Show that$ |e_n| \leq 2^{-(n+1)}(b_0 - a_0)$

Question

I would like to know if someone can shed some light on it.I'm not sure but I think Lipschitz or contraction mapping theorem is involved.

Let $x_n = \frac{a_n + b_n}{2} , r=\lim_{n \to \infty }x_n$ and $e_n =r-x_n$

Here [$a_n,b_n$] with n$\geq$0 denotes that successive intervals that arisein the bisection method when it is applied to a continuos function $f$.

Show that $|e_n| \leq 2^{-(n+1)}(b_0 - a_0)$

Answer 1

Each time you bisect, you reduce the size of the interval by a factor of two. Therefore, after $n$ bisections, the size is reduced by a factor of $2^n$.

This means that $b_n-a_n =\dfrac{b_0-a_0}{2^n} $.

To get that additional factor of two, note that $x_n$ is in the center of $[a_n, b_n]$, so wherever $r$ is, $|x_n-r| \le \dfrac{b_n-a_n}{2} = \dfrac{\frac{b_0-a_0}{2^n}}{2} = \dfrac{b_0-a_0}{2^{n+1}} $.