[seqfan] Re: need algebra proof

[Maximilian Hasler]

On Wed, Mar 17, 2010 at 10:12 PM, quantum pkt <qntmpkt at yahoo.com> wrote:
>
> Conjecture:
> Given polcoeff S(x), by way of example given A000041, the partition sequence = (1, 1, 2, 3, 5, 7, 11, 15, 22,...); then
> S(x) = (1 + x + 2x^2 + 3x^3 + ...).
>
> Then S(x) = some A(x) such that S(x) = A(x) / A(x^2).


If you replace the somehow obscure " S(x) = some A(x) " (?)
by  "there exists A(x)", then this is indeed true,
given the hypothesis that S(0) = 1.
(Else such an A cannnot exist, because A(0) / A(0) = 1 (or ill defined
if A(0)=0).)


> Our conjecture states that A(x) = S(x) * S(x^2) * S(x^4) * S(x^8) * S(x^16) * ...).


Obviously, A(x) is not uniquely defined, so you should say "one
solution is given by ...".
The choice of this solution is equivalent to the assumption (or choice)  A(0)=1.
You could multiply this A by any nonzero constant and you would again
get the same A(x) / A(x^2).


The infinite product converges in the topology of formal power series
(which means that the coefficients up to a  given order
are only affected by a finite number of terms,  cf.
http://en.wikipedia.org/wiki/Formal_power_series )
so it is perfectly well defined.
To exclude any ambiguity, you can specify
that the expression  ... / A(x^2)
means  ... * ( A(x^2) )^(-1) ,
where ( 1 - T(x) )^(-1) = 1 + T(x) + T(x)^2 + ...
(where T(x) is any power series with T(0)=0).

So you can check that the initial formula is true
(which means : coefficients of any given power are the same
on the l.h.s. and on the r.h.s.)
by writing out the relevant factors of the infinite products
and cancelling those which correspond.


> Similarly, for S(x) = B(x) / B(x^3), our conjecture states that:
>
> B(x) = S(x) * S(x^3) * S(x^9) * S(x^27) * S(x^81) * ...
>
> Generally, for any k, in S(x) = C(x) / C(x^k).
>

Yes, this is the same, the only necessary condition is again that S(0)
= 1, and assuming the choice  B(0) = C(0) = 1.


Maximilian


PS: I think Richard Mathar already pointed out roughly the same thing
in an earlier thread, mentioning the Lagrange-Bürmann formula, cf
http://en.wikipedia.org/wiki/Lagrange_inversion_theorem and references
therein.