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I'm reading Knoebel's Exponentials Reiterated, page 242 and have very hard time understanding the following:

If $z^{z^x}=x$, then $\frac{dx}{dz}\gt 0$ on the open line $$L=\{<z,x>|0\lt z\lt e^{-e}\,\text{and}\, x=e^{-1}\}.$$

  1. This doesn't make sense to me. Because if $z^{z^x}=x$, then $x=e^{-1}$ implies $z=e^{-e}$, which is not in the open interval $\left(0,e^{-e}\right)$.

  2. I know that, if $j(z,x)=z^{z^x}-x$, then $$dj(z,x)=z^{z^x}z^{x-1}(x\ln z+1)\,dz+\left(z^{z^x}z^x\ln ^2z-1\right)\,dx,$$ and there is a unique trajectory satisfying $j(z,x)=0$ when it is not true that $z=e^{-e}$ and $x=e^{-1}$ simultaneously by the Implicit function theorem, but how could that be used to prove $\frac{dx}{dz}\gt 0$ on the line?

1mik1
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2 Answers2

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Consider the implicit function $$F(x,z)=z^{z^x}-x=0$$ $$\frac{\partial F(x,z)}{\partial x}=z^{z^x+x} \log ^2(z)-1$$ $$\frac{\partial F(x,z)}{\partial z}=z^{z^x} \left(z^{x-1}+x z^{x-1} \log (z)\right)$$ $$\frac {dx}{dz}=-\frac{\frac{\partial F(x,z)}{\partial z} } {\frac{\partial F(x,z)}{\partial x} }=-\frac{z^{z^x+x-1} (x \log (z)+1)}{z^{z^x+x} \log ^2(z)-1}$$

More pleasant would be the reverse way using $$z=\left(\frac{x \log (x)}{W(x \log (x))}\right)^{\frac{1}{x}}$$ then logarithmic differentiation and chain rule with $t=x \log(x)$. We should end with $$\frac {dx}{dz}=f(x)$$ which is positive for $0 \leq x \lt e$ with a vertical asymptote at $x=\frac 1e$. For $x>e$, the derivative is always negative.

Edit

For the calculation of the derivative, start with $$z=\left(\frac{t}{W(t)}\right)^{\frac{1}{x}} \qquad t=x\log(x)$$ $$\log(z)={\frac{1}{x}} \log\left(\frac{t}{W(t)}\right)$$ $$\frac {z'} z=-\frac 1{x^2}\log\left(\frac{t}{W(t)}\right)+\frac 1x \frac d {dt}\Bigg[\log\left(\frac{t}{W(t)}\right) \Bigg]\frac {dt} {dx}$$ $$\frac d {dt}\Bigg[\log\left(\frac{t}{W(t)}\right) \Bigg]=\frac{W(t)}{t (1+W(t))}$$ All of the above gives $$z'=\frac z {x^2 t (1+W(t)) }\Bigg[x W(t) t'-t (1+W(t)) \log \left(\frac{t}{W(t)}\right) \Bigg]$$ $$\frac {dx}{dz}=\frac 1 {z'}$$

Claude Leibovici
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  • I get $\frac{d,\ln z}{dx}=\frac{W(x\ln x)-W^2(x\ln x)\ln x}{(W(x\ln x)+1)x^2\ln x}$, how do I get your result from that? – 1mik1 Jan 30 '21 at 10:26
  • The function $x\mapsto xW(x\ln x)(\ln x+1)-\left((1+W(x\ln x))\ln \frac{x\ln x}{W(x\ln x)}\right)x\ln x$ equals zero at $x=e^{-1}$, so $\frac{dx}{dz}=0$. How did you get $\frac{dx}{dz}\gt 0$? – 1mik1 Jan 30 '21 at 15:02
  • I mean, $\frac{dx}{dz}=\frac{1}{z'}$ and $\frac{1}{z'}=0$ at $x=e^{-1}$. This doesn't look good. – 1mik1 Jan 30 '21 at 15:14
  • @1mik1. I wrote that there is a vertical asymptote at $x=\frac 1e$ – Claude Leibovici Jan 30 '21 at 15:17
  • Or is it the case that $\frac{dx}{dz}$ on $L$ is nonsensical? This is the most confusing. – 1mik1 Jan 30 '21 at 15:27
  • In other words, there is no $z$ in $(0,e^{-e})$ such that $j(z,x)=0$ when $x=e^{-1}$. – 1mik1 Jan 30 '21 at 15:31
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  1. I think Knoebel meant $dx/dz>0$ where $z\in(0,e^{-e})$ and $x\in(0,e^{-1})$. Otherwise we would just be evaluating the derivative at one point, namely $(x,z)=(e^{-1},e^{-e})$.

  2. Note that $z^{z^x}=x$ is equivalent to $\ln\ln x=\ln\ln z+x\ln z$. Implicit differentiation gives $$\frac{dx}{dz}=\frac{x\ln x(1+x\ln z)}{z\ln z(1-x\ln x\ln z)}$$ after collecting terms. The derivative is positive if $\frac{1+x\ln z}{1-x\ln x\ln z}>0$ which is true following a case-by-case analysis of $1-x\ln x\ln z\in(-\infty,0)$, $(0,1/e)$, $(1/e,1)$ and $(1,\infty)$.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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