1. Problem #21: Recurrence Integral

Show that
$\int_0^1 (1 - x^2)^n~dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$

Use any technique you like. It looks, perhaps, like something using Legendre polynomials? (I didn't do it this way)

One way to attack this (and the one I'll be featuring as my solution) is to denote the integral as $I_n$. We have the recurrence relation:
$I_{k + 1} = \left ( \frac{2k + 1}{2k + 3} \right )~ I_k$

-Dan

2. Re: Problem #21: Recurrence Integral

$\int_0^1 (1-x)^n \, dx = \frac{1}{n+1}$

3. Re: Problem #21: Recurrence Integral

Originally Posted by Idea
$\int_0^1 (1-x)^n \, dx = \frac{1}{n+1}$
Whoops! Typo. I have fixed it in the problem statement. Thanks for the catch!

-Dan

4. Re: Problem #21: Recurrence Integral

$I_{k+1}=\frac{2 (k+1)}{2 k+3}I_k$

5. Re: Problem #21: Recurrence Integral

That's the last time I post a problem on two hours of sleep. Thanks (again) for the catch.

Originally Posted by topsquark
Show that
$\int_0^1 (1 - x^2)^n~dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$

Use any technique you like. It looks, perhaps, like something using Legendre polynomials? (I didn't do it this way)

One way to attack this (and the one I'll be featuring as my solution) is to denote the integral as $I_n$. We have the recurrence relation:
$I_{k + 1} = \left ( \frac{2k + 2}{2k + 3} \right )~ I_k$
-Dan

6. Re: Problem #21: Recurrence Integral

No takers this time around. Let's see if I can post the solution without any typos!

Spoiler:
Let's look at the recurrence relation first.

Let $I_{k + 1} = \int_0^1 (1 - x^2)^{k + 1}~dx$

Do this by parts:
$\int u ~ dv = uv - \int v~du$

I am choosing $u = (1 - x^2)^{k + 1}$ and $dv = dx$

This gives
$I_{k + 1} = \left . x \cdot (1 - x^2)^{k + 1} \right |_0^1 - \int_0^1 (k +1)x(1 - x^2)^k(-2x)~dx$

The first term vanishes leaving
$I_{k + 1} = (2k + 2) \int_0^1 x^2 (1 - x^2)^k~dx$

Now for a cool little trick:
$I_{k + 1} = (2k + 2) \int_0^1 \left [1 - (1 - x^2) \right ] (1 - x^2)^k~dx$

$I_{k + 1} = (2k + 2) \left [ \int_0^1 (1 - x^2)^k~dx - \int_0^1 (1 - x^2)^{k + 1}~dx \right ]$

$I_{k + 1} = (2k + 2)I_k - (2k + 2)I_{k + 1}$

So finally
$I_{k + 1} = \left ( \frac{2k + 2}{2k + 3} \right ) I_k$

Now for the induction step:

Base case: k = 0:
$\int_0^1 ( 1 - x^2)^0~dx = 1 = \frac{2^{2 \cdot 0} (0!)^2}{1!}$ (Check!)

Now let's assume the theorem is true for some integer k. We need to show that the k + 1 case works as well. So:

$I_{k +1} = \left ( \frac{2k + 2}{2k + 3} \right ) I_k = \left ( \frac{2k + 2}{2k + 3} \right ) \cdot \frac{2^{2k} ( k!)^2}{(2k + 1)!} = \frac{2(k + 1)2^{2k}(k!)^2}{(2k + 3)(2k + 1)!}$

Multiply by 1:
$I_{k + 1} = \frac{2(k + 1)}{2k + 2} \cdot \frac{2(k + 1)2^{2k}(k!)^2}{(2k + 3)(2k + 1)!}$

$I_{k + 1} = \frac{ \left [ 2(k + 1) \right ] ^2 2^{2k}(k!)^2}{(2k + 3)(2k + 2)(2k + 1)!}$

$I_{k + 1} = \frac{2^{2(k + 1)} \left [ (k + 1)! \right ]^2}{\left [ 2(k + 1) + 1 \right ]!}$

So we have that the kth case implies the k+1th case. Therefore etc.

-Dan