# uniqueness of fibonacci number extension

10 views

### bo198214

May 4, 2008, 9:30:01 PM5/4/08
to
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?

### Jannick Asmus

May 5, 2008, 8:30:02 AM5/5/08
to
On 05.05.2008 03:30, bo198214 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.

Some correction is needed here: For the definition of r^z, z complex,
there is a (branch) of the logarithm needed which does not exist on the
complex plane (but on every simply connected open subset). So (1) cannot
hold if the is meant in terms of holomorphic functions.

> 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?

--
Best wishes,
J.

### David C. Ullrich

May 5, 2008, 8:30:02 AM5/5/08
to

Some sort of growth condition would give uniqueness - see
"entire functions of exponential type" somewhere.

Given no extra conditions the solution is not unique.
For example, let g be a non-trivial entire function
with

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

for all z (for example, g(z) = a^z for a suitable constant a.)
Then

f(z) = F(z) + g(z) sin(2 pi z)

gives a second solution to (1).

David C. Ullrich

### G. A. Edgar

May 5, 2008, 3:00:00 PM5/5/08
to
In article <fvmuka\$h9\$1...@news.acm.uiuc.edu>, Jannick Asmus
<jannic...@web.de> wrote:

> On 05.05.2008 03:30, bo198214 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.
>
> Some correction is needed here: For the definition of r^z, z complex,
> there is a (branch) of the logarithm needed which does not exist on the
> complex plane (but on every simply connected open subset). So (1) cannot
> hold if the is meant in terms of holomorphic functions.

You only need a log of r and of 1-r. So choose such logarithms, once
at the start, and use F(n)=(r^n+(1-r)^n)/sqrt(5) ... No worry about
branches, F(n) is defined for n in the whole complex plane.

Actually, F(n)=(r^n+(1-r)^n)/sqrt(5) isn't the Fibonaccis. Should
be F(n)=(r^n-(1-r)^n)/sqrt(5).

Different choices of these logs will give you different solutions,
answering negatively the question of uniqueness, right?

>
> > 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?
>
> --
> Best wishes,
> J.
>

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

### Jannick Asmus

May 5, 2008, 10:00:02 PM5/5/08
to
On 05.05.2008 21:00, G. A. Edgar wrote:
> In article <fvmuka\$h9\$1...@news.acm.uiuc.edu>, Jannick Asmus
> <jannic...@web.de> wrote:
>
>> On 05.05.2008 03:30, bo198214 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.
>> Some correction is needed here: For the definition of r^z, z complex,
>> there is a (branch) of the logarithm needed which does not exist on the
>> complex plane (but on every simply connected open subset). So (1) cannot
>> hold if the is meant in terms of holomorphic functions.
>
> You only need a log of r and of 1-r. So choose such logarithms, once
> at the start, and use F(n)=(r^n+(1-r)^n)/sqrt(5) ... No worry about
> branches, F(n) is defined for n in the whole complex plane.
>
> Actually, F(n)=(r^n+(1-r)^n)/sqrt(5) isn't the Fibonaccis. Should
> be F(n)=(r^n-(1-r)^n)/sqrt(5).
>
> Different choices of these logs will give you different solutions,
> answering negatively the question of uniqueness, right?

Thanks for making this a lot clearer. I believe such a function does not
even exist by the following reasoning with brute force:

The logarithms of r are ln(r)+2iPi.Z and of 1-r they are ln(r-1)+iPi+2iPi.Z.

Now let us assume that there is a choice of the logarithms of r and 1-r
such that (1) is satisfied for all complex figures, then in particular
for all positive reals x>2. Writing down relation (1) and sorting by
terms r^x, r^(x-1), r^(x-2) and (r-1)^x,... shows that this is
impossible unless exp(iPi.x) has real values only. Contradiction.

### David C. Ullrich

May 6, 2008, 9:30:02 AM5/6/08
to
On Tue, 6 May 2008 02:00:02 +0000 (UTC), Jannick Asmus
<jannic...@web.de> wrote:

??? There is in fact no problem using (1) to define an entire
function. If you write down the details of whatever argument
you have in mind you'll find a flaw.

