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Question Number 115621 by sachin1221 last updated on 27/Sep/20

$. . . a d v a n c e d c a l c u l u s . . .$ $e v a l u a t e ::$ $s h o w t h a t {l i m}_{n \to \infty} \frac{1}{n} [{c o s}^{2 p} \frac{π}{2 n} + {c o s}^{2 p} \frac{2 π}{2 n} + {c o s}^{2 p} \frac{3 π}{2 n} . . . . . . {c o s}^{2 p} \frac{π}{2}] = \underset{r = 1}{\prod^{p}} \frac{p + r}{4 r}$
	Answered by TANMAY PANACEA last updated on 27/Sep/20

	$\lim_{n \to \infty} \frac{1}{n} \underset{r = 1}{\sum^{n}} {(c o s \frac{r π}{2 n})}^{2 p}$ $\int_{0}^{1} c o s {(\frac{x π}{2})}^{2 p} d x$ $t = \frac{x π}{2} \to \frac{d t}{d x} = \frac{π}{2}$ $\int_{0}^{\frac{π}{2}} {c o s}^{2 p} t \times \frac{2}{π} d t$ $π I = 2 \int_{0}^{\frac{π}{2}} {c o s}^{2 p} t \times {s i n}^{0} t d t$ $n o w i s h a l l u s e g a m m a f u n c t i o n$ $f o r m u l a$ $2 \int_{0}^{\frac{π}{2}} {(s i n θ)}^{2 m - 1} {(c o s θ)}^{2 n - 1} d θ = \frac{⌈ (m) ⌈ (n)}{⌈ (m + n)}$ $⌈ \to g a m m a o p e r a t o r$ $π I = 2 \int_{0}^{\frac{π}{2}} {(s i n t)}^{2 \times \frac{1}{2} - 1} {(c o s t)}^{2 \times \frac{2 p + 1}{2} - 1} d t$ $I = \frac{1}{π} \times \frac{⌈ (\frac{1}{2}) \times ⌈ (\frac{2 p + 1}{2})}{⌈ (\frac{1}{2} + \frac{2 p + 1}{2})}$
	Commented by Rasheed.Sindhi last updated on 27/Sep/20

	$S i r t a n m a y a r e y o u^{'} t a n m a y$ ${c h a u d h r y}^{'} a n o l d m e m b e r o f t h e$ $f o r u m ?$
	Commented by TANMAY PANACEA last updated on 27/Sep/20

	$y e s s i r i a m t h e o l d m e m b e r . . . g o r f e w m o n t h s i w a s i n a c t i v e$
	Commented by TANMAY PANACEA last updated on 27/Sep/20

	$r e a d f o r i n s t e d o f g o r$
	Commented by bemath last updated on 27/Sep/20

	$s a m e s i r . i^{'} m o l d m e m b e r t o o . m y$ $a g e 87 y e a r . h i h i h i h i$
	Commented by Ar Brandon last updated on 27/Sep/20
	��I now remember you Mr Tanmay. You once helped me out with Q86830. Nice to meet you again Sir��
	Commented by Ar Brandon last updated on 27/Sep/20
	��BeMath stop kidding bro. I once had a look at your profile picture and you aren't that old��. You're a young guy. <20 years old I guess. You rather look good.�� Are you gonna tell me now that it was your grandson ? ��
	Commented by bemath last updated on 27/Sep/20

	$o o h h n o b r o$
	Commented by Dwaipayan Shikari last updated on 27/Sep/20
	An e^2.8 bro I am��
	Commented by bemath last updated on 27/Sep/20

	$g o o d . y o u r i s t h e s a n t u y$
	Commented by Rasheed.Sindhi last updated on 27/Sep/20

	$T h a n k s t a n m a y s i r!$
	Answered by Ar Brandon last updated on 27/Sep/20
	$S=lim_(n→∞) (1/n)[cos^(2p) ((π/(2n)))+cos^(2p) (((2π)/(2n)))+cos^(2p) (((3π)/(2n)))+∙∙∙+cos^(2p) ((π/2))] =lim_(n→∞) (1/n)Σ_(k=1) ^n cos^(2p) (((kπ)/(2n)))=∫_0 ^1 cos^(2p) (((πx)/2))dx (Reimann′s Sum) S=∫_0 ^1 cos^(2p) (((πx)/2))dx=(2/π)∫_0 ^(π/2) cos^(2p) (u)du {setting ((πx)/2)=u} =(2/π)[(1/2)∙(3/4)∙(5/6)∙∙∙((2p−5)/(2p−4))∙((2p−3)/(2p−2))∙((2p−1)/(2p))∙(π/2)] (Walli′s method) =(2/π)∙(π/2)∙Π_(r=0) ^(p−1) ((2p−(2r+1))/(2p−2r))=Π_(r=0) ^(p−1) ((2p−(2r+1))/(2p−2r))$
	$S = \lim_{n \to \infty} \frac{1}{n} [\cos^{2 p} (\frac{π}{2 n}) + \cos^{2 p} (\frac{2 π}{2 n}) + \cos^{2 p} (\frac{3 π}{2 n}) + \cdot \cdot \cdot + \cos^{2 p} (\frac{π}{2})]$ $= \lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} \cos^{2 p} (\frac{k π}{2 n}) = \int_{0}^{1} \cos^{2 p} (\frac{π x}{2}) d x ({R e i m a n n}^{'} s S u m)$ $S = \int_{0}^{1} \cos^{2 p} (\frac{π x}{2}) d x = \frac{2}{π} \int_{0}^{\frac{π}{2}} \cos^{2 p} (u) du {s e t t i n g \frac{π x}{2} = u}$ $= \frac{2}{π} [\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \cdot \cdot \frac{2 p - 5}{2 p - 4} \cdot \frac{2 p - 3}{2 p - 2} \cdot \frac{2 p - 1}{2 p} \cdot \frac{π}{2}] ({W a l l i}^{'} s m e t h o d)$ $= \frac{2}{π} \cdot \frac{π}{2} \cdot \underset{r = 0}{\prod^{p - 1}} \frac{2 p - (2 r + 1)}{2 p - 2 r} = \underset{r = 0}{\prod^{p - 1}} \frac{2 p - (2 r + 1)}{2 p - 2 r}$
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