91影视

To sketch the graph of the given function for three periods, we need to identify the period and plot the function for three consecutive periods. The function is given as piecewise: $$ f(x)=\left\\{\begin{array}{lr}{1,} & {-L \leq x < 0,} \\\ {0,} & {0 \leq x < L}\end{array} \quad f(x+2 L)=f(x)\right. $$ The period of the function is 2L. So let's sketch the given function for three periods: 1. For the period -2L 鈮� x < 0, f(x) = 1. 2. For the period 0 鈮� x < 2L, f(x) = 0. 3. For the period 2L 鈮� x < 4L, f(x) = 1. 4. For the period 4L 鈮� x < 6L, f(x) = 0. Now, if you plot these points on a graph, you will see three periods of the given piecewise function.

To find the Fourier series representation of the given function, we have to find the Fourier coefficients 饾憥鈧�, 饾憥饾憶, and 饾憦饾憶 by computing these integrals: $$ a_{0}=\frac{1}{L} \int_{-L}^{L} f(x) d x, \quad a_{n}=\frac{1}{L} \int_{-L}^{L} f(x) \cos \left(\frac{n \pi x}{L}\right) d x, \quad b_{n}=\frac{1}{L} \int_{-L}^{L} f(x) \sin \left(\frac{n \pi x}{L}\right) d x $$ 1. Compute 饾憥鈧�: $$ a_{0}=\frac{1}{L} \int_{-L}^{L} f(x) d x $$ Since the function is given as a piecewise function, we rewrite the integral as the sum of two integrals: $$ a_{0}=\frac{1}{L} \left( \int_{-L}^{0} 1 dx + \int_{0}^{L} 0 dx \right)=\frac{1}{L}\left[L-0\right]=1 $$ 2. Compute 饾憥饾憶: $$ a_{n}=\frac{1}{L} \int_{-L}^{L} f(x) \cos \left(\frac{n \pi x}{L}\right) d x $$ Again, we rewrite the integral as the sum of two integrals: $$ a_{n}=\frac{1}{L} \left( \int_{-L}^{0} 1 \cos \left(\frac{n \pi x}{L}\right) d x + \int_{0}^{L} 0 \cos \left(\frac{n \pi x}{L}\right) d x \right)=\frac{1}{L}\int_{-L}^{0} \cos \left(\frac{n \pi x}{L}\right) d x $$ Now, integrate w.r.t x: $$ a_{n}=\frac{1}{L} \left[-\frac{L}{n \pi} \sin \left(\frac{n \pi x}{L}\right) \right]_{-L}^{0} = \frac{1}{n\pi}\left[-\sin(n\pi)+\sin(-n\pi)\right]= 0 $$ 3. Compute 饾憦饾憶: $$ b_{n}=\frac{1}{L} \int_{-L}^{L} f(x) \sin \left(\frac{n \pi x}{L}\right) d x $$ Rewrite the integral as the sum of two integrals: $$ b_{n}=\frac{1}{L} \left( \int_{-L}^{0} 1 \sin \left(\frac{n \pi x}{L}\right) d x + \int_{0}^{L} 0 \sin \left(\frac{n \pi x}{L}\right) d x \right)=\frac{1}{L}\int_{-L}^{0} \sin \left(\frac{n \pi x}{L}\right) d x $$ Integrate w.r.t x: $$ b_{n}=\frac{1}{L} \left[-\frac{L}{n \pi} \cos \left(\frac{n \pi x}{L}\right) \right]_{-L}^{0} = \frac{1}{n\pi}\left[-\cos(0) + \cos(n\pi)\right] = \frac{1}{n\pi}(1-(-1)^{n}) $$ Now we have all the Fourier coefficients: 饾憥鈧� = 1, 饾憥饾憶 = 0, and 饾憦饾憶 = 鈭玕frac{1}{n\pi}(1-(-1)^{n}). So, the Fourier series representation of the function is: $$ f(x) = \frac{1}{2} + \sum_{n=1}^{\infty} \frac{1}{n\pi} (1-(-1)^{n}) \sin \left( \frac{n\pi x}{L} \right) $$