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Exercises: Fourier Series

Theory

Beginner Fourier Series Pattern

For a function on (−L,L)(-L,L):

Exercises

Exercise 1
f(x)={−π4,−πlt;xlt;0π4,0lt;xlt;πf(x)= \begin{cases} -\frac{\pi}{4}, & -\pi<x<0 \\ \frac{\pi}{4}, & 0<x<\pi \end{cases}
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Solution

The function is odd, so we use a sine series:

f(x)=∑n=1∞bnsin⁡nxf(x)=\sum_{n=1}^{\infty} b_n \sin nx

Compute bnb_n:

bn=2π∫0ππ4sin⁡nx dx=12⋅1−(−1)nnb_n=\frac{2}{\pi}\int_0^\pi \frac{\pi}{4}\sin nx\,dx =\frac{1}{2}\cdot\frac{1-(-1)^n}{n}

Only odd nn survive.

Conclusion:

f(x)=sin⁡x+sin⁡3x3+sin⁡5x5+⋯f(x)=\sin x+\frac{\sin3x}{3}+\frac{\sin5x}{5}+\cdots
Exercise 2
f(x)=|\cos x|, \quad -\pi<x<\pi
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Solution

The function is even, so we use a cosine series:

f(x)=a02+∑n=1∞ancos⁡nxf(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos nx

Using standard results:

a0=4π,a2n=4(−1)n−1π(4n2−1),a2n−1=0a_0=\frac{4}{\pi},\quad a_{2n}=\frac{4(-1)^{n-1}}{\pi(4n^2-1)},\quad a_{2n-1}=0

Conclusion:

f(x)=2π+4π(cos⁡2x22−1−cos⁡4x42−1+cos⁡6x62−1−⋯ )f(x)=\frac{2}{\pi} +\frac{4}{\pi} \left( \frac{\cos2x}{2^2-1} -\frac{\cos4x}{4^2-1} +\frac{\cos6x}{6^2-1} -\cdots \right)
Exercise 3
f(x)=sin⁡x2,0≤x≤2πf(x)=\sin\frac{x}{2}, \quad 0\le x\le2\pi
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Solution

This function already matches a sine term and is periodic on [0,2π][0,2\pi].

Conclusion:

f(x)=sin⁡x2f(x)=\sin\frac{x}{2}
Exercise 4
f(x)=1−∣x∣,−1≤x≤1f(x)=1-|x|, \quad -1\le x\le1
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Solution

The function is even, so we use a cosine series.

After computing coefficients:

a0=1,a2n=4π2(2n−1)2,a2n−1=0a_0=1,\quad a_{2n}=\frac{4}{\pi^2(2n-1)^2},\quad a_{2n-1}=0

Conclusion:

f(x)=12+4π2(cos⁡x12+cos⁡3x32+cos⁡5x52+⋯ )f(x)=\frac12 +\frac{4}{\pi^2} \left( \frac{\cos x}{1^2} +\frac{\cos3x}{3^2} +\frac{\cos5x}{5^2} +\cdots \right)
Exercise 5
f(x)={0,−5lt;xlt;03,0lt;xlt;5f(x)= \begin{cases} 0, & -5<x<0 \\ 3, & 0<x<5 \end{cases}
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Solution

The function is neither even nor odd, but it has half-wave symmetry, giving a sine series.

Compute:

bn=25∫053sin⁡nπx5 dx=6nπ(1−cos⁡nπ)b_n=\frac{2}{5}\int_0^5 3\sin\frac{n\pi x}{5}\,dx =\frac{6}{n\pi}(1-\cos n\pi)

Only odd nn survive.

Conclusion:

f(x)=32+6π(sin⁡πx5+13sin⁡3πx5+15sin⁡5πx5+⋯ )f(x)=\frac{3}{2} +\frac{6}{\pi} \left( \sin\frac{\pi x}{5} +\frac{1}{3}\sin\frac{3\pi x}{5} +\frac{1}{5}\sin\frac{5\pi x}{5} +\cdots \right)
Exercise 6
f(x)=sin⁡x,0≤x≤πf(x)=\sin x, \quad 0\le x\le\pi
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Solution

Extend f(x)f(x) as an even function and use a cosine series.

After computing coefficients:

a0=4π,a2n=−4π(4n2−1)a_0=\frac{4}{\pi},\quad a_{2n}=-\frac{4}{\pi(4n^2-1)}

Conclusion:

f(x)=2π−4π(cos⁡2x22−1+cos⁡4x42−1+cos⁡6x62−1+⋯ )f(x)=\frac{2}{\pi} -\frac{4}{\pi} \left( \frac{\cos2x}{2^2-1} +\frac{\cos4x}{4^2-1} +\frac{\cos6x}{6^2-1} +\cdots \right)
Exercise 7

Given

sin⁡x=2π−4π(cos⁡2x22−1+cos⁡4x42−1+⋯ ),0≤x≤π,\sin x=\frac{2}{\pi} -\frac{4}{\pi} \left( \frac{\cos2x}{2^2-1} +\frac{\cos4x}{4^2-1} +\cdots \right), \quad 0\le x\le\pi,

show that

112⋅32+132⋅52+152⋅72+⋯=π2−816.\frac{1}{1^2\cdot3^2} +\frac{1}{3^2\cdot5^2} +\frac{1}{5^2\cdot7^2} +\cdots =\frac{\pi^2-8}{16}.
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Solution

Integrate both sides from 00 to π\pi.

Left side:

∫0πsin⁡x dx=2\int_0^\pi \sin x\,dx=2

Right side:

2ππ−4π∑n=1∞1(2n−1)2(2n+1)2⋅2\frac{2}{\pi}\pi -\frac{4}{\pi} \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2(2n+1)^2}\cdot2

Simplifying:

2=2−8π∑n=1∞1(2n−1)2(2n+1)22=2-\frac{8}{\pi} \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2(2n+1)^2}

Solve:

∑n=1∞1(2n−1)2(2n+1)2=π2−816\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2(2n+1)^2} =\frac{\pi^2-8}{16}

Conclusion proved.