5

We have $a_0 = x_2\in [0,A]$ and $a_{n+1} = (A−a_n)/2$. Prove that this sequence converges to $A/3$. And then prove the same for the sequences $a_{2k}$ and $a_{2k+1}$.

Andrés E. Caicedo
  • 79,201
Elena
  • 61

1 Answers1

3

Hint: write $a_n = A/3 + b_n$. Then $$a_{n+1} = (A - a_n) / 2 = A / 3 - b_n / 2 = A/3 + b_{n+1}.$$

Alternative proof: the function $x \mapsto (A - x) / 2$ is a contraction and so by Banach's fixed-point theorem has a unique fixed point $ x = (A - x) / 2$ and so $x = A/3$.

Marek
  • 6,196
  • Actually, we don't need Banach here: it's enough to observe that the graph of the function is a line not parallel with the line $y=x$ and so intersects it in a unique point. – Marek Jan 14 '13 at 17:36
  • @Marek One does not really need all that to determine the asymptotics of $b_n=b_0/2^n$... – Did Jan 14 '13 at 17:37
  • @did: what exactly are you referring to (I mentioned Banach's theorem just for fun)? Also, your expression for $b_n$ is not correct... – Marek Jan 14 '13 at 17:40
  • @Marek I referred to what you wrote in your comment (what else?). (Right, $b_n=(-1)^nb_0/2^n$.) – Did Jan 14 '13 at 17:50
  • Nice answer (+1). I wrote up a similar answer before I read yours, oh well :-) – robjohn Jan 14 '13 at 18:50