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.



Series solution for general trinomial
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 $\alpha$.


