Calculus

Fourier Series

Bessel’s Inequality and Parseval’s Theorem

Page 1
Problems 1-2
Page 2
Problems 3-4

Bessel’s Inequality

Let \(f\left( x \right)\) be a piecewise continuous function defined on the interval \(\left[ { – \pi ,\pi } \right],\) so that its Fourier series is given by

\[{f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos nx + {b_n}\sin nx} \right)} .}\]

Bessel’s inequality states that

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

From here we can conclude that the series \(\sum\limits_{n = 1}^\infty {\left( {a_n^2 + b_n^2} \right)} \) is convergent.

Parseval’s Theorem

If \(f\left( x \right)\) is a square-integrable function on the interval \(\left[ { – \pi ,\pi } \right],\) such that

\[\int\limits_{ – \pi }^\pi {{f^2}\left( x \right)dx} \le \infty,\]

then the Bessel’s inequality becomes equality. In this case we have Parseval’s formula:

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

Parseval’s Formula in Complex Form

Let again \(f\left( x \right)\) be a square-integrable function on the interval \(\left[ { – \pi ,\pi } \right]\) and let \({{c_n}}\) be complex coefficients such that

\[f\left( x \right) = \sum\limits_{n = – \infty }^\infty {{c_n}{e^{inx}}} ,\]

where

\[{{c_n} }={ \frac{1}{{2\pi }}\int\limits_{ – \pi }^\pi {f\left( x \right){e^{ – inx}}dx} .}\]

Then the Parseval’s formula can be written in the form

\[{\sum\limits_{n = – \infty }^\infty {{{\left| {{c_n}} \right|}^2}} }={ \frac{1}{{2\pi }}\int\limits_{ – \pi }^\pi {{f^2}\left( x \right)dx} .}\]

Note that the energy of a \(2\pi\)-periodic wave \(f\left( x \right)\) is

\[E = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {{f^2}\left( x \right)dx} .\]

Solved Problems

Click on problem description to see solution.

 Example 1

Apply Parseval’s formula to the function \(f\left( x \right) = x\) and find the sum of the series \(\sum\limits_{n = 1}^\infty {\large\frac{1}{{{n^2}}\normalsize}} .\)

 Example 2

Apply Parseval’s formula to the function \(f\left( x \right) = {x^2}.\)

 Example 3

Apply Parseval’s formula to the function

\[
{f\left( x \right) \text{ = }}\kern0pt
{\begin{cases}
1, & \text{if} & 0 \le \left| x \right| \le d \\
0, & \text{if} & d \le \left| x \right| \le \pi
\end{cases},}
\]

and find the sums of the series \(\sum\limits_{n = 1}^\infty {\large\frac{{{{\sin }^2}nd}}{{{n^2}}}\normalsize} \) and \(\sum\limits_{n = 1}^\infty {\large\frac{{{{\cos }^2}nd}}{{{n^2}}}\normalsize}.\)

 Example 4

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

Example 1.

Apply Parseval’s formula to the function \(f\left( x \right) = x\) and find the sum of the series \(\sum\limits_{n = 1}^\infty {\large\frac{1}{{{n^2}}\normalsize}} .\)

Solution.

Fourier series expansion of the function \(f\left( x \right) = x\) on the interval \(\left[ { – \pi ,\pi } \right]\) is given by

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

(See Example \(3\) on the page Definition of Fourier Series and Typical Examples.)

