Series solution for general trinomial Consider the equation
$x^5-2x^2+z=0$
How do you derive the Lagrange inversion theorem series solution for it? I know it exists because the answer is here for any trinomial https://arxiv.org/pdf/0910.2957.pdf
I am trying to figure out how derive the series solution.
There is this well known result, which I think is needed:
$$ h(f^{-1}(z))=h(0)+\sum_{n\geq 1}\frac{z^n}{n}\cdot [z^{n-1}]\left(h'(z)\cdot\left(\frac{z}{f(z)}\right)^n\right).$$
What functions should I use for f(z), h(z)
There is no information about this online. Most sources just give the series solution for $x^5-x+a=0$ which has $f(x)=x^5-x$ and then the bring radical solution is as follows by applying the above formula.
the best I could come up with is:
$z=2y-y^{5/2}$ taking the square root of the solution of this gives the solution to the original equation
so $h(z)$ would be a square root
$$ h(f^{-1}(z))=\sum_{n\geq 1}\frac{z^n}{n}\cdot [y^{n-1}]\left(1/2y^{-1/2}\cdot\left(\frac{1}{2-y^{3/2}}\right)^n\right).$$
This gets messy and does not generate integer y's , so it does not work. Any ideas.
 A: Addendum, for trinomials of the form $x^n+bx^m+c=0$ for integers $n>m$, we can get $n-m$ roots using a simple modification  Glasser formula, as cited in Wikipedia .
https://en.wikipedia.org/wiki/Bring_radical
What if instead of $  x = \zeta^{-\frac{1}{N-1}}\,$ we make the transformation  more general, like this:
$x = \zeta^{-\frac{k}{N-1}}$
This means that the resulting root is raised to some $k$ power as a series.
So the idea behind the modification is we:
Start by letting N be a fraction and then apply the general transformation to raise this root to the 1/N th power
so solving $x^5-2x^2-.1=0$ would be
$y^{5/2}-2y-.1=0 $
$N=5/2$
the root of original equation  is $y^{1/2}$ which uses the modification
Putting it all together, we have:
$q=(-1/b)^{1/(m-n)}$
$p_2=e^\frac{2\pi iv(m(1+w)-1}{n-m}$
$p_1=e^\frac{-2\pi iv}{n-m}$
$v$ runs through $0,1,...n-m-1$
$q\left[ p_1+\frac{1}{m-n}\sum_{w=0} \frac{  p_2q^{-n(w+1)} c^{w+1} }{w+1} {\frac{1-m-nw}{m-n} \choose w} \right] $
are n-m roots for the trinomial. this bypasses the cumbersome resolvent  method  of having to sum $n-1$  number of  hypergeometric functions, by using a single series. It would seem like something that should have been discovered by now.
Example: find a series solution for three roots of $x^5-2x^2+.1=0$
https://www.wolframalpha.com/input?i=x%5E5-2x%5E2%2B.1%3D0
one of the complex roots
first complex root
second complex root
real root
A: One pair of solutions will
be Puiseux series $ax^{1/2}+bx+cx^{3/2}+\cdots$ (where $x^{1/2}$ can
have one of two signs). Thus set
$u=x^{1/2}$ in the solution, giving a series $F(u)$ satisfying
$F(u)^5-2F(u)^2+u^2=0$. Hence $\sqrt{-F(u)^5+2F(u)^2}=u$. By ordinary
Lagrange inversion, $$ [u^n]F(u) =[u^{n-1}]\frac
1n\left(\frac{1}{\sqrt{2-u^3}}\right)^n. $$
Addendum. The series $F(x^{1/2})$ and $F(-x^{1/2})$ give two
solutions to $x^5-2x^2+z=0$. The other three solutions $G(x)$ are
given by
$$ [x^n]G(x) = \frac 1n[x^{n-1}]\left(\frac{x}{2(x+\alpha)^2
      -(x+\alpha)^5}\right)^n, $$
for $n\geq 1$, and $G(0)=\alpha$, where $\alpha=2^{1/3}$ (three
different values).
Addendum 2. The solution above can easily be generalized. Let
$P(t)\in\mathbb{C}[t]$. The fractional power series (Puiseux series)
$y$ satisfying $P(y)=x$ are given as follows: let $\alpha$ be a zero
of $P(t)$ of multiplicity $m$. Then $y=\alpha + \sum_{n\geq 1} a_n x^{n/m}$, where
$$ a_n = \frac 1n[u^{n-1}]\left(
       \frac{u}{P(u+\alpha)^{1/m}}\right)^n. $$
There are $m$ values of  $P(u+\alpha)^{1/m}$ (differing by multiplication by an $m$th root of unity), so we get $m$ solutions corresponding to $\alpha$.
