Fourier Series of Functions with an Arbitrary Period
Fourier Series Expansion on the Interval [−L , L ]
We assume that the function f (x )[−L , L ] . Using the substitution x = Ly /π (−π ≤ x ≤ π ) , we can convert it into the function
\[F\left( y \right) = f\left( {\frac{{Ly}}{\pi }} \right),\]
which is defined and integrable on [−π , π ] . Fourier series expansion of this function F (y )
\[F\left( y \right) = f\left( {\frac{{Ly}}{\pi }} \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos ny + {b_n}\sin ny} \right)} .\]
The Fourier coefficients for the function are given by
\[{a_0} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {F\left( y \right)dy} ,\]
\[{a_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {F\left( y \right)\cos nydy} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( {\frac{{Ly}}{\pi }} \right)\cos nydy} ,\]
\[{b_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {F\left( y \right)\sin nydy} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( {\frac{{Ly}}{\pi }} \right)\sin nydy} ,\;\; n = 1,2,3, \ldots \]
Returning to the initial variables, that is setting \(y = {\frac{{\pi x}}{L}},\) we obtain the following trigonometric series for \(f\left( x \right):\)
\[f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos \frac{{n\pi x}}{L} + {b_n}\sin\frac{{n\pi x}}{L}} \right)},\]
where
\[{a_0} = \frac{1}{L}\int\limits_{ - L}^L {f\left( x \right)dx} ,\;\; {a_n} = \frac{1}{L}\int\limits_{ - L}^L {f\left( x \right)\cos \frac{{n\pi x}}{L}dx} ,\;\; {b_n} = \frac{1}{L}\int\limits_{ - L}^L {f\left( x \right)\sin\frac{{n\pi x}}{L}dx} .\]
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
\[f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos \frac{{n\pi x}}{L} + {b_n}\sin\frac{{n\pi x}}{L}} \right)},\]
where \(L = \frac{{b - a}}{2}\) and Fourier coefficients are calculated as follows:
\[{a_0} = \frac{1}{L}\int\limits_{ - L}^L {f\left( x \right)dx} ,\;\; {a_n} = \frac{1}{L}\int\limits_{ - L}^L {f\left( x \right)\cos \frac{{n\pi x}}{L}dx} ,\;\; {b_n} = \frac{1}{L}\int\limits_{ - L}^L {f\left( x \right)\sin\frac{{n\pi x}}{L}dx} ,\;\; n = 1,2,3, \ldots \]
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
\[{a_0} = \frac{2}{L}\int\limits_0^L {f\left( x \right)dx} ,\;\; {a_n} = \frac{2}{L}\int\limits_0^L {f\left( x \right)\cos \frac{{n\pi x}}{L}dx}.\]
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
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 \(n = 2k,\) \(k = 1,2,3, \ldots \)
\[{b_{2k}} = \frac{A}{{2k\pi }}\left[ {1 + {{\left( { - 1} \right)}^{2k + 1}}} \right] = 0.\]
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
\[f\left( x \right) = \frac{A}{2} + \frac{{2A}}{\pi }\sum\limits_{k = 1}^\infty {\frac{1}{{2k - 1}}\sin \left( {\frac{{2k - 1}}{L}\pi x} \right)}.\]
Figure 1, A = 2, L = 2, n = 2 , n = 10
Example 2.
Find the Fourier series of the function:
\[
{f\left( x \right) }=
{\begin{cases}
0, & -1 \le x \le 0 \\
x, & 0 \lt x \le 1
\end{cases}.}
\]
Solution.
Here \(L = 1.\) Then we can write:
\[{a_0} = \frac{1}{L}\int\limits_a^b {f\left( x \right)dx} = \int\limits_{ - 1}^1 {f\left( x \right)dx} = \int\limits_0^1 {xdx} = \left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_0^1 = \frac{1}{2}.\]
Calculate the coefficients \({a_n}:\)
\[{a_n} = \frac{1}{L}\int\limits_a^b {f\left( x \right)\cos \frac{{n\pi x}}{L}dx} = \int\limits_0^1 {x\cos \left( {n\pi x} \right)dx} = \left. {\left( {\frac{1}{{n\pi }}x\sin \left( {n\pi x} \right)} \right)} \right|_0^1 - \frac{1}{{n\pi }}\int\limits_0^1 {\sin \left( {n\pi x} \right)dx} = \frac{1}{{n\pi }}\left[ {\left. {\left( {x\sin n\pi x} \right)} \right|_0^1 + \left. {\left( {\frac{{\cos n\pi x}}{{n\pi }}} \right)} \right|_0^1} \right] = \frac{1}{{n\pi }}\left[ {\sin n\pi + \frac{{\cos n\pi }}{{n\pi }} - \frac{1}{{n\pi }}} \right] = \frac{1}{{{n^2}{\pi ^2}}}\left[ {\cos n\pi - 1} \right] = \frac{1}{{{n^2}{\pi ^2}}}\left[ {{{\left( { - 1} \right)}^n} - 1} \right].\]
Determine now the coefficients \({b_n}:\)
\[{b_n} = \frac{1}{L}\int\limits_a^b {f\left( x \right)\sin\frac{{n\pi x}}{L}dx} = \int\limits_0^1 {x\sin \left( {n\pi x} \right)dx} = \left. {\left( { - \frac{1}{{n\pi }}x\cos \left( {n\pi x} \right)} \right)} \right|_0^1 + \frac{1}{{n\pi }}\int\limits_0^1 {\cos\left( {n\pi x} \right)dx} = \frac{1}{{n\pi }}\left[ {\left. { - \left( {x\cos n\pi x} \right)} \right|_0^1 + \left. {\left( {\frac{{\sin n\pi x}}{{n\pi }}} \right)} \right|_0^1} \right] = \frac{1}{{n\pi }}\left[ { - \cos n\pi + \frac{{\sin n\pi }}{{n\pi }}} \right] = \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{n\pi }}.\]
As a result, we obtain (Figure \(2\)):
\[f\left( x \right) = \frac{1}{4} + \sum\limits_{n = 1}^\infty {\left[ {\frac{{\left( {{{\left( { - 1} \right)}^n} - 1} \right)}}{{{n^2}{\pi ^2}}}\cos n\pi x + \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{n\pi }}\sin n\pi x} \right]} .\]
Figure 2, n = 5 , n = 10
See more problems on Page 2.
