prove-that-lim-n-n-2k-1-cos-k-2pi-n-pi-

Question Number 140050 by mnjuly1970 last updated on 03/May/21

p r o v e t h a t ::

ϕ := {l i m}_{n \to \infty} \frac{n}{\sqrt{2 k}} . \sqrt{1 - {c o s}^{k} (\frac{2 π}{n})} = π

\dots \dots \dots \dots \dots .

Answered by Kamel last updated on 04/May/21

Answered by Dwaipayan Shikari last updated on 03/May/21

\frac{n}{\sqrt{2 k}} \sqrt{1 - {c o s}^{k} (\frac{2 π}{n})} \approx 2 \frac{n}{\sqrt{2 k}} \sqrt{1 - (1 - \frac{2 π^{2} k}{2! n^{2}})} = \frac{\sqrt{2} n π}{\sqrt{2 k}} . \frac{\sqrt{k}}{n} = π

Answered by mnjuly1970 last updated on 04/May/21

p r o v e :: \dots \dots \dots \dots .

\frac{1}{n} = t - \Rightarrow {_{t \to 0^{+}}^{n \to \infty}

ϕ := {l i m}_{t \to 0^{+}} \frac{1}{\sqrt{2 k}} (\frac{1}{t}) \sqrt{1 - {c o s}^{k} (2 π t)}

:= {l i m}_{t \to 0^{+}} \frac{1}{\sqrt{2 k}} (\frac{1}{t}) {\sqrt{1 - c o s (2 π t)} . \sqrt{1 + c o s (2 π t) + \dots + {c o s}^{k - 1} (2 π t)}

:= {l i m}_{t \to 0^{+}} \frac{1}{\sqrt{2 k}} (\frac{1}{t}) {\sqrt{2 {s i n}^{2} (π t)} . \sqrt{1 + c o s (π t) + \dots + {c o s}^{k - 1} (2 π t)}}

:= {l i m}_{t \to 0^{+}} \frac{\sqrt{2} s i n (π t)}{\sqrt{2} \sqrt{k}} \sqrt{[1 + 1 + 1 + \dots . + 1] := k t i m e s}

\dots \dots . . ϕ : = π \dots \dots .