From 1b15f980bc16ef3abeb456fe214584f80bff88eb Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Tue, 12 Oct 2021 16:19:15 +0200
Subject: Removed appendix.

---
 ising_scaling.tex | 37 -------------------------------------
 1 file changed, 37 deletions(-)

(limited to 'ising_scaling.tex')

diff --git a/ising_scaling.tex b/ising_scaling.tex
index e6e559e..2c540e6 100644
--- a/ising_scaling.tex
+++ b/ising_scaling.tex
@@ -778,41 +778,4 @@ The successful smooth description of the Ising free energy produced in part by a
 
 \bibliography{ising_scaling}
 
-\appendix
-
-\section{Legendre series to power series}
-
-In the work, we examine a series of the form
-\[
-  h(\theta)=\left(1-\frac{\theta^2}{\theta_c^2}\right)\sum_{i=0}^\infty h_iP_{2i+1}(\theta/\theta_c),
-\]
-where $P_n$ is the $n$th Legendre polynomial. This form is useful for iteratively fitting the $h_i$, since these polynomials are orthogonal on the domain $[-\theta_c,\theta_c]$, but is not useful for evaluating the convergence of the function. For this a simple power expansion in $\theta$ is better. To make the conversion, first recall that
-\[
-  P_n(x)=2^n\sum_{k=0}^n\binom nk\binom{\frac{n+k-1}2}nx^k
-\]
-\[
-  \begin{aligned}
-    \sum_{i=0}^\infty h_iP_{2i+1}(\theta/\theta_c)
-    &=\sum_{i=0}^\infty h_i2^{2i+1}\sum_{k=0}^{2i+1}\frac{\theta^k}{\theta_c^k}\binom{2i+1}k\binom{\frac{2i+k}2}{2i+1} \\
-    &=\sum_{k=0}^\infty\theta^k\frac1{\theta_c^k}\sum_{i=(k-1)/2}^\infty h_i2^{2i+1}\binom{2i+1}k\binom{\frac{2i+k}2}{2i+1} \\
-    &=\sum_{j=0}^\infty\theta^{2j+1}\frac1{\theta_c^{2j+1}}\sum_{i=j}^\infty h_i2^{2i+1}\binom{2i+1}{2j+1}\binom{\frac{2i+2j+1}2}{2i+1}
-  \end{aligned}
-\]
-We saw in our fitting that at higher order the coefficients $h_i$ approached roughly the pattern $h_i/h_{i-1}=-b+m/i$ for positive constants $b$ and $m$. This recurrence relation can be solved to give
-\[
-  h_i\simeq\frac{(-1)^i}{i!}b^{i-1}(b-m)h_0\frac{\Gamma(2-\frac mb+i-1)}{\Gamma(2-\frac mb)}
-\]
-\[
-  \sum_{i=0}^\infty h_iP_{2i+1}(\theta/\theta_c)
-  =\sum_{j=0}^\infty\theta^{2j+1}\frac1{\theta_c^{2j+1}}
-  \frac{\Gamma(1+j-m/b)}{\Gamma(1-m/b)\Gamma(1+j)}2^{1+2j}(-b)^jh_0\binom{2j+\frac12}{2j+1}{}_2F_1(2j+\tfrac32,1+j-m/b;1+j;b)
-\]
-If we call this coefficient $H_j$, then
-\[
-  h(\theta)=\left(1-\frac{\theta^2}{\theta_c^2}\right)\sum_{i=0}^\infty H_i\theta^{2i+1}
-  =\sum_{i=0}^\infty H_i\theta^{2i+1}-\frac1{\theta_c^2}\sum_{i=0}^\infty H_i\theta^{2i+3}
-  =\sum_{i=0}^\infty(H_i-H_{i-1}/\theta_c^2)\theta^{2i+1}
-\]
-for $H_{-1}=0$.
-
 \end{document}
-- 
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