dstefan
06-05-2008, 11:58 PM
Chapter 2, Problem 4
Let f : {\mathbb R} \to {\mathbb R} given by f(x) = \frac{x^{5/2}}{1+x^2}.
(i) Use Midpoint rule with tol = 10^{-6} to compute an approximation of
I ~=~ \int_0^1 f(x) ~dx ~=~ \int_0^1 \frac{x^{5/2}}{1+x^2}
(ii) Show that f^{(4)}(x) is not bounded on the interval (0,1).
(iii) Apply Simpson's rule with n = 2^k, k=2:8, intervals to compute the integral I, and conclude that Simpson's rule converges.
Solution:
(i) The approximate value of the integral is 0.179171, and is obtained for a partition of the interval [0, 1] using 512 intervals:
\begin{tabular}{|c|c|}\hline No. Intervals & Midpoint Rule \\ \hline 4 & 0.17715737 \\ \hline 8 & 0.17867774 \\ \hline 16 & 0.17904866 \\ \hline 32 & 0.17914062 \\ \hline 64 & 0.17916354 \\ \hline 128 & 0.17916926 \\ \hline 256 & 0.17917070 \\ \hline 512 & 0.17917105 \\ \hline \end{tabular}
(ii) Without computing f^{(4)}(x), note that the denominator 1 + x^2 of f(x) is bounded away from 0, and that the fourth derivative of the numerator of f(x) is on the order of x^{-3/2}, which is not defined at 0, and is unbounded in the limit as x \to 0.
(iii) Using Simpson's rule, the following approximate values of the integral are obtained:
\begin{tabular}{|c|c|}\hline No. Intervals & Simpson's Rule \\ \hline 4 & 0.17915510 \\ \hline 8 & 0.17916982 \\ \hline 16 & 0.17917106 \\ \hline 32 & 0.17917116 \\ \hline \end{tabular}
The approximate value of the integral is 0.179171, and is obtained for a partition of the interval [0, 1] using 32 intervals.
Let f : {\mathbb R} \to {\mathbb R} given by f(x) = \frac{x^{5/2}}{1+x^2}.
(i) Use Midpoint rule with tol = 10^{-6} to compute an approximation of
I ~=~ \int_0^1 f(x) ~dx ~=~ \int_0^1 \frac{x^{5/2}}{1+x^2}
(ii) Show that f^{(4)}(x) is not bounded on the interval (0,1).
(iii) Apply Simpson's rule with n = 2^k, k=2:8, intervals to compute the integral I, and conclude that Simpson's rule converges.
Solution:
(i) The approximate value of the integral is 0.179171, and is obtained for a partition of the interval [0, 1] using 512 intervals:
\begin{tabular}{|c|c|}\hline No. Intervals & Midpoint Rule \\ \hline 4 & 0.17715737 \\ \hline 8 & 0.17867774 \\ \hline 16 & 0.17904866 \\ \hline 32 & 0.17914062 \\ \hline 64 & 0.17916354 \\ \hline 128 & 0.17916926 \\ \hline 256 & 0.17917070 \\ \hline 512 & 0.17917105 \\ \hline \end{tabular}
(ii) Without computing f^{(4)}(x), note that the denominator 1 + x^2 of f(x) is bounded away from 0, and that the fourth derivative of the numerator of f(x) is on the order of x^{-3/2}, which is not defined at 0, and is unbounded in the limit as x \to 0.
(iii) Using Simpson's rule, the following approximate values of the integral are obtained:
\begin{tabular}{|c|c|}\hline No. Intervals & Simpson's Rule \\ \hline 4 & 0.17915510 \\ \hline 8 & 0.17916982 \\ \hline 16 & 0.17917106 \\ \hline 32 & 0.17917116 \\ \hline \end{tabular}
The approximate value of the integral is 0.179171, and is obtained for a partition of the interval [0, 1] using 32 intervals.