Fourier Series of Functions with an Arbitrary Period
Solved Problems
Example 3.
Find the Fourier series of the trapezoidal wave defined by the function
\[
{f\left( x \right) }=
{\begin{cases}
x, & 0 \le x \le 1 \\
1, & 1 \lt x \le 2 \\
3-x, & 2 \lt x \le 3
\end{cases}.}
\]
Solution.
In the given case, obviously, \(L = {\frac{3}{2}}.\) Calculate the coefficients of the expansion \({a_0}\) and \({a_n}.\)
\[{a_0} = \frac{1}{L}\int\limits_a^b {f\left( x \right)dx} = \frac{2}{3}\int\limits_0^3 {f\left( x \right)dx} = \frac{2}{3}\left[ {\int\limits_0^1 {xdx} + \int\limits_1^2 {1dx} + \int\limits_2^3 {\left( {3 - x} \right)dx} } \right] = \frac{2}{3}\left[ {\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_0^1 + \left. x \right|_0^1 + \left. {\left( {3x - \frac{{{x^2}}}{2}} \right)} \right|_2^3} \right] = \frac{4}{3};\]
\[
{a_n} = \frac{1}{L}\int\limits_a^b {f\left( x \right)\cos \frac{{n\pi x}}{L}dx}
= \frac{2}{3}\int\limits_0^3 {f\left( x \right)\cos \frac{{2n\pi x}}{3}dx}
= \frac{2}{3}\left\{ {\int\limits_0^1 {x\cos \frac{{2n\pi x}}{3}dx} } \right.
+ \int\limits_1^2 {\cos \frac{{2n\pi x}}{3}dx}
+ \left. {\int\limits_2^3 {\left( {3 - x} \right)\cos \frac{{2n\pi x}}{3}dx} } \right\}
= \frac{2}{3}\left\{ {\left[ {\left. {\left( {\frac{3}{{2n\pi }}x\sin\frac{{2n\pi x}}{3}} \right)} \right|_0^1 - \int\limits_0^1 {\frac{3}{{2n\pi }}\sin\frac{{2n\pi x}}{3}dx} } \right]} \right.
+ \left. {\left( {\frac{3}{{2n\pi }}\sin\frac{{2n\pi x}}{3}} \right)} \right|_1^2
+ \left. {\left[ {\left. {\left( {\frac{3}{{2n\pi }}\left( {3 - x} \right)\sin\frac{{2n\pi x}}{3}} \right)} \right|_2^3 + \int\limits_2^3 {\frac{3}{{2n\pi }}\sin\frac{{2n\pi x}}{3}dx} } \right]} \right\}
= \frac{2}{3}\left\{ {\frac{3}{{2n\pi }}\sin\frac{{2n\pi }}{3}} \right.
+ \frac{9}{{4{n^2}{\pi ^2}}}\left( {\cos\frac{{2n\pi }}{3} - 1} \right)
+ \frac{3}{{2n\pi }}\left( {\sin\frac{{4n\pi }}{3} - \sin\frac{{2n\pi }}{3}} \right)
- \frac{3}{{2n\pi }}\sin\frac{{4n\pi }}{3}
+ \left. {\frac{9}{{4{n^2}{\pi ^2}}}\left( { \text{-}\cos 2n\pi + \cos\frac{{4n\pi }}{3}} \right)} \right\}
= \frac{2}{3}\left\{ {\frac{9}{{4{n^2}{\pi ^2}}}\left( {\cos\frac{{2n\pi }}{3} - 1} \right) + \frac{9}{{4{n^2}{\pi ^2}}}\left( {\cos\frac{{4n\pi }}{3} - 1} \right)} \right\}.\]
Since
\[\cos {\frac{{4n\pi }}{3}} = \cos \left( {2n\pi - {\frac{{2n\pi }}{3}}} \right) = \cos {\frac{{2n\pi }}{3}},\]
we obtain
\[{a_n} = \frac{2}{3} \cdot \frac{{2 \cdot 9}}{{4{n^2}{\pi ^2}}}\left( {\cos \frac{{2n\pi }}{3} - 1} \right) = \frac{3}{{{n^2}{\pi ^2}}}\left( {\cos \frac{{2n\pi }}{3} - 1} \right),\;\; n = 1,2,3, \ldots \]
The coefficients \({b_n}\) are zero because the function is even on the given interval \(\left[ {0,3} \right].\) Then the Fourier series expansion is expressed by the formula
\[f\left( x \right) = \frac{2}{3} - \frac{3}{{{\pi ^2}}} {\sum\limits_{n = 1}^\infty {\frac{{1 - \cos \frac{{2n\pi }}{3}}}{{{n^2}}}\cos \frac{{2n\pi x}}{3}}}.\]
Graphs of the given function and its Fourier approximation for \(n = 1\) and \(n = 3\) are shown in Figure \(3.\)
Example 4.
