Combining the bisection method with Newton's method

I need to code an algorithm that finds the root of a function $f$, such that $f(x)=0$. I can assume that I have identified an interval $[a,b]$ with $f(a)<0$ and $f(b)>0$ where the function is monotone and continuous, and hence I know that there is a solution to $f(x)=0$. I am also able to calculate the derivative $f'(x)$ at any point.

I also know that there are other solutions to $f(x) = 0$ outside of $[a,b]$ but I am only interested in the solution inside the interval. I could use Newton's method, but that may overshoot the interval and find the wrong solution. I could also use the bisection method but that would be too slow. How could I combine Newton's method with the bracketing such that I have a guaranteed fast convergence to the correct root?

One idea I had was to use Newton to update the point with the smallest absolute function value (e.g, update $a$ if $|f(a)|< |f(b)|$), updating the interval boundaries based on the sign of the new estimate, or use the bisection method if the updated estimate fell outside the previous interval. How would you do it?

• You should also reduce the interval with each successful Newton iteration. Overshoot the Newton step every now and then to also reduce the interval at the other side of the root. – Lutz Lehmann Nov 26 '16 at 17:31
• You might be interested in Brent's method, that combines the dissection and secant method. – Julián Aguirre Nov 27 '16 at 14:58
• Thank, I am aware of Brent's method, but I wanted to include the derivative and also Brent's method would be a bit harder to code. I coded up a combined Newton/bisection method and it was actually easier than I thought. Every iteration I have a interval [a_k,b_k] with $f(a)<0$ and $f(b)>0$ and the current best estimate $x_k$. First $x_k$ is updated to $x_{k+1}$ with Newton's, and in case this falls outside $[a,b]$ (the original interval) I use bisection instead to get $x_{k+1}$. Then update $[a_{k+1}, b_{k+1}]$ depending on the sign. – Freelunch Nov 27 '16 at 16:53