5
$\begingroup$

This question already has an answer here:

I met a question which said :

Find the value of

$\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^{.^{.^{.^{.^{.}}}}}}}}$

Now to start I declared

$y=\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^{.^{.^{.^{.^{.}}}}}}}}$

Now this implies that

$y=\sqrt2^y$

Now solving this equation we get

$y=2,4$

but then how can a single number have two values. So where am I going wrong?

Thank you :)

$\endgroup$

marked as duplicate by Simply Beautiful Art, Namaste, Did, Shailesh, Bill Dubuque Sep 26 '17 at 0:48

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.

  • $\begingroup$ Just because the question says "Find the value", doesn't mean the question was well formulated. Perhaps it should have said "Find a value** or "Find all values". $\endgroup$ – Lee Mosher Sep 25 '17 at 18:55
  • $\begingroup$ I can't exactly recall what the question said but the main problem is that how can one number give out two values. Thanks for helping :) $\endgroup$ – Mayank Mittal Sep 25 '17 at 19:00
  • 1
    $\begingroup$ see this. see also. $\endgroup$ – JMoravitz Sep 25 '17 at 19:01
  • 3
    $\begingroup$ note that $\sqrt{2}<2$ and if $0<x<2$ then $\sqrt{2}^x<2$ $\endgroup$ – Adam Sep 25 '17 at 19:04
  • 1
    $\begingroup$ Here's an analogy. Suppose I asked "what is the value of $\sqrt{25}$? It is a single number." and you reasoned. If $x = \sqrt{25}$ then $x^2 = \sqrt{25}^ 2 = 25$. But $(-5)^2= 25$ and $5^2=25$. So $x = 5$ and it equals $-5$. How can a single number have two values. The error is assuming that $x = \sqrt{2}^x$ has only one solution. Your number is one of the solutions. Hint: $\sqrt{2} < 2$ so $\sqrt{2}^{\sqrt{2}} < \sqrt{2}^2 = 2$ so by induction $x$ (if it exists) is $\le 2$. $\endgroup$ – fleablood Sep 25 '17 at 20:15
3
$\begingroup$

I’m sorry, but I have to disagree with all the previous answers except that of @JanEerland. It seems to me that the infinite expression that you have written can be interpreted in only one way, as the limit of a sequence, which we must show to be convergent. If we do this, the limit is unique.

The sequence is defined recursively as follows: \begin{align} a_0&=\sqrt2\\ a_{n+1}&=\sqrt2^{a_n}\quad\text{for }n\ge0\\ L&=\lim_{n\to\infty}a_n \end{align} One sees easily that $a_n<2$ for all $n$, and a little less easily that the sequence is increasing. Your computation gives two possible values, but only one of these is $\le2$, and hence that one is the value, to the extent that the expression is to be viewed as a limit.

$\endgroup$
  • $\begingroup$ From a logical standpoint, what you (correctly) did was to say, “If this expression means anything, it means either $2$ or $4$.” The remaining step, to find that it does mean something, is what I and others have supplied. $\endgroup$ – Lubin Sep 27 '17 at 13:48
1
$\begingroup$

The paradox arises when you assume that a solution exists and in fact there is no solution. To solve the paradox you check your proposed solution against the original equation.

Thus if a positive value of $x$ satisfies $x^{x^{x^...}}=4$ then it must also satisfy $x^4=4$, thus $x=\sqrt{2}$. If a positive value of $x$ satisfies $x^{x^{x^...}}=2$ then it must also satisfy $x^2=2$, thus $x=\sqrt{2}$. Clearly not both can be correct and whichevever one fails to check out implies that that case has no solution. We find that in fact putting in $x=\sqrt{2}$ gives $x^{x^{x^...}}=2$ so we conclude that $x^{x^{x^...}}=4$ has no solution.

$\endgroup$

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