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Re: Fibonacci
*To*: mathgroup at smc.vnet.net
*Subject*: [mg13189] Re: Fibonacci
*From*: dreissNOSPAM at nospam.earthlink.net (David Reiss)
*Date*: Mon, 13 Jul 1998 07:42:39 -0400
*Organization*: EarthLink Network, Inc.
*References*: <6nsiqc$f39@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
When using the add-on package Calculus`Limit` evaluation of
Limit[Fibonacci[n]/Fibonacci[n-1], n -> Infinity] returns 1 rather than
GoldenRatio. So, bug it is. I will pass it on to the wri bugs
folks...
In article <6nsiqc$f39 at smc.vnet.net>, BobHanlon at aol.com wrote:
> On a Macintosh running Mathematica 3.0.1, the limit returns unevaluated:
>
> Limit[Fibonacci[n]/Fibonacci[n-1], n -> Infinity]
>
> Limit[Fibonacci[n]/Fibonacci[-1 + n], n -> Infinity]
>
> fib[n_] := (((1+Sqrt[5])/2)^n - (-((1+Sqrt[5])/2))^-n)/Sqrt[5]
>
> Table[fib[n], {n, 10}]//FullSimplify
>
> {1,1,2,3,5,8,13,21,34,55}
>
> Limit[fib[n]/fib[n-1], n -> Infinity]
>
> Limit[(-(1/2*(-1 - Sqrt[5]))^(-n) + (1/2*(1 + Sqrt[5]))^n)/
> (-(1/2*(-1 - Sqrt[5]))^(1 - n) +
> (1/2*(1 + Sqrt[5]))^(-1 + n)), n -> Infinity]
>
> However, since
>
> fib[n]/fib[n-1] == (fib[n-1] + fib[n-2])/fib[n-1] == 1 +
> fib[n-2]/fib[n-1]
>
> Then, the limit must satisfy the equation (x == 1 + 1/x)
>
> Select[x /. Solve[x == 1 + 1/x, x], #>0&]
>
> {1/2*(1 + Sqrt[5])}
>
> which is the Golden Ratio
>
> %[[1]] == GoldenRatio // N
>
> True
>
> Bob Hanlon
>
> In a message dated 7/4/98 8:01:56 PM, tobi.kamke at t-online.de wrote:
>
> >I've a problem. I thought that Limit[Fibonacci[n]/Fibonacci[n-1], n ->
> >Infinity] is GoldenRatio.
> >
> >Mathematica says 1.
> >What's wrong?
--
David Reiss
dreissNOSPAM at nospam.earthlink.net
http://home.earthlink.net/~dreiss
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