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Introduction to Linear Algebra with Mathematica

 

Examples of Fourier Series


This section is a collection of Fourier series expansions for different functions.

Theorem: If a periodic function of period \( 2\ell \) is square-integrable on any finite interval, then the Fourier series converges to the function at almost every point:

\[ f(x) \,\sim \, \frac{a_0}{2} + \sum_{k\ge 0} \left[ a_k \cos \left( \frac{k \pi x}{\ell} \right) + b_k \sin \left( \frac{k \pi x}{\ell} \right) \right] , \]
where its coefficients are determined via Euler--Fourier formula:
\[ \begin{split} a_k &= \frac{1}{\ell} \, \int_{-\ell}^{\ell} f(x)\,\cos \left( \frac{k \pi x}{\ell} \right) {\text d} x , \quad k=0,1,2,\ldots , \\ b_k &= \frac{1}{\ell} \, \int_{-\ell}^{\ell} f(x)\,\sin \left( \frac{k \pi x}{\ell} \right) {\text d} x , \quad k=1,2,\ldots . \end{split} \qquad\qquad ■ \]

Expansion in a complex form (or exponential form):
\begin{equation} \label{EqFourier.1} f(x) \,\sim\, \mbox{P.V.}\sum_{n=-\infty}^{\infty} \hat{f}(n)\, e^{n{\bf j} \pi x/\ell} = \lim_{N\to \infty} \sum_{n=-N}^{N} \hat{f}(n)\, e^{n{\bf j} \pi x/\ell} , \end{equation}
where T = 2ℓ is the period and the Fourier coefficients \( \hat{f}(n) \) are evaluated according to the Euler--Fourier formula:
\begin{equation} \label{EqFourier.2} \hat{f}(n) = \frac{1}{2\ell} \int_{-\ell}^{\ell} f(x)\, e^{-n{\bf j} \pi x/\ell} \,{\text d} x = \frac{1}{T} \int_{0}^{T} f(x)\, e^{-2n{\bf j} \pi x/T} \,{\text d} x , \qquad n \in \mathbb{Z} = \left\{ 0, \pm 1, \pm 2, \ldots \right\} . \end{equation}
Here «P.V.» abbreviates the Cauchy principle value, which is a regularization of the infinite sum.

The first examples are based on well-known sum of geometric series

\[ \frac{1}{2} + z + z^2 + z^3 + \cdots = \frac{1}{2} \cdot \frac{1+z}{1-z} , \]
where \( z = r\,e^{{\bf j}x} . \) Extraction of real and imaginary parts yields the following series:
\[ P(r, x) = \frac{1}{2} + \sum_{n\ge 1} r^n \cos (nx) = \mbox{P.V.} \sum_{n = -\infty}^{\infty} r^{|n|} e^{{\bf j}nx} = \frac{1}{2} \cdot \frac{1 + r^2}{1- 2r\,\cos x + r^2} , \]
\[ Q(r, x) =\sum_{n\ge 1} r^n \sin (nx) = \frac{1}{2} \cdot \frac{r\, \sin x}{1- 2r\,\cos x + r^2} . \]
The function P(r, x) is called the Poisson kernel.

Similarly, from the formula

\[ \ln \frac{1}{1-z} = z + \frac{1}{2}\, z^2 + \frac{1}{3}\, z^3 + \frac{1}{4}\, z^4 + \cdots \qquad (0 \le r < 1), \]
we get
\[ \sum_{n\ge 1} \frac{\cos nx}{n}\,r^n = \frac{1}{2} \cdot \ln \frac{1}{1 - 2r\,\cos x + r^2} , \qquad \sum_{n\ge 1} \frac{\sin nx}{n}\, r^n = \arctan \frac{r\,\sin x}{1 - 2r\,\cos x + r^2} . \]

From these sums, we derive

\begin{eqnarray*} \sum_{n\ge 1} \frac{1}{n} \, \cos n x &=& -\frac{1}{2} \, \ln \left[ 2 \left( 1 - \cos x \right) \right] , \qquad \mbox{on interval } \ 0 < x < 2\pi ; \\ \sum_{n\ge 1} \frac{1}{n} \, \sin n x &=& \begin{cases} \phantom{-}\frac{\pi -x}{2} , & \ \mbox{for } 0< x < \pi , \\ - \frac{\pi +x}{2} , & \ \mbox{for } -\pi < x < 0 , \end{cases} \qquad \mbox{on interval } \ -\pi < x < \pi . \\ \sum_{n\ge 1} \frac{1}{n} \, \sin \frac{n\pi x}{\ell} &=& \frac{\ell-x}{2} , \qquad 0 < x < 2\ell . \\ \sum_{n\ge 1} \frac{(-1)^{n+1}}{n} \, \cos n x &=& \ln \left( 2 \, \cos \frac{x}{2} \right) , \qquad \mbox{on interval } \ |x| < \pi ; \\ \sum_{n\ge 1} \frac{(-1)^{n+1}}{n} \, \sin n x &=& \frac{x}{2} , \qquad \mbox{on interval } \ |x| < \pi . \end{eqnarray*}
\[ \frac{1}{2{\bf j}}\,\sum_{n\ne 0} \frac{e^{{\bf j}nx}}{n} = \begin{cases} - \frac{\pi}{2} - \frac{x}{2} , & \ \mbox{ if} \quad -\pi < x < 0 , \\ 0 , & \ \mbox{ if} \quad x = 0, \\ \frac{\pi}{2} - \frac{x}{2} , & \ \mbox{ if} \quad 0 < x < \pi . \end{cases} \]
S1[x_] = Sum[1/n*Cos[n*x] , {n, 1, 100}];
p1 = Plot[{S1[t]}, {t, 0, 2*\[Pi]}, PlotStyle -> {Thickness[0.01], Orange}];
p2 = Plot[{ -1/2 Log[2*(1 - Cos[t])]}, {t, 0, 2*\[Pi]}, PlotStyle -> {Thickness[0.007], Plue}];
Show[p1, p2]
S2[x_] = Sum[1/n*Sin[n*x], {n, 1, 100}];
Plot[{S2[t], Piecewise[{{(\[Pi] - t)/2, 0 < t < \[Pi]}, {-(\[Pi] + t)/ 2, -\[Pi] < t < 0}}]}, {t, -\[Pi], \[Pi]}, PlotStyle -> Thickness[0.01]]
S3[x_] = Sum[1/n*Sin[(n*\[Pi]*x)/l], {n, 1, 100}];
Manipulate[ Plot[{S3[t], (l - t)/2}, {t, 0, 20}, PlotStyle -> Thickness[0.01]], {l, 0, 20, Appearance -> "Labeled"}]
S4[x_] = Sum[(-1)^(n + 1)/n*Cos[n*x], {n, 1, 100}];
p41 = Plot[{S4[t], Log[2*Cos[t/2]]}, {t, -\[Pi], \[Pi]}, PlotStyle -> {Thickness[0.02], Blue}];
p42 = Plot[{S4[t], Log[2*Cos[t/2]]}, {t, -\[Pi], \[Pi]}, PlotStyle -> Thickness[0.007]];
Show[p41, p42]
S6[x_] = 1/(2 \[ImaginaryJ]) Sum[E^(\[ImaginaryJ]*n*x)/n, {n, 1, 100}];
p61 = Plot[ Piecewise[{{(-\[Pi] - t)/2, -\[Pi] < t < 0}, {0, t = 0}, {(\[Pi] - t)/2, 0 < t < \[Pi]}}], {t, -\[Pi], \[Pi]}, PlotStyle -> {Thickness[0.01], Blue}];
p62 = Plot[{S6[t]}, {t, -\[Pi], \[Pi]}, PlotStyle -> {Thickness[0.02], Orange}];
Show[p61, p62]
\begin{eqnarray*} \sum_{n\ge 1} \frac{1}{2n-1} \, \cos (2n-1) x &=& \frac{1}{2} \cdot \ln \left( \cot \frac{x}{2} \right) , \qquad \mbox{on interval } \ 0 < x < \pi . \\ \sum_{n\ge 1} \frac{1}{2n-1} \, \sin (2n-1) x &=& \frac{1}{2} \sum_{\nu =-\infty}^{+\infty} \frac{e^{{\bf j}\left( 2\nu -1 \right) x}}{{\bf j} \left( 2\nu -1 \right)} = \frac{\pi}{4} \cdot \mbox{sign} (x) = \frac{\pi}{4} \times \begin{cases} \phantom{-}1 , & \ \mbox{for } 0< x , \\ - 1 , & \ \mbox{for } x<0 , \end{cases} \qquad \mbox{on interval } \ -\pi < x < \pi . \\ \frac{4}{\pi}\,\sum_{n\ge 1} \frac{1}{2n-1} \, \sin \frac{(2n-1)\pi x}{\ell} &=& 2 \left[ H\left( \frac{x}{\ell} \right) - H\left( \frac{x}{\ell} -1 \right) \right] -1 . \\ \sum_{n\ge 1} \frac{(-1)^{n+1}}{2n-1} \, \cos (2n-1) x &=& \begin{cases} \phantom{-}\frac{\pi}{4} , & \ \mbox{for } |x|< \frac{\pi}{2} , \\ -\frac{\pi}{4} , & \ \mbox{for } |x|> \frac{\pi}{2} , \end{cases} \qquad \mbox{on interval } \ -\pi < x < \pi . \\ \sum_{n\ge 1} \frac{(-1)^{n+1}}{2n-1} \, \sin (2n-1) x &=& \frac{1}{2} \,\ln \left\vert \cot \left( \frac{x}{2} - \frac{\pi}{4} \right) \right\vert , \qquad \mbox{on interval } \ -\pi < x < \pi . \end{eqnarray*}
