My Math Forum Newton's Method Help
 User Name Remember Me? Password

 Calculus Calculus Math Forum

 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 $\displaystyle f(x)=x^3-2x-5$ 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: 1,636 Thanks: 569 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, $$x_{n + 1} = x_{n} - \dfrac{f(x_{n})}{f'(x_{n})}$$. Thanks from JanSolo
October 11th, 2017, 11:36 PM   #3
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,093
Thanks: 1087

Quote:
 Originally Posted by JanSolo 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 $\displaystyle f(x)=x^3-2x-5$ 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.
Newton's method will find $x$ such that $f(x)=0$

It's an iterative method such that the current approximation $x_n$ is given by

$x_n = x_{n-1} - \dfrac{f(x_{n-1})}{f^\prime(x_{n-1})}$

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 $x^6-2=0$

So apply Newton's method to $f(x) = x^6-2$

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:
 Originally Posted by romsek Newton's method will find $x$ such that $f(x)=0$ It's an iterative method such that the current approximation $x_n$ is given by $x_n = x_{n-1} - \dfrac{f(x_{n-1})}{f^\prime(x_{n-1})}$ 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 $x^6-2=0$ So apply Newton's method to $f(x) = x^6-2$

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^6-2=0 from.

Did you do this:

$\displaystyle 2^{1/6}=x$
$\displaystyle 2=x^6$
$\displaystyle 0=x^6-2$

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: 1,636
Thanks: 569

Math Focus: Yet to find out.
Quote:
 Originally Posted by JanSolo $\displaystyle 2^{1/6}=x$ $\displaystyle 2=x^6$ $\displaystyle 0=x^6-2$ Does that work? Thanks tons for your help!
The point of the exercise is to show one way how we can approximate such expressions. As romsek has said, Newton's method finds the zeros of a function (there are some other criteria as well, but ignore for the moment) and so if you can construct a function that you know has a zero at a certain value, you can find an approximation for that value.

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.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post ZMD Calculus 1 February 25th, 2017 03:07 AM helpme123 Applied Math 3 July 28th, 2015 06:06 AM cokipoon Calculus 2 February 29th, 2012 12:28 AM djo201 Calculus 6 November 15th, 2008 12:32 PM krismcqueen Algebra 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top