\documentclass{article}
\usepackage{axiom}
\begin{document}
\title{Indefinite Problems}
\author{Timothy Daly, et al.}
\maketitle
\begin{abstract}
\end{abstract}
\eject
\tableofcontents
\eject
\section{Integers}
\section{Polynomials}
\[
(x+y)^n = \sum_{k=0}^{n}
\left(
\begin{array}{c}
n\\
k
\end{array}
\right)
x^{n-k}y^k
\]
where
\[
\left(
\begin{array}{c}
n\\
k
\end{array}
\right)
= {}_nC_k
= \frac{n!}{(n-k)!k!}
\]
so
\[
\begin{array}{ccc}
(x+y)^8 & = & {}_8C_0x^{8-0}y^0 + {}_8C_1x^{8-1}y^1 + {}_8C_2x^{8-2}y^2 +\\
& & {}_8C_3x^{8-3}y^3 + {}_8C_4x^{8-4}y^4 + {}_8C_5x^{8-5}y^5 +\\
& & {}_8C_6x^{8-6}y^6 + {}_8C_7x^{8-7}y^7 + {}_8C_8x^{8-8}y^8\ \ 
\end{array}
\]
\subsection{particular cases}
\subsection{step 1 $(m^0)*(n^i)$}
The point of this exercise is to find general forms for particular
polynomials. In particular, as above, the binomial form has a 
convenient closed form. So we're looking for a closed form of the
general product
\[ (x+a)^n * (x+b)^m \]
Let $m=(x+a)$ and $m=(x+b)$, we examine particular cases:
\[
\begin{array}{ccc}
(m^0)*(n^0) &=& b^0x^0\\
(m^0)*(n^1) &=& b^0x^1+b^1x^0\\
(m^0)*(n^2) &=& b^0x^2  + 2b^1x^1 + b^2x^0\\
(m^0)*(n^3) &=& b^0x^3  + 3b^1x^2  + 3b^2x^1 + b^3x^0\\
(m^0)*(n^4) &=& b^0x^4  + 4b^1x^3  + 6b^2x^2  + 4b^3x^1 + b^4x^0\\
(m^0)*(n^5) &=& b^0x^5  + 5b^1x^4  + 10b^2x^3  + 10b^3x^2  + 5b^4x^1 + b^5x^0
\end{array}
\]
The first observation is that the coefficients follow pascal's triangle
(hardly a surprise) and could be rewritten as:
\[
\begin{array}{ccc}
(m^0)*(n^0) &=& {}_0C_0b^0x^0\\
(m^0)*(n^1) &=& {}_1C_0b^0x^1 + {}_1C_1b^1x^0\\
(m^0)*(n^2) &=& {}_2C_0b^0x^2 + {}_2C_1b^1x^1 + {}_2C_2b^2x^0\\
(m^0)*(n^3) &=& {}_3C_0b^0x^3 + {}_3C_1b^1x^2  + {}_3C_2b^2x^1 + {}_3C_3b^3x^0\\
(m^0)*(n^4) &=& {}_4C_0b^0x^4 + {}_4C_1b^1x^3  + {}_4C_2b^2x^2  + {}_4C_3b^3x^1 + {}_4C_4b^4x^0\\
(m^0)*(n^5) &=& {}_5C_0b^0x^5 + {}_5C_1b^1x^4  + {}_5C_2b^2x^3  + {}_5C_3b^3x^2  + {}_5C_4b^4x^1 + {}_5C_5b^5x^0
\end{array}
\]
and if we observe the behavor of the $b$ term we can rewrite it as
\[
\begin{array}{ccl}
(m^0)*(n^0) &=& \sum_{i=0}^0 {}_0C_ib^0x^0\\
(m^0)*(n^1) &=& \sum_{i=0}^1 {}_1C_ib^0x^1 + {}_1C_ib^1x^0\\
(m^0)*(n^2) &=& \sum_{i=0}^2 {}_2C_ib^0x^2 + {}_2C_i2b^1x^1 + {}_2C_ib^2x^0\\
(m^0)*(n^3) &=& \sum_{i=0}^3 {}_3C_ib^0x^3 + {}_3C_ib^1x^2  + {}_3C_ib^2x^1 + {}_3C_ib^3x^0\\
(m^0)*(n^4) &=& \sum_{i=0}^4 {}_4C_ib^0x^4 + {}_4C_ib^1x^3  + {}_4C_ib^2x^2  + {}_4C_ib^3x^1 + {}_4C_ib^4x^0\\
(m^0)*(n^5) &=& \sum_{i=0}^5 {}_5C_ib^0x^5 + {}_5C_ib^1x^4  + {}_5C_ib^2x^3  + {}_5C_ib^3x^2  + {}_5C_ib^4x^1 + {}_5C_ib^5x^0
\end{array}
\]
or in functional terms we can definte a [[term]] function which will
generate each term in the polynomial as:
\[
{\rm term}(\alpha,\beta) = {}_{\alpha}C{_\beta} b^{\beta} x^{\alpha-\beta}
\]
So, the Axiom code is:
<<term(alpha,beta)>>=
term(alpha,beta)==
  pascalTriangle(alpha,beta)$GALUTIL(INT)*(b**(beta))*(x**(alpha-beta))
@
\noindent
and we can compute particular terms thus:
\begin{verbatim}
term(0,0)
\end{verbatim}
\[
1 
\]
\begin{verbatim}
term(1,0)
\end{verbatim}
\[
x 
\]
\begin{verbatim}
term(1,1)
\end{verbatim}
\[
b 
\]
and the polynomial is just
\begin{verbatim}
term(1,0)+term(1,1)
\end{verbatim}
\[
x+b 
\]
with the next polynomial:
\begin{verbatim}
term(2,0)+term(2,1)+term(2,2)
\end{verbatim}
\[
{x \sp 2}+{2 \  b \  x}+{b \sp 2}  
\]

