Re: 3 functional equations is max

In article
<10975721.1224533368479.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
amy666 <tommy1729@xxxxxxxxxxx> wrote:

Robert wrote :

amy666 <tommy1729@xxxxxxxxxxx> writes:

Robert israel wrote :

amy666 <tommy1729@xxxxxxxxxxx> writes:

given an entire function f(z).

f(z) has at most 3 (independant) functional
equations.

No, it can have arbitrarily many.
Take any positive integer n.
Consider the polynomials
P(x) = product_{j=1}^{n+1} (x-j) and P_j(x) =
P(x)/(x-j)
Letting T f(x) = f(x+1), consider the functional
equations
P_j(T)(f)(z) = 0 for j = 1,...,n+1
i.e. if P_j(x) = sum_{k=0}^n c_k x^k,
P_j(T)(f)(z) = sum_{k=0}^n c_k f(z + k)

In particular f(z) = exp(r z) satisfies
P_j(T)(f)(z)=0 if and
only if P_j(exp(r)) = 0, i.e. exp(r) is one of
1,...,n+1 but not j.

These n+1 functional equations are all
independent,
since for each j=1..n+1
there is an entire function exp((ln j) z) that
satisfies all of them
except for equation number j. So an entire
function
can satisfy n
independent functional equations.
--
Robert Israel
israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics
http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver,
BC, Canada

euh ...

i dont get it :s

P_j(T)(f)(z) ??

perhaps an example with 4 functional equations
might help me understand ...

thanks for trying though :)

regards

tommy1729

For n=4, the 5 functional equations are

120*f(x)-154*f(x+1)+71*f(x+2)-14*f(x+3)+f(x+4) = 0
60*f(x)-107*f(x+1)+59*f(x+2)-13*f(x+3)+f(x+4) = 0
40*f(x)-78*f(x+1)+49*f(x+2)-12*f(x+3)+f(x+4) = 0
30*f(x)-61*f(x+1)+41*f(x+2)-11*f(x+3)+f(x+4) = 0
24*f(x)-50*f(x+1)+35*f(x+2)-10*f(x+3)+f(x+4) = 0

Each of the functions f_j(x) = j^x for j = 1, 2, 3,
4, 5 satisfies
a different four of these equations.
--
Robert Israel
israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics
http://www.math.ubc.ca/~israel
University of British Columbia Vancouver,
BC, Canada

very nice :)

but with " independant" i meant the functional equations cannot be reduced to
a lesser amount of functional equations.

( note to david ullrich : damn , ive just defined it afterall - see my reply
to david - )

since roberts examples are j^x

we can reduce to

F_j(x) satisfies :

f(x+a) = j^a * f(x)

Exactly how do you reduce those equations to this form?

You seem to be missing the point to the j^x here. The functions
j^x are not the functional equations - the functions j^x are what
show that the given functional equations are independent.

(Using a very reasonable notion of "independent": a set of
functional equations is independent of no one of the equations
follows from the others.)

another example of functional equations that can be reduced ( NOT independant
) :

f(x) = f(x+4)

f(x) = f(x+2)

Right, these are not independent, since the second obviously
implies the first.

yet another example of functional equations that can be reduced ( NOT
independant ) :

f(x) = f(x^2)

f(x) = f(x^4)

an example of independant :

f(x) = f(x+3)

f(x) = f(x+i)

And these are independent. Right. Now exactly how is it
that the equations Robert gave are not independent?

regards

tommy1729

--
David C. Ullrich
.

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