Diagonals capture behaviour of multivariate generating functions and are data structures for univariate sequences.

Rational diagonals capture algebraic behaviour too.
They can be studied using D-finite techniques.

Complex analysis in several variables is more complicated than the univariate theory, but many results generalize.

It is hard to picture things (even \(\mathbb{C}^2\) has 4 real dimensions) but polynomial amoebas are a useful tool.

We first overview the general argument with the function

$$ F(x,y) = \frac{1}{1-x-y}.$$

There are 3 Laurent expansions of \(F\) around the origin.
We consider the power series expansion

$$ F(x,y) = \sum_{i,j \geq 0} \binom{i+j}{i} x^i y^k $$
with domain of convergence \(\mathcal{D} = \{(x,y) \in \mathbb{C}^2 : |x|+|y|<1\}\).

Set \(\mathbb{r} = (1,1)\) so the Cauchy integral formula implies

$$ f_{n\mathbf{r}} = \binom{2n}{n} = \frac{1}{(2\pi i)^2} \int_{T(a,b)} \frac{1}{1-x-y} \; \frac{dxdy}{x^{n+1}y^{n+1}} $$

for all \((a,b) \in \mathcal{D}\).

Set \(\mathbb{r} = (1,1)\) so the Cauchy integral formula implies

$$ f_{n\mathbf{r}} = \binom{2n}{n} = \frac{1}{(2\pi i)^2} \int_{T(a,b)} \frac{1}{1-x-y} \; \frac{dxdy}{x^{n+1}y^{n+1}} $$

for all \((a,b) \in \mathcal{D}\).

The singular variety \(\mathcal{V}=\mathcal{V}(1-x-y)\) is now infinite, but we will try to find a finite set of singularities determining diagonal asymptotics.

As a rational diagonal, we know \(f_n\) has an asymptotic expansion as a sum of terms of the form \(C \, n^\alpha \, \zeta ^n \, (\log n)^\ell\).

Like the univariate case, the first step is to characterize the exponential growth

$$ \rho = \limsup _{n\rightarrow\infty} |f_{n,n}|^{1/n}.$$
 

As a rational diagonal, we know \(f_n\) has an asymptotic expansion as a sum of terms of the form \(C \, n^\alpha \, \zeta ^n \, (\log n)^\ell\).

Like the univariate case, the first step is to characterize the exponential growth

$$ \rho = \limsup _{n\rightarrow\infty} |f_{n,n}|^{1/n}.$$
Univariate: Reciprocal of modulus of dominant singularities
 

As a rational diagonal, we know \(f_n\) has an asymptotic expansion as a sum of terms of the form \(C \, n^\alpha \, \zeta ^n \, (\log n)^\ell\).

Like the univariate case, the first step is to characterize the exponential growth

$$ \rho = \limsup _{n\rightarrow\infty} |f_{n,n}|^{1/n}.$$
Univariate: Reciprocal of modulus of dominant singularities
 

Multivariate: Bounded by minimal points \(\partial \mathcal{D} \cap \mathcal{V}\)

In our case the maximum modulus bound implies

$$\binom{2n}{n}  = \left| \frac{1}{(2\pi i)^2} \int_{T(a,b)} \frac{1}{1-x-y} \; \frac{dxdy}{x^{n+1}y^{n+1}} \right| \leq \frac{|ab|^{-n}}{1-|a|-|b|}$$

for all \((a,b) \in \mathcal{D}\). 

In our case the maximum modulus bound implies

$$\binom{2n}{n}  = \left| \frac{1}{(2\pi i)^2} \int_{T(a,b)} \frac{1}{1-x-y} \; \frac{dxdy}{x^{n+1}y^{n+1}} \right| \leq \frac{|ab|^{-n}}{1-|a|-|b|}$$

for all \((a,b) \in \mathcal{D}\). Thus we get a family of bounds

$$ \limsup_{n \rightarrow\infty} \binom{2n}{n}^{1/n} \leq |ab|^{-1}$$
for all \((a,b)\in\mathcal{D}\).

In our case the maximum modulus bound implies

$$\binom{2n}{n}  = \left| \frac{1}{(2\pi i)^2} \int_{T(a,b)} \frac{1}{1-x-y} \; \frac{dxdy}{x^{n+1}y^{n+1}} \right| \leq \frac{|ab|^{-n}}{1-|a|-|b|}$$

for all \((a,b) \in \mathcal{D}\). Thus we get a family of bounds

$$ \limsup_{n \rightarrow\infty} \binom{2n}{n}^{1/n} \leq |ab|^{-1}$$
for all \((a,b)\in{\color{red}\mathcal{D}}\).

In our case the maximum modulus bound implies

$$\binom{2n}{n}  = \left| \frac{1}{(2\pi i)^2} \int_{T(a,b)} \frac{1}{1-x-y} \; \frac{dxdy}{x^{n+1}y^{n+1}} \right| \leq \frac{|ab|^{-n}}{1-|a|-|b|}$$

for all \((a,b) \in \mathcal{D}\). Thus we get a family of bounds

$$ \limsup_{n \rightarrow\infty} \binom{2n}{n}^{1/n} \leq |ab|^{-1}$$
for all \((a,b)\in{\color{red}\overline{\mathcal{D}}}\).

In our case the maximum modulus bound implies

$$\binom{2n}{n}  = \left| \frac{1}{(2\pi i)^2} \int_{T(a,b)} \frac{1}{1-x-y} \; \frac{dxdy}{x^{n+1}y^{n+1}} \right| \leq \frac{|ab|^{-n}}{1-|a|-|b|}$$

for all \((a,b) \in \mathcal{D}\). Thus we get a family of bounds

$$ \limsup_{n \rightarrow\infty} \binom{2n}{n}^{1/n} \leq |ab|^{-1}$$
for all \((a,b)\in{\color{red}\partial\mathcal{D}}\).

Since \(\partial \mathcal{D} = \{(a,b):|a|+|b|=1\}\) the best upper bound in

$$ \limsup_{n \rightarrow\infty} \binom{2n}{n}^{1/n} \leq |ab|^{-1}$$
is given by minimizing \(1/{\Big(}|a|(1-|a|){\Big)}\) for \(0<|a|<1\).

Since \(\partial \mathcal{D} = \{(a,b):|a|+|b|=1\}\) the best upper bound in

$$ \limsup_{n \rightarrow\infty} \binom{2n}{n}^{1/n} \leq |ab|^{-1}$$
is given by minimizing \(1/{\Big(}|a|(1-|a|){\Big)}\) for \(0<|a|<1\).

The minimum of \(4\) is achieved when \(|a|=|b|=1/2\).

Since \(\partial \mathcal{D} = \{(a,b):|a|+|b|=1\}\) the best upper bound in

$$ \limsup_{n \rightarrow\infty} \binom{2n}{n}^{1/n} \leq |ab|^{-1}$$
is given by minimizing \(1/{\Big(}|a|(1-|a|){\Big)}\) for \(0<|a|<1\).

The minimum of \(4\) is achieved when \(|a|=|b|=1/2\).

Note: \(\displaystyle\binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}}\) so the bound is tight.

The only singularity with \(|x|=|y|=1/2\) is \(\boldsymbol{\sigma} = (1/2,1/2)\).

In fact, we characterized \(\partial\mathcal{D} \cap \mathcal{V} = \{(x,1-x):0<x<1\}\).

The only singularity with \(|x|=|y|=1/2\) is \(\boldsymbol{\sigma} = (1/2,1/2)\).

In fact, we characterized \(\partial\mathcal{D} \cap \mathcal{V} = \{(x,1-x):0<x<1\}\).

The only minimal point where the Cauchy integrand

$$\frac{1}{1-x-y} \; \frac{1}{x^{n+1}y^{n+1}} $$
has exponential growth \(4^n\) is \(\boldsymbol{\sigma}\).

The only singularity with \(|x|=|y|=1/2\) is \(\boldsymbol{\sigma} = (1/2,1/2)\).

In fact, we characterized \(\partial\mathcal{D} \cap \mathcal{V} = \{(x,1-x):0<x<1\}\).

