Answer to Question #198839 in Calculus for Hanata yuji

Question

Indicate the function of f(x) belowin fourier series

0,-0,5<x<0

F(x) ={

x, 0<x<0, 5

Answer 1

$$a_0=\dfrac{1}{L}\displaystyle\int_{-L}^{L}f(x)dx$$

$$=\dfrac{1}{0.5}\displaystyle\int_{0}^{0.5}xdx=[x^2]\begin{matrix} 0.5 \\ 0 \end{matrix}=0.25$$

$$a_n=\dfrac{1}{L}\displaystyle\int_{-L}^{L}f(x)\cos\dfrac{n\pi x}{L} dx$$

$$=\dfrac{1}{0.5}\displaystyle\int_{0}^{0.5}x\cos\dfrac{n\pi x}{0.5}dx=2\displaystyle\int_{0}^{0.5}x\cos(2\pi nx)dx$$

$$=2[\dfrac{x\sin(2\pi n x)}{2\pi n}+\dfrac{\cos(2\pi n x)}{4\pi^2n^2}]\begin{matrix} 0.5\\ 0 \end{matrix}$$

$$=\dfrac{(-1)^n-1}{2\pi^2n^2}=-\dfrac{1}{\pi^2(2k+1)^2}, k\in Z^+$$

$$b_n=\dfrac{1}{L}\displaystyle\int_{-L}^{L}f(x)\sin\dfrac{n\pi x}{L} dx$$

$$=\dfrac{1}{0.5}\displaystyle\int_{0}^{0.5}x\sin\dfrac{n\pi x}{0.5}dx=2\displaystyle\int_{0}^{0.5}x\sin(2\pi nx)dx$$

$$=2[-\dfrac{x\cos(2\pi n x)}{2\pi n}-\dfrac{\sin(2\pi n x)}{4\pi^2n^2}]\begin{matrix} 0.5\\ 0 \end{matrix}$$

$$=\dfrac{(-1)^n}{2\pi n}=-\dfrac{(-1)^k}{2\pi (k+1)}, k\in Z^+$$

$$f(x)=\dfrac{1}{8}-\displaystyle\sum_{k=0}^{\infin}(\dfrac{\cos(2\pi (2k+1) x)}{\pi^2(2k+1)^2}+\dfrac{(-1)^k\sin(2\pi (k+1) x)}{2\pi (k+1)})$$