From: hagman on
On 5 Jun., 15:30, hanrahan...(a)yahoo.co.uk wrote:
> On Jun 5, 6:20 am, Stephen Montgomery-Smith
>
> <step...(a)math.missouri.edu> wrote:
> > hanrahan...(a)yahoo.co.uk wrote:
> > > On Jun 4, 4:12 pm, hanrahan...(a)yahoo.co.uk wrote:
> > >> Does f(x,y) have to be a polynomial function of x and y, if for
> > >> constant x it is a polynomial in y and for constant y it is a
> > >> polynomial in x? I haven't been able to prove this.
>
> > > I should have specified: x and y are real, so is f(x,y), and the
> > > function is a polynomial in y for EVERY constant x, and also in x for
> > > EVERY constant y.
>
> > > Michael
>
> > I solved this problem many years ago.  It isn't totally straightforward.
> >   You have to use the fact that the reals are uncountable.
>
> Hi Stephen. Thanks for this. I haven't yet worked out how the
> uncountability of the reals can be used, but do you remember whether
> the answer is yes or no to whether f(x,y) must be a polynomial?
>
> Michael

yes. And the answer is "no" for a similar function Q x Q -> RR.

Assume f: R x R -> R is a function such that
for all x in R ,the function y |-> f(x,y) is a polynomial and
for all y in R ,the function x |-> f(x,y) is a polynomial.

If n in N let
X_n := { x in R | deg( y |-> f(x,y) ) < n} and
Y_n := { y in R | deg( x |-> f(x,y) ) < n}.
Since U X_n = R, the X_n cannot all be finite.
(This is where "uncountable" is used)
And of course if X_n is infinite, so are all X_k with k>n.
The same holds for the sets Y_n.
Hence there exists n in N such that X_n and Y_n are infinite.
Select an n-element subset A of X_n and an n-element subset B of Y_n.
There exists a (unique) polynomial
g(x,y) = sum_{i<n} sum {j<n} a_{i,j} x^i y^j
that agrees with f for all (x,y) in A x B.
For x in A, the functions y |-> g(x,y) and y |-> f(x,y) agree
on at least n points (namely for y in B).
Since both are polynomials of degree <n, they are the same.
Thus for y in Y_n, the functions x |-> g(x,y) and x |-> f(x,y) agree
on at least n points (namely for x in A).
Since both are polynomials of degree <n, they are the same.
Thus for x in R, the funtions y |-> g(x,y) and y |-> f(x,y) agree
on infinitely many points (namely for y in Y_n).
Since both are polynomials, they are the same.
In total f(x,y) = g(x,y) for all (x,y) in R x R.

Next I define a counterexample for the countable case:
Let F: N -> R be a function that is not a polynomial, e.g. F(n) = 2^n.
Define f: N x N -> R recursively as follows:
On thediagonal, let f(n,n) := F(n).
If n<m, let f(n,m) = p(m) where p is the unique polynomial of degree
<n
satisfying p(k) = f(k,m) for k in {0, ..., n-1}
If n>m, let f(n,m) = p(n) where p is the unique polynomial of degree
<m
satisfying p(k) = f(n,k) for k in {0, ..., m-1}
The resulting function f: N x N -> R has the following properties:
For n in N, the function m |-> f(n,m) is a poylnomial of degree <n
For m in N, the function n |-> f(n,m) is a poylnomial of degree <m
f is not a polynomial becaus n |-> f(n,n) is not polynomial.

The same construction is possible for Q x Q -> R using an enumeration
of Q
and again an arbitrary function on the diagonal.

