Question-80139

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

$$\left[{Q}\mathrm{80131}\:{Reposted}\right] \\ $$

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

if something is ∞, you may not treat it as a “normal”value, otherwise we would get things like 1×∞=∞ 2×∞=∞ ⇒1×∞=2×∞ ⇒((1×∞)/∞)=((2×∞)/∞) ⇒1=2

$${if}\:{something}\:{is}\:\infty,\:{you}\:{may}\:{not} \\ $$$${treat}\:{it}\:{as}\:{a}\:“{normal}''{value},\:{otherwise} \\ $$$${we}\:{would}\:{get}\:{things}\:{like} \\ $$$$\mathrm{1}×\infty=\infty \\ $$$$\mathrm{2}×\infty=\infty \\ $$$$\Rightarrow\mathrm{1}×\infty=\mathrm{2}×\infty \\ $$$$\Rightarrow\frac{\mathrm{1}×\infty}{\infty}=\frac{\mathrm{2}×\infty}{\infty} \\ $$$$\Rightarrow\mathrm{1}=\mathrm{2} \\ $$

Commented by MJS last updated on 31/Jan/20

try this: (e^(1/e) )^((e^(1/e) )^((e^(1/e) )^(...) ) ) =?

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

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

$$−{j} \\ $$

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

(e^(1/e) )^x =x lnx−(x/e)=0 f(x)=lnx−(x/e) f′(x)=(1/x)−(1/e)=0 ⇒x=e f′′(x)=−(1/x^2 )<0 e is the maximum value 0 ∴(e^(1/e) )^((e^(1/e) )^(...) ) =e

$$\left({e}^{\frac{\mathrm{1}}{{e}}} \right)^{{x}} ={x} \\ $$$${lnx}−\frac{{x}}{{e}}=\mathrm{0} \\ $$$${f}\left({x}\right)={lnx}−\frac{{x}}{{e}} \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{e}}=\mathrm{0} \\ $$$$\Rightarrow{x}={e} \\ $$$${f}''\left({x}\right)=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }<\mathrm{0} \\ $$$${e}\:{is}\:{the}\:{maximum}\:{value}\:\mathrm{0} \\ $$$$\therefore\left({e}^{\frac{\mathrm{1}}{{e}}} \right)^{\left({e}^{\frac{\mathrm{1}}{{e}}} \right)^{…} } ={e} \\ $$

Commented by MJS last updated on 31/Jan/20

I claim for x>0:y=x^x^x^(...) <∞ ⇔ x≤e^(1/e)

$$\mathrm{I}\:\mathrm{claim}\:\mathrm{for}\:{x}>\mathrm{0}:{y}={x}^{{x}^{{x}^{…} } } <\infty\:\Leftrightarrow\:{x}\leqslant\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{e}}} \\ $$$$ \\ $$

Answered by MJS last updated on 31/Jan/20

x=(e^(π/2) )^(...) ln x =(π/2)x ⇔ (π/2)x−ln x =0 f(x)=(π/2)x−ln x f′(x)=0 ⇒ x=(2/π) f′′((2/π))>0 ⇒ absolute minimum at (((2/π)),((1+ln (π/2))) ) ⇒ no zero ⇒ no solution

$${x}=\left(\mathrm{e}^{\frac{\pi}{\mathrm{2}}} \right)^{…} \\ $$$$\mathrm{ln}\:{x}\:=\frac{\pi}{\mathrm{2}}{x}\:\Leftrightarrow\:\frac{\pi}{\mathrm{2}}{x}−\mathrm{ln}\:{x}\:=\mathrm{0} \\ $$$${f}\left({x}\right)=\frac{\pi}{\mathrm{2}}{x}−\mathrm{ln}\:{x} \\ $$$${f}'\left({x}\right)=\mathrm{0}\:\Rightarrow\:{x}=\frac{\mathrm{2}}{\pi} \\ $$$${f}''\left(\frac{\mathrm{2}}{\pi}\right)>\mathrm{0}\:\Rightarrow\:\mathrm{absolute}\:\mathrm{minimum}\:\mathrm{at}\:\begin{pmatrix}{\frac{\mathrm{2}}{\pi}}\\{\mathrm{1}+\mathrm{ln}\:\frac{\pi}{\mathrm{2}}}\end{pmatrix} \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{zero}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$

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