Prove existence of divergent sequence in Newton's method

Question

Given $ f(x)=x^3-1$, how to prove that there exists a sequence of initial values $x_{0,1}>x_{0,2}>x_{0,3}>...$ where $x_{0,1}=0,x_{0,2}=-2^{1/3}$, such that the sequence produced by Newton's method is divergent?

Answer 1

If we start at $x_0=0$, then the next member of the sequence is $-\frac{f(x_0)}{f'(x_0)}=-\infty$, so the method fails.

The line $y=kx$ is tangent to the curve at $x$ if $k=3x^2$ and $kx=x^3-1$, so $x=-\left(\frac{1}{2}\right)^{1/3}$. So if we take $x_0=-\left(\frac{1}{2}\right)^{1/3}$ we are also in trouble because the next approximation will be $x=0$ and then $-\infty$.

This process can be continued indefinitely, having found a bad point $x=-h$ we find a point on the curve whose tangent passes through the point $(-h,0)$. So we need a point $(-H,-H^3-1)$ such that the tangent there passes through $(-h,0)$. The gradient at $x=-H$ is $3H^2$, so the tangent is $y+H^3+1=3H^2(x+H)$. For this to pass through $(-h,0)$ we need $H^3+1=3H^2(H-h)$ or $2H^3-3hH^2-1=0$.

Put $g(x)=2x^3-3hx^2-1$. Then $g(h)=-h^3-1<0$ but the leading coefficient is positive, so there must be a root $H>h$. So we can find an infinite sequence of bad starting points.