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Question Number 161964 by HongKing last updated on 24/Dec/21
Prove that: (a series inspired Knopp Konrad) (√e^𝛑 ) = Σ_(k=0) ^∞ ((sin(((kπ)/4)))/((k!) (√2^k ))) π^k
$$\mathrm{Prove}\:\mathrm{that}:\:\left(\mathrm{a}\:\mathrm{series}\:\mathrm{inspired}\:\mathrm{Knopp}\:\mathrm{Konrad}\right) \\ $$$$\sqrt{\mathrm{e}^{\boldsymbol{\pi}} }\:\:=\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{sin}\left(\frac{{k}\pi}{\mathrm{4}}\right)}{\left(\mathrm{k}!\right)\:\sqrt{\mathrm{2}^{\boldsymbol{\mathrm{k}}} }}\:\:\pi^{\boldsymbol{\mathrm{k}}} \\ $$
Answered by mindispower last updated on 25/Dec/21
sin(a)=Ime^(ia) Σ_(k≥0) ((sin(((kπ)/4)))/(k!.(√2^k ))).π^k =ImΣ_(k≥0) (((πe^(i(π/4)) )^k )/(k!((√2))^k )) =Ime^((π/( (√2))) e^(i(π/4)) ) =Im e^((π/2)(1+i)) =e^(π/2) =(√e^π )
$${sin}\left({a}\right)={Ime}^{{ia}} \\ $$$$\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{sin}\left(\frac{{k}\pi}{\mathrm{4}}\right)}{{k}!.\sqrt{\mathrm{2}^{{k}} }}.\pi^{{k}} ={Im}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(\pi{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{{k}} }{{k}!\left(\sqrt{\mathrm{2}}\right)^{{k}} } \\ $$$$={Ime}^{\frac{\pi}{\:\sqrt{\mathrm{2}}}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} } ={Im}\:{e}^{\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+{i}\right)} \\ $$$$={e}^{\frac{\pi}{\mathrm{2}}} =\sqrt{{e}^{\pi} } \\ $$
Commented by HongKing last updated on 25/Dec/21
cool my dear Sir thank you so much
$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Answered by Lordose last updated on 25/Dec/21
A = Σ_(k=0) ^∞ ((sin(((k𝛑)/4)))/(k!(√2^k )))𝛑^k = ImΣ_(k=0) ^∞ (e^((i𝛑k)/4) /(k!(√2^k )))𝛑^k A = ImΣ_(k=0) ^∞ (((((𝛑e^((i𝛑)/4) )/( (√2))))^k )/(k!)) = Σ_(k=0) ^∞ ((((𝛑/2))^k )/(k!)) e^x = Σ_(k=0) ^∞ (x^k /(k!)), Im(e^((i𝛑)/4) ) = (1/( (√2))) A = e^(𝛑/2) = (√e^𝛑 )
$$\mathrm{A}\:=\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}\left(\frac{\mathrm{k}\boldsymbol{\pi}}{\mathrm{4}}\right)}{\mathrm{k}!\sqrt{\mathrm{2}^{\mathrm{k}} }}\boldsymbol{\pi}^{\mathrm{k}} \:=\:\boldsymbol{\mathfrak{Im}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{e}^{\frac{\mathrm{i}\boldsymbol{\pi\mathrm{k}}}{\mathrm{4}}} }{\mathrm{k}!\sqrt{\mathrm{2}^{\mathrm{k}} }}\boldsymbol{\pi}^{\mathrm{k}} \\ $$$$\mathrm{A}\:=\:\boldsymbol{\mathfrak{Im}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\boldsymbol{\pi}\mathrm{e}^{\frac{\mathrm{i}\boldsymbol{\pi}}{\mathrm{4}}} }{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{k}} }{\mathrm{k}!}\:=\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)^{\mathrm{k}} }{\mathrm{k}!} \\ $$$$\mathrm{e}^{\mathrm{x}} \:=\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{x}^{\mathrm{k}} }{\mathrm{k}!},\:\boldsymbol{\mathfrak{Im}}\left(\mathrm{e}^{\frac{\boldsymbol{\mathrm{i}\pi}}{\mathrm{4}}} \right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{A}\:=\:\boldsymbol{\mathrm{e}}^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:=\:\sqrt{\boldsymbol{\mathrm{e}}^{\boldsymbol{\pi}} } \\ $$

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