# Math Help - Newton's Method

1. ## 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.

2. $f(x)=x^3-17x-34$
$f'(x)=3x^2-17$

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

3. 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...

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

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

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

$x_1=x_0-\frac{f\left(x_0\right)}{f'\left(x_0\right)}$

4. 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

5. $x_n-\frac{x^3_n-17x_n-34}{3x_n^2-17}=\frac{x_n\cdot (3x_n^2-17)-(x_n^3-17x_n-34)}{3x_n^2-17}=\frac{3x_n^3-17x_n-x_n^3+17x_n+34}{3x_n^2-17}=\frac{2x_n^3+34}{3x_n^2-17}$

6. Isn't that the same thing as Xn - Xn^3 - 17xn - 34 / 3xn^2 - 17 ?

7. Originally Posted by lilwayne
Isn't that the same thing as Xn - Xn^3 - 17xn - 34 / 3xn^2 - 17 ?
Yes,

it's a matter of choice.
You may prefer to substitute the x values into the single fraction (which has only 2 terms in x)
or into the direct Newton's method formula.
Either way, done correctly will give the same result for the next approximation to the root.