Dean Hanafy

Posted on Wednesday, 28 May, 2003 - 03:29 pm:

Reduction formula:

I_{n} = \int_{0}^{1} (1 x^{2})^{- n} dx

show that for

n \geq 2

2 (n - 1) I_{n} = 2^{1 - n} + (2 n - 3) I_{n - 1}

Now, I have managed to prove an alternative reduction formula relating

I_{n}

and

I_{n - 1}

. And I have got so far and I just need a little push, so hints only please :-)

and secondly

$I_{n} = \int_{0}^{π / 2} \sin^{(} x) dx$

show that $(n + 2) I_{n + 2} = (n + 1) I_{n}$ , which I can do, but the next part(which I'm close to getting) is

show that $I_{2 n} = (2 n)! π / (2^{2 n + 1} (n!)^{2})$
cheers
Deano

David Loeffler

Posted on Wednesday, 28 May, 2003 - 03:36 pm:

For the first one, would you mind telling us how far you have got?

For the second one, use induction. Let's say a_2n = (2n)!p / (2²ⁿ⁺¹ (n!)² ).

You can easily evaluate I₀ , and check that it agrees with a₀ .

Now show that a_2n satisfies the same recurrence as the I's, that is that (2n+2)a_2n+2 = (2n+1)a_2n .

Now if a_2n = I_2n , this shows that a_2n+2 = I_2n+2 . And since you know a₀ = I₀ , it follows that a_2n = I_2n for all n.

David

Dean Hanafy

Posted on Thursday, 29 May, 2003 - 12:35 pm:

Cheers David, induction! Should try and learn to spot when to use it to my advantage, thanks!

as for the reduction formula i get

(2n)I_n+1 =(2n-1)I_n + 2^-n if my memory serves me correctly (not looking at my solution at the minute), which seems a valid alternative but not what is being asked to be proved.....again hints would be great....

Cheers
Deano

Matthew Buckley

Posted on Thursday, 29 May, 2003 - 01:27 pm:

If you want a hint for the first question, try thinking of 1/(1+x² )ⁿ as 1*1/(1+x² )ⁿ . Now integrate by parts. Post back if you still need help.

Andre Rzym

Posted on Thursday, 29 May, 2003 - 09:11 pm:

Dean, in the reduction formula as per your "memory", just substitute n-1 for n and you have the result you were looking to prove.

You have done it!

Andre

Matthew Buckley

Posted on Thursday, 29 May, 2003 - 09:18 pm:

So you have! Well done - the way I suggested does give you exactly what you got originally. Then, as Andre said, sub in n-1 for n, and you are done!

Philip Ellison

Posted on Friday, 30 May, 2003 - 01:07 pm:

Matthew, could you please post another hint, as I am stuck on the first integral posed. Thanks

Matthew Buckley

Posted on Friday, 30 May, 2003 - 01:48 pm:

No probs Philip:

Start with

$I_{n} = \int_{0}^{1} 1 * 1 / (1 + x^{2})^{n} dx,$

and integrate this by parts. You can do this because 1 is easy to integrate. This should give:

Philip Ellison

Posted on Friday, 30 May, 2003 - 01:51 pm:

My problem was with evaluating the last integral! It looks like it should be solveable by using a trig substitution, but this would be really messy, so I'm hoping there's a nicer way (all the other questions in this exercise came out fairly easily, so I'm hoping that I've just missed something obvious!).

Kerwin Hui

Posted on Friday, 30 May, 2003 - 01:58 pm:

Use x² =(1+x² )-1 to get the last integral in terms of I_n and I_n+1 .

Philip Ellison

Posted on Friday, 30 May, 2003 - 02:17 pm:

D'oh! Thanks very much.

Matthew Buckley

Posted on Friday, 30 May, 2003 - 02:19 pm:

Sorry Philip - didn't realise that it was that last bit that was causing problems!!!! As Kerwin has said, think of the x² on the top as x² +1-1, and then split it up as suggested.

Just to extend it a bit more, this is a very useful trick for evaluating integrals that look a little nasty at first glance. For example, if you had to evaluate

\int x^{4} / (1 + x^{2}) dx

, you could add on

x^{2}

and subtract it, to give you

\int [x^{2} - x^{2} / (1 + x^{2})] dx .

Then, apply the trick again, but this time add and subtract 1 to get

\int [x^{2} - 1 + 1 / (1 + x^{2})] dx .

Now you have something that is integrable. The first two bits are easy to do, and the last one can be done using trig identities.< /font>

Just thought you should know that - another trick up your sleeve for your exams

Philip Ellison

Posted on Friday, 30 May, 2003 - 03:33 pm:

I've seen this technique before, but haven't used it much, and so I don't have much experience with it. It's something I'll have to look out for in future.
Thanks for your help.