Given a complex number r <> 0 choose a complex number
L such that exp(L) = a. Now define p(z) = exp(Lz). This
is the function commonly denoted by r^z. There are infinitely
many such finctions, of course, one for each possible value
of L.

But regardless of the choice of L, it's easy to verify that
p(z+w) = p(z) p(w) for all complex z and w. And it
follows that p(n) = r^n for all integers n. And hence that
p(z+n) = p(z) r^n for all complex z and integral n.

And it follows that if F(z)=(r^z+(1-r)^z)/sqrt(5)
(or whatever the correct version is) then
F(z) = F(z-1) + F(z-2).

>>>> 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?

David C. Ullrich

### G. A. Edgar

May 6, 2008, 10:00:03 AM5/6/08
to
See formula (9) and the graph just above at
http://mathworld.wolfram.com/FibonacciNumber.html
This is an entire function F with the required properties. This
particular one has the advantage that F(r) is real when r is real.

### Jannick Asmus

May 8, 2008, 11:00:02 AM5/8/08
to

Right - G.A. provided a link to an explicit solution. And you will give
a family of those below.

Meanwhile I have thrown the piece of paper with my notes away, so that I
will not write it up again to find out where I went wrong. Dust bin was
just the right place for it anyway.

> Given a complex number r <> 0 choose a complex number
> L such that exp(L) = a. Now define p(z) = exp(Lz). This
> is the function commonly denoted by r^z. There are infinitely
> many such finctions, of course, one for each possible value
> of L.
>
> But regardless of the choice of L, it's easy to verify that
> p(z+w) = p(z) p(w) for all complex z and w. And it
> follows that p(n) = r^n for all integers n. And hence that
> p(z+n) = p(z) r^n for all complex z and integral n.
>
> And it follows that if F(z)=(r^z+(1-r)^z)/sqrt(5)
> (or whatever the correct version is) then
> F(z) = F(z-1) + F(z-2).

Right, too. I see now that the p's just separate leaving the
characteristic equation of r and 1-r.

>>>>> 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?

For this question consider the vector space V over the complex numbers
with entire functions f(z) satisfying (1) and f(0)=f(1)=0. This implies
that f(Z)=0.

Then r^z.q(z)^n and (1-r)^z.q(z)^m with q(z)=exp(2iPi.z), m,n integers,
are in V and linearly independent. Hence there is no uniqueness.

But this was already pointed out by David basically, including the
reference to Carlson's theorem on uniqueness if f is of exponential type.

--
Best wishes,
J.

### bo198214

May 8, 2008, 11:00:03 AM5/8/08
to
First of course I have to correct my oversight, the Fibonacci formula
is:
F(n)=(r^n-(1-r)^n)/sqrt(5)

Then though I dont understand the condition

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

in David C. Ullrich's post this triggered some remembering/
recognition:

Whenever a real analytic function f satisfies

(1*) f(x)=U(f(x-1),f(x-2),...,f(x-k)) for all x, for a fixed k, with
some initial conditions f(i)=a_i, 0<=i<k,

then also the real analytic function g(x)=f(x+phi(x)) satisifies this
equation, for any 1-periodic real analytic function phi with phi(0)=0,
i.e. for example g(x)=f(x+sin(2*pi*x)).

In the particular case of the Fibonacci function, where U is linear,
we have even some more possibilities to variate; for every solution
f(x) of (1*), also f(x)*phi(x) is a solution for every 1-periodic
function phi with phi(0)=1.
For example phi(x)=1+sin(2*pi*x), or phi(x)=cos(2*pi*x).

Regarding uniqueness conditions I didnt sorted out yet the literature
found by the keyword "entire functions of exponential type".

### David C. Ullrich

May 9, 2008, 6:53:04 AM5/9/08
to
On Thu, 8 May 2008 15:00:03 +0000 (UTC), bo198214

>First of course I have to correct my oversight, the Fibonacci formula
>is:
>F(n)=(r^n-(1-r)^n)/sqrt(5)
>
>Then though I dont understand the condition
>g(z) = -g(z-1) + g(z-2)
>in David C. Ullrich's post

I think there was a little error there. When I said
we wanted g(z) = -g(z-1) + g(z-2) what I had in
mind was

f(z) = F(z) + g(z) sin(pi z),

although what I'd actually written was

f(z) = F(z) + g(z) sin(2 pi z).

