\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{showlabels}
\newtheorem{theorem}{Theorem}
\begin{document}
\title{theorem}
\author{john}
\date{}
\maketitle
%\section*{Note on notation}
%In several parts of the paper we need to build some sort of approximation based on an existing approximation.
%The first example is when we introduce $B(z)$ to generalise from the Hermite to the Birkhoff case. Another
%example is in what I am proposing below where induction on $m$ is used.
%I think things would be better if we always used $\alpha$ and $\gamma$ for the approximation we are constructing
%and $\widehat\alpha$ and $\widehat\gamma$ for an existing case we already know about.
%This would mean that the Hermite case is unchanged but the construction of the Birkhoff case reverses the use of
%the $\widehat{\phantom{a}}$ symbols.
%{\bf Having written this note I am not convinced by it myself and I have written the result below without assuming
%and change in notation in the rest of the paper. Hence, please feel free to ignore this note.}
%\section*{Proposed result}
\begin{theorem}
Write the order $p=-1+\sum_{i=1}^n$ Birkhoff interpolation formula in the form
\begin{equation}\label{eq:poly1}
f(t) = \sum_{i=1}^n \sum_{j\in S_i} \alpha_{ij}(t) f^{(j)}(\tau_i),
\end{equation}
where
\begin{equation*}
-\frac{B(t,z)w(t)}{B(t,t)(z-t)w(z)} = -\frac1{z-t} + \sum_{i=1}^n \sum_{j\in S_i}\frac{ \gamma_{ij}(t)}{(z-\tau_i)^{j+1}}
\end{equation*}
and $\alpha_{ij}(t) =\gamma_{ij}(t)/j!$
then $\alpha_{ij}(t)$ is a polynomial in $t$.
\end{theorem}
{\bf Proof} The proof is by induction on $m$, starting with $m=0$, so that $B(t,z)=1$. For given $i\in \{1,2,\dots,n\}$,
the values of $\gamma_{ij}(t)$, $j=0,1,\dots,s_i-1$ are equal to the coefficients of $(z-\tau_i)^{-j-1}$ in the Laurent expansion
of $-w(t)/(z-t)w(z)$ about $z=\tau_i$. This is equal to the product of the Taylor series of
$-w(t)/(z-t)$ and the Laurent series of $1/w(z)$. This product is $w(t) A B$, where $B$ is the Laurent expansion of $1/w(z)$ and
\begin{equation*}
A=\frac1{t-\tau_i} + \frac{z-\tau_i}{(t-\tau_i)^2} + \cdots + \frac{(z-\tau_i)^{s_i-1}}{(t-\tau_i)^{s_i}}
\end{equation*}
is the Taylor series for $1/(t-z)$. Note that in $A$, it is necessary to include only $s_i$ terms because only terms of negative
exponent in $w(t) A B$ are relevant. The proof for $m=0$ concludes by observing that all the terms in $A$ cancel with $w(z)$
and that the Laurent coefficients in $B$ are independent of $t$.
If $m>1$, there exists $I\in \{1,2,\dots,n\}$ and $J