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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.

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closed as unclear what you're asking by Chappers, Claude Leibovici, Mario Carneiro, Empty, wythagoras Oct 3 '15 at 8:25

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Draw an example. By $a_k$ I presume you mean the bisection interval $[a_k,b_k]$. $\endgroup$ – copper.hat Oct 3 '15 at 2:33
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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 $f(x)$.

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