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I was just playing around in desmos and I think I found something that approximates Lambert W for whole numbers.

$$f(x)=a(\int_{0}^{1}(\sum_{n=1}^{x}t^n-1)dt)+b$$

Where $a\approx0.765424$ And $b\approx0.944602$

$a$ and $b$ were achieved through linearization, plugging $xe^x$ in for x.

Please point me in the right direction for proofing or disproving that as $x \to \infty$ $f(x) \to W(x)$

Daniel Fischer
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CoderBoy
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    You're effectively saying that the Lambert W function can be well approximated by a linear function of $\ln{(x)}$. But it's well known that $W(x)\sim\ln{(x)}$ so this is not surprising. – Peter Foreman Nov 24 '19 at 13:58
  • @Peter sorry I didn’t actually know that. Does ln(x) get closer to W(x) the higher x gets? – CoderBoy Nov 24 '19 at 13:59
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    @CoderBoy It depends on what you mean by closer. The difference $\ln{(x)}-W(x)\to\infty$ but the ratio $W(x)/\ln{(x)}\to1$. – Peter Foreman Nov 24 '19 at 14:06
  • @Peter I’m actually interested in the difference and not the ratio – CoderBoy Nov 24 '19 at 14:09

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This is an interesting post.

In fact $$f(x)=\int_{0}^{1}\sum_{n=1}^{x}(t^n-1)dt=\psi (x+2)+\gamma -2$$

Looking at the empirical model $$W(x)=a + b f(x)$$ for $0 \leq x \leq 10^6$ by steps of $10$, we obtain with $(R^2=0.999875)$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.174690 & 0.000366 & \{0.173973, 0.175408\} \\ b & 0.903778 & 0.000032 & \{0.903715, 0.903840\} \end{array}$$

Claude Leibovici
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  • Woah! Maybe better constants could get that R^2 much better. I assume I should compare (e^x+2)+−2 to W(xe^x) to get those better constants, and then see if anything interesting arises that could help with a proof. Thanks! – CoderBoy Nov 24 '19 at 15:52
  • @CoderBoy. I think that you are forgetting that $W(x e^x)=x$. – Claude Leibovici Nov 25 '19 at 05:05
  • no, sorry, might’ve been badly worded. I was going to use linearization to get a new a and b and see if they are a ratio of or are any known constants to help with a proof. – CoderBoy Nov 26 '19 at 10:31