October 11th, 2017, 11:19 PM  #1 
Newbie Joined: Sep 2017 From: Australia Posts: 4 Thanks: 0  Newton's Method Help
Hi, I really need help with this!!! Using Newton’s Method: a) Sketch the function and then find the third approximation $\displaystyle x_3$, starting with $\displaystyle x_1=2$, for $\displaystyle f(x)=x^32x5$ b) Find $\displaystyle \sqrt[6]{2}$ (6th root of 2) correct to eight decimal places. Hint: let $\displaystyle x$ be that number. Thanks tons to whomever can help me; I'm struggling. Last edited by skipjack; October 12th, 2017 at 01:42 AM. 
October 11th, 2017, 11:29 PM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,597 Thanks: 546 Math Focus: Yet to find out. 
What have you tried so far? Have you written down the formula? For a) I would suggest first finding the derivative of $f(x)$, then set $n = 1$ and apply the following, $$x_{n + 1} = x_{n}  \dfrac{f(x_{n})}{f'(x_{n})}$$. 
October 11th, 2017, 11:36 PM  #3  
Senior Member Joined: Sep 2015 From: USA Posts: 1,979 Thanks: 1027  Quote:
It's an iterative method such that the current approximation $x_n$ is given by $x_n = x_{n1}  \dfrac{f(x_{n1})}{f^\prime(x_{n1})}$ here $f(x) = x^3  2x 5$ $f^\prime(x) = 3x^2  2$ $x_1 = 2$ $x_2 = 2  \dfrac{2^3  2(2) 5}{3(2^2)2} = 2  \dfrac{1}{10}=\dfrac{21}{10}$ I leave it to you to compute $x_3$ b) You want to find $2^{1/6}$ This is the solution to $x^62=0$ So apply Newton's method to $f(x) = x^62$ Last edited by skipjack; October 12th, 2017 at 01:44 AM.  
October 12th, 2017, 12:01 AM  #4  
Newbie Joined: Sep 2017 From: Australia Posts: 4 Thanks: 0  Quote:
So is this right: $\displaystyle x_3=2.1−\dfrac{2.1^3−2(2.1)−5}{3(2.1)^2−2}$ $\displaystyle x_3=2.094568121$ I don't see where you got x^62=0 from. Did you do this: $\displaystyle 2^{1/6}=x$ $\displaystyle 2=x^6$ $\displaystyle 0=x^62$ Does that work? Thanks tons for your help! Last edited by JanSolo; October 12th, 2017 at 12:10 AM.  
October 12th, 2017, 12:09 AM  #5  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,597 Thanks: 546 Math Focus: Yet to find out.  Quote:
So yes, $2^{1/6}$ is a solution to the equation $x^6  2 = 0$. Last edited by skipjack; October 12th, 2017 at 02:07 AM.  

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derivative, graph calculus algebra., method, newton, newtons 
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