Editing Lambert W function (section)

== Identities ==
[[File:Plot of Lambert W function of x exp x.svg|thumb|upright=1.3|A plot of {{math|''W''<sub>''j''</sub>(''xe''<sup>''x''</sup>)}} where blue is for {{math|1=''j'' = 0}} and red is for {{math|1=''j'' = −1}}. The diagonal line represents the intervals where {{math|1=''W''<sub>''j''</sub>(''xe''<sup>''x''</sup>) = ''x''}}.]]
[[File:The product logarithm Lambert W function W 2(z) plotted in the complex plane from -2-2i to 2+2i.svg|upright=1.3|alt=The product logarithm Lambert W function W 2(z) plotted in the complex plane from -2-2i to 2+2i|thumb|The product logarithm Lambert {{math|''W''}} function {{math|''W''<sub>2</sub>(''z'')}} plotted in the complex plane from {{math|−2 − 2''i''}} to {{math|2 + 2''i''}}]]
A few identities follow from the definition:
: <math>\begin{align}
W_0(x e^x) &= x & \text{for } x &\geq -1,\\
W_{-1}(x e^x) &= x & \text{for } x &\leq -1.
\end{align}</math>
Note that, since {{math|1=''f''(''x'') = ''xe<sup>x</sup>''}} is not [[injective]], it does not always hold that {{math|1=''W''(''f''(''x'')) = ''x''}}, much like with the [[inverse trigonometric functions]]. For fixed {{math|''x'' < 0}} and {{math|''x'' ≠ −1}}, the equation {{math|1=''xe<sup>x</sup>'' = ''ye<sup>y</sup>''}} has two real solutions in {{mvar|y}}, one of which is of course {{math|1=''y'' = ''x''}}. Then, for {{math|1=''i'' = 0}} and {{math|''x'' < −1}}, as well as for {{math|1=''i'' = −1}} and {{math|''x'' ∈ (−1, 0)}}, {{math|1=''y'' = ''W<sub>i</sub>''(''xe<sup>x</sup>'')}} is the other solution.

Some other identities:<ref>{{cite web | url=http://functions.wolfram.com/ElementaryFunctions/ProductLog/17/01/0001/ | title=Lambert function: Identities (formula 01.31.17.0001)}}</ref>
: <math>
\begin{align}
& W(x)e^{W(x)} = x, \quad\text{therefore:}\\[5pt]
& e^{W(x)} = \frac{x}{W(x)}, \qquad e^{-W(x)} = \frac{W(x)}{x}, \qquad e^{n W(x)} = \left(\frac{x}{W(x)}\right)^n.
\end{align}
</math>
: <math>\ln W_0(x) = \ln x - W_0(x) \quad \text{for } x > 0.</math><ref>{{cite web | url=http://mathworld.wolfram.com/LambertW-Function.html | title=Lambert W-Function}}</ref>
: <math>W_0\left(x \ln x\right) = \ln x \quad\text{and}\quad e^{W_0\left(x \ln x\right)} = x \quad \text{for } \frac1e \leq x . </math>
: <math>W_{-1}\left(x \ln x\right) = \ln x \quad\text{and}\quad e^{W_{-1}\left(x \ln x\right)} = x \quad \text{for } 0 < x \leq \frac1e . </math>
: <math>
\begin{align}
& W(x) = \ln \frac{x}{W(x)} &&\text{for } x \geq -\frac1e, \\[5pt]
& W\left( \frac{nx^n}{W\left(x\right)^{n-1}} \right) = n W(x) &&\text{for } n, x > 0
\end{align}
</math>
:: (which can be extended to other {{mvar|n}} and {{mvar|x}} if the correct branch is chosen).
: <math>W(x) + W(y) = W\left(x y \left(\frac{1}{W(x)} + \frac{1}{W(y)}\right)\right) \quad \text{for } x, y > 0.</math>

Substituting {{math|−ln ''x''}} in the definition:<ref>https://isa-afp.org/entries/Lambert_W.html Note: although one of the assumptions of the relevant lemma states that ''x'' must be > 1/''e'', inspection of said lemma reveals that this assumption is unused. The lower bound is in fact x > 0.<!--If you're so inclinded, you could load the file in Isabelle, replace the assumptions with "x \<in> {0..exp 1} \<and> x > 0" and the first part of the proof with 'have "x > 0" using assms by simp' and it will prove it for you.--> The reason for the branch switch at ''e'' is simple: for ''x'' > 1 there are always two solutions, −ln&nbsp;''x'' and another one that you'd get from the ''x'' on the other side of ''e'' that would feed the same value to ''W''; these must crossover at ''x'' = ''e'': [https://wolframalpha.com/input/?i=plot+LambertW%280%2C-ln%28x%29%2Fx%29%3BLambertW%28-1%2C-ln%28x%29%2Fx%29x%3D0...5] W<sub>n</sub> cannot distinguish a value of ln x/x from an ''x'' < ''e'' from the same value from the other ''x'' > ''e'', so it cannot flip the order of its return values.</ref>
: <math>\begin{align}
W_0\left(-\frac{\ln x}{x}\right) &= -\ln x &\text{for } 0 &< x \leq e,\\[5pt]
W_{-1}\left(-\frac{\ln x}{x}\right) &= -\ln x &\text{for } x &> e.
\end{align}</math>

With Euler's iterated exponential {{math|''h''(''x'')}}:
: <math>\begin{align}h(x) & = e^{-W(-\ln x)}\\
 & = \frac{W(-\ln x)}{-\ln x} \quad \text{for } x \neq 1.
\end{align}</math>