The only minimal point where the Cauchy integrand

$$\frac{1}{1-x-y} \; \frac{1}{x^{n+1}y^{n+1}} $$
has exponential growth \(4^n\) is \(\boldsymbol{\sigma}\).

Thus \(\boldsymbol{\sigma}\) is the only minimal point where local behaviour could determine diagonal asymptotics.

Let

$$ \quad\qquad I = \frac{1}{(2\pi i)^2} \int_{|x|=1/2} \left( \int_{|y| = 1/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right) \frac{dx}{x^{n+1}} $$

Let

$$ \quad\qquad I = \frac{1}{(2\pi i)^2} \int_{|x|=1/2} \left( \int_{|y| = 1/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right) \frac{dx}{x^{n+1}} $$
and

$$ \;\; I_\text{loc} = \frac{1}{(2\pi i)^2} \int_{\mathcal{N}} \left( \int_{|y| = 1/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right) \frac{dx}{x^{n+1}}$$

where \(\mathcal{N} = \left\{|x| = 1/2 : \arg(x) \in \left(-\frac{\pi}{4},\frac{\pi}{4}\right)\right\}\).

Let \(\mathcal{N}' = \{|x|=1/2\}\setminus\mathcal{N}\) 

Let \(\mathcal{N}' = \{|x|=1/2\}\setminus\mathcal{N}\) and

\(\rho = |1-e^{i\pi/4}/2| = 0.736\dots\).

When \(x \in \mathcal{N}'\),

$$ \begin{aligned}\left| \frac{1}{(2\pi i)} \int_{|y| = 1/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right|  &= \left| \frac{1}{(2\pi i)} \int_{|y| = 1/4} \frac{1/(1-x)}{1-\frac{y}{1-x}} \frac{dy}{y^{n+1}} \right| \\[+4mm] &= \left| [y^n] \sum_{j \geq 0}(1-x)^{-(j+1)}y^j \right| \\[+4mm] &= |1-x|^{-(n+1)} \\[+4mm] &\leq \rho^{-(n+1)} \end{aligned}$$

When \(x \in \mathcal{N}'\),

$$ \begin{aligned}\left| \frac{1}{(2\pi i)} \int_{|y| = 1/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right|  &= \left| \frac{1}{(2\pi i)} \int_{|y| = 1/4} \frac{1/(1-x)}{1-\frac{y}{1-x}} \frac{dy}{y^{n+1}} \right| \\[+4mm] &= \left| [y^n] \sum_{j \geq 0}(1-x)^{-(j+1)}y^j \right| \\[+4mm] &= |1-x|^{-(n+1)} \\[+4mm] &\leq \rho^{-(n+1)} \end{aligned}$$

When \(x \in \mathcal{N}'\),

$$ \begin{aligned}\left| \frac{1}{(2\pi i)} \int_{|y| = 1/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right|  &= \left| \frac{1}{(2\pi i)} \int_{|y| = 1/4} \frac{1/(1-x)}{1-\frac{y}{1-x}} \frac{dy}{y^{n+1}} \right| \\[+4mm] &= \left| [y^n] \sum_{j \geq 0}(1-x)^{-(j+1)}y^j \right| \\[+4mm] &= |1-x|^{-(n+1)} \\[+4mm] &\leq \rho^{-(n+1)} \end{aligned}$$

When \(x \in \mathcal{N}'\),

$$ \begin{aligned}\left| \frac{1}{(2\pi i)} \int_{|y| = 1/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right|  &= \left| \frac{1}{(2\pi i)} \int_{|y| = 1/4} \frac{1/(1-x)}{1-\frac{y}{1-x}} \frac{dy}{y^{n+1}} \right| \\[+4mm] &= \left| [y^n] \sum_{j \geq 0}(1-x)^{-(j+1)}y^j \right| \\[+4mm] &= |1-x|^{-(n+1)} \\[+4mm] &\leq \rho^{-(n+1)} \end{aligned}$$

When \(x \in \mathcal{N}'\),

$$ \begin{aligned}\left| \frac{1}{(2\pi i)} \int_{|y| = 1/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right|  &= \left| \frac{1}{(2\pi i)} \int_{|y| = 1/4} \frac{1/(1-x)}{1-\frac{y}{1-x}} \frac{dy}{y^{n+1}} \right| \\[+4mm] &= \left| [y^n] \sum_{j \geq 0}(1-x)^{-(j+1)}y^j \right| \\[+4mm] &= |1-x|^{-(n+1)} \\[+4mm] &\leq \rho^{-(n+1)} \\[+4mm] \end{aligned}$$

When \(x \in \mathcal{N}'\),

$$ \begin{aligned}\left| \frac{1}{(2\pi i)} \int_{|y| = 1/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right|  &= \left| \frac{1}{(2\pi i)} \int_{|y| = 1/4} \frac{1/(1-x)}{1-\frac{y}{1-x}} \frac{dy}{y^{n+1}} \right| \\[+4mm] &= \left| [y^n] \sum_{j \geq 0}(1-x)^{-(j+1)}y^j \right| \\[+4mm] &= |1-x|^{-(n+1)} \\[+4mm] &\leq \rho^{-(n+1)} \\[+4mm] &= (1.35\dots)^{n+1} \end{aligned}$$

Thus

$$\begin{aligned} \left|I-I_\text{loc}\right| &= \left|\frac{1}{(2\pi i)^2} \int_{\mathcal{N}'} \left( \int_{|y| = 1/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right) \frac{dx}{x^{n+1}}\right| \\[+7mm]& \leq \frac{\text{length}(\mathcal{N}')}{2\pi} \, \rho^{-(n+1)} \, 2^{n+1} \\[+5mm]&= \frac{3}{8 \rho \pi}\left(\frac{2}{\rho}\right)^n \\[+5mm]&\leq \frac{3}{8 \rho \pi}\left(2.72\right)^n \end{aligned}$$

grows exponentially slower than \(I = f_{n,n}\).

Thus

$$\begin{aligned} \left|I-I_\text{loc}\right| &= \left|\frac{1}{(2\pi i)^2} \int_{\mathcal{N}'} \left( \int_{|y| = 1/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right) \frac{dx}{x^{n+1}}\right| \\[+7mm]& \leq \frac{\text{length}(\mathcal{N}')}{2\pi} \, \rho^{-(n+1)} \, 2^{n+1} \\[+5mm]&= \frac{3}{8 \rho \pi}\left(\frac{2}{\rho}\right)^n \\[+5mm]&\leq \frac{3}{8 \rho \pi}\left(2.72\right)^n \end{aligned}$$

grows exponentially slower than \(I = f_{n,n}\).

Thus

$$\begin{aligned} \left|I-I_\text{loc}\right| &= \left|\frac{1}{(2\pi i)^2} \int_{\mathcal{N}'} \left( \int_{|y| = 1/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right) \frac{dx}{x^{n+1}}\right| \\[+7mm]& \leq \frac{\text{length}(\mathcal{N}')}{2\pi} \, \rho^{-(n+1)} \, 2^{n+1} \\[+5mm]&= \frac{3}{8 \rho \pi}\left(\frac{2}{\rho}\right)^n \\[+5mm]&\leq \frac{3}{8 \rho \pi}\left(2.72\right)^n \end{aligned}$$

grows exponentially slower than \(I = f_{n,n}\).

Thus

$$\begin{aligned} \left|I-I_\text{loc}\right| &= \left|\frac{1}{(2\pi i)^2} \int_{\mathcal{N}'} \left( \int_{|y| = 1/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right) \frac{dx}{x^{n+1}}\right| \\[+7mm]& \leq \frac{\text{length}(\mathcal{N}')}{2\pi} \, \rho^{-(n+1)} \, 2^{n+1} \\[+5mm]&= \frac{3}{8 \rho \pi}\left(\frac{2}{\rho}\right)^n \\[+5mm]&\leq \frac{3}{8 \rho \pi}\left(2.72\right)^n \end{aligned}$$

grows exponentially slower than \(I = f_{n,n}\).

