Re: conditions for this polynomial to be in Q[X]

On 09.10.2006 23:51, ben_r_jackson@xxxxxxxxxxx wrote:
hi,

I'd like to get conditions for when
F(X) = a1*(X-b1)^n+...+ak*(X-bk)^n is a rational polynomial.

Ideally, for n sufficiently large with respect to k, I'd like to show
that the ai's and bi's are algebraic, must lie in number fields whose
degree depends on k in some way,...

When k=2, I have been able to do this for n>2 by looking at the
coefficients of X^n, X^{n-1},
X^{n-2} and X^{n-3}. However, this proof was by brute force, explicitly
calculating a1, a2, b1 and b2 in terms of these coefficients.

This produces a nice result, but trying to do the same for larger
values of k would get increasingly complicated and not generalize.
There must be a way to do this more generally. Possibly using symmetric
functions?

Also "my" brute-force method only works if n>2k-2 (as I need at least
2k coefficients, one for each of the ai's and bi's). Is this condition
necessary to get a "nice" result?

By the following technique you can reduce to the case where n>2k-2. If
you brute force proof is correct, there you are. :-)

Use primitives of F to sufficiently increase n. More precisely, the
primitive FF(X) = [a1*(X-b1)^(n+1)+...+ak*(X-bk)^(n+1)]/(n+1) of F(X) is
in Q[X] *iff* F(X) is in Q[X]. Note that F(0) = FF(0) up to a rational
constant.

Now the question is: Does your proof for n>2k-2 really work? Let's see ...

For k=2, the condition n>2k-2=2 is needed, as I have an example where
F(X) has rational coefficients, but the ai's and bi's are expressions
involving pi.

Any suggestions would be greatly appreciated.

Ben


J.
.

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