User:Dom walden/Multivariate Analytic Combinatorics/Cauchy-Hadamard Theorem and Exponential Bounds

Cauchy-Hadamard theorem in several complex variables

Theorem

Let $\alpha$ be an n-dimensional vector of natural numbers ($\alpha =(\alpha _{1},\cdots ,\alpha _{n})\in \mathbb {N} ^{n}$) with $||\alpha ||=\alpha _{1}+\cdots +\alpha _{n}$, then $f(z)$ converges with radius of convergence $\rho =(\rho _{1},\cdots ,\rho _{n})\in \mathbb {R} ^{n}$ with $\rho ^{\alpha }=\rho _{1}^{\alpha _{1}}\cdots \rho _{n}^{\alpha _{n}}$ if and only if

$\limsup _{||\alpha ||\to \infty }{\sqrt[{||\alpha ||}]{|c_{\alpha }|\rho ^{\alpha }}}=1$

where

$f(z)=\sum _{\alpha \geq 0}c_{\alpha }(z-a)^{\alpha }:=\sum _{\alpha _{1}\geq 0,\ldots ,\alpha _{n}\geq 0}c_{\alpha _{1},\ldots ,\alpha _{n}}(z_{1}-a_{1})^{\alpha _{1}}\cdots (z_{n}-a_{n})^{\alpha _{n}}$

Proof

Set $z=a+t\rho$ $(z_{i}=a_{i}+t\rho _{i})$, then

$\sum _{\alpha \geq 0}c_{\alpha }(z-a)^{\alpha }=\sum _{\alpha \geq 0}c_{\alpha }\rho ^{\alpha }t^{||\alpha ||}=\sum _{\mu \geq 0}\left(\sum _{||\alpha ||=\mu }|c_{\alpha }|\rho ^{\alpha }\right)t^{\mu }$

This is a power series in one variable $t$ which converges for $|t|<1$ and diverges for $|t|>1$. Therefore, by the Cauchy-Hadamard theorem for one variable

$\limsup _{\mu \to \infty }{\sqrt[{\mu }]{\sum _{||\alpha ||=\mu }|c_{\alpha }|\rho ^{\alpha }}}=1$

Setting $|c_{m}|\rho ^{m}=\max _{||\alpha ||=\mu }|c_{\alpha }|\rho ^{\alpha }$ gives us an estimate

$|c_{m}|\rho ^{m}\leq \sum _{||\alpha ||=\mu }|c_{\alpha }|\rho ^{\alpha }\leq (\mu +1)^{n}|c_{m}|\rho ^{m}$

Because ${\sqrt[{\mu }]{(\mu +1)^{n}}}\to 1$ as $\mu \to \infty$

${\sqrt[{\mu }]{|c_{m}|\rho ^{m}}}\leq {\sqrt[{\mu }]{\sum _{||\alpha ||=\mu }|c_{\alpha }|\rho ^{\alpha }}}\leq {\sqrt[{\mu }]{|c_{m}|\rho ^{m}}}\implies {\sqrt[{\mu }]{\sum _{||\alpha ||=\mu }|c_{\alpha }|\rho ^{\alpha }}}={\sqrt[{\mu }]{|c_{m}|\rho ^{m}}}\qquad (\mu \to \infty )$

Therefore

$\limsup _{||\alpha ||\to \infty }{\sqrt[{||\alpha ||}]{|c_{\alpha }|\rho ^{\alpha }}}=\limsup _{\mu \to \infty }{\sqrt[{\mu }]{|c_{m}|\rho ^{m}}}=1$

Example

For the central diagonal of our example, $\Delta {\frac {1}{1-x-y}}=\sum _{n\geq 0}f_{n,n}x^{n}y^{n}$:

$\limsup _{n\to \infty }{\sqrt[{n}]{|f_{n,n}|x^{n}y^{n}}}=1\implies \limsup _{n\to \infty }|f_{n,n}|={\frac {1}{(xy)^{n}}}$

$x^{n}y^{n}$ is at its largest when $x=y={\frac {1}{2}}$ so that $\limsup _{n\to \infty }|f_{n,n}|=4^{n}$.

