{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 4 121 - SECTION 4.9 NEWTON'S METHOD 383 2 2...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: SECTION 4.9 NEWTON'S METHOD 383 2 2 15.sinm=$.sof(m)=sinx—m => f’(m):cosac—2w 2 - 2 smm — :c . . . mn+1 : acn — ———-"—". From the figure. the p0s1tive root of cos mn — 2:13,. sin a: : $2 is near 1. $1 : 1 => $2 % 0.891396. :33 e 0.876985. 334 z 0.876726 z :35. So the positive root is 0.876726. to six decimal places. 16.2cosatt:x4.sof($)=2cosm—at:4 => f/(w)=—2sincc—4w3 4 2cosxn i 9:" .—3. From the figure. the positive root —2 Sin 3:" — 4:0,. : mn+1 : $11. 7 4 . of2cosar: : m is near 1. 3:1 :1 => mg % 1.014184. mg % 1.013958 2 m4. So the positive root is 1.013958. to six decimal places. 17. From the graph, we see that there appear to be points of intersection near as : 70.7 and m : 1.2. Solving $4 = 1 + m is the same as solving f(a:)=m4imi1:0.f(:c)=a:4—ac—1 :5 f’(x)=4m3—1. mfi—mn—l —2 SO $n+1 : £13" — 4$§171 331 = —0.7 :31 :12 :62 % 70.725253 $2 % 1.221380 m3 R3 —0.724493 $3 R 1.220745 374 m ~0.724492 % :35 1:4 2 1.220744 % 035 To six decimal places. the roots of the equation are ~0.724492 and 1.220744. 18. 3 From the graph. there appears to be a point of intersection near ac : 0.6. Solving em : 3 — 230 is the same as solving f(a:) 2 ex + 2x — 3 : 0. f(ac) :ez+2x—3 => f’(m) :em+2.so ex" + 23: — 3 mn+1 = 30” ~ fi. Now 121 : 0.6 :> 362 % 0.594213. ‘ x3 x 0.594205 :3 2:4. So to six decimal places. the root of the equation _1 2 is 0.594205. 0 19. From the graph. there appears to be a point of intersection near a: = 0.5. 3"; Solving tan’1 :3 = l i an is the same as solving f(.r) =tan_lm+$— 1 :0. f(;c) :tan71x+$—1 => , 1 tan—1xn+$n — 1 Z 1‘ : n , ——. ':“3 f($) 1+9:2+ 8037"“ “PC 1/(1+a:%)+1 m1 : 0.5 => 9:2 % 0.520196. 303 % 0.520269 % 3:4. So to six decimal places. the root of the equation is 0.520269. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern