Calculus

Fourier Series

Fourier Series Logo

Bessel’s Inequality and Parseval’s Theorem

Solved Problems

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Example 3

Apply Parseval's formula to the function

\[f\left( x \right) = \begin{cases} 1, & \text{if} & 0 \le \left| x \right| \le d \\ 0, & \text{if} & d \le \left| x \right| \le \pi \end{cases},\]

and find the sums of the series

\[\sum\limits_{n = 1}^\infty {\frac{{{{\sin }^2}nd}}{{{n^2}}}} \;\;\text{and}\;\; \sum\limits_{n = 1}^\infty {\frac{{{{\cos }^2}nd}}{{{n^2}}}}.\]

Example 4

Calculate the sum of the series \[\sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {2k + 1} \right)}^2}}}}.\]

Example 3.

Apply Parseval's formula to the function

\[f\left( x \right) = \begin{cases} 1, & \text{if} & 0 \le \left| x \right| \le d \\ 0, & \text{if} & d \le \left| x \right| \le \pi \end{cases},\]

and find the sums of the series

\[\sum\limits_{n = 1}^\infty {\frac{{{{\sin }^2}nd}}{{{n^2}}}} \;\;\text{and}\;\; \sum\limits_{n = 1}^\infty {\frac{{{{\cos }^2}nd}}{{{n^2}}}}.\]

Solution.

The Fourier series of this function is given by (try to find it yourself):

\[f\left( x \right) = \frac{d}{\pi } + \frac{2}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\sin nd}}{n}\cos nx} .\]

The Fourier coefficients in this expansion are

\[{a_0} = \frac{{2d}}{\pi },\;\; {a_n} = \frac{{2\sin nd}}{{n\pi }},\;\; {b_n} = 0.\]

Applying Parseval's identity

\[\frac{{{a_0^2}}}{2} + \sum\limits_{n = 1}^\infty {\left( {a_n^2 + b_n^2} \right)} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {{f^2}\left( x \right)dx},\]

we obtain

\[\frac{1}{2}{\left( {\frac{{2d}}{\pi }} \right)^2} + \sum\limits_{n = 1}^\infty {{{\left[ {\frac{{2\sin nd}}{{n\pi }}} \right]}^2}} = \frac{2}{\pi }\int\limits_0^\pi {{f^2}\left( x \right)dx} ,\;\; \Rightarrow \frac{{2{d^2}}}{{{\pi ^2}}} + \frac{4}{{{\pi ^2}}}\sum\limits_{n = 1}^\infty {\frac{{{{\sin }^2}nd}}{{{n^2}}}} = \frac{2}{\pi }\int\limits_0^d {dx} ,\;\; \Rightarrow \frac{{{d^2}}}{\pi } + \frac{2}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\sin }^2}nd}}{{{n^2}}}} = d,\;\; \Rightarrow \frac{2}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\sin }^2}nd}}{{{n^2}}}} = d - \frac{{{d^2}}}{\pi },\;\; \Rightarrow \sum\limits_{n = 1}^\infty {\frac{{{{\sin }^2}nd}}{{{n^2}}}} = \frac{{d\left( {\pi - d} \right)}}{\pi } \cdot \frac{\pi }{2},\;\; \Rightarrow \sum\limits_{n = 1}^\infty {\frac{{{{\sin }^2}nd}}{{{n^2}}}} = \frac{{d\left( {\pi - d} \right)}}{2}.\]

It is also easy to find the sum \(\sum\limits_{n = 1}^\infty {e\frac{{{{\cos }^2}nd}}{{{n^2}}}}:\)

\[\sum\limits_{n = 1}^\infty {\frac{{{\cos^2}nd}}{{{n^2}}}} = \sum\limits_{n = 1}^\infty {\frac{{1 - {\sin^2}nd}}{{{n^2}}}} = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} - \sum\limits_{n = 1}^\infty {\frac{{{\sin^2}nd}}{{{n^2}}}} .\]

Here \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} = \zeta \left( 2 \right) = {\frac{{{\pi ^2}}}{6}}\) (See Example \(1\)). Hence,

\[\sum\limits_{n = 1}^\infty {\frac{{{\cos^2}nd}}{{{n^2}}}} = \frac{{{\pi ^2}}}{6} - \frac{{d\left( {\pi - d} \right)}}{2} = \frac{{{\pi ^2} - 3\pi d + 3{d^2}}}{6}.\]

Example 4.

Calculate the sum of the series \[\sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {2k + 1} \right)}^2}}}}.\]

Solution.

We have found in the previous Example \(3\) that

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

Put \(d = {\frac{\pi }{2}}\) to obtain

\[\sum\limits_{n = 1}^\infty {\frac{{{{\sin }^2}\frac{{n\pi }}{2}}}{{{n^2}}}} = \frac{{\frac{\pi }{2}\left( {\pi - \frac{\pi }{2}} \right)}}{2}\;\; \text{or}\;\; \sum\limits_{n = 1}^\infty {\frac{{{{\sin }^2}\frac{{n\pi }}{2}}}{{{n^2}}}} = \frac{{{\pi ^2}}}{8}.\]

One can note that

\[ \frac{{n\pi }}{2} = \begin{cases} 0, & n = 2k \\ 1, & n = 4k + 1 \\ -1, & n = 4k + 3 \end{cases},\;\; \text{where}\;\; k = 0,1,2,3,\ldots\]

Hence,

\[ {{\sin }^2}\frac{{n\pi }}{2} = \begin{cases} 0, & n = 2k \\ 1, & n = 2k + 1 \end{cases},\;\; \text{where}\;\; k = 0,1,2,3,\ldots\]

Then the sum of the series is

\[\sum\limits_{k = 0}^\infty {\frac{1}{{{{\left( {2k + 1} \right)}^2}}}} = \frac{{{\pi ^2}}}{8}.\]
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