> this triggered some remembering/
>recognition:
>
>Whenever a real analytic function f satisfies
>
>(1*) f(x)=U(f(x-1),f(x-2),...,f(x-k)) for all x, for a fixed k, with
>some initial conditions f(i)=a_i, 0<=i<k,
>
>then also the real analytic function g(x)=f(x+phi(x)) satisifies this
>equation, for any 1-periodic real analytic function phi with phi(0)=0,
>i.e. for example g(x)=f(x+sin(2*pi*x)).
>
>In the particular case of the Fibonacci function, where U is linear,
>we have even some more possibilities to variate; for every solution
>f(x) of (1*), also f(x)*phi(x) is a solution for every 1-periodic
>function phi with phi(0)=1.
>For example phi(x)=1+sin(2*pi*x), or phi(x)=cos(2*pi*x).
>
>Regarding uniqueness conditions I didnt sorted out yet the literature
>found by the keyword "entire functions of exponential type".

It's a big subject, most of which I no longer remember. A crude
version of the basic fact is that sin(pi z) has as many zeroes as
is possible for a function that grows as slowly as sin(pi z).
In particular, I'm pretty sure that if, say, f is an entire function,
alpha < pi,

|f(z)| <= c exp(alpha |z|)

for all z and f(n) = 0 for every integer n then f vanishes
identically.

The actual results are much more subtle in various ways,
but unless I'm getting senile that's an easily stated consequence.
(Unless I got the numbers wrong - you can check that by
noting that sin(pi z) _is_ extremal in at least a crude sense.)

Getting back to uniqueness: Although F is not unique, it
seems like it _could_ be that F is the unique solution
of minimal growth (in the sense of exponential type,
for example).

David C. Ullrich

### Dan Asimov

May 9, 2008, 8:06:15 PM5/9/08
to

[original poster's correction incorporated]

On May 4, 6:30 pm, bo198214 <bo198...@googlemail.com> 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.

But if we just assume the form of f(z) is

f(z) = A exp(az) + B exp(bz)

for complex constants a,b,A,B,
then it's easy to show that

{exp(a), exp(b)} = {(1+sqrt(5))/2, (1-sqrt(5))/2},

so:

---------------------------------------------------------
If we set exp(a) = (1+sqrt(5))/2, exp(b) = (1-sqrt(5))/2

it follows that A = 1/sqrt(5), B = -1/sqrt(5), giving

(**) f(z) = (exp(az) - exp(bz)) / sqrt(5).
---------------------------------------------------------

I.e., *any* choice of complex logarithms a and b of
(1+sqrt(5))/2 and (1-sqrt(5))/2, respectively
will work in (**) to give an entire f(z)
satisfying (*) above (and hence giving
the Fibonacci numbers for n = 0,1,2,...).

Dan Asimov

_____________________________________________________________________
"It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

### bo198214

May 10, 2008, 4:57:31 AM5/10/08
to
On May 9, 12:53 pm, "David C. Ullrich" <dullr...@sprynet.com> wrote:
> It's a big subject, most of which I no longer remember. A crude
> version of the basic fact is that sin(pi z) has as many zeroes as
> is possible for a function that grows as slowly as sin(pi z).
> In particular, I'm pretty sure that if, say, f is an entire function,
> alpha < pi,
>
>   |f(z)| <= c exp(alpha |z|)
>
> for all z and f(n) = 0 for every integer n then f vanishes
> identically.

I also just looked up Carlson's theorem, which was mentioned by
Jannick Asmus and the accompanying Phragmén-Lindelöf theorem

http://en.wikipedia.org/wiki/Carlson's_theorem
http://en.wikipedia.org/wiki/Phragmén-Lindelöf_theorem

which is perhaps on the line what you were just saying.
I think also somehow connected to those theorems is the following
uniqueness theorem for the gamma function which I was not aware of
before:

Let F be holomorphic on the right half plane, and let F(z+1)=zF(z),
F(1)=1, further let F be bounded on the strip 1<=Re(z)<2, then F is
the gamma function.

