Commit 27ac2843 authored by Jim Hefferon's avatar Jim Hefferon

jc2 edits

parent ff791769
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......@@ -349,7 +349,7 @@
\renewcommand{\section}{\@startsection%
{section}{1}{0em}{-12ex plus1ex minus2ex}{1em}%
{\raggedright\Large\bfseries}}
%\renewcommand{\subsection}[1]{\gdef\sectioningname{#1}\@startsection%
% \renewcommand{\subsection}[1]{\gdef\sectioningname{#1}\@startsection%
% {subsection}{2}{0em}{10ex plus1ex minus1ex}{1em}%
% {\raggedright\large\bfseries}}
\renewcommand{\subsection}{%
......@@ -362,11 +362,12 @@
\newcommand{\subseccmdb}[1]{\subsecheader{#1}}
\usepackage{trimspaces}
\newcommand{\subsecheader}[1]{\pagebreak[3]%
\vspace{10ex plus1ex minus1ex}%
\gdef\sectioningname{#1}% grab the subsec's name
{\raggedright\large\bfseries \thesection.\thesubsection\ #1}%
\nopagebreak\vspace*{1em}\par\noindent}
{\raggedright\large\bfseries\thesection.\thesubsection\ #1}%
\nopagebreak\vspace*{1em}\par\noindent\trim@spaces}
% --------OPTIONAL SUBSECTION
%
......
......@@ -118,9 +118,9 @@ beginfig(4) % equiv relation; row-equiv mats from ch1.27, further split
path p[]; partition; % dotlabels.ulft(5,6,7,8,9,10,11,12,13);
label(btex {\scriptsize \ldots} etex,z13);
x14=.8[x0,x9]; y14=.7[y10,y9]; drawpoint(z14);
label.rt(btex {\scriptsize $A$} etex,z14);
label.rt(btex {\scriptsize $S$} etex,z14);
x15=.75[x0,x9]; y15=.35[y10,y9]; drawpoint(z15);
label.rt(btex {\scriptsize $B$} etex,z15);
label.rt(btex {\scriptsize $T$} etex,z15);
pickup pencircle scaled line_width_light;
% split part in center
......
......@@ -146,7 +146,7 @@ then \( m(x) \) is a \definend{factor\/} of \( c(x) \).
Any \definend{root\/} of the factor (any \( \lambda\in\Re \) such that
\( m(\lambda)=0 \)) is a root of \( c(x) \) since
\( c(\lambda)=m(\lambda)\cdot q(\lambda)=0 \).
The prior corollary immediately yields the following converse.
% The prior corollary immediately yields the following converse.
\begin{corollary}
If \( \lambda \) is a root of the polynomial \( c(x) \)
......@@ -154,6 +154,12 @@ then \( x-\lambda \) divides \( c(x) \) evenly, that is,
$x-\lambda$ is a factor of $c(x)$.
\end{corollary}
\begin{proof}
By the above corollary $c(x)=(x-\lambda)\cdot q(x)+c(\lambda)$.
Since $\lambda$ is a root,
$c(\lambda)=0$ so $x-\lambda$ is a factor.
\end{proof}
Finding the roots and factors of a high-degree polynomial can be hard.
But for second-degree polynomials we have the quadratic formula:~the
roots of \( ax^2+bx+c \) are
......@@ -198,7 +204,7 @@ we know that \( q(x)=(x-3)(x+2) \).
While \( x^2+1 \) has no real roots and so doesn't factor over the real
numbers, if we imagine a root\Dash traditionally denoted \( i \)
so that \( i^2+1=0 \)\Dash then \( x^2+1 \) factors into a product of linears
so that \( i^2+1=0 \)\Dash then \( x^2+1 \) factors into a product of linears
\( (x-i)(x+i) \).
So we adjoin this root \( i \) to the reals and close the new system with
......
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