We know by Stirling's approximation that this is a good estimate.

But what about a diagonal along an arbitrary ray, like the above example $\Delta ^{(2,1)}{\frac {1}{1-x-y}}$?

$\limsup _{|n{\textbf {r}}|\to \infty }{\sqrt[{|n{\textbf {r}}|}]{|f_{2n,n}|x^{2n}y^{n}}}=1\implies \limsup _{|n{\textbf {r}}|\to \infty }|f_{2n,n}|={\frac {1}{(x^{2}y)^{n}}}$

If we keep $x=y={\frac {1}{2}}$ then $\limsup _{|n{\textbf {r}}|\to \infty }|f_{2n,n}|=8^{n}$

This isn't a good estimate.

Better to use $x={\frac {2}{3}},y={\frac {1}{3}}$ then $\limsup _{|n{\textbf {r}}|\to \infty }|f_{2n,n}|=\left({\frac {27}{4}}\right)^{n}=(6.75)^{n}$

Convex optimisation

In the below, the function we are interested in is $F({\textbf {z}})={\frac {G({\textbf {z}})}{H({\textbf {z}})}}$.

We therefore want to find the ${\textbf {w}}$ on the domain of convergence of $F({\textbf {z}})$ that minimises ${\textbf {w}}^{-{\textbf {r}}}$.

The subject of convex optimisation already has the tools for this, but in order to use it we need to transform the domain of convergence to be a convex set and ${\textbf {w}}^{-{\textbf {r}}}$ to be a convex function.

Give examples of how useful convex is...

Fortunately, the logarithmic image of the domain of convergence of a power series of a complex function is convex.

Therefore, we define

$Relog({\textbf {z}})=(\log |z_{1}|,\cdots ,\log |z_{d}|)$

$amoeba(H)=\{Relog({\textbf {z}}):H({\textbf {z}})=0\}$

The domain of convergence of our function can now be defined as the complement of this amoeba

$amoeba(H)^{c}=\mathbb {R} ^{d}\setminus amoeba(H)$

This may leave us with multiple unconnected components, each one for a different Laurent series expansion. Denote the component we are interested in as $B$

${\mathcal {D}}=Relog^{-1}(B)$

The logarithmic image of ${\textbf {w}}^{-{\textbf {r}}}$ is $h({\textbf {w}})=-{\textbf {r}}\cdot Relog({\textbf {w}})$. Because $\log$ is a concave function, $-\log$ is convex.

So we now have a problem of minimising a convex function over a convex set.

We want to find the supporting hyperplane to ${\bar {B}}$ with outward-facing normal $-\nabla h({\textbf {w}})$.

Critical point equations

This happens when the supporting hyperplane defined above coincides with the tangent plane with normal $\nabla H({\textbf {w}})$.

This means they are not linearly independent and therefore the matrix

${\begin{pmatrix}{\frac {\partial H}{\partial z_{1}}}({\textbf {w}})&\cdots &{\frac {\partial H}{\partial z_{d}}}({\textbf {w}})\\r_{1}/w_{1}&\cdots &r_{d}/w_{d}\end{pmatrix}}$

is rank deficient, or its 2 x 2 submatrices have zero determinants. This is equivalent to a system of equations referred to as the critical point equations

$H({\textbf {w}})=0\quad r_{j}w_{1}{\frac {\partial H}{\partial z_{1}}}({\textbf {w}})-r_{1}w_{j}{\frac {\partial H}{\partial z_{j}}}({\textbf {w}})=0\quad (2\leq j\leq d).$

References

• Melczer, Stephen (2021). An Invitation to Analytic Combinatorics: From One to Several Variables (PDF). Springer Texts & Monographs in Symbolic Computation.
• Pemantle, Robin; Wilson, Mark C.; Melczer, Stephen (2024). Analytic Combinatorics in Several Variables (PDF) (2nd ed.). Cambridge University Press.
• Shabat, B. V. (1992). Introduction to Complex Analysis. Part II: Functions of Several Variables. American Mathematical Society, Providence, Rhode Island.

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