Find the Fourier series of the function \[f\left( x \right) = {\cos ^2}x.\]
Solution.
This function is even with period \(\pi\) \(\left( {L = {\frac{\pi }{2}}} \right).\) Therefore, \({b_n} = 0.\) Find the coefficients \({a_0}\) and \({a_n}.\)
\[{a_0} = \frac{2}{L}\int\limits_0^L {f\left( x \right)dx} = \frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}xdx} = \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {\left( {1 + \cos 2x} \right)dx} = \frac{2}{\pi }\left[ {\left. {\left( {x + \frac{{\sin 2x}}{2}} \right)} \right|_0^{\frac{\pi }{2}}} \right] = \frac{2}{\pi }\left[ {\frac{\pi }{2} + \frac{{\sin \pi }}{2}} \right] = 1.\]
\[{a_n} = \frac{2}{L}\int\limits_0^L {f\left( x \right)\cos \frac{{n\pi x}}{L}dx} = \frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}x\cos 2nxdx} = \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {\left( {1 + \cos 2x} \right)\cos 2nxdx} = \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {\left( {\cos 2nx + \cos 2x\cos 2nx} \right)dx} = \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {\left[ {2\cos 2nx + \cos \left( {2n - 2} \right)x + \cos \left( {2n + 2} \right)x} \right]dx} = \frac{1}{\pi }\left. {\left[ {\sin \frac{{2nx}}{n} + \sin \frac{{\left( {2n - 2} \right)x}}{{2n - 2}} + \sin \frac{{\left( {2n + 2} \right)x}}{{2n + 2}}} \right]} \right|_0^{\frac{\pi }{2}} = \frac{1}{\pi }\left[ {\frac{{\sin n\pi }}{n} + \frac{{\sin \left( {n - 1} \right)\pi }}{{2n - 2}} + \frac{{\sin \left( {n + 1} \right)\pi }}{{2n + 2}}} \right] = 0.\]
However, this result is valid only for \(n \ge 2.\) Therefore we calculate the coefficient \({a_1}\) separately:
\[{a_1} = \frac{4}{\pi }\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}x\cos 2xdx} = \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {\left( {1 + \cos 2x} \right)\cos 2xdx} = \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {\left( {\cos 2x + {{\cos }^2}2x} \right)dx} = \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {\left( {\cos 2x + \frac{{1 + \cos 4x}}{2}} \right)dx} = \frac{1}{\pi }\int\limits_0^{\frac{\pi }{2}} {\left( {2\cos 2x + 1 + \cos 4x} \right)dx} = \frac{1}{\pi }\left[ {\left. {\left( {\sin 2x + x + \frac{{\sin 4x}}{4}} \right)} \right|_0^{\frac{\pi }{2}}} \right] = \frac{1}{\pi }\left( {\sin \pi + \frac{\pi }{2} + \frac{{\sin 2\pi }}{4}} \right) = \frac{1}{2}.\]
Thus, the Fourier series of the function \(f\left( x \right) = {\cos ^2}x\) is given by
\[f\left( x \right) = {\cos ^2}x = \frac{1}{2} + \frac{1}{2}\cos 2x.\]
This result is the well-known trigonometric identity.