Question 10.2.2: Let f(x) = {0, −3 < x < −1, 1, −1 < x < 1, 0, 1 ......

Let

f(x)=\begin{cases}0, & -3\lt x\lt-1, \\1, & -1\lt x\lt1,\\0, & 1\lt x\lt3\end{cases} (21)

and suppose that f(x + 6) = f(x). Graph three periods of y = f(x). Find the coefficients in the Fourier series for f.

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

The graph of y = f(x) is shown in Figure 10.2.3.

Note that f(x) is not assigned a value at the points of discontinuity, such as x = −1 and x = 1, or at the ends of the period, such as x = −9, x = −3, x = 3, and x = 9. This has no effect on the values of the Fourier coefficients, because they result from the evaluation of integrals, and the value of an integral is not affected by the value of the integrand at a single point or at a finite number of points. Thus the coefficients are the same regardless of what value, if any, f(x) is assigned at a point of discontinuity.

Since f has period 6, it follows that L = 3 in this problem. Consequently, the Fourier series for f has the form

f(x)={\frac{a_{0}}{2}}+\sum_{n-1}^{\infty}\left(a_{n}\cos\left({\frac{n\pi x}{3}}\right)+b_{n}\sin\left({\frac{n\pi x}{3}}\right)\right), (22)

where the coefficients a_{n} and b_{n} are given by equations (13)

a_{n}=\frac{1}{L}\int_{-L}^{L}f(x)\cos\Bigl(\frac{n\pi x}{L}\Bigr)d x,\quad n=0,1,2,\dots\,. (13)

and (14)

b_{n}=\frac{1}{L}\int_{-L}^{L}f(x)\sin\Bigl(\frac{n\pi x}{L}\Bigr)d x,\quad n=1,2,3,\dots\,. (14)

with L = 3. We have

a_{0}=\frac{1}{3}\int_{-3}^{3}f(x)d x=\frac{1}{3}\int_{-1}^{1}1\,d x=\frac{2}{3}. (23)

Similarly,

a_{n}=\frac{1}{3}\int_{-1}^{1}\cos\left(\frac{n\pi x}{3}\right)d x=\frac{1}{n\pi}\sin\left(\frac{n\pi x}{3}\right)\biggr|_{-1}^{1}=\frac{2}{n\pi}\sin\left(\frac{n\pi}{3}\right),\;\;\;n=1,2,\ldots, (24)

and

b_{n}=\frac{1}{3}\int_{-1}^{1}\sin\left(\frac{n\pi x}{3}\right)d x=-\frac{1}{n\pi}\cos\left(\frac{n\pi x}{3}\right)\biggr|_{-1}^{1}=0,\;\;\;n=1,2,\ldots, (25)

Thus the Fourier series for f is

f(x)={\frac{1}{3}}+\sum_{n=1}^{\infty}{\frac{2}{n\pi}}\sin\left({\frac{n\pi}{3}}\right)\cos\left({\frac{n\pi x}{3}}\right)\\=\frac{1}{3}+\frac{\sqrt{3}}{\pi}\left(\cos\left(\frac{\pi x}{3}\right)+\frac{1}{2}\cos\left(\frac{2\pi x}{3}\right)-\frac{1}{4}\cos\left(\frac{4\pi x}{3}\right)-\frac{1}{5}\cos\left(\frac{5\pi x}{3}\right)+\cdots\right). (26)

FIGURE 10.2.3
Loading more images...

Related Answered Questions

Question: 10.1.4

Verified Answer:

The general solution is given by equation (11), [l...
Question: 10.1.3

Verified Answer:

The general solution of the differential equation ...
Question: 10.1.2

Verified Answer:

The general solution of this differential equation...
Question: 10.1.1

Verified Answer:

The general solution of the differential equation ...