Infinity again

[bonanova]

A friend pointed this one out to me. Interested in any comments.
Consider the equation which involves an infinite string of exponentiation x to the x to the x to the x ...:

[1] x ^x... = 2.

Noting that the exponent of the first x equals the whole expression gives x ² = 2, or x = sqrt(2).
Then consider the equation

[2] x ^x... = 4.

Now x ⁴ = 4, or x = 4throot(4) = sqrt(2).

Thus the expression sqrt(2) ^sqrt(2)... is multivalued, or perhaps indeterminate.
Why?
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[ralphmerridew]

You've gotten ahead of yourself. What you showed is that if there exists an x for which the power converges to 2 or 4, then that x must be sqrt(2). You did not show that a tower of sqrt(2) converges to 2 or 4, or even that it converges at all.

[bonanova]

[Swepsie]

X ² = 2 has 2 solutions.

Playing around in Excel, shows that sqrt(2) = 1.41421...
is indeed a solution that works.


X ⁴ = 4 has 4 solutions.
1.41421...,
-1.41421...,
i * 1.41421...
and i * -1.41421...

One of the last 3 should be the solution for [2].
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