Does $^{\frac12}x=e^{W(\ln x)}$, or not?

Asked Aug 10 '13 at 17:56

Active Aug 10 '13 at 18:36

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Whenever I see tetration discussed here, I inevitably see it asserted that there's no consistent continuous definition for tetration. However, it seems to me that

If we restrict ourselves to positive real values, $^{\frac12}x=e^{W(\ln x)}$ is the only consistent definition for $^{\frac12}x$; and
This definition is sufficient to define tetration to any real height, if we further say that $\displaystyle^tx = {\lim_{a \to t}}\; ^ax$. So, what gives? I don't especially think I'm smarter than professional mathematicians, so if there's an inconsistency I'd like to be able to recognize it. Is there?

edited Aug 10 '13 at 18:36

Ben Grossmann

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asked Aug 10 '13 at 17:56

Jim Butler

1

Could you expand on how you would define ${}^tx$ using only integer values and the fact that $^{\frac12}x=e^{W(\ln x)}$? – Ben Grossmann Aug 10 '13 at 18:43
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so $e^{W(W(\ln(x)))}=e^x $ ? Try W|A http://www.wolframalpha.com/input/?i=e^[W[W[ln[x]]]] – Gottfried Helms Aug 20 '13 at 20:27
How are we supposed to know this given we can't really evaluate $^{1/2}x$ well? – Simply Beautiful Art Dec 31 '15 at 17:03

Does $^{\frac12}x=e^{W(\ln x)}$, or not?

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