<p><em>Here is a detailed outline, not entirely made rigorous.</em></p>

<p>Setting $b_n \stackrel{\rm def}{=} \frac{1}{6}-Y_n$, with $b_0 = 1/6$, we have
$$
b_{n+1} = \frac{4}{3}\left( \sqrt{1+\frac{3}{2} b_n - \frac{9}{2} b_n^2} - 1\right)\tag{1}
$$
(I like to set things near zero.) By induction, $b_n \geq 0$ for every $n$, and a simple computation shows that $b_{n+1} - b_n \leq 0$ for all $n$. By monotone convergence, $(b_n)_n$ converges, and the fixed point being zero we have $$\lim_{n\to \infty}b_n = 0\,.\tag{2}$$</p>

<p>We can do a Taylor expansion of (1) to get
$$
b_{n+1} = b_n - \frac{27}{8}b_n^2 + o(b_n^2)\tag{3}
$$
i.e., summing, $$b_{n} = b_0 - \frac{27}{8}\sum_{k=0}^{n-1} \left(b_k^2 + o(b_k^2)\right)\,\tag{4}$$</p>

<p>Solving the recurrence
$$
\tilde{b}_{n} = \frac{1}{6} - \frac{27}{8}\sum_{k=0}^{n-1} \tilde{b}_{k}^2$$
yields $\tilde{b}_{n} = \frac{1+o(1)}{6+\frac{27}{8}n} \displaystyle\operatorname*{\sim}_{n\to\infty} \frac{8}{27n}$. (For instance, solving the continuous version $h' = -\frac{27}{8}h^2$ with $h(0)=1/6$.) "Therefore", $b_n \displaystyle\operatorname*{\sim}_{n\to\infty} \tilde{b}_{n}$ and
$$
Y_n = \frac{1}{6} - \frac{8}{27n} + o\left(\frac{1}{n}\right)\,.\tag{5}
$$</p>

<hr>

<p>Here is a plot (in Mathematica) to show how they match:</p>

<pre><code>t1 := RecurrenceTable[{a[1 + n] == 3/2 - Sqrt[2 - 8 a[n]^2], 
   a[0] == 0.0}, a[n], {n, 1, 10000, 100}]
t2 := Table[ 1/6 - 8/(27 n),  {n, 1, 10000, 100}]
ListPlot[{t1, t2}, PlotLegends -&gt; {"Recurrence", "Asymptotics"}, 
 PlotMarkers -&gt; Automatic]
</code></pre>

<p><a href="https://i.sstatic.net/2OVGQ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/2OVGQ.png" alt="enter image description here"></a></p>

CC BY-SA 3.0
2018-04-20T12:59:16.683Z
2018-04-20T14:01:59.610Z
2018-04-20T14:01:59.610Z
Clement C.
<p>Let $f(x) = 3/2 - \sqrt{2 - 8x^2}$. I am interested in the recurrence $Y_0 = 0$ and $Y_{i+1} = f(Y_i) = f(f(Y_{i-1})) = ... = f(f(f(f(... f(0) ...))))$. A quick plot shows that $Y_i$ converges to $1/6$ (also $f(1/6) = 1/6$). I am now interested in how quickly the sequence converges, i.e., for a given $\delta &gt; 0$, how can I determine the smallest $i$ such that $Y_i \ge 1/6 - \delta$? The function $f$ is rather unusual for these type of recurrences and I don't know how to deal with such a function.</p>

CC BY-SA 3.0
2018-04-20T12:20:40.927Z
Not quite sure this is entirely in the scope of this site, but by a Taylor expansion of $1/6-Y_n$ around $0$ one can show $$Y_n = \frac{1}{6} - \frac{8}{27n} + o\left(\frac{1}{n}\right)\,.$$
Clement C.
<fixed-points>
Solving recurrence
Chris
# Solving recurrence

(Asked by: username_0 on 2018-04-20)

Let $f(x) = 3/2 - \sqrt{2 - 8x^2}$. I am interested in the recurrence $Y\_0 = 0$ and $Y\_{i+1} = f(Y\_i) = f(f(Y\_{i-1})) = ... = f(f(f(f(... f(0) ...))))$. A quick plot shows that $Y\_i$ converges to $1/6$ (also $f(1/6) = 1/6$). I am now interested in how quickly the sequence converges, i.e., for a given $\delta > 0$, how can I determine the smallest $i$ such that $Y\_i \ge 1/6 - \delta$? The function $f$ is rather unusual for these type of recurrences and I don't know how to deal with such a function.

**Tags:** fixed-points
> **Comments:**
> **username_1**: Not quite sure this is entirely in the scope of this site, but by a Taylor expansion of $1/6-Y_n$ around $0$ one can show $$Y_n = \frac{1}{6} - \frac{8}{27n} + o\left(\frac{1}{n}\right)\,.$$  

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## 1 answers

### Answer by username_1 on 2018-04-20. Score: 1

Here is a detailed outline, not entirely made rigorous.
Setting $b_n \stackrel{\rm def}{=} \frac{1}{6}-Y_n$, with $b_0 = 1/6$, we have
$$
b_{n+1} = \frac{4}{3}\left( \sqrt{1+\frac{3}{2} b_n - \frac{9}{2} b_n^2} - 1\right)\tag{1}
$$
(I like to set things near zero.) By induction, $b_n \geq 0$ for every $n$, and a simple computation shows that $b_{n+1} - b_n \leq 0$ for all $n$. By monotone convergence, $(b_n)_n$ converges, and the fixed point being zero we have $$\lim_{n\to \infty}b_n = 0\,.\tag{2}$$
We can do a Taylor expansion of (1) to get
$$
b_{n+1} = b_n - \frac{27}{8}b_n^2 + o(b_n^2)\tag{3}
$$
i.e., summing, $$b_{n} = b_0 - \frac{27}{8}\sum_{k=0}^{n-1} \left(b_k^2 + o(b_k^2)\right)\,\tag{4}$$
Solving the recurrence
$$
\tilde{b}_{n} = \frac{1}{6} - \frac{27}{8}\sum_{k=0}^{n-1} \tilde{b}_{k}^2$$
yields $\tilde{b}_{n} = \frac{1+o(1)}{6+\frac{27}{8}n} \displaystyle\operatorname*{\sim}_{n\to\infty} \frac{8}{27n}$. (For instance, solving the continuous version $h' = -\frac{27}{8}h^2$ with $h(0)=1/6$.) "Therefore", $b_n \displaystyle\operatorname*{\sim}_{n\to\infty} \tilde{b}_{n}$ and
$$
Y_n = \frac{1}{6} - \frac{8}{27n} + o\left(\frac{1}{n}\right)\,.\tag{5}
$$

Here is a plot (in Mathematica) to show how they match:
t1 := RecurrenceTable[{a[1 + n] == 3/2 - Sqrt[2 - 8 a[n]^2], 
   a[0] == 0.0}, a[n], {n, 1, 10000, 100}]
t2 := Table[ 1/6 - 8/(27 n),  {n, 1, 10000, 100}]
ListPlot[{t1, t2}, PlotLegends -> {"Recurrence", "Asymptotics"}, 
 PlotMarkers -> Automatic]


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