Does anyone think this can be conveyed to the case of the Fibonacci
function?

### David C. Ullrich

May 10, 2008, 11:27:30 AM5/10/08
to
On Fri, 9 May 2008 17:06:15 -0700 (GMT-07:00), Dan Asimov

>
>[original poster's correction incorporated]
>
>On May 4, 6:30 pm, bo198214 <bo198...@googlemail.com> 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

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...

>But if we just assume the form of f(z) is
>
> f(z) = A exp(az) + B exp(bz)
>
>for complex constants a,b,A,B,
>then it's easy to show that
>
> {exp(a), exp(b)} = {(1+sqrt(5))/2, (1-sqrt(5))/2},
>
>so:
>
>---------------------------------------------------------
>If we set exp(a) = (1+sqrt(5))/2, exp(b) = (1-sqrt(5))/2
>
>it follows that A = 1/sqrt(5), B = -1/sqrt(5), giving
>
>(**) f(z) = (exp(az) - exp(bz)) / sqrt(5).
>---------------------------------------------------------
>
>I.e., *any* choice of complex logarithms a and b of
>(1+sqrt(5))/2 and (1-sqrt(5))/2, respectively
>will work in (**) to give an entire f(z)
>satisfying (*) above (and hence giving
>the Fibonacci numbers for n = 0,1,2,...).
>
>Dan Asimov
>
>
>_____________________________________________________________________
>"It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

David C. Ullrich

### Robert Israel

May 11, 2008, 1:33:18 AM5/11/08
to

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.

--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

### David C. Ullrich

May 11, 2008, 11:51:15 AM5/11/08
to

Something clicked when I read that - I saw how to make what I
was trying yesterday work.

The general solution to (**) above is

g(z) = alpha^z p_1(z) + beta^z p_2(z),

where alpha = (1 + sqrt(3)i)/2, beta = (1 - sqrt(3)i)/2, and
p_1, p_2 are entire functions of period 1. At least so it seems;
there are probably errors below, but it seems clear that the
approach works. (So the solution space _is_ "two-dimensional"
in a sense, maybe in a certain module where the "constants"
are the periodic entire functions. Which makes sense in
retrospect, since p is periodic if and only if Dp = 0
(D as below), just as p is constant if and only if p' = 0.)

Ok. Define

Dg(z) = g(z) - g(z-1),

and for convenience in the notation define

e_a(z) = e^(az)

(just because we need to give the function e_a a name.)

One calculates that

D(e_a g) = e^(-a) e_a (Dg - (1-e^a) g).

First we want to solve the equation

(D - alpha) g = 0

where alpha is a constant. Note that if alpha = 1
(which doesn't come up below) this is trivial, and
has only the trivial solution. Assume that alpha <> 1.
Let a = log(1 - alpha). Multiplying the equation by
e^(-a) e_a and applying the above it becomes

D(e_a g) = 0,

so the solution is e_a g = p, where p has period 1,
or:

(1) The general solution to (D - alpha) g = 0
is g = e_{-a} p, where a = log(1-alpha) and
p has period 1.

Now note that (**) above is equivalent to

(***) (D^2 - D + 1) g = 0.

(Not because the coefficients in (**) and (***) are
the same! That's just a coincidence; work out what
(***) says and you get (**).)

Factoring that polynomial it becomes

(D - alpha)(D - beta) g = 0,

where alpha and beta are as above.

Now applying (1) we see that this is equivalent to

(D - beta) g = e_{-a} p_1,

where p_1 is periodic and a = log(1-alpha).
We also set b = log(1-beta) and note that
a + b = 0 and also that a = log(beta) and
b = log(alpha).

If you multiply the last equation by e^(-b) e_b
and then replace p_1 by a certain constant times
p_1 it becomes

(****) D(e_b g) = e_{b-a} p_1.