S7[x_] = Sum[1/(2*n - 1)*Cos[(2*n - 1)*x], {n, 1, 100}];
p71 = Plot[{S7[t]}, {t, 0, \[Pi]}, PlotStyle -> {Thickness[0.02], Purple}];
p72 = Plot[{1/2*Log[Cot[t/2]]}, {t, 0, \[Pi]}, PlotStyle -> {Thickness[0.01], Green}];
Show[p71, p72]
S8[x_] = Sum[1/(2*n - 1)*Sin[(2*n - 1)*x], {n, 1, 100}];
S81 = Plot[\[Pi]/4*Sign[t], {t, -\[Pi], \[Pi]}, PlotStyle -> {Thickness[0.007], Blue}];
S82 = Plot[{S8[t]}, {t, -\[Pi], \[Pi]}, PlotStyle -> {Thickness[0.007], Orange}];
Show[S82, S81]
Manipulate[ Plot[{4/\[Pi]*Sum[Sin[((2 n - 1)*\[Pi]*t)/l]/(2*n - 1), {n, 1, 100}], 2*(HeavisideTheta[t/l] - HeavisideTheta[t/l - 1]) - 1}, {t, -10, 10}, PlotStyle -> {{Thickness[0.01], Orange}, {Thickness[0.01], Blue}}], {l, 1, 5}]
S10[x_] = Sum[(-1)^(n + 1)/(2 n - 1)*Cos[(2 n - 1)*x], {n, 1, 100}];
p101 = Plot[S10[t], {t, -\[Pi], \[Pi]}, PlotStyle -> {Thickness[0.015], Orange}];
p102 = Plot[{Piecewise[{{\[Pi]/4, Abs[t] < \[Pi]/2}, {-\[Pi]/4, Abs[t] > \[Pi]/2}}]}, {t, -\[Pi], \[Pi]}, PlotStyle -> {Thickness[0.007], Blue}];
Show[p101, p102]
S11[x_] = Sum[(-1)^(n + 1)/(2 n - 1)*Sin[(2 n - 1)*x], {n, 1, 100}];
p111 = Plot[{S11[t]}, {t, -\[Pi], \[Pi]}, PlotStyle -> {Thickness[0.02], Purple}];
p112 = Plot[Log[Cot[t/2 - Pi/4]]/2, {t, -\[Pi], \[Pi]}, PlotStyle -> {Thickness[0.007], Blue}];
Show[p111, p112]
\begin{eqnarray*} \frac{1}{2} + \sum_{n\ge 0} \frac{(-1)^n}{1+n^2} \left( \cos nx -n\,\sin nx \right) &=& \frac{\pi}{2\,\sinh \pi}\, e^x , \qquad \mbox{on interval } \ |x|< \pi . \\ \frac{2}{\pi} - \frac{4}{\pi} \,\sum_{n\ge 1} \frac{1}{4n^2 -1} \, \cos 2nx &=& \sin x , \qquad \mbox{on interval } \ 0\le x < \pi . \end{eqnarray*}
S12[x_] = 0.5 + Sum[(-1)^n/(1 + n^2)*(Cos[n*x] - n*Sin[n*x]), {n, 1, 100}];
Plot[{S12[t], (\[Pi]*E^t)/(2*Sinh[\[Pi]])}, {t, -\[Pi], \[Pi] + 0.2}, PlotStyle -> {{Thickness[0.01], Orange}, {Thickness[0.01], Blue}}]
S13[x_] = 2/\[Pi] - 4/\[Pi]*Sum[1/(4*n^2 - 1)*Cos[2*n*x], {n, 1, 100}];
p131 = Plot[{S13[t]}, {t, -0.2, \[Pi] + 0.2}, PlotStyle -> {Thickness[0.02], Orange}];
p132 = Plot[{Sin[t]}, {t, -0.2, \[Pi] + 0.2}, PlotStyle -> {Thickness[0.007], Blue}];
Show[p131, p132]
\begin{eqnarray*} \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^2} \, \cos n x &=& - \frac{x^2}{4} + \frac{\pi^2}{12} , \qquad \mbox{on interval } \ |x| < \pi . \\ \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^2} \, \sin n x &=& ????, \qquad \mbox{on interval } \ |x| < \pi . \\ \frac{\pi^2}{12} + \sum_{n\ge 1} \frac{1}{n^2} \, \cos n x &=& \frac{(x-\pi )^2}{4} , \qquad \mbox{on interval } \ 0< x < 2\pi . \\ \sum_{n\ge 1} \frac{1}{n^2} \, \sin n x &=& 1.63498\,\sin x , \qquad \mbox{on interval } \ |x| < \pi . \end{eqnarray*}
S14[x_] = Sum[(-1)^(n + 1)/n^2*Cos[n*x], {n, 1, 100}];
p141 = Plot[{S14[t]}, {t, -\[Pi], \[Pi] + 0.2}, PlotStyle -> {Thickness[0.02], Orange}];
p142 = Plot[{-(0.5 t)^2 + \[Pi]^2 /12}, {t, -\[Pi], \[Pi] + 0.2}, PlotStyle -> {Thickness[0.007], Blue}];
Show[p141, p142]
S15[x_] = Sum[(-1)^(n + 1)/n^2*Sin[n*x], {n, 1, 100}];
Plot[{S15[t] }, {t, 0, \[Pi]}, PlotStyle -> {Thickness[0.01], Orange}]
S16[x_] = Pi^2/12 + Sum[Cos[n*x]/n^2, {n, 1, 20}] ;
Plot[{S16[t], (Pi - t)^2/4}, {t, 0, 2*Pi}, PlotStyle -> Thickness[0.006]]

