Answer to Question #198325 in Calculus for hanata yuji

Question

indicate the function of f(x) below in fourier series

$0,-1<x<0$

$f(x)=$ {

$1,0<x<3$

Answer 1

$T=4$

$$a_0=\dfrac{1}{4}\displaystyle\int_{-1}^{3}f(x)dx=\dfrac{1}{4}\displaystyle\int_{0}^{3}dx$$

$$=\dfrac{1}{4}[x]\begin{matrix} 3 \\ 0 \end{matrix}=\dfrac{3}{4}$$

$$a_n=\dfrac{1}{4}\displaystyle\int_{-1}^{3}f(x)\cos(\dfrac{n\pi x}{2})dt=\dfrac{1}{4}\displaystyle\int_{0}^{3}\cos(\dfrac{n\pi x}{2})dx$$

$$=\dfrac{1}{2n\pi}\big[\sin(\dfrac{n\pi x}{2})\big]\begin{matrix} 3 \\ 0 \end{matrix}=\dfrac{1}{2n\pi}\sin\big(\dfrac{3n\pi }{2}\big)$$

$n=1, a_{1}=-\dfrac{1}{2(1)\pi}$

$n=2, a_{2}=0$

$n=3, a_{3}=\dfrac{1}{2(3)\pi}$

$...$

$$a_{2k+1}=\dfrac{(-1)^{k+1}}{2(2k+1)\pi}, k=0,1,2,...$$

$$b_n=\dfrac{1}{4}\displaystyle\int_{-1}^{3}f(x)\sin(\dfrac{n\pi x}{2})dx=\dfrac{1}{4}\displaystyle\int_{0}^{3}\sin(\dfrac{n\pi x}{2})dx$$

$$=-\dfrac{1}{2n\pi}\big[\cos(\dfrac{n\pi x}{2})\big]\begin{matrix} 3 \\ 0 \end{matrix}=\dfrac{1}{2n\pi}\big(1-\cos\big(\dfrac{3n\pi }{2}\big)\big)$$

$n=1, b_{1}=\dfrac{1}{2(1)\pi}$

$n=2, b_{2}=\dfrac{2}{2(2)\pi}$

$n=3, b_{3}=\dfrac{1}{2(3)\pi}$

$n=4, b_{4}=0$

$n=5,b_{5}=\dfrac{1}{2(5)\pi}$

$b_{6}=\dfrac{2}{2(6)\pi}$

$...$

$$b_{2k+1}=\dfrac{1}{2(2k+1)\pi}, k=0,1,2,...$$

$$b_{4m+2}=\dfrac{2}{2(4m+2)\pi}, m=0,1,2,..$$

$$f(x)=\dfrac{3}{8}+\displaystyle\sum_{k=0}^{\infin}\dfrac{(-1)^{k+1}}{2(2k+1)\pi}\cos(\dfrac{(2k+1)\pi x}{2})$$

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

$$+\displaystyle\sum_{m=0}^{\infin}\dfrac{1}{2(2m+1)\pi}\sin((2m+1)\pi x)$$