Re: Re: idiom for recurrence relations
- To: mathgroup at smc.vnet.net
- Subject: [mg27100] Re: [mg27081] Re: [mg26714] idiom for recurrence relations
- From: BobHanlon at aol.com
- Date: Sun, 4 Feb 2001 02:58:26 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Or more generally,
Sum[Fibonacci[n]/x^(n+1), {n, Infinity}]//Simplify // ExpandAll // Simplify
1/(x^2 - x - 1)
% /. x -> 10
1/89
Sum[Fibonacci[n]*x^n, {n, Infinity}]//Simplify // ExpandAll // Simplify
-(x/(x^2 + x - 1))
And @@ Table[
Fibonacci[n] == (D[x/(1-x-x^2), {x, n}]/n! /. x -> 0), {n, 20}]
True
Bob Hanlon
In a message dated 2001/2/3 5:30:21 AM, spiralcenter314 at my-deja.com writes:
>I have done some research on the Golden ratio, though not probably as
>extensive as you.
>
>But this may be pertinent, 1 / 89 = .011235955...
>This has all the fibonacci in it. Which is:
>
>0/(10^1)+
>1/(10^2)+
>1/(10^3)+
>2/(10^4)+
>3/(10^5)+
>5/(10^6)+
>etc...
>