S16[x_] = \[Pi]^2/12 + Sum[1/n^2*Cos[n*x], {n, 1, 100}];
p161 = Plot[{S16[t]}, {t, 0, 2*\[Pi]+ 0.3}, PlotStyle -> {Thickness[0.02], Orange}];
p162 = Plot[{(t - \[Pi])^2/4}, {t, 0, 2*\[Pi] +0.3}, PlotStyle -> {Thickness[0.007], Blue}];
Show[p161, p162]
S17[x_] = Sum[1/n^2*Sin[x], {n, 1, 100}];
s171 = Plot[{S17[t]}, {t, -\[Pi], \[Pi] + 0.3}, PlotStyle -> {Thickness[0.02], Orange}];
s172 = Plot[{1.63498 Sin[t]}, {t, -\[Pi], \[Pi] + 0.3}, PlotStyle -> {Thickness[0.007], Blue}];
Show[s171, s172]
\begin{eqnarray*} \sum_{n\ge 1} \frac{1}{(2n-1)^2} \, \cos (2n-1) x &=& \frac{5}{4} - \frac{\pi |x|}{4} , \qquad \mbox{on interval } \ |x| < \pi ; \\ \sum_{n\ge 1} \frac{1}{(2n-1)^2} \, \sin (2n-1) x &=& ?? , \qquad \mbox{on interval } \ |x| < \pi . \end{eqnarray*}
S18[x_] = Sum[Cos[(2*n - 1)*x]/(2*n - 1)^2, {n, 1, 100}];
p181 = Plot[{S18[t]}, {t, -\[Pi], \[Pi] + 0.3}, PlotStyle -> {Thickness[0.02], Orange}];
p182 = Plot[5/4 - \[Pi]/4*Abs[t], {t, -\[Pi], \[Pi] + 0.3}, PlotStyle -> {Thickness[0.007], Blue}];;
Show[p181, p182]
S19[x_] = Sum[Sin[(2 n - 1)*x]/(2 n - 1)^2, {n, 1, 100}];
p191 = Plot[{S19[t]}, {t, -\[Pi], \[Pi]}, PlotStyle -> {Thickness[0.02], Orange}];
\[ \begin{split} \sum_{n\ge 1} \frac{1}{n^3} \, \cos n x &= ?? , \qquad \mbox{on interval } \ 0< x < < 2\pi ; \\ \sum_{n\ge 1} \frac{(-1)^n}{n^3} \, \cos n x &= ?? , \qquad \mbox{on interval } \ 0< x < < 2\pi ; \\ \sum_{n\ge 1} \frac{1}{n^3} \, \sin n x &= \frac{\pi^2 x}{6} - \frac{\pi x^2}{4} + \frac{x^3}{12} , \qquad \mbox{on interval } \ 0 < x < 2\pi . \\ \sum_{n\ge 1} \frac{(-1)^n}{n^3} \, \sin n x &= \frac{x^3 - \pi^2 x}{12} . \\ \sum_{n\ge 1} \frac{1}{n^3} \, \sin (2n x) &= \frac{1}{6} \times \begin{cases} 4 x^3 + 6\pi x^2 + 2 \pi^2 x , & \quad\mbox{for} \quad x \in (-\pi ,0) , \\ 4 x^3 - 6\pi x^2 + 2 \pi^2 x , & \quad\mbox{for} \quad x \in (0, \pi) . \end{cases} \end{split} \]

