The surgery approach to ACSV requires finite minimality but uses only univariate residues and the saddle-point method.

When \(\mathcal{V}\) is smooth and the contributing points are nondegenerate then we get computable asymptotics.

$$f_{n\mathbf{r}} = \mathbf{w}^{-n\mathbf{r}} n^{(1-d)/2} \frac{(2\pi)^{(1-d)/2}}{\sqrt{\det(r_d\mathcal{H})}} \, \left(\frac{- G(\mathbf{w})}{w_dH_{z_d}(\mathbf{w})} + O\left(n^{-1}\right)\right).$$
 

The surgery approach to ACSV requires finite minimality but uses only univariate residues and the saddle-point method.

When \(\mathcal{V}\) is smooth and the contributing points are nondegenerate then we get computable asymptotics

$$f_{n\mathbf{r}} = \mathbf{w}^{-n\mathbf{r}} n^{(1-d)/2} \frac{(2\pi)^{(1-d)/2}}{\sqrt{\det(r_d\mathcal{H})}} \, \left(\frac{- G(\mathbf{w})}{w_dH_{z_d}(\mathbf{w})} + O\left(n^{-1}\right)\right).$$
 

Hardest part of the analysis: finding minimal points.

Assume that \(F(\mathbf{z}) = \sum_{\mathbf{i}\in\mathbb{Z}^d}f_\mathbf{i}\mathbf{z}^\mathbf{i}\) is combinatorial so \(f_\mathbf{i} \geq 0\).
 

Vivanti-Pringsheim: If \(\mathbf{w} \in \mathcal{V}\) is a minimal point then \((|w_1|,\dots,|w_d|)\in \mathcal{V}\) and is minimal.

 

Assume that \(F(\mathbf{z}) = \sum_{\mathbf{i}\in\mathbb{Z}^d}f_\mathbf{i}\mathbf{z}^\mathbf{i}\) is combinatorial so \(f_\mathbf{i} \geq 0\).
 

Vivanti-Pringsheim: If \(\mathbf{w} \in \mathcal{V}\) is a minimal point then \((|w_1|,\dots,|w_d|)\in \mathcal{V}\) and is minimal.

Corollary: A point \(\mathbf{w} \in \mathcal{V}\) with positive coordinates is minimal if and only if \(H(tw_1,\dots,tw_d) \neq 0\) for all \(0<t<1\).

Special Case

If \(H(\mathbf{z}) = 1-P(\mathbf{z})\) where \(p_\mathbf{i}\geq 0\) and the integer span of the exponents in \(P\) is \(\mathbb{Z}^d\) then every minimal point is strictly minimal and has positive coordinates.

$$\frac{1}{1-x-y}$$

$$\frac{1}{2-e^{x+y}}$$

$$\frac{1}{1-t(x+y)}$$

$$\frac{1}{1-x+y}$$

A lonesum matrix is a \(0-1\) matrix that is uniquely determined by its row and column sums.

$$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \quad \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$

$$\begin{array}{ccccc} \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} & \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} & \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} & \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} & \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \\[3ex] \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} & \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} & \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} & \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} & \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \\[3ex] & \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} & \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} & \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} & \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \end{array}$$

A lonesum matrix is a \(0-1\) matrix that is uniquely determined by its row and column sums.

$$F(x,y) = \sum_{n,k\geq0}\frac{B_{n,k}}{n!k!}x^ny^k = \frac{e^{x+y}}{e^{x}+e^{y}-e^{x+y}}$$

Noncommutative Biology: Sequential Regulation of Complex Networks. Letsou and Cai,
PLOS Computational Biology, 2016.

A lonesum matrix is a \(0-1\) matrix that is uniquely determined by its row and column sums.

$$F(x,y) = \sum_{n,k\geq0}\frac{B_{n,k}}{n!k!}x^ny^k = \frac{e^{x+y}}{e^{x}+e^{y}-e^{x+y}}$$

Since

$$ H(x,y) = e^{x}+e^{y}-e^{x+y} = 1 - \sum_{k \geq 0}\sum_{j=1}^{k-1}\frac{x^jy^{k-j}}{j!(k-j)!}$$
we know all minimal points are positive and strictly minimal.

Since

$$ H(x,y) = e^{x}+e^{y}-e^{x+y} = 1 - \sum_{k \geq 0}\sum_{j=1}^{k-1}\frac{x^jy^{k-j}}{j!(k-j)!}$$
we know all minimal points are positive and strictly minimal.

Furthermore, if \(H(x,y)=0\) then \(y=x-\log(e^x-1)\) decreases with \(x>0\) so every point with positive coordinates is minimal.

The function \(f(t) = t{\Big/}(1-e^t)\log(1-e^{-t})\) is a bijection on \(\mathbb{R}_{>0}.\)

The critical point equations in direction \(\mathbf{r}=(r,s)\) have unique positive solution \(x = f^{-1}(r/s)\) and \(y=f^{-1}(s/r)\).
 

The function \(f(t) = t{\Big/}(1-e^t)\log(1-e^{-t})\) is a bijection on \(\mathbb{R}_{>0}.\)

The critical point equations in direction \(\mathbf{r}=(r,s)\) have unique positive solution \(x = f^{-1}(r/s)\) and \(y=f^{-1}(s/r)\).

Theorem (Khera, Lundberg, Melczer 2019)
If \(k,n\rightarrow\infty\) such that \(n/k\rightarrow\lambda>0\) then
 

$$ B_{rn,sn} = \frac{a^{-rn}b^{-sn}}{\sqrt{n}}\frac{(rn)!(sn)!}{\sqrt{2\pi sae^{-a}(be^{-b}+ae^{-a}-ab)}}\left(1+O\left(n^{-1}\right)\right)$$
where \(a=f^{-1}(\lambda)\) and \(b = f^{-1}(1/\lambda)\).

Univariate case: Asymptotics given by dominant singularities.

Multivariate case: Asymptotics always given by minimal points?

Univariate case: Asymptotics given by dominant singularities.

Multivariate case: Asymptotics always given by minimal points?

NO!

Univariate case: Asymptotics given by dominant singularities.

Multivariate case: Asymptotics always given by minimal points?

NO!

It turns out critical points are the ones influencing asymptotics.

Minimality just helps determine which critical points matter.

Considering minimal points makes the proof (relatively) easy,
but makes checking the conditions difficult.

Problem: Because \(\mathcal{V}\) is infinite, finding the points in \(T(\mathbf{w}) \cap \mathcal{V}\) is expensive. It can also contain spurious points.

