How to Initialize Newton- Raphson's method for a nonnegative function

Question

The root solving method Newton-Raphson converges quickly to the estimated root value but requires a 'close' enough initial guess to converge. I have read that an initial value is often chosen by use of the bisection method, where it iterates until a low level of tolerance and it is fed as an initial guess into Newton's methods. However, the bisection method requires a change of signs along the function.

My question is what other method could you use to feed into Newton's if the function is never negative on its domain?

Answer 1

You describe a situation where your function $f$ is defined on $\Omega \subseteq \mathbb{R}$ and is nonnegative, i.e., $f : \Omega \rightarrow [0,\infty)$.

As you correctly observe, you cannot hope to apply the bisection method to $f$ in order to narrow the search for an initial guess.

However, there are at least two options. Any root of $f$ is necessarily a global minimum, of $f$, hence a root of $f'$.

1. If $f$ is at least three times differentiable and $f'''$ is continuous, then you can hope to apply Newton's method to the equation $f'(x) = 0$ and have quadratic convergence. In this case, you may be able to detect a sign change of $f'$ and apply bisection to narrow the search interval. This procedure will give you a list of candidates for roots of $f$, but you will of course have to verify them one by one.
2. Alternatively, you can attempt to locate a minimum of $f$ using a dedicated algorithm, such as the golden section search. Again, you will have to verify if the candidates are in fact roots of $f$. Restrictions apply to the application of the golden section search and while the convergence to a local minimum is assured it is only linear.

Answer 2

Taylor series might help. Suppose you what to find the root for $f(x)=x-\cos(x)$. $\cos(x)\approx 1-x^2/2$, so a root of $f(x)$ is likely close to the roots of $x^2/2+x-1\implies x=\frac{-1\pm \sqrt{3}}{1}\approx 0.73205$ The real root is closer to $0.73908$.