Re: 3 functional equations is max

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
.

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