Considering minimal points makes the proof (relatively) easy,
but makes checking the conditions difficult.

Problem: Because \(\mathcal{V}\) is infinite, finding the points in \(T(\mathbf{w}) \cap \mathcal{V}\) is expensive. It can also contain spurious points.

Example

$$F(x,y) = \frac{1}{(1+2y)(1-x-y)}$$
still has critical point \(\mathbf{w}=(1/2,1/2)\) but it is no longer finitely minimal as \(T(\mathbf{w}) \cap \mathcal{V} \supset \left\{(e^{i\theta}/2,-1/2) : \theta \in (-\pi,\pi]\right\}\).

Considering minimal points makes the proof (relatively) easy,
but makes checking the conditions difficult.

Problem: Because \(\mathcal{V}\) is infinite, finding the points in \(T(\mathbf{w}) \cap \mathcal{V}\) is expensive. It can also contain spurious points.

Solution: Reduce importance of minimality. Only critical points really matter for asymptotics.

Considering minimal points makes the proof (relatively) easy,
but makes checking the conditions difficult.

Problem: Because \(\mathcal{V}\) is infinite, finding the points in \(T(\mathbf{w}) \cap \mathcal{V}\) is expensive. It can also contain spurious points.

Solution: Reduce importance of minimality. Only critical points really matter for asymptotics.

Generically there are finite set of critical points, even though there are an infinite set of minimal points.

Theorem (Baryshnikov-Pemantle 2011): Suppose that
 

$$ H = 0 , \qquad r_jz_1H_{z_1} = r_1z_jH_{z_j} \qquad\quad (2 \leq j \leq d) $$
admits a finite set of solutions. If
 

• there is exactly one minimal solution \(\mathbf{w} \in \mathbb{C}_*^d\) and
   
• \(H_{z_d}(\mathbf{w})\) and \(\det \mathcal{H}_\mathbf{w}\) are non-zero

then \(\displaystyle f_{n\mathbf{r}} = \mathbf{w}^{-n\mathbf{r}} n^{(1-d)/2} \frac{(2\pi)^{(1-d)/2}}{\sqrt{\det(r_d\mathcal{H})}} \, \left(\frac{- G(\mathbf{w})}{w_dH_{z_d}(\mathbf{w})} + O\left(n^{-1}\right)\right)\).

Only critical points matter now.

We get the same asymptotic expansions as the finitely minimal case, including for higher order poles.

Only critical points matter now.

We get the same asymptotic expansions as the finitely minimal case, including for higher order poles.

If there are a finite number of critical points satisfying these conditions then one adds their asymptotic contributions.

Only critical points matter now.

We get the same asymptotic expansions as the finitely minimal case, including for higher order poles.

If there are a finite number of critical points satisfying these conditions then one adds their asymptotic contributions.

Algorithm for Asymptotics: Characterize the (hopefully) finite set of critical points, and verify which are minimal.

Proof Idea (Baryshnikov-Pemantle 2011)

The surgery method deforms each \(|z_i|\) independently. We need to construct multivariate deformations.

Proof Idea (Baryshnikov-Pemantle 2011)

The surgery method deforms each \(|z_i|\) independently. We need to construct multivariate deformations.

Every minimal point has a cone of hyperbolicity. 

Proof Idea (Baryshnikov-Pemantle 2011)

The surgery method deforms each \(|z_i|\) independently. We need to construct multivariate deformations.

Every minimal point has a cone of hyperbolicity. 

At a smooth point \(\mathbf{w}\) we have \((\nabla_{\log} H)(\mathbf{w}) = \lambda \mathbf{v}\) for \(\mathbf{v}\in\mathbb{R}^d\) and the cone is a real halfspace with normal \(\mathbf{v}\).

Proof Idea (Baryshnikov-Pemantle 2011)

The surgery method deforms each \(|z_i|\) independently. We need to construct multivariate deformations.

Every minimal point has a cone of hyperbolicity. 

At a smooth point \(\mathbf{w}\) we have \((\nabla_{\log} H)(\mathbf{w}) = \lambda \mathbf{v}\) for \(\mathbf{v}\in\mathbb{R}^d\) and the cone is a real halfspace with normal \(\mathbf{v}\).

If \(\mathbf{v} \neq \mathbf{r}\) then we can locally deform the domain of integration to lower height. Glue the local deformations together.

Do we always have minimal critical points when \(\mathcal{V}\) is smooth?

Do we always have minimal critical points when \(\mathcal{V}\) is smooth?

No, if \(h_\mathbf{r}\) decreases without bound on \(\mathcal{D}\).

Do we always have minimal critical points when \(\mathcal{V}\) is smooth?

No, if \(h_\mathbf{r}\) decreases without bound on \(\mathcal{D}\).

What if \(h_{\mathbf{r}}\) achieves its infimum on \(\overline{\mathcal{D}}\)?

Do we always have minimal critical points when \(\mathcal{V}\) is smooth?

No, if \(h_\mathbf{r}\) decreases without bound on \(\mathcal{D}\).

What if \(h_{\mathbf{r}}\) achieves its infimum on \(\overline{\mathcal{D}}\)?

Example: Let \(\displaystyle F(x,y) = \frac{1}{2+y-x(1+y)^2}.\)

Where is the
critical point?

The branches
\((x,y_1(x))\) and \((x,y_2(x))\) don't intersect in \(\mathbb{R}^2\)

\((|x|,|y_1(x)|)\) and \((|x|,|y_2(x)|)\) do intersect in \(\mathbb{R}^2\)

\((\log|x|,\log|y_1(x)|)\) and \((\log|x|,\log|y_2(x)|)\) do intersect in \(\mathbb{R}^2\)

“Amoebas are shadows 
of algebraic varieties...” 
-- Timo de Wolff (2017)

Such ghost intersections happen when the amoeba in \(\mathbb{R}^d\) does not reflect \(\mathcal{V}\) in \(\mathbb{C}^d\).

This does not happen in the combinatorial case, when \(\mathcal{V} \cap \partial \mathcal{D}\) is captured by the points with positive real coordinates.

Such ghost intersections happen when the amoeba in \(\mathbb{R}^d\) does not reflect \(\mathcal{V}\) in \(\mathbb{C}^d\).

This does not happen in the combinatorial case, when \(\mathcal{V} \cap \partial \mathcal{D}\) is captured by the points with positive real coordinates.

Proposition

For a combinatorial Laurent expansion, if a smooth point
\(\mathbf{w}\in\mathbb{R}_{>0}^d\) is a local minimizer (or maximizer) of \(h_\mathbf{r}\) on \(\overline{\mathcal{D}}\) then it is a critical point in the direction \(\mathbf{r}\). 

