\newcommand{\titlesize}{30pt}  % for large title
\input{m211oh.tex}


\begin{center}
\tcr{\rmH Finding rem(f,g): an Example}   
\end{center}\vspace{1pt}

\LARGE
\newcommand{\rem}{\mbox{rem}}
\begin{description} 
\item[\tcr{Example:}] Let \mc{g(t)=(t-1)^2(t-2)}.  Find a \tcg{formula} for 
\mc{\rem(f,g)} which is valid for every \mc{f \in \C[t]}. 


\item[\tcr{Solution:}] Write \mc{r(t) = \mbox{rem}(f,g) = a_0+a_1t+a_2t^2}. Then the \mc{a_i}'s 
satisfy the following system of (linear) equations: %(and its row-reductions):
$$
\tcb{\begin{array}{rclc|rcl}
r(1) & = & f(1)  & & a_0 + a_1  + a_2 & = & f(1)\\
r'(1)& = & f'(1) & &  a_1 +2a_2 & = & f'(1) \\
r(2) & = & f(2)  & & a_0 + 2a_1 + 4a_2 & = & f(2) 
\end{array}}
$$ 
\tcg{Row-reducing} the associated augmented matrix yields:
%\exprow{1.1}\expcol{-2pt}
%$\hspace{1.1cm} 
$$
\tcb{
\left(\begin{array}{ccc|c}
1&1&1&f(1)\\
0&1&2&f'(1)\\
1&2&4&f(2)
\end{array} \right )
\stackrel{R3-R1}{\leftrightarrow}  
\left(\begin{array}{ccc|c}
1&1&1&f(1)\\
0&1&2&f'(1)\\
0&1&3&f(2)-f(1)
\end{array}\right)} 
$$
$$
\tcb{\stackrel{R3-R2}{\leftrightarrow} 
\left(\begin{array}{ccc|c}
1&1&1&f(1)\\
0&1&2&f'(1)\\
0&0&1&f(2)-f(1)-f'(1)
\end{array}\right).}
$$
%\iffalse
\exprow{1.0}\expcol{2pt}
\exprow{1.3}\expcol{-4pt} 
Thus, by \tcg{back-substitution} we obtain: 
$$
\tcb{\begin{array}[t]{ccccl}
a_2& = & \multicolumn{2}{l}{f(2)-f(1)-f'(1)}\\
a_1& = & f'(1)-2a_2& = &f'(1)-2f(2)+2f(1)+2f'(1)\\
   &   &           & = &3f'(1)-2f(2)+2f(1),\\
a_0& = & f(1)-a_1-a_2 & = & f(1)-3f'(1)+2f(2)-2f(1) \\
   &   &              &   & -f(2)+f(1)+f'(1)\\
   &   &           & = & -2f'(1)+f(2),
\end{array}}
$$
\newpage

and so we have
\begin{eqnarray}
\tcb{r(t)} &\tcb{=} & \tcb{\tcr{(}f(2)-2f'(1)\tcr{)} + \tcr{(}2f(1)+3f'(1)-2f(2)\tcr{)}t} 
\label{1}\\
& & \tcb{+ \tcr{(}f(2)-f(1)-f'(1)\tcr{)}t^2.} \nonumber
\end{eqnarray}

\item[\tcr{Note:}] The above formula exhibits the \tcg{coefficients} of \mc{r(t)} explicitly.
However, it is frequently useful to re-write the above formula in a way that exhibits its 
dependence on \mc{f} more clearly. 

\item[\tcr{Example:}] If \mc{g(t) = (t-1)^2(t-2)},  find a formula for \mc{\rem(f, g)} 
in terms of \mc{f(1),\, f'(1)} and \mc{f(2)}.
 
\item[\tcr{Solution:}] Looking at the result \sref{1} of the previous example, 
we shall separate out and collect the terms involving \mc{f(1), f'(1)} and \mc{f(2):}
\begin{eqnarray*}
\tcb{r(t)} &\tcb{\stackrel{\mbox{\large\sref{1}}}{=}} 
     & \tcb{(f(2)-2f'(1))+(2f(1)+3f'(1)-2f(2))t} \\
     & & \tcb{+(f(2)-f(1)-f'(1))t^2} \\  
     &\tcb{=}& \tcb{f(1)(2t-t^2)+f'(1)(-2+3t-t^2)} \\
     & & \tcb{+f(2)(1-2t+t^2).}
\end{eqnarray*}
In other words, we have for every \mc{f \in \C[t]:}
$$
\tcb{\rem(f,g)=f(1)e_{10}(t)+f'(1)e_{11}(t)+f(2)e_{20}(t)}
$$
where
\exprow{1.15}
\tcb{$\begin{array}[t]{rclcl}
e_{10}(t)&=&2t-t^2&=&t(2-t)\\
e_{11}(t)&=&-2+3t-t^2&=&(1-t)(t-2)\\
e_{20}(t)&=&1-2t+t^2&=&(1-t)^2.
\end{array}$}

\noi Note that these polynomials \tcr{do not depend} on \mc{f.}
\end{description}
\end{document}
