Pade’ approximants. Extracting critical points

A Pade’ approximant is a rational approximation to a function $f$ of the form

$[L,M](x)={\sum_{j=0}^M p_jx^j\over 1+\sum_{j=1}^L q_jx^j}$

It can be computed very algorithmically from a Taylor series
expansion of the function

$f(x)\rightarrow \sum_{n=0}^\infty a_n x^n$

First cut the Taylor series off at $n_{max}=L+M$ and multiply it out,

$\sum_{j=0}^M p_jx^j-\Big(\sum_{n=0}^{L+M} a_n x^n\Big)\Big(1+\sum_{j=1}^L q_jx^j}\Big)=0$

Equate like coefficients on each side, a process that results in two
sets of equations

(1) $\begin{eqnarray*} p_0&=& a_0\\ p_1&=& a_1+a_0q_1\\ p_2&=& a_2+a_1q_1+a_0q_2\\ \vdots &=& \vdots\\ p_M&=& a_M+a_{M-1}q_1+a_{M-2}q_2+\cdots+a_0 q_M\end{eqnarray*}$

(2) $\begin{eqnarray*} -a_{M+1}&=& a_Mq_1+a_{M-1}q_2+a_{M-2}q_3+\cdots+a_0 q_{M+1}\\ -a_{M+2}&=& a_{M+1}q_1+a_{M}q_2+a_{M-1}q_3+\cdots+a_0 q_{M+2}\\ -a_{M+3}&=& a_{M+2}q_1+a_{M+1}q_2+a_{M}q_3+\cdots+a_0 q_{M+3}\\ \vdots &=& \vdots\\ -a_{M+L}&=& a_{M+L-1}q_1+a_{M+L-2}q_2+a_{M+L-3}q_3+\cdots+a_0 q_{M+L}\end{eqnarray*}$

however $q_k=0$ for $k>L$ on the right-hand side.

Solve the second set for $\{q_1,q_2,\cdots,q_L\}$ and use it in the first
set to get the $\{p_1,p_2,\cdots,p_M\}$ .

We can do this in REDUCE
operator p,q,a,eqtn,rhs;
q(0):=1;
!M:=3;
!L:=5;
matrix !T(!L,!L);
matrix inhom(!L,1);
Numerator:=for j:=0:!M sum p(j)*x^j$
Denominator:=for j:=0:!L sum q(j)*x^j$
Func:=for j:=0:(!M+!L) sum a(j)*x^j$
zilch:=Denominator*Func-Numerator$
for j:=0:(!M+!L) do eqtn(j):=coeffn(zilch,x,j);
for j:=1:!L do for k:=1:!L do !T(k,j):=df(eqtn(k+!M),q(j));
for j:=1:!L do inhom(j,1):=eqtn(j+!M)-for k:=1:!L sum !T(j,k)*q(k);
soln:=-1/!T*inhom$
for j:=1:!L do let q(j)=soln(j,1)$
for j:=0:!M do let solve(eqtn(j),p(j))$
a(0):=1; a(1):=1; a(2):=1/2; a(3):=1/6; a(4):=1/24; a(5):=1/120; a(6):=1/720;a(7):=1/5040;a(8):=1/40320;
pade:=numerator/denominator;

As you can see I set this up so that we will get the $[3,3]$ approximant to $f=e^x$ . Let’s run it:

pade:=numerator/denominator;

             3       2
          - x  - 12*x  - 60*x - 120
pade := ----------------------------
           3       2
          x  - 12*x  + 60*x - 120

In a previous post on recursions, we worked out the Miller algorithm for computing Bessel functions using a descending three-term recursion. This type of algorithm, in which the final step is a”renormalization” using an addition identity, leads to a Pade’ approximant. For example we start with $J_6(x)=0$ and $J_5(x)=x$ , the descending recursion $J_{n-1}(x)={2(n-1)\over x}J_{n-1}(x)-J_n(x)$ terminating in $J_0(x)$ together with $1=J_0+2J_2+2J_4+2J_6+\cdots$ leads to