Under the assumptions of the Main Theorem of Smooth ACSV (simple pole case), for any \(M \in \mathbb{N}\) there is an expansion

$$ f_{n\mathbf{r}} = \mathbf{w}^{-n\mathbf{r}} n^{(1-d)/2} \frac{(2\pi)^{(1-d)/2}}{\sqrt{\det(r_d\mathcal{H})}} \left(\sum_{j=0}^M C_j (r_dn)^{-j} + O\left(n^{-M-1}\right)\right) $$
with computable constants \(C_0,\dots,C_M\).


 

Under the assumptions of the Main Theorem of Smooth ACSV (simple pole case), for any \(M \in \mathbb{N}\) there is an expansion

$$ f_{n\mathbf{r}} = \mathbf{w}^{-n\mathbf{r}} n^{(1-d)/2} \frac{(2\pi)^{(1-d)/2}}{\sqrt{\det(r_d\mathcal{H})}} \left(\sum_{j=0}^M C_j (r_dn)^{-j} + O\left(n^{-M-1}\right)\right) $$
with computable constants \(C_0,\dots,C_M\).


Hörmander (1990): Explicit expression for the \(C_j\).

$$ C_j = (-1)^j \sum_{0 \le \ell \le 2j} \left. \frac{\mathcal{E}^{\ell+j} {\Big(} P(\theta) \psi(\theta)^\ell {\Big)}}{2^{\ell+j} \ell! (\ell+j)!} \right|_{\theta=\mathbf{0}}$$
where
$$\begin{aligned}\mathcal{E} &= - \sum_{1 \le i,j \le k} (\mathcal{H}^{-1})_{ij} \partial_i \partial_j \\[+6mm] P(\theta) &= \frac{-G \left( \widehat{\mathbf{w}}e^{i\theta}, g \left( \widehat{\mathbf{w}}e^{i\theta} \right) \right)}{g \left( \widehat{\mathbf{w}}e^{i\theta} \right) H_{z_d} \left( \widehat{\mathbf{w}}e^{i\theta}, g \left( \widehat{\mathbf{w}}e^{i\theta} \right) \right)} \\[+6mm]\psi(\theta) &= \log \left( \frac{g \left( \widehat{\mathbf{w}}e^{i\theta} \right)}{g(\widehat{\mathbf{w}})} \right) + i(\hat{\mathbf{r}} \cdot \theta)/r_d - (1/2)\theta \cdot \mathcal{H} \cdot \theta^T\end{aligned}$$

$$ C_j = (-1)^j \sum_{0 \le \ell \le 2j} \left. \frac{{\color{red}\mathcal{E}^{\ell+j}} {\Big(} P(\theta) \psi(\theta)^\ell {\Big)}}{2^{\ell+j} \ell! (\ell+j)!} \right|_{\theta=\mathbf{0}}$$
where
$$\begin{aligned}{\color{red}\mathcal{E}} &= - \sum_{1 \le i,j \le k} (\mathcal{H}^{-1})_{ij} \partial_i \partial_j \\[+6mm] P(\theta) &= \frac{-G \left( \widehat{\mathbf{w}}e^{i\theta}, g \left( \widehat{\mathbf{w}}e^{i\theta} \right) \right)}{g \left( \widehat{\mathbf{w}}e^{i\theta} \right) H_{z_d} \left( \widehat{\mathbf{w}}e^{i\theta}, g \left( \widehat{\mathbf{w}}e^{i\theta} \right) \right)} \\[+6mm]\psi(\theta) &= \log \left( \frac{g \left( \widehat{\mathbf{w}}e^{i\theta} \right)}{g(\widehat{\mathbf{w}})} \right) + i(\hat{\mathbf{r}} \cdot \theta)/r_d - (1/2)\theta \cdot \mathcal{H} \cdot \theta^T\end{aligned}$$

$$ C_j = (-1)^j \sum_{0 \le \ell \le 2j} \left. \frac{\mathcal{E}^{\ell+j} {\Big(} {\color{red}P(\theta)} \psi(\theta)^\ell {\Big)}}{2^{\ell+j} \ell! (\ell+j)!} \right|_{\theta=\mathbf{0}}$$
where
$$\begin{aligned}\mathcal{E} &= - \sum_{1 \le i,j \le k} (\mathcal{H}^{-1})_{ij} \partial_i \partial_j \\[+6mm] {\color{red}P(\theta)} &= \frac{-G \left( \widehat{\mathbf{w}}e^{i\theta}, g \left( \widehat{\mathbf{w}}e^{i\theta} \right) \right)}{g \left( \widehat{\mathbf{w}}e^{i\theta} \right) H_{z_d} \left( \widehat{\mathbf{w}}e^{i\theta}, g \left( \widehat{\mathbf{w}}e^{i\theta} \right) \right)} \\[+6mm]\psi(\theta) &= \log \left( \frac{g \left( \widehat{\mathbf{w}}e^{i\theta} \right)}{g(\widehat{\mathbf{w}})} \right) + i(\hat{\mathbf{r}} \cdot \theta)/r_d - (1/2)\theta \cdot \mathcal{H} \cdot \theta^T\end{aligned}$$

$$ C_j = (-1)^j \sum_{0 \le \ell \le 2j} \left. \frac{\mathcal{E}^{\ell+j} {\Big(} P(\theta) {\color{red}\psi(\theta)^\ell} {\Big)}}{2^{\ell+j} \ell! (\ell+j)!} \right|_{\theta=\mathbf{0}}$$
where
$$\begin{aligned}\mathcal{E} &= - \sum_{1 \le i,j \le k} (\mathcal{H}^{-1})_{ij} \partial_i \partial_j \\[+6mm] P(\theta) &= \frac{-G \left( \widehat{\mathbf{w}}e^{i\theta}, g \left( \widehat{\mathbf{w}}e^{i\theta} \right) \right)}{g \left( \widehat{\mathbf{w}}e^{i\theta} \right) H_{z_d} \left( \widehat{\mathbf{w}}e^{i\theta}, g \left( \widehat{\mathbf{w}}e^{i\theta} \right) \right)} \\[+6mm]{\color{red}\psi(\theta)} &= \log \left( \frac{g \left( \widehat{\mathbf{w}}e^{i\theta} \right)}{g(\widehat{\mathbf{w}})} \right) + i(\hat{\mathbf{r}} \cdot \theta)/r_d - (1/2)\theta \cdot \mathcal{H} \cdot \theta^T\end{aligned}$$

