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Homework 1 Solutions

Homework 1 Solutions - 4.2 use the stopping criterion.2...

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Unformatted text preview: 4.2 use the stopping criterion: .2; = {15): 152-255 = 5.555 Tmevalue: 5111(3513}: U.3fifi025___ zero order: sin[£] =5 = 1514:5193 3 3 E: : UBfifiUEE —l_fl4?193 x 103%: 2.332% {1.355525 first Dfdfll'. sing] =1.54:r'195 — {K :53? = 5.555551 5, =1.15% an =‘fl'355f‘fifiéflfiwg x100%=22_36% Md 01115:: 5 5' 5 =5_555551+ (”T ”r 3] =5_355255 3 125 5, =5.53155 a, = —5_355295—5_555351 x155%=1.211% 5.555295 11de urdar: T s' 5 =5555295—W3] =5_555521 3 55-15 flBfififlll — 0.866295 {1.3661121 5} = fi_flflfl4?’?% Ed =| x1flfl% = fl_fl3lfi% Since the appmflmate mm is below 0.5%, the compmatiun can be terminated 4.5 T111»: value: 313} = 554. zero order: f8)=f{l}=-EE s,=‘554;;523x1oo=ya=111_19m first ordm'. f{3)=-fil+f'(l){3—l)=—52+T0{2)=TB £,=85.9‘21% second order: f8)=?3+f2fl) [3—1)2 =T3+¥4=354 £,=36_1fi1% flaird mflar: ‘3} 150 f8)=354+f fifl}(3_1}3 2354+?82554 53.20% Thus, 1h: fl'fird—ordcr result is pflfect because the original flmctiun is a thiId—order polynomial. 4.6 True value: f' {x} = ?5252 —12x+2 f' (2) = 25(2)2 —12{2)+ 2 = 233 fimction values xH = 1.8 fin-.1) = 58.96 x!- = 2 fin) = 102 xm = 2.2 firm} 2 164.56 forward: 16456—102 283—3128 f'[2)=—=312.8 at: x100%=1l).53% 0.2 283 backward: f'[2)=w=255_2 5!: fl x1m=93239¢ 0.2 283 centered: 16456—5036 283—234 f'(2)=—=284 a}: X100€6=0.353% 2(fl.2} 283 Bod: the forward and backward have errors that can be approximated by (recall 1321.415), ‘E | g f"(x.-) h ' 2 f"(2} = lfiflx— 12 =150(2}—12 = 288 |Et| 22—3822 =2s.s Thisis very closetothe actual errorthat occurred inthe approximations forward: |E,| s5 |233—312.3| =29.3 backward: |E,| a pas—2552‘ 222.3 The centered approximation has an error that can be approximated by (3: E, as —f :2) 2:2 = —%022 = —1 which is exact: E = 283 — 284 = —1. This result occurs because the original fimction is a cubic equation which has zero fourth and higher derivatives. 4.? True value: f"(x) =15'Dx —l2 f"(2) 2150(2) —12= 283 fi=02i f"(2}=w=w=lgg ”-25 0.25 h =o_125: we) =W zw = 233 {11252 0.1252 Buflnresults ammm becausethe mars aIeafimction of4fi’ andhigher defivafives whicharezemfin a Bed—order polynomial. 4.12 The condition number is computed as CNZEf'GE) ft?) J—‘ ] 1mm — (a) CH: 2 1.00001—1 =1.nmm(153_1139)=m_6” «.Jl.00001—1+1 1.003162 The result is ill—conditioned because the derivative is large nearx = 1. —10 —5 a.) .93.;wa = —1o 3"“ 454x10—5 The result is ill—conditioned because A: is large. 31:] fl_l J3m2 +1 _ 300{—5.555556x10_6) {c} (EN = — = —0.99999444 430:)? +1_3m 0.001556? The result is well—conditioned x —xs_x — 3—: +1 2 ([1) CN 2 x = 0.031(049'9657) = _0_ 000 5 g“ _1 — 0.9995 x The result is well—conditioned x (1+cos x)cosx+sinx(sinx} 2 [9) EN = {1 +_cos x) = 3.14190? [20,204,233 2 —10,001 smx — 63 65.2 l+cosx The result is ill—conditioned because, as in the following plug the fimction has a singularity at x = Fr. 2000 1000 0 400013 3.1 3.12 3.1 3.16 3.13 3.2 -2000 ...
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