Prove-that-a-series-inspired-Knopp-Konrad-e-k-0-sin-kpi-4-k-2-k-pi-k-

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

Prove that : (a series inspired Knopp Konrad)

\sqrt{e^{π}} = \underset{k = 0}{\sum^{\infty}} \frac{\sin (\frac{k π}{4})}{(k!) \sqrt{2^{k}}} π^{k}

Answered by mindispower last updated on 25/Dec/21

s i n (a) = {I m e}^{i a}

\sum_{k ⩾ 0} \frac{s i n (\frac{k π}{4})}{k! . \sqrt{2^{k}}} . π^{k} = I m \sum_{k ⩾ 0} \frac{{(π e^{i \frac{π}{4}})}^{k}}{k! {(\sqrt{2})}^{k}}

= {I m e}^{\frac{π}{\sqrt{2}} e^{i \frac{π}{4}}} = I m e^{\frac{π}{2} (1 + i)}

= e^{\frac{π}{2}} = \sqrt{e^{π}}

Commented by HongKing last updated on 25/Dec/21

cool my dear Sir thank you so much

Answered by Lordose last updated on 25/Dec/21

A = \underset{k = 0}{\sum^{\infty}} \frac{\sin (\frac{k π}{4})}{k! \sqrt{2^{k}}} π^{k} = Im \underset{k = 0}{\sum^{\infty}} \frac{e^{\frac{i π k}{4}}}{k! \sqrt{2^{k}}} π^{k}

A = Im \underset{k = 0}{\sum^{\infty}} \frac{{(\frac{π e^{\frac{i π}{4}}}{\sqrt{2}})}^{k}}{k!} = \underset{k = 0}{\sum^{\infty}} \frac{{(\frac{π}{2})}^{k}}{k!}

e^{x} = \underset{k = 0}{\sum^{\infty}} \frac{x^{k}}{k!}, Im (e^{\frac{i π}{4}}) = \frac{1}{\sqrt{2}}

A = e^{\frac{π}{2}} = \sqrt{e^{π}}

Facebook Tweet Pin