# Calculus

## Fourier Series # Even and Odd Extensions

• Suppose that a function $$f\left( x \right)$$ is piecewise continuous and defined on the interval $$\left[ {0,\pi } \right].$$ To find its Fourier series, we first extend this function to the interval $$\left[ {-\pi,\pi } \right].$$ This can be done in two ways:

• We can construct the even extension of $$f\left( x \right):$$
${{f_\text{even}}\left( x \right) \text{ = }}\kern0pt {\begin{cases} f\left( {-x} \right), & -\pi \le x \lt 0 \\ f\left( {x} \right), & 0 \le x \le \pi \end{cases},}$
• or the odd extension of $$f\left( x \right):$$
${{f_\text{odd}}\left( x \right) \text{ = }}\kern0pt {\begin{cases} -f\left( {-x} \right), & -\pi \le x \lt 0 \\ f\left( {x} \right), & 0 \le x \le \pi \end{cases}.}$

For the even function, the Fourier series is called the Fourier Cosine series and is given by

${{f_\text{even}}\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {{a_n}\cos nx} ,}$

where

${{a_n} }={ \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\cos nxdx} ,\;\;}\kern-0.3pt {n = 0,1,2,3, \ldots }$

Respectively, for the odd function, the Fourier series is called the Fourier Sine series and is given by

${{f_\text{odd}}\left( x \right) }={ \sum\limits_{n = 1}^\infty {{b_n}\sin nx} ,}$

where the Fourier coefficients are

${{b_n} }={ \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\sin nxdx} ,\;\;}\kern-0,3pt {n = 1,2,3, \ldots }$

We can also define the Fourier Sine and Cosine series for a function with an arbitrary period $$2L.$$ Let $$f\left( x \right)$$ be defined on the interval $$\left[ {0,L } \right].$$ Using even extension of the function to the interval $$\left[ {-L,L } \right],$$ we obtain

${{f_\text{even}}\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {{a_n}\cos \frac{{n\pi x}}{L}} ,}$

where

${{a_n} }={ \frac{2}{L}\int\limits_0^L {f\left( x \right)\cos \frac{{n\pi x}}{L}dx} ,\;\;}\kern-0.3pt {n = 0,1,2,3, \ldots }$

For the odd extension, we have

${{f_\text{odd}}\left( x \right) }={ \sum\limits_{n = 1}^\infty {{b_n}\sin \frac{{n\pi x}}{L}} ,}$

where the coefficients $${b_n}$$ are

${{b_n} }={ \frac{2}{L}\int\limits_0^L {f\left( x \right)\sin \frac{{n\pi x}}{L}dx} ,\;\;}\kern-0.3pt {n = 1,2,3, \ldots }$

• ## Solved Problems

Click a problem to see the solution.

### Example 1

Find the Fourier Cosine series of the function
$f\left( x \right) = \begin{cases} 1, & 0 \le x \le d \\ 0, & d \lt x \le \pi \end{cases}.$

### Example 2

Find the Fourier Cosine series of the function
${f\left( x \right) \text{ = }}\kern0pt {\begin{cases} 1 – \frac{x}{d}, & 0 \le x \le d \\ 0, & d \lt x \le \pi \end{cases}.}$

### Example 3

Find the Fourier Sine series of the function $$f\left( x \right) = \cos x$$ defined on the interval $$\left[ {0,\pi } \right].$$

### Example 4

Find the Fourier Sine series of the function $$f\left( x \right) = x\sin x,$$ defined on the interval $$\left[ {0,\pi } \right].$$

### Example 1.

Find the Fourier Cosine series of the function
$f\left( x \right) = \begin{cases} 1, & 0 \le x \le d \\ 0, & d \lt x \le \pi \end{cases}.$

Solution.

We construct even extension of the given function. The corresponding Fourier series has the form:

$f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos nx} .$

Calculate the Fourier coefficients $${a_0}$$ and $${a_n}:$$

${{a_0} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)dx} }={ \frac{2}{\pi }\int\limits_0^d {dx} }={ \frac{{2d}}{\pi },}$

${{a_n} }={ \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\cos nxdx} } = {\frac{2}{\pi }\int\limits_0^d {\cos nxdx} } = {\frac{2}{{n\pi }}\left[ {\left. {\left( {\sin nx} \right)} \right|_0^d} \right] } = {\frac{2}{{n\pi }}\sin nd.}$

Thus, the Fourier series of the step function is given by

${f\left( x \right) = \frac{d}{\pi } }+{ \frac{2}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\sin nd}}{n}\cos nx} .}$

Graphs of the function and its Fourier approximation for $$n = 5$$ and $$n = 50$$ are shown in Figure $$1.$$