Re: uniqueness of fibonacci number extension

Robert Israel <israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> writes:

"David C. Ullrich" <dullrich@xxxxxxxxxxx> writes:

On Fri, 9 May 2008 17:06:15 -0700 (GMT-07:00), Dan Asimov
<dasimov@xxxxxxxxxxxxx> wrote:


[original poster's correction incorporated]

On May 4, 6:30 pm, bo198214 <bo198...@xxxxxxxxxxxxxx> wrote:

<<
We know that there is a closed form for the Fibonacci function

(1) F(n)=F(n-1)+F(n-2), F(0)=0, F(1)=1

given by

F(n)=(r^n-(1-r)^n)/sqrt(5), where r=(1+sqrt(5))/2

Actually this is an extension from the domain of definition being the
natural numbers to real or even complex numbers. The extension is
analytic
and satisfies equations (1) for all complex n.

My question:
Are there research results which give conditions that makes this
extension unique under all analytic extensions that also satisfy
equations (1) for all complex n? Or is this extension already unique
be these requirements?


[The following post repeats some things already mentioned,
but I think clarifies some loose ends.]

I agree -- I'd like to know all entire functions f(z)
satisfying

(*) f(0) = 0, f(1) = 1, and f(z) = f(z-1) + f(z-2) for all z.

So you want them all, eh? There were some hints in
my original comments:

If F is one solution to (*), for example the one
given, then the general solution to (*) is

f = F + G,

where G is a solution to

G(z) = G(z-1) + G(z-2), G(0) = G(1) = 0.

That's clearly equivalent to

G(z) = G(z-1) + G(z-2), G(n) = 0 (n in Z).

And now it follow that there is an entire function
g with

G(z) = sin(pi z) g(z),

and then the recursion becomes

(**) g(z) = - g(z-1) + g(z-2).

That seems like some progress, the initial conditions
have become irrelevant.

Oops. I was about to try to show that the set of solutions
to (**) is two-dimensional. But of course as you point
out below that's not so. Never mind...

Suppose g is any meromorphic function (not identically 0) with
g(z) = -g(z-1) + g(z-2).
Then u(z) = g(z)/g(z-1) is a meromorphic function satisfying

(***) u(z) = -1 + 1/u(z-1)

Let r1 = (-1+sqrt(5))/2 and r2= (-1-sqrt(5))/2, the constant solutions
of (***).
Writing v(z) = (u(z)-r1)/(u(z)-r2) we get

(****) v(z) = r2/r1 v(z-1)

Now if s is any logarithm of r2/r1, the meromorphic solutions of
(****) are v(z) = exp(s z) p(z) where p is any periodic meromorphic
function with period 1. This corresponds to

u(z) = r2 - sqrt(5)/(exp(sz) p(z) - 1)

So that reduces the problem to solving

g(z) = (r2 - sqrt(5)/(exp(sz) p(z) - 1)) g(z-1)

... which I don't know how to do, but perhaps it counts as progress.

After a peek at Ullrich's solution, of course I can solve this:
one solution is

g(z) = r1 exp(t z) - r2 p(z) exp((t+s) z)

where t = ln(r1).

and then the general solution is g(z) = P(z) r1^z + Q(z) r2^z
where P(z) and Q(z) are arbitrary periodic meromorphic functions
with period 1.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

.

Relevant Pages

  • Re: uniqueness of fibonacci number extension
    ... Actually this is an extension from the domain of definition being the ... Then u= g/gis a meromorphic function satisfying ... which I don't know how to do, but perhaps it counts as progress. ...
    (sci.math.research)
  • Re: uniqueness of fibonacci number extension
    ... We know that there is a closed form for the Fibonacci function ... Actually this is an extension from the domain of definition being the natural numbers to real or even complex numbers. ... extension unique under all analytic extensions that also satisfy ... satisfying above (and hence giving ...
    (sci.math.research)
  • Re: uniqueness of fibonacci number extension
    ... We know that there is a closed form for the Fibonacci function ... Actually this is an extension from the domain of definition being the natural numbers to real or even complex numbers. ... satisfying above (and hence giving ...
    (sci.math.research)