Fourier series practice problems with solutions

Solution to part 1:

Since \(f\) is odd, \(a_0 = 0\) and \(a_n = 0\) for all \(n \geq 1\). For \(b_n\):

\[b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx)\,dx = \frac{2}{\pi}\int_0^{\pi}\sin(nx)\,dx = \frac{2}{\pi}\left[-\frac{\cos(nx)}{n}\right]_0^{\pi}\]
\[= \frac{2}{n\pi}\bigl(1 – \cos(n\pi)\bigr) = \frac{2}{n\pi}\bigl(1 – (-1)^n\bigr)\]
Thus \(b_n = \dfrac{4}{n\pi}\) when \(n\) is odd, and \(b_n = 0\) when \(n\) is even.
Solution to part 2:
\[f(x) \sim \frac{4}{\pi}\sum_{k=0}^{\infty}\frac{\sin\!\bigl((2k+1)x\bigr)}{2k+1} = \frac{4}{\pi}\!\left(\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right)\]
The series contains only odd-indexed sine terms because \(f\) is odd (no cosines) and the even-indexed sines cancel due to the specific jump symmetry.

Solution to part 1:

Since \(f\) is odd, only sine terms survive. Using integration by parts:

\[b_n = \frac{2}{\pi}\int_0^{\pi} x\sin(nx)\,dx = \frac{2}{\pi}\!\left[-\frac{x\cos(nx)}{n} + \frac{\sin(nx)}{n^2}\right]_0^{\pi} = \frac{2(-1)^{n+1}}{n}\]
\[f(x) \sim 2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(nx) = 2\!\left(\sin x – \frac{\sin 2x}{2} + \frac{\sin 3x}{3} – \cdots\right)\]
Solution to part 2:

At \(x = \pi\) the periodic extension has a jump discontinuity: \(f(\pi^-) = \pi\) and \(f(\pi^+) = -\pi\). By the Dirichlet theorem the series converges to

\[\frac{f(\pi^-) + f(\pi^+)}{2} = \frac{\pi + (-\pi)}{2} = 0\]
which is consistent because every \(\sin(n\pi) = 0\).

Solution to part 1:
\[a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi} x^2\,dx = \frac{2\pi^2}{3}\]
\[a_n = \frac{2}{\pi}\int_0^{\pi} x^2\cos(nx)\,dx = \frac{4(-1)^n}{n^2}\]
\[f(x) \sim \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(nx)\]
Solution to part 2:

Since \(f\) is continuous everywhere on \([-\pi,\pi]\) and the periodic extension is continuous, the series converges to \(f(\pi) = \pi^2\) at \(x = \pi\):

\[\pi^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}(-1)^n = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{1}{n^2}\]
\[\Longrightarrow \quad \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}\]

Solution to part 1:
\[a_0 = \frac{1}{2}\int_{-2}^{2}f(x)\,dx = \frac{1}{2}\int_0^2 x\,dx = 1\]
\[a_n = \frac{1}{2}\int_0^2 x\cos\!\left(\frac{n\pi x}{2}\right)dx = \frac{2}{n^2\pi^2}\bigl((-1)^n – 1\bigr)\]
So \(a_n = -\dfrac{4}{n^2\pi^2}\) for odd \(n\), and \(a_n = 0\) for even \(n \geq 2\).
\[b_n = \frac{1}{2}\int_0^2 x\sin\!\left(\frac{n\pi x}{2}\right)dx = \frac{-2(-1)^n}{n\pi}\]
\[f(x) \sim \frac{1}{2} – \frac{4}{\pi^2}\sum_{k=0}^{\infty}\frac{\cos\!\left(\frac{(2k+1)\pi x}{2}\right)}{(2k+1)^2} + \sum_{n=1}^{\infty}\frac{-2(-1)^n}{n\pi}\sin\!\left(\frac{n\pi x}{2}\right)\]
Solution to part 2:

At \(x = 0\): the left limit (approaching from below in the periodic extension) gives \(f(0^-) = 0\) and \(f(0^+) = 0\), so the series converges to \(0\).

At \(x = 2\): the periodic extension has \(f(2^-) = 2\) and \(f(2^+) = 0\) (start of next period), so the series converges to \(\dfrac{2+0}{2} = 1\).

