Calculus

Fourier Series

Fourier Series Logo

Complex Form of Fourier Series

Let the function \(f\left( x \right)\) be defined on the interval \(\left[ { - \pi ,\pi } \right].\) 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

Click or tap a problem to see the solution.

Example 1

Using complex form, find the Fourier series of the function

\[ f\left( x \right) = \text{sign}\,x = \begin{cases} -1, & -\pi \le x \le 0 \\ 1, & 0 \lt x \le \pi \end{cases}.\]

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 \(1.\)

Fourier series of the function f(x)=sign(x)
Figure 1, n = 5, n = 50

See more problems on Page 2.

Page 1 Page 2