Question 10.4.1: Let f(x) = x, −L < x < L, and let f(−L) = f(L) = 0. Le......
Let f(x) = x, −L < x < L, and let f(−L) = f(L) = 0. Let f be defined elsewhere so that it is periodic of period 2L. The function defined in this manner is known as a sawtooth wave. Graph three periods of y = f(x). Find the Fourier series for this function.
The graph of y = f(x) on [−L , L] and one period to the left and one period to the right is shown in Figure 10.4.2.
Since f is an odd function, its Fourier coefficients are, according to equation (8),
a_{n}=0,\quad\quad\quad n=0,1,2,\,\dots\,,\\b_{n}=\frac{1}{L}\int_{-L}^{L}f(x)\sin\Bigl(\frac{n\pi x}{L}\Bigr)d x=\frac{2}L\int_{0}^{L}f(x)\sin\Bigl(\frac{n\pi x}L\Bigr)d x,\ \ \ n=1,2,\ \dots\,, (8)
a_{n}=0,\quad n=0,1,2,\,\dots\,;\\b_{n}=\frac{2}{L}\int_{0}^{L}x\sin\Bigl(\frac{n\pi x}{L}\Bigr)d x\\=\left.\frac{2}L\left(\frac{L}{n\pi}\right)^{2}\left(\sin\left(\frac{n\pi x}{L}\right)-\frac{n\pi x}{L}\cos\left(\frac{n\pi x}{L}\right)\right)\right|^{L}_{0}\\\frac{2L}{n\pi}(-1)^{n+1}\quad n=1,2,\ \dots\,.
Hence the Fourier series for f , the sawtooth wave, is
f(x)={\frac{2L}{\pi}}\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{n}}\sin\left({\frac{n\pi x}{L}}\right)\\={\frac{2L}{\pi}}\biggl(\sin\biggl({\frac{\pi x}{L}}\biggr)-{\frac{1}{3}}\sin\biggl({\frac{3\pi x}{L}}\biggr)+{\frac{1}{5}}\sin\biggl({\frac{5\pi x}{L}}\biggr)-\cdots\biggr). (9)
Observe that the periodic function f is discontinuous at the points ±L, ±3L, . . . , as shown in Figure 10.4.2. At these points the series (9) converges to the mean value of the left and right limits, namely, to zero. The partial sum of the series (9) for n = 9 is shown in Figure 10.4.3. The Gibbs phenomenon (mentioned in Section 10.3) again occurs near the points of discontinuity.