Solution to part 1:
\[b_n = \frac{2}{\pi}\int_0^{\pi} x\sin(nx)\,dx = \frac{2(-1)^{n+1}}{n}\]
\[f(x) \sim 2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(nx),\quad 0 < x < \pi\] Solution to part 2:

The odd periodic extension \(f_\text{odd}\) satisfies \(f_\text{odd}(-x) = -f_\text{odd}(x)\) and has period \(2\pi\). It has jump discontinuities at \(x = 0, \pm\pi, \pm 2\pi, \ldots\) where the one-sided limits are \(\pm\pi\). At each such point the series converges to \(0\) (the average of \(\pi\) and \(-\pi\)), while the original function values are \(0\) or \(\pi\).

Solution to part 1:
\[a_0 = 2\int_0^1 x^2\,dx = \frac{2}{3}, \qquad a_n = 2\int_0^1 x^2\cos(n\pi x)\,dx = \frac{4(-1)^n}{n^2\pi^2}\]
\[f(x) \sim \frac{1}{3} + \frac{4}{\pi^2}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(n\pi x), \quad 0 \leq x \leq 1\]
Solution to part 2:

At \(x = 0\) the even extension is continuous and equals \(f(0) = 0\):

\[0 = \frac{1}{3} + \frac{4}{\pi^2}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2} \implies \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2} = -\frac{\pi^2}{12}\]
\[\therefore \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2\pi^2} = \frac{1}{12}\]

Solution to part 1 (Sine series):
\[b_n = \frac{2}{\pi}\int_0^{\pi}(\pi – x)\sin(nx)\,dx = \frac{2}{n}\]
\[f(x) \sim 2\sum_{n=1}^{\infty}\frac{\sin(nx)}{n} = 2\!\left(\sin x + \frac{\sin 2x}{2} + \frac{\sin 3x}{3} + \cdots\right)\]
Solution to part 2 (Cosine series):
\[a_0 = \frac{2}{\pi}\int_0^{\pi}(\pi-x)\,dx = \pi,\qquad a_n = \frac{2}{\pi}\int_0^{\pi}(\pi-x)\cos(nx)\,dx = \frac{2}{n^2\pi}\bigl(1 – (-1)^n\bigr)\]
So \(a_n = \dfrac{4}{n^2\pi}\) for odd \(n\), and \(0\) for even \(n\).
\[f(x) \sim \frac{\pi}{2} + \frac{4}{\pi}\sum_{k=0}^{\infty}\frac{\cos\!\bigl((2k+1)x\bigr)}{(2k+1)^2}\]
Solution to part 3:

At \(x = \pi/2\) both extensions are continuous. Each series converges to \(f(\pi/2) = \pi – \pi/2 = \pi/2\). One may verify this for the cosine series by noting \(\cos\!\bigl((2k+1)\pi/2\bigr) = 0\) for all \(k\), leaving only the constant term \(\pi/2\). ✓

Solution to part 1:
\[\frac{1}{\pi}\int_{-\pi}^{\pi}|f(x)|^2\,dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty}\bigl(a_n^2 + b_n^2\bigr)\]
Solution to part 2:

For \(f(x) = x^2\): \(a_0 = \frac{2\pi^2}{3}\), \(a_n = \frac{4(-1)^n}{n^2}\), \(b_n = 0\).

\[\text{LHS} = \frac{1}{\pi}\int_{-\pi}^{\pi}x^4\,dx = \frac{2\pi^4}{5}\]
\[\text{RHS} = \frac{1}{2}\cdot\frac{4\pi^4}{9} + \sum_{n=1}^{\infty}\frac{16}{n^4} = \frac{2\pi^4}{9} + 16\sum_{n=1}^{\infty}\frac{1}{n^4}\]
Setting LHS = RHS:
\[\frac{2\pi^4}{5} = \frac{2\pi^4}{9} + 16\sum_{n=1}^{\infty}\frac{1}{n^4} \implies 16\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{2\pi^4}{5} – \frac{2\pi^4}{9} = \frac{8\pi^4}{45}\]
\[\therefore\quad \sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{\pi^4}{90}\]

Solution to part 1:
\[\frac{1}{\pi}\int_{-\pi}^{\pi}x^2\,dx = \frac{2\pi^2}{3}\]
\[\sum_{n=1}^{\infty} b_n^2 = \sum_{n=1}^{\infty}\frac{4}{n^2}\]
Parseval: \(\dfrac{2\pi^2}{3} = 4\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2} \implies \displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}\). ✓
Solution to part 2:

In signal processing, \(\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^2\,dx\) is the total power (energy per period) of the signal. The right-hand side \(\frac{a_0^2}{2} + \sum(a_n^2 + b_n^2)\) is the sum of the powers carried by each harmonic component. Parseval’s identity states that total power equals the sum of powers of all harmonics — energy is conserved in the Fourier representation.

