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Question Number 80131 by M±th+et£s last updated on 31/Jan/20

Commented by MJS last updated on 31/Jan/20

+∞  e^(π/2) ≈4.8  4.8^(4.8) ≈1913  ...

$$+\infty \\ $$$$\mathrm{e}^{\frac{\pi}{\mathrm{2}}} \approx\mathrm{4}.\mathrm{8} \\ $$$$\mathrm{4}.\mathrm{8}^{\mathrm{4}.\mathrm{8}} \approx\mathrm{1913} \\ $$$$... \\ $$

Commented by mr W last updated on 31/Jan/20

do you mean e^(−(π/2)) ? then:  (e^(−(π/2)) )^(....) =x  (e^(−(π/2)) )^x =x  e^(−((πx)/2)) =x  ((πx)/2)e^((πx)/2) =(π/2)  ((πx)/2)=W((π/2))  ⇒x=(2/π)W((π/2))≈((2×0.7454)/π)=0.4745

$${do}\:{you}\:{mean}\:{e}^{−\frac{\pi}{\mathrm{2}}} ?\:{then}: \\ $$$$\left({e}^{−\frac{\pi}{\mathrm{2}}} \right)^{....} ={x} \\ $$$$\left({e}^{−\frac{\pi}{\mathrm{2}}} \right)^{{x}} ={x} \\ $$$${e}^{−\frac{\pi{x}}{\mathrm{2}}} ={x} \\ $$$$\frac{\pi{x}}{\mathrm{2}}{e}^{\frac{\pi{x}}{\mathrm{2}}} =\frac{\pi}{\mathrm{2}} \\ $$$$\frac{\pi{x}}{\mathrm{2}}={W}\left(\frac{\pi}{\mathrm{2}}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}}{\pi}{W}\left(\frac{\pi}{\mathrm{2}}\right)\approx\frac{\mathrm{2}×\mathrm{0}.\mathrm{7454}}{\pi}=\mathrm{0}.\mathrm{4745} \\ $$

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