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);


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


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.

  • $\begingroup$ What do you mean by nth x and $\sqrt(1st \; x$? $\endgroup$ – DanZimm May 30 '14 at 22:01
  • $\begingroup$ oh, i wasn't allowed to write in subscript $\endgroup$ – CuriousPaths May 30 '14 at 22:03
  • $\begingroup$ See this: math.stackexchange.com/questions/431594/… $\endgroup$ – Ron Gordon May 31 '14 at 2:48
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    $\begingroup$ Because $\sqrt{n.n}=n.$ $\endgroup$ – Yves Daoust Jun 3 '14 at 7:42

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 $xy = y^2 \implies x = y$.

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    $\begingroup$ Mh, be careful with that reasoning. $X = 1-1+1-1+1-1+...$, so $X=1-(1+1-1+1-...) = 1-X$, therefore $X=1/2$ $\endgroup$ – AnalysisStudent0414 May 30 '14 at 22:06
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    $\begingroup$ 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. $\endgroup$ – Slade May 30 '14 at 22:15
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    $\begingroup$ AnalysisStudent0414's reasoning actually computes the Cesàro sum. These kinds of questions are always tied up with giving the definition of your expression. $\endgroup$ – Slade May 30 '14 at 22:17
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    $\begingroup$ @Arkamis Is there an easy way to exclude $y=0$ as a solution, apart from the form of DanielVs answer $\endgroup$ – Ragnar May 30 '14 at 22:30
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    $\begingroup$ @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? $\endgroup$ – Emily May 30 '14 at 23:01

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

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    $\begingroup$ Oh, this way I understand the most, because it is basically rewriting the 1/2 + 1/4 + 1/8... sequence as exponents, I just looked at it as a radical. $\endgroup$ – CuriousPaths May 30 '14 at 22:07
  • $\begingroup$ Wow, this is brilliant. $\endgroup$ – jeremy radcliff May 7 '15 at 19:57

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 $$ \begin{align} \log\left(\sqrt{n\sqrt{n\sqrt{n\dots}}}\right) &=\frac12\left(\log(n)+\frac12\left(\log(n)+\frac12\left(\log(n)+\vphantom{\frac12}\dots\right)\right)\right)\\ &=\frac12\log(n)+\frac14\log(n)+\frac18\log(n)+\dots\\ &=\log(n)\left(\frac12+\frac14+\frac18+\dots\right)\\[6pt] &=\log(n) \end{align} $$

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  • $\begingroup$ 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. $\endgroup$ – Emily May 30 '14 at 23:08
  • $\begingroup$ I might have made a few algebra mistakes along the way... editing in the comment box on a laptop sucks. $\endgroup$ – Emily May 30 '14 at 23:09
  • $\begingroup$ @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. $\endgroup$ – robjohn May 31 '14 at 0:57
  • $\begingroup$ @Arkamis: BTW, I did not downvote your answer. $\endgroup$ – robjohn May 31 '14 at 0:59
  • $\begingroup$ The k is the x part of the expansion, after factoring out n -m fs. $\endgroup$ – Emily May 31 '14 at 1:03

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 $x$

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    $\begingroup$ The limits of the sum in your last line don't look right. Do you mean $x^{\sum_{k=1}^n{\frac{1}{2^k}}}$? $\endgroup$ – Théophile May 30 '14 at 22:09
  • $\begingroup$ Yup. Changed sentence mid-way, thanks a lot $\endgroup$ – AnalysisStudent0414 May 30 '14 at 22:16

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

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

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  • $\begingroup$ This is the steady state approach, assume $a_k$ = $a_{k-1}$ and solving for that value, nice application. $\endgroup$ – DanielV Jun 2 '14 at 16:53
  • $\begingroup$ There are two approaches: knowing the limit, show that it is the limit; or not knowing the limit, find it. $\endgroup$ – Yves Daoust Jun 2 '14 at 18:03

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 $y=x$.

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