next up previous
Next: Dirac's delta function and Up: No Title Previous: Borel Summation

Convergence of the Fourier series

Here we shall show why the Fourier series (4-1) converges to $[f(\theta+)+f(\theta-)]/2$. Let us consider the sum

\begin{displaymath}S_N(\theta)\equiv A_0/2+\sum_{n=1}^N (A_n \cos n\theta+B_n\sin n\theta),
\end{displaymath} (68)

where An and Bn are defined by eqs. (4-2). Inserting these definitions we get
 
$\displaystyle S_N(\theta)$ = $\displaystyle (1/\pi)\int_0^{2\pi}f(\theta')[1/2+\sum_{n=1}^N(\cos n\theta
\cos n\theta'+\sin n\theta\sin n\theta')]d\theta'$  
  = $\displaystyle (1/\pi)\int_0^{2\pi}f(\theta')
[1/2+\sum_{n=1}^N\cos n(\theta-\theta')]d\theta'.$ (69)

The sum over the cosines can be performed,
 
$\displaystyle \sum_{n=1}^N\cos n(\theta-\theta')$ = $\displaystyle (1/2)\sum_{n=1}^N
[\exp (in(\theta-\theta'))
+\exp (-in(\theta-\theta'))]$  
  = $\displaystyle \frac{1}{2}e^{i(\theta-\theta')}\frac{e^{iN(\theta-\theta')}-1}
{e^{i(\theta-\theta')}-1}+{\rm complex ~conjugate}$  
  = $\displaystyle \frac{1}{2}\frac{\sin ((N+\frac{1}{2})(\theta-\theta'))-\sin
\frac{1}{2}(\theta-\theta')}{\sin \frac{1}{2}(\theta-\theta')}.$ (70)

Inserting this result we get by use of the substitution

\begin{displaymath}\theta'=\theta+u/(N+\frac{1}{2}),~~u=(N+\frac{1}{2})(\theta-\theta'),
\end{displaymath} (71)

that the sum SN reduces to

\begin{displaymath}S_N(\theta)=\frac{1}{2\pi}\int_{(N+\frac{1}{2})(\theta-2\pi)}...
...rac{\sin u}{(N+\frac{1}{2})\sin
\left(\frac{u}{2N+1}\right)}.
\end{displaymath} (72)

. Taking $0<\theta<2\pi$ we can rewrite this integral as
$\displaystyle S_N(\theta)$ = $\displaystyle \frac{1}{2\pi}\int_0^{(N+
\frac{1}{2})\theta}du f\left(\theta+\frac{u}{N+\frac{1}{2}}\right)
\frac{\sin u}{(N+
\frac{1}{2})\sin \frac{u}{2N+1}}$  
  + $\displaystyle \frac{1}{2\pi}\int_0^{(N+
\frac{1}{2})(2\pi-\theta)}du f\left(\th...
...rac{u}{N+\frac{1}{2}}\right)
\frac{\sin u}{(N+\frac{1}{2})\sin \frac{u}{2N+1}}.$ (73)

Remembering that $0<\theta<2\pi$ and taking N to be large, we arrive at the result

\begin{displaymath}S_N(\theta)\rightarrow [f(\theta+)+f(\theta-)]\frac{1}{\pi}\i...
...u}{u}=\frac{1}{2}[f(\theta+)+f(\theta-)],~~N\rightarrow\infty.
\end{displaymath} (74)

Here we used that $(N+\frac{1}{2})\sin u/(2N+1)\rightarrow (1/2)u$ for $N\rightarrow\infty$ as well as the value $\int_0^\infty du \sin u/u=\pi/2$ obtained in eq. (3-11) in the book.

The result is thus that the sum $S_\infty$ is equal to $(1/2)[f(\theta+)+
f(\theta-)]$, as mentioned in the book.


next up previous
Next: Dirac's delta function and Up: No Title Previous: Borel Summation
Mette Lund
2000-04-27