Re: sum of zetas

Am 08.09.05 14:52 schrieb Jon Slaughter:
>
> this is how you get the identity that
>
> 2Z(n,n) = 1/2*Z(n)^2 - Z(n)
>
Hmmm, so I got it right. I looked at this from the
idea of elementary symmetrical polynoms and the
solutions, that the Newton-Girard-formula provides.

This formula provides you with the sum of powers of
elements P(x) (if you have the sum of the crossproducts)
or with the sum of crossproducts C(x) if you have the sum
of powers (plus the related lower-exponent items, of course).

But you cannot get both from the lower-exponent items
only; for the exponent of 3 you get (with the sum-of-powers
p(x,1) and p(x,2), and the sum of crossproducts c(x,1), c(x,2))

p(x,3) + 6*c(x,3) = ... polynom-of (p1) of deg 2
= known-solution from lower degrees

so it behaves somehow like if the related P() and C()-functions
have the same modulus base polynom-of 2 - and we cannot
know, whether they are algebraically independent of the
lower-degree-values.

Seems to be no way out of this - the known value of the
lower-degree polynom cannot be separated into the
two parameters.
The only idea I can think of to get rid of the 6*c(x,3)-
term would be to substitute x by a complex value based
on the 3rd root of unity x'. But then the lower-degree-
polynomials had also to be computed for that complex
values, and I don't know, whether that can be done
without introducing the new modulus in the computations
of p(x') and c(x') implicitely ... and even all this is
just only a vague idea.

Gottfried Helms
.