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Possible Duplicate:
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$

how can one solve for $x$, $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}}}}$

we know, if $x=\sqrt[]{2+\sqrt{2}}$, then, $x^2=2+\sqrt{2}$

now, if $x=\sqrt[]{2+\sqrt{2}}$, then, $(x-\sqrt{2})(x+\sqrt{2})=\sqrt{2}$

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  • $\begingroup$ This may be relevant: math.stackexchange.com/questions/179981/… $\endgroup$ Aug 25, 2012 at 6:59
  • $\begingroup$ What can you say about $x^2$? $\endgroup$
    – user14082
    Aug 25, 2012 at 7:00
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    $\begingroup$ Is that expression even well-defined? What does an infinite series of square roots actually mean? Is this some kind of limit as $n\to\infty$? It's easy to show that if this limit exists, then it must equal 2 (see the answers below); but a full solution would have to show that there is some kind of convergence going on. $\endgroup$
    – user22805
    Aug 25, 2012 at 7:25
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    $\begingroup$ Sorry, @Mathemagician1234, but I don't see how your argument proves anything. I think the best way to go is to show that the sequence of expressions with a finite number of square roots is monotonically increasing and bounded above. You need to do something like this to show that the limit is defined. $\endgroup$
    – user22805
    Aug 25, 2012 at 8:04
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    $\begingroup$ @Arjang Then such a, universal, reminder might not be completely relevant as an answer to David Wallace's comment which, as far as I can see, quite properly reminds some very basic aspects of limits, in the most classical sense of the term, seemingly overlooked by some on this page. In the end, "the limit is defined" is perfect and I would not "replace" it by "the limit is classically defined". $\endgroup$
    – Did
    Aug 25, 2012 at 10:19

6 Answers 6

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Define a sequence $\{a_n\}_{n \geq 1}$ such that $a_1 = \sqrt 2$ and $a_{n+1} = \sqrt{2 + a_n}$. It should be clear that $x$ is the limit of this sequence as $n$ goes to infinity, i.e:

$$x = \lim_{n \rightarrow \infty} a_n $$

To prove that this limit exists, it is sufficient to show that the sequence is bounded above, and monotonically increasing. Both of these facts may be proved by induction.

The fact that $a_n$ is bounded above follows since $a_1 < 2$ and $a_{n+1} = \sqrt{2+a_n}$ which is less than $2$ if $a_n$ is, because then $a_{n+1} < \sqrt{2+2} = 2$.

To prove that $a_n$ is monotonically increasing note that $a_1 > a_2$ and $a_{n+1} > \sqrt{2+a_{n-1}} = a_n$, assuming of course that $a_n > a_{n-1}$.

Every monotonically increasing sequence that is bounded from above converges, so the other solutions are justified in concluding that $x=2$.

Edit: As pointed out by did, the argument above only applies when $a_1 < 2$ as there is no reason to choose $a_1 = \sqrt{2}$ specifically, as $a_1$ occurs in the "$\dots$" portion of the nested radicals.

If $a_1>2$ we may show that the sequence $\{a_n\}$ is bounded below by 2 and monotonically decreasing, by a similar argument. Finally, it is simple to show that $x$ converges if $a_1 = 2$. Thus, $x$ converges to $2$ regardless of the starting value $a_1$.

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  • $\begingroup$ Why did you choose $a_1=\sqrt2$? The value of $a_1$ is hidden in the $\cdots$ at the right end of the formula, so to speak, hence it would seem wiser to check that the limit of the iterations you consider is $2$ for every starting point. $\endgroup$
    – Did
    Aug 25, 2012 at 8:30
  • $\begingroup$ Indeed! The above argument only applies when $a_1 < 2$. If however, $a_1 > 2$ we may show that the sequence $a_n$ is bounded below by 2 and monotonically decreasing, by a similar argument. $\endgroup$ Aug 25, 2012 at 9:04
  • $\begingroup$ Daniel: The argument is definitely salvageable... If $a_1\gt2$, then $a_n\gt2$ for every $n$ and $(a_n)_n$ is decreasing, hence everything works fine. Provided one does not forget the trivial case $a_1=2$, the proof is complete. $\endgroup$
    – Did
    Aug 25, 2012 at 9:05
  • $\begingroup$ Daniel: I suggest you include in your post the argument when $a_1\geqslant2$. (Upvoted anyway.) $\endgroup$
    – Did
    Aug 25, 2012 at 9:14
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put $x = \sqrt{2+x} \implies x^2 = 2+x \implies x^2 -x-2=0$

Now just solve the quadratic equation.

