Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

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

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)))

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com