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\begin{center}{\large \bf \sc Mathematics 180
\hspace{1in}Calculus I \\

Newton's Method Program for the TI 81 and TI 82}
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This program calculates the solution to  $f(x)=0$ using
Newton's method.

Enter the function $f(x)$  as $Y1$  and graph it to find the

approximate zeros of $f$, i.e. the initial guesses.  Then
enter

$df/dx$ as $Y2$. Then run the Newton program. It will ask you

for your initial guess and for the number of iterations
($N=10$ is

usually sufficient).  The output of each iteration is then
displayed.



\bigskip\noindent
Program: NEWTON\\
:Disp "INITIAL"\\
:Input $A$\\
:Disp "$N$"\\
:Input $N$\\
:$A \to X$\\
:$1 \to J$\\
:Lbl $1$\\
:Disp X\\
:$X-Y1/Y2 \to X$\\
:$IS>(J,N)$  \hskip 0.5truein({\tt punch PRGM CTRL ALPHA A}) \\
:GoTo $1$\\
:Disp "AGAIN?"\\

\bigskip
\noindent
{\bf EXAMPLES:}\\


\noindent
1)  Find the zeros of $x^2-3$.  Here $Y1 = x^2-3$ and $Y2 =
2x$.

\bigskip
\noindent
2) Find the zeros of $x^3-4x^2 - 4$. Here $Y1 = x^3-4x^2 - 4$
and $Y2 =
3x^2-8x$.  Your initial guess should be $4$.  Try this initial
guess with

$N=10$ and see how fast it converges.  Then try it with
$N=2,0, \mbox{

and }-2$

and see what happens.  Can you give an explanation of why it
takes so

much longer to converge?


\bigskip
\noindent
3)  Find the critical points of $x^3 - e^x$.  Here we want to
find the

zeros of the {\bf derivative} of this function, so $Y1$ will
be the derivative

of $x^3 - e^x$, that is $Y1 = 3x^2-e^x$ and $Y2 = 6x-e^x$.

\bigskip
\noindent
4)  Try it on some other functions, say some polynomials of
degree $4$,

or find the point $x$ where $\cos x = x$.


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