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 October 12th, 2017, 12:19 AM #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 02:42 AM.
 October 12th, 2017, 12:29 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,519 Thanks: 506 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 12th, 2017, 12:36 AM   #3
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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 02:44 AM.

October 12th, 2017, 01:01 AM   #4
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 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?

Last edited by JanSolo; October 12th, 2017 at 01:10 AM.

October 12th, 2017, 01:09 AM   #5
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 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 03:07 AM.

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