$$ C_j = (-1)^j \sum_{0 \le \ell \le 2j} \left. \frac{\mathcal{E}^{\ell+j} {\Big(} P(\theta) \psi(\theta)^\ell {\Big)}}{2^{\ell+j} \ell! (\ell+j)!} \right|_{\theta=\mathbf{0}}$$
where
$$\begin{aligned}\mathcal{E} &= - \sum_{1 \le i,j \le k} (\mathcal{H}^{-1})_{ij} \partial_i \partial_j \\[+6mm] P(\theta) &= \frac{-G \left( \widehat{\mathbf{w}}e^{i\theta}, g \left( \widehat{\mathbf{w}}e^{i\theta} \right) \right)}{g \left( \widehat{\mathbf{w}}e^{i\theta} \right) H_{z_d} \left( \widehat{\mathbf{w}}e^{i\theta}, g \left( \widehat{\mathbf{w}}e^{i\theta} \right) \right)} \\[+6mm]\psi(\theta) &= \log \left( \frac{g \left( \widehat{\mathbf{w}}e^{i\theta} \right)}{g(\widehat{\mathbf{w}})} \right) + i(\hat{\mathbf{r}} \cdot \theta)/r_d - (1/2)\theta \cdot \mathcal{H} \cdot \theta^T\end{aligned}$$

If \(G(\mathbf{w})\neq 0\) then higher-order terms give more accuracy.

$$ [x^n y^n] \frac{1}{1 - x - y} = 4^n \left( \frac{1}{\sqrt{\pi} n^{1/2}} - \frac{1}{8\sqrt{\pi} n^{3/2}} + \frac{1}{128\sqrt{\pi} n^{5/2}} + O\left(n^{-7/2}\right) \right) $$
 

If \(G(\mathbf{w})\neq 0\) then higher-order terms give more accuracy.

$$ [x^n y^n] \frac{1}{1 - x - y} = 4^n \left( \frac{1}{\sqrt{\pi} n^{1/2}} - \frac{1}{8\sqrt{\pi} n^{3/2}} + \frac{1}{128\sqrt{\pi} n^{5/2}} + O\left(n^{-7/2}\right) \right) $$
If \(G(\mathbf{w})=0\) then higher order terms may give dominant asymptotics.

$$[x^n y^n] \frac{x - 2y^2}{1 - x - y} = 4^n \left( \frac{1}{4\sqrt{\pi}n^{3/2}} + \frac{3}{32\sqrt{\pi}n^{5/2}} + O\left(n^{-7/2}\right) \right)$$
 

If \(G(\mathbf{w})\neq 0\) then higher-order terms give more accuracy.

$$ [x^n y^n] \frac{1}{1 - x - y} = 4^n \left( \frac{1}{\sqrt{\pi} n^{1/2}} - \frac{1}{8\sqrt{\pi} n^{3/2}} + \frac{1}{128\sqrt{\pi} n^{5/2}} + O\left(n^{-7/2}\right) \right) $$
If \(G(\mathbf{w})=0\) then higher order terms may give dominant asymptotics.

$$[x^n y^n] \frac{x - 2y^2}{1 - x - y} = 4^n \left( \frac{1}{4\sqrt{\pi}n^{3/2}} + \frac{3}{32\sqrt{\pi}n^{5/2}} + O\left(n^{-7/2}\right) \right)$$
In the worst case, all terms may be zero!

$$[x^n y^n] \frac{x - y}{1 - x - y} = O \left( \frac{4^n}{n^M} \right) \quad \text{for all } M > 0$$

The number of walks in \(\mathbb{N}^2\) starting at the origin and taking \(n\) steps in \(\{\text{NE},\text{NW},\text{SE},\text{SW}\} \) is

$${\Big[}(xyt)^n{\Big]}\frac{(1+x)(1+y)}{1-txyS(x,y)} \sim \frac{2}{\pi} \cdot \frac{4^n}{n}$$
where \(S(x,y) = xy + \frac{x}{y} + \frac{y}{x} + \frac{1}{xy}\).

The number of walks in \(\mathbb{N}^2\) starting at the origin and taking \(n\) steps in \(\{\text{NE},\text{NW},\text{SE},\text{SW}\} \) is

$${\Big[}(xyt)^n{\Big]}\frac{(1+x)(1+y)}{1-txyS(x,y)} \sim \frac{2}{\pi} \cdot \frac{4^n}{n}$$
where \(S(x,y) = xy + \frac{x}{y} + \frac{y}{x} + \frac{1}{xy}\). The critical points are

$${\color{white}\left(1,1,\frac{1}{4}\right),\quad } {\color{white}\left(1,-1,\frac{1}{4}\right),\quad }{\color{white}\left(-1,1,\frac{1}{4}\right),\quad } {\color{white}\left(-1,-1,\frac{1}{4}\right)} $$

The number of walks in \(\mathbb{N}^2\) starting at the origin and taking \(n\) steps in \(\{\text{NE},\text{NW},\text{SE},\text{SW}\} \) is

$${\Big[}(xyt)^n{\Big]}\frac{(1+x)(1+y)}{1-txyS(x,y)} \sim \frac{2}{\pi} \cdot \frac{4^n}{n}$$
where \(S(x,y) = xy + \frac{x}{y} + \frac{y}{x} + \frac{1}{xy}\). The critical points are

Numerator Vanishes

$${\color{white}\left(1,1,\frac{1}{4}\right),\quad } {\color{paleturquoise}\left(1,-1,\frac{1}{4}\right),\quad }{\color{paleturquoise}\left(-1,1,\frac{1}{4}\right),\quad } {\color{paleturquoise}\left(-1,-1,\frac{1}{4}\right)} $$

The number of walks in \(\mathbb{N}^2\) starting at the origin and taking \(n\) steps in \(\{\text{NE},\text{NW},\text{SE},\text{SW}\} \) is

$${\Big[}(xyt)^n{\Big]}\frac{(1+x)(1+y)}{1-txyS(x,y)} \sim \frac{2}{\pi} \cdot \frac{4^n}{n}$$
where \(S(x,y) = xy + \frac{x}{y} + \frac{y}{x} + \frac{1}{xy}\). The critical points are

Numerator Vanishes

$${\color{white}\left(1,1,\frac{1}{4}\right),\quad } {\color{paleturquoise}\left(1,-1,\frac{1}{4}\right),\quad }{\color{paleturquoise}\left(-1,1,\frac{1}{4}\right),\quad } {\color{paleturquoise}\left(-1,-1,\frac{1}{4}\right)} $$

The number of walks in \(\mathbb{N}^2\) starting at the origin and taking \(n\) steps in \(\{\text{NE},\text{NW},\text{SE},\text{SW}\} \) and ending on the \(x\)-axis is

$${\Big[}(xyt)^n{\Big]}\frac{(1-x^2)(1+y)}{1-txyS(x,y)} \sim \frac{2\left(1 + (-1)^n\right)}{\pi} \cdot \frac{4^n}{n^2}$$
The critical points are still

$${\color{white}\left(1,1,\frac{1}{4}\right),\quad } {\color{white}\left(1,-1,\frac{1}{4}\right),\quad }{\color{white}\left(-1,1,\frac{1}{4}\right),\quad } {\color{white}\left(-1,-1,\frac{1}{4}\right)} $$

The number of walks in \(\mathbb{N}^2\) starting at the origin and taking \(n\) steps in \(\{\text{NE},\text{NW},\text{SE},\text{SW}\} \) and ending on the \(x\)-axis is

$${\Big[}(xyt)^n{\Big]}\frac{(1-x^2)(1+y)}{1-txyS(x,y)} \sim \frac{2\left(1 + (-1)^n\right)}{\pi} \cdot \frac{4^n}{n^2}$$
The critical points are still

Numerator Vanishes
to Second Order

Numerator Vanishes
to First Order

$${\color{red}\left(1,1,\frac{1}{4}\right),\quad } {\color{red}\left(1,-1,\frac{1}{4}\right),\quad }{\color{paleturquoise}\left(-1,1,\frac{1}{4}\right),\quad } {\color{paleturquoise}\left(-1,-1,\frac{1}{4}\right)} $$

The number of walks in \(\mathbb{N}^2\) starting at the origin and taking \(n\) steps in \(\{\text{NE},\text{NW},\text{SE},\text{SW}\} \) and ending on the \(x\)-axis is

$${\Big[}(xyt)^n{\Big]}\frac{(1-x^2)(1+y)}{1-txyS(x,y)} \sim \frac{2\left(1 + (-1)^n\right)}{\pi} \cdot \frac{4^n}{n^2}$$
The critical points are still

Numerator Vanishes
to Second Order

Numerator Vanishes
to First Order

$${\color{red}\left(1,1,\frac{1}{4}\right),\quad } {\color{red}\left(1,-1,\frac{1}{4}\right),\quad }{\color{paleturquoise}\left(-1,1,\frac{1}{4}\right),\quad } {\color{paleturquoise}\left(-1,-1,\frac{1}{4}\right)} $$

The number of walks in \(\mathbb{N}^2\) starting at the origin and taking \(n\) steps in \(\{\text{NE},\text{NW},\text{SE},\text{SW}\} \) and ending at the origin is

$${\Big[}(xyt)^n{\Big]}\frac{(1-x^2)(1-y^2)}{1-txyS(x,y)} \sim \frac{4\left(1 + (-1)^n\right)}{\pi} \cdot \frac{4^n}{n^3}$$
The critical points are still

$${\color{white}\left(1,1,\frac{1}{4}\right),\quad } {\color{white}\left(1,-1,\frac{1}{4}\right),\quad }{\color{white}\left(-1,1,\frac{1}{4}\right),\quad } {\color{white}\left(-1,-1,\frac{1}{4}\right)} $$

The number of walks in \(\mathbb{N}^2\) starting at the origin and taking \(n\) steps in \(\{\text{NE},\text{NW},\text{SE},\text{SW}\} \) and ending at the origin is

$${\Big[}(xyt)^n{\Big]}\frac{(1-x^2)(1-y^2)}{1-txyS(x,y)} \sim \frac{4\left(1 + (-1)^n\right)}{\pi} \cdot \frac{4^n}{n^3}$$
The critical points are still

$${\color{paleturquoise}\left(1,1,\frac{1}{4}\right),\quad } {\color{paleturquoise}\left(1,-1,\frac{1}{4}\right),\quad }{\color{paleturquoise}\left(-1,1,\frac{1}{4}\right),\quad } {\color{paleturquoise}\left(-1,-1,\frac{1}{4}\right)} $$

Numerator Vanishes to Second Order

The number of walks in \(\mathbb{N}^2\) starting at the origin and taking \(n\) steps in \(\{\text{NE},\text{NW},\text{SE},\text{SW}\} \) and ending at the origin is

$${\Big[}(xyt)^n{\Big]}\frac{(1-x^2)(1-y^2)}{1-txyS(x,y)} \sim \frac{4\left(1 + (-1)^n\right)}{\pi} \cdot \frac{4^n}{n^3}$$
The critical points are still

Numerator Vanishes to Second Order

$${\color{paleturquoise}\left(1,1,\frac{1}{4}\right),\quad } {\color{paleturquoise}\left(1,-1,\frac{1}{4}\right),\quad }{\color{paleturquoise}\left(-1,1,\frac{1}{4}\right),\quad } {\color{paleturquoise}\left(-1,-1,\frac{1}{4}\right)} $$

Suppose that our expansion of \(F=G/H\) is combinatorial. We form the extended critical point system

$$\begin{aligned} H(\mathbf{z})&=0 \\ z_1H_{z_1}(\mathbf{z}) - r_1\lambda &= 0 \\ &\;\;\vdots \\ z_dH_{z_d}(\mathbf{z}) - r_d\lambda &= 0 \\ H(t\mathbf{z}) &= 0\end{aligned}$$
 

Suppose that our expansion of \(F=G/H\) is combinatorial. We form the extended critical point system

$$\begin{aligned} H(\mathbf{z})&=0 \\ z_1H_{z_1}(\mathbf{z}) - r_1\lambda &= 0 \\ &\;\;\vdots \\ z_dH_{z_d}(\mathbf{z}) - r_d\lambda &= 0 \\ H(t\mathbf{z}) &= 0\end{aligned}$$
Positive Minimal Points: Solutions \((\mathbf{w}, 1, \lambda)\) with \(\mathbf{w} \in \mathbb{R}_{>0}\) such that there is no solution \((\mathbf{w}, s, \lambda)\) with \(0<s<1\).

Suppose that our expansion of \(F=G/H\) is combinatorial. We form the extended critical point system

$$\begin{aligned} H(\mathbf{z})&=0 \\ z_1H_{z_1}(\mathbf{z}) - r_1\lambda &= 0 \\ &\;\;\vdots \\ z_dH_{z_d}(\mathbf{z}) - r_d\lambda &= 0 \\ H(t\mathbf{z}) &= 0\end{aligned}$$
Then find points with the same coordinate-wise modulus.

