Power of sine!

Calculus Level 5

I ( n ) = ∫ 0 π 2 sin ⁡ 2 n x d x X = ∑ r = 1 ∞ I ( r ) ( 2 r r ) I(n)= \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin ^{ 2n }{ x } } \ dx \qquad \qquad X=\sum _{ r=1 }^{ \infty }{ \frac { I(r) }{ { \left( \begin{matrix} 2r \\r \end{matrix} \right) } } }

Given the above, find ⌊ 10 4 X ⌋ \left\lfloor { 10 }^{ 4 }X \right\rfloor

Notation:

  • ( n m ) = n ! n ! ( n − m ) ! \displaystyle \binom{n}{m} = \frac{n!}{n! (n-m)!} denotes the binomial coefficient .
  • ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .
This is part of my set Powers of the ordinary .


The answer is 5235.

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Raghav Vaidyanathan by Raghav Vaidyanathan , 23, USA

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1 solution

Using reduction formula, we get: I ( n ) = 2 n − 1 2 n I ( n − 1 ) , I ( 0 ) = π 2 ⇒ I ( n ) = ( 2 n − 1 ) ( 2 n − 3 ) ( 2 n − 5 ) . . . . ( 1 ) ( 2 n ) ( 2 n − 2 ) ( 2 n − 4 ) . . . ( 2 ) × π 2 ⇒ I ( n ) = ( 2 n n ) × π 2 × 2 − 2 n ⇒ I ( n ) ( 2 n n ) = π 2 × 2 − 2 n ⇒ ∑ r = 1 ∞ I ( r ) ( 2 r r ) = π 2 ( 1 4 + 1 16 + 1 64 + . . . . ) = π 6 = X ⇒ ⌊ 10 4 X ⌋ = 5235 I(n)=\frac { 2n-1 }{ 2n } I(n-1)\quad ,\quad I(0)=\quad \frac { \pi }{ 2 } \\ \\ \Rightarrow I(n)\quad =\frac { (2n-1)(2n-3)(2n-5)....(1) }{ (2n)(2n-2)(2n-4)...(2) } \times \frac { \pi }{ 2 } \\ \\ \Rightarrow I(n)\quad =\left( \begin{matrix} 2n \\ n \end{matrix} \right) \times \frac { \pi }{ 2 } \times { 2 }^{ -2n }\quad \\ \Rightarrow \frac { I(n) }{ \left( \begin{matrix} 2n \\ n \end{matrix} \right) } =\quad \frac { \pi }{ 2 } \times { 2 }^{ -2n }\\ \Rightarrow \sum _{ r=1 }^{ \infty }{ \frac { I(r) }{ { \left( \begin{matrix} 2r \\ r \end{matrix} \right) } } } =\frac { \pi }{ 2 } (\frac { 1 }{ 4 } +\frac { 1 }{ 16 } +\frac { 1 }{ 64 } +....)=\frac { \pi }{ 6 } =X\\ \Rightarrow \left\lfloor { 10 }^{ 4 }X \right\rfloor =\boxed { 5235 }

15 Helpful 0 Interesting 0 Brilliant 0 Confused

Did exactly the same

Rohit Shah - 6 years, 3 months ago

Oh! That is something new to me! Can anyone explain me about this?

I rather did using Beta function!

Kartik Sharma - 6 years, 3 months ago

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I would like to add another hint , if i may :

∫ 0 π 2 s i n m x ⋅ c o s n x d x = Γ ( m + 1 2 ) ⋅ Γ ( n + 1 2 ) 2 ⋅ Γ ( m + n + 2 2 ) \int_{0}^{\frac{\pi}{2}} sin^{m} x \cdot cos^{n} x dx \\= \dfrac{\Gamma{ ( \dfrac{m+1}{2} ) }\cdot \Gamma{ ( \dfrac{n+1}{2} ) }}{2\cdot \Gamma{ ( \dfrac{m+n+2}{2} ) }}

Did anyone consider this approach , by substituting m → 2 n , n → 0 m\rightarrow 2n , n\rightarrow 0 in the formula provided above ?

A Former Brilliant Member - 6 years, 3 months ago

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Yeah, my friend used this method to solve it.

Raghav Vaidyanathan - 6 years, 3 months ago

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@Raghav Vaidyanathan – Yes , it is a nice method .

Best of Luck Raghav for JEE

A Former Brilliant Member - 6 years, 3 months ago

Reduction formula is just recurrence using IBP. Just google Wallis reduction formula

Rohit Shah - 6 years, 3 months ago

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Oh k. Yes it is Wallis reduction formula. I know about it a little.

Kartik Sharma - 6 years, 3 months ago

Actually you know what , using Walli's Reduction formulae is the best approach for this question since most JEE aspirants know these formulae by heart . I am not saying that using Beta function won't do any good, just that use it for more complex problems .

But it's still your choice of using it or not :)

A Former Brilliant Member - 6 years, 3 months ago

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Oh, is it so? Thanks for that! I will keep that in mind!

Kartik Sharma - 6 years, 3 months ago

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