Re: 3 functional equations is max
- From: amy666 <tommy1729@xxxxxxxxxxx>
- Date: Tue, 21 Oct 2008 18:43:29 EDT
David wrote :
In article
<10975721.1224533368479.JavaMail.jakarta@xxxxxxxxxxxxx
forum.org>,
amy666 <tommy1729@xxxxxxxxxxx> wrote:
Robert wrote :functional
amy666 <tommy1729@xxxxxxxxxxx> writes:
Robert israel wrote :
amy666 <tommy1729@xxxxxxxxxxx> writes:
given an entire function f(z).equations.
f(z) has at most 3 (independant) functional
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
ofequations
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
thatindependent,1,...,n+1 but not j.
These n+1 functional equations are all
since for each j=1..n+1
there is an entire function exp((ln j) z)
israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxfunctionsatisfies all of them
except for equation number j. So an entire
can satisfy n
independent functional equations.
--
Robert Israel
0Vancouver,Department of Mathematics
http://www.math.ubc.ca/~israel
University of British Columbia
might help me understand ...BC, Canada
euh ...
i dont get it :s
P_j(T)(f)(z) ??
perhaps an example with 4 functional equations
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) =
3,60*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,
Vancouver,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
equations cannot be reduced toBC, Canada
very nice :)
but with " independant" i meant the functional
a lesser amount of functional equations.it afterall - see my reply
( note to david ullrich : damn , ive just defined
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?
how i do it doesnt matter.
what matters is
but with " independant" i meant the functional
equations cannot be reduced to
a lesser amount of functional equations.
which can be done with roberts examples.
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.
yes i get the example but my notion of independant was
but with " independant" i meant the functional
equations cannot be reduced to
a lesser amount of functional equations
for clarity , not the functional equations that robert gave must be independant , just that there is a smaller amount of functional equations that j^x satisfies ( for each j seperately )
(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 bereduced ( 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 thatcan 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?
point is less functional equations suffice for each given j individually.
regards
tommy1729
--
David C. Ullrich
regards
tommy1729
.
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