Dean Hanafy
Posted on Wednesday, 28 May, 2003 - 03:29 pm:
Reduction formula:
In = ó
õ 		1

0 
 (1x2)-n  dx
show that for n ³ 2
2(n-1) In = 21-n + (2n-3) In-1
Now, I have managed to prove an alternative reduction formula relating In and In-1. And I have got so far and I just need a little push, so hints only please :-)

and secondly
In = ó
õ 		p/2

0 
 sin(x)  dx
show that (n+2) In+2 = (n+1)In, which I can do, but the next part(which I'm close to getting) is

show that I2n = (2n)! p/(22n+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 a2n = (2n)!p / (22n+1 (n!)2 ).

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

Now show that a2n satisfies the same recurrence as the I's, that is that (2n+2)a2n+2 = (2n+1)a2n .

Now if a2n = I2n , this shows that a2n+2 = I2n+2 . And since you know a0 = I0 , it follows that a2n = I2n 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)In+1 =(2n-1)In + 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
In = ó
õ 		1

0 
 1*1/(1+x2)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 In and In+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 òx4/(1+x2) dx, you could add on x2 and subtract it, to give you
ó
õ 		[x2 - x2 /(1+x2 )]  dx.
Then, apply the trick again, but this time add and subtract 1 to get
ó
õ 		[x2 - 1 + 1/(1+x2 )] 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.