Lambert W of a constant multiple

Question

$ \ln(cx) $ can be expressed as $ f(c) + \ln(x) $, where $ f(c)=\ln(c) $.

Does the lambert W function have a similar property? (Can $ W(cx) $ be expressed as $ f(c)+W(x) $ for some function $ f $).

Answer 1

Short answer: no. But here's what happens if you try (along one particular path of analogous reasoning).

Every identity about a function is equivalent to an identity about its inverse. For the logarithm, $$ \ln(x\,y) = \ln x + \ln y \label{1}\tag{1}$$ is equivalent to $$ \exp X \, \exp Y = \exp(X+Y), $$ which you can see by setting $x = \exp X$ and $y = \exp Y$. (I'm suppressing domain considerations for concision.)

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Now, Lambert's $W$ function is the inverse of the product-exponential, i.e. $$ X = W(x) $$ means that $$ x = X \, \exp X. $$ (Again, we should be careful of domains, but we're trying get intuition here first.) This is a more complicated relationship, so we should expect any identity to be more complicated than the analogous one for the logarithm.

Here's a derivation of one possible identity that involves a sum of $W$ values: \begin{align} (X + Y) \, \exp(X+Y) &= (X+Y) \, ( \exp X \, \exp Y ) \\ &= (X \exp X) \, \exp Y + \exp X \, (Y \exp Y) \\ &= (X \exp X) \, \frac{Y \, \exp Y}{Y} + \frac{X \, \exp X}{X} \, (Y \exp Y) \\ &= (X \exp X) \, (Y \exp Y) \, \biggl( \frac{1}{X} + \frac{1}{Y} \biggr). \end{align}

Applying $W$ to both sides, where $X = W(x)$ and $Y = W(y)$ yields $$ W(x) + W(y) = W \Biggl( xy \, \biggl( \frac{1}{W(x)} + \frac{1}{W(y)} \biggr) \Biggr). $$

If we try to mimic the left side of Equation $(\ref{1})$ instead, it's even worse: $$ (X \, \exp X) \, (Y \, \exp Y) = X \, Y \, \exp(X + Y) = \frac{XY}{X + Y} \, (X + Y) \, \exp(X + Y), $$ but this doesn't lead anywhere useful that I can see. Hence, this shows that $W(xy)$ is equal to a mess.