Menu Close

n-1-n-n-cos-pin-5-




Question Number 127793 by Dwaipayan Shikari last updated on 02/Jan/21
Σ_(n=1) ^∞ (n/(n!))cos(((πn)/5))
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{{n}!}{cos}\left(\frac{\pi{n}}{\mathrm{5}}\right) \\ $$
Commented by Dwaipayan Shikari last updated on 03/Jan/21
I have found  e^(((√5)+1)/4) cos(((4π+5(√(2(5−(√5)))))/(20)))=e^(√ϕ) cos(((4π+5(√(2(5−(√5)))))/(20)))
$${I}\:{have}\:{found} \\ $$$${e}^{\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}} {cos}\left(\frac{\mathrm{4}\pi+\mathrm{5}\sqrt{\mathrm{2}\left(\mathrm{5}−\sqrt{\mathrm{5}}\right)}}{\mathrm{20}}\right)={e}^{\sqrt{\varphi}} {cos}\left(\frac{\mathrm{4}\pi+\mathrm{5}\sqrt{\mathrm{2}\left(\mathrm{5}−\sqrt{\mathrm{5}}\right)}}{\mathrm{20}}\right) \\ $$
Answered by mnjuly1970 last updated on 02/Jan/21
Σ_(n=1) ^∞ (1/((n−1)!))cos(((nπ)/5))  =Re(Σ_(n=0) ^∞ (e^((i(n−1)π)/5) /(n!)))=Re(e^((−iπ)/5) Σ(e^((inπ)/5) /(n!)))  =cos((π/5))Re[Σ_(n=0) ^∞ ((e^((inπ)/5) /(n!)) )]  =cos((π/5))Re(e^(i(π/5)) )=cos^2 ((π/5))  =(((1+(√5) )/4))^2 =((6+2(√5) )/(16))=((3+(√5))/8)
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}{cos}\left(\frac{{n}\pi}{\mathrm{5}}\right) \\ $$$$={Re}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{e}^{\frac{{i}\left({n}−\mathrm{1}\right)\pi}{\mathrm{5}}} }{{n}!}\right)={Re}\left({e}^{\frac{−{i}\pi}{\mathrm{5}}} \Sigma\frac{{e}^{\frac{{in}\pi}{\mathrm{5}}} }{{n}!}\right) \\ $$$$={cos}\left(\frac{\pi}{\mathrm{5}}\right){Re}\left[\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{e}^{\frac{{in}\pi}{\mathrm{5}}} }{{n}!}\:\right)\right] \\ $$$$={cos}\left(\frac{\pi}{\mathrm{5}}\right){Re}\left({e}^{{i}\frac{\pi}{\mathrm{5}}} \right)={cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{5}}\right) \\ $$$$=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}\:}{\mathrm{4}}\right)^{\mathrm{2}} =\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\:}{\mathrm{16}}=\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 02/Jan/21
S =Σ_(n=1) ^∞  (1/((n−1)!))cos(((nπ)/5)) ⇒S =Re(Σ_(n=1) ^∞ (1/((n−1)!))e^((inπ)/5) ) we have  Σ_(n=1) ^∞  (1/((n−1)!)) (e^((iπ)/5) )^n  =Σ_(n=0) ^∞  (1/(n!))(e^((iπ)/5) )^(n+1)  =e^((iπ)/5)  Σ_(n=0) ^∞  (((e^((iπ)/5) )^n )/(n!))  =e^((iπ)/5)  ×e^e^((iπ)/5)   =e^(((iπ)/5)+cos((π/5))+isin((π/5)))  =e^(cos((π/5))+i((π/5) +sin((π/5)))   =e^(cos((π/5))) ×{cos((π/5)+sin((π/5)))+isin((π/5) +sin((π/5)))} ⇒  S =e^(cos((π/5)))  .cos((π/5)+sin((π/5)))  with cos((π/5))=((1+(√5))/4)
$$\mathrm{S}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{n}−\mathrm{1}\right)!}\mathrm{cos}\left(\frac{\mathrm{n}\pi}{\mathrm{5}}\right)\:\Rightarrow\mathrm{S}\:=\mathrm{Re}\left(\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{n}−\mathrm{1}\right)!}\mathrm{e}^{\frac{\mathrm{in}\pi}{\mathrm{5}}} \right)\:\mathrm{we}\:\mathrm{have} \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{n}−\mathrm{1}\right)!}\:\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{5}}} \right)^{\mathrm{n}} \:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}!}\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{5}}} \right)^{\mathrm{n}+\mathrm{1}} \:=\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{5}}} \:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{5}}} \right)^{\mathrm{n}} }{\mathrm{n}!} \\ $$$$=\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{5}}} \:×\mathrm{e}^{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{5}}} } \:=\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{5}}+\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right)+\mathrm{isin}\left(\frac{\pi}{\mathrm{5}}\right)} \:=\mathrm{e}^{\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right)+\mathrm{i}\left(\frac{\pi}{\mathrm{5}}\:+\mathrm{sin}\left(\frac{\pi}{\mathrm{5}}\right)\right.} \\ $$$$=\mathrm{e}^{\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right)} ×\left\{\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}+\mathrm{sin}\left(\frac{\pi}{\mathrm{5}}\right)\right)+\mathrm{isin}\left(\frac{\pi}{\mathrm{5}}\:+\mathrm{sin}\left(\frac{\pi}{\mathrm{5}}\right)\right)\right\}\:\Rightarrow \\ $$$$\mathrm{S}\:=\mathrm{e}^{\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right)} \:.\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}+\mathrm{sin}\left(\frac{\pi}{\mathrm{5}}\right)\right)\:\:\mathrm{with}\:\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *