# Algorithmic time complexity of Newton's method vs bisection method

In the context of root finding, it is often stated that the bisection method is slower than Newton's method due to linear convergence. However, I am trying to understand why this is the case from an algorithmic time complexity viewpoint.

My understanding is that the time complexity of Newton's method is $O(log(N)F(N))$ where $F(N)$ is the cost of calculating $\frac{f(x)}{f'(x)}$ with $N$-digit precision. But given the architecture of the bisection method, which halves the search interval at each iteration, I was under the impression that its time complexity was also logarithmic. I was therefore wondering whether anyone could shed some light on why the bisection method is slower than Newton's method from a complexity point of view? Any comments or references to literature would be greatly appreciated.

• "I was under the impression that its time complexity was also logarithmic": just an impression, it is linear in the number of digits (exactly one bit per function evaluation). – Yves Daoust Feb 19 '16 at 8:00

Per every bit you need one bisection step, so the cost is $\sim N·eval(f)$, while Newton is, as you wrote, $\sim 3·\log(N)·eval(f)$, where it is used that $eval(f,f')=3·eval(f)$ if automatic/algorithmic differentiation is used.