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 February 26th, 2012, 10:11 PM #1 Newbie   Joined: Feb 2012 Posts: 10 Thanks: 0 Newton's Method? Use Newton's method to approximate the root of the equation x^3=33x+66 that belongs to the interval (4,. Start with x0=8 and perform three iterations, i.e., find x1, x2, and x3. Calculate lx0?x1l lx1?x2l , and lx3?x2l . Answers: 1. Use Newton's method xn+1=xn? f(xn) / f (xn) where the function f has a positive leading coefficient so that f(x)= x^3-33*x-66 . x1=? lx1?x0l=? x2=? lx2?x1l= ? x3=? lx3?x2l=?
 February 26th, 2012, 10:50 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Newton's Method? First, express the equation as a function equal to zero: $f(x)=x^3-33x-66=0$ Calculate the function's first derivative: $f'(x)=3x^2-33=3$$x^2-11$$$ Newton's method gives us the recursion: $x_{n+1}=x_n-\frac{f$$x_n$$}{f#39;$$x_n$$}$ With $x_0=8$ we find: $x_1=8-\frac{8^3-33(-66}{3$$8^2-11$$}=\frac{1090}{159}" /> $x_2=\frac{1090}{159}-\frac{$$\frac{1090}{159}$$^3-33$$\frac{1090}{159}$$-66}{3$$\(\frac{1090}{159}$$^2-11\)}=\frac{2855356814}{434074293}$ $x_3=\frac{2855356814}{434074293}-\frac{$$\frac{2855356814}{434074293}$$^3-33$$\frac{2855356814}{434074293}$$-66}{3$$\(\frac{2855356814}{434074293}$$^2-11\)}=\frac{51957845787745398688939458250}{7918084 338467701533185157903}$ $\|x_0-x_1\|=\|8-\frac{1090}{159}\|=\frac{182}{159}$ $\|x_1-x_2\|=\|\frac{1090}{159}-\frac{2855356814}{434074293}\|=\frac{120372616}{43 4074293}$ $\|x_2-x_3\|=\|\frac{2855356814}{434074293}-\frac{51957845787745398688939458250}{7918084338467 701533185157903}\|=\frac{1276069429705248446812893 44}{7918084338467701533185157903}$
 February 29th, 2012, 01:28 AM #3 Newbie   Joined: Feb 2012 Posts: 10 Thanks: 0 Re: Newton's Method? Thank you!

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### where the function f has a positive leading coefficient so that f(x)

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