Suppose that our expansion of \(F=G/H\) is combinatorial. We form the extended critical point system

$$\begin{aligned} H(\mathbf{z})&=0 \\ z_1H_{z_1}(\mathbf{z}) - r_1\lambda &= 0 \\ &\;\;\vdots \\ z_dH_{z_d}(\mathbf{z}) - r_d\lambda &= 0 \\ H(t\mathbf{z}) &= 0\end{aligned}$$
How do we do this in practice?

Let's look at an example.

A polyomino is a collection of unit squares stacked vertically or horizontally.

A polyomino is horizontally convex if each row has no holes.

Delest (1988): The generating function for the number of horizontally convex polyominoes \(f_{n,k}\) with \(n\) boxes and \(k\) rows is
 

$$ F(x,y) = \sum_{n,k\geq0}f_{n,k}x^ny^k = \frac{xy(1 - x)^3}{(1 - x)^4 - xy(1 - x - x^2 + x^3 + x^2y)}. $$

Let's try to encode the real solutions to the extended critical point system for the direction \(\mathbf{r}=(2,1)\).

There are a finite set of solutions to the extended system.

First Idea: Encode them using triangular systems of the form
 

$$\begin{aligned}P_0(t)&=0 \\ P_1(t,z_1)&=0 \\ &\;\;\vdots \\ P_d(t,z_1,\dots,z_d)&=0\end{aligned}$$
 

There are a finite set of solutions to the extended system.

First Idea: Encode them using triangular systems of the form
 

$$\begin{aligned}P_0(t)&=0 \\ P_1(t,z_1)&=0 \\ &\;\;\vdots \\ P_d(t,z_1,\dots,z_d)&=0\end{aligned}$$
Problems: Might need generic change of variables (destroying structure), hard to detect the real solutions, etc.

Introduce a new variable \(u\) that is a linear form

$$ u = \kappa_1z_1 + \cdots + \kappa_dz_d + \kappa_{d+1}t + \kappa_{d+2}\lambda \qquad (\kappa_i \in \mathbb{Z})$$

Introduce a new variable \(u\) that is a linear form

$$ u = \kappa_1z_1 + \cdots + \kappa_dz_d + \kappa_{d+1}t + \kappa_{d+2}\lambda \qquad (\kappa_i \in \mathbb{Z})$$

If \(u\) is generic then a reduced Gröbner basis with respect to an elimination ordering with \(u\) smallest has the form

$$P(u) = 0, \qquad z_1 = R_1(u), \; \dots,\; \lambda = R_{d+2}(u)$$

Introduce a new variable \(u\) that is a linear form

$$ u = \kappa_1z_1 + \cdots + \kappa_dz_d + \kappa_{d+1}t + \kappa_{d+2}\lambda \qquad (\kappa_i \in \mathbb{Z})$$

If \(u\) is generic then a reduced Gröbner basis with respect to an elimination ordering with \(u\) smallest has the form

$$P(u) = 0, \qquad z_1 = R_1(u), \; \dots,\; \lambda = R_{d+2}(u)$$

Geometrically: Parameterize solutions by level sets of the hyperplane with normal \(\boldsymbol{\kappa}\).

Benefit: Easy to compute, and to find real solutions.

Benefit: Easy to compute, and to find real solutions.

Problem: Coefficients of the \(R_j\) are huge.

Benefit: Easy to compute, and to find real solutions.

Problem: Coefficients of the \(R_j\) are huge.

Benefit: Easy to compute, and to find real solutions.

Problem: Coefficients of the \(R_j\) are huge.

Benefit: Easy to compute, and to find real solutions.

Problem: Coefficients of the \(R_j\) are huge.

Parameterize the solutions by introducing generic

$$ u = \kappa_1z_1 + \cdots + \kappa_dz_d + \kappa_{d+1}t + \kappa_{d+2}\lambda \qquad (\kappa_i \in \mathbb{Z})$$

and encoding the solutions as

$$P(u) = 0, \qquad z_1 = \frac{Q_1(u)}{P'(u)}, \; \dots,\; \lambda = \frac{Q_{d+2}(u)}{P'(u)}.$$

This is known as a Kronecker Representation or Rational Univariate Representation (RUR).

Dates back to work of Kronecker (1882) and Macaulay (1916) on solving zero-dimensional polynomial systems.

The height of a polynomial is the maximum bitsize of its coefficients.

Schost, 2001 (Specialized): If \(H\) has degree \(\delta\) and height \(h\) then there is a Kronecker representation of the extended critical point system with polynomials of degree at most \(\delta^d\) and heights \(\tilde{O}(h\delta^d)\).

Dates back to work of Kronecker (1882) and Macaulay (1916) on solving zero-dimensional polynomial systems.

The height of a polynomial is the maximum bitsize of its coefficients.

Schost, 2001 (Specialized): If \(H\) has degree \(\delta\) and height \(h\) then there is a Kronecker representation of the extended critical point system with polynomials of degree at most \(\delta^d\) and heights \(\tilde{O}(h\delta^d)\).

Compare to heights in \(\tilde{O}(h\delta^{2d})\) for univariate representation.

Giusti, Lecerf, Salvy (2001): Gröbner free method of computing Kronecker representations.

Safey El Din and Schost (2018): Multihomogeneous bounds.

Hauenstein, Safey El Din, Schost, Vu (2018): Homotopy algorithm to compute them for determinantal systems.

Giusti, Lecerf, Salvy (2001): Gröbner free method of computing Kronecker representations.

Safey El Din and Schost (2018): Multihomogeneous bounds.

Hauenstein, Safey El Din, Schost, Vu (2018): Homotopy algorithm to compute them for determinantal systems.

In our package we simply use a Gröbner basis computation.

Even for a small example we go from 42 digit numbers in the univariate representation to 16 digit numbers in the RUR.

M. and Salvy (2016)

Under generic assumptions (including smoothness and nondegeneracy of critical points) it is possible to find all minimal critical points in \(\tilde{O}(h\delta^{4d+3})\) bit operations for a combinatorial expansion.

M. and Salvy (2016)

Under generic assumptions (including smoothness and nondegeneracy of critical points) it is possible to find all minimal critical points in \(\tilde{O}(h\delta^{4d+3})\) bit operations for a combinatorial expansion.

In the general case, we split all variables into real and imaginary parts. We use critical point techniques to form a square system whose real solutions encode critical points.