Solution to part 1:
\[b_n = \frac{1}{2}\int_{-2}^{2}t\sin\!\left(\frac{n\pi t}{2}\right)dt = \frac{4(-1)^{n+1}}{n\pi}\]
\[f(t) \sim \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin\!\left(\frac{n\pi t}{2}\right)\]
Solution to part 2:

At \(t = 2\) the function has a jump: \(f(2^-) = 2\) and \(f(2^+) = -2\) (start of next period). By the Dirichlet theorem the series converges to \(\frac{2 + (-2)}{2} = 0\). Substituting \(t = 2\) confirms this since \(\sin(n\pi) = 0\) for all integers \(n\).

Solution to part 1:

Since period \(T = \pi\), the fundamental angular frequency is \(\omega_0 = 2\pi/T = 2\). All terms involve harmonics \(2, 4, 6, \ldots\)

\[a_0 = \frac{2}{\pi}\int_0^{\pi}\sin x\,dx = \frac{4}{\pi}\]
\[a_n = \frac{2}{\pi}\int_0^{\pi}\sin x\cos(2nx)\,dx = \frac{-4}{\pi(4n^2-1)}\]
\[f(x) \sim \frac{2}{\pi} – \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\cos(2nx)}{4n^2 – 1}\]
Solution to part 2:
\[\text{At } x = \tfrac{\pi}{2}: \quad \frac{2}{\pi} – \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\cos(n\pi)}{4n^2-1} = \frac{2}{\pi} – \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^n}{4n^2-1}\]
Using the known result \(\sum_{n=1}^{\infty}\frac{(-1)^n}{4n^2-1} = \frac{1-\pi/2}{2} \cdot \frac{2}{\pi}\), the sum evaluates to \(1 = f(\pi/2)\). ✓

Solution to part 1:
\[a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}e^x\,dx = \frac{2\sinh\pi}{\pi}\]
\[a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}e^x\cos(nx)\,dx = \frac{2(-1)^n\sinh\pi}{\pi(1+n^2)}\]
\[b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}e^x\sin(nx)\,dx = \frac{-2n(-1)^n\sinh\pi}{\pi(1+n^2)}\]
\[e^x \sim \frac{\sinh\pi}{\pi}\!\left(1 + 2\sum_{n=1}^{\infty}\frac{(-1)^n}{1+n^2}\bigl(\cos(nx) – n\sin(nx)\bigr)\right)\]
Solution to part 2:

Apply Parseval: \(\frac{1}{\pi}\int_{-\pi}^{\pi}e^{2x}\,dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty}(a_n^2 + b_n^2)\).

\[\text{LHS} = \frac{e^{2\pi} – e^{-2\pi}}{2\pi} = \frac{\sinh(2\pi)}{\pi}\]
\[a_n^2 + b_n^2 = \frac{4\sinh^2\!\pi}{\pi^2}\cdot\frac{1+n^2}{(1+n^2)^2} = \frac{4\sinh^2\!\pi}{\pi^2(1+n^2)}\]
\[\text{RHS} = \frac{2\sinh^2\!\pi}{\pi^2} + \frac{4\sinh^2\!\pi}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{1+n^2} = \frac{2\sinh^2\!\pi}{\pi^2}\!\left(1 + 2\sum_{n=1}^{\infty}\frac{1}{1+n^2}\right)\]
Setting equal and using \(\sinh(2\pi) = 2\sinh\pi\cosh\pi\):
\[\pi\coth\pi = 1 + 2\sum_{n=1}^{\infty}\frac{1}{1+n^2} = \sum_{n=-\infty}^{\infty}\frac{1}{1+n^2}\]

Solution to part 1:

For \(n = 0\): \(c_0 = \frac{1}{2\pi}\int_{-\delta}^{\delta}dx = \frac{\delta}{\pi}\).

For \(n \neq 0\):

\[c_n = \frac{1}{2\pi}\int_{-\delta}^{\delta}e^{-inx}dx = \frac{1}{2\pi}\left[\frac{e^{-inx}}{-in}\right]_{-\delta}^{\delta} = \frac{e^{in\delta} – e^{-in\delta}}{2\pi\,in} = \frac{\sin(n\delta)}{n\pi}\]
Solution to part 2:
\[c_n = \frac{\sin(n\delta)}{n\pi} = \frac{\delta}{\pi}\cdot\frac{\sin(n\delta)}{n\delta} = \frac{\delta}{\pi}\,\mathrm{sinc}(n\delta)\]
This expression is also valid at \(n = 0\) since \(\lim_{u\to 0}\mathrm{sinc}(u) = 1\), giving \(c_0 = \delta/\pi\). The coefficients therefore form a sampled sinc envelope, which is the hallmark of a rectangular pulse spectrum in Fourier analysis.

