Complex Form of Fourier Series
Let the function f (x) be defined on the interval [−π, π]. Using the well-known Euler's formulas
\[\cos \varphi = \frac{{{e^{i\varphi }} + {e^{ - i\varphi }}}}{2},\;\; \sin \varphi = \frac{{{e^{i\varphi }} - {e^{ - i\varphi }}}}{{2i}},\]
we can write the Fourier series of the function in complex form:
\[f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos nx + {b_n}\sin nx} \right)} = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\frac{{{e^{inx}} + {e^{ - inx}}}}{2} + {b_n}\frac{{{e^{inx}} - {e^{ - inx}}}}{{2i}}} \right)} = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\frac{{{a_n} - i{b_n}}}{2}{e^{inx}}} + \sum\limits_{n = 1}^\infty {\frac{{{a_n} + i{b_n}}}{2}{e^{ - inx}}} = \sum\limits_{n = - \infty }^\infty {{c_n}{e^{inx}}} .\]
Here we have used the following notations:
\[{c_0} = \frac{{{a_0}}}{2},\;\; {c_n} = \frac{{{a_n} - i{b_n}}}{2},\;\; {c_{ - n}} = \frac{{{a_n} + i{b_n}}}{2}.\]
The coefficients \({c_n}\) are called complex Fourier coefficients. They are defined by the formulas
\[{c_n} = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {f\left( x \right){e^{ - inx}}dx} ,\;\; n = 0, \pm 1, \pm 2, \ldots\]
If necessary to expand a function \(f\left( x \right)\) of period \(2L,\) we can use the following expressions:
\[f\left( x \right) = \sum\limits_{n = - \infty }^\infty {{c_n}{e^{\frac{{in\pi x}}{L}}}} ,\]
where
\[{c_n} = \frac{1}{{2L}}\int\limits_{ - L}^L {f\left( x \right){e^{ - \frac{{in\pi x}}{L}}}dx} ,\;\; n = 0, \pm 1, \pm 2, \ldots \]
The complex form of Fourier series is algebraically simpler and more symmetric. Therefore, it is often used in physics and other sciences.
Solved Problems
Example 1.
Using complex form, find the Fourier series of the function
\[
f\left( x \right) = \text{sign}\,x =
\begin{cases}
-1, & -\pi \le x \le 0 \\
1, & 0 \lt x \le \pi
\end{cases}.\]
Solution.
We calculate the coefficients \({c_0}\) and \({c_n}\) for \(n \ne 0:\)
\[{c_0} = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {f\left( x \right)dx} = \frac{1}{{2\pi }}\left[ {\int\limits_{ - \pi }^0 {\left( { - 1} \right)dx} + \int\limits_0^\pi {dx} } \right] = \frac{1}{{2\pi }}\left[ {\left. {\left( { - x} \right)} \right|_{ - \pi }^0 + \left. x \right|_0^\pi } \right] = \frac{1}{{2\pi }}\left( { - \cancel{\pi} + \cancel{\pi }} \right) = 0,\]
\[{c_n} = \frac{1}{{2\pi }}\int\limits_{ - \pi }^\pi {f\left( x \right){e^{ - inx}}dx} = \frac{1}{{2\pi }}\left[ {\int\limits_{ - \pi }^0 {\left( { - 1} \right){e^{ - inx}}dx} + \int\limits_0^\pi {{e^{ - inx}}dx} } \right] = \frac{1}{{2\pi }}\left[ { - \frac{{\left. {\left( {{e^{ - inx}}} \right)} \right|_{ - \pi }^0}}{{ - in}} + \frac{{\left. {\left( {{e^{ - inx}}} \right)} \right|_0^\pi }}{{ - in}}} \right] = \frac{i}{{2\pi n}}\left[ { - \left( {1 - {e^{in\pi }}} \right) + {e^{ - in\pi }} - 1} \right] = \frac{i}{{2\pi n}}\left[ {{e^{in\pi }} + {e^{ - in\pi }} - 2} \right] = \frac{i}{{\pi n}}\left[ {\frac{{{e^{in\pi }} + {e^{ - in\pi }}}}{2} - 1} \right] = \frac{i}{{\pi n}}\left[ {\cos n\pi - 1} \right] = \frac{i}{{\pi n}}\left[ {{{\left( { - 1} \right)}^n} - 1} \right].\]
If \(n = 2k,\) then \({c_{2k}} = 0.\) If \(n = 2k - 1,\) then \({c_{2k - 1}} = - {\frac{{2i}}{{\left( {2k - 1} \right)\pi }}}.\)
Hence, the Fourier series of the function in complex form is
\[f\left( x \right) = \text{sign}\,x = - \frac{{2i}}{\pi }\sum\limits_{k = - \infty }^\infty {\frac{1}{{2k - 1}}{e^{i\left( {2k - 1} \right)x}}} .\]
We can transform the series and write it in the real form. Rename: \(n = 2k - 1,\) \(n = \pm 1, \pm 2, \pm 3, \ldots \) Then
\[f\left( x \right) = \text{sign}\,x = - \frac{{2i}}{\pi }\sum\limits_{k = - \infty }^\infty {\frac{1}{{2k - 1}}{e^{i\left( {2k - 1} \right)x}}} = - \frac{{2i}}{\pi }\sum\limits_{n = - \infty }^\infty {\frac{{{e^{inx}}}}{n}} = - \frac{{2i}}{\pi }\sum\limits_{n = 1}^\infty {\left( {\frac{{{e^{ - inx}}}}{{ - n}} + \frac{{{e^{inx}}}}{n}} \right)} = \frac{4}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{e^{inx}} - {e^{ - inx}}}}{{2in}}} = \frac{4}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\sin nx}}{n}} = \frac{4}{\pi }\sum\limits_{k = 1}^\infty {\frac{{\sin \left( {2k - 1} \right)x}}{{2k - 1}}} .\]
Graph of the function and its Fourier approximation for \(n = 5\) and \(n = 50\) are shown in Figure \(1.\)
Figure 1, n = 5, n = 50
See more problems on Page 2.
