# Calculus 1 Newton's Method

posted by Olivia

Using Newton's method, approximate the root of the equation x^3+x+2=0 with the initial guess x1=-1 gives: x2=? and x3=? answers are not 0 or -1

1. Reiny

let f(x) = y = x^3 + x +2
dy/dx = 3x^2 + 1

newton's formula

x new = x old - f(xold/f '(xold)

x2 = -1 - (-1-1+2)/(3+1) = 1-0 = -1

since x1 = x1 exactly, x = -1 is a root

2. MathMate

Check for typo's in the question.
I believe there is only one real root for the given equation at -1.

So your initial guesses have to be complex, such as 1+i. Convergence will depend on the form of the iteration equation.
You can try
f(x)=(-2-x)^(1/3)
f(x)=-2-x^3
and all kinds of other ones.
The one that seems to converge best is
first make
x^3=-2-x
divide by x and take the square root to get
f(x)=sqrt(-1-2/x)
With a starting value of 1+i, ou should converge quite well, as the iterations alternate between the targeted root of (1/2)±sqrt(7)/2.

Once you have converged to one, you can take the conjugate for the other, without having to do the same things all over again.

3. MathMate

the targeted root of (1/2)±sqrt(7)/2i.

4. Reiny

My last line should say:

since x2 = x1 exactly, x = -1 is a root

5. MathMate

I'll get it right this time!
the targeted root of (1/2)±isqrt(7)/2.

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