Answer to Question #198325 in Calculus for hanata yuji

Question #198325

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

0,1<x<00,-1<x<0

f(x)=f(x)= {

1,0<x<31,0<x<3


1
Expert's answer
2021-05-26T02:20:06-0400

T=4T=4


a0=1413f(x)dx=1403dxa_0=\dfrac{1}{4}\displaystyle\int_{-1}^{3}f(x)dx=\dfrac{1}{4}\displaystyle\int_{0}^{3}dx

=14[x]30=34=\dfrac{1}{4}[x]\begin{matrix} 3 \\ 0 \end{matrix}=\dfrac{3}{4}

an=1413f(x)cos(nπx2)dt=1403cos(nπx2)dxa_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

=12nπ[sin(nπx2)]30=12nπsin(3nπ2)=\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,a1=12(1)πn=1, a_{1}=-\dfrac{1}{2(1)\pi}

n=2,a2=0n=2, a_{2}=0

n=3,a3=12(3)πn=3, a_{3}=\dfrac{1}{2(3)\pi}

......


a2k+1=(1)k+12(2k+1)π,k=0,1,2,...a_{2k+1}=\dfrac{(-1)^{k+1}}{2(2k+1)\pi}, k=0,1,2,...


bn=1413f(x)sin(nπx2)dx=1403sin(nπx2)dxb_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

=12nπ[cos(nπx2)]30=12nπ(1cos(3nπ2))=-\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,b1=12(1)πn=1, b_{1}=\dfrac{1}{2(1)\pi}

n=2,b2=22(2)πn=2, b_{2}=\dfrac{2}{2(2)\pi}

n=3,b3=12(3)πn=3, b_{3}=\dfrac{1}{2(3)\pi}

n=4,b4=0n=4, b_{4}=0

n=5,b5=12(5)πn=5,b_{5}=\dfrac{1}{2(5)\pi}

b6=22(6)πb_{6}=\dfrac{2}{2(6)\pi}

......


b2k+1=12(2k+1)π,k=0,1,2,...b_{2k+1}=\dfrac{1}{2(2k+1)\pi}, k=0,1,2,...

b4m+2=22(4m+2)π,m=0,1,2,..b_{4m+2}=\dfrac{2}{2(4m+2)\pi}, m=0,1,2,..


f(x)=38+k=0(1)k+12(2k+1)πcos((2k+1)πx2)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})

+k=012(2k+1)πsin((2k+1)πx2)+\displaystyle\sum_{k=0}^{\infin}\dfrac{1}{2(2k+1)\pi}\sin(\dfrac{(2k+1)\pi x}{2})




+m=012(2m+1)πsin((2m+1)πx)+\displaystyle\sum_{m=0}^{\infin}\dfrac{1}{2(2m+1)\pi}\sin((2m+1)\pi x)


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