Differentiation and Integration of Fourier Series

Solved Problems

Example 3.

Find the Fourier series of the function \(f\left( x \right) = {x^3}\) knowing that

\[{x^2} = \frac{{{\pi ^2}}}{3} + 4\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}\cos nx}\]

for \(- \pi \le x \le \pi.\)

Solution.

Integrating this series term by term, we have

\[\int\limits_{ - \pi }^x {{t^2}dt = \int\limits_{ - \pi }^x {\frac{{{\pi ^2}}}{3}dt} + 4\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\int\limits_{ - \pi }^x {\frac{{\cos nt}}{{{n^2}}}dt} } ,\;\; \Rightarrow \left. {\left( {\frac{{{t^3}}}{3}} \right)} \right|_{ - \pi }^\pi = \left. {\left( {\frac{{{\pi ^2}}}{3}t} \right)} \right|_{ - \pi }^x + 4\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\left[ {\left. {\left( {\frac{{\sin nt}}{{{n^3}}}} \right)} \right|_{ - \pi }^x} \right]} ,\;\; \Rightarrow \frac{{{x^3}}}{3} + \cancel{\frac{{{\pi ^3}}}{3}} = \frac{{{\pi ^2}x}}{3} + \cancel{\frac{{{\pi ^3}}}{3}} + 4\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{\sin nx}}{{{n^3}}}} ,\;\;} \Rightarrow {x^3} = {\pi ^2}x + 12\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^3}}}\sin nx} .\]

Substitute the Fourier series expansion for \(x\) from Example \(3\) of the section Definition of Fourier Series and Typical Examples into the last expression

\[x = 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{n}\sin nx} .\]

Then the complete answer is

\[{x^3} = 2{\pi ^2}\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{n}\sin nx} + 12\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^3}}}\sin nx} = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\left( {\frac{{12}}{{{n^3}}} - \frac{{2{\pi ^2}}}{n}} \right) \sin nx}.\]

Example 4.

Investigate the process of differentiating term by term the Fourier series expansion of the function \(f\left( x \right) = x\) defined on the interval \(\left[ { - \pi ,\pi } \right].\)

Solution.

We write the Fourier series expansion for the linear function:

\[x = 2\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\frac{{\sin nx}}{n}} .\]

By formal differentiation, we find that

\[1 \sim 2\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\cos nx} = 2\left( {\cos x - \cos 2x + \cos 3x - \ldots } \right).\]

We obtain a contradiction since the Fourier series for \(1\) should consist just of a single constant term. To explain this paradox, we introduce the Dirac delta function or the unit impulse function \(\delta \left( x \right).\) The so-called "weak" definition of \(\delta \left( x \right)\) implies that

\[ \delta \left( x \right) = \begin{cases} 0, & x \ne 0 \\ \infty, & x = 0 \end{cases} \]

assuming that the total area under the graph of the function is \(1:\)

\[\int\limits_{ - \infty }^\infty {\delta \left( x \right)dx} = 1.\]

The delta function can also be defined by the limit as \(n \to \infty\) in the following way:

\[\delta \left( x \right) = \lim\limits_{n \to \infty } \frac{1}{{2\pi }}\frac{{\sin \left( {n + \frac{1}{2}} \right)x}}{{\sin \frac{x}{2}}}.\]

The graph of the delta function for \(n = 5\) and \(n = 20\) is shown in Figure \(2.\)

The Fourier series of the delta function is given by

\[\delta \left( x \right) = \frac{1}{{2\pi }} + \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\cos nx} = \frac{1}{{2\pi }} + \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\left( {\cos x + \cos 2x + \cos 3x + \ldots } \right)}\]

This series has only cosines since the delta function is even.

Now we consider the periodic extension \({f_1}\left( x \right)\) of the original function \(f\left( x \right)\) (Figure \(3\)).

A periodic function with jump discontinuities — Figure 3.

This function has jump discontinuities at the points \(x = \left( {2m + 1} \right)\pi ,\) \(m = 0, \pm 1, \pm 2, \ldots \) The derivative \({f'_1}\left( x \right)\) of the periodic extension has an additional delta function concentrated at each jump discontinuity, so that

\[{f'_1}\left( x \right) = 1 - 2\pi \sum\limits_{m = - \infty }^\infty {\delta \left[ {x - \left( {2m + 1} \right)\pi } \right]} = 1 - 2\pi \bar \delta \left( {x - \pi } \right),\]

where \(\bar \delta \left( {x - \pi } \right)\) denotes the \(2\pi\)-periodic extension of the delta function.

It is seen from here that

\[\delta \left[ {x - \left( {2m + 1} \right)\pi } \right] = \frac{1}{{2\pi }} + \frac{1}{\pi }\sum\limits_{n = 1}^\infty {\cos n\left[ {x - \left( {2m + 1} \right)\pi } \right]} = \frac{1}{{2\pi }} + \frac{1}{\pi }\left\{ {\cos \left[ {x - \left( {2m + 1} \right)\pi } \right] + \cos 2\left[ {x - \left( {2m + 1} \right)\pi } \right] + \ldots } \right\} = \frac{1}{{2\pi }} + \frac{1}{\pi }\left\{ { - \cos x + \cos 2x - \cos 3x + \cos 4x - \ldots } \right\} = \frac{1}{{2\pi }} - \frac{1}{\pi }\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\cos nx} .\]

Hence, the Fourier series expansion of the derivative \({f'_1}\left( x \right)\) is given by

\[{f_1}^\prime \left( x \right) = 1 - 2\pi \bar \delta \left( {x - \pi } \right) = 2\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\cos nx} \sim 1.\]

Thus, the function \({f_1}^\prime \left( x \right) =\) \( 1 - 2\pi \bar \delta \left( {x - \pi } \right)\) is the Fourier series expansion for the constant term \(1.\) The graph of this function is shown in Figure \(4.\)