# Fourier Series of Functions with an Arbitrary Period

### 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, that is 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 or tap a problem to see the 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 $$n = 2k,$$ $$k = 1,2,3, \ldots$$

${{b_{2k}} }={ \frac{A}{{2k\pi }}\left[ {1 + {{\left( { – 1} \right)}^{2k + 1}}} \right] }={ 0.}$

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

${f\left( x \right) = \frac{A}{2} \text{ + }}\kern0pt{ \frac{{2A}}{\pi }\sum\limits_{k = 1}^\infty {\frac{1}{{2k – 1}}\sin \left( {\frac{{2k – 1}}{L}\pi x} \right)}}$

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