Thus

$$\begin{aligned} \left|I-I_\text{loc}\right| &= \left|\frac{1}{(2\pi i)^2} \int_{\mathcal{N}'} \left( \int_{|y| = 1/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right) \frac{dx}{x^{n+1}}\right| \\[+7mm]& \leq \frac{\text{length}(\mathcal{N}')}{2\pi} \, \rho^{-(n+1)} \, 2^{n+1} \\[+5mm]&= \frac{3}{8 \rho \pi}\left(\frac{2}{\rho}\right)^n \\[+5mm]&\leq \frac{3}{8 \rho \pi}\left(2.72\right)^n \end{aligned}$$

Note: This is exponentially slower than \(I = f_{n,n}\).

Now we introduce

$$I_\text{out} = \frac{1}{(2\pi i)^2} \int_{\mathcal{N}} \left( \int_{{\color{red}|y| = 3/4}} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right) \frac{dx}{x^{n+1}}. $$

Now we introduce

$$I_\text{out} = \frac{1}{(2\pi i)^2} \int_{\mathcal{N}} \left( \int_{{\color{red}|y| = 3/4}} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right) \frac{dx}{x^{n+1}}. $$

The maximum modulus bound implies

$$\left| I_\text{out} \right| \leq C \; 2^n \; \left(\frac{4}{3}\right)^n = O\left(\left(\frac{8}{3}\right)^n \right) $$

Now we introduce

$$I_\text{out} = \frac{1}{(2\pi i)^2} \int_{\mathcal{N}} \left( \int_{\color{red}|y| = 3/4} \frac{1}{1-x-y} \frac{dy}{y^{n+1}} \right) \frac{dx}{x^{n+1}}. $$

The maximum modulus bound implies

$$\left| I_\text{out} \right| \leq C \; 2^n \; \left(\frac{4}{3}\right)^n = O\left(\left(\frac{8}{3}\right)^n \right) $$

Note: This is exponentially slower than \(I-I_\text{out}\).

Finally, we set

$$\chi = I_\text{loc}-I_\text{out}$$

so that

$$\begin{aligned} \left|\binom{2n}{n} - \chi \right| &= \left|I-\left(I_\text{loc} - I_\text{out} \right) \right| \\[+5mm] &\leq \left|I-I_\text{loc} \right| + \left|I_\text{out} \right| \\[+5mm] &= O\left(2.72^n\right)\end{aligned}$$

For a rough comparison, using Sage to compute the integrals numerically we get

$$\begin{aligned} \binom{20}{10} &= {\color{red}18}4756 \\[+5mm] \left[\chi\right] &= {\color{red}18}5572 \end{aligned} $$

$$\begin{aligned} \binom{40}{20} &= {\color{red}1378}46528820 \\[+5mm] \left[\chi\right] &= {\color{red}1378}37443285 \end{aligned} $$

$$\begin{aligned} \binom{200}{100} &= {\color{red}90548514656103281}165404177077484163874504589675413336841320 \\[+5mm] \left[\chi\right] &= {\color{red}90548514656103281}244226514758732425815049173009882161476501 \end{aligned} $$

For a rough comparison, using Sage to compute the integrals numerically we get

$$\begin{aligned} \binom{20}{10} &= {\color{red}18}4756 \\[+5mm] \left[\chi\right] &= {\color{red}18}5572 \end{aligned} $$

$$\begin{aligned} \binom{40}{20} &= {\color{red}1378}46528820 \\[+5mm] \left[\chi\right] &= {\color{red}1378}37443285 \end{aligned} $$

$$\begin{aligned} \binom{200}{100} &= {\color{red}90548514656103281}165404177077484163874504589675413336841320 \\[+5mm] \left[\chi\right] &= {\color{red}90548514656103281}244226514758732425815049173009882161476501 \end{aligned} $$

The exp. growth of \(\binom{2n}{n} - \chi\) numerically approaches \(2/\rho\approx2.72\).

For fixed \(x \in \mathcal{N}\) the function \(1/(1-x-y)\) has a unique pole with \(1/4<|y|<3/4\).

The univariate residue theorem implies
 

$$ \frac{1}{2\pi i}\left(\int_{|y|=3/4}\frac{1}{1-x-y}\,\frac{dy}{y^{n+1}} - \int_{|y|=1/4}\frac{1}{1-x-y}\,\frac{dy}{y^{n+1}} \right) = -(1-x)^{-(n+1)}. $$

For fixed \(x \in \mathcal{N}\) the function \(1/(1-x-y)\) has a unique pole with \(1/4<|y|<3/4\).

The univariate residue theorem implies
 

$$ \frac{1}{2\pi i}\left(\int_{|y|=3/4}\frac{1}{1-x-y}\,\frac{dy}{y^{n+1}} - \int_{|y|=1/4}\frac{1}{1-x-y}\,\frac{dy}{y^{n+1}} \right) = -(1-x)^{-(n+1)}. $$

Thus our diagonal sequence is well-approximated by

$$ \chi = \frac{1}{2\pi i} \int_\mathcal{N} \frac{dx}{x^{n+1}(1-x)^{n+1}}.$$

Note: In fact

$$ \binom{2n}{n} = \frac{1}{2\pi i} \int_{|x|=1/2}\frac{dx}{x^{n+1}(1-x)^{n+1}}.$$

Our (exponentially negligble) error comes from localizing to \(\mathcal{N}\).

Note: In fact

$$ \binom{2n}{n} = \frac{1}{2\pi i} \int_{|x|=1/2}\frac{dx}{x^{n+1}(1-x)^{n+1}}.$$

Our (exponentially negligble) error comes from localizing to \(\mathcal{N}\).

We localize to avoid singularities in \(I_\text{out}\).

Note: In fact

$$ \binom{2n}{n} = \frac{1}{2\pi i} \int_{|x|=1/2}\frac{dx}{x^{n+1}(1-x)^{n+1}}.$$

Our (exponentially negligble) error comes from localizing to \(\mathcal{N}\).

We localize to avoid singularities in \(I_\text{out}\).

In this case there is no need to localize if we take \(|y|>3/2\) for the domain of integration in \(I_\text{out}\). This does not work in general.

The integral

$$ \chi = \frac{1}{2\pi i} \int_\mathcal{N} \frac{dx}{x^{n+1}(1-x)^{n+1}}$$

is perfect for the saddle-point method. Set \(x=e^{i\theta}/2\) so

$$ \chi = \frac{4^n}{2 \pi} \int_{-\pi/4}^{\pi/4} A(\theta)e^{-n\phi(\theta)}d\theta$$
where

$$A(\theta) = \frac{1}{1-e^{i\theta}/2} \quad\text{and}\quad \phi(\theta) = \log\left(2-e^{i\theta}\right) + i\theta.$$

Since

$$ A(\theta) = \frac{1}{1-e^{i\theta}/2} = 2 + 2i\theta + \cdots \quad\text{and}\quad \phi(\theta) = \theta^2 + i\theta^3 + \cdots$$

and some other conditions we can approximate

$$ \chi \;=\; \frac{4^n}{2 \pi} \int_{-\pi/4}^{\pi/4} A(\theta)e^{-n\phi(\theta)}d\theta \;\sim\; \frac{4^n}{2\pi} \int_{-\infty}^\infty 2 e^{-n\theta^2}d\theta \;=\; \frac{4^n}{\sqrt{\pi n}}.$$

The other conditions needed are

• \(\phi(0)=\phi'(0)=0\) while \(\phi''(0)\neq0\) (quadratic near the origin)
   
• real part of \(\phi\) is non-negative on domain of integration
   
• \(\phi'(\theta)\neq0\) on the domain of integration unless \(\theta=0\)

The other conditions needed are

• \(\phi(0)=\phi'(0)=0\) while \(\phi''(0)\neq0\) (quadratic near the origin)
   
• real part of \(\phi\) is non-negative on domain of integration
   
• \(\phi'(\theta)\neq0\) on the domain of integration unless \(\theta=0\)

The "magic" of ACSV: By looking for optimal bounds on exponential growth we get saddle-point integrals.

Now suppose we want asymptotics of the \((r,s)\)-diagonal

$$ f_{rn,sn} = \frac{1}{(2\pi i)^2} \int_{T(a,b)} \frac{1}{1-x-y} \, \frac{dxdy}{x^{rn+1}y^{sn+1}}.$$

Now suppose we want asymptotics of the \((r,s)\)-diagonal

$$ f_{rn,sn} = \frac{1}{(2\pi i)^2} \int_{T(a,b)} \frac{1}{1-x-y} \, \frac{dxdy}{x^{rn+1}y^{sn+1}}.$$

The singular variety \(\mathcal{V}\) and domain of convergence \(\mathcal{D}\) are unchanged.

