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Question Number 50702 by pooja24 last updated on 19/Dec/18
write the fourier series of   f(x)=x                    0≤x≤2
writethefourierseriesoff(x)=x0x2
Commented by maxmathsup by imad last updated on 19/Dec/18
i think  0≤x≤2π   if f is 2π periodic odd we have  f(x)=Σ_(n=1) ^∞  a_n sin(nx)  with a_n =(2/T) ∫_([T])  f(x) sin(nx)dx  =(2/(2π)) ∫_(−π) ^π  x sin(nx)dx =(2/π) ∫_0 ^π  xsin(nxdx ⇒(π/2)a_n =∫_0 ^π  xsin(nx)dx  by parts ∫_0 ^π  x sin(nx)dx =[−(x/n)cos(nx)]_0 ^π  +∫_0 ^π  (1/n) cos(nx)dx  =−(π/n)(−1)^n   +(1/n)[(1/n)sin(nx)]_0 ^π  =((π(−1)^(n−1) )/n) ⇒(π/2)a_n =(π/n)(−1)^(n−1)  ⇒    a_n =(2/n)(−1)^(n−1)  ⇒ f(x) =Σ_(n=1) ^∞  (2/n)(−1)^(n−1) sin(nx).
ithink0x2πiffis2πperiodicoddwehavef(x)=n=1ansin(nx)withan=2T[T]f(x)sin(nx)dx=22πππxsin(nx)dx=2π0πxsin(nxdxπ2an=0πxsin(nx)dxbyparts0πxsin(nx)dx=[xncos(nx)]0π+0π1ncos(nx)dx=πn(1)n+1n[1nsin(nx)]0π=π(1)n1nπ2an=πn(1)n1an=2n(1)n1f(x)=n=12n(1)n1sin(nx).

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