# Newtons Method

• January 11th 2010, 01:55 PM
bearhug
Newtons Method
I am working on problems using Newton's Method in which i'm given an equation and am told to find a root between 2 numbers.

For this particular question there is no 'x' in the equation so I'm not quite sure what to do:
Estimate $^4\sqrt{19}$ to 2 decimal place accuracy. Use 2 as your initial estimate.
• January 11th 2010, 01:59 PM
pickslides
Make $f(x) = x^{1\over 4}$
• January 11th 2010, 02:13 PM
bearhug
• January 11th 2010, 02:21 PM
pickslides
I understand Newton's Method to be

$x_{n+1} = x_{n}-\frac{f(x_{n})}{f'(x_{n})}$

Now find $f'(x)$ and start iterating.
• January 11th 2010, 04:19 PM
bearhug
So I'm using $f(x) = x^\frac{1}{4}$
and the derivative $f'(x) = \frac{1}{4}x^{-\frac{3}{4}}$ and iterating until it equals $^4\sqrt{19}$ ?
• January 11th 2010, 04:49 PM
No!

Use $f(x)=x^4.$

Now you are looking for the root of $x^4-19=0$

$f'(x)=4x^3$

Take a first shot at the answer, call it $x_n$

Say $x_n=4$

Then $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$
• January 15th 2010, 11:57 AM
I just want to clean this up,
as i answered in a very general way...

$f(x)=x^4-19$

$f'(x)=4x^3$

This is a very steep curve and an initial guess at the solution
can be within a range of x.
Simplest is to take the square root of the square root of 16,
or the fourth root of 16 which is 2.

$x_n=2$

The next approximation to the 4th root of 19 is then found from

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=2-\frac{f(2)}{f'(2)}$

$x_{n+1}=2-\frac{2^4-19}{(4)2^3}=2-\frac{-3}{32}$

Note... $(2+\frac{3}{32})^4=19.2176$