# Solve X=sqrt(A)^sqrt(A)^sqrt(A)^…infinty? [duplicate]

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 $X^2-e^{1/X}$ ?

## marked as duplicate by Winther, dantopa, Daniel W. Farlow, JMP, user370967 Jun 19 '17 at 7:52

• What is $X$? ${}{}$ – kingW3 Jun 18 '17 at 23:28
• – Winther Jun 18 '17 at 23:39

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

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

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

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

• Sir,i got this result already but my ans should come A=x^2-e^1/x . i think i should edit my question upto this. or the answer given is wrong? – Rohit Jun 18 '17 at 23:53
• @Rohit: this gives a value for $A$, which could also be written $A=X^{\frac 2X}$. Your question asks for the value of an expression in $X$. Getting $X=$ some expression in $A$ looks difficult. It could be handled numerically. This only makes sense if $A \lt e^{\frac 2e}$ so the power tower converges. – Ross Millikan Jun 19 '17 at 0:10
• @RossMillikan Don't forget $e^{-2e}\le A$. Also, it converges when $A=e^{2/e}$ I believe... – Simply Beautiful Art Jun 19 '17 at 1:00

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.