# Tag Info

2

One way to write it is to let $$f(x) = \frac{z \log (1-x)+(1-2 z) \log (x)+(z-1) \log (x z-x+z)}{z (2 z-1)}$$ and write $f^{(-1)}\left(\frac{z \log (1-{a_0})-2 z \log ({a_0})+z \log ({a_0} z-{a_0}+z)-\log ({a_0} z-{a_0}+z)+\log ({a_0})}{z (2 z-1)}-\frac{t}{2}\right)$ for InverseFunction[..][..]. So in regular mathematical notation, the ...

4

The equation x Exp[x] == y has multiple solutions for x. For example, evaluating tab = Table[{x -> ProductLog[i, 1]}, {i, 0, 5}] Exp[x] x /. tab N[tab] gives {1, 1, 1, 1, 1, 1} and {{x -> 0.567143}, {x -> -1.53391 + 4.37519 I}, {x -> -2.40159 + 10.7763 I}, {x -> -2.85358 + 17.1135 I}, {x -> -3.16295 + 23.4277 I}, {x -> -3....

9

The identity does not hold for x < -1: Plot[ProductLog[x*Exp[x]], {x, -5, 5}] FullSimplify[ProductLog[x*Exp[x]], x >= -1] x (* result in 10.1.0 under Windows *)

4

PowerExpand[ProductLog[x Exp[x]]] x This assumes $x\ge0$

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