SOLUTION: Find the Fourier expansion on [pi, pi] of the function f(x)= x
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Question 30437: Find the Fourier expansion on [pi, pi] of the function f(x)= x
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Find the Fourier expansion on [pi, pi] of the function f(x)= x
I THINK YOU MEAN -PI TO PI
F(X)=X...THIS AN ODD FUNCTION..SO
F(X)= SIGMA[BN*SIN(NX)]
BN=(1/PI)*INTEGRAL[-PI TO PI {F(X)SIN(NX)DX}]
=(1/PI)*INTEGRAL[-PI TO PI {X*SIN(NX)DX}]
=(1/PI)
=(1/PI)
=(1/PI)*{(-PI*COS(N*PI)/N)-(PI*COS(-N*PI)/N + SIN(N*PI)/N^2 - SIN(-N*PI)/N }
=(1/PI)*(-2PI*COS(N*PI)/N)..SINCE SIN(N*PI)=0..FOR ALL N AND
COS(-N*PI)=COS(N*PI)
BN=(1/PI)*{-2PI*(-1)^N}/N...SINCE COS(N*PI)=-1 WHEN N IS ODD AND COS(N*PI)=1
WHEN N IS EVEN.
BN=-2(-1)^N /N...HENCE THE SERIES IS
X= 2{SIN(X)/1 - SIN(2X)/2 + SIN(3X)/3 - SIN(4X)/4 +............}
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