Now, I'm embarassed to say that in a half hour I
couldn't figure out exactly when

Dg = h

has a solution; if one series converges it gives
a solution, if another series converges it gives
a solution, if one assumed that every series
always conveged one could easily give a
necessary and sufficient condition on h for
a solution to exist, etc. But in general I don't
know the answer, and the somewhat general
cases where I got solutions don't apply to
(****).

But based on the obvious solutions to the problem
we pull a solution to (****) out of our ass: One
solution is given by

e_b g = e_{b-a} p_1/(1 - e^(a-b)).

So the general solution is e_b g = that + p_2;
multiplying everything by e_{-b} and
absorbing another constant factor into
p_1 we get

g = e_{-a} p_1 + e_{-b} p_2.

Now note that e_{-a}(z) = e_b(z) = beta^z,
similarly for e_{-b}, swap p_1 and p_2 and
this becomes

g = alpha^z p_1(z) + beta^z p_2(z),

as claimed. (Exercise: explain how this
can be right; why doesn't choosing
different logarithms for alpha and beta
give yet _more_ solutions?)

So, if I've put it all together properly,
it looks like the general solution to the
original Fibonacci problem is

f(z) = F(z) + sin(pi z) g(z),

where g is as above.
David C. Ullrich

### David C. Ullrich

May 11, 2008, 12:15:42 PM5/11/08
to
On Sun, 11 May 2008 00:33:18 -0500, Robert Israel
<isr...@math.MyUniversitysInitials.ca> wrote:

Aargh. I just posted a reply to this that was totally
wrong (I'd post this as a reply to that instead of a
sibling except for the delay in posts appearing
here because of the approval mechanism.)

It seemed a little funny that (1 +- sqrt(3)i)/2
showed up instead of (-1 +- sqrt(5))/2. The
reason is that at the very start I changed

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

to

g(z) - g(z-1) + g(z-2) = 0.

I haven't worked out the details, but I _bet_
that exactly the method I used in that wrong
post shows that the general solution to the
original problem is

f(z) = F(z) + sin(pi z) g(z)

where

g(z) = alpha^z p_1(z) + beta^z p_2(z),

alpha, beta = (-1 +- sqrt(5))/2 and p_1
and p_2 are entire functions of period 1.

David C. Ullrich

### Waldek Hebisch

May 11, 2008, 12:42:06 PM5/11/08
to
>
> Let F be holomorphic on the right half plane, and let F(z+1)=zF(z),
> F(1)=1, further let F be bounded on the strip 1<=Re(z)<2, then F is
> the gamma function.
>
> Does anyone think this can be conveyed to the case of the Fibonacci
> function?
>

Yes, by direct argument. Namely, on integers general solution
of Fibonacci equation has form: F(n) = c_1*exp(a_1*z) + c_2*exp(a_2*z)
Now, consider G_z(n) = F(n+z). Applying the integer result to G_z
(z beeing a parameter) we see that:

F(z) = P_1(z)*exp(a_1*z) + P_2(z)*exp(a_2*z)

where P_1 and P_2 are holomorphic and 1-periodic (that is
P_i(z+1) = P_i(z) for i=1,2). If F is bounded in the strip 0<= Re(z) <= 1
then also P_i are bounded in this strip. But then, P_i must be
bounded on the whole complex plane, so P_i must be constant.

It is easy to see that any holomorphic, 1-periodic P_i give solution
to Fibonacci equation. holomorphic, 1-periodic P can be expanded
in the Fourier series, so this argument also give description of
general solution. In particular one can write the solution as a sum
of convergent series of exponentials where each term solves
Fibonacci equation.

Note: in the paragraph above I ignored initial conditions -- of course
one have to choose P_i to satisfy them (say fixing values at 0).

--
Waldek Hebisch
heb...@math.uni.wroc.pl

### Robert Israel

May 12, 2008, 12:40:39 PM5/12/08
to
Robert Israel <isr...@math.MyUniversitysInitials.ca> writes:

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.

### bo198214

May 20, 2008, 11:24:17 AM5/20/08
to

On 11 Mai, 18:42, Waldek Hebisch <hebi...@math.uni.wroc.pl> wrote:
> Yes, by direct argument.

