Complex Form of Fourier Series

Solved Problems

Example 2.

Using complex form find the Fourier series of the function \(f\left( x \right) = {x^2},\) defined on the interval \(\left[ { - 1,1} \right].\)

Solution.

Here the half-period is \(L = 1.\) Therefore, the coefficient \({c_0}\) is

\[ {c_0} = \frac{1}{{2L}}\int\limits_{ - L}^L {f\left( x \right)dx} = \frac{1}{2}\int\limits_{ - 1}^1 {{x^2}dx} = \frac{1}{2}\left[ {\left. {\left( {\frac{{{x^3}}}{3}} \right)} \right|_{ - 1}^1} \right] = \frac{1}{6}\left[ {{1^3} - {{\left( { - 1} \right)}^3}} \right] = \frac{1}{3}.\]

For \(n \ne 0,\)

\[{c_n} = \frac{1}{{2L}}\int\limits_{ - L}^L {f\left( x \right){e^{ - \frac{{in\pi x}}{L}}}dx} = \frac{1}{2}\int\limits_{ - 1}^1 {{x^2}{e^{ - {in\pi x}}}dx} .\]

Integrating by parts twice, we obtain

\[ {c_n} = \frac{1}{2}\left[ {\left. {\left( {\frac{{{x^2}{e^{ - in\pi x}}}}{{ - in\pi }}} \right)} \right|_{ - 1}^1 - \int\limits_{ - 1}^1 {\frac{{2x{e^{ - in\pi x}}}}{{ - in\pi }}dx} } \right] = \frac{1}{2}\left[ {\left. {\left( {\frac{{{x^2}{e^{ - in\pi x}}}}{{ - in\pi }}} \right)} \right|_{ - 1}^1 + \frac{2}{{in\pi }}\int\limits_{ - 1}^1 {x{e^{ - in\pi x}}dx} } \right] = - \frac{1}{{2in\pi }}\left[ {{e^{ - in\pi }} + \frac{2}{{in\pi }}{e^{ - in\pi }} + \frac{2}{{{{\left( {in\pi } \right)}^2}}}{e^{ - in\pi }} + \frac{2}{{in\pi }}{e^{in\pi }} - \frac{2}{{{{\left( {in\pi } \right)}^2}}}{e^{in\pi }}} \right] = \frac{1}{{2in\pi }}\left[ {{e^{in\pi }} - {e^{ - in\pi }} - \frac{2}{{in\pi }}\left( {{e^{in\pi }} + {e^{ - in\pi }}} \right) + \frac{2}{{{{\left( {in\pi } \right)}^2}}}\left( {{e^{in\pi }} - {e^{ - in\pi }}} \right)} \right] = \frac{1}{{n\pi }} \cdot \frac{{{e^{in\pi }} - {e^{ - in\pi }}}}{{2i}} + \frac{2}{{{n^2}{\pi ^2}}} \cdot \frac{{{e^{in\pi }} + {e^{ - in\pi }}}}{2} - \frac{2}{{{n^3}{\pi ^3}}} \cdot \frac{{{e^{in\pi }} - {e^{ - in\pi }}}}{{2i}} = \frac{1}{{n\pi }} \sin n\pi + \frac{2}{{{n^2}{\pi ^2}}}\cos n\pi - \frac{2}{{{n^3}{\pi ^3}}}\sin n\pi .\]

Substituting \(\sin n\pi = 0\) and \(\cos n\pi = {\left( { - 1} \right)^n},\) we get the compact expression for the coefficients \({c_n}:\)

\[{c_n} = \frac{2}{{{n^2}{\pi ^2}}}{\left( { - 1} \right)^n}.\]

Thus, the Fourier extension in complex form is given by

\[f\left( x \right) = {x^2} = \frac{1}{3} + \sum\limits_{n = 1}^\infty {\frac{2}{{{n^2}{\pi ^2}}}{{\left( { - 1} \right)}^n}{e^{in\pi x}}} + \sum\limits_{n = 1}^\infty {\frac{2}{{{{\left( { - n} \right)}^2}{\pi ^2}}}{{\left( { - 1} \right)}^{ - n}}{e^{ - in\pi x}}} .\]

Taking into account that \(\left( { - 1} \right)^{ - n} = \left( { - 1} \right)^n,\) we can finally write:

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

Graphs of the function and the Fourier series expansions are shown in Figure \(2.\)

Fourier series of the function f(x)=x^2 — Figure 2, n = 2, n = 5

Example 3.

