2

Greetings with utmost respect, everyone!

Today, I found a fascinating math question online. I seem to be stuck while solving it, however. I did not find any relevant solution to the problem yet. Please have a look at it.

Question

Find the value of $\frac{1}{2}^{\frac{1}{3}^{\frac{1}{4}^{\cdots}}}$.

My Approach

As it seems to relate to infinity, I applied logarithms to see if it works. It didn't.

\begin{gather*} y=\left(\frac{1}{2}\right)^{\left(\frac{1}{3}\right)^{\left(\frac{1}{4}\right) \cdots ^{}}} \Longrightarrow \ln y=\left(\frac{1}{3}\right)^{\left(\frac{1}{4}\right){^{\cdots }}^{}}\ln\frac{1}{2}\\ \Longrightarrow \ln^{2} y=\left(\frac{1}{4}\right){^{\cdots }}^{}\ln\frac{1}{3} +\ln^{2}\frac{1}{2} \end{gather*}

Note: $\ln^n(x)$ means $\ln(\ln\cdots n \text{ times} (x)))$

I'm not quite sure how to proceed from here. Or, if necessary, change the approach?

P.S. This is not a homework question.

Rohan Bari
  • 393
  • 3
    Whatever $L=(1/3)^{(1/4)^{(1/5)^{\cdots}}}$ is, necessarily $L\ln\frac12\le 0$, so you can't consider $\ln\ln y$. – Sassatelli Giulio Nov 19 '23 at 02:50
  • 1
    I think this does not converge. It approaches two limit points, $0.658$ and $0.690$, approximately. – GEdgar Nov 19 '23 at 02:53
  • 6
    I suspect I can guess how you have defined your infinite tower of exponents. But you have not stated how it is defined. For one thing, the symbol ∞ should have no place in this expression. For another thing, I suspect you want to define it as the limit of something, but it has not been stated what it is hoped to be the limit of. Finally, the numbers in this expression do not have parentheses telling us in what order the exponentiations are meant to be performed. – Dan Asimov Nov 19 '23 at 02:53
  • 2
    Question on same constants. There is likely no closed form – Тyma Gaidash Nov 19 '23 at 02:54
  • @SassatelliGiulio Oh, I can imagine. $\ln{0.5}$ must be a negative value and a logarithm is defined if and only if it is a nonzero positive. You are correct... – Rohan Bari Nov 19 '23 at 03:01
  • @DanAsimov Actually, that $\infty$ symbol is meant to denote that the tower goes on and on indefinitely. – Rohan Bari Nov 19 '23 at 03:02
  • @ТymaGaidash So there is no direct approach to get this done? – Rohan Bari Nov 19 '23 at 03:07
  • 6
    As written, the question doesn't mean anything, because the expression doesn't mean anything. – Thomas Andrews Nov 19 '23 at 03:14
  • 4
    Infinity is not a number, and expressions that insert the symbol $\infty$ where a numbe should go don't have meaning. Learning to use mathematical notation in a standard way is part of communicating with others about mathematics. – Greg Martin Nov 19 '23 at 04:42
  • 1
    @RohanBari the use of the inifinity symbol is unnecessary. The ellipsis "$...$" alone conveys that the tower goes on and on indefinitely. – RyRy the Fly Guy Nov 19 '23 at 06:00
  • @GregMartin Noted. – Rohan Bari Nov 19 '23 at 06:24
  • @RyRytheFlyGuy Yeah, thanks. Greg mentioned the reason... – Rohan Bari Nov 19 '23 at 06:25
  • 2
    Remember that $a^{b^c}$ means $a^{(b^c)}$, not $(a^b)^c$, which is related to why your expression doesn't make sense: it's not a computation that never ends (which might approach a limit), it's a computation that never begins. Is your expression supposed to denote the limit of a sequence of values you can compute? If so, what are the first few values in that sequence? – Karl Nov 19 '23 at 06:43

0 Answers0