8
$\begingroup$

Why is $\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\cdots}}}}}}$ equal to 2? Does this work for other numbers?

$\endgroup$

marked as duplicate by kingW3, user91500, mlc, B. Mehta, Xam Jun 2 '17 at 18:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    $\begingroup$ By any chance did you mean ${\sqrt 2}^{{{\sqrt 2}^{\sqrt 2}}^{\dots}} = 2$? If so, look at math.stackexchange.com/questions/233290. Or maybe you meant $\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}$? If so, look at math.stackexchange.com/questions/11945. $\endgroup$ – sdcvvc Jun 28 '13 at 13:11
  • 2
    $\begingroup$ With the reformulated question you have $x^2=2x$ on the assumption of convergence. $\endgroup$ – Mark Bennet Jun 28 '13 at 13:46
  • 1
    $\begingroup$ I cleared up some comments from before the question was edited into present form. That should include the comment of @sdcvvc, but said comment includes some useful links to related, but not identical, questions. $\endgroup$ – Willie Wong Jun 28 '13 at 13:59
  • $\begingroup$ The dots should be on the left side of the expression... in fact one starts the operation from the inner part $\endgroup$ – Emanuele Paolini Jun 28 '13 at 14:03
  • 1
    $\begingroup$ I personally have great difficulty with such "infinite expressions" because of the inherent ambiguity. Although I can understand the remark by @EmanuelePaolini that expression must be evaluated from the inside outward, that would not be possible for the variation $\sqrt {1\sqrt {2\sqrt {3\sqrt {4\sqrt{5\sqrt\cdots}}}}}$ which nevertheless might be given some meaning. My real problem with the question is the final "$2$"; as there are infinitely many operations around in, one never gets to the point of inserting it. I would say, if you mean to take a limit, write a limit. $\endgroup$ – Marc van Leeuwen Jun 28 '13 at 14:20
35
$\begingroup$

The number in question is simply

$$2^{1/2+1/4+1/8+\cdots} = 2^1 = 2$$

Yes, this works for other numbers. More interesting is if the numbers are not equal inside the radicals. For example, say you have a positive sequence element $a_n$ inside the $n$th radical. If we assume the expression converges to some value $P$, then

$$\log{P} = \sum_{k=1}^{\infty} \frac{\log{a_n}}{2^n}$$

$\endgroup$
  • $\begingroup$ Perhaps it is more precise to say that the number in question is the limit of the sequence $a_k = 2^{1/2+1/4+\cdots+1/2^k}$ for $k\geq 1$. $\endgroup$ – Álvaro Lozano-Robledo Jun 28 '13 at 13:57
  • 11
    $\begingroup$ @ÁlvaroLozano-Robledo: perhaps. But I will sleep well with what I have posted. $\endgroup$ – Ron Gordon Jun 28 '13 at 14:00
15
$\begingroup$

A general way to ascribe a value to this non rigorously defined formula and to many others similar to it, is to consider that they describe the "infinite" iteration of a given function. Thus, one considers a sequence $(x_n)$ such that $x_{n+1}=\sqrt{2x_n}$ for every $n\geqslant0$ and $x_0\geqslant0$. Then, indeed the value of the formula is $2$ in the following sense:

For every $x_0\gt0$, $x_n\to2$.

To see this, note that $x_{n+1}=u(x_n)$, where the function $u:x\mapsto\sqrt{2x}$ is such that $x\lt u(x)\lt2$ for every $x$ in $(0,2)$, $u(2)=2$, and $2\lt u(x)\lt x$ if $x\gt2$. Thus $(x_n)$ is increasing if $0\lt x_0\lt2$ and decreasing if $x_0\gt2$ and converges to $2$ in both cases (and I will let you solve the case $x_0=2$).

Does this work for other numbers?

Indeed, if one means by this the fact that:

$\sqrt{a\sqrt{a\sqrt{a\sqrt{a\sqrt{a\sqrt{\cdots}}}}}}=a$, for every $a\geqslant0$.

To see this, consider $a\gt0$ and $x_{n+1}=u_a(x_n)$, where $u_a$ is the function $u_a:x\mapsto\sqrt{ax}$. Note that $x\lt u_a(x)\lt a$ for every $x$ in $(0,a)$, $u_a(a)=a$, and $a\lt u_a(x)\lt x$ if $x\gt a$. Thus $(x_n)$ is increasing if $0\lt x_0\lt a$ and decreasing if $x_0\gt a$ and converges to $a$ in both cases (and I will let you solve the case $x_0=a$).

$\endgroup$
  • $\begingroup$ For completeness, it shouldn't be too hard to answer OP's follow up question ("Does this work for other numbers?") as well. $\endgroup$ – Lord_Farin Jun 28 '13 at 13:56
  • $\begingroup$ @Lord_Farin Done. Thanks. $\endgroup$ – Did Jun 28 '13 at 14:04
  • 9
    $\begingroup$ Yes! A downvote... Halelujah! $\endgroup$ – Did Jun 28 '13 at 14:07
11
$\begingroup$

Let $x=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{\cdots2}}}}}}$ then

$x^{2}=2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{\cdots2}}}}}}=2x$

so $x(x-2)=0$ since $x\neq 0$

$x=2$.

$\endgroup$
  • 16
    $\begingroup$ You should prove the convergence first. $\endgroup$ – eccstartup Jun 28 '13 at 13:53
2
$\begingroup$

In the case of $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$, this is a limit of the short-chord of a polygon $2^n \cdot x$ as $n$ goes large. The short-chord is the chord of a regular polygon that spans two edges. There is a triangle formed by two edges and the short-chord. As the polygon gets more sides, this triangle approaches a straight line, and the short-chord approaches two edges in length.

$\endgroup$
0
$\begingroup$

$$k=\sqrt{a\sqrt{a\sqrt{a\cdots}}}$$ $$k^2/a=k$$

$$k/a=1$$

$$k=a$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.