Thus the more general conclusion of
\[
(n^k) = \sum_{i=0}^k term(k,i)
\]
which in math notation is:
\[
(n^k) = \sum_{i=0}^k {}_kC_i b^ix^{k-i}
\]
\subsection{step 2  $(m^1)*(n^i)$}
\[
\begin{array}{ccl}
(m^1)*(n^0) &=& x^1+ax^0\\
(m^1)*(n^1) &=& x^2 + (b + a)x^1 + abx^0\\
(m^1)*(n^2) &=& x^3 + (2b + a)x^2  + (b^2 + 2ab)x^1 + ab^2x^0\\
(m^1)*(n^3) &=& x^4 + (3b + a)x^3  + (3b^2  + 3ab)x^2  + (b^3  + 3ab^2)x^1 + ab^3x^0\\
(m^1)*(n^4) &=& x^5 + (4b + a)x^4  + (6b^2  + 4ab)x^3  + (4b^3  + 6ab^2)x^2\\
&&  + (b^4 + 4ab^3)x^1 + ab^4x^0\\
(m^1)*(n^5) &=& x^6  + (5b + a)x^5  + (10b^2  + 5ab)x^4 \\
&&+ (10b^3  + 10ab^2 )x^3  + (5b^4 + 10ab^3 )x^2 + (b^5 + 5ab^4 )x^1 + ab^5x^0
\end{array}
\]
\subsection{step 3  $(m^2)*(n^i)$}
\[
\begin{array}{ccl}
(m^2)*(n^0) &=& x^2 + 2ax^1 + a^2x^0\\
(m^2)*(n^1) &=& x^3 + (b + 2a)x^2  + (2ab + a^2)x + abx^0\\
(m^2)*(n^2) &=& x^4 + (2b + 2a)x^3  + (b^2  + 4ab + a^2)x^2\\
&&+ (2ab^2  + 2a^2b)x^1 + a^2b^2x^0\\
(m^2)*(n^3) &=& x^5 + (3b + 2a)x^4  + (3b^2  + 6ab + a^2)x^3 + \\
&&(b^3  + 6ab^2 + 3a^2 b)x^2 + (2ab^3  + 3a^2b^2)x^1 + a^2b^3x^0\\
(m^2)*(n^4) &=& x^6 + (4b + 2a)x^5  + (6b^2  + 8ab + a^2)x^4 + \\
&&(4b^3  + 12ab^2  + 4a^2b)x^3 + (b^4 + 8ab^3  + 6a^2 b^2 )x^2  \\
&&+ (2a b^4  + 4a^2 b^3 )x^1 + a^2 b^4x^0\\
(m^2)*(n^5) &=& x^7 + (5b + 2a)x^6  + (10b^2  + 10a b + a^2 )x^5  \\
&&+ (10b^3  + 20a b^2  + 5a^2 b)x^4 + (5b^4  + 20a b^3  + 10a^2 b^2 )x^3  \\
&&+ (b^5  + 10a b^4  + 10a^2 b^3 )x^2  + (2a b^5  + 5a^2 b^4 )x^1 + a^2 b^5x^0
\end{array}
\]
\subsection{step 4  $(m^3)*(n^i)$}
\[
\begin{array}{ccl}
(m^3)*(n^0) &=& x^3 + 3a x^2  + 3a^2 x^1 + a^3x^0\\
(m^3)*(n^1) &=& x^4 + (b + 3a)x^3 + (3a b + 3a^2)x^2  + (3a^2 b + a^3)x^1 + a^3 bx^0 \\
(m^3)*(n^2) &=& x^5  + (2b + 3a)x^4  + (b^2  + 6a b + 3a^2 )x^3  \\
&&+ (3a b^2  + 6a^2 b + a^3 )x^2 + (3a^2 b^2  + 2a^3 b)x^1 + a^3 b^2x^0 \\
(m^3)*(n^3) &=& x^6  + (3b + 3a)x^5  + (3b^2  + 9a b + 3a^2 )x^4  + \\
