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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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HOLYBIBLETHE
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  • This may be relevant: http://math.stackexchange.com/questions/179981/how-can-one-express-sqrt2-sqrt2-without-using-the-square-root-of-a-squar/180000#180000 – Bombyx mori Aug 25 '12 at 06:59
  • What can you say about $x^2$? –  Aug 25 '12 at 07:00
  • See also: this thread, this thread, this thread and the links therein. – t.b. Aug 25 '12 at 07:09
  • If I wanted to be quirky, I'd say that $x$ is already solved for. – Bruno Joyal Aug 25 '12 at 07:18
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    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. –  Aug 25 '12 at 07:25
  • I first saw this in a problem solving seminar for math majors-it's really a trick question because all you have to realize is that no matter how complicated the expression is, since x is an undetermined variable, you can simply assign x to the whole mess on the right side of the first square root. Then high school algebra gets you a quadratic equation and the rest is history. : ) – Mathemagician1234 Aug 25 '12 at 07:25
  • @David That's a very good point you've made,but if we replace the square roots with rational exponents, the resulting infinite sum of (2^(1/(2^n)) converges to 2 since if we square both sides and manipulate the sum, we obtain 2 + (2^(1/(2^(n-1))). The right hand sum clearly converges to 0 as n goes to infinity. Simple calculus. – Mathemagician1234 Aug 25 '12 at 07:43
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    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. –  Aug 25 '12 at 08:04
  • @DavidWallace : replace "the limit defined" with "the limit is classically defined". There are plenty of examples of non bounded series that converge but not in the classical sense. – jimjim Aug 25 '12 at 08:30
  • @Arjang There are plenty of examples of non bounded series that converge but not in the classical sense... No idea what you are talking about here. Care to elaborate? – Did Aug 25 '12 at 08:32
  • @did : Just want to remind that classical convergence is not the only game in town. – jimjim Aug 25 '12 at 10:07
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    @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". – Did Aug 25 '12 at 10:19
  • @did : that is true. – jimjim Aug 25 '12 at 10:49
  • I really don't see how this question is a duplicate of the linked one. While similar proofs work, there are solutions which work for $2$ but not for the linked one..... After all $2 \neq 7$. – N. S. May 31 '13 at 14:22

6 Answers6

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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 $a_1$.

Daniel Pietrobon
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  • 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. – Did Aug 25 '12 at 08:30
  • 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. – Daniel Pietrobon Aug 25 '12 at 09:04
  • 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. – Did Aug 25 '12 at 09:05
  • Daniel: I suggest you include in your post the argument when $a_1\geqslant2$. (Upvoted anyway.) – Did Aug 25 '12 at 09:14
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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.

celtschk
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Rahul
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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, $x=2$

HOLYBIBLETHE
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    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. – Mathemagician1234 Aug 25 '12 at 07:46
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    Let $$x=1-1+1-1+1 ...$$ then substracting 1 from each side: $$x-1 = -1 +1 -1 +1 ...=-x$$ $$2x =1$$ $$x=1/2$$ – leonbloy Jan 19 '13 at 04:24
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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 $x_0>x_*$.

timur
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    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. – Did Aug 25 '12 at 08:30
  • @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. – timur Aug 25 '12 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

n0nChun
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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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NoChance
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