Question

Newton's Method Help!!!

Answer 1

Use Newton's method with the specific initial value x1=−1 to approximate the root of the equation

Answer 2

x^3+2=0

Answer 3

I assume you know the formula ?

Answer 4

nope

Answer 5

well the x2=x1-f(x1)/f(x2)?

Answer 6

i mean f(x1)/f'(x1)

Answer 7

Okay, so now do it. You have an initial value. Off you go!

Answer 8

This has the equation, and if you scroll down, an animation of the process
http://en.wikipedia.org/wiki/Newton%27s_method

Answer 9

You begin with
\[ x_{n+1} = x_n -\frac{f(x_n)}{f'(x_n)} \]

so you need the derivative of your function...

Answer 10

the derivative would be 3x

Answer 11

try again.

Answer 12

\[ \frac{d}{dx} x^n = n x^{n-1} \]

Answer 13

oh sorry. 3x^2

Answer 14

so the basic equation is
\[ x_{n+1} = x_n -\frac{f(x_n)}{f'(x_n)} \\
x_{n+1} = x_n -\frac{x_n^3+2}{3x_n^2} \]
we start with \(n=0, x_0 = -1\)
\[ x_1= -1 - \frac{(-1)^3+2}{3(-1)^2} \]

Answer 15

i got -2/3 for x1 and -35/18 for x2

Answer 16

A calculator (or better a computer) is helpful at this point.

-1 - (-1+2)/3 = -1 - 1/3 = -4/3
for the 1st iteration.

Answer 17

oh ok.

Answer 18

with my calculator, I can use a "variable" and type in the calculation as one line that saves its answer into the variable.
Initialize the variable to -1, and keep hitting = sign .

Answer 19

so i now have x2=-91/72

Answer 20

yes

Answer 21

i got it x2=-4/3 x3=-91/72 and x4=-1126819/894348

Answer 22

yes

Answer 23

if you cube x4 you get about -2.00006