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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 :)

Simply Beautiful Art
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Mayank Mittal
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  • 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". – Lee Mosher Sep 25 '17 at 18:55
  • 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 :) – Mayank Mittal Sep 25 '17 at 19:00
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    see this. see also. – JMoravitz Sep 25 '17 at 19:01
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    note that $\sqrt{2}<2$ and if $0<x<2$ then $\sqrt{2}^x<2$ – Adam Sep 25 '17 at 19:04
  • You've written down mere symbols. Only if you assume that has a value, you get your equation. So just start with a reasonable definition. –  Sep 25 '17 at 19:05
  • @Adam That doesn't necessarily prove anything; what if the limit is $2$? – Théophile Sep 25 '17 at 19:06
  • @Théophile it proves that it certainly cannot be 4 – Adam Sep 25 '17 at 19:06
  • @Adam I see; I misunderstood your point. – Théophile Sep 25 '17 at 19:07
  • If you sketch the curves $y=2^x$ and $y=x^2$, then you'll see that they cross in three places. https://www.wolframalpha.com/input/?i=2%5Ex%3Dx%5E2 – Fly by Night Sep 25 '17 at 19:11
  • https://math.stackexchange.com/questions/2421293/is-sqrt2-sqrt2-sqrt2-sqrt2-sqrt2-4-correct is worth a look. – Barry Cipra Sep 25 '17 at 19:48
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    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 $\le 2$. – fleablood Sep 25 '17 at 20:15
  • $x^2 = 4 $ has two solutions... – amcalde Sep 26 '17 at 09:38

2 Answers2

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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.

Lubin
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  • 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. – Lubin Sep 27 '17 at 13:48
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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.

Oscar Lanzi
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