hw3_solutions - Homework 3 Solution Sketches Math 471 Fall...

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Homework 3, Solution Sketches Math 471, Fall 2007 Assigned: Friday, September 24, 2007 Due: Friday, October 1, 2007 (1) (Newton versus Secant) Bradie, p. 113, #12. The sequence of iterates generated by the secant method follows. ======================================== n p(n) |e(n)| ======================================== 0 -2.00000000000000 0.91222917848440 1 -3.00000000000000 0.08777082151560 2 -2.83333333333333 0.07889584515106 3 -2.90792838874680 0.00430078973759 4 -2.91244964042237 0.00022046193798 5 -2.91222858559119 0.00000059289320 6 -2.91222917840283 0.00000000008157 7 -2.91222917848440 0.00000000000000 ======================================== It takes five iterations and six function evaluations to obtain p 6 , which has an error of - 8 . 16 · 10 - 11 . In this case, Secant is comparable with Newton. 1
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2 (2) (Finding Multiple Roots) (a) Bradie, p. 105, #11 =============================================================================== n p(n) |e(n)| ln e(n)/ln e(n-1) e(n)/e(n-1) =============================================================================== 0 0 0.33333333333333 1 0.11290322580645 0.22043010752688 1.37644064480852 0.66129032258065 2 0.18714686945667 0.14618646387666 1.27159410916128 0.66318737270876 3 0.23620832719698 0.09712500613635 1.21264234763671 0.66439124089009 4 0.26872882688378 0.06460450644956 1.17485304339133 0.66516862154799 5 0.29032765253196 0.04300568080137 1.14855128324725 0.66567617593288 6 0.30469112286118 0.02864221047215 1.12917865475386 0.66600993027961 7 0.31425102136460 0.01908231196874 1.11430737959058 0.66623042196037 8 0.32061732842082 0.01271600491251 1.10252616151281 0.66637653410887 9 0.32485845235361 0.00847488097972 1.09295869129396 0.66647355344927 10 0.32768450259607 0.00564883073726 1.08503204972648 0.66653806121626 =============================================================================== The apparent order of convergence is linear with asymptotic rate constant 0 . 666. The Bisection method converges faster. (Why?) (b) Bradie, p. 125, #14 (i) The Secant Method appears to be converging with order α 1 . 6. The lack of accuracy reflects the fact that the root has higher multiplicity. ============================================================= n p(n) e(n) ln e(n)/ln e(n-1) ============================================================= 0 0.50000000000000 0.50000000000000 1 2.00000000000000 1.00000000000000 2 0.85907511736897 0.14092488263103 3 0.97121826967156 0.02878173032844 1.81064723424204 4 1.00157397663295 0.00157397663295 1.81908784573958 5 0.99998454815926 0.00001545184074 1.71638131481937 6 0.99999999190723 0.00000000809277 1.68195171220410 7 0.99999999190723 0.00000000809277 ============================================================= (ii) The Secant Method appears to be converging to the root. The data on the order of convergence are not really conclusive.
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