\hfill \thepage} %} \input{tcilatex} \begin{document} \subsection*{Ma116 \hspace{2.0in} Project 2 \hfill 3/03/00} \subsection*{{\protect\small Name: \protect\underline{\hspace{2.5in}} \hfill ID: \protect\underline{\hspace{2.0in}}}} \subsection*{{\protect\small E-Mail: \protect\underline{\hspace{2.5in}} \hfill T.A./Recitation: \protect\underline{\hspace{2.0in}}}} \hfill {\small \textit{I pledge my honor that I have abided by the Stevens Honor System.}\hfill\underline{\hspace{2.3in}}}\\[10pt] \textbf{Please enter your final answers within the underlined regions along the right-hand side of the page. Show your work in the space below the question.} \hfill \textbf{Introduction}\\[2pt] In this project we approximate the infinite series \[ S \,=\, \dsum\limits_{k=2}^{\infty}\dfrac{1}{k\ln^{2}k} \] accurate to $10^{-4}$. We first try to make clear why achieving such accuracy is a non-trivial problem. The point here is that one can ``never'' achieve this accuracy by simply truncating the series at the $N$th partial sum. Then, using the fact that the terms in the series are positive and decreasing, an estimate for $S$ is derived with a much sharper bound on the maximum error, enabling us to easily estimate the sum to the specified accuracy. We begin by writing the above series as a sum of two pieces, the partial sum $S_N$ up to and including $k=N\,$, and the remainder $R_N$. Thus \[ \dsum\limits_{k=2}^{\infty}\dfrac{1}{k\ln ^{2}k} \,=\, \dsum\limits_{k=2}^{N}\dfrac{1}{k\ln ^{2}k} \,+\, \dsum\limits_{k=N+1}^{\infty}\dfrac{1}{k\ln^{2}k} \,=\,S_{N} + R_{N}. \] \hfill \textbf{1.\hspace{2pt} An upper bound on $R_N$.}\\[2pt] As shown in Figure 1, the graph of the continuous function $f(x) = 1\,/\,(x\ln^{2}x)$ makes it clear that the remainder term is bounded from above by an improper integral, \begin{equation} R_{N} \,=\, \dsum\limits_{N+1}^{\infty}\dfrac{1}{k\ln^{2}k} \,\leq\, \dint_{N}^{\infty}\dfrac{1}{x\ln^{2}x}\,dx\,=\,R_N^{\,+}\,. \tag{(1)} \end{equation} \[ \FRAME{itbpF}{4.21in}{1.84in}{0in}{\Qcb{{\protect\small \mathbf{Figure 1. \hspace{5pt} $R_N\,=\,\dsum\limits_{N+1}^{\infty}\,f\,(k) \,\leq\,\dint_{N}^{\infty}\,f\,(x)\,dx\,=\,R_N^{\,+}$.}}}}{} {proj2_fig1.gif% } {\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.21in;height 1.84in;depth 0in;original-width 4.21in;original-height 1.84in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'proj2_fig1.gif';file-properties "XNPEU";}} \] The upper bound in (1) just follows from $f(x)$ being a \emph{\textit{% decreasing}} function on $2 \leq x < \infty$. In fact, this is the same argument used when applying the \emph{\textit{integral test}} to prove convergence (see section 8.3 of the Stewart text). \hfill \textbf{a) }\hspace{1pt} Carry out the integration in equation (1) to determine the upper bound on the remainder $R_N$, as a function of $N$. \hfill\left.