Now suppose we want asymptotics of the \((r,s)\)-diagonal

$$ f_{rn,sn} = \frac{1}{(2\pi i)^2} \int_{T(a,b)} \frac{1}{1-x-y} \, \frac{dxdy}{x^{rn+1}y^{sn+1}}.$$

The singular variety \(\mathcal{V}\) and domain of convergence \(\mathcal{D}\) are unchanged.

Again we get a bound
 

$$ \limsup_{n \rightarrow \infty} |f_{rn,sn}|^{1/n} \leq |a|^{-r}|b|^{-s} \quad \text{ for all } (a,b) \in \mathcal{D}.$$

To minimize this upper bound we define the height function

$$h_{r,s}(x,y) = -r\log|x|-s\log|y|.$$

Working on \(\partial \mathcal{D}\) we set \(|x|=t\) and solve \(g'(t)=0\) where

$$ g(t) = -r\log t - s \log (1-t).$$

To minimize this upper bound we define the height function

$$h_{r,s}(x,y) = -r\log|x|-s\log|y|.$$

Working on \(\partial \mathcal{D}\) we set \(|x|=t\) and solve \(g'(t)=0\) where

$$ g(t) = -r\log t - s \log (1-t).$$

This gives the minimum \((|x|,|y|) = \boldsymbol{\sigma}_{r,s} = \left(\frac{r}{r+s},\frac{s}{r+s}\right)\).

To minimize this upper bound we define the height function

$$h_{r,s}(x,y) = -r\log|x|-s\log|y|.$$

Working on \(\partial \mathcal{D}\) we set \(|x|=t\) and solve \(g'(t)=0\) where

$$ g(t) = -r\log t - s \log (1-t).$$

This gives the minimum \((|x|,|y|) = \boldsymbol{\sigma}_{r,s} = \left(\frac{r}{r+s},\frac{s}{r+s}\right)\).

The point \(\boldsymbol{\sigma}_{r,s}\) is the only singularity on \(T(\boldsymbol{\sigma}_{r,s})\).

Note: If \(p=\log|x|\) and \(q=\log|y|\) then we want to minimize the linear function \(\tilde{h}_{r,s}(p,q) = -(r,s) \cdot (p,q)\) on \(B = \text{Relog}(\mathcal{D})\).

\(B\)

\(r_1\)

\(r_2\)

\(r_3\)

Note: If \(p=\log|x|\) and \(q=\log|y|\) then we want to minimize the linear function \(\tilde{h}_{r,s}(p,q) = -(r,s) \cdot (p,q)\) on \(B = \text{Relog}(\mathcal{D})\).

\(B\)

A similar analysis shows

$$ f_{rn,sn} \sim \left(\frac{r+s}{r}\right)^{rn}\left(\frac{r+s}{s}\right)^{rn} \cdot \frac{1}{2\pi} \int_{-\delta}^{\delta}A(\theta)e^{-n\phi(\theta)}d\theta$$

where

$$A(\theta) = \frac{r+s}{s} + O(\theta) \quad\text{ and }\quad \phi(\theta) = \frac{r(r+s)}{2s}\theta^2 + O(\theta^3). $$

A similar analysis shows

$$ f_{rn,sn} \sim \left(\frac{r+s}{r}\right)^{rn}\left(\frac{r+s}{s}\right)^{rn} \cdot \frac{1}{2\pi} \int_{-\delta}^{\delta}A(\theta)e^{-n\phi(\theta)}d\theta$$

where

$$A(\theta) = \frac{r+s}{s} + O(\theta) \quad\text{ and }\quad \phi(\theta) = \frac{r(r+s)}{2s}\theta^2 + O(\theta^3). $$
Thus \(\displaystyle f_{rn,sn} \sim \left(\frac{r+s}{r}\right)^{rn}\left(\frac{r+s}{s}\right)^{rn} \frac{1}{2\pi} \int_{-\infty}^\infty e^{-n\left(\frac{r(r+s)}{2s}\right)\theta^2}d\theta\).

A similar analysis shows

$$ f_{rn,sn} \sim \left(\frac{r+s}{r}\right)^{rn}\left(\frac{r+s}{s}\right)^{rn} \cdot \frac{1}{2\pi} \int_{-\delta}^{\delta}A(\theta)e^{-n\phi(\theta)}d\theta$$

where

$$A(\theta) = \frac{r+s}{s} + O(\theta) \quad\text{ and }\quad \phi(\theta) = \frac{r(r+s)}{2s}\theta^2 + O(\theta^3). $$
Thus \(\displaystyle \binom{rn+sn}{rn} \sim \left(\frac{r+s}{r}\right)^{rn}\left(\frac{r+s}{s}\right)^{rn} \frac{\sqrt{r+s}}{\sqrt{2rs\pi n}}\).

Consider the Laurent expansion

$$\frac{1}{1-x-y} = \sum_{i,j \geq 0}\binom{i}{j}(-1)^{j+1}y^jx^{-i-1}$$

with domain of convergence \(\mathcal{D}_2 = \{1+|y|<|x|\}\).

Consider the Laurent expansion

$$\frac{1}{1-x-y} = \sum_{i,j \geq 0}\binom{i}{j}(-1)^{j+1}y^jx^{-i-1}$$

with domain of convergence \(\mathcal{D}_2 = \{1+|y|<|x|\}\).

It now makes sense to consider directions with negative coordinates.

Consider the Laurent expansion

$$\frac{1}{1-x-y} = \sum_{i,j \geq 0}\binom{i}{j}(-1)^{j+1}y^jx^{-i-1}$$

with domain of convergence \(\mathcal{D}_2 = \{1+|y|<|x|\}\).

It now makes sense to consider directions with negative coordinates.

Which directions yield interesting (non-zero) asymptotics?

A bound on exponential growth is still given by

$$h_\mathbf{r}(x,y) = -r \log |x| - s\log|y|.$$

Parameterizing \(\mathcal{V} \cap \mathcal{D}_2\) by \((x,y) = (1+t,t)\) for \(t>0\), we want to minimize the function
 

$$ g(t) = -r \log(1+t) - s \log(t).$$
 

A bound on exponential growth is still given by

$$h_\mathbf{r}(x,y) = -r \log |x| - s\log|y|.$$

Parameterizing \(\mathcal{V} \cap \mathcal{D}_2\) by \((x,y) = (1+t,t)\) for \(t>0\), we want to minimize the function
 

$$ g(t) = -r \log(1+t) - s \log(t).$$
If \(g\) has no minimum then \(f_{n\mathbf{r}}\) decays super-exponentially. This implies it is eventually zero.

The series expansions
 

$$ \begin{aligned} g(t) &= s \log (t^{-1})-rt + rt^2/2 + \cdots &&(t \rightarrow 0^+) \\ g(t) &= -(r+s) \log (t)-rt^{-1} + rt^{-2}/2 + \cdots &&(t \rightarrow \infty) \end{aligned}$$
show \(g\) is unbounded below if \(s<0\) or \(r+s>0\).

The series expansions
 

$$ \begin{aligned} g(t) &= s \log (t^{-1})-rt + rt^2/2 + \cdots &&(t \rightarrow 0^+) \\ g(t) &= -(r+s) \log (t)-rt^{-1} + rt^{-2}/2 + \cdots &&(t \rightarrow \infty) \end{aligned}$$
show \(g\) is unbounded below if \(s<0\) or \(r+s>0\).

Note: The bounded directions have nonpositive dot product with all elements of the recession cone of \(B_2 = \text{Relog}(\mathcal{D}_2)\).

They form the polar cone to the recession cone of \(B_2\).

