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# MATRIX AND POWER SERIES METHODS MTH 306

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This 5 page Class Notes was uploaded by Mrs. Dedric Little on Monday October 19, 2015. The Class Notes belongs to MTH 306 at Oregon State University taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/224445/mth-306-oregon-state-university in Mathematics (M) at Oregon State University.

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Date Created: 10/19/15

Matrix and Power Series Methods Sample Problems Set 1 Bent E Petersen Winter 2005 Most perhaps all of the problems on the tests will be formulated as multiplechoice prob lems However I have not bothered to do so for these sample problems Note that logw means the natural logarithm of x If ever I need the logarithm base 10 of w I will denote it by log10w Problem 1 Determine if the improper integral O0 div 0 e 0 e x converges or not If it converges determine its value Problem 2 Sum the series i lt73 2 2 710 2 n Problem 3 Test for convergence i 4 0 2n 3n Problem 4 Test for convergence i 3 710 2 4n B E Petersen Version 20050130 Matrix and Power Series Methods Problem 5 Test for convergence 00 23710104 2 2701 710 Problem 6 Use Stirling s formula to estimate 10 Problem 7 Test for convergence 1 Z n10gngt239 n2 Problem 8 Test for convergence sinz For what values of w does the series 0 anw 710 Problem 9 converge Problem 10 Test for convergence Problem 11 Y 39 the or quot of the improper integral 39 the or quot of the series 71gtn1n71nr 1 Problem 12 n Mth 306 Winter 2005 B E Petersen Version 20050130 Matrix and Power Series Methods Problem 13 Y 39 the or quot of the series X 2671 0 3 3n1 Problem 14 Y 39 the or quot of the series iv 7 2n 3 2 3n 2 710 Problem 15 Given 1 TL 42 m lt7 gt e 7 271711 how many terms do we need to sum to approximate e 5 X 10 12 12 with an error no larger than Problem 16 The TaylorMaclaurin polynomial of degree 4 for a certain function g is given by 1 Pw 1 232 7 3 7 3w4 Suppose we also know 95gtwl 3 36 for S l Ifwe use P05 to estimate 905 compute the estimate and nd a good upper bound for the error Here are some proposed answers Note that the answers are not unique in many cases that is there is more than one way to do it Note also that I have been known to make a mistake every now and then so if we disagree check both your work and mine In fact part of your job is to check everything Answer 1 The integrand is dominated by e z substitution u ea yields the value 7r 4 Hence the integral converges The Answer 2 g Answer 3 Diverges by ratio test limiting ratio is 43 Alternater you can use the simple comparison 4 1 4 n gt 7 7 2 3 7 2 3 Answer 4 Converges by ratio test limiting ratio is 34 Alternater you can use the simple comparison 3 3 n 7 lt 7 2 4 lt4gt Mth 306 3 Winter 2005 B E Petersen Version 20050130 Matrix and Power Series Methods Answer 5 Converges by root test limiting root is l 2 Use Stirling s formula to compute the limiting root You can also use the ratio test the limiting ratio is 12 Answer 6 According to Stirling l l 5 10910011 i10g1027rngt 71 gt 10mm n10g10 gt m 10g10 gtr Replacing 71 by 101 3628800 we obtain log10101 2222810395674425 X 107 2222810395674425 It follows 101 905199 gtlt 1022928403 Answer 7 Converges by integral test since substituting u logw yields O dw iltgt0 dui l 2 log2 2 Answer 8 Since 0 lt sinw lt w ifO lt w lt 7r2 we have sin2ln lt ln2 ifn gt1 Thus we have convergence by comparison with the convergent p series with p 2 Answer 9 If at 7 0 then the absolute value of ratio of the n lSt term to the mm term is n l ml which tends to 00 ifw 7 0 Thus we have convergence if and only ifw 0 Answer 10 We know 33 7 l 10gw S 7 S P P for any at gt O andp gt O In particular logn 3 27112 Thus w logn 1 n2 3 7132 39 Thus we have convergence by comparison with the p series with p 32 Answer 11 b d 734lt17b 14gtH4 1 at so the integral converges Answer 12 1 10917171 7 a O Mth 306 4 Winter 2005 B E Petersen Version 20050130 Matrix and Power Series Methods Thus n ln a l and so we have divergence by the 0test 64 Answer 13 Converges by the ratio test The limiting ratio is a 4 Answer 14 Converges absolutely by the root test The limiting root is 5 Answer 15 The series is an alternating series with terms of decreasing magnitude with limit 0 Hence the error is bounded by the magnitude of the rst term omitted Thus we want to nd 71 such that 1 712 1 11 mgwm thatrs 2 n1122gtlt10 Now 211111 82 X1010 212121 20 X1012 Hence we need to take 71 l 12 that is n 11 Since the terms are numerated from O we need 12 terms Answer 16 The estimate is PO5 079167 and the error in this estimate is bounded by 5 E 1 000094 51 2 Note it is certainly also true that the error is bounded by 10300 but in problems such as this one you are expected to nd a an error estimate by making a fairly close estimate of the Taylor remainder Copyright 2004 Bent E Petersen Permission is granted to duplicate this document for nonipro t educational purposes provided that no alterations are made andprovided that this copyright notice is preserved on all copies Bent E Petersen 24 hour phone numbers Department of Mathematics o l ce 541 7375163 Oregon State University Corvallis OR 973314605 fax 541 7370517 email petersenmathoregonstateedu web httporegonstateeduNpeterseb Mth 306 5 Winter 2005

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