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by: Burnice Ratke


Burnice Ratke
GPA 3.72


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Class Notes
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This 7 page Class Notes was uploaded by Burnice Ratke on Tuesday October 20, 2015. The Class Notes belongs to MATH693B at San Diego State University taught by Staff in Fall. Since its upload, it has received 16 views. For similar materials see /class/225280/math693b-san-diego-state-university in Mathematics (M) at San Diego State University.




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Date Created: 10/20/15
Corrections to Second Printing of Numerical Optimization Last updated February 13 2006 To make it easier for the typesetter we have included in many cases an explicit correction of the offending passages marked by the tag corrected sentence77 or similar so that he she can cut and paste from the source latex le of this document into the latex le for the book Note that in this document the labels of equations appearing in the corrected sentences may not print or may print with wrong equation numbers because they are using labels de ned elsewhere in the book le When inserted in the book le they will produce the correct labels WE DE r gt 01 page 26 line 9 77 should be 7 in this equation page 47 line 75 Replace gk by ka in this formula page 49 Theorem 34 The theorem needs to be corrected to include a slightly larger multiplier on the right hand side of the inequality and to all the proviso that the inequality holds for all k su iciently large The corrected statement of the theorem is as follows Suppose that f 1R 7 R is twice continuously differentiable and that the iterates generated by the steepest descent method with exact line searches converge to a point ac where the Hessian matrix V2fw is positive de nite Let r be any scalar satisfying 1 A 7 A T E lt n 1 where A1 lt S An are the eigenvalues of V2fw Then for all k su i n17 ciently large we have fwk1 7 W S r2fwk 7 fwlr pages 50 51 In Theorems 35 and 36 we only need to assume that f is twice contin uously differentiable not three times page 53 lines 1 5 This proof needs to be corrected so that we do not assume one of the things we are trying to prove namely that wk 7 ac Replace the sentence Since V2fw is nonsingular 7 on line 1 by the following Since V2fw is nonsingular there is a radius r gt 0 such that HVkaAH g 2HV2fw 1H for all sick with 7 wi H S r 77 Replace the sentence Using this inequality on line 5 by the following Choosing am so that lle 7 wi H S minr12I we can use this inequality inductively to deduce that the sequence converges to w and the rate of convergence is quadratic N93 0000 H H H I H 9quot H 01 H O H T 58 in the line before 344 it should read O kaTilpkA page 68 The object A is used in a different sense here from in the proof of Theorem 47 on pages 90 91 To avoid confusion replace A with A on page 68 lines 2 and 10 of Algorithm 41 and on the line immediately following Algorithm 41 page 68 line 7 of Algorithm 41 should read Ak iAk page 73 line 13 It should read follows from Exercise 4677 page 71 line 16 before to simplify the notation77 insert and replace ka by g Corrected phrase so we drop the subscript k from the quantities Ak pk and mk and replace ka by g to simplify the notation page 74 line 6 The sentence immediately after 414 should read This is a problem in two variables that is computationally inexpensive to solve After some algebraic manipulation one observes that it can be solved by nding the roots of a fourth degree polynomial page 75 line 77 There should be a 7 before 1 The corrected formula is dj1 THI j1dj page 80 Figure 45 and page 82 Figure 46 In both gures the 7V along the horizontal axis should be N page 87 last line The right hand side of this expression should be mk iAkakHka fk Corrected equation is mpg WHO mkPAkakWfkll fk page 89 lines 77 to 75 The phrase and that the level set is bounded should be replaced by and that f is bounded below on the level set 7 The formula 438 remains the same The new text