# Using a Reduction Formula

• Dec 6th 2009, 02:51 PM
Using a Reduction Formula
I want to prove equation (1). Note that the equation (2) can be proven by the reduction formula for $\int sin^n(x)dx$, but I'll just write the formula down since I know it's true.

I want to show that:
----------------------------------------------------
$\int_0^{\frac{\pi}{2}}sin^{2n+1}xdx=\frac{(2)(4)(6 )...(2n)}{(3)(5)(7)...(2n+1)}$
----------------------------------------------------

This seems rather difficult, and I'm not entirely sure where to begin. I'm an amateur, so I don't have much experience with proofs like this. I'll write the equation like this:

(1) $\int_0^{\frac{\pi}{2}}sin^{2n+1}xdx=\prod_{i=1}^n\ frac{2i}{2i+1}$

Using the following reduction formula:

(2) $\int_0^{\frac{\pi}{2}}sin^n(x)dx=\frac{n-1}{n}\int_0^{\frac{\pi}{2}}sin^{n-2}xdx$

I can write the hypothesis for $n=k$

$\int_0^{\frac{\pi}{2}}sin^{2k+1}(x)dx=\frac{(2k+1)-1}{2k+1}\int_0^{\frac{\pi}{2}}sin^{(2k+1)-2}(x)dx$

So now I have:

$\frac{2k}{2k+1}\int_0^{\frac{\pi}{2}}sin^{2k-1}(x)dx=\prod_{i=1}^k\frac{2i}{2i+1}$

Factoring the left side:

$\prod_{i=1}^k\frac{2i}{2i+1}=\frac{2k}{2k+1}\prod_ {i=2}^k\frac{2(i-1)}{2i-1}$

(3) $\int_0^{\frac{\pi}{2}}sin^{2k-1}(x)dx=\prod_{i=2}^k\frac{2(i-1)}{2i-1}$

All I've managed to do is write the inductive hypothesis. It's really frustrating that I can't proceed. I can't figure out how to connect one side to the other. Wring the next term for n=k+1 doesn't seem to get me anywhere. Can anyone offer some insight? (Headbang)
• Dec 7th 2009, 01:26 PM
Jester
Quote:

I want to prove equation (1). Note that the equation (2) can be proven by the reduction formula for $\int sin^n(x)dx$, but I'll just write the formula down since I know it's true.

I want to show that:
----------------------------------------------------
$\int_0^{\frac{\pi}{2}}sin^{2n+1}xdx=\frac{(2)(4)(6 )...(2n)}{(3)(5)(7)...(2n+1)}$
----------------------------------------------------

This seems rather difficult, and I'm not entirely sure where to begin. I'm an amateur, so I don't have much experience with proofs like this. I'll write the equation like this:

(1) $\int_0^{\frac{\pi}{2}}sin^{2n+1}xdx=\prod_{i=1}^n\ frac{2i}{2i+1}$

Using the following reduction formula:

(2) $\int_0^{\frac{\pi}{2}}sin^n(x)dx=\frac{n-1}{n}\int_0^{\frac{\pi}{2}}sin^{n-2}xdx$

I can write the hypothesis for $n=k$

$\int_0^{\frac{\pi}{2}}sin^{2k+1}(x)dx=\frac{(2k+1)-1}{2k+1}\int_0^{\frac{\pi}{2}}sin^{(2k+1)-2}(x)dx$

So now I have:

$\frac{2k}{2k+1}\int_0^{\frac{\pi}{2}}sin^{2k-1}(x)dx=\prod_{i=1}^k\frac{2i}{2i+1}$

Factoring the left side:

$\prod_{i=1}^k\frac{2i}{2i+1}=\frac{2k}{2k+1}\prod_ {i=2}^k\frac{2(i-1)}{2i-1}$

(3) $\int_0^{\frac{\pi}{2}}sin^{2k-1}(x)dx=\prod_{i=2}^k\frac{2(i-1)}{2i-1}$

All I've managed to do is write the inductive hypothesis. It's really frustrating that I can't proceed. I can't figure out how to connect one side to the other. Wring the next term for n=k+1 doesn't seem to get me anywhere. Can anyone offer some insight? (Headbang)

It seems to me like you have all the pieces. Just assemby them.

Assume true for $n = k$ so

$\int_0^{\frac{\pi}{2}}sin^{2k+1}x\,dx=\prod_{i=1}^ k\frac{2i}{2i+1}$

and prove true for $n = k+1$, i.e.

$\int_0^{\frac{\pi}{2}}sin^{2k+3}x\,dx=\prod_{i=1}^ {k+1}\frac{2i}{2i+1}$.

$\int_0^{\frac{\pi}{2}}sin^{2k+3} x\,dx = \frac{2k+2}{2k+3}\int_0^{\frac{\pi}{2}}sin^{2k+1} x\,dx$ (from your reduction formula)

$= \frac{2k+2}{2k+3} \prod_{i=1}^k\frac{2i}{2i+1}$

$= \prod_{i=1}^{k+1}\frac{2i}{2i+1}$

as required.
• Dec 7th 2009, 01:43 PM
Quote:

Originally Posted by Danny
It seems to me like you have all the pieces. Just assemby them.

Assume true for $n = k$ so

$\int_0^{\frac{\pi}{2}}sin^{2k+1}x\,dx=\prod_{i=1}^ k\frac{2i}{2i+1}$

and prove true for $n = k+1$, i.e.

$\int_0^{\frac{\pi}{2}}sin^{2k+3}x\,dx=\prod_{i=1}^ {k+1}\frac{2i}{2i+1}$.

$\int_0^{\frac{\pi}{2}}sin^{2k+3} x\,dx = \frac{2k+2}{2k+3}\int_0^{\frac{\pi}{2}}sin^{2k+1} x\,dx$ (from your reduction formula)
$= \frac{2k+2}{2k+3} \prod_{i=1}^k\frac{2i}{2i+1}$
$= \prod_{i=1}^{k+1}\frac{2i}{2i+1}$