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Dirichlet's Theorem

Theory

Dirichlet’s Theorem gives conditions under which a Fourier series converges.

If f(x)f(x) is a periodic function with period 2L2L and is piecewise continuous on (−L,L)(-L,L), then:

12[lim⁡x→x0−f(x)+lim⁡x→x0+f(x)].\frac{1}{2}\left[\lim_{x\to x_0^-}f(x)+\lim_{x\to x_0^+}f(x)\right].

Exercises

Exercise 1

Write the Fourier series of f(x)=xf(x)=x on (−π,π)(-\pi,\pi).

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Solution

The function f(x)=xf(x)=x is an odd function.
Therefore,

a0=0,an=0.a_0=0,\qquad a_n=0.

We compute only bnb_n:

bn=1π∫−ππxsin⁡(nx) dx.b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}x\sin(nx)\,dx.

Since xsin⁡(nx)x\sin(nx) is even,

bn=2π∫0πxsin⁡(nx) dx.b_n=\frac{2}{\pi}\int_{0}^{\pi}x\sin(nx)\,dx.

Using integration by parts,

bn=2(−1)n+1n.b_n=\frac{2(-1)^{n+1}}{n}.

Thus, the Fourier series is

x=\sum_{n=1}^{\infty}\frac{2(-1)^{n+1}}{n}\sin(nx),\quad -\pi<x<\pi.
Exercise 2

Write the Fourier series of the periodic function

f(x)={0,−π≤x≤0,1,0lt;xlt;π2,0,π2lt;xlt;π,f(x+2π)=f(x),f(x)= \begin{cases} 0, & -\pi\le x\le 0,\\ 1, & 0<x<\frac{\pi}{2},\\ 0, & \frac{\pi}{2}<x<\pi, \end{cases} \quad f(x+2\pi)=f(x),

and compute the convergent values at x=0,π4,π2,3π4,πx=0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi.

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Solution

The period is 2π2\pi, so L=πL=\pi.

Step 1: Compute a0a_0

a0=1π∫−ππf(x) dx=1π∫0π21 dx=12.a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\,dx =\frac{1}{\pi}\int_{0}^{\frac{\pi}{2}}1\,dx =\frac{1}{2}.

Step 2: Compute ana_n

an=1π∫0π2cos⁡(nx) dx=1nπsin⁡nπ2.a_n=\frac{1}{\pi}\int_{0}^{\frac{\pi}{2}}\cos(nx)\,dx =\frac{1}{n\pi}\sin\frac{n\pi}{2}.

Step 3: Compute bnb_n

bn=1π∫0π2sin⁡(nx) dx=1nπ(1−cos⁡nπ2).b_n=\frac{1}{\pi}\int_{0}^{\frac{\pi}{2}}\sin(nx)\,dx =\frac{1}{n\pi}\left(1-\cos\frac{n\pi}{2}\right).

Fourier series

f(x)=14+∑n=1∞sin⁡nπ2nπcos⁡(nx)+∑n=1∞1−cos⁡nπ2nπsin⁡(nx).f(x)=\frac{1}{4} +\sum_{n=1}^{\infty}\frac{\sin\frac{n\pi}{2}}{n\pi}\cos(nx) +\sum_{n=1}^{\infty}\frac{1-\cos\frac{n\pi}{2}}{n\pi}\sin(nx).

Convergent values (Dirichlet’s Theorem)

  • x=0x=0: 12(0+1)=12\displaystyle \frac{1}{2}(0+1)=\frac{1}{2}
  • x=π4x=\frac{\pi}{4}: f(π4)=1f(\frac{\pi}{4})=1
  • x=π2x=\frac{\pi}{2}: 12(1+0)=12\displaystyle \frac{1}{2}(1+0)=\frac{1}{2}
  • x=3π4x=\frac{3\pi}{4}: f(3π4)=0f(\frac{3\pi}{4})=0
  • x=πx=\pi: f(π)=0f(\pi)=0
Exercise 3

Write the Fourier series of

f(x)={0,−2lt;xlt;−1,k,−1lt;xlt;1,0,1lt;xlt;2,f(x+4)=f(x),f(x)= \begin{cases} 0, & -2<x<-1,\\ k, & -1<x<1,\\ 0, & 1<x<2, \end{cases} \quad f(x+4)=f(x),

and compute the convergent values at x=−1,0,1.5x=-1,0,1.5.

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Solution

The period is 2L=42L=4, so L=2L=2.

Using Dirichlet’s Theorem:

  • x=−1x=-1: 12(0+k)=k2\displaystyle \frac{1}{2}(0+k)=\frac{k}{2}
  • x=0x=0: f(0)=kf(0)=k
  • x=1.5x=1.5: f(1.5)=0f(1.5)=0
Exercise 4

Suppose

f(x)={π,0lt;xlt;π2,0,otherwise in (−π,π),f(x+2π)=f(x).f(x)= \begin{cases} \pi, & 0<x<\frac{\pi}{2},\\ 0, & \text{otherwise in }(-\pi,\pi), \end{cases} \quad f(x+2\pi)=f(x).
  1. Find the Fourier series of f(x)f(x) and its sum at x=−π,−π2,0,π2x=-\pi,-\frac{\pi}{2},0,\frac{\pi}{2}.
  2. Find the sum of ∑n=1∞(−1)n2n−1\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{2n-1}.
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Solution

By Dirichlet’s Theorem, at x=π2x=\frac{\pi}{2} the Fourier series converges to

12(π+0)=π2.\frac{1}{2}(\pi+0)=\frac{\pi}{2}.

Evaluating the Fourier series at x=π2x=\frac{\pi}{2} leads to

π2=π4−(−1+13−15+⋯ ).\frac{\pi}{2} =\frac{\pi}{4}-\left(-1+\frac{1}{3}-\frac{1}{5}+\cdots\right).

Hence,

∑n=1∞(−1)n2n−1=−π4.\sum_{n=1}^{\infty}\frac{(-1)^n}{2n-1} =-\frac{\pi}{4}.