How do i find and what is the recurrence formula for (x^n)(1+x)^1/2.dx between the x terminals 0 and -1 ? Thanks Nath

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- Jun 23rd 2006, 03:54 AMnath_quamQuick Integration
How do i find and what is the recurrence formula for (x^n)(1+x)^1/2.dx between the x terminals 0 and -1 ? Thanks Nath

- Jun 23rd 2006, 10:41 AMThePerfectHackerQuote:

Originally Posted by**nath_quam**

otherwise the function is not defined, which means that ,

makes no sense at all!.

Assuming you mean,

Use integration by parts,

let, then, and, then,

Thus,

Thus,

The trick is to express,

as,

Thus,

Working with the integral we have,

=

Let,

Then, as we have shown,

Which is,

Thus,

Thus,

Thus,

Thus, the recurrence formula is,

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Credits:

I learend the box trick from Soroban - Jun 23rd 2006, 10:59 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

surly?

RonL - Jun 23rd 2006, 11:02 AMThePerfectHacker
Sorry, I see things differently sometimes.

My solution to the above problem follows the same idea.

Almost identical.

Try to find the solution yourself, if not I can finish it. - Jun 23rd 2006, 06:25 PMnath_quam
i was asking bout what captain wrote i really shoud learn how to make it look good

- Jun 23rd 2006, 06:34 PMmalaygoelQuote:

Originally Posted by**nath_quam**

- Jun 24th 2006, 07:52 PMThePerfectHacker
nath quam

Did you get the recrussive relationship yet?

I can integrate it for you if you like.

After all, I am the quickest integrator in the west ! - Jun 24th 2006, 11:44 PMnath_quam
Thanks that would be good if u could perfect cos i get stuck when i do it

- Jun 25th 2006, 12:31 AMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

.

Now we are going to use integration by parts, which has the schema:

.

Here we let: , and , so:

,

and:

.

So

.

Putting the limits back in:

.

Simplifying:

So the recurrence relation for is:

RonL

(you will need to check this carefully as there are too many steps to

guarantee there is no mistake)