Calculus

Fourier Series

Fourier Series Logo

Definition of Fourier Series and Typical Examples

Baron Jean Baptiste Joseph Fourier \(\left( 1768-1830 \right) \) introduced the idea that any periodic function can be represented by a series of sines and cosines which are harmonically related.

Baron Jean Baptiste Joseph Fourier (1768−1830)
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 \(\left[ { - \pi ,\pi } \right].\)

  1. 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:
    \[\int\limits_{ - \pi }^\pi {\left| {f\left( x \right)} \right|dx} \lt \infty ;\]
  2. 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

Click or tap a problem to see the solution.

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 \({{b_n}}.\)

Example 2

Find the Fourier series for the square \(2\pi\)-periodic wave defined on the interval \(\left[ { - \pi ,\pi } \right]:\)

\[ f\left( x \right) = \begin{cases} 0, & \text{if} & - \pi \le x \le 0 \\ 1, & \text{if} & 0 < x \le \pi \end{cases}.\]

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 \(2.\)

Fourier series for the square  2pi-periodic wave
Figure 2, n = 10

See more problems on Page 2.

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