fiddle around with lambert w function

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These days I was fiddling around with one particular parameterization for Lambert curve. Although it was not that hard, I was happy that finally I made it clear.

The problem is like this: a Lambert curve is a curve given by [C = {(z,w) ; | ; z = w e^w } subset mathbb{C}^* times mathbb{C}^*.] Note that the above equation defines (w) as an implicit function of (z). Such functions are not unique, we call them the Lambert W function .

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By inverse function theorem, (w) can be written locally as a function of (z). Note that there are two branches, the branches we care about is called (w_0). From wikipedia, one could find the asymptotic expansion of (w_0) w.r.t. (z), which is [w0(z) = sum{kge 1}frac{(-k)^{k-1}}{k!}x^k.] Now if we redefine the Lambert curve to be (x = y e^{-y}), and define [t = 1 + sum_{k=1}^infty frac{k^k}{k!}x^k,] how the expression (x=y e^{-y}) may be changed when we substitute (y) by (t)?

Notice that the Taylor expansions of (w_0(z)) and (t(x)) are alike, it is not so hard to establish a functional relationship between the two. So lets start by finding a Taylor series of (y) w.r.t. (x). It is easy to see two definitions of Lambert curves are identified if we introduce the transform (y(x) = -w0(x)). So (y) can be expand to a Taylor series: [y(x)=sum{kge 1}frac{k^{k-1}}{k!}x^k.] By term-by-term differentiation, we know (x y’(x) = t-1). But from another point of view, by implicit differentiation, we find (1 = xy’(y^{-1} - 1)). So (y = 1-t^{-1}). And, if written in the new parameter (t), Lambert curve can be formatted as [x = (1-t^{-1}) e^{-(1-t^{-1})}.]