# Newtons Method

• Jan 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 $\displaystyle ^4\sqrt{19}$ to 2 decimal place accuracy. Use 2 as your initial estimate.
• Jan 11th 2010, 01:59 PM
pickslides
Make $\displaystyle f(x) = x^{1\over 4}$
• Jan 11th 2010, 02:13 PM
bearhug
• Jan 11th 2010, 02:21 PM
pickslides
I understand Newton's Method to be

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

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

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

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

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

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

Say $\displaystyle x_n=4$

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

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

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

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.

$\displaystyle x_n=2$

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

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

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

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