Series solution for general trinomial

Question

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.

Answer 1

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 $\alpha$.

Answer 2

I.- Euler / Lambert trinomial equations. Proof based on

Wang, Fei, Proof of a series solution for Euler’s trinomial equation, ACM Commun. Comput. Algebra 50, No. 4, 136-144 (2016). ZBL1370.33022.,

using Lagrange-Bürmann Theorem (LBT) to derive a series solution. (Technique is used below to solve the equation asked for)

The trinomial equation $$y = q + y^m$$ was solved in series form by Lambert, in 1758. Few years later, 1779, Euler transformed Lambert’s equation into the more symmetrical form with $y=x^{-\beta}$, $m=\alpha/\beta$ and $q=(\alpha-\beta)\cdot v$ as $$x^\alpha-x^\beta = (\alpha-\beta)\cdot v\cdot x^{\alpha+\beta}$$ so that $$v=f(x)=\frac{x^{-\beta}-x^{-\alpha}}{\alpha-\beta}$$ where $f(1)=0$ and $f'(1)=1$, therefore applying LBT $$x = 1+\sum_{n=1}^{\infty}\frac{v^n}{n!}\cdot[ \lim_{x\rightarrow1}\frac{d^{n-1}}{dx^{n-1}}\frac{(x-1)^n}{f(x)^n}] $$ $$x = 1+\sum_{n=1}^{\infty}\frac{v^n(\alpha-\beta)^n}{n}\cdot \mathrm{res}_{x=1}\lbrack(x^{-\beta}-x^{-\alpha})^{-n}\rbrack$$ residues can be computed using some CAS (Wolfram or Maple). This gives $$x = 1+\sum_{n=1}^{\infty}\frac{v^n(\alpha-\beta)^n}{n}\cdot \frac{\prod_{j=1}^{n-1}[1+j\alpha+(n-j)\beta]}{(n-1)!(\alpha-\beta)^n}$$ finally, this expression can be written using Gamma functions or Binomials $$x = 1+\sum_{n=1}^{\infty}\frac{v^n(\alpha-\beta)^n}{n\alpha+1}\cdot \binom{(n\alpha+1)/(\alpha-\beta)}{n}$$ Lambert series is recovered with $\beta = -1$, $\alpha=-m$ and $x=y$, giving $$y = 1-\sum_{n=1}^{\infty}\frac{q^n}{nm-1}\cdot \binom{(nm-1)/(m-1)}{n}$$This series can be obtained directly in the same way using $q=g(y)=y-y^m$ and

computing the summand residues $\mathrm{res}_{y=1}[(y-y^m)^{-n}]$.

Both series can be represented as a special case of the generalized Fox-Wright Function $_k\Psi_ℓ^*(A;B;ζ)$ normalized such that $_k\Psi_ℓ^*(A;B;ζ)|_{ζ=0}=1$ (See this MO Answer here for details). $y$ series for $m\in \mathbb{R} \land m>1$ can be put as $$y = -\frac{1}{m-1}\sum_{n=0}^{\infty}\frac{q^n}{n!}\cdot \frac{\Gamma(-\frac{1}{m-1}+n\cdot\frac{m}{m-1})}{\Gamma(1-\frac{1}{m-1}+n\cdot\frac{1}{m-1})}$$ which converges for $|q|<(m-1)m^{-\frac{m}{m-1}}$ giving $$y=\ _1\Psi_1^*([\frac{-1}{m-1},\frac{m}{m-1}];[1-\frac{1}{m-1},\frac{1}{m-1}];q)$$ A similar solution can be found in Section 4 of

Miller, A. R.; Moskowitz, I. S., Reduction of a class of Fox-Wright Psi functions for certain rational parameters, Comput. Math. Appl. 30, No. 11, 73-82 (1995). ZBL0839.33003.

and

Belkić, Dževad, All the trinomial roots, their powers and logarithms from the Lambert series, Bell polynomials and Fox-Wright function: illustration for genome multiplicity in survival of irradiated cells, J. Math. Chem. 57, No. 1, 59-106 (2019). ZBL1406.92284.

as well.

II.- A slight variation of this procedure allows to find the solution series for the specific problem asked for ($\gamma=2,\ n=2,\ m=5$) assuming $\gamma>0,\ m,n \in \mathbb{N},\ m>n$ $$z=\gamma\cdot\zeta^n-\zeta^m$$ We start from Ch 3 Bürmann's Theorem in

Sofo, Anthony, Computational techniques for the summation of series, New York, NY: Kluwer Academic/Plenum Publishers (ISBN 0-306-47805-6/hbk). xvi, 189 p. (2003). ZBL1059.65002.

Let $\phi(\zeta)$ be a simple function in a domain $\mathbb{D}$, zero at a point $a$ of $\mathbb{D}$, and let $$\vartheta(\zeta)=\frac{\zeta-a}{\phi(\zeta)},\ \ \vartheta(a)=\frac{1}{\phi'(a)}$$ If $f(\zeta)$ is analytic in $\mathbb{D}$ then $\forall \zeta \in \mathbb{D}$

$$f(\zeta)=f(a)+\sum_{k=1}^n \frac{[\phi(\zeta)]^k}{k!}\cdot \frac{d^{k-1}}{dt^{k-1}}[f'(t)\cdot[\vartheta(t)]^k]_{t=a}+R_{n+1}$$ $$f(\zeta)=f(a)+\sum_{k=1}^n \frac{[\phi(\zeta)]^k}{k}\cdot\mathrm{res}_{\,t=a}[f'(t)\cdot[\phi(t)]^{-k}]+R_{n+1}$$ This provides an expansion of a function in terms of series of powers of another function. Conditions for convergence can be seen in the given references. For now we can take it as a formal series. Using these relationships $$z=\phi(\zeta)=\gamma\cdot\zeta^n-\zeta^m,\ \ \ f(\zeta)=\zeta$$ The zeroes of $\phi(\zeta)$ are given by $m-n$ different roots $\zeta_\ell=\gamma^{\frac{1}{m-n}}\cdot\omega_{m-n}^\ell,\ \ \ell=0,1,..,m-n-1$ being $\omega_{k}$ the $k$-th root of unity and $\zeta_{m-n}=0$ with multiplicity $n$. Thus $$\zeta=\zeta_j+\sum_{k=1}^\infty \frac{z^k}{k}\cdot \mathrm{res}_{t=\zeta_j}[(\gamma\cdot t^n-t^m)^{-k}]$$ For example by taking $\gamma=2,\ n=2,\ m=5$. The root $\zeta_\mathrm{o}=2^{1/3}$ provides $$\zeta=2^\frac{1}{3}+\sum_{k=1}^\infty c_k z^k$$ whose first 10 terms $c_k$ are given by

The same procedure can be followed with every zero of $\phi(\zeta)$, giving different expansions.

III.- Another approach is to consider a solution based on Fox Wright series function. This MO link provides the inversion series that corresponds to a $_1\Psi_1^*$ function. For $\omega\in\mathbb{C},\ \alpha,\beta\in\mathbb{R}$ the general (standard) trinomial equation $$u-\omega\cdot u^\alpha-1=0$$ has the principal solution for $\alpha>1$ $$u =\,_{1}\Psi_{1}^{*}([1,\alpha];[2,\alpha-1];\omega)$$ whose powers satisfy $$u^\beta=\,_{1}\Psi_{1}^{*}([\beta,\alpha];[\beta+1,\alpha-1];\omega)$$ giving these series converging for $|w|<(\alpha-1)^{\alpha-1}\alpha^{-\alpha}$ $$u=\sum_{k=0}^{\infty}\frac{\omega^k}{k!}\cdot\frac{\Gamma(1+\alpha\cdot k)}{\Gamma(2+(\alpha-1)\cdot k)}$$ $$u^\beta=\beta\cdot\sum_{k=0}^{\infty}\frac{\omega^k}{k!}\cdot\frac{\Gamma(\beta+\alpha\cdot k)}{\Gamma(\beta+1+(\alpha-1)\cdot k)}$$ respectively. Since $\log{u}=\lim_{\beta\rightarrow 0}\frac{u^\beta-1}{\beta}$, logarithm series is obtained from this last series as well$$\log{u}=\sum_{k=1}^{\infty}\frac{\omega^k}{k!}\cdot\frac{\Gamma(\alpha\cdot k)}{\Gamma(1+(\alpha-1)\cdot k)}$$

The original problem $z=\gamma\cdot\zeta^n-\zeta^m$ can be worked now using simple algebra. Set $$u\cdot z =\gamma\cdot\zeta^n,\ \alpha=m/n,\ \omega=z^{-1}(z/\gamma)^\alpha\ \mathrm{and}\ \beta=1/n$$ This produces the Puiseux series (although it is also valid for $m,n\in\mathbb{R},\ \mathrm{s.t.}\ m>n>0$)$$\zeta=\frac{1}{n}(z/\gamma)^\frac{1}{n}\sum_{k=0}^{\infty}\frac{[z^{-1}(z/\gamma)^{\frac{m}{n}}]^k}{k!}\cdot\frac{\Gamma(\frac{1}{n}+\frac{m}{n}\cdot k)}{\Gamma(1+\frac{1}{n}+(\frac{m}{n}-1)\cdot k)}$$ Case $[m,n,\gamma]=[5,2,2]$, gives $$\zeta=\frac{1}{2}\sqrt{z/2}\sum_{k=0}^{\infty}\frac{(z/2)^{\frac{3}{2}k}}{2^k\,k!}\cdot\frac{\Gamma(\frac{1}{2}+\frac{5}{2}\cdot k)}{\Gamma(\frac{3}{2}+\frac{3}{2}\cdot k)}$$ converges for $0\le z\le (\frac{m}{n}-1)(\frac{\gamma\,n}{m})^\frac{m}{m-n}=1.034128651...$

Answer 3

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

This question is pretty popular. The explicit formulas for reference purposes are as follows:

Series solution for powers $k$ of the large roots for the trinomial: $x^v-jx^u+z=0$:

\begin{equation} R_I^k(z) = j^{\frac{k}{v-u}}\left[e^{\frac{2kI\pi i}{u-v}}+\frac{k}{u-v}\sum_{n=0} \frac{z^{1+n}}{n+1} j^{\frac{v(n+1)}{u-v}} e^{\frac{2I\pi i(u(1+n)-k)}{v-u}} {\frac{vn+u-k}{v-u} \choose n}\right] \end{equation}

for $I=0,1,..v-u-1$

Powers $k$ of the $u$ quantity of small roots of the trinomial $x^v-jx^u+z=0$:

\begin{equation} f^{k}(z)=r_I^k=k\sum_{n=0} \frac{e^{2i{\pi}I( (v-u)n+k)/u} z^{(n(v-u)+k)/u}}{j^{(nv+k)/u} (n(v-u)+k)} {nv/u +k/u-1 \choose n} \end{equation}

for $I=0,1,...u-1$

Source: https://arxiv.org/pdf/2212.09919.pdf