I saw this formula which inverts a series
https://encyclopediaofmath.org/wiki/B%C3%BCrmann-Lagrange_series
It's straightforward about a=0, for roots close to zero, or for non-composite functions. Its harder for larger solutions. I am trying to compute one of the large roots of the polynomial and then square this result as a single series.
$$ F(w)= f(a) + \sum^\infty_{n=1} \frac{(w-b)^n}{n!}\frac{d^{n-1}}{dz^{n-1}}[f'(z)\left[\left( \frac{z-a}{g(z) - b} \right)^n \right]]_{z = a} $$
Consider the polynomial $$g(z)=-(z^3+3z^2-5z)=.15=w$$ which is equal to $-z(z-r_1)(z-r_2)$ for the two roots of the ancillary quadratic. Hence we have:
$$r_1=a,g(r_1)=0=b$$
So I am trying to find a series for the square of one of the roots, which for the large ones is around $17.63$. This means $$f(a)=z^2$$
It works for $f(a)=z$ but not squared
Here is $$ \frac{.15^n}{n!} \frac{d^{n-1}}{dz^{n-1}} \frac{2x}{(x(x-r_2))^n}$$
about $x=-r_1$
Taking the first three terms shows it gets close but does not work
The first two terms get very close, 17.6335 vs 17.633343, but fails on the third.
What mistake is made?
Edit I seem to have made an error regarding substitutions in wolfram. Updated, gives 17.63334294:
which is much closer to the correct value. Dealing with sign changes and substitutions is tricky.