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Get an approximation formula for the following integral: $$ \sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy $$

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Evaluating the integral could be a start. Use the Beta function. –  Ragib Zaman Dec 27 '11 at 1:02
    
Thak you. Also a small question: how to accept answers here? thank you. –  David Dec 27 '11 at 1:06
    
@David Did you mean $\cos^{2(n-k)+1}\left( \sin^{2(k-1)}(y) \right) \mathrm{d}y$ or $\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \cdot \mathrm{d}y$ ? –  Sasha Dec 27 '11 at 1:09
    
$\cos^{2(n-k)+1}\cdot\sin^{2(k-1)}ydy$ –  David Dec 27 '11 at 1:14
    
@Patrick To the OP's credit, they do ask in a comment how to accept answers in this site. :) // David: In each answer for your questions, under the vote count will be a grey V-shaped tick mark. Clicking it will accept the answer. [You can also unaccept answers subsequently; just click the tick mark once again.] –  Srivatsan Dec 27 '11 at 2:05

2 Answers 2

$$\sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy $$

First observe the powers of $cos$ and $sin$

$$ cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) = cos^{2n-2k}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y) = (1-sin^{2})^{(n-k)}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y)$$

$$ \int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy = \int_0^{\frac{\pi}{2}} \left( (1-sin^{2})^{(n-k)}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y) \right) dy $$

If we use the substitution $t = sin(y) \Rightarrow dt = cos(y) \hspace{4pt} dy$

$t = 0$ when $y=0$ and $t=1$ when $y=\frac{\pi}{2}$

$$ \int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy = \int_0^{\frac{\pi}{2}} \left( (1-sin^2)^{(n-k)}(y) \cdot \sin^{2(k-1)}(y)\cdot \cos(y) \right) dy $$

$$ = \int_0^1 t^{(2k-2)} (1-t^{2})^{(n-k)} dt $$

The expression

$$ \sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy = \sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1} \left( \int_0^1 t^{(2k-2)} (1-t^{2})^{(n-k)} dt \right) $$

$$ = \left( \frac{1}{35} \right)^{k-1}\int_0^1 \left( \sum_{k=1}^{n} t^{(2k-2)} (1-t^{2})^{(n-k)} \right) dt $$

Notice that the sum is a geometric series with first term $\frac{1}{35}(1-t^{2})^{n}$ and the common ratio $\frac{t^{2}}{35(1-t^{2})^{2}}$

The expression therefore simplifies to

$$ = \int_0^1 \left( \sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1} t^{(2k-2)} (1-t^2)^{(n-k)} \right) dt $$

$$ \sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy = \left( \frac{1}{35} \right)^{k-1} \sum_{k=1}^n \left( \int_0^1 t^{(2k-2)} (1-t^{2})^{(n-k)} dt \right) $$

$$ = \int_0^1 \left( \sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1} t^{(2k-2)} (1-t^{2})^{(n-k)} \right) dt $$

Notice that the sum is a geometric series with first term $(1-t^{2})^{(n-1)}$ and the common ratio $\frac{t^{2}}{35(1-t^{2})}$

The expression therefore simplifies to

$$ = \int_0^1 \left( \sum_{k=1}^{n} \left( \frac{1}{35} \right)^{k-1} \left( t^{(2k-2)} (1-t^{2})^{(n-k)} \right) \right) dt $$

This expression turns out to be the summation of constant multiple of Beta function

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$$ \sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy $$

We have that

$$B(x,y) = 2 \int_0^{\pi/2} \sin^{2x-1} \theta \cos^{2y-1} \theta d\theta$$

So we can write your expression as

$$\sum\limits_{k = 1}^n {{{\left( {\frac{1}{{35}}} \right)}^{k - 1}}} \frac{1}{2}B\left( {k - \frac{1}{2},n - k + 1} \right)$$

But we also know that

$$B\left( {x,y} \right) = \frac{{\Gamma \left( x \right)\Gamma \left( y \right)}}{{\Gamma \left( {x + y} \right)}}$$

so that

$$\frac{1}{2}B\left( {k - \frac{1}{2},n - k + 1} \right) = \frac{1}{2}\frac{{\Gamma \left( {k - \frac{1}{2}} \right)\Gamma \left( {n - k + 1} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}}$$

And we have closed formulas for two of the three $\Gamma$ functions there, namely:

$$\eqalign{ & \Gamma \left( {n + \frac{1}{2}} \right) = \frac{{\left( {2n - 1} \right)!!}}{{{2^n}}}\sqrt \pi \cr & \Gamma \left( {k - \frac{1}{2}} \right) = \Gamma \left( {k - 1 + \frac{1}{2}} \right) = \frac{{\left( {2k - 3} \right)!!}}{{{2^{k - 1}}}}\sqrt \pi \cr} $$

For the last one, we simply put

$$\Gamma \left( {n - k + 1} \right) = \left( {n - k} \right)!$$

So we get

$$\frac{1}{2}B\left( {k - \frac{1}{2},n - k + 1} \right) = \frac{{\left( {2k - 3} \right)!!\left( {n - k} \right)!}}{{{2^{k - n}}\left( {2n - 1} \right)!!}}$$

And then

$$\sum\limits_{k = 1}^n {{{\left( {\frac{1}{{35}}} \right)}^{k - 1}}} \frac{{\left( {2k - 3} \right)!!\left( {n - k} \right)!}}{{{2^{k - n}}\left( {2n - 1} \right)!!}}$$

or

$$\frac{{{2^n}}}{{\left( {2n - 1} \right)!!}}\sum\limits_{k = 1}^n {{{\left( {\frac{1}{{35}}} \right)}^{k - 1}}} \frac{{\left( {2k - 3} \right)!!\left( {n - k} \right)!}}{{{2^k}}}$$

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