Fourier series: coefficients, even and odd extensions, convergence and Parseval

Worked solutions

Problem 1

1. $\sin3x\sin5x = \tfrac12(\cos2x - \cos8x)$The product-to-sum identity $\sin A\sin B = \tfrac12\big(\cos(A - B) - \cos(A + B)\big)$.
2. $\int_{-\pi}^{\pi}\cos kx\,dx = 0$ for $k \ne 0$Over a whole number of periods a cosine has as much area above the axis as below.
3. $\sin^24x = \tfrac12(1 - \cos8x)$The same identity with $A = B$: the $\cos 0 = 1$ term survives.
4. $\int_{-\pi}^{\pi}\tfrac12\,dx = \pi$Only the constant contributes.
5. $0$, $\pi$ and $0$The last integrand $\cos2x\sin7x$ is odd, so its integral over $[-\pi, \pi]$ is $0$ without any work. These three facts are why integrating against $\sin nx$ picks out $b_n$ alone.

Problem 2

1. $a_n = 0$ for every $n$$f$ is odd, so $f(x)\cos nx$ is odd and integrates to $0$; that includes $a_0$.
2. $b_n = \tfrac2\pi\int_0^\pi\sin nx\,dx$For an odd function, twice the integral over $[0, \pi]$.
3. $= \tfrac{2}{n\pi}\big(1 - \cos n\pi\big) = \tfrac{2}{n\pi}\big(1 - (-1)^n\big)$$\big[-\tfrac{\cos nx}{n}\big]_0^\pi$.
4. $b_n = \tfrac{4}{n\pi}$ for odd $n$, $b_n = 0$ for even $n$$1 - (-1)^n$ is $2$ or $0$.
5. $f(x) \sim \tfrac4\pi\big(\sin x + \tfrac13\sin3x + \tfrac15\sin5x + \cdots\big)$The coefficients shrink only like $1/n$, the mark of a jump. Near each jump the partial sums peak at about $1.179$ instead of $1$: an overshoot of about $0.179$, or $9\%$ of the jump of $2$, which moves closer to the jump but does not shrink as terms are added.

Problem 3

1. $a_n = 0$$x$ is odd.
2. $b_n = \tfrac2\pi\int_0^\pi x\sin nx\,dx$Twice the half-period integral.
3. $\int_0^\pi x\sin nx\,dx = \Big[-\tfrac{x\cos nx}{n}\Big]_0^\pi + \tfrac1n\int_0^\pi\cos nx\,dx$Parts with $u = x$, $dv = \sin nx\,dx$. The last integral is $0$.
4. $= -\tfrac{\pi(-1)^n}{n}$, so $b_n = \tfrac{2(-1)^{n+1}}{n}$$\cos n\pi = (-1)^n$.
5. $x \sim 2\big(\sin x - \tfrac12\sin2x + \tfrac13\sin3x - \cdots\big)$; it converges to $\tfrac\pi2$ at $x = \tfrac\pi2$ and to $0$ at $x = \pi$At $\pi$ the periodic extension jumps from $\pi$ down to $-\pi$, and the series takes the average, $0$: every term has $\sin n\pi = 0$ (Mistake 3). At $\pi/2$ the series reads $2(1 - \tfrac13 + \tfrac15 - \cdots) = \tfrac\pi2$, which is Leibniz's $1 - \tfrac13 + \tfrac15 - \cdots = \tfrac\pi4$.

Problem 4

1. $b_n = 0$ and $a_0 = \tfrac2\pi\int_0^\pi x\,dx = \pi$$\lvert x\rvert$ is even. The constant term is $\tfrac{a_0}{2} = \tfrac\pi2$, its average.
2. $a_n = \tfrac2\pi\int_0^\pi x\cos nx\,dx = \tfrac2\pi\Big[\tfrac{x\sin nx}{n} + \tfrac{\cos nx}{n^2}\Big]_0^\pi$Integration by parts.
3. $a_n = \tfrac{2\big((-1)^n - 1\big)}{\pi n^2}$The $\sin$ term vanishes at both ends; $\cos n\pi = (-1)^n$ and $\cos0 = 1$.
4. $a_n = -\tfrac{4}{\pi n^2}$ for odd $n$, $0$ for even $n$Mistaking $\cos n\pi$ for $0$ gives $-\tfrac{2}{\pi n^2}$ for every $n$ (Mistake 4).
5. $0 = \tfrac\pi2 - \tfrac4\pi\sum_{k\ge1}\tfrac{1}{(2k - 1)^2}$At $x = 0$ the function is continuous, so the series equals $\lvert0\rvert = 0$.
6. $\lvert x\rvert \sim \tfrac\pi2 - \tfrac4\pi\big(\cos x + \tfrac19\cos3x + \tfrac{1}{25}\cos5x + \cdots\big)$, and $\sum\tfrac{1}{(2k - 1)^2} = \tfrac{\pi^2}{8} \approx 1.2337$The coefficients fall like $1/n^2$, faster than the square wave's: $\lvert x\rvert$ has corners but no jumps.