If the minimum exists it is again given by

$$ \boldsymbol{\sigma} = \left(\frac{r}{r+s},\frac{s}{r+s}\right).$$

The same arguments apply (after care with localizing) to give

$$ \binom{-nr-1}{ns}(-1)^{ns+1} = f_{rn,sn} \sim \left(\frac{r+s}{r}\right)^{rn}\left(\frac{r+s}{s}\right)^{sn}\left(\frac{r+s}{s}\right)\sqrt{\frac{s}{2r(r+s)\pi n}}.$$

 

If the minimum exists it is again given by

$$ \boldsymbol{\sigma} = \left(\frac{r}{r+s},\frac{s}{r+s}\right).$$

The same arguments apply (after care with localizing) to give

$$ \binom{-nr-1}{ns}(-1)^{ns+1} = f_{rn,sn} \sim \left(\frac{r+s}{r}\right)^{rn}\left(\frac{r+s}{s}\right)^{sn}\left(\frac{r+s}{s}\right)\sqrt{\frac{s}{2r(r+s)\pi n}}.$$

Note: Need to be careful with signs when simplifying.

Fix a rational function \(F(\mathbf{z}) = \frac{G(\mathbf{z})}{H(\mathbf{z})}\) with \(G\) and \(H\) coprime.

Let \(\mathcal{D}\) denote the domain of convergence of an expansion

$$ F(\mathbf{z}) = \sum_{\mathbf{i}\in\mathbb{Z}^d}f_\mathbf{i}\mathbf{z}^\mathbf{i}.$$

We determine asymptotics of \(f_{n\mathbf{r}}\) as \(n\rightarrow\infty\) where \(\mathbf{r}\in\mathbb{R}_*^d\).

Statements with \(f_{n\mathbf{r}}\) are interpreted to hold when \(n\mathbf{r}\in\mathbb{Z}^d\).

The singularities of \(F\) form the singular variety \(\mathcal{V}=\mathcal{V}(H)\).

We assume \(H\) is square-free until noted. (If not replace \(H\) with the product of its distinct irreducible factors.)

The singularities of \(F\) form the singular variety \(\mathcal{V}=\mathcal{V}(H)\).

We assume \(H\) is square-free until noted. (If not replace \(H\) with the product of its distinct irreducible factors.)

Under this assumption, a smooth point of \(\mathcal{V}\) is a point where at least one partial derivative of \(H\) does not vanish.

The singularities of \(F\) form the singular variety \(\mathcal{V}=\mathcal{V}(H)\).

We assume \(H\) is square-free until noted. (If not replace \(H\) with the product of its distinct irreducible factors.)

Under this assumption, a smooth point of \(\mathcal{V}\) is a point where at least one partial derivative of \(H\) does not vanish.

Note: The implicit function theorem implies \(\mathcal{V}\) is a complex manifold near any smooth point.

Again we start with the representation 

$$f_{n\mathbf{r}} = \frac{1}{(2\pi i)^d} \int_{T(\mathbf{w})} F(\mathbf{z}) \frac{d\mathbf{z}}{\mathbf{z}^{n\mathbf{r}+\mathbf{1}}}$$

for \(\mathbf{w} \in \mathcal{D}\) and define the height function

$$h_\mathbf{r}(\mathbf{z}) = - \sum_{j=1}^d r_j \log|z_j|,$$

so that \(\displaystyle\limsup_{n\rightarrow\infty}|f_{n\mathbf{r}}|^{1/n} \leq e^{h_\mathbf{r}(\mathbf{w})}\) for all \(\mathbf{w} \in \mathcal{D}\).

If \(h_\mathbf{r}\) is unbounded below on \(\overline{\mathcal{D}}\) then \(f_{n\mathbf{r}}\) is eventually zero.

If \(h_\mathbf{r}\) is unbounded below on \(\overline{\mathcal{D}}\) then \(f_{n\mathbf{r}}\) is eventually zero.

If \(h_\mathbf{r}\) is bounded below but the minimum is not achieved then there is an amoeba limit direction normal to \(\mathbf{r}\).

If \(h_\mathbf{r}\) is unbounded below on \(\overline{\mathcal{D}}\) then \(f_{n\mathbf{r}}\) is eventually zero.

If \(h_\mathbf{r}\) is bounded below but the minimum is not achieved then there is an amoeba limit direction normal to \(\mathbf{r}\).

Otherwise the minimum is achieved on \(\partial \mathcal{D}\).
For every \(\mathbf{w} \in \partial \mathcal{D}\) there is a minimal point in \(T(\mathbf{w})\).

If \(h_\mathbf{r}\) is unbounded below on \(\overline{\mathcal{D}}\) then \(f_{n\mathbf{r}}\) is eventually zero.

If \(h_\mathbf{r}\) is bounded below but the minimum is not achieved then there is an amoeba limit direction normal to \(\mathbf{r}\).

Otherwise the minimum is achieved on \(\partial \mathcal{D}\).
For every \(\mathbf{w} \in \partial \mathcal{D}\) there is a minimal point in \(T(\mathbf{w})\).

Recall: If \(\mathbf{w}\) is a minimal point then \((\nabla_\text{log} H)(\mathbf{w})\) is a scalar multiple of a real vector.

Proposition: If \(\mathbf{w}\in\mathcal{V} \cap \partial\mathcal{D}\) and \((\nabla_\text{log} H)(\mathbf{w}) = \lambda \mathbf{r}\) then \(\mathbf{w}\) is either a minimizer or a maximizer of \(h_\mathbf{r}\) on \(\overline{\mathcal{D}}\).

Proof: The log-gradient is the normal to a supporting hyperplane of \(B = \text{Relog}(\mathcal{D})\) at \(\text{Relog}(\mathbf{w})\).

If \((\nabla_\text{log} H)(\mathbf{w}) = \lambda \mathbf{r}\) then the matrix

$$ M = \begin{pmatrix} (\nabla_\text{log} H)(\mathbf{w}) \\ \mathbf{r} \end{pmatrix} = \begin{pmatrix} z_1H_{z_1}(\mathbf{w}) &\cdots &z_dH_{z_d}(\mathbf{w}) \\ r_1 &\cdots &r_d \end{pmatrix}$$
is rank deficient, so all \(2 \times 2\) minors vanish.

If \((\nabla_\text{log} H)(\mathbf{w}) = \lambda \mathbf{r}\) then the matrix

$$ M = \begin{pmatrix} (\nabla_\text{log} H)(\mathbf{w}) \\ \mathbf{r} \end{pmatrix} = \begin{pmatrix} z_1H_{z_1}(\mathbf{w}) &\cdots &z_dH_{z_d}(\mathbf{w}) \\ r_1 &\cdots &r_d \end{pmatrix}$$
is rank deficient, so all \(2 \times 2\) minors vanish.

Smooth Critical Point Equations

$$H(\mathbf{z})=0, \qquad \frac{z_1}{r_1}H_{z_1}(\mathbf{z}) = \frac{z_2}{r_2}H_{z_2}(\mathbf{z}) = \cdots = \frac{z_d}{r_d}H_{z_d}(\mathbf{z})$$

Another Perspective

Let \(\mathcal{V}_* = \mathcal{V} \cap \mathbb{C}_*^d\) and define the map

$$ \phi_\mathbf{r}(\mathbf{z}) = -\sum_{j=1}^d r_j \log z_j$$from \(\mathcal{V}_*\) to \(\mathbb{C}\).

The critical points of \(\phi_\mathbf{r}\) as a map of complex manifolds are the points where \(\nabla \phi_\mathbf{r}\) is parallel to \(\nabla H\).

These are the solutions of the critical point equations.

Yet Another Perspective

In algebraic statistics this is solving the logarithmic derivative of the likelihood function \(\ell_\mathbf{u}\) with observed data \(\mathbf{u}=\mathbf{r}\).

Algebraic statistics is interested in characterizing properties of the critical points (for example, counting them based on topological properties of \(\mathcal{V}\)).

Yet Another Perspective

In algebraic statistics this is solving the logarithmic derivative of the likelihood function \(\ell_\mathbf{u}\) with observed data \(\mathbf{u}=\mathbf{r}\).

Algebraic statistics is interested in characterizing properties of the critical points (for example, counting them based on topological properties of \(\mathcal{V}\)).

There is potential to better exploit such results in ACSV.
 

In ACSV we have to find special critical points and then do further analysis to find asymptotics.

