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\markright{Chapter 2: Answers 4}
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\Bf{Chapter 2: Answers 4 \hfill Jack K. Cohen \hfill Colorado School of Mines}

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\item Divide the equation by $a$ to get an equation of the form $x^3 + p x^2 + q x + r = 0$.  Let $f(x) = x^3 + p x^2 + q x + r $.   Note that for sufficiently large positive $x$, $f(x)$ is as large as you like---hence we can find $B > 0$ such that $f(B) > 0$.  Similarly, considering  large negative $x$, we see that we can find $A < 0$ such that $f(A) < 0$.  Since $f(x)$ is continuous in the closed interval $[A, B]$ and $f(A) < 0 < f(B)$, the intermediate value theorem guarantees that there is a $C$ in $[A, B]$ such that $f(C) = 0$.

\noindent Discussion: the division by $a$ is not essential, it just lets us avoid considering two cases (positive $a$ and negative $a$).   Likewise, insisting on  a \Em{positive} endpoint $B$ and a \Em{negative} endpoint $A$ was just one way of guaranteeing $A < B$ to avoid having to consider the possibility that the interval is really $[B, A]$.    Most importantly, note that a rough sketch of $f(x)$ makes clear the existence of the root and is a guide to the formal proof.

\item Each bisection cuts the interval in half.  So 10 bisections reduces it to $2^{-10}$-th of its original size.  The approximation $2^{10} = 1024 \approx 10^3$ justifies the folklore.  Ahem.  Remembering that $2^{10} \approx 10^3$ may well be more important than anything else on this worksheet!

\item $r \approx 50008.332$.

\item $r \approx 50008.331912$.  An appropriate input is:
\begin{verbatim}
Bisect[Cos[1000/r] - r/(r + h), {r, 50008.3215, 50008.3325},  0.0000005]
\end{verbatim}

\item Our interval has length on the order of $10^6$, so after about 30 iterations  we'd have 3 places after the decimal point (do you know why?).  This \Em{might} be considered practical by some, but expert opinion is that bisection alone is an inadequate equation solving method.  Stay tuned.


\end{enumerate}
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