Problem 5

1. $a_0 = \tfrac2\pi\int_0^\pi x^2\,dx = \tfrac{2\pi^2}{3}$Even function, so $b_n = 0$. The constant term is $\tfrac{a_0}{2} = \tfrac{\pi^2}{3}$.
2. $a_n = \tfrac2\pi\int_0^\pi x^2\cos nx\,dx = \tfrac{4(-1)^n}{n^2}$Integrate by parts twice; only the term $\tfrac{2x\cos nx}{n^2}$ survives at the ends.
3. $x^2 \sim \tfrac{\pi^2}{3} + 4\sum_{n\ge1}\tfrac{(-1)^n}{n^2}\cos nx$Assemble the series.
4. At $x = \pi$: $\pi^2 = \tfrac{\pi^2}{3} + 4\sum\tfrac{1}{n^2}$$(-1)^n\cos n\pi = 1$. The periodic extension of $x^2$ is continuous at $\pi$, so the series equals $\pi^2$ there.
5. At $x = 0$: $0 = \tfrac{\pi^2}{3} - 4\sum\tfrac{(-1)^{n+1}}{n^2}$$(-1)^n\cos0 = (-1)^n$.
6. $\sum\tfrac{1}{n^2} = \tfrac{\pi^2}{6} \approx 1.6449$ and $\sum\tfrac{(-1)^{n+1}}{n^2} = \tfrac{\pi^2}{12} \approx 0.8225$The Basel sum. Writing $a_0$ instead of $\tfrac{a_0}{2}$ as the constant term breaks both evaluations (Mistake 1).

Problem 6

1. $L = 2$, so the terms are $\cos\tfrac{n\pi x}{2}$ and $\sin\tfrac{n\pi x}{2}$The period is $2L = 4$.
2. $a_0 = \tfrac12\int_0^2 x\,dx = 1$$\tfrac1L\int_{-L}^{L}$, and $f = 0$ on the left half. The constant term $\tfrac{a_0}{2} = \tfrac12$ is the average of $f$ over a period.
3. $a_n = \tfrac12\int_0^2 x\cos\tfrac{n\pi x}{2}\,dx = \tfrac{2\big((-1)^n - 1\big)}{n^2\pi^2}$Parts with $k = \tfrac{n\pi}{2}$: $\big[\tfrac{x\sin kx}{k} + \tfrac{\cos kx}{k^2}\big]_0^2 = \tfrac{(-1)^n - 1}{k^2}$.
4. $b_n = \tfrac12\int_0^2 x\sin\tfrac{n\pi x}{2}\,dx = \tfrac{2(-1)^{n+1}}{n\pi}$$\tfrac12\big[-\tfrac{x\cos kx}{k}\big]_0^2 = -\tfrac{(-1)^n}{k}$; the $\sin kx$ term vanishes.
5. $f \sim \tfrac12 - \tfrac{4}{\pi^2}\sum_{n\text{ odd}}\tfrac{1}{n^2}\cos\tfrac{n\pi x}{2} + \tfrac2\pi\sum_{n\ge1}\tfrac{(-1)^{n+1}}{n}\sin\tfrac{n\pi x}{2}$The cosine terms are the even part of $f$, the sine terms the odd part.
6. At $x = 2$ the series converges to $1$$f$ jumps from $2$ down to $0$ there, and $\tfrac12(2 + 0) = 1$. Check: $\cos n\pi = -1$ for odd $n$ gives $\tfrac12 + \tfrac{4}{\pi^2}\cdot\tfrac{\pi^2}{8} = 1$, using Problem 4. Using $\tfrac1\pi$ in place of $\tfrac1L$ gives the wrong average (Mistake 2).

Problem 7

1. Extend $f$ to $[-\pi, 0]$ as an odd functionThen the series has sines only, and on $[0, \pi]$ it represents $f$ itself.
2. $b_n = \tfrac2\pi\int_0^\pi x(\pi - x)\sin nx\,dx$The odd-function formula.
3. $b_n = \tfrac{4\big(1 - (-1)^n\big)}{\pi n^3}$Parts three times, or split as $\pi x - x^2$ and reuse the integrals of Problems 3 and 5.
4. $b_n = \tfrac{8}{\pi n^3}$ for odd $n$, $0$ for even $n$The odd extension is smooth enough that its coefficients fall like $1/n^3$.
5. $\tfrac{\pi^2}{4} = \tfrac8\pi\big(1 - \tfrac{1}{3^3} + \tfrac{1}{5^3} - \cdots\big)$At $x = \tfrac\pi2$, $f = \tfrac{\pi^2}{4}$, and $\sin\tfrac{n\pi}{2}$ is $1, -1, 1, \dots$ for $n = 1, 3, 5, \dots$
6. $x(\pi - x) \sim \tfrac8\pi\big(\sin x + \tfrac{1}{27}\sin3x + \tfrac{1}{125}\sin5x + \cdots\big)$, and the sum is $\tfrac{\pi^3}{32} \approx 0.9689$The first term alone already has nearly the right height: its peak $\tfrac8\pi \approx 2.546$ is close to the true peak $\tfrac{\pi^2}{4} \approx 2.467$.

