Posted: Mon Oct 25, 2004 7:27 pm    Post subject: Recurrence relation

Consider a recurrence satisfying T1=1, and Tn=T1T_(n-1)+T2T_(n-2)+…+T_(n-1)T1 I consider f(x)=T1x+T2x^2+…. (suppose f(x) is convergent) then i get f(x)^2 = f(x)-x f(x) = [1-sqrt(1-4x)]/2.

But then how to find out the general term of T_n?

Posted: Tue Oct 26, 2004 7:14 pm    Post subject: Re: Recurrence relation

This would be useful: http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A000108

Edit:

Since f is a g.f. of , we could use the binomial expansion of to get the coefficients of each term.

Posted: Tue Oct 26, 2004 7:49 pm    Post subject:

Thanks

Posted: Tue Oct 26, 2004 9:17 pm    Post subject:

How about a variation?
Tn=T0T_(n-1)+T1T_(n-2)+…+T_(n-1)T0

Posted: Tue Oct 26, 2004 9:19 pm    Post subject: Re: Recurrence relation

may i know what does the binonmial theorem with non-integer power look like?

Posted: Wed Oct 27, 2004 7:08 pm    Post subject:

Let’s say T0=1. For taylor’s series, do you mean using this formula,

f(x)=f(0)+f'(0)x+f”(0)x^2+… and so on?

Posted: Thu Oct 28, 2004 7:30 pm    Post subject:

It is

—

If we define , then T’ would be same as the original T.

BTW, if (in the original def.) , then . (C(n) is the function described in here)

—

Also, we can define the binomial coefficent for natural k as (not using Gamma function):

Posted: Thu Oct 28, 2004 9:04 pm    Post subject:

ah yes…. i typed taylor’s formula wrongly…. missing out all those factorials in the denominator.. sorry.
Thanks anyway.

Posted: Thu Oct 28, 2004 9:50 pm    Post subject:

This is very interesting. I was working on the following:

with , , and

with . Is there anything that you can say about them?