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Let $0<x_0<1$. Given the recursive defined sequence $x_{n+1}=\frac{1}{3}(x_n^2+2)$ for $n \in \mathbb{N}$. Show that this sequence converges and calculate it's value.

Show that it's bounded above with $x_n <1$ Base Case: $x_1=\frac{1}{3}(x_0^2+2)<1$ Induction Hypothesis: Be $x_k<1$ Induction Step: $n\rightarrow n+1$

$x_{n+1}=\frac{1}{3}(x_n^2+2)<1$.

Show that it's monotonically nondecreasing: $x_{n+1}-x_n=\frac{1}{3}(x_n^2+2) -x_n=...$ I've made a few steps more, but i can't see why this is in the end $>0$...

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  • $\begingroup$ Another method is to argue inductively: $x_{n+2}-x_{n+1}=\frac13(x_{n+1}^2-x_n^2)=\frac13(x_{n+1}-x_n)(x_{n+1}+x_n)$. $\endgroup$ – Andrés E. Caicedo Feb 14 '14 at 8:07
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Hint:\begin{align*} \frac 1 3 (x_n^2 + 2) - x_n &= \frac 1 3 (x_n^2 - 3x_n + 2) \\ &= \frac 1 3 (x_n - 2)(x_n - 1) \end{align*}

Now use the previous part.

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  • $\begingroup$ Since $x_n<1$, $(x_n - 2)(x_n - 1)$ are bouth negativ, so their product is positive. $\endgroup$ – fear.xD Feb 10 '14 at 19:16
  • $\begingroup$ @fear.xD Exactly. And that's exactly what you need. $\endgroup$ – user61527 Feb 10 '14 at 19:16
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In what follows we show that $x_n\to 1$.

Set $y_n=x_n-1$, then $y_0\in (-1,0)$ and $$ y_n=\frac{y_n+2}{3}\cdot y_n. $$

We shall show that: $y_{n}\in \left(-\dfrac{2^n}{3^n},0\right)$.

Indeed, $y_0\in (-1,0)$. Assume that $y_{k}\in \left(-\dfrac{2^k}{3^k},0\right)$. Then $$ \frac{2y_k+2}{3} \in \left(\frac{1}{3},\frac{2}{3}\right), $$ and hence $$ y_{k+1}=y_k\cdot \frac{2y_k+2}{3} \in \left(-\dfrac{2^{k+1}}{3^{k+1}},0\right) $$

Once this is proved, this implies that $y_n\to 0$, and hence $x_n\to 1$.

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  • $\begingroup$ Very interesting approach(as usual)! +1 $\endgroup$ – user119228 Feb 13 '14 at 0:52
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Let be $f(x)={1\over 3}(x^2+2)$. $f'(x)={2\over 3}x$ and we have: $$ x\in[0,1]\implies 0\le f(x)={1\over 3}(x^2+2)\le{1\over 3}(1+2)=1, $$ $$ x\in[0,1]\implies 0\le f'(x)={2\over 3}x\le{2\over 3}<1. $$ So $f:[0,1]\longrightarrow[0,1]$ is contractive and the limit of the sequence is the only fixed point of $f$.

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