&&(b^3  + 9a b^2  + 9a^2 b + a^3 )x^3 + (3a b^3  + 9a^2 b^2  + 3a^3 b)x^2  \\
&&+ (3a^2 b^3  + 3a^3 b^2 )x^1 + a^3 b^3x^0\\
(m^3)*(n^4) &=& x^7 + (4b + 3a)x^6  + (6b^2  + 12a b + 3a^2 )x^5  \\
&&+ (4b^3  + 18a b^2  + 12a^2 b + a^3 )x^4 + (b^4  + 12a b^3  + 18a^2 b^2  + 4a^3 b)x^3\\
&& + (3a b^4  + 12a^2 b^3  + 6a^3 b^2 )x^2 + (3a^2 b^4  + 4a^3 b^3 )x^1 + a^3 b^4x^0\\
(m^3)*(n^5) &=& x^8 + (5b + 3a)x^7  + (10b^2  + 15a b + 3a^2)x^6  \\
&&+ (10b^3  + 30a b^2  + 15a^2 b + a^3)x^5 + (5b^4  + 30a b^3  + 30a^2 b^2  + 5a^3 b)x^4 \\
&&+ (b^5  + 15a b^4  + 30a^2 b^3  + 10a^3 b^2)x^3 + (3a b^5  + 15a^2 b^4  + 10a^3 b^3 )x^2\\
&&  + (3a^2 b^5  + 5a^3 b^4 )x^1 + a^3 b^5x^0
\end{array}
\]
\subsection{step 5  $(m^4)*(n^i)$}
\[
\begin{array}{ccl}
(m^4)*(n^0) &=& x^4 + 4a x^3  + 6a^2 x^2  + 4a^3 x^1 + a^4x^0\\
(m^4)*(n^1) &=& x^5 + (b + 4a)x^4  + (4a b + 6a^2 )x^3  + (6a^2 b + 4a^3 )x^2\\
&&  + (4a^3 b + a^4 )x^1 + a^4 bx^0\\
(m^4)*(n^2) &=& x^6  + (2b + 4a)x^5  + (b^2  + 8a b + 6a^2 )x^4  + (4a b^2  + 12a^2 b + 4a^3 )x^3\\
&& + (6a^2 b^2  + 8a^3 b + a^4 )x^2  + (4a^3 b^2  + 2a^4 b)x^1 + a^4 b^2x^0\\
(m^4)*(n^3) &=&  x^7  + (3b + 4a)x^6  + (3b^2  + 12a b + 6a^2 )x^5  + \\
&&(b^3  + 12a b^2  + 18a^2 b + 4a^3 )x^4 + (4a b^3  + 18a^2 b^2  + 12a^3 b + a^4 )x^3  \\
&&+ (6a^2 b^3  + 12a^3 b^2  + 3a^4 b)x^2 + (4a^3 b^3  + 3a^4 b^2 )x^1 + a^4 b^3x^0\\
(m^4)*(n^4) &=& x^8  + (4b + 4a)x^7  + (6b  + 16a b + 6a )x^6  + (4b  + 24a b  + 24a b + 4a )x^5\\
&& + (b  + 16a b  + 36a b  + 16a b + a )x^4  + (4a b  + 24a b  + 24a b  + 4a b)x^3\\
&& + (6a b  + 16a b  + 6a b )x^2  + (4a b  + 4a b )x^1 + a bx^0\\
(m^4)*(n^5) &=& x^9  + (5b + 4a)x^8  + (10b^2  + 20a b + 6a^2 )x^7\\
&&  + (10b^3  + 40a b^2  + 30a^2 b + 4a^3 )x^6 \\
&&+ (5b^4  + 40a b^3  + 60a^2 b^2  + 20a^3 b + a^4 )x^5 \\
&&+ (b^5  + 20a b^4  + 60a^2 b^3  + 40a^3 b^2  + 5a^4 b)x^4 \\
&&+ (4a b^5  + 30a^2 b^4  + 40a^3 b^3  + 10a^4 b^2 )x^3  \\
