Compute four steps of Newton's method to solve $x^3-2x+2=0$ with $x_0=0$. Explain what is happening.
What I have so far:
Since $f(x)=x^3-2*x+2$, we have $f'(x)=3x^2-2$
Applying Newton's method $x_{n+1}=x_n-\frac{x_n^3-2x_n^2+2}{3x_n^2-2}$
Since $x_0=0$, we can get $x_1=1$, $x_2=0$, $x_3=1$, $x_4=0$
I am having trouble explaining what is happening partially because I am not sure what the problem is asking in this latter part.
My guess is that it is asking me to explain why x here is jumping from 0 to 1 and this may somehow have something to do with the convergence?
Best Answer
The goal of the Newton algorithm is to find a sequence that converge to the roots of an equation (to approximate it). Here as your sequence doesn't converge, this is obviously a fail.
Now, you did prove earlier in your lessons (at least I hope so) a theorem that assure that under some hypothesis, the sequence converge : what hypothesis is missing here?