I know about the convergence of the bisection method. I was wondering if we could find an equality or an inequality between x_n-x_{n-1} (successive sequences made while using the method) and $(a,b)$ where $(a,b)$ is the interval we need to find the root in. the approximate error is 0.5. so what can we say about x_n-x_{n-1}, when we know that xn - p <{\frac {|b-a|}{2^{n}}}.
This formula can be ? (p is the root for example)
On
$|x_n-p| \leq 2^{-1} (b_n-a_n)=2^{-1-n} (b-a)$. That's an inequality for the error, which follows from the fact that $p \in [a_n,b_n]$.
On the other hand, for bisection the step size satisfies an equality: $|x_n-x_{n-1}|=2^{-2}(b_{n-1}-a_{n-1})=2^{-1-n}(b-a)$. This is because $x_{n-1}$ is the midpoint of $[a_{n-1},b_{n-1}]$ and $x_n$ is the midpoint of one of the two halves of that interval, so your next guess is always a quarter-bracket away from your current guess.
(I assume in my notation that $[a_0,b_0]=[a,b]$ and $x_n=\frac{a_n+b_n}{2}$.)
The step sizes being known in advance is particular to bisection, and is arguably one of the reasons why it's so slow.
For the bisection method, $\lvert x_n-x_{n-1}\rvert=2^{-n}(b-a)$ (either sign is possible at any step, depending on the function). That's kind of the point of the algorithm. You can prove this by induction using the fact that $x_n$ is the mid-point of a segment of length $2^{1-n}(b-a)$ having $x_{n-1}$ as one of the endpoints.