2

Let $f : [a_{0}; b_{0}] \to \mathbb{R}$ be continuous with $f(a_{0})f(b_{0})< 0$. A method for the approximation of $\alpha \in (a_{0}; b_{0})$ such that $f(\alpha) = 0$ is defined which is identical to the Bisection Method except that instead of testing the midpoint of the current interval $[a_{k}; b_{k}]$, the point $$c =\frac{a_{k}f(b_{k})-b_{k}f(a_{k})}{f(b_{k})-f(a_{k})}$$ is chosen. Draw a sketch to illustrate the geometric description of this method in the case where f has no inflection points in $[a_{0}; b_{0}]$. Would it be reasonable to use $|b_{k}-a_{k}]<\text{tol}$ as a convergence criterion with this method?

I was able to do all the parts except the last one. What is meant by a reasonable convergence criterion here? My thinking is that clearly this convergence criterion is better suited to Bisection method. But since the false position converges this would be reasonable convergence criterion although i am aware that the exact convergence criterion for the false position method would be much better.

Btw this is a question from online problem sheets of Oxford (i am doing these to improve my understanding) .

Asim Mahmood
  • 23

1 Answers1

1

For a sufficiently smooth function with a non-zero second derivative around the root the regula falsi develops the unwanted characteristic that the secant roots or new midpoints eventually all fall on the same side of the root. The other interval end stalls, remains unchanged. Thus the interval length has a lower bound as the distance from the root to the stalling interval end. The convergence of the other interval end reduces to linear and can be arbitrarily slow. I've done a model examination of this in How to show that regula falsi has linear rate of convergence?

A suitable convergence test for the unchanged method could check if the midpoint sequence has become single-sided, and then use an extrapolation of the limit point and the distance to it as error measure.

Lutz Lehmann
  • 131,652