M. and Salvy (2016)

Under generic assumptions (including smoothness and nondegeneracy of critical points) it is possible to find all minimal critical points in \(\tilde{O}(h\delta^{4d+3})\) bit operations for a combinatorial expansion.

In the general case, we split all variables into real and imaginary parts. We use critical point techniques to form a square system whose real solutions encode critical points.

One can find minimal critical points in the general case in \(\tilde{O}(h2^{3d}\delta^{9d+5})\) bit operations.

Can install with 

Hackl, Luo, M., Selover, and Wong (2023): Smooth combinatorial case.

Hackl, Luo, M., and Schost (2025): Extension to more complicated geometries (and other improvements).

Can install with 

Hackl, Luo, M., Selover, and Wong (2023): Smooth combinatorial case.

Hackl, Luo, M., and Schost (2025): Extension to more complicated geometries (and other improvements).

Lee, M., Smolčić (2022): Package for (possibly) non-combinatorial expansions using HomotopyContinuation.jl.

Example: Asymptotics of

$$ \binom{2n}{n} = [x^ny^n]\left(\frac{1}{1-x-y}\right) $$

$$ \binom{2n}{n} =\frac{4^n}{\sqrt{\pi n}} + O\left(\frac{4^n}{n^{3/2}}\right) $$

Example: Asymptotics of

$$ \binom{2n}{n} = [x^ny^n]\left(\frac{1}{1-x-y}\right) $$

Example: Asymptotics of

$$ \binom{2n}{n} = [x^ny^n]\left(\frac{1}{1-x-y}\right) $$

Example: Asymptotics of

$$ \binom{2n}{n} = [x^ny^n]\left(\frac{1}{1-x-y}\right) $$

Recall the generating function
 

$$ F(x,y) = \sum_{n,k\geq0}f_{n,k}x^ny^k = \frac{xy(1 - x)^3}{(1 - x)^4 - xy(1 - x - x^2 + x^3 + x^2y)} $$
for the number \(f_{n,k}\) of horizontally convex polyominoes with \(n\) boxes and \(k\) rows.

$$b_n = \sum_{k=0}^n \binom{n}{k}^2 \binom{n+k}{k}^2 = [(xyzt)^n] \frac{1}{1 - t(1+x)(1+y)(1+z)(1+y+z+yz+xyz)}$$

A. van der Poorten, A proof that Euler missed... Apéry’s proof  
 of the irrationality of ζ(3). Math. Intelligencer, 1978.

$$b_n = \sum_{k=0}^n \binom{n}{k}^2 \binom{n+k}{k}^2 = [(xyzt)^n] \frac{1}{1 - t(1+x)(1+y)(1+z)(1+y+z+yz+xyz)}$$

A. van der Poorten, A proof that Euler missed... Apéry’s proof  
 of the irrationality of ζ(3). Math. Intelligencer, 1978.

$$b_n = \sum_{k=0}^n \binom{n}{k}^2 \binom{n+k}{k}^2 = [(xyzt)^n] \frac{1}{1 - t(1+x)(1+y)(1+z)(1+y+z+yz+xyz)}$$

A. van der Poorten, A proof that Euler missed... Apéry’s proof  
 of the irrationality of ζ(3). Math. Intelligencer, 1978.

If the smooth critical point equations
 

• have exactly one minimal solution \(\mathbf{w} \in \mathbb{C}_*^d\) and
   
• \(H_{z_d}(\mathbf{w})\) and \(\det \mathcal{H}_\mathbf{w}\) are non-zero

then \(\displaystyle f_{n\mathbf{r}} = \mathbf{w}^{-n\mathbf{r}} n^{(1-d)/2} \frac{(2\pi)^{(1-d)/2}}{\sqrt{\det(r_d\mathcal{H})}} \, \left(\frac{- G(\mathbf{w})}{w_dH_{z_d}(\mathbf{w})} + O\left(n^{-1}\right)\right)\).

If the smooth critical point equations
 

• have exactly one minimal solution \(\mathbf{w} \in \mathbb{C}_*^d\) and
   
• \(H_{z_d}(\mathbf{w})\) and \(\det \mathcal{H}_\mathbf{w}\) are non-zero

then \(\displaystyle f_{n\mathbf{r}} = \mathbf{w}^{-n\mathbf{r}} n^{(1-d)/2} \frac{(2\pi)^{(1-d)/2}}{\sqrt{\det(r_d\mathcal{H})}} \, \left(\frac{- G(\mathbf{w})}{w_dH_{z_d}(\mathbf{w})} + O\left(n^{-1}\right)\right)\).

If \(\mathbf{r}\) varies in some compact set where the above conditions are satisfied then the error term in the asymptotic estimate can be uniformly bounded.

Recall that

$$ \left[x^{rn}y^{sn}\right]\frac{1}{1-x-y} \sim \left(\frac{r+s}{r}\right)^{rn}\left(\frac{r+s}{s}\right)^{sn}\frac{\sqrt{r+s}}{\sqrt{2rs\pi n}}$$

Recall that

$$ {\color{paleturquoise}\left[x^{rn}y^{sn}\right]\frac{1}{1-x-y}} \sim {\color{orange}\left(\frac{r+s}{r}\right)^{rn}\left(\frac{r+s}{s}\right)^{sn}\frac{\sqrt{r+s}}{\sqrt{2rs\pi n}}}$$

Only makes sense
if \(rn,sn \in\mathbb{N}\)

Valid for any \(r,s>0\)

Recall that

$$ \left[x^{rn}y^{sn}\right]\frac{1}{1-x-y} \sim \left(\frac{r+s}{r}\right)^{rn}\left(\frac{r+s}{s}\right)^{sn}\frac{\sqrt{r+s}}{\sqrt{2rs\pi n}}$$

What if we put \((r,s)=(\pi,1)\) into the approximation?

What does this correspond to?

Recall that

$$ \left[x^{rn}y^{sn}\right]\frac{1}{1-x-y} \sim \left(\frac{r+s}{r}\right)^{rn}\left(\frac{r+s}{s}\right)^{sn}\frac{\sqrt{r+s}}{\sqrt{2rs\pi n}}$$

What if we put \((r,s)=(\pi,1)\) into the approximation?

What does this correspond to?

Compare to \(f_{[n\pi],n}\) where \([z]=\) closest integer to \(z\).