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    $\begingroup$ If you want to draw $\Rightarrow$ (or better yet, $\implies$) use \Rightarrow (or better yet \implies). Don't use $=>$ (=>). $\endgroup$
    – Asaf Karagila
    Aug 25, 2012 at 7:08
  • $\begingroup$ @asaf : Thanks for the suggestion $\endgroup$
    – Rahul
    Aug 25, 2012 at 7:16
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    $\begingroup$ Most upvoted post, at the moment, and not even a solution, for reasons explained elsewhere on the page. $\endgroup$
    – Did
    Aug 25, 2012 at 8:34
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$x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\ldots }}}}}}$

$x^2 =2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\ldots }}}}}$

$x^2 =2+x$

$x^2-x-2=0$

$x^2-2x+x-2=0$

$(x-2)(x+1)=0$

we have, $x=2$

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    $\begingroup$ Easy peasy. And a good example of how important it is to realize when the simple methods are applicable in situations where a problem looks harder then it really is. But there is the technical question of convergence raised by David above,so this isn't really a complete solution unless we assume the convergence. $\endgroup$ Aug 25, 2012 at 7:46
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    $\begingroup$ Let $$x=1-1+1-1+1 ...$$ then substracting 1 from each side: $$x-1 = -1 +1 -1 +1 ...=-x$$ $$2x =1$$ $$x=1/2$$ $\endgroup$
    – leonbloy
    Jan 19, 2013 at 4:24
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This is to address the convergence issue. Assume that $x_0$ is given, and with $c>0$ and $0<\alpha<1$, let me define $$ x_{n+1}=(c+x_n)^\alpha\quad\textrm{for }n=1,2,\ldots,\quad \textrm{and}\quad x=\lim_{n\to\infty}x_n, $$ if the latter limit exists. So the problem is about the fixed point iteration $x_{n+1}=\phi(x_n)$ with $\phi(x)=(c+x)^\alpha$. It is well known that a fixed point $x$ of this iteration is attracting if $|\phi'(x)|<1$. There is only one fixed point $x_*>0$ with $x_*=\phi(x_*)$. We check $$ \phi'(x_*)=\frac{\alpha(c+x_*)^\alpha}{c+x_*}=\frac{\alpha x_*}{c+x_*}<1, $$ so $x_*$ is an attracting fixed point. From here it is easy to see that any iteration with initial point $x_0\geq-c$ will converge to $x_*$.

A more direct way to see the convergence is to note that for $-c\leq x<x_*$, we have $(c+x)^\alpha<x_*$ and $(c+x)^\alpha>x$, and for $x>x_*$, we have $(c+x)^\alpha>x_*$ and $(c+x)^\alpha<x$. Hence the sequence is increasing and bounded from above if $-c\leq x_0<x_*$, and is decreasing and bounded from below if $x_0>x_*$.

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    $\begingroup$ Why did you choose $x_0=\sqrt2$? The value of $x_0$ is hidden in the $\cdots$ at the right end of the formula, so to speak, hence it would seem wiser to check that the limit of the iterations you consider is $2$ for every starting point. $\endgroup$
    – Did
    Aug 25, 2012 at 8:30
  • $\begingroup$ @did: That is a god idea. I incorporated it into the answer. But we still have the restriction $x_0\geq-2$, because otherwise it will become something different. $\endgroup$
    – timur
    Aug 25, 2012 at 15:43
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Well, $x = (2 + x)^.5 $ from the expression of x. Implying $x^2 = 2 + x$. And now you can solve the quadratic for x, giving x = -1 and x = 2. X can't be negative, so x = 2

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In general, the function $f(x)=\displaystyle\bigg(\small\sqrt{\normalsize x+\small\sqrt{\normalsize x+\sqrt{ x+\sqrt{x}}}}\;\normalsize\bigg)$ is:

$0.5(1+\sqrt[]{1+4x})$

Ref: Math-Integration of nested square roots of x

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