Two solutions to one number.

Question

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 :)

Answer 1

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.

Answer 2

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.