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How to solve

$$x^{x^{x^x}} = \frac{1}{3^{\sqrt{48}}}$$

Attempt :

Let $x^{x^{\cdots}} = y$ $$\begin{align} x^y &=y\\ y\ln(x) &= \ln(y)\\ -\ln(x) &= -\ln(y)e^{-\ln(y)}\\ -\ln(y) &= W(-\ln(x))\\ y &= e^{-W(-\ln(x))} \end{align}$$

I'll stop this. Am i doing this right? I know the original question is not continuous power. By the way i'm assuming $x^{x^{x^x}} = x^{x^{\cdots}}$ But is this allowed?

Some advice and help are needed.

Thanks

user376343
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user516076
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  • Is it $x^{x^{x^x}}=3^{-\sqrt{48}}$, or is it $x^{x^{\cdots}}=3^{-\sqrt{48}}$? –  Jul 19 '20 at 06:32
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    @Gae.S. the first one. There is no typo – user516076 Jul 19 '20 at 06:33
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    You can't say $x^y=y$ if this is not an infinite power tower. – VIVID Jul 19 '20 at 06:46
  • $\frac{1}{3^\sqrt{48}}$ is less than the minimum of the function $x^{x^{x^x}}$ when $x > 0$, so $x$ must be negative if it exists. – Toby Mak Jul 19 '20 at 06:52
  • @TobyMak. Using $x<0$ could make some problems, don't you think ? I hav the feeling that it could be funny in the complex domain. Cheers :-) – Claude Leibovici Jul 19 '20 at 08:00
  • I am aware of that, but I wasn't sure if the OP was looking for just real-valued solutions or complex-valued solutions. It would be very strange to look for solutions in the complex numbers, as you have mentioned. – Toby Mak Jul 19 '20 at 08:02
  • I have a pair of solutions for the original equation. See my edit. – Claude Leibovici Jul 19 '20 at 08:25

2 Answers2

2

There is no real-valued solution. We first show that for $x > 0$, $x^x \ge x$. If $x \ge 1$, then $x^x \ge x^1 = x$. If $0 < x < 1$, then $$\log (x^x) = x \log x > 1 \log x = \log x,$$ since $\log x < 0$. Therefore, $$x^{x^{x^x}} \ge x^{x^x} \ge x^x \ge x.$$ Next, we show $x^x \ge 1/2$. For the reason above, we need only consider $0 < x < 1$. Then by differentiation, if $f(x) = x^x$, then $$f'(x) = f(x) (\log f(x))' = x^x (1 + \log x),$$ and since we established that $f(x) \ge x$ for $0 < x < 1$ the only critical point is when $x = e^{-1}$. The second derivative $$f''(x) = x^x \left(x^{-1} + (1 + \log x)^2\right)$$ shows that $f$ is a convex function on the same interval. Thus $x = e^{-1}$ is the absolute minimum and the minimum value attained is $f(e^{-1}) = e^{-1/e} > 1/2$. Consequently $$x^{x^{x^x}} > x^{x^{1/2}} = \left((x^{1/2})^{x^{1/2}}\right)^2 > (1/2)^2 > 1/4 > 3^{-4 \sqrt{3}}.$$

heropup
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2

As said in comment, in the real domain, there is no zero for the function $$f(x)=x^{x^{x^x}}-3^{-\sqrt{48}}$$ the first derivative $$f'(x)=x^{x^x+x^{x^x}-1} \left(x^x \log (x) (x \log (x) (\log (x)+1)+1)+1\right)$$ cancels close to $x_*\sim 0.275$ (this is a minimum by the second erivative test) and $f(x_*) \sim 0.593$.

If the problem was $$g(x)=x^{x^{x^x}}-3^{\sqrt{48}}$$ it would be a very different story. Plotting $$h(x)=\log \left(\log \left(\log \left(x^{x^{x^x}}\right)\right)\right)-\log \left(\log \left(4 \sqrt{3} \log (3)\right)\right)$$ shows almost a straight line around $x=2$.

Newton method will work like a charm $$\left( \begin{array}{cc} n & x_n \\ 0 & 2.0000000 \\ 1 & 1.9447990 \\ 2 & 1.9466308 \\ 3 & 1.9466333 \end{array} \right)$$

Edit

Back to the original equation, there are at least two complex roots which are $$x_\pm=-0.332844\pm 0.291254\, i$$

In comments, I have asked how I found these roots. In a preliminary step, I looked at function $$F(a)=\Im(f(-a(1+i)))^2+\Re(f(-a(1+i)))^2$$ and noticed that for $a \sim \frac \pi {10}$ the result was very small $(F\left(\frac{\pi }{10}\right)\sim 1.37 \times 10^{-6}$).

Now, Newton iterations $$\left( \begin{array}{cc} n & x_n \\ 0 & -0.31415927-0.31415927\, i \\ 1 & -0.32787054-0.30029210\, i \\ 2 & -0.33258317-0.29293379\, i \\ 3 & -0.33287291-0.29129953\, i \\ 4 & -0.33284380-0.29125412\, i \\ 5 & -0.33284375-0.29125414\, i \end{array} \right)$$

Claude Leibovici
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