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Question Number 80139 by M±th+et£s last updated on 31/Jan/20

Commented by M±th+et£s last updated on 31/Jan/20

$$\left[Q80131Reposted\right]$$

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

$$ifsomethingis\mathrm{\infty },youmaynot$$$$treatitasa‘‘{normal}^{″}value,otherwise$$$$wewouldgetthingslike$$$$1\times \mathrm{\infty }=\mathrm{\infty }$$$$2\times \mathrm{\infty }=\mathrm{\infty }$$$$\Rightarrow 1\times \mathrm{\infty }=2\times \mathrm{\infty }$$$$\Rightarrow \frac{1\times \mathrm{\infty }}{\mathrm{\infty }}=\frac{2\times \mathrm{\infty }}{\mathrm{\infty }}$$$$\Rightarrow 1=2$$

Commented by MJS last updated on 31/Jan/20

$$\mathrm{try}\mathrm{this}:$$$${\left({\mathrm{e}}^{\frac{1}{\mathrm{e}}}\right)}^{{\left({\mathrm{e}}^{\frac{1}{\mathrm{e}}}\right)}^{{\left({\mathrm{e}}^{\frac{1}{\mathrm{e}}}\right)}^{...}}}=?$$

Commented by M±th+et£s last updated on 31/Jan/20

$$-j$$

Commented by Tony Lin last updated on 31/Jan/20

$${\left(e^{\frac{1}{e}}\right)}^x=x$$$$lnx-\frac{x}{e}=0$$$$f\left(x\right)=lnx-\frac{x}{e}$$$$f^{\prime }\left(x\right)=\frac{1}{x}-\frac{1}{e}=0$$$$\Rightarrow x=e$$$$f^{″}\left(x\right)=-\frac{1}{x^2}<0$$$$eisthemaximumvalue0$$$$\therefore {\left(e^{\frac{1}{e}}\right)}^{{\left(e^{\frac{1}{e}}\right)}^{...}}=e$$

Commented by MJS last updated on 31/Jan/20

$$\mathrm{I}\mathrm{claim}\mathrm{for}x>0:y=x^{x^{x^{...}}}<\mathrm{\infty }\iff x⩽{\mathrm{e}}^{\frac{1}{\mathrm{e}}}$$$$$$

Answered by MJS last updated on 31/Jan/20

$$x={\left({\mathrm{e}}^{\frac{\pi }{2}}\right)}^{...}$$$$\ln x=\frac{\pi }{2}x\iff \frac{\pi }{2}x-\ln x=0$$$$f\left(x\right)=\frac{\pi }{2}x-\ln x$$$$f^{\prime }\left(x\right)=0\Rightarrow x=\frac{2}{\pi }$$$$f^{″}\left(\frac{2}{\pi }\right)>0\Rightarrow \mathrm{absolute}\mathrm{minimum}\mathrm{at}\left(\begin{array}{c}\frac{2}{\pi }\\ 1+\ln \frac{\pi }{2}\end{array}\right)$$$$\Rightarrow \mathrm{no}\mathrm{zero}\Rightarrow \mathrm{no}\mathrm{solution}$$

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