%$Header$ \chapter[Solutions: \ccfrzeroxrefcomma{}Chapter \ref{ccfr0}] {Solutions: \ccfrzeroxrefcomma{}Chapter \ref{ccfr0}, \ccfrzerolongtitle{}} \label{ccfs0} \vworkexercisechapterheader{} \begin{vworkexercisesolution}{\ref{exe:cfr0:sexe0:a01}} Placeholder. \end{vworkexercisesolution} \vworkexerciseseparator \begin{vworkexercisesolution}{\ref{exe:cfr0:sexe0:b01}} Note that $1.609344 > 255/255$ (i.e. beyond the corner point'' in the sense suggested by Fig. \cfryzeroxrefhyphen{}\ref{fig:cfry0:ili0:00}), so it is necessary to find the Farey neighbors of $1.609344^{-1}$ in $F_{255}$ (Algorithm \ccfrzeroxrefhyphen{}\ref{alg:ccfr0:scba0:cfenclosingneighborsfab}), and then invert and re-order the results. Table \ref{tbl:ccfs0:exe:cfr0:sexe0:b01:01} shows the application of Algorithm \ccfrzeroxrefhyphen{}\ref{alg:ccfr0:scrn0:akgenalg} to form the continued fraction partial quotients and convergents of $1.609344^{-1}$ = 1,000,000/1,609,344. \begin{table} \caption{Continued Fraction Partial Quotients And Convergents Of $1.609344^{-1}$, The Reciprocal Of The Conversion Factor From Miles To Kilometers (Solution To Exercise \ccfrzeroxrefhyphen{}\ref{exe:cfr0:sexe0:b01})} \label{tbl:ccfs0:exe:cfr0:sexe0:b01:01} \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|} \hline \small{Index} & \small{Dividend} & \small{Divisor} & $a_k$ & \small{Remainder} & $p_k$ & $q_k$ \\ \small{(k)} & & & & & & \\ \hline \hline \small{-1} & \small{N/A} & \small{1,000,000} & \small{N/A} & \small{1,609,344} & \small{1} & \small{0} \\ \hline \small{0} & \small{1,000,000} & \small{1,609,344} & \small{0} & \small{1,000,000} & \small{0} & \small{1} \\ \hline \small{1} & \small{1,609,344} & \small{1,000,000} & \small{1} & \small{609,344} & \small{1} & \small{1} \\ \hline \small{2} & \small{1,000,000} & \small{609,344} & \small{1} & \small{390,656} & \small{1} & \small{2} \\ \hline \small{3} & \small{609,344} & \small{390,656} & \small{1} & \small{218,688} & \small{2} & \small{3} \\ \hline \small{4} & \small{390,656} & \small{218,688} & \small{1} & \small{171,968} & \small{3} & \small{5} \\ \hline \small{5} & \small{218,688} & \small{171,968} & \small{1} & \small{46,720} & \small{5} & \small{8} \\ \hline \small{6} & \small{171,968} & \small{46,720} & \small{3} & \small{31,808} & \small{18} & \small{29} \\ \hline \small{7} & \small{46,720} & \small{31,808} & \small{1} & \small{14,912} & \small{23} & \small{37} \\ \hline \small{8} & \small{31,808} & \small{14,912} & \small{2} & \small{1,984} & \small{64} & \small{103} \\ \hline \small{9} & \small{14,912} & \small{1,984} & \small{7} & \small{1,024} & \small{471} & \small{758} \\ \hline \small{10} & \small{1,984} & \small{1,024} & \small{1} & \small{960} & \small{535} & \small{861} \\ \hline \small{11} & \small{1,024} & \small{960} & \small{1} & \small{64} & \small{1,006} & \small{1,619} \\ \hline \small{12} & \small{960} & \small{64} & \small{15} & \small{0} & \small{15,625} & \small{25,146} \\ \hline \end{tabular} \end{center} \end{table} From the table, the highest-ordered convergent with a denominator not greater than 255 is 64/103. Applying Theorem \ccfrzeroxrefhyphen{}\ref{thm:ccfr0:scba0:convergentbetterappthanlesserdenominator} yields 151/243 as the other neighbor to $1.609344^{-1}$ in $F_{255}$. Inverting and ordering these fractions yields $$\frac{243}{151} < 1.609344 < \frac{103}{64}.$$ Either rational approximation is quite good, but 103/64 is closer to 1.609344 than 243/151, and so we choose 103/64 as the best rational approximation under the constraints. \end{vworkexercisesolution} \vworkexerciseseparator \begin{vworkexercisesolution}{\ref{exe:cfr0:sexe0:b02}} To find the best rational approximation to 1.609344 with a numerator and denominator no larger than 65,535, first consider the possibility that 1.609344 is already a rational number that can be expressed under these constraints. Examining the final convergent\footnote{For a rational number, the final convergent is the rational number in lowest terms. This is established by Theorem \ccfrzeroxrefhyphen{}\ref{thm:ccfr0:scnv0:evenslessthanoddsgreaterthan}.} of Table \ref{tbl:ccfs0:exe:cfr0:sexe0:b01:01} verifies that 1.609344 = 25,146/15,625. Thus, the best rational approximation under the constraints is 25,146/15,625, which is precisely equal to the number to be approximated. \end{vworkexercisesolution} \vworkexercisechapterfooter %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \vfill \noindent\begin{figure}[!b] \noindent\rule[-0.25in]{\textwidth}{1pt} \begin{tiny} \begin{verbatim} $RCSfile: c_cfs0.tex,v$ $Source: /home/dashley/cvsrep/e3ft_gpl01/e3ft_gpl01/dtaipubs/esrgubka/c_cfs0/c_cfs0.tex,v$ $Revision: 1.2$ $Author: dtashley$ $Date: 2001/07/01 21:10:59$ \end{verbatim} \end{tiny} \noindent\rule[0.25in]{\textwidth}{1pt} \end{figure} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % $Log: c_cfs0.tex,v$ % Revision 1.2 2001/07/01 21:10:59 dtashley % Safety check-in after major re-org. % % Revision 1.1.1.1 2001/07/01 00:14:34 dtashley % Initial import. % % %End of file C_CFS0.TEX