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Question Number 161964 by HongKing last updated on 24/Dec/21

$$\mathrm{Prove}\mathrm{that}:\left(\mathrm{a}\mathrm{series}\mathrm{inspired}\mathrm{Knopp}\mathrm{Konrad}\right)$$$$\sqrt{{\mathrm{e}}^{\mathit{\pi }}}=\underset{\mathbf{k}=0}{\sum ^{\mathrm{\infty }}}\frac{\sin \left(\frac{k\pi }{4}\right)}{(\mathrm{k}!)\sqrt{2^{\mathbf{k}}}}{\pi }^{\mathbf{k}}$$

Answered by mindispower last updated on 25/Dec/21

$$sin\left(a\right)={Ime}^{ia}$$$$\sum _{k⩾0}\frac{sin\left(\frac{k\pi }{4}\right)}{k!.\sqrt{2^k}}.{\pi }^k=Im\sum _{k⩾0}\frac{{\left(\pi e^{i\frac{\pi }{4}}\right)}^k}{k!{\left(\sqrt{2}\right)}^k}$$$$={Ime}^{\frac{\pi }{\sqrt{2}}{e}^{i\frac{\pi }{4}}}=Im{e}^{\frac{\pi }{2}(1+i)}$$$$=e^{\frac{\pi }{2}}=\sqrt{e^{\pi }}$$

Commented by HongKing last updated on 25/Dec/21

$$\mathrm{cool}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

Answered by Lordose last updated on 25/Dec/21

$$\mathrm{A}=\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{\sin \left(\frac{\mathrm{k}\mathit{\pi }}{4}\right)}{\mathrm{k}!\sqrt{2^{\mathrm{k}}}}{\mathit{\pi }}^{\mathrm{k}}=\mathfrak{Im}\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{e}}^{\frac{\mathrm{i}\mathit{\pi }\mathbf{k}}{4}}}{\mathrm{k}!\sqrt{2^{\mathrm{k}}}}{\mathit{\pi }}^{\mathrm{k}}$$$$\mathrm{A}=\mathfrak{Im}\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{\mathit{\pi }{\mathrm{e}}^{\frac{\mathrm{i}\mathit{\pi }}{4}}}{\sqrt{2}}\right)}^{\mathrm{k}}}{\mathrm{k}!}=\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{\mathit{\pi }}{2}\right)}^{\mathrm{k}}}{\mathrm{k}!}$$$${\mathrm{e}}^{\mathrm{x}}=\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{x}}^{\mathrm{k}}}{\mathrm{k}!},\mathfrak{Im}\left({\mathrm{e}}^{\frac{\mathbf{i}\mathit{\pi }}{4}}\right)=\frac{1}{\sqrt{2}}$$$$\mathrm{A}={\mathbf{e}}^{\frac{\mathit{\pi }}{2}}=\sqrt{{\mathbf{e}}^{\mathit{\pi }}}$$

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