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

\[
{\cos \varphi = \frac{{{e^{i\varphi }} + {e^{ – i\varphi }}}}{2},\;\;\;}\kern0pt
{\sin \varphi = \frac{{{e^{i\varphi }} – {e^{ – i\varphi }}}}{{2i}},}
\]

we can write the Fourier series of the function in complex form:

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

Here we have used the following notations:

\[
{{c_0} = \frac{{{a_0}}}{2},\;\;\;}\kern0pt
{{c_n} = \frac{{{a_n} – i{b_n}}}{2},\;\;\;}\kern0pt
{{c_{ – n}} = \frac{{{a_n} + i{b_n}}}{2}.}
\]

The coefficients \({c_n}\) are called complex Fourier coefficients. They are defined by the formulas

\[{{c_n} = \frac{1}{{2\pi }}\int\limits_{ – \pi }^\pi {f\left( x \right){e^{ – inx}}dx} ,\;\;}\kern0pt{n = 0, \pm 1, \pm 2, \ldots }\]

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

\[{{c_n} = \frac{1}{{2L}}\int\limits_{ – L}^L {f\left( x \right){e^{ – \frac{{in\pi x}}{L}}}dx} ,\;\;}\kern0pt{n = 0, \pm 1, \pm 2, \ldots }\]

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 2

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

Example 3

Using complex form find the Fourier series of the function
\[{f\left( x \right) = \frac{{a\sin x}}{{1 – 2a\cos x + {a^2}}},\;\;}\kern-0.3pt{\left| a \right| \lt 1.}\]

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 \(n \ne 0:\)

\[\require{cancel}
{{c_0} = \frac{1}{{2\pi }}\int\limits_{ – \pi }^\pi {f\left( x \right)dx} }
= {\frac{1}{{2\pi }}\left[ {\int\limits_{ – \pi }^0 {\left( { – 1} \right)dx} + \int\limits_0^\pi {dx} } \right] }
= {\frac{1}{{2\pi }}\left[ {\left. {\left( { – x} \right)} \right|_{ – \pi }^0 + \left. x \right|_0^\pi } \right] }
= {\frac{1}{{2\pi }}\left( { – \cancel{\pi} + \cancel{\pi }} \right) }={ 0,}
\]

\[
{{c_n} = \frac{1}{{2\pi }}\int\limits_{ – \pi }^\pi {f\left( x \right){e^{ – inx}}dx} }
= {{\frac{1}{{2\pi }}\left[ {\int\limits_{ – \pi }^0 {\left( { – 1} \right){e^{ – inx}}dx} }\right.}+{\left.{ \int\limits_0^\pi {{e^{ – inx}}dx} } \right] }}
= {{\frac{1}{{2\pi }}\left[ { – \frac{{\left. {\left( {{e^{ – inx}}} \right)} \right|_{ – \pi }^0}}{{ – in}} }\right.}+{\left.{ \frac{{\left. {\left( {{e^{ – inx}}} \right)} \right|_0^\pi }}{{ – in}}} \right] }}
= {{\frac{i}{{2\pi n}}\left[ { – \left( {1 – {e^{in\pi }}} \right) }\right.}+{\left.{ {e^{ – in\pi }} – 1} \right] }}
= {\frac{i}{{2\pi n}}\left[ {{e^{in\pi }} + {e^{ – in\pi }} – 2} \right] }
= {\frac{i}{{\pi n}}\left[ {\frac{{{e^{in\pi }} + {e^{ – in\pi }}}}{2} – 1} \right] }
= {\frac{i}{{\pi n}}\left[ {\cos n\pi – 1} \right] }
= {\frac{i}{{\pi n}}\left[ {{{\left( { – 1} \right)}^n} – 1} \right].}
\]

If \(n = 2k,\) then \({c_{2k}} = 0.\) If \(n = 2k – 1,\) then \({c_{2k – 1}} \) \(= – {\large\frac{{2i}}{{\left( {2k – 1} \right)\pi }}\normalsize}.\)

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
Page 1
Problem 1
Page 2
Problems 2-3