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let-f-x-n-1-sin-nx-n-x-n-with-1-lt-x-lt-1-1-find-a-explicite-form-of-f-x-2-find-the-value-of-n-1-1-n2-n-sin-n-2-

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Question Number 42089 by maxmathsup by imad last updated on 17/Aug/18
let f(x) =Σ_(n=1) ^∞ ((sin(nx))/n) x^n with −1<x<1 1) find a explicite form of f(x) 2) find the value of Σ_(n=1) ^∞ (1/(n2^n ))sin((n/2))
letf(x)=∑n=1∞sin(nx)nxnwith−1<x<11)findaexpliciteformoff(x)2)findthevalueof∑n=1∞1n2nsin(n2)
Commented by maxmathsup by imad last updated on 06/Nov/18
1) we have f(x)=Im(Σ_(n=1) ^∞ ((e^(inx) x^n )/n)) =Im(Σ_(n=1) ^∞ (((xe^(ix) )^n )/n)) let z =x e^(ix) and W(z) =Σ_(n=1) ^∞ (z^n /n) ⇒(dW/dz)(z)=Σ_(n=1) ^∞ z^(n−1) =Σ_(n=0) ^∞ z^n =(1/(1−z)) ⇒ W(z)=−ln(1−z) +c but c=W(0)=0 ⇒W(z)=−ln(1−z) =−ln(1−xe^(ix) )=−ln(1−xcosx −ix sinx)=a+ib ⇒ e^(a+ib) = (1/(1−xcosx−ixsinx)) =((1−xcosx+ixsinx)/((1−xcosx)^2 +x^2 sin^2 x)) =((1−xcosx+ixsnx)/(1−2xcosx +x^2 cos^2 x +x^2 sin^2 x)) =((1−xcosx+ixsinx)/(1−2xcosx +x^2 )) ⇒ e^a (cosb +isinb) =((1−xcosx)/(1−2xcosx +x^2 )) +i((xsinx)/(1−2xcosx +x^2 )) ⇒ e^a cosb =((1−xcosx)/(1−2xcosx +x^2 )) and e^a sinb =((x sinx)/(1−2xcosx +x^2 )) ⇒ tanb =((xsinx)/(1−xcosx)) ⇒b =arctan(((xsinx)/(1−x cosx))) ⇒f(x)=Im(W(z)) =arctan(((xsinx)/(1−xcosx))) 2) Σ_(n=1) ^∞ (1/(n2^n ))sin((n/2)) =f((1/2)) =arctan(((sin((1/2)))/(2(1−(1/2)cos((1/2))))) =arctan{ ((sin(2^(−1) ))/(2−cos(2^(−1) )))} .
1)wehavef(x)=Im(∑n=1∞einxxnn)=Im(∑n=1∞(xeix)nn)letz=xeixandW(z)=∑n=1∞znn⇒dWdz(z)=∑n=1∞zn−1=∑n=0∞zn=11−z⇒W(z)=−ln(1−z)+cbutc=W(0)=0⇒W(z)=−ln(1−z)=−ln(1−xeix)=−ln(1−xcosx−ixsinx)=a+ib⇒ea+ib=11−xcosx−ixsinx=1−xcosx+ixsinx(1−xcosx)2+x2sin2x=1−xcosx+ixsnx1−2xcosx+x2cos2x+x2sin2x=1−xcosx+ixsinx1−2xcosx+x2⇒ea(cosb+isinb)=1−xcosx1−2xcosx+x2+ixsinx1−2xcosx+x2⇒eacosb=1−xcosx1−2xcosx+x2andeasinb=xsinx1−2xcosx+x2⇒tanb=xsinx1−xcosx⇒b=arctan(xsinx1−xcosx)⇒f(x)=Im(W(z))=arctan(xsinx1−xcosx)2)∑n=1∞1n2nsin(n2)=f(12)=arctan(sin(12)2(1−12cos(12))=arctan{sin(2−1)2−cos(2−1)}.

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