A smooth critical point \(\mathbf{w}\in\mathbb{C}^d\) such that \(\mathbf{r}\) points away from \(B\) at \(\text{Relog}(\mathbf{w})\) is called a smooth contributing point.

A smooth minimal contributing point is a minimizer of \(h_\mathbf{r}\) on \(\overline{\mathcal{D}}\).

A smooth critical point \(\mathbf{w}\in\mathbb{C}^d\) such that \(\mathbf{r}\) points away from \(B\) at \(\text{Relog}(\mathbf{w})\) is called a smooth contributing point.

A smooth minimal contributing point is a minimizer of \(h_\mathbf{r}\) on \(\overline{\mathcal{D}}\).

Note: For a power series expansion any minimal smooth critical point is a smooth contributing point.

A smooth critical point \(\mathbf{w}\in\mathbb{C}^d\) such that \(\mathbf{r}\) points away from \(B\) at \(\text{Relog}(\mathbf{w})\) is called a smooth contributing point.

A smooth minimal contributing point is a minimizer of \(h_\mathbf{r}\) on \(\overline{\mathcal{D}}\).

Note: For a power series expansion any minimal smooth critical point is a smooth contributing point.

Note: Bounded components of the amoeba complement will always have minimizers and maximizers.

\(\mathcal{C}(H)\)

$$1 - x - y - 6xy - x^2y^2$$

$$\mathbf{r}=(1,1)$$

\(\mathcal{C}(H)\)

$$1 - x - y - 6xy - x^2y^2$$

$$\mathbf{r}=(1,1)$$

$$1 - x - y - 6xy - x^2y^2$$

\(\mathcal{C}(H)\)

$$\mathbf{r}=(1,1)$$

$$1 - x - y - 6xy - x^2y^2$$

\(\mathcal{C}(H)\)

$$\mathbf{r}=(1,1)$$

$$1 - x - y - 6xy - x^2y^2$$

\(\mathcal{C}(H)\)

$$\mathbf{r}=(1,1)$$

$$1 - x - y - 6xy - x^2y^2$$

\(\mathcal{C}(H)\)

$$\mathbf{r}=(1,1)$$

$$1 - x - y - 6xy - x^2y^2$$

\(\mathcal{C}(H)\)

$$\mathbf{r}=(1,1)$$

$$1 - x - y - 6xy - x^2y^2$$

\(\mathcal{C}(H)\)

$$\mathbf{r}=(1,1)$$

$$1 - x - y - 6xy - x^2y^2$$

\(\mathcal{C}(H)\)

$$\mathbf{r}=(1,1)$$

A minimal point \(\mathbf{w}\) is called finitely minimal if \(T(\mathbf{w}) \cap \mathcal{V}\) is finite and strictly minimal if \(T(\mathbf{w}) \cap \mathcal{V} = \{\mathbf{w}\}\).

Because of smoothness, some partial derivative \(H_{z_j}(\mathbf{w})\neq0\). Without loss of generality, we assume that \(H_{z_d}(\mathbf{w})\neq0\).

Thus we can parameterize \(z_d = g(z_1,\dots,z_{d-1})\) on \(\mathcal{V}\) near \(\mathbf{w}\).

A minimal point \(\mathbf{w}\) is called finitely minimal if \(T(\mathbf{w}) \cap \mathcal{V}\) is finite and strictly minimal if \(T(\mathbf{w}) \cap \mathcal{V} = \{\mathbf{w}\}\).

Because of smoothness, some partial derivative \(H_{z_j}(\mathbf{w})\neq0\). Without loss of generality, we assume that \(H_{z_d}(\mathbf{w})\neq0\).

Thus we can parameterize \(z_d = g(z_1,\dots,z_{d-1})\) on \(\mathcal{V}\) near \(\mathbf{w}\).

To simplify notation we write \(\widehat{\mathbf{z}}=(z_1,\dots,z_{d-1})\).

A minimal point \(\mathbf{w}\) is called finitely minimal if \(T(\mathbf{w}) \cap \mathcal{V}\) is finite and strictly minimal if \(T(\mathbf{w}) \cap \mathcal{V} = \{\mathbf{w}\}\).

Because of smoothness, some partial derivative \(H_{z_j}(\mathbf{w})\neq0\). Without loss of generality, we assume that \(H_{z_d}(\mathbf{w})\neq0\).

Thus we can parameterize \(z_d = g(z_1,\dots,z_{d-1})\) on \(\mathcal{V}\) near \(\mathbf{w}\).

Start with
$$f_{n\mathbf{r}} = I = \frac{1}{(2\pi i)^d} \int_{T(\widehat{\mathbf{w}})} \left(\int_{|w_d|\pm\epsilon} F(\mathbf{z})\frac{dz_d}{z_d^{nr_d+1}} \right)\frac{d\mathbf{z}}{\widehat{\mathbf{z}}^{n\widehat{\mathbf{r}}+\mathbf{1}}}$$

A minimal point \(\mathbf{w}\) is called finitely minimal if \(T(\mathbf{w}) \cap \mathcal{V}\) is finite and strictly minimal if \(T(\mathbf{w}) \cap \mathcal{V} = \{\mathbf{w}\}\).

Because of smoothness, some partial derivative \(H_{z_j}(\mathbf{w})\neq0\). Without loss of generality, we assume that \(H_{z_d}(\mathbf{w})\neq0\).

Thus we can parameterize \(z_d = g(z_1,\dots,z_{d-1})\) on \(\mathcal{V}\) near \(\mathbf{w}\).

Start with
$$f_{n\mathbf{r}} = I = \frac{1}{(2\pi i)^d} \int_{T(\widehat{\mathbf{w}})} \left(\int_{|w_d|{\color{red}\pm}\epsilon} F(\mathbf{z})\frac{dz_d}{z_d^{nr_d+1}} \right)\frac{d\mathbf{z}}{\widehat{\mathbf{z}}^{n\widehat{\mathbf{r}}+\mathbf{1}}}$$

A minimal point \(\mathbf{w}\) is called finitely minimal if \(T(\mathbf{w}) \cap \mathcal{V}\) is finite and strictly minimal if \(T(\mathbf{w}) \cap \mathcal{V} = \{\mathbf{w}\}\).

Because of smoothness, some partial derivative \(H_{z_j}(\mathbf{w})\neq0\). Without loss of generality, we assume that \(H_{z_d}(\mathbf{w})\neq0\).

Thus we can parameterize \(z_d = g(z_1,\dots,z_{d-1})\) on \(\mathcal{V}\) near \(\mathbf{w}\).

Start with
$$f_{n\mathbf{r}} = I = \frac{1}{(2\pi i)^d} \int_{T(\widehat{\mathbf{w}})} \left(\int_{|w_d|{\color{red}-}\epsilon} F(\mathbf{z})\frac{dz_d}{z_d^{nr_d+1}} \right)\frac{d\mathbf{z}}{\widehat{\mathbf{z}}^{n\widehat{\mathbf{r}}+\mathbf{1}}}$$

Assume \(\mathbf{w}\) is strictly minimal.

Again we define \(I_\text{loc}\) by restricting the torus \(T(\widehat{\mathbf{w}})\) to a neighbourhood of \((w_1,\dots,w_d)\).

Assume \(\mathbf{w}\) is strictly minimal.

Again we define \(I_\text{loc}\) by restricting the torus \(T(\widehat{\mathbf{w}})\) to a neighbourhood of \((w_1,\dots,w_d)\).

Again we define \(I_\text{out}\) by moving \(|z_d|=|w_d|-\epsilon\) to \(|z_d|=|w_d|+\epsilon\).

Assume \(\mathbf{w}\) is strictly minimal.

Again we define \(I_\text{loc}\) by restricting the torus \(T(\widehat{\mathbf{w}})\) to a neighbourhood of \((w_1,\dots,w_d)\).

Again we define \(I_\text{out}\) by moving \(|z_d|=|w_d|-\epsilon\) to \(|z_d|=|w_d|+\epsilon\).

When \(\mathbf{w}\) is a strictly minimal smooth contributing point then \(I\) equals \(\chi = I_\text{loc} - I_\text{out}\) up to an exponentially negligible error.

