Give an example in which $a_0=a_1<a_2=a_3<a_4=a_5<a_6=\cdots$ for the bisection method.
I don't see how to make an example of this form. I'm sure there is something simple/elegant that does the trick, but I don't see it. Any hints or solutions are greatly appreciated.
We solve the equation $3x=1$ using the Bisection Method, with initial values $a_0=0$ and $b_0=1$. The root is given by $x=1/3$.
The root is in $[0,1/2]$, so $a_1=0=a_0$ and $b_1=1/2$.
The root is in $[1/4,1/2]$, so $a_2=1/4\gt a_1$ and $b_2=1/2$.
Now look at the points $1/4,1/3,1/2$. The distance from $1/4$ to the root is $1/12$ and the distance from the root to $1/2$ is $1/6$, the same $1$ to $2$ ratio we started with. So the pattern will repeat.
Remark: The above example is in a (weak) sense the only one. Suppose we start with an interval $[a_0,b_0]$ and a continuous $f(x)$ that has different sign at $a_0$ than at $b_0$. Apply the Bisection Method. Suppose we get the pattern in the OP. Then the root of $f(x)=0$ that we converge to is one-third of the way from $a_0$ to $b_0$. However, there is a fair bit of latitude in the choice of $f(x)$.