Definition of Fourier Series and Typical Examples
Baron Jean Baptiste Joseph Fourier (1768 − 1830) introduced the idea that any periodic function can be represented by a series of sines and cosines which are harmonically related.
Fig.1 Baron Jean Baptiste Joseph Fourier (1768−1830) To consider this idea in more detail, we need to introduce some definitions and common terms.
Basic Definitions
A function \(f\left( x \right)\) is said to have period \(P\) if \(f\left( {x + P} \right) = f\left( x \right)\) for all \(x.\) Let the function \(f\left( x \right)\) has period \(2\pi.\) In this case, it is enough to consider behavior of the function on the interval \(\left[ { - \pi ,\pi } \right].\)
Suppose that the function \(f\left( x \right)\) with period \(2\pi\) is absolutely integrable on \(\left[ { - \pi ,\pi } \right]\) so that the following so-called Dirichlet integral is finite:
\[\int\limits_{ - \pi }^\pi {\left| {f\left( x \right)} \right|dx} \lt \infty ;\]
Suppose also that the function \(f\left( x \right)\) is a single valued, piecewise continuous (must have a finite number of jump discontinuities), and piecewise monotonic (must have a finite number of maxima and minima).
If the conditions \(1\) and \(2\) are satisfied, the Fourier series for the function \(f\left( x \right)\) exists and converges to the given function (see also the Convergence of Fourier Series page about convergence conditions.)
At a discontinuity \({x_0}\), the Fourier Series converges to
\[\lim\limits_{\varepsilon \to 0} \frac{1}{2}\left[ {f\left( {{x_0} - \varepsilon } \right) - f\left( {{x_0} + \varepsilon } \right)} \right].\]
The Fourier series of the function \(f\left( x \right)\) is given by
\[f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}} ,\]
where the Fourier coefficients \({{a_0}},\) \({{a_n}},\) and \({{b_n}}\) are defined by the integrals
\[{a_0} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)dx} ,\;\; {a_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\cos nx dx} ,\;\; {b_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\sin nx dx} .\]
Sometimes alternative forms of the Fourier series are used. Replacing \({{a_n}}\) and \({{b_n}}\) by the new variables \({{d_n}}\) and \({{\varphi_n}}\) or \({{d_n}}\) and \({{\theta_n}},\) where
\[{d_n} = \sqrt {a_n^2 + b_n^2} ,\;\; \tan {\varphi _n} = \frac{{{a_n}}}{{{b_n}}},\;\; \tan {\theta _n} = \frac{{{b_n}}}{{{a_n}}},\]
we can write:
\[
f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{d_n}\sin \left( {nx + {\varphi _n}} \right)} \;\; \text{or}\;\;
f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{d_n}\cos\left( {nx + {\theta _n}} \right)} .\]
Fourier Series of Even and Odd Functions
The Fourier series expansion of an even function \(f\left( x \right)\) with the period of \(2\pi\) does not involve the terms with sines and has the form:
\[f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos nx} ,\]
where the Fourier coefficients are given by the formulas
\[{a_0} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)dx} ,\;\; {a_n} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\cos nxdx} .\]
Accordingly, the Fourier series expansion of an odd \(2\pi\)-periodic function \(f\left( x \right)\) consists of sine terms only and has the form:
\[f\left( x \right) = \sum\limits_{n = 1}^\infty {{b_n}\sin nx} ,\]
where the coefficients \({{b_n}}\) are
\[{b_n} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\sin nxdx} .\]
Below we consider expansions of \(2\pi\)-periodic functions into their Fourier series assuming that these expansions exist and are convergent.
Solved Problems
Example 1.
Let the function \(f\left( x \right)\) be \(2\pi\)-periodic and suppose that it is presented by the Fourier series:
\[f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}}.\]
Calculate the coefficients \({{a_0}},\) \({{a_n}},\) and \({{b_n}}.\)
Solution.
To define \({{a_0}},\) we integrate the Fourier series on the interval \(\left[ { - \pi ,\pi } \right]:\)
\[\int\limits_{ - \pi }^\pi {f\left( x \right)dx} = \pi {a_0} + \sum\limits_{n = 1}^\infty {\left[ {{a_n}\int\limits_{ - \pi }^\pi {\cos nxdx} + {b_n}\int\limits_{ - \pi }^\pi {\sin nxdx} } \right]}.\]
For all \(n \gt 0\),
\[
\int\limits_{ - \pi }^\pi {\cos nxdx} = \left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_{ - \pi }^\pi = 0\;\; \text{and}\;\;\; \int\limits_{ - \pi }^\pi {\sin nxdx} = \left. {\left( { - \frac{{\cos nx}}{n}} \right)} \right|_{ - \pi }^\pi = 0.\]
Therefore, all the terms on the right of the summation sign are zero, so we obtain
