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Let $ \left\{ a_n \right\} $ be a recursive sequence such that $$a_{n+1}=\frac{1}{4-3a_n}\quad,n\ge1 $$ Determine for which $a_1$ the sequence converges and in case of convergence find its limit.

The problem is from the book 'Problems in Mathematical Analysis I by W.J.Kaczor'.

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  • $\begingroup$ $$ {\sqrt{x^2+y}} = {x+{\cfrac {y}{2x+{\cfrac {y}{2x+{\cfrac {y}{2x+{\ddots }}}}}}}} $$ $\endgroup$ Sep 30, 2018 at 20:09

2 Answers 2

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The Moebius transformation $$T:\quad\bar{\mathbb C}\to\bar{\mathbb C},\qquad x\mapsto T(x):={1\over 4-3x}$$ has the two fixed points $1$ and ${1\over3}$. We therefore introduce a new complex projective coordinate $z$ via $$z:={x-{1\over3}\over x-1},\qquad{\rm resp.},\qquad x={z-{1\over3}\over z-1}\ .$$ In terms of this coordinate $T$ appears as ${\displaystyle \hat T(z)={z\over3}}$ (with fixed points $0$ and $\infty$), so that $$\bigl(\hat T\bigr)^{\circ n}(z)={z\over 3^n}\ .$$ It follows that for all initial points $z\ne\infty$ we have $$\lim_{n\to\infty}\bigl(\hat T\bigr)^{\circ n}(z)=0\ .$$ In terms of the original variable $x$ this means that for all initial points $x\ne1$ we have $$\lim_{n\to\infty}T^{\circ n}(x)={1\over3}\ .$$ There is, however, the following caveat: The above argument refers to the domain $\bar{\mathbb C}$; but maybe you want to exclude $x=\infty$ as a generic point. In terms of the coordinate $z$ this is the point $z_*=1$. For all initial values $z_k=3^k$ $(k\geq1)$ we have $\bigl(\hat T\bigr)^{\circ k}z_k=z_*$. This implies that in the original formulation of the problem you have $T^{\circ k}(x_k)=\infty$ (i.e., you "accidentally" hit $\infty$ after finitely many steps) for all initial points $x_k=\bigl(3^k-{1\over3}\bigr)/(3^k-1)$ $(k\geq1)$.

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  • $\begingroup$ Nice solution :-) (+1) upvote $\endgroup$
    – user798113
    Dec 24, 2020 at 20:05
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Suppose that this sequence does converge to A. Then we must have $A= \frac{1}{4- 3A}$. Then $A(4- 3A)= 4A- 3A^2= 1$. $3A^2- 4A+ 1= (3A- 1)(A- 1)= 0$. A is either 1 or 1/3.

If $a_1> 1$ the sequence clearly converges to 1. If $a_1\le 1/3$ if clearly converges to $\frac{1}{3}$. It's a little harder to show, but still true, that if $1/3< a_1< 1$ then the sequence converges to 1/3: if $\frac{1}{3}< a< 1$ then $1< 3a< 3$ so that 0< 3a-1< 2. But for $\frac{1}{3}< a< 1$, $a- 1< 0$. That is, 3a-1 is positive while a- 1 is negative so that $(3a- 1)(a- 1)= 3a^2- 4a+ 1< 0$. Then $3a^2- 4a= a(3a- 4)> 1$ and $a> \frac{1}{3a- 4}$. That is, for $\frac{1}{3}< a_1< 1$ the sequence is decreasing to $\frac{1}{3}$.

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    $\begingroup$ what if $a_1 = 4/3$? $\endgroup$
    – Brian Shin
    Sep 30, 2018 at 16:31
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    $\begingroup$ Moreover, what if $a_1 =2$? I'm finding the sequence converges to $1/3$. $\endgroup$
    – Brian Shin
    Sep 30, 2018 at 16:52
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    $\begingroup$ Let $\,a_1\in\mathbb{R}\,$. 1) How to show it converge for all $\,a_1\,$. 2) "If $\,a_1$... the sequence clearly converges to ..."; is it that clear?! Thanks. $\endgroup$ Sep 30, 2018 at 16:53

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