hagman
From: hagman on
On 5 Jun., 23:37, hagman <goo...(a)von-eitzen.de> wrote:
> On 5 Jun., 15:30, hanrahan...(a)yahoo.co.uk wrote:
>
>
>
> > On Jun 5, 6:20 am, Stephen Montgomery-Smith
>
> > <step...(a)math.missouri.edu> wrote:
> > > hanrahan...(a)yahoo.co.uk wrote:
> > > > On Jun 4, 4:12 pm, hanrahan...(a)yahoo.co.uk wrote:
> > > >> Does f(x,y) have to be a polynomial function of x and y, if for
> > > >> constant x it is a polynomial in y and for constant y it is a
> > > >> polynomial in x? I haven't been able to prove this.
>
> > > > I should have specified: x and y are real, so is f(x,y), and the
> > > > function is a polynomial in y for EVERY constant x, and also in x for
> > > > EVERY constant y.
>
> > > > Michael
>
> > > I solved this problem many years ago.  It isn't totally straightforward.
> > >   You have to use the fact that the reals are uncountable.
>
> > Hi Stephen. Thanks for this. I haven't yet worked out how the
> > uncountability of the reals can be used, but do you remember whether
> > the answer is yes or no to whether f(x,y) must be a polynomial?
>
> > Michael
>
> yes. And the answer is "no" for a similar function Q x Q -> RR.
>
> Assume f: R x R -> R is a function such that
> for all x in R ,the function y |-> f(x,y) is a polynomial and
> for all y in R ,the function x |-> f(x,y) is a polynomial.
>
> If n in N let
> X_n := { x in R | deg( y |-> f(x,y) ) < n} and
> Y_n := { y in R | deg( x |-> f(x,y) ) < n}.
> Since U X_n = R, the X_n cannot all be finite.
> (This is where "uncountable" is used)
> And of course if X_n is infinite, so are all X_k with k>n.
> The same holds for the sets Y_n.
> Hence there exists n in N such that X_n and Y_n are infinite.
> Select an n-element subset A of X_n and an n-element subset B of Y_n.
> There exists a (unique) polynomial
> g(x,y) = sum_{i<n} sum {j<n} a_{i,j} x^i y^j
> that agrees with f for all (x,y) in A x B.
> For x in A, the functions y |-> g(x,y) and y |-> f(x,y) agree
> on at least n points (namely for y in B).
> Since both are polynomials of degree <n, they are the same.
> Thus for y in Y_n, the functions x |-> g(x,y) and x |-> f(x,y) agree
> on at least n points (namely for x in A).
> Since both are polynomials of degree <n, they are the same.
> Thus for x in R, the funtions y |-> g(x,y) and y |-> f(x,y) agree
> on infinitely many points (namely for y in Y_n).
> Since both are polynomials, they are the same.
> In total f(x,y) = g(x,y) for all (x,y) in R x R.
>
> Next I define a counterexample for the countable case:
> Let F: N -> R be a function that is not a polynomial, e.g. F(n) = 2^n.
> Define f: N x N -> R recursively as follows:
> On thediagonal, let f(n,n) := F(n).
> If n<m, let f(n,m) = p(m) where p is the unique polynomial of degree
> <n
> satisfying p(k) = f(k,m) for k in {0, ..., n-1}
Oops, this should have been:
If n<m, let f(n,m) = p(m) where p is the unique polynomial of
degree <n satisfying p(k) = f(n,k) for k in {0, ..., n-1}

> If n>m, let f(n,m) = p(n) where p is the unique polynomial of degree
> <m
> satisfying p(k) = f(n,k) for k in {0, ..., m-1}
and here
If n>m, let f(n,m) = p(n) where p is the unique polynomial of
degree <m satisfying p(k) = f(k,m) for k in {0, ..., m-1}

In other words:
Once we have defined f(x,y) for the case min(x,y)<n, we define f(n,n)
arbitrarily with the aim "non-polynomial" in mind and then
define f(x,n) by a polynomial suiting the finitely
many prescribed values at x=0, ..., n.
Same for f(n,y)

> The resulting function f: N x N -> R has the following properties:
> For n in N, the function m |-> f(n,m) is a poylnomial of degree <n
> For m in N, the function n |-> f(n,m) is a poylnomial of degree <m
> f is not a polynomial becaus n |-> f(n,n) is not polynomial.
>
> The same construction is possible for Q x Q -> R using an enumeration
> of Q
> and again an arbitrary function on the diagonal.
>
> hagman

From: hanrahan398 on
On Jun 6, 9:24 am, hagman <goo...(a)von-eitzen.de> wrote:
> On 5 Jun., 23:37, hagman <goo...(a)von-eitzen.de> wrote:

> > On 5 Jun., 15:30, hanrahan...(a)yahoo.co.uk wrote:
>
> > > On Jun 5, 6:20 am, Stephen Montgomery-Smith
>
> > > <step...(a)math.missouri.edu> wrote:
> > > > hanrahan...(a)yahoo.co.uk wrote:
> > > > > On Jun 4, 4:12 pm, hanrahan...(a)yahoo.co.uk wrote:
> > > > >> Does f(x,y) have to be a polynomial function of x and y, if for
> > > > >> constant x it is a polynomial in y and for constant y it is a
> > > > >> polynomial in x? I haven't been able to prove this.

Many thanks for this. I don't quite follow these three lines:

> > There exists a (unique) polynomial
> > g(x,y) = sum_{i<n} sum {j<n} a_{i,j} x^i y^j
> > that agrees with f for all (x,y) in A x B.

Why does such a polynomial exist?

Many thanks!

Michael
From: Dave L. Renfro on
Michael wrote:

>> Does f(x,y) have to be a polynomial function of x and y, if for
>> constant x it is a polynomial in y and for constant y it is a
>> polynomial in x? I haven't been able to prove this.

Stephen Montgomery-Smith wrote:

> I solved this problem many years ago. It isn't totally straightforward.
> You have to use the fact that the reals are uncountable.

If anyone is still interested in this, and for those who might
later stumble on this thread through a search of the sci.math
archives in the future, a solution is given by E. Grassmann
(University of Calgary) in Canadian Mathematical Bulletin
Volume 19, Number 2 (June 1976), pp. 252-253 (solution to
Problem 238).

The solution makes use of these two facts (that hopefully I've
stated correctly -- I'm verbalizing a lot of math expressions):

(1) There exists an integer m and an infinite subset E of the
reals R such that the (m+1)'st difference operator with
respect to x vanishes on R x E.

(2) There exists an integer n and an infinite subset F of the
reals R such that the (n+1)'st difference operator with
respect to y vanishes on F x R.

Reference is also made to the following paper:

Stephen A. Andrea, "Algebraic addition theorems", Advances
in Mathematics 13 (1974), 20-30.

Dave L. Renfro