\[\int\limits_{ - \pi }^\pi {f\left( x \right)dx} = \pi {a_0}\;\;\text{or}\;\; {a_0} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)dx} .\]
In order to find the coefficients \({{a_n}},\) we multiply both sides of the Fourier series by \(\cos mx\) and integrate term by term:
\[
\int\limits_{ - \pi }^\pi {f\left( x \right)\cos mxdx}
= \frac{{{a_0}}}{2}\int\limits_{ - \pi }^\pi {\cos mxdx}
+ \sum\limits_{n = 1}^\infty {\left[ {{a_n}\int\limits_{ - \pi }^\pi {\cos nx\cos mxdx} + {b_n}\int\limits_{ - \pi }^\pi {\sin nx\cos mxdx} } \right]} .\]
The first term on the right side is zero. Then, using the well-known Product-to-Sum Identities , we have
\[\int\limits_{ - \pi }^\pi {\sin nx\cos mxdx} = \frac{1}{2}\int\limits_{ - \pi }^\pi {\left[ {\sin{\left( {n + m} \right)x} + {\sin \left( {n - m} \right)x}} \right]dx} = 0,\]
\[\int\limits_{ - \pi }^\pi {\cos nx\cos mxdx} = \frac{1}{2}\int\limits_{ - \pi }^\pi {\left[ {\cos {\left( {n + m} \right)x} + {\cos \left( {n - m} \right)x}} \right]dx} = 0,\]
if \(m \ne n.\)
In case when \(m = n\), we can write:
\[\int\limits_{ - \pi }^\pi {\sin nx\cos mxdx} = \frac{1}{2}\int\limits_{ - \pi }^\pi {\left[ {\sin 2mx + \sin 0} \right]dx} ,\;\; \Rightarrow \int\limits_{ - \pi }^\pi {{\sin^2}mxdx} = \frac{1}{2}\left[ {\left. {\left( { - \frac{{\cos 2mx}}{{2m}}} \right)} \right|_{ - \pi }^\pi } \right] = \frac{1}{{4m}}\left[ { - \cancel{\cos \left( {2m\pi } \right)} + \cancel{\cos \left( {2m\left( { - \pi } \right)} \right)}} \right] = 0;\]
\[\int\limits_{ - \pi }^\pi {\cos nx\cos mxdx} = \frac{1}{2}\int\limits_{ - \pi }^\pi {\left[ {\cos 2mx + \cos 0} \right]dx} ,\;\; \Rightarrow \int\limits_{ - \pi }^\pi {{\cos^2}mxdx} = \frac{1}{2}\left[ {\left. {\left( {\frac{{\sin 2mx}}{{2m}}} \right)} \right|_{ - \pi }^\pi + 2\pi } \right] = \frac{1}{{4m}}\left[ {\sin \left( {2m\pi } \right) - \sin \left( {2m\left( { - \pi } \right)} \right)} \right] + \pi = \pi .\]
Thus,
\[\int\limits_{ - \pi }^\pi {f\left( x \right)\cos mxdx} = {a_m}\pi ,\;\; \Rightarrow {a_m} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\cos mxdx} ,\;\; m = 1,2,3, \ldots\]
Similarly, multiplying the Fourier series by \(\sin mx\) and integrating term by term, we obtain the expression for \({{b_m}}:\)
\[{b_m} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\sin mxdx} ,\;\; m = 1,2,3, \ldots\]
Rewriting the formulas for \({{a_n}},\) \({{b_n}},\) we can write the final expressions for the Fourier coefficients:
\[{a_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\cos nxdx} ,\;\; {b_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\sin nxdx} .\]
Example 2.
Find the Fourier series for the square \(2\pi\)-periodic wave defined on the interval \(\left[ { - \pi ,\pi } \right]:\)
\[
f\left( x \right) = \begin{cases}
0, & \text{if} & - \pi \le x \le 0 \\
1, & \text{if} & 0 < x \le \pi
\end{cases}.\]
Solution.
First we calculate the constant \({{a_0}}:\)
\[{a_0} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)dx} = \frac{1}{\pi }\int\limits_0^\pi {1dx} = \frac{1}{\pi } \cdot \pi = 1.\]
Find now the Fourier coefficients for \(n \ne 0:\)
\[{a_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\cos nxdx} = \frac{1}{\pi }\int\limits_0^\pi {1 \cdot \cos nxdx} = \frac{1}{\pi }\left[ {\left. {\left( {\frac{{\sin nx}}{n}} \right)} \right|_0^\pi } \right] = \frac{1}{{\pi n}} \cdot 0 = 0,\]
\[{b_n} = \frac{1}{\pi }\int\limits_{ - \pi }^\pi {f\left( x \right)\sin nxdx} = \frac{1}{\pi }\int\limits_0^\pi {1 \cdot \sin nxdx} = \frac{1}{\pi }\left[ {\left. {\left( { - \frac{{\cos nx}}{n}} \right)} \right|_0^\pi } \right] = - \frac{1}{{\pi n}} \cdot \left( {\cos n\pi - \cos 0} \right) = \frac{{1 - \cos n\pi }}{{\pi n}}.\]
Since \(\cos n\pi = {\left( { - 1} \right)^n},\) we can write:
\[{b_n} = \frac{{1 - {{\left( { - 1} \right)}^n}}}{{\pi n}}.\]
Thus, the Fourier series for the square wave is
\[f\left( x \right) = \frac{1}{2} + \sum\limits_{n = 1}^\infty {\frac{{1 - {{\left( { - 1} \right)}^n}}}{{\pi n}}\sin nx} .\]
We can easily find the first few terms of the series. By setting, for example, \(n = 5,\) we get
\[
f\left( x \right) = \frac{1}{2} + \frac{{1 - \left( { - 1} \right)}}{\pi }\sin x
+ \frac{{1 - {{\left( { - 1} \right)}^2}}}{{2\pi }}\sin 2x
+ \frac{{1 - {{\left( { - 1} \right)}^3}}}{{3\pi }}\sin 3x
+ \frac{{1 - {{\left( { - 1} \right)}^4}}}{{4\pi }}\sin 4x
+ \frac{{1 - {{\left( { - 1} \right)}^5}}}{{5\pi }}\sin 5x + \ldots
= \frac{1}{2} + \frac{2}{\pi }\sin x
+ \frac{2}{{3\pi }}\sin 3x
+ \frac{2}{{5\pi }}\sin 5x + \ldots\]
The graph of the function and the Fourier series expansion for \(n = 10\) is shown below in Figure \(2.\)
Figure 2, n = 10
See more problems on Page 2.
