# Bessel’s Inequality and Parseval’s Theorem

### 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 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 $$\sum\limits_{n = 1}^\infty {\large\frac{1}{{{n^2}}\normalsize}}.$$

### 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