Condition for convergence of Newton-Raphson method.

Let $$f :[a,b]\to\mathbb{R}$$ be any function which is twice differentiable in $$(a,b)$$ with only one root $$\alpha$$ in $$(a,b)$$. Let $$f'(x)$$ and $$f''(x)$$ denote the first and second order derivatives of $$f(x)$$ with respect to $$x$$. If $$\alpha$$ is a simple root and is computed by the Newton-Raphson method, then the method converges if

$$|f(x)f''(x)|<|f'(x)|^2.$$

How to show this argument?

I'll offer only a sketch proof. We can show the error terms $$\varepsilon_n:=\alpha-x_n$$ satisfy $$\varepsilon_{n+1}=-\frac{f^{\prime\prime}(\xi_n)}{f^\prime(\xi_n)}\varepsilon_n^2$$ for some $$\xi_n$$ between $$x_n$$ and $$\alpha$$. We're assured of convergence if $$\left|\frac{f^\prime(\xi_n)}{f^{\prime\prime}(\xi_n)}\right|>|\varepsilon_n|.$$But the given inequality may be rearranged, viz.$$\left|\frac{f^\prime(\xi_n)}{f^{\prime\prime}(\xi_n)}\right|>\left|\frac{f(\xi_n)}{f^\prime(\xi_n)}\right|\approx|\xi_n-\alpha|\sim|\epsilon_n|.$$
• But the given relation is $\epsilon_{n+1}=-\frac{f''(\zeta_n)}{2f'(\zeta_n)}\epsilon_n^2$. – Mittal G Mar 5 '19 at 17:26
• @MittalG The aim is to force $\left|\frac{\varepsilon_{n+1}}{\varepsilon_n}\right|<1$. – J.G. Mar 5 '19 at 17:36
Look at $$g(x)=x-\frac{f(x)}{f'(x)}.$$ Then $$g'(x)=\frac{f(x)f''(x)}{f'(x)^2}$$ and by assumption $$|g^\prime(x)|<1$$. Now apply the fixed-point formulation of your choice.
• In the derivative, power $2$ should be in the denominator. – Mittal G Mar 6 '19 at 8:57