8
$\begingroup$

$$\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}=\sqrt{a}$$

Just for fun. I would like to read the proof of this if it exists. Any references would be appreciated.

$\endgroup$
  • 1
    $\begingroup$ As a first step, try to define the sequence rigorously, i.e., give the explicit recursive definition. Then you should be able to show the sequence is monotonic, bounded, and then figure out which equation does the limit satisfy. $\endgroup$ – Ittay Weiss Apr 9 '15 at 23:41
  • 2
    $\begingroup$ So this is equivalent to $$\prod_{i=1}^\infty a^{\frac 1{3^i}}$$ $\endgroup$ – abiessu Apr 9 '15 at 23:42
  • $\begingroup$ @abiessu hahah oh wow I did not see that. It's easy then. Thank you! $\endgroup$ – Ivan Apr 9 '15 at 23:53
  • 1
    $\begingroup$ Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Martin Sleziak Apr 10 '15 at 11:09
  • $\begingroup$ This is almost the same question as math.stackexchange.com/questions/815418/… $\endgroup$ – DanielV Jun 3 '15 at 1:56
6
$\begingroup$

Taking the "redefinition" as $\prod_{i=1}^\infty a^{\frac 1{3^i}}$, we immediately have $\sum_{i=1}^\infty \frac 1{3^i}=\frac 12$, which is the effect of summing the exponents as would occur with the product described. Therefore, we have $$\prod_{i=1}^\infty a^{\frac 1{3^i}}=\sqrt a$$

$\endgroup$
16
$\begingroup$

Let $x_n = \sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\sqrt[k]{a\cdots}}}}}}}}}$, where the $\sqrt[k]{\cdot}$ appears $n$ times. Prove that this sequence is monotone and bounded. Now Setting $\lim_{n \to \infty} x_n = x$, we obtain $$x = \sqrt[k]{ax} \implies x^k = ax \implies x^{k-1} = a \implies x = \sqrt[k-1]{a}$$ In your case, $k=3$.

$\endgroup$
  • 4
    $\begingroup$ $$x_{n+1}=\sqrt[k]{ax_n},$$ $$x=\lim_{n\to \infty} x_n= \lim_{n\to \infty} x_{n+1} =\lim_{n\to \infty} \sqrt[k]{ax_n}=\sqrt[k]{a \lim_{n\to \infty} x_n}=\sqrt[k]{ax}$$ $\endgroup$ – Leonhardt von M Apr 10 '15 at 0:00
11
$\begingroup$

Let $\displaystyle b=\sqrt[3]{a\cdot\ \underbrace{\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}_{\text{This is $b$, which is allegedly $\sqrt a$.}}}$.

Then $\displaystyle b = \sqrt[3]{ab}$, so $b^3 = ab$, and then $b^2 = a$. Thus $b = \sqrt a$.

If we put $\sqrt{a}$ in place of the expression said to be $b=\sqrt{a}$, we get $\sqrt[3]{a\sqrt{a}}$. And it is easy to see that that is indeed $\sqrt a$.

As to convergence: let $g(x) = (ax)^{1/3}$. The question is the behavior of the sequence $$ a,\ g(a),\ g(g(a)),\ g(g(g(a))),\ \ldots\ . $$ For $x$ between $\sqrt{a}$ and $a$, we have $0<g'(x)<1/3$, so $g$ is a contraction and thus has a unique attractive fixed point.

$\endgroup$
  • 1
    $\begingroup$ Not sure why this got down votes... $\endgroup$ – abiessu Apr 10 '15 at 1:20
  • $\begingroup$ I think people misread "allegedly" $\endgroup$ – k_g Apr 10 '15 at 2:20
  • $\begingroup$ From $b^3=ab$, how do you rule out the possibility $b=0$? $\endgroup$ – Barry Cipra Apr 10 '15 at 11:37
  • $\begingroup$ @BarryCipra : Interesting question. I'll see if I can show (maybe by induction?) that $\sqrt a$ is a lower bound, at least if $a>1$. And $a$ itself if $a<1$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Apr 10 '15 at 15:56
  • $\begingroup$ @BarryCipra : OK, how about this argument: Iteration of a contraction moves you toward the attractive fixed point, so the sequence will go to $\sqrt a$, not to $0$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Apr 10 '15 at 16:09
4
$\begingroup$

Another way to go is to show that $x_n=a^{b_n}$ where $b_n$ satisfies

$$b_1=\frac13\quad\text{and}\\ b_{n+1}=(1+b_n)\frac13$$

then show $\lim b_n=\frac12$ in a manner similar to user17762's answer.

$\endgroup$
1
$\begingroup$

What we have is the $lim_{n\to\infty}$ $a^{\frac{1}{3}}a^{\frac{1}{9}}a^{\frac{1}{27}}..a^{\frac{1}{3^n}}$.

Using the property $(a^b)(a^c)=a^(b+c)$

We get:

$a^{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...\frac{1}{3^n}}$

The exponent is in fact is geometric series which has a value of $(1/(1-(1/3))-1=\frac{1}{2}$ so:

We get your expression to be $a^{\frac{1}{2}}=\sqrt{a}$

$\endgroup$
  • 2
    $\begingroup$ You should write your answer in $\rm\LaTeX$. I think you might have the right idea more or less and is quite similar in nature to @abiessu's answer. However it's hard to tell when it's written like this. $\endgroup$ – Cameron Williams Jun 3 '15 at 0:30
  • $\begingroup$ May I ask you to join a chat room of mine? The chat room is at the bottom of my profile description. $\endgroup$ – Simply Beautiful Art Feb 2 '17 at 1:27
0
$\begingroup$

this might not be entirely rigorous, but here is a simple intuitive way to think of it: \begin{align*}x &= \sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}\\ x^3&=a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}\\ \frac{x^3}{a}&=\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}\\ \frac{x^3}{a} &= x\\ x^2&=a\\ x&=\sqrt{a}\\ \sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}&=\sqrt{a} \end{align*}

$\endgroup$
  • $\begingroup$ This assumes that the sequence converges. $\endgroup$ – robjohn Jun 3 '15 at 2:03
  • $\begingroup$ Yes, so like I said, not entirely rigorous, just a solid intuition $\endgroup$ – nosyarg Jun 3 '15 at 2:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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