S21[x_] = Sum[Cos[n*x]/n^3, {n, 1, 100}];
Plot[{S21[t]}, {t, 0, 2 \[Pi]+0.3}, PlotStyle -> {Thickness[0.02], Orange}]
S22[x_] = Sum[(-1)^n *Cos[n*x]/n^3, {n, 1, 100}];
Plot[{S22[t]}, {t, 0, 2 \[Pi] + 0.3}, PlotStyle -> {Thickness[0.02], Orange}]
S23[x_] = Sum[Sin[n*x]/n^3, {n, 1, 100}];
p231 =Plot[{S23[t]}, {t, 0, 2 \[Pi] + 0.3}, PlotStyle -> {Thickness[0.02], Orange}];
p232 = Plot[{(\[Pi]^2*t)/6 - (\[Pi]*t^2)/4 + t^3/12}, {t, 0, 2 \[Pi]}, PlotStyle -> {Thickness[0.007], Blue}];
Show[p231, p232]
S23a[x_] = Sum[Sin[2*n*x]/n^3, {n, 1, 100}];
f23a[x_] = Piecewise[{{4*x^3 + 6*Pi*x^2 + 2*Pi^2 *x, -Pi < x <= 0}, {4*x^3 - 6*Pi*x^2 + 2*Pi^2 *x, 0 < x < Pi}}]
Plot[{S23a[x]*6, f23a[x]}, {x, -Pi, Pi}, PlotStyle -> Thick]
S24[x_] = Sum[(-1)^n *Sin[n*x]/n^3, {n, 1, 100}];
p241=Plot[{S24[t]}, {t, 0, 2 \[Pi] + 0.3}, PlotStyle -> {Thickness[0.02], Orange}];
p242 = Plot[{(t^3 - \[Pi]^2*t)/12}, {t, -\[Pi], \[Pi]}, PlotStyle -> {Thickness[0.007], Blue}];
Show[p241, p242]