Solution to part 1:

The general conversion formulas are \(c_0 = \frac{a_0}{2}\), and for \(n \geq 1\):

\[c_n = \frac{a_n – ib_n}{2}, \qquad c_{-n} = \frac{a_n + ib_n}{2}\]
Here \(a_0 = 1\), \(a_n = 1/n\), \(b_n = 2/n^2\). Therefore:
\[c_0 = \frac{1}{2}, \qquad c_n = \frac{1}{2n} – \frac{i}{n^2}\quad (n \geq 1), \qquad c_{-n} = \frac{1}{2n} + \frac{i}{n^2}\quad (n \geq 1)\]
Solution to part 2:

Taking the complex conjugate of \(c_n\) for \(n \geq 1\):

\[\overline{c_n} = \overline{\frac{1}{2n} – \frac{i}{n^2}} = \frac{1}{2n} + \frac{i}{n^2} = c_{-n} \quad \checkmark\]
This conjugate symmetry \(c_{-n} = \overline{c_n}\) is a universal property of the complex Fourier coefficients of any real-valued function, since \(\overline{f(x)} = f(x)\) implies \(\overline{c_n} = c_{-n}\) directly from the coefficient formula.

Solution to part 1 (general solution):

Substituting \(u = X(x)T(t)\) gives \(X^{\prime\prime} + \lambda X = 0\) with \(X(0)=X(\pi)=0\), yielding:

\[X_n(x) = \sin(nx),\quad \lambda_n = n^2,\quad T_n(t) = e^{-n^2 t}\]
\[u(x,t) = \sum_{n=1}^{\infty} B_n e^{-n^2 t}\sin(nx)\]
Solution to part 2:

At \(t = 0\): \(u(x,0) = \sum_{n=1}^{\infty}B_n\sin(nx) = \sin(2x) – \frac{1}{2}\sin(5x)\).

By orthogonality, \(B_2 = 1\), \(B_5 = -\frac{1}{2}\), and all other \(B_n = 0\). Therefore:

\[u(x,t) = e^{-4t}\sin(2x) – \tfrac{1}{2}e^{-25t}\sin(5x)\]

As \(t \to \infty\), the higher-frequency term decays faster (factor \(e^{-25t}\) vs. \(e^{-4t}\)), leaving the rod temperature exponentially approaching zero, consistent with the zero boundary conditions.

Solution to part 1 (Fourier sine coefficients):
\[B_n = \frac{2}{\pi}\int_0^{\pi}x(\pi – x)\sin(nx)\,dx\]
Integrating by parts twice:
\[B_n = \frac{2}{\pi}\cdot\frac{2\bigl(1-(-1)^n\bigr)}{n^3} \cdot \pi = \frac{4(1-(-1)^n)}{n^3\pi}\cdot\pi\]
More precisely: \(B_n = \dfrac{8}{\pi n^3}\) for odd \(n\), and \(B_n = 0\) for even \(n\).
Solution to part 2:
\[u(x,t) = \frac{8}{\pi}\sum_{k=0}^{\infty}\frac{e^{-(2k+1)^2 t}\sin\!\bigl((2k+1)x\bigr)}{(2k+1)^3}\]
Solution to part 3:

At \(x = \pi/2\), \(\sin\!\bigl((2k+1)\pi/2\bigr) = (-1)^k\). As \(t \to \infty\) the \(k = 0\) term dominates:

\[u\!\left(\frac{\pi}{2}, t\right) \approx \frac{8}{\pi}e^{-t}\sin\!\left(\frac{\pi}{2}\right) = \frac{8}{\pi}e^{-t}\]
The dominant decay rate is \(e^{-t}\) (eigenvalue \(\lambda_1 = 1^2 = 1\)). Higher modes decay at rates \(e^{-9t}, e^{-25t}, \ldots\), becoming negligible quickly.

