Math Question and Answers


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 144311 by mnjuly1970 last updated on 24/Jun/21

$. . . . . . Nice . . . . Calculus . . . . . .$ $Find the value of ::$ $Θ := \underset{n = 1}{\sum^{\infty}} \frac{1}{4^{n} {c o s}^{2} (\frac{π}{2^{n + 2}})} = ?$ $. . . . . . . . . .$
Commented by Kamel last updated on 24/Jun/21

$S_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{4^{k} {c o s}^{2} (\frac{π}{2^{k + 2}})}, T_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{4^{k} {s i n}^{2} (\frac{π}{2^{k + 2}})}$ $S o : S_{n} + T_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{4^{k} {(s i n (\frac{π}{2^{k + 2}}) c o s (\frac{π}{2^{k + 2}}))}^{2}} = \underset{k = 1}{\sum^{n}} \frac{1}{4^{k - 1} {s i n}^{2} (\frac{π}{2^{k + 1}})}$ $\overset{p = k - 1}{=} \underset{p = 0}{\sum^{n - 1}} \frac{1}{4^{p} {s i n}^{2} (\frac{π}{2^{p + 2}})} = 2 + T_{n} - \frac{1}{4^{n} {s i n}^{2} (\frac{π}{2^{n + 2}})}$ $T h e n : S_{n} = 2 - \frac{1}{4^{n} {s i n}^{2} (\frac{π}{2^{n + 2}})}$ $∴ \underset{n = 1}{\sum^{+ \infty}} \frac{1}{4^{n} {c o s}^{2} (\frac{π}{2^{n + 2}})} = 2 - \underset{n \to + \infty}{l i m} {(\frac{\overset{}{1}}{2^{n} s i n (\frac{π}{2^{n + 2}})})}^{2} \overset{t = \frac{π}{2^{n + 2}}}{=} 2 - \underset{t \to 0^{+}}{l i m} {(\frac{4}{π} \frac{t}{s i n (t)})}^{2}$ $∴ \underset{n = 1}{\sum^{+ \infty}} \frac{1}{4^{n} {c o s}^{2} (\frac{π}{2^{n + 2}})} = 2 - \frac{16}{π^{2}}$
Commented by mnjuly1970 last updated on 24/Jun/21

$b r a v o m r k a m e l t a s h a k o r . .$
	Answered by ArielVyny last updated on 24/Jun/21

	$Θ = \underset{n = 1}{\sum^{\infty}} \frac{1}{4^{n}} [1 + {t g}^{2} (\frac{π}{2^{n + 2}})] = \underset{n = 1}{\sum^{\infty}} {(\frac{1}{4})}^{n} + \underset{n = 1}{\sum^{\infty}} \frac{{t g}^{2} (\frac{π}{2^{n + 2}})}{4^{n}}$ $Θ = \frac{4}{3} + \underset{n = 1}{\sum^{\infty}} {(\frac{1}{4})}^{n} {t g}^{2} (\frac{π}{4 \times 2^{n}})$ $p o s o n s {(\frac{1}{2})}^{n} = t$ $Θ = \frac{4}{3} + \underset{t = \frac{1}{2}}{\sum^{0}} t^{2} {t g}^{2} (\frac{π}{4} t) = \frac{4}{3} - \underset{t = 0}{\sum^{\frac{1}{2}}} t^{2} {t g}^{2} (\frac{π}{4} t)$ $Θ = \frac{4}{3} - \frac{1}{2} \underset{t = 0}{\sum^{1}} t^{2} {t g}^{2} (\frac{π}{4} t)$ $Θ = \frac{4}{3} - \frac{1}{2}$ $Θ = \frac{5}{6}$
	Commented by mathmax by abdo last updated on 25/Jun/21

	$not correct!$
	Answered by abdullahoudou last updated on 24/Jun/21

	$1 - {(2 / π)}^{2}$
	Commented by ArielVyny last updated on 24/Jun/21

	$y o u c a n s h o w t h a t p l e a s e ?$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum