# Newton's Method

• Oct 21st 2010, 12:32 AM
lilwayne
Newton's Method
x^3=17x+34

= x^3 - 17x - 34 = 0

x^3 - 17x - 34 (that is f(x))

X0 = 6

Find X1 and |X1-X0|

So Xn+1 = Xn - f(Xn) / f'(Xn)

So Xn+1 = Xn - Xn^3 - 17xn - 34 / 3xn^2 - 17n?

how do i derive that 17xn? I'm confused at that part basically, because i dont know what to do with 17n (if i did do that right).

thanks.
• Oct 21st 2010, 01:27 AM
MathoMan
$f(x)=x^3-17x-34$
$f'(x)=3x^2-17$

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x^3_n-17x_n-34}{3x_n^2-17}=\frac{3x_n^3-17x_n-x_n^3+17x_n+34}{3x_n^2-17}=\frac{2x_n^3+34}{3x_n^2-17}$
• Oct 21st 2010, 01:32 AM
Quote:

Originally Posted by lilwayne
x^3=17x+34

= x^3 - 17x - 34 = 0

x^3 - 17x - 34 (that is f(x))

X0 = 6

Find X1 and |X1-X0|

So Xn+1 = Xn - f(Xn) / f'(Xn)

So Xn+1 = Xn - Xn^3 - 17xn - 34 / 3xn^2 - 17n?

how do i derive that 17xn? I'm confused at that part basically, because i dont know what to do with 17n (if i did do that right).

thanks.

Unfortunately, you didn't do it right...

$f(x)=x^3-17x-34$

$f'(x)=3x^2-17$

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

$x_1=x_0-\frac{f\left(x_0\right)}{f'\left(x_0\right)}$
• Oct 21st 2010, 01:41 AM
lilwayne
Quote:

Originally Posted by MathoMan
$f(x)=x^3-17x-34$
$f'(x)=3x^2-17$

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x^3_n-17x_n-34}{3x_n^2-17}=\frac{3x_n^3-17x_n-x_n^3+17x_n+34}{3x_n^2-17}=\frac{2x_n^3+34}{3x_n^2-17}$

Your answer was right but i dont get how you turned Xn - xn^3 ..... to 3Xn^3 .....

how did you do that :S
• Oct 21st 2010, 01:47 AM
MathoMan
$x_n-\frac{x^3_n-17x_n-34}{3x_n^2-17}=\frac{x_n\cdot (3x_n^2-17)-(x_n^3-17x_n-34)}{3x_n^2-17}=\frac{3x_n^3-17x_n-x_n^3+17x_n+34}{3x_n^2-17}=\frac{2x_n^3+34}{3x_n^2-17}$
• Oct 21st 2010, 01:51 AM
lilwayne
Isn't that the same thing as Xn - Xn^3 - 17xn - 34 / 3xn^2 - 17 ?
• Oct 21st 2010, 02:38 AM