Calculus

Fourier Series

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Differentiation and Integration of Fourier Series

  • Differentiation of Fourier Series

    Let \(f\left( x \right)\) be a \(2\pi\)-periodic piecewise continuous function defined on the closed interval \(\left[ { – \pi ,\pi } \right].\) As we know, the Fourier series expansion of such a function exists and is given by

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

    If the derivative \(f’\left( x \right)\) of this function is also piecewise continuous and the function \(f\left( x \right)\) satisfies the periodicity conditions

    \[{f\left( { – \pi } \right) = f\left( \pi \right),\;\;\;}\kern-0.3pt{f’\left( { – \pi } \right) = f’\left( \pi \right),}\]

    then the Fourier series expansion of the derivative \(f’\left( x \right)\) is expressed by the formula

    \[{f’\left( x \right) \text{ = }}\kern0pt{ \sum\limits_{n = 1}^\infty {\left( {n{b_n}\cos nx – n{a_n}\sin nx} \right)} }.\]

    Integration of Fourier Series

    Let \(g\left( x \right)\) be a \(2\pi\)-periodic piecewise continuous function on the interval \(\left[ { – \pi ,\pi } \right].\) Then this function can be integrated term by term on this interval. The Fourier series for \(g\left( x \right)\) is given by

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

    Consider the function

    \[
    {G\left( x \right) = \int\limits_0^x {g\left( t \right)dt} }
    \sim {\frac{{{A_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {\left( {{A_n}\cos nx + {B_n}\sin nx} \right)}}
    \]

    where \({A_n} = – {\large\frac{{{b_n}}}{n}\normalsize},\) \({B_n} = {\large\frac{{{a_n}}}{n}\normalsize}.\)

    By setting \(x = 0,\) we see that

    \[
    {G\left( 0 \right) = 0 }={ \frac{{{A_0}}}{2} + \sum\limits_{n = 1}^\infty {{A_n}} }={ \frac{{{A_0}}}{2} – \sum\limits_{n = 1}^\infty {\frac{{{b_n}}}{n}} \;\;\text{or}\;\;}\kern-0.3pt
    {\frac{{{A_0}}}{2} = \sum\limits_{n = 1}^\infty {\frac{{{b_n}}}{n}} .}
    \]

    Hence, the Fourier series expansion of the function \(G\left( x \right)\) is defined by

    \[
    {G\left( x \right) = \int\limits_0^x {g\left( t \right)dt} }
    = {{\int\limits_0^x {\frac{{{a_0}}}{2}dx} \text{ + }}}\kern0pt{{ \sum\limits_{n = 1}^\infty {\int\limits_0^x {\left( {{a_n}\cos nx + {b_n}\sin nx} \right)dx} } }}
    = {{\frac{{{a_0}x}}{2} \text{ + }}\kern0pt{ \sum\limits_{n = 1}^\infty {\frac{{{a_n}\sin nx + {b_n}\left( {1 – \cos nx} \right)}}{n}}}}
    \]

    where the series on the right-hand side is obtained by the formal term-by-term integration of the Fourier series for \(g\left( x \right).\)

    Because of the presence of the term depending on \(x\) on the right-hand side, this is not clearly a Fourier series expansion of the integral of \(g\left( x \right).\) The result can be rearranged to be a Fourier series expansion of the function

    \[{\Phi \left( x \right) }={ \int\limits_0^x {g\left( t \right)dt} – \frac{{{a_0}x}}{2}.}\]

    The Fourier series of the function \(\Phi\left( x \right)\) is given by the expression

    \[
    {\Phi \left( x \right) = \int\limits_0^x {g\left( t \right)dt} – \frac{{{a_0}x}}{2} }
    = {\frac{{{A_0}}}{2} \text{ + }}\kern0pt{ \sum\limits_{n = 1}^\infty {\left( {{A_n}\cos nx + {B_n}\sin nx} \right)} ,}
    \]

    where the Fourier coefficients are defined by the relationships:

    \[
    {\frac{{{A_0}}}{2} = \sum\limits_{n = 1}^\infty {\frac{{{b_n}}}{n}} ,\;\;\;}\kern0pt
    {{A_n} = – \frac{{{b_n}}}{n},\;\;\;}\kern0pt
    {{B_n} = \frac{{{a_n}}}{n}.}
    \]


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Find the Fourier series of the sign function
    \[ {f\left( x \right) = \text{sign}\,x }= {\begin{cases} -1, & -\pi \le x \le 0 \\ 1, & 0 \lt x \le \pi \end{cases},} \]
    knowing that the Fourier series expansion of the function \(F\left( x \right) = \left| x \right|\) on the interval \(\left[ { – \pi ,\pi } \right]\) is given by
    \[{F\left( x \right) = \left| x \right| }={ \frac{\pi }{2} – \frac{4}{\pi }\sum\limits_{n = 0}^\infty {\frac{{\cos \left( {2n + 1} \right)x}}{{{{\left( {2n + 1} \right)}^2}}}}.\;}\]

