請用公式幫我解以下這題:
求下面周期為2π函數的Fourier級數展開式。
f(x)= 3, 0≦x<π
f(x)= x, -π≦x<0
2006-06-17 09:09:08 · 1 個解答 · 發問者 小薯 2 in 科學 ➔ 數學
f(x)= { 3, 0≦x<π , 週期 p=2π { x, -π≦x<0a0=(1/2π){∫-π0 xdx +∫0π 3dx} =(1/2π){(1/2)x2|-π0 +3x|0π} =(1/2π){-(π2/2) +3π} = -(π/4) + (3/2)an=(1/π){∫-π0 xcosnxdx +∫0π 3cosnxdx} =(1/π){[x(1/n)sinnx-(-1/n2)cosnx]|-π0 +[(3/n)sinnx]|0π} =(1/n2π)(1-cosnπ) = { 2/(n2π), n=1,3,5,... { 0 , n=2,4,6,... bn=(1/π){∫-π0 xsinnxdx +∫0π 3sinnxdx} =(1/π){[x(-1/n)cosnx-(-1/n2)sinnx]|-π0 +[(-3/n)cosnx]|0π} =(1/π){-(π/n)cosnπ + (3/n)(1-cosnπ)} = -(1/n)cosnπ + (3/nπ)(1-cosnπ) = { (1/n)+ 6/(nπ), n=1,3,5,... { -(1/n) , n=2,4,6,... Fourier級數展開式 f(x)=-(π/4) + (3/2) + (2/π){cosx+(1/32)cos3x+(1/52)cos5x+...} +[(π+6)/π]{sinx+(1/3)sin3x+(1/5)sin5x+...} -{(1/2)sin2x+(1/4)sin4x+(1/6)sin6x+...}
2006-06-17 11:24:52 · answer #1 · answered by 蔡春益 7 · 0⤊ 0⤋