Re: Game: find the series

[Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>]

On Tue, Jun 11, 2024 at 06:37:12AM -0500, Kurt Foster wrote:
> On Jun 8, 2024, at 11:41 AM, Max Alekseyev wrote:
> 
> > Raising to the power x, we have S = (1+S)^x.
> > Letting x -> 0, we conclude that S = 1 + x*T for some power series x.
> > Taking logarithm, we then have
> > log(1 + x*T) = x*log(2 + x*T) = x*log(2) + x*log(1+x*T/2)
> > That is, we have the power series identity
> > x*log(2) + Sum_{k>=1} (-1)^(k-1) * x^k * T^k * (x - 2^k) / k / 2^k = 0.
> <snip>
> Nice analysis!
> 
> > %3 = [L, 1/2*L^2 + 1/2*L, 1/6*L^3 + 5/8*L^2 + 1/4*L, 1/24*L^4 + 3/8*L^3
> > + 9/16*L^2 + 1/8*L, 1/120*L^5 + 9/64*L^4 + 17/32*L^3 + 7/16*L^2 +
> > 1/16*L, 1/720*L^6 + 7/192*L^5 + 55/192*L^4 + 115/192*L^3 + 5/16*L^2 +
> > 1/32*L]
> 
> 
> Here is a slight variation of the implementation, with c = log(2) and S = 1
> + z.  Pari-GP has a built-in series reversion routine serreverse().  I
> truncated to a polynomial to extract the coefficients.
> 
> ? xinS=log(1+z)/(c+log(1+z/2));
> ? Sinx=serreverse(xinS);
> ? P=truncate(Sinx);

Sorry, I forgot to post my answer, but it is similar:

S=serreverse(log(1+x)/(log2+log(1+x/2)))+1;
2*exp(log(S)/x-log2) == S+1
%3 = 1

(Your formula is missing the +1, but this does not change the polynomials in log2).

Cheers,
Bill.