Solve X=sqrt(A)^sqrt(A)^sqrt(A)^..............infinty?

Question

If $X= \newcommand{\W}{\operatorname{W}}\sqrt{A}^{\sqrt{A}^{\sqrt{A}^{\sqrt{A}^{\sqrt{A}^{\sqrt{A}^{\sqrt{A}^{\sqrt{A}^{.{^{.^{\dots}}}}}}}}}}} $ then what is the value of $X^2-e^{1/X}$ ?

Answer 1

You have $X= \sqrt{A}^X.$ So

$$\ln X = X \ln \sqrt{A} = \frac{X}{2}\ln A$$

$$\frac{2\ln X}{X} = \ln A$$

$$ A = \exp\left(\frac{2\ln X}{X}\right).$$

Answer 2

Firstly, this has to converge, which occurs when $e^{-2e}\le A\le e^{2e^{-1}}$. More elaboration on the convergence is discussed in this question.

$$X=A^{X/2}=e^{\frac12\ln(A)X}$$

$$Xe^{-\frac12\ln(A)X}=1$$

$$-\frac12\ln(A)Xe^{-\frac12\ln(A)X}=-\frac12\ln(A)$$

$$-\frac12\ln(A)X=W\left(-\frac12\ln(A)\right)$$

$$X=\frac{W\left(-\frac12\ln(A)\right)}{-\frac12\ln(A)}=e^{-W\left(-\frac12\ln(A)\right)}$$

Where I used the Lambert W function. Now it's easy to compute the rest.