\[ \frac{8}{\pi} \,\sum_{n\ge 1} \frac{1}{(2n-1)^3}\,\sin\left( 2n-1 \right) x = \begin{cases} x \left( \pi - x \right) , & \quad 0 \le x \le \pi , \\ x \left( \pi + x \right) , & \quad -\pi \le x \le 0. \end{cases} \]
S25[x_] = 8/\[Pi]*Sum[Sin[(2 n - 1)*x]/(2 n - 1)^3, {n, 1, 100}];
p251 = Plot[S25[t], {t, -\[Pi], \[Pi]}, PlotStyle -> {Thickness[0.02], Orange}];
f[x_] = Piecewise[{{x*(Pi-x), 0 p252 = Plot [f[t], {t, -\[Pi], Pi}, PlotStyle -> {Thickness[0.007], Blue}];
Show[p251, p252]
\[ \sum_{n\ge 1} \frac{1}{(2n-1)^3}\,\cos\left( 2n-1 \right) x = ??? \]
S26[x_] = Sum[Cos[(2 n - 1)*x]/(2 n - 1)^3, {n, 1, 100}];
p261 = Plot[S26[t], {t, -\[Pi], \[Pi]*2}, PlotStyle -> {Thickness[0.02], Orange}];
\[ \begin{split} \sum_{n\ge 1} \frac{(-1)^n}{(2n-1)^3}\,\sin\left( 2n-1 \right) x &= ?? , \qquad \mbox{on interval } \ 0< x < < 2\pi ; \\ \sum_{n\ge 1} \frac{(-1)^n}{(2n-1)^3}\,\cos\left( 2n-1 \right) x &= ?? , \qquad \mbox{on interval } \ 0< x < < 2\pi ; \end{split} \]
S27[x_] = Sum[(-1)^n/(2*n - 1)^3*Sin[(2*n - 1)*x], {n, 1, 100}];
p271 = Plot[{S27[t]}, {t, 0, 2*\[Pi]}, PlotStyle -> {Thickness[0.02], Orange}];
S27[x_] = Sum[(-1)^n/(2*n - 1)^3*Sin[(2*n - 1)*x], {n, 1, 100}];
p271 = Plot[{S27[t]}, {t, 0, 2*\[Pi]}, PlotStyle -> {Thickness[0.02], Orange}]
\[ 48 \sum_{n\ge 1} \frac{(-1)^{n-1}}{n^4}\,\cos\left( n\,x \right) = x^4 - 2\pi^2 x^2 + \frac{7 \pi^4}{15} , \qquad |x| < \pi . \]
S[x_] = -48*Sum[(-1)^n *Cos[n*x]/n^4, {n, 1, 100}];
p = Plot[S[t], {t, -\[Pi], \[Pi]*2}, PlotStyle -> {Thickness[0.02], Orange}];
q = Plot [t^4 -2*Pi^2 *t^2 +7*Pi^4/15, {t, -\[Pi], Pi}, PlotStyle -> {Thickness[0.007], Blue}];
Show[p, q]

Example 1: Consider the following square wave functions on interval \( (0, 2\ell ) : \)

