My Math Forum Newton's Method

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 October 21st, 2010, 12:30 AM #1 Member   Joined: Sep 2010 Posts: 55 Thanks: 0 Newton's Method x^3=17x+34 = x^3 - 17x - 34 = 0 x^3 - 17x - 34 (that is f(x)) X0 = 6 Find X1 and |X1-X0| So Xn+1 = Xn - f(Xn) / f'(Xn) So Xn+1 = Xn - Xn^3 - 17Xn - 34 / 3Xn^2 - [color=#BF0000]17n[/color]? How do I derive that 17Xn? I'm confused at that part basically, because I don't know what to do with 17n (if I did do that right). Thanks.
 October 21st, 2010, 01:36 AM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Newton's Method You can just let that be 17. So you are working with: $x_{n+1}=x_n-\frac{{(x_n)}^3- 17(x_n) - 34}{3(x_n)^2-17}$ The brackets aren't necessary, it's for your understanding. Does the exercise require you to use Newton's method? Hoempa
October 21st, 2010, 01:46 AM   #3
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Re: Newton's Method

Quote:
 Originally Posted by Hoempa You can just let that be 17. So you are working with: $x_{n+1}=x_n-\frac{{(x_n)}^3- 17(x_n) - 34}{3(x_n)^2-17}$ The brackets aren't necessary, it's for your understanding. Does the exercise require you to use Newton's method? Hoempa
Yeah we have to use Newton's method... I think that may be it, one minute

 October 21st, 2010, 01:50 AM #4 Member   Joined: Sep 2010 Posts: 55 Thanks: 0 Re: Newton's Method Yeah it is, thanks a lot!
 October 21st, 2010, 01:51 AM #5 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Newton's Method You're welcome, good luck! Can you go from here? Hoempa

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