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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What is $X$? ${}{}$ – kingW3 Jun 18 '17 at 23:28
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5Possible duplicate of What is the maximum convergent $x$ in the power tower $x^{x^{x^{x\cdots}}}$? ; See also https://math.stackexchange.com/questions/890319/convergence-of-tetration-sequence ; https://math.stackexchange.com/questions/1089458/how-can-i-prove-the-convergence-of-a-power-tower ; https://math.stackexchange.com/questions/108288/infinite-tetration-convergence-radius – Winther Jun 18 '17 at 23:39
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You have $X= \sqrt{A}^X.$ So
$$\ln X = X \ln \sqrt{A} = \frac{X}{2}\ln A$$
$$\frac{2\ln X}{X} = \ln A$$
$$ A = \exp\left(\frac{2\ln X}{X}\right).$$
Andrei
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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
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@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 00:10
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@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 01:00
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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.
Simply Beautiful Art
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