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Topic: Which Polynomial?
Replies: 28   Last Post: Jul 27, 2006 3:13 PM

 Messages: [ Previous | Next ]
 luciepoirier Posts: 59 Registered: 1/30/05
Re: Which Polynomial?
Posted: Jul 31, 2005 1:53 PM

>> No, no, Kirby. Look at the sequence f1^n+f2^n. Every
>> recursively-generated sequence has a generating
>> polynomial. f1 and f2 are the two roots.

>
> No really, the usual thing is to approach phi using two successive terms of
> the sequence. That's not to say your exercise is irrelevant though --
> although the sequence I'm getting from your instructions isn't
> 0,1,1,2,3,5,8,13... but 1,3,4,7,11,18,29...
>

f[n] = q^n satisfies the Fibonacci recurrence relation f[n] - f[n-1] -
f[n-2] = 0 (n=0,1,2,...) iff q^n - q^(n-1) - q^(n-2) = 0 (with q non zero)
that is, q^2 - q - 1 = 0.

The roots are then q(1) = (1+SQRT5)/2 and q(2) = (1-SQRT5)/2 both solutions
of the recurrence relation.

Since the relation is linear and homogeneous, the general solution is f[n] =
c(1)q(1)^n + c(2)q(2)^n for any choice of c(1) and c(2). In order to satisfy
the (two) initial values f(0) = 0 and f(1) = 1, we determine c(1) = 1/SQRT5
and c(2) = -1/SQRT5 which gives the correct formula for f[n] corresponding
to the initial values 0 and 1.

JP

Date Subject Author
7/30/05 Kirby Urner
7/30/05 talmanl@mscd.edu
7/30/05 Victor Steinbok
7/30/05 Kirby Urner
7/31/05 Victor Steinbok
7/31/05 Kirby Urner
7/31/05 luciepoirier
7/31/05 Kirby Urner
7/31/05 luciepoirier
7/31/05 Victor Steinbok
8/1/05 Kirby Urner
8/1/05 Kirby Urner
8/1/05 talmanl@mscd.edu
8/2/05 Wayne Bishop
8/2/05 Victor Steinbok
8/2/05 Wayne Bishop
8/2/05 Victor Steinbok
8/4/05 luciepoirier
8/2/05 talmanl@mscd.edu
8/2/05 Kirby Urner
7/30/05 Pam
7/30/05 talmanl@mscd.edu
7/31/05 luciepoirier
7/30/05 Kirby Urner
7/31/05 Victor Steinbok
7/31/05 Kirby Urner
9/20/05 Kirby Urner
3/14/06 Kirby Urner
7/27/06 Kirby Urner