Solution to part 1:
\[u(x,t) = \sum_{n=1}^{\infty}\sin(nx)\bigl(A_n\cos(cnt) + B_n\sin(cnt)\bigr)\]
Solution to part 2 (\(c = 1\)):

From \(u_t(x,0) = 0\): \(B_n = 0\) for all \(n\). From \(u(x,0)\): \(A_1 = 1\), \(A_3 = \frac{1}{3}\), all other \(A_n = 0\). Thus:

\[u(x,t) = \sin x\cos t + \frac{1}{3}\sin(3x)\cos(3t)\]
Solution to part 3:

The fundamental frequency corresponds to \(n = 1\): angular frequency \(\omega_1 = c\cdot 1 = 1\,\text{rad/s}\), frequency \(f_1 = \frac{1}{2\pi}\,\text{Hz}\). The second harmonic present (note: \(n=2\) has zero coefficient so the next mode is \(n=3\)) has angular frequency \(\omega_3 = 3\,\text{rad/s}\), which is the third harmonic. The amplitude ratio of the two modes is \(3\!:\!1\).

Solution to part 1:
\[a_0 = \frac{1}{\pi}\int_0^{\pi}x\,dx = \frac{\pi}{2}\]
\[a_n = \frac{1}{\pi}\int_0^{\pi}x\cos(nx)\,dx = \frac{(-1)^n – 1}{n^2\pi}\]
\[b_n = \frac{1}{\pi}\int_0^{\pi}x\sin(nx)\,dx = \frac{(-1)^{n+1}}{n}\]
\[f(x) \sim \frac{\pi}{4} – \frac{2}{\pi}\sum_{k=0}^{\infty}\frac{\cos\!\bigl((2k+1)x\bigr)}{(2k+1)^2} + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(nx)\]
Solution to part 2:

At \(x = 0\): in the periodic extension, \(f(0^-) = 0\) (end of preceding period where \(f \to 0\)) and \(f(0^+) = 0\). Both limits are \(0\), so the series converges to \(\dfrac{0+0}{2} = 0\). (The constant term alone is \(\pi/4 \neq 0\), but the alternating cosine sum cancels it at \(x=0\).)

At \(x = \pi\): \(f(\pi^-) = \pi\) and \(f(\pi^+) = 0\) (next period starts). Series converges to \(\dfrac{\pi + 0}{2} = \dfrac{\pi}{2}\).

Solution to part 3:

At \(x = \pi/2\), the series converges to \(f(\pi/2) = \pi/2\):

\[\frac{\pi}{2} = \frac{\pi}{4} – \frac{2}{\pi}\sum_{k=0}^{\infty}\frac{\cos\!\bigl((2k+1)\pi/2\bigr)}{(2k+1)^2} + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin\!\left(\frac{n\pi}{2}\right)\]
Since \(\cos\!\bigl((2k+1)\pi/2\bigr) = 0\) for all \(k\), only the sine sum survives. The non-zero terms occur at odd \(n = 2k+1\):
\[\frac{\pi}{2} = \frac{\pi}{4} + \sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1} \implies \frac{\pi}{4} = \sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1} = 1 – \frac{1}{3} + \frac{1}{5} – \cdots \quad\checkmark\]

Solution to part 1:
\[S_N(x) = \frac{4}{\pi}\sum_{k=0}^{N}\frac{\sin\!\bigl((2k+1)x\bigr)}{2k+1} = \frac{4}{\pi}\!\left(\sin x + \frac{\sin 3x}{3} + \cdots + \frac{\sin((2N+1)x)}{2N+1}\right)\]
Solution to part 2:

Differentiating and using the closed form \(S_N'(x) = \frac{4}{\pi}\cdot\frac{\sin((2N+2)x)}{2\sin x}\) (derived via telescoping product), setting \(S_N'(x^*) = 0\) gives \(\sin((2N+2)x^*) = 0\), so \(x^* = \frac{\pi}{2N+2} = \frac{\pi}{2(N+1)}\). This is the first positive maximum of \(S_N\).

Solution to part 3:

Substituting \(x^* \approx \frac{\pi}{2N+2}\) into \(S_N\) and taking the Riemann sum limit as \(N \to \infty\):

\[\lim_{N\to\infty} S_N(x^*) = \frac{2}{\pi}\int_0^{\pi}\frac{\sin u}{u}\,du \approx \frac{2}{\pi}\cdot 1.8519 \approx 1.1789\]
The function value at the jump is \(f(0^+) = 1\), so the overshoot is \(1.1789 – 1 \approx 0.179\). As a fraction of the jump size (which is \(2\)), the relative overshoot is \(\approx 8.9\%\). This is the Gibbs constant: the partial sums of any Fourier series always overshoot by approximately \(9\%\) of the total jump at a discontinuity, regardless of how many terms are taken.