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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 $a_{2k+1}$.

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Elena, I edited your question. Please check and make sure that I didn't change the meaning of the question. –  Thomas Jan 14 '13 at 16:32
    
@Elena: Welcome to MSE! Please use MathJax to format your questions. It would help if you showed us what you have tried so we can help correct that. Is this homework? If so, please tag is as such. Regards –  Amzoti Jan 14 '13 at 16:32
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Is "these sequents" this sequence? Is "congruent" converges? Is "$an$" $a_n$? –  Andres Caicedo Jan 14 '13 at 16:34
    
@amWhy: Yeah, I missed that one. Thanks. –  Thomas Jan 14 '13 at 16:40
    
@Elena : I think you meant "Prove that this sequence converges to $A/3$." –  Michael Hardy Jan 14 '13 at 16:42
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1 Answer

up vote 3 down vote accepted

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 $x = A/3$.

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Thank youu very much :) –  Elena Jan 14 '13 at 17:35
    
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
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