\underline{\,R_N^{\,+}\,=\,\hspace{1.0in}}\right. \hfill Let $S_e$ denote our estimate for $S$ and $E = S - S_e$ the \emph{\textit{% error}} in our estimate. One might first try to estimate $S$ by simply taking $S_e = S_N$ for sufficiently large $N$. In this case the upper bound on $R_N$ serves as our \emph{\textit{error bound}}, \[ \left|\, E \,\right| \,=\, \left|\,S - S_e\,\right| \,=\, \left|\,S - S_N\,\right| \,=\, \left|\,R_N\,\right| \,\leq\, R_N^{\,+}\,. \] \hfill \textbf{b) }\hspace{1pt} Using the upper bound from part (a) as our error estimate, how big should $N\,$ be to ensure that the error is less than $% 10^{-4}$? \hfill\left.\underline{\,N\,\geq\,\hspace{1.0in}}\right. \hfill The number $N\,$ just calculated is so large that summing the series to obtain $S_{N}$, at a rate of one term per second would require $% 2.8\times10^{4333}$ centuries. Rather than just truncating the series at the $N\,$th partial sum we need to make a more precise estimate of the remainder $R_N$\/. \hfill \textbf{2.\hspace{2pt} A lower bound on $R_N$.}\\[2pt] Returning to the graph of $f(x)$ in Figure 2, you see that by drawing boxes starting at $N+1\,$, instead of $N\,$, you get a \emph{\textit{lower bound}} for $R_{N}\,$, vis. \begin{equation} R_N^{\,-} \,=\, \dint_{N+1}^{\infty }\dfrac{1}{x\ln^{2}x}\,dx \,\leq\, \dsum\limits_{N+1}^{\infty}\dfrac{1}{k\ln^{2}k} \,=\, R_{N}\,. \tag{(2)} \end{equation} \[ \FRAME{itbpF}{4.21in}{1.84in}{0in}{\Qcb{{\protect\small \mathbf{Figure 2. \hspace{5pt} $R_N^{\,-}\,=\,\dint_{N+1}^{\infty}f(x)\,dx\,\leq\, \dsum\limits_{N+1}^{\infty}f(k)\,=\,R_N$.}}}}{}{proj2_fig2.gif} {\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.21in;height 1.84in;depth 0in;original-width 4.21in;original-height 1.84in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'proj2_fig2.gif';file-properties "XNPEU";}} \] Again it is worth noting that the inequality in (2) is the same argument used in applying the integral test to prove divergence of an infinite series. \hfill \textbf{a) }\hspace{1pt} Carry out the integration in equation (2) to determine the lower bound on the remainder $R_N$, as a function of $N$. \hfill\left.\underline{\,R_N^{\,-}\,=\,\hspace{1.0in}}\right. \hfill \textbf{3.\hspace{2pt} A sharper bound on the error.}\\[2pt] From Figures 1 and 2, it is clear that the bounds on $R_N$ in equations (1) and (2) hold for any infinite series of the form $\sum f(k)$, where $f(x) > 0 $ is a decreasing function. Combining the two inequalities from (1) and (2), we have the result that the remainder $R_N$ must lie in the range, \begin{equation} R_N^{\,-} \,=\, \dint_{N+1}^{\infty }\,f\,(x)\,dx \,\leq\, R_{N} \,\leq\, \dint_{N}^{\infty}\,f\,(x)\,dx \,=\, R_N^{\,+}. \tag{(3)} \end{equation} Adding the partial sum $S_N$ to each term in the previous inequality gives upper and lower bounds on the the exact value of the infinite series $% S=\sum\,f\,(k)$, \[ S_N + \dint_{N+1}^{\infty }\,f\,(x)\,dx \,\leq\, \dsum\limits_{2}^{\infty}\,f\,(k) \,\leq\, S_N + \dint_{N}^{\infty}\,f\,(x)\,dx\,. \] Taking our estimate $S_e$ to lie somewhere between these lower and upper bounds means that the absolute value of the error in our estimate is no larger than the distance between the upper and lower bounds on $R_N$. Choosing the estimate $S_e$ to be the midpoint of these upper and lower bounds, \begin{equation} S_e = S_N + 0.5\,\left(R_N^{\,-} + R_N^{\,+}\right)\,, \tag{(4)} \end{equation} reduces the maximum error by another factor of two. \begin{equation} \left|\, E \,\right| \,=\, \left|\,\dsum\limits_{2}^{\infty}\,f\,(k) - S_e\,\right| \,\leq\, 0.5\,\left(R_N^{\,+} - R_N^{\,-}\right) \tag{(5)} \end{equation} \hfill \textbf{a) }\hspace{1pt} For our specific example with $f(x) = 1/(xln^{2}x)$% , express the error bound in equation (5) as a function of $N$. \hfill\left.% \underline{\,\left|\, E \,\right|\,\leq\,\hspace{1.0in}}\right. \hfill \textbf{4.\hspace{2pt} A more convenient error bound.}\\[2pt] Returning to the bounds on the remainder $R_N$ in equation (3), observe that the lower bound can be rewritten as two integrals, \[ R_N^{\,-}\,=\, \dint_{N}^{\infty }\,f\,(x)\,dx - \dint_{N}^{N+1}\,f\,(x)\,dx \,\leq\, R_{N} \,\leq\, \dint_{N}^{\infty}\,f\,(x)\,dx \,=\, R_N^{\,+}\,. \] As shown in Figure 3, the integral $\int_N^{N+1} \,f\,(x)\,dx$ is bounded from above by $f\,(N)$ and we can replace the estimate in (3) with a slightly weaker estimate, \[ \dint_{N}^{\infty }\,f\,(x)\,dx - f\,(N) \,\leq\, R_{N} \,\leq\, \dint_{N}^{\infty}\,f\,(x)\,dx\,. \] \[ \FRAME{itbpF}{4.21in}{1.84in}{0in}{\Qcb{{\protect\small \mathbf{Figure 3. \hspace{5pt}$\dint_{N}^{N+1}\,f(x)\,dx\,\leq\,f(N)$.}}}}{} {proj2_fig3.gif} {% \special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.21in;height 1.84in;depth 0in;original-width 4.21in;original-height 1.84in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'proj2_fig3.gif';file-properties "XNPEU";}} \] As in equation (4), we use the midpoint of this interval, $\int_N^{\infty} \,f\,(x)\,dx - 0.5f\,(N)\,$, as our estimate for $R_N$, giving us a new estimate for $S$, \begin{equation} S_* \,=\, S_N + \dint_{N}^{\infty}\,f\,(x)\,dx - 0.5f\,(N)\,. \tag{(6)} \end{equation} One advantage is that we now have a very simple means of estimating the error. Using either of the estimates $S_e$ or $S_*$, the error can be no larger than $f\,(N)\,/\,2$, \begin{equation} \left|\, E \,\right| \,\leq\, 0.5\,f\,(N)\,. \tag{(7)} \end{equation} \hfill \textbf{a) }\hspace{1pt} Use the error bound in (7) to determine how large $N $ must be to estimate our infinite series to within $10^{-4}$\,. \hfill\left.% \underline{\,N\,\geq\,\hspace{1.0in}}\right. \hfill \textbf{b) }\hspace{1pt} Complete the estimate for $S=\sum_{k=2}^{\infty}\,f% \,(k)$ accurate to $10^{-4}$\,. \hfill\left.\underline{\,S_*\,=\,\hspace{% 1.0in}}\right. \hfill \textbf{5.\hspace{2pt} $\left|\,E\,\right| < 10^{-5}$.}\\[2pt] \textbf{a) }\hspace{1pt} Use the error bound in (7) to determine how large $N $ must be to estimate our infinite series to within $10^{-5}$\,. \hfill\left.% \underline{\,N\,\geq\,\hspace{1.0in}}\right. \hfill \textbf{b) }\hspace{1pt} Estimate $S=\sum_{k=2}^{\infty}\,f\,(k)$ accurate to $10^{-5}$\,. \hfill\left.\underline{\,S_*\,=\,\hspace{1.0in}}\right. 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