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Ok, so I've been playing around with radical graphs and such lately, and I discovered that if the 

nth x = √(1st x √ 2nd x ... √nth x);

Then

$$\text{the "infinith" } x = x$$

Example:

$$\sqrt{4\sqrt{4\sqrt{4\sqrt{4\ldots}}}}=4$$
Try it yourself, type calc in Google search, hit then a number, such as $4$, and repeat, ending with $4$, (or press the buttons instead).

I'm a math-head, not big enough though, I think this sequence is divergent or convergent or whatever, too lazy to search up the difference.

However, can this be explained to me? Like how the Pythagorean Theorem can be explained visually.

Martin Sleziak
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CuriousPaths
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6 Answers6

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$$\begin{align} X & = \sqrt{n \cdot \sqrt {n \cdot \sqrt{n \dots} } } \\ & = \sqrt{n} \cdot \sqrt{\sqrt{n}} \cdot \sqrt{\sqrt{\sqrt{n}}} \cdot \dots \\ &= n^{1/2} \cdot n^{1/4} \cdot n^{1/8} \dots \\ &= n^{1/2 + 1/4 + 1/8 \dots} \\ &= n^1 \\ &= n \\ \end{align}$$

DanielV
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Suppose $y = \sqrt{x\sqrt{x\sqrt{x\cdots}}}$.

Multiply both sides by $x$ and take the square root:

$$\sqrt{xy} = \sqrt{x\sqrt{x\sqrt{x\cdots}}} = y$$

Therefore, $\sqrt{xy} = y$, and solving we have $xy = y^2 \implies x = y$.

Emily
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    Mh, be careful with that reasoning.

    $X = 1-1+1-1+1-1+...$, so $X=1-(1+1-1+1-...) = 1-X$, therefore

    $X=1/2$

     – AnalysisStudent0414 May 30 '14 at 22:06
  • Yes, my edits got all messed up when my internet crashed. I have to fix it. – Emily May 30 '14 at 22:07
  • There, it makes more sense when I use the right symbols and don't beg the question. – Emily May 30 '14 at 22:11
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    You are begging the question—which is fine, but it should be a little more explicit. The point is to suppose that $\sqrt{x \sqrt{x \sqrt{x \cdots}}}$ has a definite value $y$, then compute what it would have to be. – Andrew Dudzik May 30 '14 at 22:15
  • ^exactly what you-sir-33433 is saying. Once you know it converges to some value, it works. But if it doesn't, that reasoning could be tricky – AnalysisStudent0414 May 30 '14 at 22:15
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    AnalysisStudent0414's reasoning actually computes the Cesàro sum. These kinds of questions are always tied up with giving the definition of your expression. – Andrew Dudzik May 30 '14 at 22:17
  • You could use this reasoning to find that $y = 3^0 + 3^1 + 3^2+ \dots \Rightarrow y = -1/2$, but +1 anyway since this is a good enough approach for beginners. Convergence can be explained another time. – DanielV May 30 '14 at 22:17
  • No, I am not begging the question. Yes, I am assuming it converges to a finite value. However, that proof is pretty elementary. – Emily May 30 '14 at 22:18
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    @you-sir-33433 "The point is to suppose that..." I wrote, literally, "Suppose that $y =$ ..." I don't know how much more suppose-y I can get. – Emily May 30 '14 at 22:22
  • @Arkamis I wasn't criticizing the proof, I was clarifying what it shows and what it doesn't. The omitted part may be "pretty elementary" to you, but it is still an omission. I suppose I should have said that you are begging half the question. – Andrew Dudzik May 30 '14 at 22:29
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    @Arkamis your "suppose" can be easily interpreted as "let $y=...$ ". You can probably rephrase it better. +1, by the way. – Ant May 30 '14 at 22:30
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    @Arkamis Is there an easy way to exclude $y=0$ as a solution, apart from the form of DanielVs answer – Ragnar May 30 '14 at 22:30
  • @you-sir-33433 Maybe, but we're not going to answer every question down to set-theoretic construction of numbers. The OP identified a pattern and asked about why that pattern holds. I attempted to provide nothing more than an answer to his question. In fact, I changed my answer because my first attempt would have been a contrapositive, which I felt might have been over his head. – Emily May 30 '14 at 22:43
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    @Ragnar We have to define the thing we're talking about first. The most standard would be to say that it's the limit of the sequence $\sqrt{x}, \sqrt{x\sqrt{x}}, \cdots$. It is not hard to show that if $x>1$, then this sequence is decreasing, and every term is $>x$, so it has a limit $\geq x$. Similarly, if $0<x<1$, the sequence is increasing, and every term is $<x$, so it has a positive limit $\leq x$. – Andrew Dudzik May 30 '14 at 22:44
  • @Ragnar Suppose $x \neq \sqrt{x\sqrt{x\cdots}}$. Then, $\sqrt{x \cdot x} \neq \sqrt{x \cdot \sqrt{x\sqrt{x\cdots}}}$. Hence $x \neq \sqrt{x\sqrt{x\cdots}}$, a contradiction. – Emily May 30 '14 at 22:47
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    @Arkamis Okay, This has nothing to do with set theory, this is about the very real possibility that an expression can have no reasonable interpretations, or multiple ones. Saying that we're only allowed to draw attention to this question if we're literally prepared to construct set theory from the ground up is unreasonable. – Andrew Dudzik May 30 '14 at 22:48
  • Well, how else do you justify $\sqrt{n\sqrt{n\cdots}} = \sqrt{n}\cdot \sqrt{\sqrt{n}}\cdots$ without first defining rules of exponentiation? And how do you generalize that to non-integer values of $n$? And so forth... You can make the argument that the $k$th $n$ is under $k$ radicals, but likewise does that argument extend to infinity? At some point you just need to answer the OPs direct question. – Emily May 30 '14 at 22:49
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    @Arkamis Axioms and propositions are not the same thing. I repeat that you are being unreasonable. – Andrew Dudzik May 30 '14 at 22:51
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    And I repeat that an answer to an OP who, in the question itself explicitly admits to not caring about convergence, need not address convergence. – Emily May 30 '14 at 22:54
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    @Arkamis Asserting it does not make it so. – Andrew Dudzik May 30 '14 at 22:56
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    @you-sir-33433 This is a site to answer people's questions, not to slowly assemble a textbook. I simply answered his question. He doesn't understand convergence; is this the appropriate place to show that $1/2^n$ converges, as well? Should we begin with the definition of convergence of infinite series? What's the right place to begin? – Emily May 30 '14 at 23:01
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    Anyone is free to provide an answer as «complete» as his/her will will provide. There is no point in this discussion (and, IMHO, this is a perfectly reasonable answer to the OP's question...) so it would probably be a good idea not to continue. – Mariano Suárez-Álvarez May 31 '14 at 03:09
  • Despite not needing to, I addressed convergence in an edit, although after brief self-reflection I decided I will clean it up and re-post it after I have ensured it is correct. – Emily Jun 02 '14 at 20:00
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    So, this answer gets attacked in the comments for assuming it converges...yet the other answer gets away with distributing a square root $\sqrt{ab} = \sqrt{a}\sqrt{b}$ in the case that $b$ is a limit expression, moreover applying the distribution of the square root infinitely many times, and gets no such attacks? Good grief. – Caleb Stanford Jul 17 '16 at 17:52
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It is important to show that the limit exists. Let define the sequence $$ a_k=\sqrt{\vphantom{A}na_{k-1}} $$ Since $\dfrac{a_k}{a_{k-1}}=\sqrt{\dfrac{n}{a_{k-1}}}$ and $\dfrac{a_k}{n}=\sqrt{\dfrac{a_{k-1}}{n}}$, we have