fragment is and that f is bounded below on the level set wfw Sflt7 0r 1 page 90 equation 444 Delete 2 from the denominator Corrected equation is 1 lt YAk8 YAk2 04 cleminAk 39 2 lpzc page 91 line 6 Delete 2 from the denominator of the middle expression and replace by in the rightmost expression Corrected equation is moleav 1 7 1 lt lpk I 7 01 Ak 2 H 3390 0 9310 r gt 5197 PO 3 0 NH 973 page 91 line 7 replace pk gt g by pk gt page 109 line 10 6 should be C page 115 display equation on line 19 me 7 wl H should be meH 7 lly Corrected equation llwm1 7 new a eHwo 7 new page 118 equation 537 and also line 9 I ll should be C Tb in both places page 118 line 4 Should read Set 130 730 k e 0 page 119 line 5 it should read we will discuss it brie y in Chapter 6 page 125 line 8 Missing in the equation for t Corrected equation t dSf 1gt1 gt page 129 line 8 The 3 should be page 129 line 11 and also equation 565 In both places X3 should be cg page 139 line 5 Replace the phrase we complete the rst CG iteration compute the new iterate sea and stop by the phrase we set mm b 7ka and return page 140 lines 7 8 Replace pk is the steepest descent direction 7ka by pk is set to the steepest descent direction 7ka Also delete the sentence This is the reason for choosing the initial estimate in the CG iteration as mm 0 page 140 two lines before Algorithm 61 it should read 71k min05 ka ie gradient not Hessian page 140 line 6 The vector 13 should be pk Corrected sentence V2fwkpk 7ka starting from sea 0 page 141 line 1 remove minor in minor weakness In line 11 remove slightly page 148 lines 5 and 6 of Algorithm 65 These lines Find index 1 and lnterchange row should be moved below the next line that starts with for j 12 n 8 lines below the statement ifj S n should read ifj lt n page 170 line 2 Remove set after setting page 171 line after 716 and page 172 line 6 Hessian should be Jacobian page 181 Figure 73 This gure should contain p64 62 rather than p6 4 e 77 page 183 4 lines after 734 we saw in Chapter 9 should be we will see in Chapter 9 T wweo m r gt U1 9 905 3 page 190 exercise 77 Should read expressing the intermediate derivatives Van i 4 9 in terms of quantities available at their parent nodes and then in terms of the independent variables ail mg mg page 190 exercise 79 It should read for the evaluation point cc 12 7r2T77 page 236 line 4 The terms 12903 in both expressions should read 1290 7 4w1 page 249 line 17 i 12q 1 should be i 12 q 1 ellipsis missing page 269 line 711 The indices in the two summations should be 9 page 316 line 78 Replace 71000 by O 7100 page 322 displayed formula on line 710 The denominator should contain an extra factor Corrected formula reads Vfw d 61 Hwltwgtlwvc1ltwgtw page 329 formula 1232 The fraction should be Corrected formula should read 17461in 3 2 14 17w1w2 mzin 17 ltw27 gt st 15 in 20 3 1w1w2 page 334 equation 1236 should read 2 20T 1kw page 336 line 12 The last word in the line should be 77constraint77 rather than 77constrained The line should read One frequently used constraint quali cation is the linear independence con straint quali cation LICQ page 339 line 14 The phrase should be holds if VfwTd 2 O for all d 6 F1 page 346 line 15 Immediately after the sign add fc Corrected equation fltZkgt 2 y an 7 wTsz w7gtZk 06 00le KHZ 1 y 51le wllszsz w7 4961 01le i 001112 page 354 1 line below 1275 mi should be xi Corrected sentence We simply take M E ac in De nition 77 5 O 5 5 I 5 03 5 5 H r gt 01 page 371 equation 1314 The lefthand side of this equation should be AP page 377 line 13 The 7 before the nal term should be replaced by Corrected equation cTw cng 7 cq 7 slaw cqacqJr cng 5115c page 377 equation 1324 The 7 before the nal term should be replaced by Corrected equation cTw cTw 511546 4 page 390 line 6 After such that77 append t gt 0 t gt 0 and Corrected passage we have a tie if there are two indices il E B such that ti gt 0 t gt 0 and77 page 391 line 7 Append t gt 0 for all i E B and77 Corrected sentence Given a set of tied indices 8 C B with ti gt 0 for all i E B and77 page 391 lines 13 18 Replace these lines