advanced-calculus-evaluate-show-that-lim-n-1-n-cos-2p-pi-2n-cos-2p-2pi-2n-cos-2p-3pi-2n-cos-2p-pi-2-r-1-p-p-r-4r-

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

... advanced calculus... evaluate :: show that lim_(n→∞) (1/n)[cos^(2p) (π/(2n))+cos^(2p) ((2π)/(2n))+cos^(2p) ((3π)/(2n))......cos^(2p) (π/2)] =Π_(r=1) ^p ((p+r)/(4r))

\dots a d v a n c e d c a l c u l u s \dots

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} \dots \dots {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→∞) (1/n)Σ_(r=1) ^n (cos((rπ)/(2n)))^(2p) ∫_0 ^1 cos(((xπ)/2))^(2p) dx t=((xπ)/2)→(dt/dx)=(π/2) ∫_0 ^(π/2) cos^(2p) t×(2/π)dt πI=2∫_0 ^(π/2) cos^(2p) t×sin^0 t dt now i shall use gamma function formula 2∫_0 ^(π/2) (sinθ)^(2m−1) (cosθ)^(2n−1) dθ=((⌈(m)⌈(n))/(⌈(m+n))) ⌈→gamma operator πI=2∫_(0 ) ^(π/2) (sint)^(2×(1/2)−1) (cost)^(2×((2p+1)/2)−1) dt I=(1/π)×((⌈((1/2))×⌈(((2p+1)/2)))/(⌈((1/2)+((2p+1)/2))))

\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

Sir tanmay are you ′tanmay chaudhry′ an old member of the forum?

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

yes sir i am the old member ...gor few months i was inactive

y e s s i r i a m t h e o l d m e m b e r \dots 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

same sir. i′m old member too. my age 87 year. hihihihi

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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