Re: help working with functions (follow-up)
- To: mathgroup at smc.vnet.net
- Subject: [mg60035] Re: help working with functions (follow-up)
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Mon, 29 Aug 2005 01:38:39 -0400 (EDT)
- Organization: The University of Western Australia
- References: <dero92$s7j$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <dero92$s7j$1 at smc.vnet.net>, Bob Hanlon <hanlonr at cox.net>
wrote:
> Amplifying on my previous response
>
> f1[i_,j_,n_]:= Module[{k=0},
> Nest[
> Append[#,#[[-2]]+#[[-1]]+i*j^(k++)]&,
> {0,1}, n-2]];
>
> f1[0,4,10]
>
> {0,1,1,2,3,5,8,13,21,34}
>
> You can also solve for the closed form
>
> expr=FullSimplify[a[k]/.
> RSolve[{a[k]==a[k-1]+a[k-2]+i*j^(k-2),a[0]==0,a[1]==1}, a[k], k][[1]],
> Element[k, Integers]]
>
> (2^(-k - 1)*(2*Sqrt[5]*(-(1 - Sqrt[5])^k + (1 + Sqrt[5])^k)*((j - 1)*j - 1) +
> i*(5*2^(k + 1)*j^k + 2*Sqrt[5]*((1 - Sqrt[5])^k - (1 + Sqrt[5])^k)*j +
> (-5 +
> Sqrt[5])*(1 + Sqrt[5])^k -
> (1 - Sqrt[5])^k*(5 + Sqrt[5]))))/(5*((j - 1)*j - 1))
>
> Consequently,
>
> f2[i_,j_,n_]:=Table[(2^(-k-1)*(2*Sqrt[5]*(-(1-Sqrt[5])^k+(1+
> Sqrt[5])^k)*((j-1)*j-1)+i*(5*2^(k+1)*j^k+2*Sqrt[5]*((
> 1-Sqrt[5])^k-(1+Sqrt[5])^k)*j+(-5+Sqrt[5])*(
> 1+Sqrt[5])^
> k-(1-Sqrt[
> 5])^k*(5+Sqrt[5]))))/(5*((j-1)*j-1)),{
> k,0,n-1}]//Simplify;
>
> Checking that f1 and f2 are equivalent
>
> And@@Flatten@Table[f1[i,j,n]==f2[i,j,n],{i,5},{j,5},{n,2,10}]
>
> True
>
> For i=0 the sequence is the Fibonacci sequence
You can use this observation to rewrite expr as
(i (-((1/2)(1 + Sqrt[5]))^k + j^k))/(-1 - j + j^2) +
(1 + (i (1 + Sqrt[5] - 2j))/(2 (-1 - j + j^2))) Fibonacci[k]
which is somewhat simpler, and clearly equal to Fibonacci[k] when i = 0.
Cheers,
Paul
_______________________________________________________________________
Paul Abbott Phone: 61 8 6488 2734
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