I would like to find a solution $x$ for $x^x = c$ where $c$ is a positive constant.
Firstly I'm looking for an approximative solution when $c$ tends to infinity.
Thank you in advance
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Use this formula $$ x = \frac{\log c}{W_k(\log c)}, k \in \mathbb{Z} $$ where $W_k$ is the $k$th branch of the product logarithm. This is the inverse of $f(x) = x^x$.
Henricus V.
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the product logarithm ? the Lambert W function ? – reuns Mar 17 '16 at 05:26
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@user1952009 Yes. They are the same thing. – Henricus V. Mar 17 '16 at 05:27
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and why would you consider some particular branch of $W(z)$ ? – reuns Mar 17 '16 at 05:29
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@user1952009 The solution is not unique. – Henricus V. Mar 17 '16 at 05:29
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yes, we can consider the branch we'd like, that's what you meant, ok. – reuns Mar 17 '16 at 05:29
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and wikipedia doesn't give details on the $\mathbb{Z}$ different branches (one for each solution of $z e^{z} = 1 \implies z + \ln z = 0 \implies$ one for each branch of the logarithm), how do we know what looks like the set ${ z \in \mathbb{C} \ \mid \ z e^z = u }$ ? – reuns Mar 17 '16 at 05:32
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@user1952009 Sorry, what do you mean by "what looks like"? – Henricus V. Mar 17 '16 at 05:34
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if I draw the set, is it on a line, a curve, doesn't it have accumulation points, does it go to $\infty$, is it contained on a half-plane, etc... ? – reuns Mar 17 '16 at 05:35
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@user1952009 It is a countably infinite set. I don't know about the other properties though. – Henricus V. Mar 17 '16 at 05:37
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yes that it is (at most) countably infinite, I can prove it because $W(z)$ has as many branches as the logarithm has – reuns Mar 17 '16 at 05:38
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Thank you very much for your answer. – Dingo13 Mar 17 '16 at 21:39