with the following material BilE BTIE T tgt21ltlttgtn forauzezsaa z 239 l we choose i as the leaving index Otherwise we retain in B the indices i that tie for the smallest value of B lE 1t and evaluate the second column of B lE If just one value of B lE 2t achieves the minimum we choose it as the leaving index Otherwise we consider coe icients B lEMc for successively higher k until just one candidate index remains page 406 equation 1422 The second occurrence of 04 should be 043 Corrected equation Mag cc aggAwa Ts a galAsa Vn page 406 line 73 delete the comma between AXa and Asa e Corrected fragment AXa ASa e page 407 equations 1424a and 1424b The algorithm is slightly better if the min with 1 is not performed Corrected equations 16 def pH 7 2 04 7 min 7 max iAzilt0 7 k d f i 5 a a 2 min 7 2 iAsiltO A52 FWPE Q 9 390 0 H 10 aw 0301 page 416 lines 6 and 8 Remove w from these equations page 435 below 1525 remove the phrase denotes the 2 norm page 436 line 10 it should read 1 lt M instead of the other way round page 437 lines 7 and 8 KI should be replaced with 45 page 444 lines 8 and 9 Replace and that the constraints by so that the constraints page 454 5 6 lines below 1626d Should read active constraint gradients are lin early independent at the solution page 456 line 10 Should read First linear dependence page 476 last line add space after to page 483 equation 1653 Replace As by AA and AA by Ay Also replace ASe by AYe Co Tracth aquatic 7L5 G 7AT 0 Aw 7rd A 0 7 AA in 0 Y A Ay iAYe Glue page 483 equation 1654 Replace As by AA page 484 lines 6 7 Replace all three occurrences of s in these lines by A Also replace S by A page 485 equation 1656a replace maxz A by maX minz Also it should simply read 1656 Without the a page 487 exercise 1610 Replace the rst sentence by Consider the problem 163a 163b and assume that A has full row rank and that Z is the matrix Whose columns span the null space of A Also replace A by AT in the last line of page 487 page 487 last line Replace A by AT Corrected inline equation reads GX d Tva page 498 line 9 Replace full column rank by full row rank page 499 3 lines before 1717 ill conditioning of Q should be ill conditioning of v297 page 501ln gure 173 label the curves by value of u page 515 line 8 Replace and terminating when H Aw A 3 7 by satisfying H Awk A S 779 This change is not necessary but it improves clarity 8 8 8 8 8 9 9 01 O 00 J O H K page 515 add a new line after line 12 Choose new tolerance Tk1 E On 19 Page 517 Equation 1753 should read 7 akkf ie a is missing the subscript k Same typo in 1754 and 1755 page 536 line 5 A comma is missing in kak page 537 line 3 AT should be Ag page 537 line 6 for for should be for page 542 line 71 Omit akpz page 543 line 3 Should read pk kaz Ykpy page 548 line 8 The term D wkpk should read D wk Iapk The same correction should be made in page 552 line 3 page 551 line 7 the two B should be M Corrected passage Set Mk1 Mk page 560 line 76 A should be Ak Corrected phrase where Zk is a basis for the null space of Ak page 574 Exercise 182 The starting point and solution are transposed The two sentences should read Use the starting point 30 7181719708708 The solution is approximately ac 71711591827076370763 page 575 Exercise 1810 Equation 1869 should read I 7 ATAAT1Aw page 579 The de nition of a cone should have 04 gt 0 instead of 04 2 O The corrected de ntion should read as follows we awef forallagt0 5 page 579 The de nition of a ine hull in A3 and the example are wrong The sentence containing the formula A3 and the sentence that follows should be replaced by the following Corrected passage An a ine set in R is a the set of all vectors x y where w is given and y is any vector in a given subspace of R Given f C R the a ine hall of f denoted by afff is the smallest a ine set containing f For instance when f is the ice cream cone de ned as below by 7 we have afff R3 while if f is the set of two isolated points f 100 O20 we have afff 100 047120 for all a e R page 612 The year for reference 23 should be 1995


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