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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))
$${let}\:{f}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{sin}\left({nx}\right)}{{n}}\:{x}^{{n}} \:\:\:\:\:{with}\:\:−\mathrm{1}<{x}<\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{find}\:\:{a}\:{explicite}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}\mathrm{2}^{{n}} }{sin}\left(\frac{{n}}{\mathrm{2}}\right) \\ $$
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) )))} .
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)={Im}\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{inx}} {x}^{{n}} }{{n}}\right)\:={Im}\left(\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left({xe}^{{ix}} \right)^{{n}} }{{n}}\right)\:{let}\: \\ $$$${z}\:={x}\:{e}^{{ix}} \:\:{and}\:{W}\left({z}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{z}^{{n}} }{{n}}\:\Rightarrow\frac{{dW}}{{dz}}\left({z}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:{z}^{{n}−\mathrm{1}} =\sum_{{n}=\mathrm{0}} ^{\infty} \:{z}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{z}}\:\Rightarrow \\ $$$${W}\left({z}\right)=−{ln}\left(\mathrm{1}−{z}\right)\:+{c}\:\:{but}\:{c}={W}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow{W}\left({z}\right)=−{ln}\left(\mathrm{1}−{z}\right) \\ $$$$=−{ln}\left(\mathrm{1}−{xe}^{{ix}} \right)=−{ln}\left(\mathrm{1}−{xcosx}\:−{ix}\:{sinx}\right)={a}+{ib}\:\Rightarrow \\ $$$${e}^{{a}+{ib}} \:=\:\frac{\mathrm{1}}{\mathrm{1}−{xcosx}−{ixsinx}}\:=\frac{\mathrm{1}−{xcosx}+{ixsinx}}{\left(\mathrm{1}−{xcosx}\right)^{\mathrm{2}} \:+{x}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{1}−{xcosx}+{ixsnx}}{\mathrm{1}−\mathrm{2}{xcosx}\:+{x}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}\:+{x}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}\:=\frac{\mathrm{1}−{xcosx}+{ixsinx}}{\mathrm{1}−\mathrm{2}{xcosx}\:+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${e}^{{a}} \left({cosb}\:+{isinb}\right)\:=\frac{\mathrm{1}−{xcosx}}{\mathrm{1}−\mathrm{2}{xcosx}\:+{x}^{\mathrm{2}} }\:+{i}\frac{{xsinx}}{\mathrm{1}−\mathrm{2}{xcosx}\:+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${e}^{{a}} {cosb}\:=\frac{\mathrm{1}−{xcosx}}{\mathrm{1}−\mathrm{2}{xcosx}\:+{x}^{\mathrm{2}} }\:{and}\:{e}^{{a}} {sinb}\:=\frac{{x}\:{sinx}}{\mathrm{1}−\mathrm{2}{xcosx}\:+{x}^{\mathrm{2}} }\:\:\Rightarrow \\ $$$${tanb}\:=\frac{{xsinx}}{\mathrm{1}−{xcosx}}\:\Rightarrow{b}\:={arctan}\left(\frac{{xsinx}}{\mathrm{1}−{x}\:{cosx}}\right)\:\Rightarrow{f}\left({x}\right)={Im}\left({W}\left({z}\right)\right) \\ $$$$={arctan}\left(\frac{{xsinx}}{\mathrm{1}−{xcosx}}\right) \\ $$$$\left.\mathrm{2}\right)\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}\mathrm{2}^{{n}} }{sin}\left(\frac{{n}}{\mathrm{2}}\right)\:={f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:={arctan}\left(\frac{{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right.}\right) \\ $$$$={arctan}\left\{\:\frac{{sin}\left(\mathrm{2}^{−\mathrm{1}} \right)}{\mathrm{2}−{cos}\left(\mathrm{2}^{−\mathrm{1}} \right)}\right\}\:. \\ $$

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