How is false position method used when the root is a double root?

Question

How do we use it for such cases?

For example, how to find $3$ with the method?

$f(x)=(x-3)^2(x-1)$

Answer 1

The root $x=3$ can not be found with the false position method. Any root found by a bracketing method has to be a zero crossing, with values of both signs in every neighborhood of the root. You also need to start the method with a bracketing interval, which here has to contain the other single root $x=1$. The method will converge to that root, possibly involving a longer initial phase where one interval end point moves towards $x=3$.

For a polynomial function, you can eliminate the multiple roots by computing the GCD with its derivative and dividing it out. This depends on the coefficients being integers or given as rationals.

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Starting the regula falsi method for the given polynomial on the interval $[-2,5]$, the method in some sense "finds" the root at $x=3$ in that the active interval end moves toward that point, the lower end $-2$ never changing. However, the bracketing interval does not change and the convergence is very slow, for instance $x[ 5] = 3.50$, $x[ 8] = 3.38$, $x[ 12] = 3.30$, $x[22] = 3.20$, $x[ 55] = 3.10$.

Changing to the Illinois modification does not help much, as this anti-stalling measure is geared towards simple roots in the standard situation of a convex increasing function (or any of its flipped variants) over the interval. The double root still leads to long stalling segments. Enhancing this variant by a stalling count and an over-relaxed Aitkens delta-squared formula restores fast convergence to the root $-1$.