Question-196303

Question Number 196303 by sonukgindia last updated on 22/Aug/23

Answered by sniper237 last updated on 22/Aug/23

Σ_(n=1) ^∞ Im((e^(iπ/4) /2))^n =Im(Σ_(n=1) ^∞ (e^(iπ/4) /2)^n ) = Im (((e^(iπ/4) /2)/(1−e^(iπ/4) /2)))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:{Im}\left(\frac{{e}^{{i}\pi/\mathrm{4}} }{\mathrm{2}}\right)^{{n}} ={Im}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({e}^{{i}\pi/\mathrm{4}} /\mathrm{2}\right)^{{n}} \right) \\ $$$$=\:{Im}\:\left(\frac{{e}^{{i}\pi/\mathrm{4}} /\mathrm{2}}{\mathrm{1}−{e}^{{i}\pi/\mathrm{4}} /\mathrm{2}}\right) \\ $$

Commented by sonukgindia last updated on 22/Aug/23

$${explain}\:{this} \\ $$

Commented by JDamian last updated on 22/Aug/23

1. sin α = Im{e^(iα) } 2. Σ(Im{z_i }) = Im{Σz_i }

$$\mathrm{1}.\:\:\:\mathrm{sin}\:\alpha\:=\:\mathrm{Im}\left\{\mathrm{e}^{{i}\alpha} \right\} \\ $$$$\mathrm{2}.\:\:\:\Sigma\left(\mathrm{Im}\left\{{z}_{{i}} \right\}\right)\:\:=\:\:\mathrm{Im}\left\{\Sigma{z}_{{i}} \right\} \\ $$

Answered by Mathspace last updated on 22/Aug/23

s(x)=Σ_(n=1) ^∞ ((sin(nx))/2^n ) ⇒ s(x)=Im(Σ_(n=1) ^∞ (e^(inx) /2^n )) but Σ_(n=1) ^∞ (e^(inx) /2^n )=Σ_(n=1) ^∞ ((1/2)e^(ix) )^n (1/(1−(1/2)e^(ix) )) (look that ∣(1/2)e^(ix) ∣<1) =(2/(2−cosx−isinx)) =((2(2−cosx+isinx))/((2−cosx)^2 +sin^2 x)) ⇒s(x)=((2sinx)/(4−4cosx +1)) =((2sinx)/(5−4cosx)) and Σ_(n=1) ^∞ ((sin(((nπ)/4)))/2^n ) =s((π/4))=((2sin((π/4)))/(5−4cos((π/4)))) =((2×(1/( (√2))))/(5−4.((√2)/2)))=((√2)/(5−2(√2)))

$${s}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{sin}\left({nx}\right)}{\mathrm{2}^{{n}} }\:\Rightarrow \\ $$$${s}\left({x}\right)={Im}\left(\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{e}^{{inx}} }{\mathrm{2}^{{n}} }\right)\:{but} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{e}^{{inx}} }{\mathrm{2}^{{n}} }=\sum_{{n}=\mathrm{1}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{2}}{e}^{{ix}} \right)^{{n}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{e}^{{ix}} }\:\:\left({look}\:{that}\:\mid\frac{\mathrm{1}}{\mathrm{2}}{e}^{{ix}} \mid<\mathrm{1}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}−{cosx}−{isinx}} \\ $$$$=\frac{\mathrm{2}\left(\mathrm{2}−{cosx}+{isinx}\right)}{\left(\mathrm{2}−{cosx}\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} {x}} \\ $$$$\Rightarrow{s}\left({x}\right)=\frac{\mathrm{2}{sinx}}{\mathrm{4}−\mathrm{4}{cosx}\:+\mathrm{1}} \\ $$$$=\frac{\mathrm{2}{sinx}}{\mathrm{5}−\mathrm{4}{cosx}} \\ $$$${and}\:\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{sin}\left(\frac{{n}\pi}{\mathrm{4}}\right)}{\mathrm{2}^{{n}} } \\ $$$$={s}\left(\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{4}}\right)}{\mathrm{5}−\mathrm{4}{cos}\left(\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\frac{\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\mathrm{5}−\mathrm{4}.\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{2}}} \\ $$

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