Exercise 7.15

If we start with ${\eta }_1<{\eta }_2<{\eta }_3$, in first iteration, we have ${\overline{\eta }}_1=\frac{1}{2}({\eta }_1+{\eta }_3)$, if we end up with the U-arrangement of ${\overline{\eta }}_1,{\eta }_2,{\eta }_3$, the interval is halfed. Otherwise, we have the U-arrangement of ${\eta }_1,{\overline{\eta }}_1,{\eta }_2$, which in the second iteration will end up with either ${\eta }_1,{\overline{\eta }}_2,{\overline{\eta }}_1$ or ${\overline{\eta }}_2,{\overline{\eta }}_1,{\eta }_3$, where ${\overline{\eta }}_2=\frac{1}{2}({\eta }_1+{\overline{\eta }}_1)$. both of them are at least half of the original interval $|{\eta }_3-{\eta }_1|$.

So after two iterations, the original interval is at least halfed, Keep doing these, we see that it decreases exponentially in the bisection algorithm.