User:Balabiot/LambertW

The function $$f(z) = W(e^z) - 1\,$$ satisfies the equation


 * $$1 + f(z) + \ln (1 + f(z)) = z.\,$$

Then $$z + \ln (1 + z)\,$$ can be expanded into a power series and inverted. This gives a series for $$f(z+1) = W(e^{z+1})-1\,$$:


 * $$W(e^{1+z}) = 1 + \frac{z}{2} + \frac{z^2}{16}

- \frac{z^3}{192} - \frac{z^4}{3072} + \frac{13 z^5}{61440} - \frac{47 z^6}{1474560} - \frac{73 z^7}{41287680} + \frac{2447 z^8}{1321205760} + O(z^9).$$

$$W(x)\,$$ can be computed by substituting $$\ln x - 1\,$$ for z in the above series. For example, substituting −1 for z gives the value of $$W(1) = 0.567143\,$$. Expanding $$W(e^(1+z))$$ in series, one can find the coefficients to be


 * $$a_n = \frac{1}{n!}\sum_{i=1}^n \left\{ {n \atop i} \right\} e^i W^{(i)}(e)$$

where $$\lbrace{n\atop i}\rbrace$$ are the Stirling numbers of the second kind and all $$b_n = e^n W^{(n)}(e)$$ are rational. The initial values are:


 * $$b_1 = 1/2$$
 * $$b_2 = -3/8$$
 * $$b_3 = 19/32$$
 * $$b_4 = -185/128$$
 * $$b_5 = 2437/512$$
 * $$b_6 = -40523/2048$$
 * $$b_7 = 814355/8192$$