# (20點)用傅立葉Fourier級數做展開expand

expand f(x) = / 1-x (0<x<1)

\ 1 (-1<x=<0)

(a)by Fourier series

(b)by Fourier integral

Update:

f(x)在0<1是1-x

Update 2:

f(x)在0<1是1-x

Update 3:

-1<0 怎麼打不出來

Update 4:

f(x)大於0小於等於1時是1-x

f(x)大於-1小於等於0時是1

Update 5:

f(x)大於0小於1時是1-x

f(x)大於-1小於等於0時是1

Rating

expand

f(x)=

_____{1-x ;(0<x<1)

_____{1 ;(-1<x<=0)

(a)by Fourier series

(b)by Fourier integral

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sol:

(a)by Fourier series:

令F(x)=f(x)且F(x)=F(x-2)即週期T=2

F.S.expand{F(x)}= a[0]+Σ(1~∞) a[n]cos(nπx)+b[n]sin(nπx)

其中

a[0]={1+(1*1)/2}/2=3/4

a[n]=∫(-1~1)F(x)cos(nπx)dx

___=∫(-1~0)cos(nπx)dx+∫(0~1)(1-x)cos(nπx)dx

___=[sin(nπx)/nπ](-1~0)

___+[(1-x)sin(nπx)/(nπ)-cos(nπx)/(nπ)^2](0~1)

___=[1-(-1)^n]/(nπ)^2

b[n]=∫(-1~1)F(x)sin(nπx)dx

___=∫(-1~0)sin(nπx)dx+∫(0~1)(1-x)sin(nπx)dx

___=[-cos(nπx)/nπ](-1~0)

___+[(x-1)cos(nπx)/(nπ)-sin(nπx)/(nπ)^2](0~1)

___=(-1)/(nπ)

F.S.expand{F(x)}= 3/4+Σ(1~∞) {[1-(-1)^n]/(nπ)^2}cos(nπx)

___________________________+[(-1)/(nπ)]sin(nπx)

F(x)=

{F.S.expand{F(x)};當x為連續點

{[F(0)+F(1)]/2=0.5;當x為不連續點

(b)by Fourier integral:

F.I.expand{F(x)}= ∫(0~∞) A(ω)cos(ωx)+B(ω)sin(ωx) dω

其中

A(ω)=∫(-∞~∞)f(x)cos(ωx)dx

____=∫(-1~0)cos(ωx)dx+∫(0~1)(1-x)cos(ωx)dx

____=[-sin(ωx)/(ω)](-1~0)

____+[(1-x)sin(ωx)/(ω)-cos(ωx)/(ω)^2](0~1)

____=-sin(ω)/(ω)+[1-cos(ω)]/(ω)^2

B(ω)=∫(-∞~∞)f(x)sin(ωx)dx

____=∫(-1~0)sin(ωx)dx+∫(0~1)(1-x)sin(ωx)dx

____=[-cos(ωx)/(ω)](-1~0)

____+[(x-1)cos(ωx)/(ω)-sin(ωx)/(ω)^2](0~1)

____=[-1+cos(ω)]/(ω)+[1/(ω)-sin(ω)/(ω)^2]

____=[cos(ω)]/(ω)-[sin(ω)]/(ω)^2

F.I.expand{f(x)}=∫(0~∞){-sin(ω)/ω+[1-cos(ω)]/ω^2}cos(ωx)

____________________+{[cosω]/ω-[sinω]/ω^2}sin(ωx) dω

f(x)=

{F.I.expand{f(x)};當x為連續點

{[F(0)+F(1)]/2=0.5;當x為不連續點

Source(s):