Problem 8

1. $\tfrac1\pi\int_{-\pi}^{\pi}x^4\,dx = \tfrac{2\pi^4}{5}$The left side of Parseval: $\tfrac1\pi\cdot\tfrac{2\pi^5}{5}$.
2. $\tfrac{a_0^2}{2} = \tfrac12\Big(\tfrac{2\pi^2}{3}\Big)^2 = \tfrac{2\pi^4}{9}$The constant term's share, with $a_0 = \tfrac{2\pi^2}{3}$.
3. $\sum a_n^2 = \sum\tfrac{16}{n^4}$$a_n = \tfrac{4(-1)^n}{n^2}$ and $b_n = 0$.
4. $16\sum\tfrac{1}{n^4} = \tfrac{2\pi^4}{5} - \tfrac{2\pi^4}{9} = \tfrac{8\pi^4}{45}$Solve for the sum.
5. $\sum\tfrac{1}{n^4} = \tfrac{\pi^4}{90} \approx 1.0823$Leaving out the $\tfrac{a_0^2}{2}$ term gives $\tfrac{\pi^4}{40} \approx 2.435$ (Mistake 5).

Problem 9

1. $c_n = \tfrac{1}{2\pi}\int_{-\pi}^{\pi}e^{(1 - in)x}\,dx$Combine $e^x$ and $e^{-inx}$.
2. $= \tfrac{e^{(1 - in)\pi} - e^{-(1 - in)\pi}}{2\pi(1 - in)}$The exponential rule with a complex constant.
3. $e^{\mp in\pi} = (-1)^n$So the numerator is $(-1)^n(e^\pi - e^{-\pi}) = 2(-1)^n\sinh\pi$.
4. $c_n = \tfrac{(-1)^n\sinh\pi}{\pi(1 - in)}$For every integer $n$, including $n = 0$, where $c_0 = \tfrac{\sinh\pi}{\pi}$ is the average of $e^x$.
5. $c_n + c_{-n} = \tfrac{(-1)^n\sinh\pi}{\pi}\cdot\tfrac{2}{1 + n^2}$At $x = 0$ every $e^{inx} = 1$; pair $n$ with $-n$, since $\tfrac{1}{1 - in} + \tfrac{1}{1 + in} = \tfrac{2}{1 + n^2}$.
6. $1 = \tfrac{\sinh\pi}{\pi}\Big(1 + 2\sum_{n\ge1}\tfrac{(-1)^n}{1 + n^2}\Big)$$e^x$ is continuous at $0$, so the series equals $e^0 = 1$.
7. $c_n = \tfrac{(-1)^n\sinh\pi}{\pi(1 - in)}$, and $\sum\tfrac{(-1)^n}{1 + n^2} = \tfrac12\big(\tfrac{\pi}{\sinh\pi} - 1\big) \approx -0.3640$At $x = \pi$ the series converges instead to $\cosh\pi$, the average of $e^{-\pi}$ and $e^\pi$.

Problem 10

1. Try $y = \sum_{n\text{ odd}}B_n\sin nx$The forcing has only odd sine terms, and the equation maps $\sin nx$ to a multiple of itself.
2. $(\sin nx)'' + 4\sin nx = (4 - n^2)\sin nx$Differentiate twice.
3. $(4 - n^2)B_n = \tfrac{4}{n\pi}$ for odd $n$Match the coefficients of $\sin nx$ on both sides.
4. $B_n = \tfrac{4}{n\pi(4 - n^2)}$No odd $n$ makes $4 - n^2 = 0$: the forcing has no $\sin2x$ term, so there is no resonance with the natural frequency $2$.
5. $B_1 = \tfrac{4}{3\pi} \approx 0.4244$, $B_3 = -\tfrac{4}{15\pi} \approx -0.0849$, $B_5 = -\tfrac{4}{105\pi} \approx -0.0121$$4 - n^2$ is $3$, $-5$, $-21$.
6. $y = \tfrac{4}{3\pi}\sin x - \tfrac{4}{15\pi}\sin3x - \tfrac{4}{105\pi}\sin5x - \cdots$The $B_n$ fall like $1/n^3$, so the response is far smoother than the square wave that drives it. Terms below the natural frequency stay in phase with the forcing; those above it come out with the opposite sign.