&&+ (6a^2 b^5  + 20a^3 b^4  + 10a^4 b^3 )x^2 \\
&&+ (4a^3 b^5  + 5a^4 b^4 )x^1 + a^4 b^5x^0
\end{array}
\]
which could be more insightfully written as
\[
\begin{array}{ccc}
(m^4)*(n^0) &=& x^4 \\
&&+ 4a x^3  \\
&&+ 6a^2 x^2  \\
&&+ 4a^3 x^1 \\
&&+ a^4x^0\\
(m^4)*(n^1) &=& x^5 \\
&&+ (b + 4a)x^4  \\
&&+ (4a b + 6a^2 )x^3  \\
&&+ (6a^2 b + 4a^3 )x^2\\
&&+ (4a^3 b + a^4 )x^1 \\
&&+ a^4 bx^0\\
(m^4)*(n^2) &=& x^6  \\
&&+ (2b + 4a)x^5  \\
&&+ (b^2  + 8a b + 6a^2 )x^4  \\
&&+ (4a b^2  + 12a^2 b + 4a^3 )x^3\\
&&+ (6a^2 b^2  + 8a^3 b + a^4 )x^2  \\
&&+ (4a^3 b^2  + 2a^4 b)x^1 \\
&&+ a^4 b^2x^0\\
(m^4)*(n^3) &=& x^7 \\
&&+ (3b + 4a)x^6  \\
&&+ (3b^2  + 12a b + 6a^2 )x^5  \\
&&+ (b^3  + 12a b^2  + 18a^2 b + 4a^3 )x^4 \\
&&+ (4a b^3  + 18a^2 b^2  + 12a^3 b + a^4 )x^3\\
&&+ (6a^2 b^3  + 12a^3 b^2  + 3a^4 b)x^2 \\
&&+ (4a^3 b^3  + 3a^4 b^2 )x^1 \\
&&+ a^4 b^3x^0\\
(m^4)*(n^4) &=& x^8 \\
&&+ (4b + 4a)x^7  \\
&&+ (6b  + 16a b + 6a )x^6  \\
&&+ (4b  + 24a b  + 24a b + 4a )x^5\\
&&+ (b  + 16a b  + 36a b  + 16a b + a )x^4 \\
&&+ (4a b  + 24a b  + 24a b  + 4a b)x^3\\
&&+ (6a b  + 16a b  + 6a b )x^2 \\
&&+ (4a b  + 4a b )x^1 \\
&&+ a bx^0\\
(m^4)*(n^5) &=& x^9 \\
&&+ (5b + 4a)x^8  \\
&&+ (10b^2  + 20a b + 6a^2 )x^7\\
&&+ (10b^3  + 40a b^2  + 30a^2 b + 4a^3 )x^6\\
&&+ (5b^4  + 40a b^3  + 60a^2 b^2  + 20a^3 b + a^4 )x^5 \\
&&+ (b^5  + 20a b^4  + 60a^2 b^3  + 40a^3 b^2  + 5a^4 b)x^4 \\
&&+ (4a b^5  + 30a^2 b^4  + 40a^3 b^3  + 10a^4 b^2 )x^3  \\
&&+ (6a^2 b^5  + 20a^3 b^4  + 10a^4 b^3 )x^2 \\
&&+ (4a^3 b^5  + 5a^4 b^4 )x^1 \\
&&+ a^4 b^5x^0
\end{array}
\]
which could be more insightfully written as
\[
\begin{array}{ccc}
(m^4)*(n^0) &=& x^4 \\
&&+ 4a x^3  \\
&&+ 6a^2 x^2  \\
&&+ 4a^3 x^1 \\
&&+ a^4x^0\\
(m^4)*(n^1) &=& x^5 \\
&&+ (b + 4a)x^4  \\
&&+ (4a b + 6a^2 )x^3  \\
&&+ (6a^2 b + 4a^3 )x^2\\
&&+ (4a^3 b + a^4 )x^1 \\
&&+ a^4 bx^0\\
(m^4)*(n^2) &=& x^6  \\
&&+ (2b + 4a)x^5  \\