Using complex form find the Fourier series of the function

\[f\left( x \right) = \frac{{a\sin x}}{{1 - 2a\cos x + {a^2}}},\; \left| a \right| \lt 1.\]

Solution.

We apply the formulas

\[\cos x = \frac{{{e^{ix}} + {e^{ - ix}}}}{2},\;\; \sin x = \frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}.\]

This results in the following expression:

\[f\left( x \right) = \frac{{a \cdot \frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}}}{{1 - 2a \cdot \frac{{{e^{ix}} + {e^{ - ix}}}}{2} + {a^2}}} = \frac{1}{{2i}} \cdot \frac{{a\left( {{e^{ix}} - {e^{ - ix}}} \right)}}{{1 - a\left( {{e^{ix}} + {e^{ - ix}}} \right) + {a^2}}} = \frac{1}{{2i}} \cdot \frac{{a\left( {{e^{ix}} - {e^{ - ix}}} \right)}}{{1 - a{e^{ix}} - a{e^{ - ix}} + {a^2}{e^{ix}}{e^{ - ix}}}} = \frac{1}{{2i}} \cdot \frac{{a\left( {{e^{ix}} - {e^{ - ix}}} \right)}}{{\left( {1 - a{e^{ix}}} \right) - a{e^{ - ix}}\left( {1 - a{e^{ix}}} \right)}} = \frac{1}{{2i}} \cdot \frac{{a\left( {{e^{ix}} - {e^{ - ix}}} \right)}}{{\left( {1 - a{e^{ix}}} \right)\left( {1 - a{e^{ - ix}}} \right)}}.\]

Using partial decomposition, we can write:

\[f\left( x \right) = \frac{1}{{2i}} \cdot \frac{{a\left( {{e^{ix}} - {e^{ - ix}}} \right)}}{{\left( {1 - a{e^{ix}}} \right)\left( {1 - a{e^{ - ix}}} \right)}} = \frac{1}{{2i}}\left( {\frac{A}{{1 - a{e^{ix}}}} + \frac{B}{{1 - a{e^{ - ix}}}}} \right).\]

Calculate the coefficients \(A, B:\)

\[A\left( {1 - a{e^{ - ix}}} \right) + B\left( {1 - a{e^{ix}}} \right) = a{e^{ix}} - a{e^{ - ix}},\;\; \Rightarrow A - aA{e^{ - ix}} + B - aB{e^{ix}} = a{e^{ix}} - a{e^{ - ix}},\;\; \Rightarrow A = 1,\;B = - 1.\]

As a result, the function \(f\left( x \right)\) can be written in the form

\[f\left( x \right) = \frac{1}{{2i}}\left( {\frac{1}{{1 - a{e^{ix}}}} - \frac{1}{{1 - a{e^{ - ix}}}}} \right).\]

We see that

\[\left| {a{e^{ix}}} \right| = \left| a \right|\left| {{e^{ix}}} \right| = \left| a \right|\sqrt {{{\cos }^2}x + {{\sin }^2}x} = \left| a \right| \lt 1.\]

For conjugate, we have the same result:

\[\left| {a{e^{ - ix}}} \right| = \left| {a{e^{ix}}} \right| = \left| a \right| \lt 1.\]

Expanding the fractions into power series, we get

\[\frac{1}{{1 - a{e^{ix}}} = \left( {1 - a{e^{ix}}} \right)^{ - 1}} = \sum\limits_{n = 0}^\infty {{a^n}{e^{inx}}} ,\]

\[\frac{1}{{1 - a{e^{ - ix}}} = \left( {1 - a{e^{ - ix}}} \right)^{ - 1}} = \sum\limits_{n = 0}^\infty {{a^n}{e^{ - inx}}} .\]

Thus, the Fourier series of the function \(f\left( x \right)\) is

\[f\left( x \right) = \frac{1}{{2i}}\sum\limits_{n = 0}^\infty {{a^n}\left( {{e^{inx}} - {e^{ - inx}}} \right)} = \sum\limits_{n = 0}^\infty {{a^n}\sin nx} .\]

Since \(\sin nx = 0,\) the final answer is

\[f\left( x \right) = \sum\limits_{n = 1}^\infty {{a^n}\sin nx} .\]