Plot of \(\displaystyle \frac{f_{[n\pi],n}}{\text{asymptotics when \((r,s)=(\pi,1)\)}}\)

Plot of \(\displaystyle \frac{f_{[n\pi],n}}{\text{asymptotics when \((r,s)=(\pi,1)\)}}\)

Recall that

$$ \left[x^{rn}y^{sn}\right]\frac{1}{1-x-y} \sim {\color{orange}\left(\frac{r+s}{r}\right)^{rn}\left(\frac{r+s}{s}\right)^{sn}\frac{\sqrt{r+s}}{\sqrt{2rs\pi n}}}$$

If \(A_n\) is this approximation with \((r,s) =(\pi,1)\) then

$$ f_{[n\pi],n} \sim \underbrace{\left(\frac{\pi+1}{\pi}\right)^{[n\pi]-n\pi}}_\text{bounded factor} \; \cdot \; {\color{orange}A_n}$$

Plot of \(\displaystyle \frac{f_{[n\pi],n}}{\left(\frac{\pi+1}{\pi}\right)^{[n\pi]-n\pi} \; A_n}\)

Fix \(\mathbf{m}=(\widehat{\mathbf{m}},1)\) and suppose that for \(\mathbf{r}=(\widehat{\mathbf{r}},1)\) near \(\mathbf{m}\) there is a smoothly varying strictly minimal critical point \(\mathbf{w}=\mathbf{w}(\mathbf{r})\) with

• \(H_{z_d}(\mathbf{w})\) and \(G(\mathbf{w})\) are non-zero
   
• the matrix \(\mathcal{H}=\mathcal{H}_{\mathbf{w}}\) is nonsingular
   

If \(\widehat{\mathbf{s}}=\widehat{\mathbf{s}}(n)\) is a sequence in \(\mathbb{N}^{d-1}\) with \(|{\color{red}\widehat{\mathbf{m}}}-{\color{paleturquoise}\widehat{\mathbf{s}}}/n| = o(n^{-1/3})\) then

$$\begin{aligned}f_{\hat{\mathbf{s}},n} &\sim {\color{red}\mathbf{w}^{-n\mathbf{m}} n^{(1-d)/2} \left( \frac{-G(\mathbf{w})(2\pi)^{(1-d)/2}}{w_d H_{z_d}(\mathbf{w})\sqrt{\det \mathcal{H}}} \right)} \\[+5mm]&\qquad\qquad{\color{paleturquoise}\cdot\;\widehat{\mathbf{w}}^{-(\hat{\mathbf{s}} - n\widehat{\mathbf{m}})} \exp \left[ -\frac{(\hat{\mathbf{s}} - n\widehat{\mathbf{m}})^T \mathcal{H}^{-1}(\hat{\mathbf{s}} - n\widehat{\mathbf{m}})}{2n} \right].}\end{aligned}$$

The first works on multivariate asymptotics via generating functions dealt with limit theorems.

Bender (1973): CLTs from specific bivariate behaviour.

Bender and Richmond (1983): Arbitrary dimension.

Richmond and Gao (1992): Different types of singularities.

These are essentially multivariate transfer theorems.

Suppose that in some direction \((\mathbf{m},1)\) there is a strictly minimal critical point \(\mathbf{w}=(\mathbf{1},t)\) with \(t>0\) such that

• \(H_{z_d}(\mathbf{w})\) and \(G(\mathbf{w})\) are non-zero
   
• the matrix \(\mathcal{H}=\mathcal{H}_{\mathbf{w}}\) is nonsingular

Then as \(n\rightarrow\infty\) the coefficients of \([z_d^n]F(\mathbf{z})\) approach a multivariate normal distribution

$$F(x,y,z) = \square + \cdots + {\color{red}\left(\square + \square x + \square y + \square xy + \cdots\right)z^n} + \cdots$$

Suppose that in some direction \((\mathbf{m},1)\) there is a strictly minimal critical point \(\mathbf{w}=(\mathbf{1},t)\) with \(t>0\) such that

• \(H_{z_d}(\mathbf{w})\) and \(G(\mathbf{w})\) are non-zero
   
• the matrix \(\mathcal{H}=\mathcal{H}_{\mathbf{w}}\) is nonsingular

Then

$$ \sup_{\mathbf{s} \in \mathbb{Z}^{d-1}} n^{(d-1)/2} {\Big|} t^n f_{\mathbf{s},n} - \nu_n(\mathbf{s}) {\Big|} \to 0 $$
where $$\nu_n(\mathbf{s}) = \frac{-G(\mathbf{w})}{w_d H_{z_d}(\mathbf{w})} \frac{(2\pi n)^{(1-d)/2}}{\sqrt{\det \mathcal{H}}} \exp \left[ -\frac{(\mathbf{s} - n\mathbf{m})^T \mathcal{H}^{-1}(\mathbf{s} - n\mathbf{m})}{2n} \right].$$

Theorem (Hosten-Sullivant 2010)

If \(\text{ML}(n,k) = \) ML-degree for the multinomial missing data problem with random variables \(X_1\in\{1,\dots,n\}\) and
\(X_2 \in \{1,\dots,k\}\) then

$$\sum_{n,k \geq 0}\text{ML}(n,k)\frac{x^ny^k}{n!k!} = \frac{1}{e^x + e^y - e^{x+y}}.$$

Theorem (Hosten-Sullivant 2010)

If \(\text{ML}(n,k) = \) ML-degree for the multinomial missing data problem with random variables \(X_1\in\{1,\dots,n\}\) and
\(X_2 \in \{1,\dots,k\}\) then

$$\sum_{n,k \geq 0}\text{ML}(n,k)\frac{x^ny^k}{n!k!} = \frac{1}{e^x + e^y - e^{x+y}}.$$

Theorem (Khera, Lundberg, M. 2021): For all fixed \(K>0\),

$$ \sup_{|k-n/2| \le K\sqrt{n}} \left| \frac{(2 \log 2)^n}{n!} \mathrm{ML}(n-k, k) - \frac{2^{-\frac{2(k-n/2)^2}{n(1-\log 2)}}}{(4 \log 2)\sqrt{1 - \log 2}} \right| \to 0$$

GF tracking divisions
and polynomial degree
in Euclidean Algorithm
over \(\mathbb{F}_p[x]\) is

$$F(u,z) = \frac{1}{1-pz-p(p-1)uz}$$
(Flajolet and Sedgewick Ex. IX.15)

Under generic assumptions, the methods of ACSV give explicit asymptotics.

When \(F\) is combinatorial, these methods are rigorously implemented in Sage, along with simple limit theorems.

In the combinatorial case the amoeba picture captures relevant information for asymptotics.

In general, you need to be careful about behaviour in \(\mathbb{C}^d\).