\begin{align*} f(x) &= 2 \left[ H(x/\ell ) - H(x/\ell -1) \right] -1 = \begin{cases} \phantom{-}1, & \ 0< x< \ell , \\ -1, & \ \ell < x < 2\ell ; \end{cases} \\ g(x) &= H(x/\ell ) - H(x/\ell -1) = \begin{cases} 1, & \ 0< x< \ell , \\ 0, & \ \ell < x < 2\ell ; \end{cases} \\ h(x) &= H(x/\ell -1) - H(x/\ell -2) = \begin{cases} 0, & \ 0< x< \ell , \\ 1, & \ \ell < x< 2\ell ; \end{cases} \end{align*}
where H(t) is the Heaviside function. Since the function g is an odd function, all coefficients ak are zeroes and we get sine Fourier series (setting \( \ell =1 \) for simplicity):
\[ b_k = \int_0^2 f(x) \,\sin \left( k\pi x \right) {\text d}x = - \frac{4}{k\pi} \,(-1)^k \sin^2 \frac{k\pi}{2} = \frac{4}{k\pi} \times \begin{cases} 1 , & \ \mbox{if $k$ is odd}, \\ 0, & \ \mbox{if $k$ is even}. \end{cases}. \]
Therefore, we get the following Fourier series for f(x):
\[ f(x) = \frac{4\,\ell}{\pi}\, \sum_{n\ge 0} \,\frac{1}{2n+1} \, \sin \left( \frac{(2n+1)\pi x}{\ell} \right) . \]
Using Mathematica,
g[x_,L_]=HeavisideTheta[x/L] - HeavisideTheta[-1 + x/L]
ak = Assuming[L > 0, Integrate[g[x, L]*Cos[k*Pi*x/L], {x, 0, 2*L}]]
bk = Assuming[L > 0, Integrate[g[x, L]*Sin[k*Pi*x/L], {x, 0, 2*L}]]
h[x_, L_] = -HeavisideTheta[-2 + x/L] + HeavisideTheta[-1 + x/L]
ak = Assuming[L > 0, Integrate[h[x, L]*Cos[k*Pi*x/L], {x, 0, 2*L}]]
a0 = Assuming[L > 0, Integrate[h[x, L]*Cos[0*Pi*x/L], {x, 0, 2*L}]]
bk = Assuming[L > 0, Integrate[h[x, L]*Sin[k*Pi*x/L], {x, 0, 2*L}]]
we find other Fourier series:
\begin{align*} g(x) &= \frac{\ell}{2} + \frac{2\ell}{\pi} \sum_{n\ge 0} \frac{1}{2n+1}\, \sin \frac{(2n+1)\,\pi x}{\ell} , \\ h(x) &= \frac{\ell}{2} - \frac{2\ell}{\pi} \sum_{n\ge 0} \frac{1}{2n+1}\, \sin \frac{(2n+1)\,\pi x}{\ell} . \end{align*}
Then we plot partial sums with 10 terms (for simplicity setting \( \ell =1 \) ):
ff[x_] = Sum[4/Pi/(2*n + 1)*Sin[(2*n + 1)*Pi*x], {n, 0, 10}]
Plot[ff[x], {x, -2, 2}, PlotStyle -> Thick]
gg[x_] = 1/2 + 2/Pi*Sum[1/(2*n+1)*Sin[(2*n+1)*Pi*x],{n,0,10}]
hh[x_] = 1/2 - 2/Pi*Sum[1/(2*n+1)*Sin[(2*n+1)*Pi*x],{n,0,10}]
Plot[gg[x], {x, -2, 2}, PlotStyle -> Thick]

The Fourier series for the characteristic function is

\[ \frac{b-a}{\ell} + \sum_{n\ge 1} \frac{1}{n\pi} \left( \sin \frac{n\pi b}{\ell} - \sin \frac{n\pi a}{\ell} \right) \cos \frac{n\pi x}{\ell} + \sum_{n\ge 1} \frac{1}{n\pi} \left( \cos \frac{n\pi a}{\ell} - \cos \frac{n\pi b}{\ell} \right) \sin \frac{n\pi x}{\ell} = \chi_{[a,b]} = \begin{cases} 1, & \ \mbox{ for }\ x \in [a,b] , \\ 0, & \ \mbox{ otherwise.} \end{cases} \]
   ■

Example 2: On the interval [-ℓ, ℓ], consider three saw-tooth functions

\[ |x| = \begin{cases} \phantom{-}x, & \ \mbox{ for} \quad 0 < x < \ell , \\ -x, & \ \mbox{ for} \quad -\ell < x 0; \ell \end{cases} \]
\[ f(x) = \begin{cases} \frac{\ell - x}{2} , & \ \mbox{ for } \ 0 < x < \ell , \\ \frac{\ell + x}{2} , & \ \mbox{ for } \ -\ell < x < 0 . \end{cases} \]
and

