Calculus

Fourier Series

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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 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