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This question already has an answer here:

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

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

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

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

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

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  • $\begingroup$ 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? $\endgroup$ – Rohit Jun 18 '17 at 23:53
  • $\begingroup$ @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. $\endgroup$ – Ross Millikan Jun 19 '17 at 0:10
  • $\begingroup$ @RossMillikan Don't forget $e^{-2e}\le A$. Also, it converges when $A=e^{2/e}$ I believe... $\endgroup$ – Simply Beautiful Art Jun 19 '17 at 1:00
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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.

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