            4        2
         3*x  - 128*x  + 640
J_0(x) := ---------------------
           4       2
          x  + 32*x  + 640

So what is the application to critical phenomena? Suppose that a function $f(z)$ has singular behaviour at some critical value $f(z)\sim (z-z_c)^{\beta}\, R(z)$ , in which $R(z)$ is regular. Then its logarithmic derivative has a pole there

${d\over dz}\ln (f(z))={\beta\over z-z_c}\Big(1+\mathcal{O}(z-z_c)^1\Big)$

so that the residue of the logarithmic derivative at $z=z_c$ is $\beta$

$Res(z_c)=\lim_{z\rightarrow z_c} (z-z_c){d\ln f(z)\over dz}=\beta$

One of the most influential models of strong interactions in two dimensions is to consider a triangular lattice of $N$ lattice points on which non-overlapping hexagons will be placed. The activity of a single hexagon is $z=e^{\mu/kT}$ , and the system has a critical point at a density below the close-packing density of $1/3$ .

A series expansion for ${d\over dz}\ln (f(z))$ can be approximated by a Pade’ approximant $[N,M]$

${d\over dz}\ln(f(z))=\sum_n a_nz^n\approx [N,M]={P_M(z)\over Q_N(z)}$

$={b_0+b_1z+\cdots +b_M z^M\over 1+c_1z+\cdots+c_Nz^N}={P_M(z)\over c_N\prod_{j}^N(z-z_j)}$

which is a ratio of an $M^{th}$ order polynomial and an $N^{th}$ order polynomial, in which the zeroes of $Q_N$ are the poles/singularities of $f(z)$ . All computer algebra systems have a Pade’ approximant function and a real root finding function.
A triangular lattice of $N$ can be divided into three sublattices “1”, “2”, “3”. Consider a completely filled up or covered lattice in which every site of type “1” has a hexagon on it (we could start with all “2” sites or all “3” sites covered). Now we can create a “hole” or anti-hexagon with activity $z'=1/z=e^{-\beta\mu}$ by removing a hexagon from a site. Doing so we find high-density formulas for the densities each of the three sublattices

$\rho_1=1-z'-5z'^2-34z'^3-267z'^4-2037z'^5+\cdots$

$\rho_2=\rho_3=z'^2+9z'^3+80z'^4+965z'^5+\cdots$

The total density is $\rho={1\over 3}(\rho_1+\rho_2+\rho_3)={z\over N}{d\over dz}\ln Q$ .
An order parameter for the phase transition (solid-fluid) is

$R=\rho_1-\rho_2$

Use your favorite computer CAS to find Pade’ approximants $[2,2]$ and $[3,3]$ for ${d\over dz'}\ln R$ in variable $z'$ . One root of $Q_2$ will persist to a root of $Q_3$ , use it to find the critical activity $z'_c=1/z_c$ . The order parameter critical exponent is $\beta$ , $R\sim t^\beta$ with $t=z-z_c$ , find $\beta$ . Let’s use REDUCE, and load its own pade library (not use our DIY above)…

!R:=1-z-6*z^2-43*z^3-347*z^4-3002*z^5;
load(pade);
load(roots);
dlogR:=taylortostandard(taylor(df(log(!R),z),z,0,7));
!Q:=pade(dlogR,z,0,3,3);
realroots(den(!Q));     
#  {z= - 0.0119137,z=0.0142508,z=0.0877545} last is persistent
!Q:=pade(dlogR,z,0,2,2);
realroots(den(!Q));
#  {z=0.0899435,z=0.300713}  see?
# so z_c=11.11809, very close to the exact value, the model is exactly solvable
sub(z=0.0899435, num(!Q)/(z-0.300713));
# 0.108764605959  Exact value is 1/9=beta, you are very close

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Pade’ approximants. Extracting critical points

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Pade’ approximants. Extracting critical points

Published by Jeff on September 9, 2021September 9, 2021

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