Calculus

Fourier Series

Fourier Series of Functions with an Arbitrary Period

Page 1
Problem 1
Page 2
Problems 2-4

Fourier Series Expansion on the Interval \(\left[ { – L,L} \right]\)

We assume that the function \(f\left( x \right)\) is piecewise continuous on the interval \(\left[ { – L,L} \right].\) Using the substitution \(x = \) \({\large\frac{{Ly}}{\pi }\normalsize}\) \(\left( { – \pi \le x \le \pi } \right)\), we can transform it into the function
\[F\left( y \right) = f\left( {\frac{{Ly}}{\pi }} \right),\] which is defined and integrable on \(\left[ { – \pi ,\pi } \right].\) Fourier series expansion of this function \(F\left( y \right)\) can be written as
\[
{F\left( y \right) = f\left( {\frac{{Ly}}{\pi }} \right) }
= {\frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos ny + {b_n}\sin ny} \right)} .}
\] The Fourier coefficients for the function are given by
\[{a_0} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {F\left( y \right)dy} ,\] \[
{{a_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {F\left( y \right)\cos nydy} }
= {\frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( {\frac{{Ly}}{\pi }} \right)\cos nydy} ,}
\] \[
{{b_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {F\left( y \right)\sin nydy} }
= {\frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( {\frac{{Ly}}{\pi }} \right)\sin nydy} ,\;\;}\kern-0.3pt
{n = 1,2,3, \ldots }
\] Returning to the initial variables, i.e. setting \(y = {\large\frac{{\pi x}}{L}\normalsize},\) we obtain the following trigonometric series for \(f\left( x \right):\)
\[
{f\left( x \right) = \frac{{{a_0}}}{2} \text{ + }}\kern0pt{ \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos \frac{{n\pi x}}{L} + {b_n}\sin\frac{{n\pi x}}{L}} \right)}}
\] where
\[
{{{a_0} }={ \frac{1}{L}\int\limits_{ – L}^L {f\left( x \right)dx} ,\;\;}}\kern-0.3pt
{{{a_n} }={ \frac{1}{L}\int\limits_{ – L}^L {f\left( x \right)\cos \frac{{n\pi x}}{L}dx} ,\;\;}}\kern-0.3pt
{{{b_n} }={ \frac{1}{L}\int\limits_{ – L}^L {f\left( x \right)\sin\frac{{n\pi x}}{L}dx} .}}
\]

Fourier Series Expansion on the Interval \(\left[ { a,b} \right]\)

If the function \(f\left( x \right)\) is defined on the interval \(\left[ { a,b} \right],\) then its Fourier series representation is given by the same formula
\[{f\left( x \right) = \frac{{{a_0}}}{2} \text{ + }}\kern0pt{ \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos \frac{{n\pi x}}{L} + {b_n}\sin\frac{{n\pi x}}{L}} \right)}}\] where \(L = \large\frac{{b – a}}{2}\normalsize\) and Fourier coefficients are calculated as follows:
\[
{{{a_0} }={ \frac{1}{L}\int\limits_{ – L}^L {f\left( x \right)dx} ,\;\;}}\kern-0.3pt
{{{a_n} }={ \frac{1}{L}\int\limits_{ – L}^L {f\left( x \right)\cos \frac{{n\pi x}}{L}dx} ,\;\;}}\kern-0.3pt
{{{b_n} }={ \frac{1}{L}\int\limits_{ – L}^L {f\left( x \right)\sin\frac{{n\pi x}}{L}dx} ,\;\;}}\kern-0.3pt
{n = 1,2,3, \ldots }
\]

Even and Odd Functions

The Fourier series expansion of an even function, defined on the interval \(\left[ { – L,L} \right]\) has the form:
\[{f\left( x \right) }={ \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos \frac{{n\pi x}}{L}} ,}\] where
\[
{{a_0} = \frac{2}{L}\int\limits_0^L {f\left( x \right)dx} ,\;\;}\kern-0.3pt
{{a_n} = \frac{2}{L}\int\limits_0^L {f\left( x \right)\cos \frac{{n\pi x}}{L}dx}.}
\] The Fourier series expansion of an odd function defined on the interval \(\left[ { – L,L} \right]\) is expressed by the formula
\[f\left( x \right) = \sum\limits_{n = 1}^\infty {{b_n}\sin\frac{{n\pi x}}{L}} ,\] where the Fourier coefficients are
\[{{b_n} }={ \frac{2}{L}\int\limits_0^L {f\left( x \right)\sin\frac{{n\pi x}}{L}dx} .}\]

Solved Problems

Click on problem description to see solution.