\[ g(x) = \begin{cases} 1 - |x|/\delta , & \ \mbox{ if } \ |x| \le \delta , \\ 0, & \ \mbox{ if } \ |x| > \delta . \end{cases} \]
Here δ is some small positive number. We expand these functions into Fourier series:
\begin{align*} |x| &= 2 - \frac{4\ell}{\pi^2}\, \sum_{k\ge 1} \frac{1}{(2k-1)^2}\,\cos \frac{(2k-1)\pi x}{\ell} , \\ f(x) &= \frac{\ell}{4} - \frac{2\ell}{\pi^2} \, \sum_{k\ge 1} \frac{(-1)^k}{(2k-1)^2} \,\cos \frac{(2k-1) \pi x}{\ell} \\ g(x) &= \frac{\delta}{2\ell} + \frac{2\ell}{\delta \pi^2} \,\sum_{n\ge 1} \frac{1}{n^2} \left[ 1 - \cos \frac{\delta n\pi}{\ell} \right] \cos \frac{n\pi x}{\ell} . \end{align*}
Integrate[x*Cos[n*x*Pi/L], {x, 0, L}]*2/L
(2 L (-1 + Cos[n \[Pi]] + n \[Pi] Sin[n \[Pi]]))/(n^2 \[Pi]^2)
Integrate[(L - x)/2*Cos[n*x*Pi/L], {x, 0, L}]*2/L
(2 L Sin[(n \[Pi])/2]^2)/(n^2 \[Pi]^2)
Assuming[ d > 0, Integrate[(1 - Abs[x]/d)*Cos[n*x*Pi/L], {x, 0, d}]*2/L]
(2 L (-1 + Cos[(d n \[Pi])/L]))/(d n^2 \[Pi]^2)
We plot these functions.

Fig.1: Graph of |x|
     
Fig.2: Graph of f(x)
     
Fig.3: Graph of g(x)

s20[x_] = 2 - (4/Pi^2)*Sum[Cos[Pi*(2*k - 1) *x]/(2*k - 1)^2 , {k, 1, 20}];
Plot[s20[x], {x, -2.5, 2.5}, PlotStyle -> Thickness[0.01]]
s40[x_] = 1/4 + (2/Pi^2)* Sum[ Sin[(n \[Pi])/2]^2 *Cos[Pi*n*x]/(n)^2 , {n, 1, 40}];
Plot[s40[x], {x, -2.5, 2.5}, PlotStyle -> Thickness[0.01]]
s30[x_] = 1/6 + (6/Pi^2)* Sum[ (1 - Cos[n*Pi/3]) *Cos[Pi*n*x]/(n)^2 , {n, 1, 30}];
Plot[s30[x], {x, -2.5, 2.5}, PlotStyle -> Thickness[0.01]]
   ■

 

Monomial Fourier Expansions


\begin{align*} x &= - \frac{2\ell}{\pi} \sum_{n\ge 1} \frac{(-1)^n}{n}\,\sin \left( \frac{n\pi x}{\ell} \right) , \qquad 0 < x < \ell , \\ &= - \frac{4\ell}{\pi^2} \sum_{k\ge 1} \frac{1}{(2k-1)^2}\,\cos \left( \frac{\left( 2k-1 \right) \pi x}{\ell} \right) , \qquad 0 < x < \ell . \end{align*}
2*Integrate[x*Sin[Pi*n*x/L], {x, 0, L}]/L
(2 L (-n \[Pi] Cos[n \[Pi]] + Sin[n \[Pi]]))/(n^2 \[Pi]^2)
\begin{align*} x^2 &= \frac{2\ell^2}{\pi^3} \sum_{n\ge 1} \frac{-2 +(-1)^n \left( 2- n^2 \pi^2 \right)}{n^3}\,\sin \left( \frac{n\pi x}{\ell} \right) , \qquad 0 < x < \ell , \\ &= \frac{\ell^2}{3} + \frac{4\ell^2}{\pi^2} \sum_{n\ge 1} \frac{(-1)^n}{n^2}\,\cos \left( \frac{n\pi x}{\ell} \right) , \qquad 0 < x < \ell . \end{align*}
2*Integrate[x^2 *Cos[n*Pi*x/L], {x, 0, L}]/L
(2 L^2 (2 n \[Pi] Cos[n \[Pi]] + (-2 + n^2 \[Pi]^2) Sin[ n \[Pi]]))/(n^3 \[Pi]^3)

 

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