# Approximation for Lambert W function near zero

I am looking for a good approximation for the $W_0$ branch of the Lambert $W$ function. I am looking for values $0 < x < e$ only, so I expect something simpler than the general Taylor expansion. Thanks.

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@Moshen: $0 < x < e$ or $0 < x < 1/e$? – Shai Covo Mar 13 '11 at 10:01
The first one is correct. – Helium Mar 13 '11 at 10:29
Which approximation do you know for $0 < x < e$? – Shai Covo Mar 13 '11 at 10:34
I'm extremely amused that we posted very similar questions about the same special function (albeit at different ends): math.stackexchange.com/q/27355/1778 – Charles Mar 16 '11 at 5:58

I don't know how simple you need it, and since you never said anything on how accurate you want your approximant to be (i.e., to how many correct decimal places should the approximant match the Lambert function?),

$$W_0(z)\approx\ln(1+z)\frac{1+\frac{123}{40}z+\frac{21}{10}z^2}{1+\frac{143}{40}z+\frac{713}{240}z^2}$$

should be good enough, which has a maximum error of around $1.6\times 10^{-4}$ for $z\in[0,e]$.

The rational portion here is a Padé approximant; probably one might do better with a minimax rational approximation, but I don't have the patience and inclination to derive it since your question's rather vague to begin with.

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In fact, I am not a mathematician but an engineer. So I am looking for something as simple as possible. I'm trying to find a Nash equilibrium of a game. To that end, I need to solve a system of 2 equations with Lambert $W_0$ function. I have already solved the problem numerically, but an analytical solution is greatly appreciated. The system is $$x = 1 - W_0 (\frac{C_1 e^{y + 1}}{y + 1})$$ $$y = 1 - W_0 (\frac{C_2 e^{x + 1}}{x + 1})$$ To me, your given approximation looks wonderful but it's still too complex to solve my problem – Helium Mar 13 '11 at 10:45
If I'm not mistaken, your system reduces to something like $y = \frac{k_1 - k_2x}{k_1x-k_2}$ (a hyperbola) where $k_1 = c_2-c_1, k_2 = c_2+c_1$. – Eelvex Mar 13 '11 at 14:44
@Mohsen: See comment above (forgot to @ you). – Eelvex Mar 13 '11 at 20:35
@Eelvex: Thanks a lot man(lady? :D). The point I already computed numerically wonderfully fits in the hyperbola you gave. So it must be correct. But I am a little bit confused. Actually, $x$ and $y$ are curves. right? So the intersection must be a point, not a hyperbola. In fact, when I plot the functions, I can easily see that the intersection is a point. Note that I am only interested in real values, not imaginaries. – Helium Mar 13 '11 at 22:00
@Mohsen:(man :)) Yes there are two curves. Use the fact that $f = W_0(f)e^{W_0(f)}$ to simplify your system, or open another question for details :) – Eelvex Mar 13 '11 at 22:28