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For the natural exponential function,

$$ f(x)=e^x \to f(x)=f'(x). $$

Is there a natural tetration (tetral?) function?

$$ f(x)={{^x}b} \to f(x)=f'(x) $$

Is it base $e$?

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    $\begingroup$ You seem to be confusing two things - the differential equation $f'(x)=f(x)$ has the solution $f(x)=ae^x$ for a constant $a$, so you won't get anything essentially new from that. (consider $e^{-x}f(x)$ and differentiate this to obtain $-e^{-x}f(x)+e^{-x}f'(x)=0$ so $e^{-x}f(x)=a$ and $f(x)=ae^x$) $\endgroup$ – Mark Bennet Jun 1 at 7:54
  • $\begingroup$ I see it now. Over morning coffee I did feel it would be nice if a natural tetration function did exist. But thanks for clarifying. $\endgroup$ – lukejanicke Jun 1 at 8:20
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    $\begingroup$ Is there anything natural about tetration? $\endgroup$ – darij grinberg Jun 1 at 9:18
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$f(x)=f'(x)$ implies that $f(x)=ce^x$, so an exponential function, not tetration.

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  • $\begingroup$ If the natural exponential function is the only function that equals its derivative, then this answers my question: there is no natural tetration function. I think. $\endgroup$ – lukejanicke Jun 1 at 8:08
  • $\begingroup$ Which apparently is true, so marking your answer as correct. math.stackexchange.com/questions/644879/… $\endgroup$ – lukejanicke Jun 1 at 8:18

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