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Question Number 38942 by math khazana by abdo last updated on 01/Jul/18
find Σ_(n=1) ^∞ (((−1)^n )/n)cos(nx) and Σ_(n=1) ^∞ (((−1)^n )/n)sin(nx)
$${find}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{cos}\left({nx}\right)\:{and}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{sin}\left({nx}\right) \\ $$
Commented by prof Abdo imad last updated on 09/Jul/18
let f(x)=Σ_(n=1) ^∞ (((−1)^n )/n) cos(nx) f(x) =Re( Σ_(n=1) ^∞ (((−1)^n )/n) e^(inx) )=Re(w(x)) w(x)=Σ_(n=1) ^∞ (((−e^(ix) )^n )/n) =−ln(1+e^(ix) ) ( Σ (z^n /n) =−ln(1−z)) but ln(1+e^(ix) ) =ln( 1 +cosx +isinx) =ln(2cos^2 ((x/2)) +2i sin((x/2))cos((x/2))) =ln(2) +ln( cos((x/2))e^((ix)/2) ) =ln(2) +ln(cos((x/2))) +((ix)/2) ⇒ w(x)=−ln(2cos((x/2)))−((ix)/2) ⇒ f(x)=−ln(2cos((x/2))) also we get Σ_(n=1) ^∞ (((−1)^n )/n)sin(nx)=−(x/2) .
$${let}\:{f}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:{cos}\left({nx}\right) \\ $$$${f}\left({x}\right)\:={Re}\left(\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:{e}^{{inx}} \right)={Re}\left({w}\left({x}\right)\right) \\ $$$${w}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−{e}^{{ix}} \right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{1}+{e}^{{ix}} \right) \\ $$$$\left(\:\Sigma\:\frac{{z}^{{n}} }{{n}}\:=−{ln}\left(\mathrm{1}−{z}\right)\right)\:\:{but} \\ $$$${ln}\left(\mathrm{1}+{e}^{{ix}} \right)\:={ln}\left(\:\mathrm{1}\:+{cosx}\:+{isinx}\right) \\ $$$$={ln}\left(\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\:+\mathrm{2}{i}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$={ln}\left(\mathrm{2}\right)\:+{ln}\left(\:{cos}\left(\frac{{x}}{\mathrm{2}}\right){e}^{\frac{{ix}}{\mathrm{2}}} \right) \\ $$$$={ln}\left(\mathrm{2}\right)\:+{ln}\left({cos}\left(\frac{{x}}{\mathrm{2}}\right)\right)\:+\frac{{ix}}{\mathrm{2}}\:\Rightarrow \\ $$$${w}\left({x}\right)=−{ln}\left(\mathrm{2}{cos}\left(\frac{{x}}{\mathrm{2}}\right)\right)−\frac{{ix}}{\mathrm{2}}\:\Rightarrow \\ $$$${f}\left({x}\right)=−{ln}\left(\mathrm{2}{cos}\left(\frac{{x}}{\mathrm{2}}\right)\right)\:{also}\:{we}\:{get} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{sin}\left({nx}\right)=−\frac{{x}}{\mathrm{2}}\:\:. \\ $$

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