&&+ (b^2  + 8a b + 6a^2 )x^4  \\
&&+ (4a b^2  + 12a^2 b + 4a^3 )x^3\\
&&+ (6a^2 b^2  + 8a^3 b + a^4 )x^2  \\
&&+ (4a^3 b^2  + 2a^4 b)x^1 \\
&&+ a^4 b^2x^0\\
(m^4)*(n^3) &=& x^7 \\
&&+ (3b + 4a)x^6  \\
&&+ (3b^2  + 12a b + 6a^2 )x^5  \\
&&+ (b^3  + 12a b^2  + 18a^2 b + 4a^3 )x^4 \\
&&+ (4a b^3  + 18a^2 b^2  + 12a^3 b + a^4 )x^3\\
&&+ (6a^2 b^3  + 12a^3 b^2  + 3a^4 b)x^2 \\
&&+ (4a^3 b^3  + 3a^4 b^2 )x^1 \\
&&+ a^4 b^3x^0\\
(m^4)*(n^4) &=& x^8 \\
&&+ (4b + 4a)x^7  \\
&&+ (6b  + 16a b + 6a )x^6  \\
&&+ (4b  + 24a b  + 24a b + 4a )x^5\\
&&+ (b  + 16a b  + 36a b  + 16a b + a )x^4 \\
&&+ (4a b  + 24a b  + 24a b  + 4a b)x^3\\
&&+ (6a b  + 16a b  + 6a b )x^2 \\
&&+ (4a b  + 4a b )x^1 \\
&&+ a bx^0\\
(m^4)*(n^5) &=& x^9 \\
&&+ (5b + 4a)x^8  \\
&&+ (10b^2  + 20a b + 6a^2 )x^7\\
&&+ (10b^3  + 40a b^2  + 30a^2 b + 4a^3 )x^6\\
&&+ (5b^4  + 40a b^3  + 60a^2 b^2  + 20a^3 b + a^4 )x^5 \\
&&+ (b^5  + 20a b^4  + 60a^2 b^3  + 40a^3 b^2  + 5a^4 b)x^4 \\
&&+ (4a b^5  + 30a^2 b^4  + 40a^3 b^3  + 10a^4 b^2 )x^3  \\
&&+ (6a^2 b^5  + 20a^3 b^4  + 10a^4 b^3 )x^2 \\
&&+ (4a^3 b^5  + 5a^4 b^4 )x^1 \\
&&+ a^4 b^5x^0
\end{array}
\]
\subsection{step 6  $(m^5)*(n^i)$}
\[
\begin{array}{ccl}
(m^5)*(n^0) &=& \\
(m^5)*(n^1) &=& \\
(m^5)*(n^2) &=& \\
(m^5)*(n^3) &=& \\
(m^5)*(n^4) &=& \\
(m^5)*(n^5) &=& 
\end{array}
\]
\section{Groups}
Notes from the meeting:
\[
G = <a,b ; [a,b]^n=1>
\]
The dihedral group:
\[
D=<a,b; a^2=1, b^{2n}=1>
\]
\[
is |D| = 4n?
\]
\[
{\rm\ G\ maps\ to\ D\ under\ }\rho
\]
\[
a \rightarrow a
\]
\[
b \rightarrow b
\]
\[
[a,b] = a^{-1}b^{-1}ab
\]
\[
\rho([a,b])=b^2
\]
\[
\rho([a,b]^n)=b^{2n}
\]
So $\rho$ is a homomorphism.
Let $K={\rm kernel}\rho$ and $G/K \equiv D$.
Thre is an algorithm to rewrite elements in $K$ in terms of coset
representatives. For example, let
\[
\alpha=(a^{-1}b^{-1}ab)^n
\]
\section{Matrices}
\section{Series}
\subsection{Generalized Binomial Expansion}
From \cite{1} we read:

By 1665, Issac Newton had found a simple way to expand -- his word was 
``reduce''-- binomial expressions into series. For him, such reductions
would be a means of recasting binomials in alternate form as well as
an entryway into the method of fluxions. This theorem was the starting
point for much of Newton's mathematical innovation.

As described in the {\sl epistola prior}, the issue as hand was to reduce
the binomial $(P+PQ)^{m/n}$ and to do so whether $m/n$ ``is integral or
(so to speak) fractional, whether positive or negative''. $\ldots$

Newton discovered a pattern for expanding not only elementary
binomials like $(1+x)^5$ but more sophisticated ones like
$\frac{1}{\sqrt[3]{(1+x)^5}} = (1+x)^{5/3}$. The reduction,
as Newton explained to Leibniz, obeyed the rule
\[
(P+PQ)^{m/n}  =  p^{m/n} + \frac{m}{n}AQ + \frac{m-n}{2n}BQ 
+\frac{m-2n}{3n}CQ + \frac{m-3n}{4n}DQ +{\rm etc.},
\]
\noindent
where each of $A$, $B$, $C$, $\ldots$ represents the previous term,
as will be illustrated below. This is his famous binomial expansion,
although perhaps in an unfamiliar guise.

Newton provided the example of
$\sqrt{c^2+x^2}=[c^2+c^2(x^2)/c^2]^{1/2}$.
Here, $P=c^2$, $Q={x^2}/{c^2}$, $m=1$, and $n=2$. Thus,
\[
\sqrt{c^2+x^2} = (c^2)^{1/2} + \frac{1}{2}A\frac{x^2}{c^2}
-\frac{1}{4}B\frac{x^2}{c^2}-\frac{1}{2}C\frac{x^2}{c^2} 
-\frac{5}{8}D\frac{x^2}{c^2} - \ldots
\]

To identify $A$, $B$, $C$, and the rest, we recall that each is the
immediately preceding term. Thus, $A=(c^2)^{1/2}=c$, giving us
\[
\sqrt{c^2+x^2} =  c 
+\frac{x^2}{2c}
-\frac{1}{4}B\frac{x^2}{c^2}
-\frac{1}{2}C\frac{x^2}{c^2}
-\frac{5}{8}D\frac{x^2}{c^2} - \ldots
\]

Likewise $B$ is the previous term -- i.e., $B={x^2}/{2c}$ --
so at this stage we have
\[
\sqrt{c^2+x^2} =  c 
+\frac{x^2}{2c}
-\frac{x^4}{8c^3}
-\frac{1}{2}C\frac{x^2}{c^2}
-\frac{5}{8}D\frac{x^2}{c^2} - \ldots
\]