    Example 2

    Find the Fourier series expansion of the function \(f\left( x \right) = {x^2}\) knowing that
    \[ {x = 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { – 1} \right)}^{n + 1}}}}{n}\sin nx} \;\;\;}\kern-0.3pt {\text{for}\; – \pi \le x \le \pi .} \]

    Example 3

    Find the Fourier series of the function \(f\left( x \right) = {x^3}\) knowing that
    \[ {{x^2} = \frac{{{\pi ^2}}}{3} }+{ 4\sum\limits_{n = 1}^\infty {\frac{{{{\left( { – 1} \right)}^n}}}{{{n^2}}}\cos nx} \;\;\;}\kern-0.3pt {\text{for}\; – \pi \le x \le \pi .} \]

    Example 4

    Investigate the process of differentiating term by term the Fourier series expansion of the function \(f\left( x \right) = x\) defined on the interval \(\left[ { – \pi ,\pi } \right].\)

    Example 1.

    Find the Fourier series of the sign function
    \[ {f\left( x \right) = \text{sign}\,x }= {\begin{cases} -1, & -\pi \le x \le 0 \\ 1, & 0 \lt x \le \pi \end{cases},} \]
    knowing that the Fourier series expansion of the function \(F\left( x \right) = \left| x \right|\) on the interval \(\left[ { – \pi ,\pi } \right]\) is given by
    \[{F\left( x \right) = \left| x \right| }={ \frac{\pi }{2} – \frac{4}{\pi }\sum\limits_{n = 0}^\infty {\frac{{\cos \left( {2n + 1} \right)x}}{{{{\left( {2n + 1} \right)}^2}}}}.\;}\]

    Solution.

    As \(f\left( x \right) = F’\left( x \right)\) for all \(x \ne 0,\) we obtain

    \[{f\left( x \right) = \frac{d}{{dx}}\Big[ {\frac{\pi }{2} \text{ − }}}\kern0pt{{ \frac{4}{\pi }\sum\limits_{n = 0}^\infty {\frac{{\cos \left( {2n + 1} \right)x}}{{{{\left( {2n + 1} \right)}^2}}}} } \Big] }\]

    or

    \[f\left( x \right) = \frac{4}{\pi }\sum\limits_{n = 0}^\infty {\frac{{\sin\left( {2n + 1} \right)x}}{{2n + 1}}} .\]

    The graphs of the given function and its Fourier approximation are shown in Figure \(1.\)

    Fourier approximation of the sign function
    Figure 1, n = 5, n = 50

    Example 2.

    Find the Fourier series expansion of the function \(f\left( x \right) = {x^2}\) knowing that
    \[ {x = 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { – 1} \right)}^{n + 1}}}}{n}\sin nx} \;\;\;}\kern-0.3pt {\text{for}\; – \pi \le x \le \pi .} \]

    Solution.

    As the function \(f\left( x \right)\) is piecewise continuous on the interval \(\left[ { – \pi ,\pi } \right],\) we may integrate its Fourier series term by term to obtain

    \[{\int\limits_{ – \pi }^x {tdt} \text{ = }}\kern0pt{ 2\sum\limits_{n = 1}^\infty {\int\limits_{ – \pi }^x {\frac{{{{\left( { – 1} \right)}^{n + 1}}}}{n}\sin nt\,dt} } .}\]

    Consequently,

    \[ {\frac{{{x^2}}}{2} – \frac{{{\pi ^2}}}{2} \text{ = }}\kern0pt{ 2\sum\limits_{n = 1}^\infty {{{\left( { – 1} \right)}^{n + 1}}\left[ {\left. {\left( { – \frac{{\cos nt}}{{{n^2}}}} \right)} \right|_{ – \pi }^x } \right]} ,\;\;}\Rightarrow {{\frac{{{x^2}}}{2} – \frac{{{\pi ^2}}}{2} \text{ = }}}\kern0pt{{ 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { – 1} \right)}^n}}}{{{n^2}}}\left[ {\cos nx }\right.}-{\left.{ \cos \left( { – \pi n} \right)} \right]} ,\;\;}}\Rightarrow {{\frac{{{x^2}}}{2} – \frac{{{\pi ^2}}}{2} }={ 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { – 1} \right)}^n}}}{{{n^2}}}\cos nx} }}-{{ 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { – 1} \right)}^n}{{\left( { – 1} \right)}^n}}}{{{n^2}}}} ,\;\;}}\Rightarrow {{\frac{{{x^2}}}{2} – \frac{{{\pi ^2}}}{2} }={ 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { – 1} \right)}^n}}}{{{n^2}}}\cos nx} }}-{{ 2\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} .}} \]

    We know from Example \(1\) on the page Bessel’s Inequality and Parseval’s Theorem that \(\zeta \left( 2 \right) = \sum\limits_{n = 1}^\infty {\large\frac{1}{{{n^2}}}\normalsize} \) \(= {\large\frac{{{\pi ^2}}}{6}\normalsize}.\) Then we get

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

    or

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

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