That s striking. Let me rephrase whether I ve got everything right so
far:

Every function F that satisfies

(a) F(z+1)=F(z)+F(z-1)

can (by induction) be shown to be described by
F(n) = c1*b1^n + c2*b2^n
for integer n, for suitable b1, b1 (i.e. just the solutions of
b^2=b^1+1) and for
suitable c1 and c2 depending on F(0) and F(1) (i.e. c1+c2=F(0) and
c1*b1+c2*b2=F(1)).

If we now consider G_z(n) = F(z+n) than every G_z also satisfies (a)
and hence can
be described by
G_z(n) = c1(z)*b1^n + c2(z)*b2^n, where

(b) c1(z)+c2(z)=F(z) and c1(z)*b1+c2(z)*b2=F(z+1).

Hence
G_z(n) = c1(z)/b1^z * b1^(n+z) + c2(z)/b2^z * b2^(n+z)
u:=z+n
F(u) = c1(u-n)/b1^(u-n) * b1^u + c2(u-n)/b2^(u-n) * b2^u
independent of n.

So p1(z):=c1(z)/b1^z and p2(z):=c2(z)/b2^z are our candidates for
being 1-periodic.
We need to show that
ci(z+1)/bi=ci(z)

When we solve (b) we get:

c1(z)=F(z)*b2-F(z+1)/(b2-b1)
c2(z)=F(z)*b1-F(z+1)/(b1-b2)

Now in this case we have b2+b1=1 and b1*b2=-1 then
c1(z+1)/b1=-b2*(F(z+1)*b2-F(z+2))=-b2*(F(z+1)*(b2-1)-F(z))=-F(z
+1)+b2*F(z)=c1(z)
correspondingly for c2(z+1)/b2

So p1(z) and p2(z) are indeed 1-periodic.

As a test consider our previous construction of different solutions
by:

F(z)= (exp((log(b1)+2*pi*i*k1)*z) - exp((log(b2)+2*pi*i*k2)*z)/sqrt(5)

This confirms our thesis that every holomorphic solutions must be of
the form
F(z)=p1(z)*b1^z+p2(z)*b2^z (which is also in accordance with Robert
Israel's derivation)
by letting
p1(z)=exp(2*pi*i*k1*z)/sqrt(5) and p2(z)=-exp(2*pi*i*k2*z)/sqrt(5)

If now F is bounded on the strip then c1 and c2 (as solutions of (b))
are also bounded on the strip and so is p1 and p2. p1 and p2 must be
holomorphic as a consequence of F being holomorphic (is that true?).
Hence as periodic functions on the whole plane they are constants ...

Waldek, it came so fluently, did you just invent this argumentation or
is it written somewhere? If so I would really like to know where.
Thanks a lot for this insight.

### Waldek Hebisch

May 23, 2008, 10:00:02 AM5/23/08
to
>
>
> On 11 Mai, 18:42, Waldek Hebisch <hebi...@math.uni.wroc.pl> wrote:
> > Yes, by direct argument.
>
> That s striking. Let me rephrase whether I ve got everything right so
> far:
>
>
> So p1(z):=c1(z)/b1^z and p2(z):=c2(z)/b2^z are our candidates for

<snip>

>
> When we solve (b) we get:
>
> c1(z)=F(z)*b2-F(z+1)/(b2-b1)
> c2(z)=F(z)*b1-F(z+1)/(b1-b2)
>

<snip>

> If now F is bounded on the strip then c1 and c2 (as solutions of (b))
> are also bounded on the strip and so is p1 and p2. p1 and p2 must be
> holomorphic as a consequence of F being holomorphic (is that true?).

c1 and c2 are linear combinations of F(z) and F(z+1) so are holomorphic.
By the get from them p1 and p2 dividing by b1^z (or b2^z respectively),
again the result is holomorphic.

> Hence as periodic functions on the whole plane they are constants ...
>
> Waldek, it came so fluently, did you just invent this argumentation or
> is it written somewhere? If so I would really like to know where.

I invented the argumentation.

--
Waldek Hebisch
heb...@math.uni.wroc.pl