 Example 1

Find the Fourier series of the function
\[
{f\left( x \right) }=
{\begin{cases}
A, & 0 \le x \le L \\
0, & L \lt x \le 2L
\end{cases}.}
\]

 Example 2

Find the Fourier series of the function:
\[
{f\left( x \right) }=
{\begin{cases}
0, & -1 \le x \le 0 \\
x, & 0 \lt x \le 1
\end{cases}.}
\]

 Example 3

Find the Fourier series of the trapezoidal wave defined by the function
\[
{f\left( x \right) }=
{\begin{cases}
x, & 0 \le x \le 1 \\
1, & 1 \lt x \le 2 \\
3-x, & 2 \lt x \le 3
\end{cases}}
\]

 Example 4

Find the Fourier series of the function \(f\left( x \right) = {\cos ^2}x.\)

Example 1.

Find the Fourier series of the function
\[
{f\left( x \right) }=
{\begin{cases}
A, & 0 \le x \le L \\
0, & L \lt x \le 2L
\end{cases}.}
\]

Solution.

Determine the Fourier coefficients:
\[
{{a_0} }={ \frac{1}{L}\int\limits_a^b {f\left( x \right)dx} }
= {\frac{1}{L}\int\limits_0^L {Adx} }={ A,}
\] \[
{{a_n} }={ \frac{1}{L}\int\limits_a^b {f\left( x \right)\cos \frac{{n\pi x}}{L}dx} }
= {\frac{1}{L}\int\limits_a^b {A\cos \frac{{n\pi x}}{L}dx} }
= {\frac{A}{L}\left[ {\left. {\left( {\frac{L}{{n\pi }}\sin\frac{{n\pi x}}{L}} \right)} \right|_0^L} \right] }
= {\frac{A}{{n\pi }}\left( {\sin n\pi – \sin 0} \right) }={ 0,}
\] \[
{{b_n} }={ \frac{1}{L}\int\limits_a^b {f\left( x \right)\sin\frac{{n\pi x}}{L}dx} }
= {\frac{1}{L}\int\limits_a^b {A\sin \frac{{n\pi x}}{L}dx} }
= {\frac{A}{L}\left[ {\left. {\left( { – \frac{L}{{n\pi }}\cos\frac{{n\pi x}}{L}} \right)} \right|_0^L} \right] }
= {\frac{A}{{n\pi }}\left[ { – \cos n\pi + \cos 0} \right] }
= {\frac{A}{{n\pi }}\left[ {1 – {{\left( { – 1} \right)}^n}} \right] }
= {\frac{A}{{n\pi }}\left[ {1 + {{\left( { – 1} \right)}^{n + 1}}} \right].}
\] One can notice that for even \(n = 2k,\) \(k = 1,2,3, \ldots \)
\[{{b_{2k}} }={ \frac{A}{{2k\pi }}\left[ {1 + {{\left( { – 1} \right)}^{2k + 1}}} \right] }={ 0.}\]

Fourier series for a rectangular function

Figure 1, A = 2, L = 2, n = 2, n = 10

For odd \(n = 2k – 1,\) \(k = 1,2,3, \ldots \)
\[
{{b_{2k – 1}} }={ \frac{A}{{\left( {2k – 1} \right)\pi }}\left[ {1 + {{\left( { – 1} \right)}^{2k}}} \right] }
= {\frac{{2A}}{{\left( {2k – 1} \right)\pi }}.}
\] Hence, the Fourier series expansion of the given function (Figure \(1\)) is
\[{f\left( x \right) = \frac{A}{2} \text{ + }}\kern0pt{ \frac{{2A}}{\pi }\sum\limits_{k = 1}^\infty {\frac{1}{{2k – 1}}\sin \left( {\frac{{2k – 1}}{L}\pi x} \right)}}\]

Page 1
Problem 1
Page 2
Problems 2-4