# Calculus

## Fourier Series # 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

Click or tap a problem to see the solution.

### Example 1

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

### Example 2

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

### 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.