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Chapter 4 125 - SECTION4.9 NEWTON'SMETHOD El 387(b:31 = 0.6...

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Unformatted text preview: SECTION4.9 NEWTON'SMETHOD El 387 (b) :31 = 0.6. £132 2 17.9. $3 m 11.946802. $4 N 7.985520. 375 z 5.356909. x5 B 3.624996. :27 z 2.505589. .713 % 1.820129. 309 % 1.461044. (510 % 13393233311 % 1.324913. x12 2 1.324718 % $13 (C) an : 0.57. x2 2 —54.165455. .753 % ——36.114293. 3:4 z —24.082094. 1'5 x —16.063387, are % —10.721483. m7 2 —7.165534. mg % —4.801704. mg % —3.233425. 2310 % —2.193674. 2011 w —1.496867. 3312 % —0.997546. 9313 % —0.496305. $14 % —2.894162. 1315 % —1.967962. 1216 g —1.341355. —0.870187. 3318 m —0.249949. 3:19 % —1.192219.:B20 % 7073195291521 % 0.355213. 22 $17 3:22 ”~V —1.753322. $23 % ~1.189420. 1624 % —0.729123. 5625 R 0377844136 % —1.937872. $27 as —1.320350. $23 2 70.851919. 5629 R4 —0.200959. $30 RV. —1.119386. 0331 m v0.654291. $32 ANJ 1.547010. 1133 % 1.360051. $34 «"1 1.325828, (1235 % 1.324719. $36 3 1.324718 Q“; $37. From the figure, we see that the tangent line corresponding to 3:1 : 1 results in a sequence of approximations that converges quite quickly (905 m $6). The tangent line corresponding to $1 : 0.6 is close to being horizontal. so 302 is quite far from the root. But the sequence still converges — just a little more slowly (2:12 a $13). Lastly. the tangent line corresponding to 3:1 : 0.57 is very nearly horizontal. 3:2 is farther away from the root. and the sequence takes more iterations to converge ($36 a $37)- 33. For f(a:) : 1121/3.f'(w) = g 4/3 and 3 1 3 _ f (mu) 1% $n+1 — 33,. i I — sun 1 *2/3 : Inn 3% = 2m... f (33") 533" P . . . . 3 3 Therefore. each successwe approx1mation becomes tw1ce as large as the a previous one in absolute value. so the sequence of approximations fails to ‘ converge to the root. which is 0. In the figure. we have .711 : 0.5. —3 m2 : ~2(0.5) : 41. and $3 = —2(41)= 2. 34. According to Newton’s Method. for my. > 0. xn+i:$n— :mnw2wn:7:cn andforzcn<0. 1/(2,/.rn) a: m ”“3" m[2( )1 s n+1: n_———: n—~ ‘33,. 273:". owecan 1/(2./_m..,) see that after choosing any value 931 the subsequent values will alternate between 7231 and $1 and never approach the root. 35. (a) f(;v) : 3m4 4 28353 + 621:2 + 249: => f’(a:) = 123:3 4 84:32 + 12:17 + 24 : f’tcm) * a f”(~'€1) 3 $3 3 0.6455 => x4 x 0.6452 :> $5 % 0.6452. Now try $1 = 6 :> 3:2 : 7.12 => 1:3 2 6.8353 :> 204 w 6.8102 :> 365 x 6.8100. Finally try :31 : —0.5 2 $2 % —0.4571 :> $3 r”: —0.4552 => 14 x —0.4552. Therefore. an : —0.455. 6.810 and 0.645 are all critical numbers correct to three decimal places. f"($) : 36$2 — 1683: + 12. Now to solve f’(:c) : 0. try $1 — 1 5 Jig—$1 ...
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