Calculus

Fourier Series

Differentiation and Integration of Fourier Series

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

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 on problem description to see 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