The analogous substitutions yield $C=-{x^4}/{8c^3}$ and then
$D={x^6}/{16c^5}$. Working from left to right in this fashion,
Newton arrived at
\[
\sqrt{c^2+x^2} =  c 
+\frac{x^2}{2c}
-\frac{x^4}{8c^3}
-\frac{x^6}{16c^5}
-\frac{5x^8}{128c^7}
\]

Obviously, the technique has a recursive flavor: one finds the coefficient
of $x^8$ from the coefficient of $x^6$, which in turn requires the
coefficient of $x^4$, and so on. Although the modern reader is probably
accustomed to a ``direct'' statement of the binomial theorem, Newton's
recursion has an undeniable appeal, for it streamlines the arithmetic
when calculating a numerical coefficient from its predecessor.

For the record, it is a simple matter to replace $A$, $B$, $C$, $\ldots$
by their equivalent expressions in terms of $P$ and $Q$, then factor the
common $P^{m/n}$ from both sides to arrive at the result found in today's
texts:
\[
(1+Q)^{m/n} = 1 
+\frac{m}{n}Q
+\frac{\frac{m}{n}\left(\frac{m}{n}-1\right)}{2\rm{\ x\ }1}Q^2 
+\frac{\frac{m}{n}\left(\frac{m}{n}-1\right)
                     \left(\frac{m}{n}-2\right)}
         {3\rm{\ x\ }2\rm{\ x\ }1}Q^3 
\]

Newton likened such reductions to the conversion of square roots into
infinite decimals, and he was not shy in touting the benefits of the 
operation. ``It is a convenience attending infinite series,'' he
wrote in 1671,
\begin{quote}
that all kinds of complicated terms,\ldots may be reduced to the
class of simple quantities, i.e., to an infinite series of fractions
whose numerators and denominators are simple terms, which will thus
be freed from those difficulties that in their original form seem'd
almost insuperable
\end{quote}

To be sure, freeing mathematics from insuperable difficulties is a
worthy undertaking.

One additional example may be helpful. Consider the expansion of
${1}/{\sqrt{1-x^2}}$, which Newton put to good use in a result
we shall discuss later in the chapter. We first write this as
$(1-x^2)^{-1/2}$, identify $m=-1$, $n=2$, and $Q=-x^2$, to get
\[
\frac{1}{\sqrt{1-x^2}} = 1
+\left(-\frac{1}{2}\right)(-x^2)
+\frac{(-1/2)(-3/2)}{2\rm{\ x\ }1}(-x^2)^2 
\]
\[
+\frac{(-1/2)(-3/2)(-5/2)}{3\rm{\ x\ }2\rm{\ x\ }1}(-x^2)^3
\]
\[
+\frac{(-1/2)(-3/2)(-5/2)(-7/2)}{4\rm{\ x\ }3\rm{\ x\ }2\rm{\ x\ }1}(-x^2)^4
+\ldots 
\]
\[
= 1+\frac{1}{2}x^2+\frac{3}{8}x^4+\frac{5}{16}x^6+\frac{35}{128}x^8
+\ldots
\]
\subsection{Inverting Series}

Having described a method for reducing certain binomials to infinite
series of the form $z=A+Bx+Cx^2+Dx^3+\ldots$, Newton next sought a
way of finding the series for $x$ in terms of $z$. In modern terminology,
he was seeking the inverse relationship. The resulting technique involves
a bit of heavy algebraic lifting, but it warrents our attention for it too
will appear later on. As Newton did, we describe the inversion procedure by
means of a specific example.

Beginning with the series $z=x-x^2+x^3-x^4+\ldots$, we rewrite it as
\[(x-x^2+x^3-x^4+\dots)-z=0\]
and discard all powers of $x$ greater than or equal to the quadratic.
This, of course, leaves $x-z=0$, and so the inverted series begins as
$x=z$.

