Question-123823

Question Number 123823 by Dwaipayan Shikari last updated on 28/Nov/20

Answered by mindispower last updated on 29/Nov/20

{(1 + x)}^{a} = 1 + \sum_{n ⩾ 1} \frac{1}{n!} \underset{k = 0}{\prod^{n - 1}} (a - k) . x^{n}

{(1 - {k s i n}^{2} (x))}^{- \frac{1}{2}} = 1 + Σ \frac{1}{n!} \underset{k = 0}{\prod^{n - 1}} (- \frac{1}{2} - k) . {s i n}^{2 n} (x)

= 1 + \sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{n! 2^{n}} \prod_{k ⩽ n - 1} (1 + 2 k) . k^{n} {s i n}^{2 n} (x)

= 1 + \sum_{n ⩾ 1} \frac{{(- 1)}^{n} (2 n - 1)!!}{n! . 2^{n}} k^{n} {s i n}^{2 n} (x)

\int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{1 - {k s i n}^{2} (x)}} = \int_{0}^{\frac{π}{2}} (1 + \sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{n! . 2^{n}} (2 n - 1)!! . k^{n} {s i n}^{2 n} (x)) d x

S_{n} = \frac{π}{2} + Σ \frac{{(- 1)}^{n} (2 n - 1)!! . k^{n}}{2^{n} n!} \int_{0}^{\frac{π}{2}} {s i n}^{2 n} (x) d x

\int_{0}^{\frac{π}{2}} {s i n}^{a} (x) = \frac{1}{2} . 2 \int_{0}^{\frac{π}{2}} {s i n}^{2 (\frac{a}{2} + \frac{1}{2}) - 1} {c o s}^{2 \frac{1}{2} - 1} (x) d x

\frac{1}{2} β (\frac{a}{2} + \frac{1}{2}, \frac{1}{2}) = \frac{Γ (\frac{a}{2} + \frac{1}{2}) Γ (\frac{1}{2})}{2 Γ (1 + \frac{a}{2})}, v a l i d f o r R e (a) > - 1

s_{n} = \frac{π}{2} + \sum_{n ⩾ 1} \frac{{(- 1)}^{n} (2 n - 1)!!}{2^{n} . n!} . k^{n} . \frac{1}{2} \frac{Γ (n + \frac{1}{2}) Γ (\frac{1}{2})}{Γ (1 + n)}

Γ (n + \frac{1}{2}) = \underset{k = 0}{\prod^{n - 1}} (n - \frac{1}{2} - k) Γ (\frac{1}{2}), Γ (\frac{1}{2}) = \sqrt{π}

Γ (1 + n) = n!

\Leftrightarrow

S_{n} = \frac{π}{2} + \sum_{n ⩾ 1} \frac{{(- 1)}^{n} (2 n - 1)!!}{2^{n} n!} k^{n} . \frac{1}{2} \prod_{k ⩽ n - 1} (\frac{2 n - 1 - 2 k}{2}) Γ (\frac{1}{2}) Γ (\frac{1}{2}) . \frac{1}{n!}

= \frac{π}{2} + \sum_{n ⩾ 1} \frac{{(- 1)}^{n} (2 n - 1)!! . (2 n - 1)!!}{2^{n} n! . 2 . 2^{n} n!} k^{n} Γ^{2} (\frac{1}{2})

= \frac{π}{2} + \sum_{n ⩾ 1} {(\frac{(2 n - 1)!!}{n!})}^{2} \frac{{(- k)}^{n} π}{2^{2 n + 1}}

s o m t h i n g m i s s i n g

n o t, \sum_{n ⩾ 0} {(\frac{(2 n - 1)!!}{n!})}^{2} \dots .

b y c o m p a r i s o n \Rightarrow {(- k)}^{n} = 256^{n}

\Rightarrow {(- k)}^{n} = 256^{n} e^{2 i π m}

\Rightarrow - k = 256 e^{\frac{2 i π m}{n}}, \forall n t r u

n \to \infty, - k = 256 \Rightarrow k = - 256

\sqrt{k} = \underline{+} 16 i

Commented by mnjuly1970 last updated on 29/Nov/20

e x c e l l e n t s i r \dots

Commented by Dwaipayan Shikari last updated on 29/Nov/20

T h i s i s a q u e s t i o n {f r o m}^{″} B r i l l i a n t^{″}

Commented by Dwaipayan Shikari last updated on 29/Nov/20

https://brilliant.org/problems/elliptic-integral

Commented by mindispower last updated on 29/Nov/20

w e c a n u s e (x + 1)!! = (x + 1) (x - 1)!!

\Rightarrow 1!! = 1 . (- 1)!!

\Rightarrow (- 1)!! = 1

w i t h e t h a t

n = 0

{(\frac{(- 1)!!}{0!})}^{2} . \frac{256^{0} . π}{2^{2.0 + 1}} = \frac{π}{2}

w e a r e d o n n e