Posted by Olivia on Monday, October 31, 2011 at 11:54pm.
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

Calculus 1 Newton's Method  Reiny, Tuesday, November 1, 2011 at 9:14am
let f(x) = y = x^3 + x +2
dy/dx = 3x^2 + 1
newton's formula
x _{new} = x _{old}  f(x_{old}/f '(x_{old})
x2 = 1  (11+2)/(3+1) = 10 = 1
since x1 = x1 exactly, x = 1 is a root

Calculus 1 Newton's Method  MathMate, Tuesday, November 1, 2011 at 9:14am
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)=(2x)^(1/3)
f(x)=2x^3
and all kinds of other ones.
The one that seems to converge best is
first make
x^3=2x
divide by x and take the square root to get
f(x)=sqrt(12/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.

Calculus 1 Newton's Methodcorr  MathMate, Tuesday, November 1, 2011 at 9:16am
the targeted root of (1/2)±sqrt(7)/2i.

Calculus 1 Newton's Method  Reiny, Tuesday, November 1, 2011 at 9:16am
My last line should say:
since x2 = x1 exactly, x = 1 is a root

Calculus 1 Newton's Methodcorr  MathMate, Tuesday, November 1, 2011 at 9:17am
I'll get it right this time!
the targeted root of (1/2)±isqrt(7)/2.
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