Newton was aware that discarding all those higher degree terms rendered
the solution inexact. The {\sl exact} answer would have the form
$x=z+p$, where $p$ is a series yet to be determined. Substituting $z+p$
for $x$ gives
\[ 
[(z+p)-(z+p)^2+(z+p)^3-(x+p)^4+\dots] 
\]
which we then expand and rearrange to get
\[
[-z^2+z^3-z^4+z^5-\dots]
+[1-2z+3z^2-4z^3+5z^4-\dots]p
\]
\[
+[-1+3z-6z^2+10z^3-\dots]p^2
+[1-4z+10z^2-\dots]p^3
\]
\[
+[-1+5z-\dots]p^4
+\dots=0
\]

Next, jettison the quadratic, cubic, and higher degree terms in $p$ and
solve to get
\[
p=\frac{z^2-z^3+z^4-z^5+\dots}{1-2z+3z^2-4z^3+\dots}
\]

Newton now did a second round of weeking, as he tossed out all but
the lowest power of $z$ in the numerator and denominator. Hence $p$
is approximately ${z^2}/{1}$, so the inverted series at this stage
looks like $x=z+p=z+z^2$.

But $p$ is not {\sl exactly} $z^2$. Rather, we say that $p=z^2+q$, where
$q$ is a series to be determined. To do so, we substitute to get
\[
[-z^2+z^3-z^4+z^5-\dots]
+[1-2z+3z^2-4z^3+5z^4-\dots](z^2+q)
\]
\[
+[-1+3z-6z^2+10z^3-\dots](z^2+q)^2
+[1-4z+10z^2-\dots](z^2+q)^3
\]
\[
+[-1+5z-\dots](z^2+q)^4+\dots=0
\]

We expand and collect terms by powers of $q$:
\[
[-z^3+z^4-z^6+\dots]
+[1-2z+z^2+2z^3-\dots]q
\]
\[
+[-1+3z-3z^2-2z^3+\dots]q^2+\dots
\]

As before, discard terms involving powers of $q$ above the first, 
solve to get 
\[q=\frac{z^3-z^4+z^6-\dots}{1-2z+z^2+2z^3+\dots}\] and then drop
all but the lowest degree terms top and bottom to arrive at
$q={z^3}/{1}$. At this point, the series looks like
$x=z+z^2+q=z+z^2+z^3$.

The process would be continued by substituting $q=z^3+r$.
Newton, who had a remarkable tolerance for algebraic monotony, seemed
able to continue such calculations {\sl ad infinitum} (almost). But
eventually even he was ready to step back, examine the output, and seek
a pattern. Newton put it this way: ``Let it be observed here, by the bye,
that when 5 or 6 terms $\ldots$ are known, they may be continued at 
pleasure for most part, by observing the analogy of the progression''

For our example, such an examination suggests that
\[x=z+z^2+z^3+z^4+z^5+\dots\] is the inverse of the series
\[z=x-x^2+x^3-x^4+\dots\] with which we began.

In what sense can this be trusted? After all, Newton discarded most of
his terms most of the time, so what confidence remains that the answer
is correct?

Again, we take comfort in the following ``check.'' The original series
\[z=x-x^2+x^3-x^4+\dots\] is geometric with common ratio $-x$, and so in
closed form $z={x}/(1+x)$. Consequently, $x={z}/(1-z)$, which
we recognize to be the sum of the geometric series
\[z+z^2+z^3+z^4+z^5+\dots\]. This is precisely the result to which
Newton's procedure has led us. Everything seems to be in working order.
\section{Riedemeister-Schrier}
\section{The Makefile}
<<*>>=
TANGLE=/usr/local/bin/NOTANGLE
WEAVE=/usr/local/bin/NOWEAVE

all: indefinite.dvi

indefinite.dvi: indefinite.pamphlet
	@echo building indefinite.dvi from indefinite.pamphlet
	@${WEAVE} -delay indefinite.pamphlet >indefinite.tex
	@latex indefinite.tex
	@latex indefinite.tex
	@rm -f indefinite*~
	@rm -f indefinite.log
	@rm -f indefinite.aux
	@rm -f indefinite.tex
	@rm -f indefinite.toc

@
\eject
\begin{thebibliography}{99}
\bibitem{1} A Calculus Gallery
\bibitem{2} William Sit paper
\bibitem{4} Axiom, The 30 Year Horizon
\end{thebibliography}
\end{document}