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Let $x_0=\sqrt{2}+\sqrt{3}+\sqrt{6}$ and

$$\forall n \in \mathbb{N} : x_{n+1}=\dfrac{x_n^2-5}{2(x_n+2)}$$

Then $x_n=?$


My Try :

$$x_{n+1}(2x_n+4)=x_n^2-5 \\2x_{n+1}x_n+4x_{n+1}=x_n^2-5\\x_n^2-2x_{n+1}x_n-(4x_{n+1}+5)=0$$

So we have :

$$x_n=\dfrac{2x_{n+1}\pm\sqrt{2x_{n+1}^2+4(4x_{n+1}+5)}}{2}$$

Now what ?

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    $\begingroup$ for which value of $n$ do you need $x_n$? $\endgroup$ – Math-fun Nov 6 '17 at 14:35
  • $\begingroup$ In the expression for $x_{n+1}$ replace $n+1$ by $n$ (so, shift all $n$ by $-1$.) That way you can express $x_n$ in terms $x_{n-1}$. $x_{n-1}$ you can express in terms of $x_{n-2}$ (using the same trick) and so on all the way down to $x_0$. Were you looking for that? Or for the limit of this sequence? $\endgroup$ – Piotr Benedysiuk Nov 6 '17 at 14:41
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    $\begingroup$ @PiotrBenedysiuk: There is no limit since the given iteration is Newton's method applied to $x^2+4x+5$, which has no real root. $\endgroup$ – Jack D'Aurizio Nov 6 '17 at 14:48
  • $\begingroup$ @PiotrBenedysiuk And anyway, how is the approach in your comment supposed to yield a formula for $x_n$? $\endgroup$ – Did Nov 6 '17 at 14:49
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    $\begingroup$ Except that you will not be able to "express $x_n$ in terms of $x_0$" for every $n$ using this approach. $\endgroup$ – Did Nov 6 '17 at 16:28
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A crucial observation is that the substitution $x_n=a_n-2$ brings the given recurrence relation into a familiar one, namely $$ a_{n+1} = \frac{a_n^2-1}{2a_n} $$ which is associated with the duplication formula for the cotangent function. In particular, $a_0=-\cot\theta$ implies $a_n=-\cot(2^n\theta)$ and $x_n=-2-\cot(2^n\theta)$. Since $a_0=\cot\frac{\pi}{24}$ we have

$$ x_n = -2+\cot\left(\frac{\pi\cdot2^n}{24}\right) $$ and our sequence keeps oscillating between $-2-\frac{1}{\sqrt{3}}$ and $-2+\frac{1}{\sqrt{3}}$ from $x_3$ on.


Anyway, for any $x_0\in\mathbb{R}$ the given sequence is never converging, since, as remarked in the comments, it is Newton's method applied to a quadratic polynomial with a negative discriminant ($x^2+4x+5$).

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  • $\begingroup$ @thanks . why $a_0 =\cotθ $ $\endgroup$ – Almot1960 Nov 6 '17 at 15:07
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    $\begingroup$ @Almot1960: that is a harmless assumption. Any real number is the cotangent of something, here we get lucky that $a_0$ is the cotangent of a rational multiple of $\pi$. $\endgroup$ – Jack D'Aurizio Nov 6 '17 at 15:08
  • $\begingroup$ @ok now why $\cot \frac{\pi}{24}=a_0$ $\endgroup$ – Almot1960 Nov 6 '17 at 15:30
  • $\begingroup$ @Almot1960: that can be proved in a number of ways, for instance by applying the cotangent bisection formulas to $\cot\frac{\pi}{12}$ found here by Micheal. $\endgroup$ – Jack D'Aurizio Nov 6 '17 at 15:33
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    $\begingroup$ @Almot1960 See also this thread and the links in there for more related stuff. $\endgroup$ – Jyrki Lahtonen Nov 6 '17 at 21:47

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