Equation and Wave Equation by Separation of Variables.