Again we simplify \(\chi = I_\text{loc} - I_\text{out}\) by taking a 1-dimensional residue at \(z_d = g(z_1,\dots,z_{d-1})\).

What remains is a \((d-1)\)-dimensional integral over a neighbourhood \(\mathcal{N}\) of \(\widehat{\mathbf{w}}\) in \(T(\widehat{\mathbf{w}})\).

Again we simplify \(\chi = I_\text{loc} - I_\text{out}\) by taking a 1-dimensional residue at \(z_d = g(z_1,\dots,z_{d-1})\).

What remains is a \((d-1)\)-dimensional integral over a neighbourhood \(\mathcal{N}\) of \(\widehat{\mathbf{w}}\) in \(T(\widehat{\mathbf{w}})\).

We end up with an expression of the form
 

$$\chi = \frac{\mathbf{w}^{-n\mathbf{r}}}{(2\pi)^{d-1}} \int_{\mathcal{N}'} A(\boldsymbol{\theta})e^{-n\phi(\boldsymbol{\theta})} d\boldsymbol{\theta}.$$
 

Again we simplify \(\chi = I_\text{loc} - I_\text{out}\) by taking a 1-dimensional residue at \(z_d = g(z_1,\dots,z_{d-1})\).

What remains is a \((d-1)\)-dimensional integral over a neighbourhood \(\mathcal{N}\) of \(\widehat{\mathbf{w}}\) in \(T(\widehat{\mathbf{w}})\).

We end up with an expression of the form
 

$$\chi = \frac{\mathbf{w}^{-n\mathbf{r}}}{(2\pi)^{d-1}} \int_{\mathcal{N}'} A(\boldsymbol{\theta})e^{-n\phi(\boldsymbol{\theta})} d\boldsymbol{\theta}.$$
The residue depends on the multiplicities of the factors of \(H\).

When \(H\) has a simple pole at \(\mathbf{w}\) then
 

$$\chi = \frac{\mathbf{w}^{-n\mathbf{r}}}{(2\pi)^{d-1}} \int_{\mathcal{N}'} A(\boldsymbol{\theta})e^{-n\phi(\boldsymbol{\theta})} d\boldsymbol{\theta}$$

where
 

$$\begin{aligned} A(\mathbf{\theta}) &= \frac{-\text{sgn}(r_d)G(\widehat{\mathbf{z}},g(\widehat{\mathbf{z}}))}{g(\widehat{\mathbf{z}})H_{z_d}(\widehat{\mathbf{z}},g(\widehat{\mathbf{z}}))} \\[+5mm] \phi(\mathbf{\theta}) &= r_d\log\left(\frac{g(\widehat{\mathbf{z}})}{g(w_1,\dots,w_{d-1})}\right) + i(\widehat{\mathbf{r}} \cdot \boldsymbol{\theta}) \end{aligned}$$
after setting \(\widehat{\mathbf{z}} = \left(w_1e^{i\theta_1},\dots,w_{d-1}e^{i\theta_{d-1}}\right)\).

When \(H\) has a simple pole at \(\mathbf{w}\) then
 

$$\chi = \frac{\mathbf{w}^{-n\mathbf{r}}}{(2\pi)^{d-1}} \int_{\mathcal{N}'} A(\boldsymbol{\theta})e^{-n\phi(\boldsymbol{\theta})} d\boldsymbol{\theta}$$

where
 

$$\begin{aligned} A(\mathbf{\theta}) &= \frac{-\text{sgn}(r_d)G(\widehat{\mathbf{z}},g(\widehat{\mathbf{z}}))}{g(\widehat{\mathbf{z}})H_{z_d}(\widehat{\mathbf{z}},g(\widehat{\mathbf{z}}))} \\[+5mm] \phi(\mathbf{\theta}) &= r_d\log\left(\frac{g(\widehat{\mathbf{z}})}{g(w_1,\dots,w_{d-1})}\right) + i(\widehat{\mathbf{r}} \cdot \boldsymbol{\theta}) \end{aligned}$$
This is messy but explicit (and still is for higher multiplicity).

Asymptotic behaviour depends on the Hessian \(\mathcal{H}\) of \(\phi(\boldsymbol{\theta})\) at the origin.

Proposition

If \(\displaystyle U_{i,j} = \frac{w_iw_jH_{z_iz_j}(\mathbf{w})}{w_dH_{z_d}(\mathbf{w})}\) and \(\displaystyle V_i = \frac{r_i}{r_d}\) then

$$ \mathcal{H}_{i,j} = \begin{cases} V_iV_j + U_{i,j} - V_jU_{i,d} - V_iU_{j,d} + V_iV_jU_{d,d} &: i \neq j \\[+3mm] V_i + V_i^2 + U_{i,i} - 2V_iU_{i,d} + V_i^2U_{d,d} &: i=j \end{cases}$$

Our (partially skipped over) careful setup ensures that

• \(\phi(\mathbf{0})=(\nabla \phi)(\mathbf{0})=0\),
   
• the origin is the only point in \(\mathcal{N}'\) where \(\nabla \phi\) vanishes, and
   
• the real part of \(\phi(\boldsymbol{\theta})\) is non-negative in \(\mathcal{N}'\).

Our (partially skipped over) careful setup ensures that

• \(\phi(\mathbf{0})=(\nabla \phi)(\mathbf{0})=0\),
   
• the origin is the only point in \(\mathcal{N}'\) where \(\nabla \phi\) vanishes, and
   
• the real part of \(\phi(\boldsymbol{\theta})\) is non-negative in \(\mathcal{N}'\).

Thus, the saddle-point method applies whenever \(\det \mathcal{H} \neq 0.\) 
We call such a point nondegenerate.

Theorem (Simple Pole Case)

Suppose that \(F(\mathbf{z})\) admits a nondegenerate strictly minimal smooth contributing point \(\mathbf{w}\in\mathbb{C}_*^d\) in the direction \(\mathbf{r}\in\mathbb{R}_*^d\).

Theorem (Simple Pole Case)

Suppose that \(F(\mathbf{z})\) admits a nondegenerate strictly minimal smooth contributing point \(\mathbf{w}\in\mathbb{C}_*^d\) in the direction \(\mathbf{r}\in\mathbb{R}_*^d\).

If \(H_{z_d}(\mathbf{w})\neq0\) then

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

Theorem (Simple Pole Case)

Suppose that \(F(\mathbf{z})\) admits a nondegenerate strictly minimal smooth contributing point \(\mathbf{w}\in\mathbb{C}_*^d\) in the direction \(\mathbf{r}\in\mathbb{R}_*^d\).

If \(H_{z_d}(\mathbf{w})\neq0\) then for any integer \(M\geq0\) 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 \(C_0,\dots,C_M\).

Theorem (Simple Pole Case)

Suppose that \(F(\mathbf{z})\) admits a nondegenerate strictly minimal smooth contributing point \(\mathbf{w}\in\mathbb{C}_*^d\) in the direction \(\mathbf{r}\in\mathbb{R}_*^d\).

If \(H_{z_d}(\mathbf{w})\neq0\) then for any integer \(M\geq0\) there is an expansion

$$ f_{n\mathbf{r}} = \mathbf{w}^{-n\mathbf{r}} n^{(1-d)/2} \frac{(2\pi)^{(1-d)/2}}{\sqrt{\det(r_d\mathcal{H})}} \left(\sum_{j=0}^M C_j (r_dn)^{-j} + O\left(n^{-M-1}\right)\right) $$
If \(\mathbf{w}\) is finitely minimal and all points in \(\mathcal{V} \cap T(\mathbf{w})\) satisfy these conditions then one sums the expansions given by each.

Theorem (General Pole Order)

Suppose that \(F(\mathbf{z})\) admits a nondegenerate strictly minimal smooth contributing point \(\mathbf{w}\in\mathbb{C}_*^d\) in the direction \(\mathbf{r}\in\mathbb{R}_*^d\).

