How can I find extremum of $f(x) = x^{x^{…{x^x}}}$, where x is written 2n times? I tried looking at f’(x), but it didn’t get me anywhere. I also researched tetration (https://en.m.wikipedia.org/wiki/Tetration), but didn’t find anything on my problem. All ideas are welcome. Thanks!
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1LImit of $x^{x^{x^\ldots}}$ is known. – Wuestenfux Mar 10 '22 at 13:00
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@Wuestenfux only for some values of $x$- for others there is no limit. – Henry Mar 10 '22 at 13:06
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@Wuestenfux, how can it help with the extremum? If u look at the graph of, say, x^x^x^x, there’s also a local minimum somewhere between 0 and 1. I am trying to figure out where exactly. For x^x it’s at 1/e – ABlack Mar 10 '22 at 13:06
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I doubt there is an easy answer, though for $x^{x^{x^x}}$ it appears to be about $0.593237$ at when $x\approx0.274689$ – Henry Mar 10 '22 at 13:16
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See https://math.stackexchange.com/questions/534820/how-can-we-calculate-xx – Nurator Mar 10 '22 at 13:17
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@Nurator I have no problem differentiating f, I can’t find zeros of f’ – ABlack Mar 10 '22 at 13:22
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Try to use the logarithm trick for $x^{g(x)}$:
$f(x) = x^{g(x)} => \ln(f(x))= \ln(x) \cdot g(x)$
Thus, differentiating gives us
$\Rightarrow \frac{f'(x)}{f(x)}=\frac{g(x)}{x}+\ln(x)\cdot g'(x)$
$\Rightarrow f'(x)= g(x)x^{g(x)-1}+\ln(x)\cdot g'(x)$
As $g(x)$ is of order less than before, you can recursively use this formula to get to a direct formulation of $f(x)$.
Nurator
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I thought of that, too, but if we write down f’(x) we get a complicated equation on x and we can’t really find zeros of f’ for big n. – ABlack Mar 10 '22 at 13:10
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Let $f(x)=y$. The given function can be put as $y=x^y$. Or $\log y=y\log x$
Differentiating wrt $x$, we get
$\frac{1}{y}\frac{dy}{dx}=\frac{y}{x}+\frac{dy}{dx} \log x$
$\frac{dy}{dx}=\frac{\frac{y}{x}}{\frac{1}{y}-\log x}$
Can you proceed from there, noting that for extremum, the derivative vanishes?
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You had the same idea as me :D But dont you need to consider the inner derivative of y? – Nurator Mar 10 '22 at 13:08
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Still, the equation we get for n > 2 is rather complicated and I can’t get to the answer. – ABlack Mar 10 '22 at 13:16