Bessel’s Inequality and Parseval’s Theorem
Bessel's Inequality
Let f (x) be a piecewise continuous function defined on the interval [−π, π], 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
Example 1.
Apply Parseval's formula to the function \(f\left( x \right) = x\) and find the sum of the series \[\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}}.\]
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 \({b_n} = {\frac{2}{n}} {\left( { - 1} \right)^{n + 1}}.\)
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 \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^s}}}} \) 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
\[{a_0} = \frac{{2{\pi ^2}}}{3},\;\; {a_n} = \frac{4}{{{n^2}}}{\left( { - 1} \right)^n},\;\; {b_n} = 0.\]
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 \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^s}}}} \) is known as Riemann zeta function \(\zeta \left( s \right).\) Consequently,
\[\zeta \left( 4 \right) = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^4}}}} = \frac{{{\pi ^4}}}{{90}}.\]
See more problems on Page 2.
