Question

We have said that Fourier series can represent discontinuous functions although power series cannot. It might occur to you to wonder why we could not substitute the power series for $\mathbf{sin}\mathbf{nx}$and $\mathbf{cos}\mathbf{nx}$(which converge for all x) into a Fourier series and collect terms to obtain a power series for a discontinuous function. As an example of what happens if we try this, consider the series in Problem 9.5. Show that the coefficients of x, if collected, form a divergent series; similarly, the coefficients of ${\mathbf{x}}^3$form a divergent series, and so on.

Short answer

The given series is divergent $\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{(2n+1)}^2=\infty \ne 0$.

Step-by-step solution

Step 1:Determine the type of series using Fourier series:

The Fourier series of the given function is

$\frac{4}{\pi }(\sin x+\frac{1}{3}\sin 3x+\frac{1}{5}\sin 5x+\mathrm{...})\mathrm{...}\left(1\right)$

Write each sine term as a power series and the coefficients of x in the series (1).

$\frac{4}{\pi }(\sin x+\frac{1}{3}\sin 3x+\frac{1}{5}\sin 5x+\mathrm{...})=\frac{4}{\pi }(x-\frac{1}{6}{x}^3+\frac{1}{120}{x}^5-\mathrm{...}\infty +\frac{1}{3}(3x-\frac{9}{2}{x}^3+\frac{81}{40}{x}^5-\mathrm{...}\infty )+\frac{1}{5}(5x-\frac{125}{6}{x}^3+\frac{625}{24}{x}^5-\mathrm{...}\infty )+\mathrm{...})$

The coefficient of x in the above series is again a series and it can be written as follows:

$\begin{array}{c}\text{coefficient\hspace{0.17em}of\hspace{0.17em}x=}\frac{4}{\pi }(1+\frac{1}{3}\times 3+\frac{1}{5}\times 5+\mathrm{...}\infty )\\ =\frac{4}{\pi }(1+1+1+\mathrm{...}\infty )\\ =\frac{4}{\pi }\sum _{n=1}^{\infty }1\end{array}$

Obviously, this is a divergent series and it can be easily proved using limit of an nth term.

$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }1=1\ne 0$

Determine the type of series using limit of an nth term:

Write each sine term as a power series and the coefficients of $x^3$in the series (1).

$\frac{4}{\pi }(\sin x+\frac{1}{3}\sin 3x+\frac{1}{5}\sin 5x+\mathrm{...})=\frac{4}{\pi }(x-\frac{1}{6}{x}^3+\frac{1}{120}{x}^5-\mathrm{...}\infty +\frac{1}{3}(3x-\frac{9}{2}{x}^3+\frac{81}{40}{x}^5-\mathrm{...}\infty )+\frac{1}{5}(5x-\frac{125}{6}{x}^3+\frac{625}{24}{x}^5-\mathrm{...}\infty )+\mathrm{...})$

The coefficient of in the above series is again a series and it can be written as follows:

$\begin{array}{c}\text{coefficient\hspace{0.17em}of\hspace{0.17em}}{x}^3=\frac{4}{\pi }(-\frac{1}{6}+\frac{1}{3}(-\frac{9}{2})+\frac{1}{5}(-\frac{125}{6})+\mathrm{...})\\ =\frac{4}{\pi }(-\frac{1}{6}-\frac{1}{6}\left(9\right)-\frac{1}{6}\left(25\right)-\frac{1}{6}\left(49\right)-\mathrm{...})\\ =\frac{-2}{3\pi }(1^2+3^2+5^2+7^2+\mathrm{...}\infty )\\ =\frac{-2}{3\pi }\sum _{n=0}^{\infty }{(2n+1)}^2\end{array}$

Obviously, this is a divergent series and it can be easily proved using limit of an nth term.

$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{(2n+1)}^2=\infty \ne 0$

Similarly, it can be proved that coefficients of odd powers of x are again divergent series.