Example 4 (calculation of Fourier coefficients)

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Example 4 (calculation of Fourier coefficients)

Let us calculate the Fourier coefficients for the function in Example 2 whose graph is shown in Figure 1. In order to calculate

, we have to know the values of the function

when

is changing from $-\pi$ to $\pi$ . From the graph we read that

for $x \in [-\pi, 0)$ , and

for $x \in [0, \pi)$ . Note that the values of

at the endpoints of the intervals do not affect the value of the integrals. Thus,

$a_0 = \displaystyle{\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx = \ \frac{1}{... ...\pi}^{0} 0 \cdot dx + \ \frac{1}{\pi} \int_{0}^{\pi} 1 \cdot dx = 1,}$

$a_n = \displaystyle{ \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx dx = \... ...\cos nx dx + \ \frac{1}{\pi} \int_{0}^{\pi} 1 \cdot \cos nx dx = \\ }$

$= \displaystyle{ \biggl. \frac{1}{\pi} \frac{1}{n} \sin nx \biggr\vert^{\pi}_0 = \frac{1}{n\pi} (\sin n\pi - \sin 0) = 0} ,$

and

$b_n = \displaystyle{ \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx dx = \... ... \sin nx dx + \ \frac{1}{\pi} \int_{0}^{\pi} 1 \cdot \sin nx dx = \ }$

$= \displaystyle{ \biggl. \frac{1}{\pi} ( -\frac{1}{n}) \cos nx \biggr\vert^... ...= - \frac{1}{n\pi} (\cos n\pi - \cos 0) = \ \frac{1}{n\pi} (1 - (-1)^n), }$

whence

$\begin{displaymath}b_n = \displaystyle{ \left \{ \begin{array}{cl} \displaysty... ... 0 & {\mbox for} n {\mbox even.} \\ \end{array}\right . }\end{displaymath}$

The resulting Fourier series representation for on the interval $(-\pi, \pi)$ is the following:

$\displaystyle f(x) \thicksim \displaystyle{ \frac{1}{2} + \sum_{n {\mbox ... ...i} \sin x + \frac{2}{3\pi} \sin 3x + \frac{2}{5\pi} \sin 5x + \ldots}$

(8)

Note that in the equation (8) in place of the expected "=" sign we see the " $\thicksim$ " sign instead. The paragraph below provides the explanation.

Next: Points of discontinuity and Up: Fseries_1 Previous: Fourier coefficients for -periodic

Matthias Neufang 2002-09-18