Here the Fourier coefficients are \({a_0} = {a_n} \) \(= 0\) (since the function \(f\left( x \right) = x\) is odd) and \({b_n} =\) \( {\large\frac{2}{n}\normalsize} {\left( { – 1} \right)^{n + 1}}.\)

Using Parseval’s formula, we have

\[
{{\sum\limits_{n = 1}^\infty {{{\left[ {\frac{2}{n}{{\left( { – 1} \right)}^{n + 1}}} \right]}^2}} }={ \frac{1}{\pi }\int\limits_{ – \pi }^\pi {{x^2}dx} ,\;\;}}\Rightarrow
{{4\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} }={ \frac{1}{\pi }\left[ {\left. {\left( {\frac{{{x^3}}}{3}} \right)} \right|_{ – \pi }^\pi } \right],\;\;}}\Rightarrow
{{\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} }={ \frac{1}{{4\pi }}\left( {\frac{{{\pi ^3}}}{3} – \frac{{{{\left( { – \pi } \right)}^3}}}{3}} \right),\;\;}}\Rightarrow
{{\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} }={ \frac{1}{{4\pi }} \cdot \frac{{2{\pi ^3}}}{3} }={\frac{{{\pi ^2}}}{6}.}}
\]

Note that \(\sum\limits_{n = 1}^\infty {\large\frac{1}{{{n^s}}}\normalsize} \) is called Riemann zeta function \(\zeta \left( s \right).\) Thus, we have proved that

\[{\zeta \left( 2 \right) = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} }={ \frac{{{\pi ^2}}}{6}.}\]

Example 2.

Apply Parseval’s formula to the function \(f\left( x \right) = {x^2}.\)

Solution.

We have found in Example \(4\) in the section Definition of Fourier Series and Typical Examples that the Fourier series of the function \(f\left( x \right) = {x^2}\) on the interval \(\left[ { – \pi ,\pi } \right]\) is given by

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

where

\[{{a_0} = \frac{{2{\pi ^2}}}{3},\;\;}\kern0pt{{a_n} = \frac{4}{{{n^2}}}{\left( { – 1} \right)^n},\;\;}\kern0pt{{b_n} = 0.}\]

Applying Parseval’s formula to the function, we obtain

\[
{\frac{{{a_0^2}}}{2} + \sum\limits_{n = 1}^\infty {\left( {a_n^2 + b_n^2} \right)} }={ \frac{1}{\pi }\int\limits_{ – \pi }^\pi {{f^2}\left( x \right)dx} ,\;\;}\Rightarrow
{\frac{1}{2}{\left( {\frac{{2{\pi ^2}}}{3}} \right)^2} }+{ \sum\limits_{n = 1}^\infty {{{\left[ {\frac{4}{{{n^2}}}{{\left( { – 1} \right)}^n}} \right]}^2}} }={ \frac{1}{\pi }\int\limits_{ – \pi }^\pi {{x^4}dx} ,\;\;}\Rightarrow
{\frac{{2{\pi ^4}}}{9} + 16\sum\limits_{n = 1}^\infty {\frac{1}{{{n^4}}}} }={ \frac{1}{\pi }\left[ {\left. {\left( {\frac{{{x^5}}}{5}} \right)} \right|_{ – \pi }^\pi } \right],\;\;}\Rightarrow
{\frac{{2{\pi ^4}}}{9} + 16\sum\limits_{n = 1}^\infty {\frac{1}{{{n^4}}}} }={ \frac{1}{\pi } \cdot \frac{{2{\pi ^5}}}{5},\;\;}\Rightarrow
{16\sum\limits_{n = 1}^\infty {\frac{1}{{{n^4}}}} }={ \frac{{2{\pi ^4}}}{5} – \frac{{2{\pi ^4}}}{9},\;\;}\Rightarrow
{16\sum\limits_{n = 1}^\infty {\frac{1}{{{n^4}}}} = \frac{{8{\pi ^4}}}{{45}},\;\;}\Rightarrow
{\sum\limits_{n = 1}^\infty {\frac{1}{{{n^4}}}} = \frac{{{\pi ^4}}}{{90}}.}
\]

The series \(\sum\limits_{n = 1}^\infty {\large\frac{1}{{{n^s}}}\normalsize} \) is known as Riemann zeta function \(\zeta \left( s \right).\) Consequently,

\[{\zeta \left( 4 \right) = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^4}}}} }={ \frac{{{\pi ^4}}}{{90}}.}\]
Page 1
Problems 1-2
Page 2
Problems 3-4