If \(p\) is the smallest integer such that \((\partial_d^pH)(\mathbf{w})\neq0\) then for any \(M\) there are computable constants \(C_0,\dots,C_M\) such that

$$ f_{n\mathbf{r}} = \mathbf{w}^{-n\mathbf{r}} n^{{\color{violet}p-1}+(1-d)/2} \frac{(2\pi)^{(1-d)/2}}{\sqrt{\det(r_d\mathcal{H})}} \left(\sum_{j=0}^M C_j (r_dn)^{-j} + O\left(n^{-M-1}\right)\right). $$

Theorem (General Pole Order)

Suppose that \(F(\mathbf{z})\) admits a nondegenerate strictly minimal smooth contributing point \(\mathbf{w}\in\mathbb{C}_*^d\) in the direction \(\mathbf{r}\in\mathbb{R}_*^d\).

If \(p\) is the smallest integer such that \((\partial_d^pH)(\mathbf{w})\neq0\) then for any \(M\) there are computable constants \(C_0,\dots,C_M\) such that

$$ f_{n\mathbf{r}} = \mathbf{w}^{-n\mathbf{r}} n^{{\color{violet}p-1}+(1-d)/2} \frac{(2\pi)^{(1-d)/2}}{\sqrt{\det(r_d\mathcal{H})}} \left(\sum_{j=0}^M C_j (r_dn)^{-j} + O\left(n^{-M-1}\right)\right). $$

Recall univariate rational asymptotics

$$f_n = \lambda^{-n} \, n^{p-1} \, \left(\frac{p \; G(\lambda)}{(-\lambda)^{p} H^{(p)}(\lambda)}+O\left(n^{-1}\right)\right)$$

This result is valid for meromorphic functions not just rational functions.

The derivatives of \(g\) can be computed by implicitly differentiating \(H(\widehat{\mathbf{z}},g(\widehat{\mathbf{z}}))=0\).

The constants \(C_0,\dots,C_M\) can be computed solely from the partial derivatives of \(G\) at \(\mathbf{z}=\mathbf{w}\) up to order \(2M\) and the partial derivatives of \(H\) at \(\mathbf{z}=\mathbf{w}\) up to order \(2M+2\).

$$F(x,y) = \frac{1}{1-x-y} = \sum_{i,j \geq 0} \binom{i+j}{i}x^iy^j $$

Critical Point Equations in Direction \(\mathbf{r}=(1,1)\)

$$1-x-y=0 \qquad\qquad -x=-y$$

Strictly Minimal Critical Point: \((x_*,y_*) = \left(\frac{1}{2},\frac{1}{2}\right)\)

Hessian

\(g(x)=1-x\) so \(\phi(\theta) = \log(2-e^{i\theta})\) and \(\mathcal{H}=\phi''(0)=2\).

$$F(x,y) = \frac{1}{1-x-y} = \sum_{i,j \geq 0} \binom{i+j}{i}x^iy^j $$

Critical Point Equations in Direction \(\mathbf{r}=(1,1)\)

$$1-x-y=0 \qquad\qquad -x=-y$$

Strictly Minimal Critical Point: \((x_*,y_*) = \left(\frac{1}{2},\frac{1}{2}\right)\)

Asymptotics

$$ \binom{2n}{n} = [x^ny^n]F(x,y) = \frac{4^n}{\sqrt{\pi n}}\left(1+O\left(n^{-1}\right)\right)$$

$$F(x,y) = \frac{1}{1-x-y} = \sum_{i,j \geq 0} \binom{i+j}{i}x^iy^j $$

Critical Point Equations in Direction \(\mathbf{r}=(r,s)\)

$$1-x-y=0 \qquad\qquad -sx=-ry$$

Strictly Minimal Critical Point: \((x_*,y_*) = \left(\frac{r}{r+s},\frac{s}{r+s}\right)\)

Asymptotics

$$ \binom{rn+sn}{rn} = \left(\frac{r+s}{r}\right)^{rn}\left(\frac{r+s}{s}\right)^{rn} \frac{\sqrt{r+s}}{\sqrt{2rs\pi n}}\left(1+O\left(n^{-1}\right)\right)$$

Ramgoolam, Wilson and Zahabi (2020): The generating function for the chiral operators in the large \(N\) limit of the bi-clover quiver gauge theory is

$$ F(x,y) = \frac{1}{\prod_{k \geq 1}(1-x^k-y^k)}.$$

Ramgoolam, Wilson and Zahabi (2020): The generating function for the chiral operators in the large \(N\) limit of the bi-clover quiver gauge theory is

$$ F(x,y) = \frac{1}{\prod_{k \geq 1}(1-x^k-y^k)}.$$

Note: \(F(x,y)=G(x,y)/(1-x-y)\) where
 

$$ G(x,y) = \prod_{k \geq 2}\left(1-x^k-y^k\right)^{-1} $$
is analytic at the roots of \(1-x-y\) with \(|x|,|y|<1\).

Ramgoolam, Wilson and Zahabi (2020): The generating function for the chiral operators in the large \(N\) limit of the bi-clover quiver gauge theory is

$$ F(x,y) = \frac{1}{\prod_{k \geq 1}(1-x^k-y^k)}.$$

Thus,

$$ \begin{aligned}[x^{rn}y^{sn}]F(x,y) = &G\left(\frac{r}{r+s},\frac{s}{r+s}\right) \\[+4mm]& \qquad\cdot \left(\frac{r+s}{r}\right)^{rn}\left(\frac{r+s}{s}\right)^{rn} \frac{\sqrt{r+s}}{\sqrt{2rs\pi n}}\left(1+O\left(n^{-1}\right)\right)\end{aligned}$$

The generating function for the number of lattice walks on the steps NSEW starting at the origin and staying in the first quadrant is the main diagonal of

$$F(x,y,t) = \frac{(1+x)(1+y)}{1-txy(x+1/x+y+1/y)}.$$

If \(|x|,|y|\leq1\) and
 

$$4 = \left|t^{-1}\right| = \left|x^2y + y + y^2x + x\right|$$

then \(|x|=|y|=1\) and \(x\) and \(y\) are real.

If \(|x|,|y|\leq1\) and
 

$$4 = \left|t^{-1}\right| = \left|x^2y + y + y^2x + x\right|$$

then \(|x|=|y|=1\) and \(x\) and \(y\) are real.

Thus \(\boldsymbol{\sigma} = (1,1,1/4)\) and \(\boldsymbol{\tau} = (-1,-1,1/4)\) are minimal critical points and the only points on \(T(\boldsymbol{\sigma})\).

If \(|x|,|y|\leq1\) and
 

$$4 = \left|t^{-1}\right| = \left|x^2y + y + y^2x + x\right|$$

then \(|x|=|y|=1\) and \(x\) and \(y\) are real.

Thus \(\boldsymbol{\sigma} = (1,1,1/4)\) and \(\boldsymbol{\tau} = (-1,-1,1/4)\) are minimal critical points and the only points on \(T(\boldsymbol{\sigma})\).

Because the numerator of \(F\) vanishes at \(\boldsymbol{\tau}\) its asymptotic contribution is \(O(4^nn^{-2})\).

We then compute

$$f_{n,n,n} \sim 4^n n^{-1} \frac{(2\pi)^{-1}}{\sqrt{\det(\mathcal{H})}} \cdot \frac{- G(\boldsymbol{\sigma})}{(1/4)H_{t}(\boldsymbol{\sigma}) } = \frac{4^n}{n} \cdot \frac{4}{\pi}$$
 

We then compute

$$f_{n,n,n} \sim 4^n n^{-1} \frac{(2\pi)^{-1}}{\sqrt{\det(\mathcal{H})}} \cdot \frac{- G(\boldsymbol{\sigma})}{(1/4)H_{t}(\boldsymbol{\sigma}) } = \frac{4^n}{n} \cdot \frac{4}{\pi}$$
Next Lecture: Automating the smooth case (and complexity)

We then compute

$$f_{n,n,n} \sim 4^n n^{-1} \frac{(2\pi)^{-1}}{\sqrt{\det(\mathcal{H})}} \cdot \frac{- G(\boldsymbol{\sigma})}{(1/4)H_{t}(\boldsymbol{\sigma}) } = \frac{4^n}{n} \cdot \frac{4}{\pi}$$
Next Lecture: Automating the smooth case (and complexity)

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.

Amoebas (and contours) help with intuition for the analysis.

Also Next Lecture: Importance of minimal vs critical points, tools to prove minimality, limit theorems