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I have to admit that I have forget most of my limit knowledge and I would appreaciate an advice with this.

The problem is:

$a_1 := 1$

$a_{n+1} := \dfrac{a_n^2}{4} + 1$

Calculate the limit for $n \rightarrow \infty$.

My thoughts:

I would say the limit is $\infty$ and maybe rewrite the problem as $a_{n+1} := \dfrac{a_n^2 + 4}{4}$?

Also, I remember the "known limits" - i.e. $(1 + 1/x)^x$ converges to $e$ in $\infty$ etc. etc. But I cannot see anything useful to solve the problem above.

Thank you for help!

Clemens Bartholdy
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Tereza Tizkova
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  • I am not sure of the rigorous way, but I remember doing these kind of questions assuming that $a_{n+1} \approx a_n$ and solving the quadratic. The intuition is that if sequence converges, we cna make that $a_{n+1}$ arbitarily close to $a_n$ – Clemens Bartholdy Mar 22 '22 at 13:42
  • I have to admit a friend asked me this and I feel really ashamed I cannot do it rigorously. But it seems so trivial to me, that it has to go to $\infty$, or am I wrong? You see that the numerator of the fraction runs to $\infty$, as $a_n > 1$ for $n \geq 2$... – Tereza Tizkova Mar 22 '22 at 13:46
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    It depends on the initial condition basically. It's actually a deep concept. Different initial conditions give us drastically different behaviours at endpoints. Look up basin and attractors. For this particular case, I'd suggest you write down a few terms of the sequence and see if you can intuit from that – Clemens Bartholdy Mar 22 '22 at 13:57

2 Answers2

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There is a fixed point: $$x=\frac{x^2}{4}+1\implies(x-2)^2=0\implies x=2$$

Does the sequence converge to a fixed point from the starting state? Well, the map $x\mapsto x^2/4+1$ has derivative $x/2$, which is bounded in magnitude $\lt1$ on the interval $[0,2]$: as this interval is compact, in the complete metric space of $\Bbb R$, and you have a (weak) contraction by the mean value theorem, bounded derivative, it does indeed converge to the fixed point by the Banach Contraction Mapping theorem (since $a_1=1\in[0,2]$).

Why compactness matters for the case $d(Ta,Tb)\lt d(a,b)$ (as opposed to $\|T\|\le c\lt 1$ for a fixed $c$).

FShrike
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  • How can you write it as $x = \dfrac{x^2}{4} + 1$, when in fact one $x$ is $a_n+1$ and the other is $a_n$ so they are different? I dont understand how can you rewrite it to the formula as you did. – Tereza Tizkova Mar 22 '22 at 13:54
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    $x$ here represents a hypothetical fixed point @TerezaTizkova so I solved the equation (proving a fixed point does exist) then used a famous theorem to say that the sequence converges to this fixed point. If the equation had no solutions then the fixed point theorem would have been useless. – FShrike Mar 22 '22 at 13:55
  • Ah, makes sense, thank you! It concerns me that we haven´t learned this during my Math. Analysis uni courses, since this is really interesting! – Tereza Tizkova Mar 22 '22 at 14:00
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    You’re welcome. If the Banach theorem seems daunting, there is a simpler special case proof for the real numbers (just look up “fixed point convergence bounded derivative”) – FShrike Mar 22 '22 at 14:01
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    @TerezaTizkova $a_{n+1}$ and $a_n$ must converge to the same limit if it exists, which Fshrike shows must be $x = 2$. That's the intuition for considering fixed points. From there you need only justify that the limit does indeed exist. – infinitylord Mar 22 '22 at 14:06
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1)$a_n (>0)$ is an increasing sequence:

$a_{n+1}-a_n=a_n^2/4 -a_n+1=$

$(1/4)(a_n^2-4a_n)+1=$

$(1/4)(a_n-2)^2-1+1=$

$(1/4)(a_n-2)^2 \ge 0;$

2)Bounded above by $2$

By induction:

$a_1<2;$

Assume $a_n \le 2.$

$a_{n+1}=a_n^2/4 +1\le 2;$

3)Limit

$L=L^2/4+1;$

$L^2-4L+4=0;$

$(L-2)^2=0$, $L=2.$

Peter Szilas
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