prove that:: φ :=lim_(n→∞) (n/( (√(2k)))) .(√(1−cos^k (((2π)/n)))) =π ................


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

$p r o v e t h a t ::$ $ϕ := {l i m}_{n \to \infty} \frac{n}{\sqrt{2 k}} . \sqrt{1 - {c o s}^{k} (\frac{2 π}{n})} = π$ $. . . . . . . . . . . . . . . .$
	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
	$prove::............. (1/n) =t −⇒{_( t→0^+ ) ^( n→∞) φ:= lim_(t→0^+ ) (1/( (√(2k)))) ((1/t))(√(1−cos^k (2πt))) :=lim_(t→0^+ ) (1/( (√(2k)))) ((1/t)){(√(1−cos(2πt))) .(√(1+cos(2πt)+...+cos^(k−1) (2πt))) :=lim_(t→0^+ ) (1/( (√(2k))))((1/t)){(√(2sin^2 (πt))) .(√(1+cos(πt)+...+cos^(k−1) (2πt))) } :=lim_(t→0^+ ) (((√2) sin(πt))/( (√2) (√k))) (√([1+1+1+....+1]:=k times)) ........ φ : = π .......$
	$p r o v e :: . . . . . . . . . . . . .$ $\frac{1}{n} = t - \Rightarrow {_{t \to 0^{+}}^{n \to \infty}$ $ϕ := {l i m}_{t \to 0^{+}} \frac{1}{\sqrt{2 k}} (\frac{1}{t}) \sqrt{1 - {c o s}^{k} (2 π t)}$ $:= {l i m}_{t \to 0^{+}} \frac{1}{\sqrt{2 k}} (\frac{1}{t}) {\sqrt{1 - c o s (2 π t)} . \sqrt{1 + c o s (2 π t) + . . . + {c o s}^{k - 1} (2 π t)}$ $:= {l i m}_{t \to 0^{+}} \frac{1}{\sqrt{2 k}} (\frac{1}{t}) {\sqrt{2 {s i n}^{2} (π t)} . \sqrt{1 + c o s (π t) + . . . + {c o s}^{k - 1} (2 π t)}}$ $:= {l i m}_{t \to 0^{+}} \frac{\sqrt{2} s i n (π t)}{\sqrt{2} \sqrt{k}} \sqrt{[1 + 1 + 1 + . . . . + 1] := k t i m e s}$ $. . . . . . . . ϕ : = π . . . . . . .$
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