\section{WHY IS NUMERICAL ANALYSIS NECESSARY?}
Consider the quadratic equation $x^2 - 80x + 1 = (x-r_1)(x-r_2) = 0$
with roots
$$
{80 \pm \sqrt{80^2-4}\over 2}= 
\cases{\displaystyle
80\left(1-{1\over 80^2}-\cdots\right)\cr
\displaystyle
{1\over 80}\left(1+{1\over 80^2}+\cdots\right)\cr}\approx
\cases{\displaystyle 80.0\cr .0125\cr}
$$
to 3 significant digits.
Using 3-digit arithmetic, the roots are calculated as
$$
\eqalign{
{.800\cdot 10^2 \pm \sqrt{.640\cdot 10^4 - .400\cdot 10^1}
\over .200\cdot 10^1}
& = {10^2\bigl[.800 \pm \sqrt{.640 - .0004}\,\bigr]
\over .200\cdot 10^1}\cr
& \rightarrow
{10^2\bigl[.800 \pm \sqrt{.640}\,\bigr]
\over .200\cdot 10^1}=
\cases{\displaystyle 80.0\cr 0.00}.\cr}
$$
Note that
$ax^2 + bx + c = a(x-r_1)(x-r_2)=ax^2-(r_1+r_2)x+ar_1r_2
\Rightarrow r_2=c/(ar_1).$
Calculating the roots this way gives
$r_1=80.0, r_2={1/80.0}=.0125$, perfect!
In general,
$$
r_1 = {-b -\hbox{sign}(b) \sqrt{b^2-4ac} \over 2a},
\qquad r_2={c\over ar_1}.
$$

Consider using 5-digit arithmetic to calculate the dot product of
$$ \displaylines{
x=
\pmatrix{100.01\cr 102.01\cr .01\cr}
\hbox{and\ }
y=
\pmatrix{1.02\cr -1.00\cr .04\cr}.
\cr
\matrix{\underline{\hbox{calculated $x_iy_i$}}
\cr 102.01\cr -102.01\cr 0.0004\cr}
\qquad
\matrix{\underline{\hbox{exact $x_iy_i$}}
\cr 102.0102\cr -102.01\cr .0004\cr}.
\cr}
$$
The calculated dot product
$$
fl\left(\sum_{i=1}^3 x_iy_i\right)=0.004,
$$
whereas the exact $\sum_{i=1}^3 x_iy_i = 0.0006$, a 33\% relative error.
A mathematical analysis shows $$
fl\left(\sum_{i=1}^3 x_iy_i\right)=
\sum_{i=1}^3 x_iy_i (1+3.03\mu\theta_i), \hbox{ where }
\mu=10^{-4},\ |\theta_i| \leq 1.
$$

\vfil\eject
\noindent {\bf Rules of thumb}:

\item{1)} Avoid cancellation error if possible. Examples:
$$\displaylines{
\sqrt{x+1} - \sqrt{x} = {1\over\sqrt{x+1} + \sqrt{x}},\cr
\sin(x+\epsilon) - \sin x =
2 \cos\left(x+ {\epsilon\over 2}\right) \sin {\epsilon\over 2},\cr
\int_n^{n+1} {dx\over 1+x^2} = \tan^{-1} (n+1) - \tan^{-1} n =
\tan^{-1} \left({1 \over 1 + n (n+1)}\right), \cr
{1\over\sqrt{x}} - {1\over\sqrt{x+1}} =
{1\over(x+1)\sqrt{x} + x \sqrt{x+1}}.\cr}$$

\item{2)} Don't add terms with alternating signs or very different
magnitudes. e.g., write $$\sum_{k=1}^{100} {(-1)^k\over{k^4}} \qquad
\hbox{as} \qquad \sum_{i=50}^1 {(-1)^{2i}\over{(2i)^4}} + \sum_{i=49}^0
{(-1)^{2i+1}\over{(2i+1)^4}}.$$

\item{3)} Write iterative formulas in terms of corrections. e.g.,
write the iteration $$
x_{n+1} = {x_{n-1} f(x_n) - x_nf(x_{n-1}) \over f(x_n) - f(x_{n-1})}
\qquad \hbox{as} \qquad x_{n+1} = x_n - \underbrace{
f(x_n) \left({x_n - x_{n-1} \over f(x_n) - f(x_{n-1})}\right)
}_{\hbox{correction to $x_n$}}.$$

\item{4)} Don't divide by small numbers.  e.g., $$
{{1 - \cos x\over\sin x}} = \tan{x\over 2}\quad\hbox{for }x\approx0.$$
\vfil\eject