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Question Number 45960 by maxmathsup by imad last updated on 19/Oct/18

find f(x)=Σ_(n=1) ^∞ ((x^n sin(nx))/n)

$${find}\:{f}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}} {sin}\left({nx}\right)}{{n}} \\ $$

Answered by Smail last updated on 22/Oct/18

f(x)=Im(Σ_(n=1) ^∞ ((x^n e^(inx) )/n)) Σ_(n=1) ^∞ ((x^n e^(inx) )/n)=Σ_(n=1) ^∞ (((xe^(ix) )^n )/n) ln(1−x)=−Σ_(n=1) ^∞ (x^n /n) with ∣x∣<1 Σ_(n=1) ^∞ (((xe^(ix) )^n )/n)=−ln(1−xe^(ix) ) =−ln(1−xcos(x)−xisin(x))=a+ib (1/(1−xcosx−ixsinx))=e^(a+ib) e^a (cosb+isinb)=((1−xcosx+ixsinx)/(1+x^2 −2xcosx)) e^a cos(b)=((1−xcosx)/(1+x^2 −2xcosx)) e^a sin(b)=((xsinx)/(1+x^2 −2xcosx)) tan(b)=((xsinx)/(1−xcosx)) b=tan^(−1) (((xsinx)/(1−xcosx))) Σ_(n=1) ^∞ ((x^n sin(nx))/n)=tan^(−1) (((xsinx)/(1−xcosx)))

$${f}\left({x}\right)={Im}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} {e}^{{inx}} }{{n}}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} {e}^{{inx}} }{{n}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({xe}^{{ix}} \right)^{{n}} }{{n}} \\ $$$${ln}\left(\mathrm{1}−{x}\right)=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}}\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({xe}^{{ix}} \right)^{{n}} }{{n}}=−{ln}\left(\mathrm{1}−{xe}^{{ix}} \right) \\ $$$$=−{ln}\left(\mathrm{1}−{xcos}\left({x}\right)−{xisin}\left({x}\right)\right)={a}+{ib} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{xcosx}−{ixsinx}}={e}^{{a}+{ib}} \\ $$$${e}^{{a}} \left({cosb}+{isinb}\right)=\frac{\mathrm{1}−{xcosx}+{ixsinx}}{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{xcosx}} \\ $$$${e}^{{a}} {cos}\left({b}\right)=\frac{\mathrm{1}−{xcosx}}{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{xcosx}} \\ $$$${e}^{{a}} {sin}\left({b}\right)=\frac{{xsinx}}{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{xcosx}} \\ $$$${tan}\left({b}\right)=\frac{{xsinx}}{\mathrm{1}−{xcosx}} \\ $$$${b}={tan}^{−\mathrm{1}} \left(\frac{{xsinx}}{\mathrm{1}−{xcosx}}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} {sin}\left({nx}\right)}{{n}}={tan}^{−\mathrm{1}} \left(\frac{{xsinx}}{\mathrm{1}−{xcosx}}\right) \\ $$

Commented by maxmathsup by imad last updated on 23/Oct/18

correct answer thanks a lots..

$${correct}\:{answer}\:{thanks}\:{a}\:{lots}.. \\ $$

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