I have a continuous, decreasing and convex function $f$. Given an interval $[a, b]$ such that $f (a)>0 $and $f (b)<0$, if I apply the secant method twice, where the outcome point will be located? I know if we apply once it will be $r <=x < b$, where $r$ is a root of function $f$ and $x$ is the outcome point of secant method.
Any idea-?
Let $x$ be the point where the secant through $a,b$ crosses $0$. As you said, $r\le x\le b$. Actually, $r<b$ because the secant is negative at $b$. If $r=x$, then $f(x)=0$ and we are not going to use the secant method again. So, the interesting case is $r<x<b$, when $f(x)<b$.
But this case is exactly like your original setup, with $[a,x]$ instead of $[a,b]$. Therefore, the second secant will yield $x_2$ such that $r\le x_2\le x$.
Using the method again and again, we get a sequence $x_n\searrow r$.
