\magnification 2200
\parindent 0pt
\parskip 10pt
\pageno=1
\hsize 7truein
\hoffset -0.35truein
\vsize 9.2truein
\def\u{\vskip -5pt}
\def\v{\vskip 10pt}


\centerline{\bf{EXISTANCE}}


$span{1, t, t^2} = $ polynomials of degree 2

A polynomial $p(t)$ is in the span of $\{1, t, t^2\}$ if and only if there exists a solution for $a, b, c$ to the equation $$p(t) = a + bt + ct^2$$

Example 1:  $2 + t^3$ is not in the span of  $\{1, t, t^2\}$ since there does not exist 
$a, b, c$ such that $2 + t^3 = a + bt + ct^2$

Example 2:  $1 + 2t +  3t^2 $ is  in the span of  $\{1, t, t^2\}$ since there  exists 
$a, b, c$ such that $1 + 2t +  3t^2 = a + bt + ct^2$

$a = 1, b=2, c = 3$ is a solution.

\centerline{\bf{UNIQUENESS}}




${\bf b_1, ..., b_n}$ are linearly independent if 
\u

$c_1{\bf b_1} + c_2{\bf b_2} + ... + c_n{\bf b_n}  = 
d_1{\bf b_1} + d_2{\bf b_2} + ... + d_n{\bf b_n}$ 

\centerline{implies
$c_1 = d_1$, $c_2 = d_2$..., $c_n = d_n$.}

or equivalently,
 
${\bf b_1, ..., b_n}$ are linearly independent if
\u

$c_1{\bf b_1} + c_2{\bf b_2} + ... + c_n{\bf b_n}  = 0$ 
implies $c_1 = c_2 = ... c_n$.

\v\v

Example 1:
${\bf b_1} = (1, 0, 0)$, ${\bf b_2} = (0, 1, 0)$, ${\bf b_3} = (0, 0, 1)$.
\v

\centerline{$(1, 2, 3) \not= (1, 2, 4)$.}

\v
\centerline{If $(a, b, c) = (1, 2, 3)$ then $a = 1$, $b = 2$, $c = 3$.}

\v\v
Example 2:
${\bf b_1} = 1$, ${\bf b_2} = t$, ${\bf b_3} = t^2$.
\v
\centerline{$1 + 2t +  3t^2 \not= 1 + 2t +  4t^2$.}
\v

\centerline{If $a + bt +  ct^2 = 1 + 2t + 3t^2$ then $a = 1$, $b = 2$, $c = 3$.}

\eject
Application:  Partial Fractions

\centerline{${ 4\over (x^2 + 1)(x - 3)} = {Ax + B \over x^2 + 1} + {C \over x - 3}$}

If you don't like denominators, get rid of them:

$\hskip .7in  {4 = (Ax + B)(x-3) + C(x^2 + 1) }$

%Hence ${ 4} = {(Ax + B)(x-3) + C(x^2 + 1)  }$
%\v
%\centerline{Thus $4 = (Ax + B)(x-3) + C(x^2 + 1)$}
\v
\centerline{$ {4 = Ax^2 + Bx - 3Ax - 3B + Cx^2 + C}$}
\v
\centerline{${4 = (A + C)x^2 + (B - 3A)x - 3B + C }$}



I.e., $0x^2 + 0x + 4 = {(A + C)x^2 + (B - 3A)x - 3B + C }$

Thus $0 = A + C$,\hfill $0 = B - 3A$, \hfill ~~$4 = -3B + C$.  

$C = -A$, \hfil  ~$B = 3A$,\hfill 
 $4 = -3(3A) + -A$ $\Rightarrow$ $4 = -10A$. 

Hence $A = -{2 
\over 5}$, $B = 3(-{2 
\over 5}) = -{6 \over 5}$, $C = {2 \over 5}$.
\v
\centerline{Thus,
${ 4\over (x^2 + 1)(x - 3)} 
= {-{2 \over 5}x  - {6 \over 5} \over x^2 + 1} + {{2 \over 5} \over 
x - 3}$}

\hskip 1.65in = {${-{2}x - {6} \over 5(x^2 + 1)} + {{2} \over 
5(x - 3)}$}

\vfill\vfill\vfill\vfill

Alternatively, can plug in $x = 3$ to quickly find $C$ and then solve for $A$ and $B$.  Can also use matrices to solve linear eqns.

\end