  1. if $a_{k-1}\le n$, then $a_{k-1}\le a_k\le n$; that is, $a_k$ is increasing and bounded above by $n$.

  2. if $a_{k-1}\ge n$, then $a_{k-1}\ge a_k\ge n$; that is, $a_k$ is decreasing and bounded below by $n$.

In either case, $a_k$ is convergent. Using the continuity of multiplication by a constant and the continuity of square root, we get $$ \lim_{k\to\infty}a_k=\lim_{k\to\infty}\sqrt{\vphantom{A}na_{k-1}}=\sqrt{n\lim_{k\to\infty}a_k} $$ Squaring and dividing by $\lim_{k\to\infty}a_k$, we get that $$ \lim_{k\to\infty}a_k=n $$

Another Approach

Not as rigorous, but perhaps more intuitive. Take the logarithm of both sides and we get $$ \begin{align} \log\left(\sqrt{n\sqrt{n\sqrt{n\dots}}}\right) &=\frac12\left(\log(n)+\frac12\left(\log(n)+\frac12\left(\log(n)+\vphantom{\frac12}\dots\right)\right)\right)\\ &=\frac12\log(n)+\frac14\log(n)+\frac18\log(n)+\dots\\ &=\log(n)\left(\frac12+\frac14+\frac18+\dots\right)\\[6pt] &=\log(n) \end{align} $$

robjohn
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  • Convergence is even easier, really. Let $f_n$ be $n$ square root operations, e.g. $f_2 = \sqrt{x\sqrt{x}}$. Then, for $n >m$,$f_n-f_m = \sqrt{x\sqrt{x\cdots x f_m}}-f_m = k_{n-m}f_m^{1/2^{n-m}}-f_m$ which is Cauchy. – Emily May 30 '14 at 23:08
  • I might have made a few algebra mistakes along the way... editing in the comment box on a laptop sucks. – Emily May 30 '14 at 23:09
  • @Arkamis: I assume that $k_{n-m}$ is supposed to be $f_{n-m}$. However, perhaps I am missing something, but I don't see how $f_n-f_m=f_{n-m}f_m^{2^{m-n}}-f_m$ shows that $f_n$ is Cauchy. In any case, this, or another justification, should be in an answer. – robjohn May 31 '14 at 00:57
  • @Arkamis: BTW, I did not downvote your answer. – robjohn May 31 '14 at 00:59
  • The k is the x part of the expansion, after factoring out n -m fs. – Emily May 31 '14 at 01:03
  • When I'm not on my phone, I'll add the convergence to my answet – Emily May 31 '14 at 01:23
  • IMHO This is the only correct answer because 1. it shows convergence 2. it defines correctly the problem through a sequence – Bogdan Lataianu Nov 24 '23 at 00:35
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You're basically doing this:

$x_0 = \sqrt x$

$x_1 = \sqrt{x x_0}$

$\displaystyle x_n = \sqrt{x x_{n-1}} = x^{\sum_{k=1}^n \frac{1}{2^k}} $

So it's pretty obvious that converges to $x$

AnalysisStudent0414
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Assume that you iterated infinitely many times (can take a while), and observed a convergence to $n$.

One more iteration yields $\sqrt{n.n}=n$.

  • This is the steady state approach, assume $a_k$ = $a_{k-1}$ and solving for that value, nice application. – DanielV Jun 02 '14 at 16:53
  • There are two approaches: knowing the limit, show that it is the limit; or not knowing the limit, find it. –  Jun 02 '14 at 18:03
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Suppose $y = \sqrt{x\sqrt{x\sqrt{x\cdots}}}$. Then obviously $y^2=xy$, whence $y=x$ or $y=0$. But $y \gt 0$, so $y=x$.

Nicky Hekster
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