umair butt

Posted on Tuesday, 03 February, 2004 - 08:43 pm:

I need some help with this, many thanks:

given that $I_{n} = \int_{0}^{1} (x^{n / 2}) (1 - x)^{1 / 2} dx$ , show that for $n \geq 2$ , $I_{n} = (n / (n + 3)) I_{n - 2}$

Kerwin Hui

Posted on Tuesday, 03 February, 2004 - 08:49 pm:

Integrate by parts, expand (1-x)^3/2 as (1-x)^1/2 -x(1-x)^1/2 .

Kerwin

umair butt

Posted on Tuesday, 03 February, 2004 - 09:07 pm:

thanks kerwin,how would you advise me to tackle these binomial type reduction formula questions?
the expo and trigonometric ones are much easier!

Kerwin Hui

Posted on Tuesday, 03 February, 2004 - 09:19 pm:

The process is more-or-less universal in deriving reduction formulae, namely an exercise in integration by parts and rearranging your integrand. For example, in this question we don't really have much choice - either we have u=x^n/2 or u=(1-x)^1/2 . The latter choice would mean that we have another integral, which has (1-x) in its denominator and there are no easy way to get rid of it [You could try integrating by parts, but that will either make things even more complicated or reduce to the integral we started with]. So u=x^n/2 is the only sensible move. After this, the only way to rearrange the integrand x^n/2-1 (1-x)^3/2 into sums of terms in the form we started with is to expand a factor (1-x).

Kerwin

umair butt

Posted on Tuesday, 03 February, 2004 - 09:33 pm:

i get the point for which function to differentiate and integrate, but its just the end integrand i find difficult to rearrange into sums of the initial function.How would you advise me to get over this problem? Should i do as many of these type of questions as possible to get the hang of it?

Kerwin Hui

Posted on Tuesday, 03 February, 2004 - 09:59 pm:

Doing questions is one of the best way to get the hang of techniques. Try to keep your mind focus on what you want to achieve - it should help you to decide which part of the integrand needs some manipulation.

Kerwin

umair butt

Posted on Tuesday, 03 February, 2004 - 10:24 pm:

thanks a bunch kerwin,p.s when you did reduction formula did u first find it quite challenging?
Did you need to dedicate some amount of time on it?

Kerwin Hui

Posted on Tuesday, 03 February, 2004 - 10:40 pm:

I think the challenging part is not to make any mistakes going from one line to another when you manipulate the integrand - these mistakes are usually harder to spot, especially if you jump steps in your writing.

Kerwin