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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

(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?

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.

> 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.

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

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:

<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/

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?

> 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.

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:

<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

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.

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.

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.

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)

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".

May 9, 2008, 6:53:04 AM5/9/08

to

On Thu, 8 May 2008 15:00:03 +0000 (UTC), bo198214

<bo19...@googlemail.com> wrote:

<bo19...@googlemail.com> wrote:

>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

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

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.

> 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?

May 10, 2008, 11:27:30 AM5/10/08

to

On Fri, 9 May 2008 17:06:15 -0700 (GMT-07:00), Dan Asimov

<das...@earthlink.net> wrote:

<das...@earthlink.net> wrote:

>

>[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

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...

>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

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

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

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:

<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

May 11, 2008, 12:42:06 PM5/11/08

to

bo198214 <bo19...@googlemail.com> wrote:

>

> 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?

>

>

> 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

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.

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.

May 23, 2008, 10:00:02 AM5/23/08

to

bo198214 <bo19...@googlemail.com> wrote:

>

>

> 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:

>

>

>

>

> 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

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