Fourier series for even and odd functions

Fourier series: Coefficients of Fourier series

Fourier series for even and odd functions

In the Fourier series of odd and even functions about half the coefficients will be zero.

Fourier coefficients of even and odd functionsLet #L# be a positive number and let #f# be a #2L#-periodic function. Suppose that #f# has Fourier series

\[\begin{array}{rcl}\displaystyle \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos \left(\frac{\pi n x}{L}\right) + b_n \sin \left(\frac{\pi n x}{L}\right) \end{array}\]

If #f# is even, then #b_n=0# for #n=1,2,\ldots#
If #f# is odd, then #a_n=0# for #n=0,1,2\ldots#

Suppose that #f# is even. Since #\cos\left(\frac{\pi n x}{L}\right)# is an even function, the product #f(x)\cdot \cos\left(\frac{\pi n x}{L}\right)# is also even. At the same time, #\sin\left(\frac{\pi n x}{L}\right)# is odd, so #f(x)\cdot \sin\left(\frac{\pi n x}{L}\right)# is also odd. Using the Euler formulas for the coefficients, we can calculate the coefficients for #f#.

\[\begin{array}{rcl}\displaystyle a_n &=&\displaystyle\frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{\pi n x}{L}\right) \dd x = \frac{2}{L} \int_0^{L} f(x)\cos\left(\frac{\pi n x}{L}\right) \dd x \quad \text{for } n=0,1,\ldots \\ b_n &=& \displaystyle\frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{\pi n x}{L}\right) \dd x = 0 \quad \text{for } n=1,2,\ldots \end{array}\]

So the Fourier series of an even function contains only cosines and can be written as stated.

Now let #f(x)# be an odd function. Since #\cos\left(\frac{\pi n x}{L}\right)# is clearly an even function for all non-negative #n#, the product #f(x)\cos\left(\frac{\pi n x}{L}\right)# is odd. At the same time, #\sin\left(\frac{\pi n x}{L}\right)# is odd so that #f(x)\sin\left(\frac{\pi n x}{L}\right)# is even. Then, again using the Euler formulas, we can calculate the coefficients of the Fourier series for #f#:

\[\begin{array}{rcl}\displaystyle a_n &=&\displaystyle\frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{\pi n x}{L}\right) \dd x = 0 \quad \text{for } n=0,1,\ldots \\ b_n &=& \displaystyle\frac{1}{L} \int_{-L}^{L} f(x)\sin\left(\frac{\pi n x}{L}\right) \dd x = \displaystyle\frac{2}{L} \int_0^{L} f(x)\sin\left(\frac{\pi n x}{L}\right) \dd x \quad \text{for } n=1,2,\ldots \end{array}\]

Thus, the Fourier series of a periodic even function contains only terms with sines and can be written as

\[\begin{array}{c}\displaystyle \sum_{n=1}^{\infty} b_n \sin\left(\frac{\pi n x}{L}\right) \end{array}\]

Piecewise smooth

Even in the case where #f# is periodic and piecewise smooth and behaves like an even or odd function at the point where #f# and #f'# are smooth, the theorem holds.

For instance, the #2#-periodic function #f# that is #-1# on #\ivco{-1}{0}# and #1# on #\ivco{0}{1}# is odd outside the points #2k# for integers #k# and the coefficient #a_0# of the Fourier series is #0#.

Consider the #2#-periodic function that is determined by \[f(x)=\begin{cases}-x^2 &\text{if } -1 \leq x\lt 0\\ x^2 &\text{if } 0\leq x\lt 1\end{cases} \]

Its Fourier series has the form \[f(x)=\frac{a_0}{2}+\sum_{m=1}^{\infty}\left(a_m\cdot\cos\left(\frac{m\cdot\pi\cdot x}{L}\right)+b_m\cdot\sin\left(\frac{m\cdot\pi\cdot x}{L}\right)\right)\] where \(L=1\) is the half-period of \(f\).

Of which of the coefficients of the Fourier series can we tell that they are zero without further computation?

\(a_0 =0\)
\(a_m=0\) for \( m=1,2,3,\dots\)

The function is odd, so all coefficients #a_m# are zero.

The graph of the function \(f(x)\) has been plotted over three periods in the figure below.

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