# Approximation for the Lambert $W$ function from $x=5$ to $x=105$

Recently, I learnt about the Lambert $$W$$ function which is the inverse of $$f(x) = x\cdot e^{x}$$ and have been trying to solve problems related to it on the internet. After solving each problem, however, I find myself going back to WolframAlpha to calculate the result which is rather inconvenient. So I was wondering if I could find an approximation for $$W(x)$$ which I can simply plug into my standard calculator which is almost always close to me.

Playing around with the graph of $$W(x)$$ seemed like the most obvious method to me and so I began to try out different functions to match its graph. After a while, I settled upon $$f(x) = 1.006 \log _{3.96} (x+1)$$ from $$x=5$$ to $$x=105$$. From the graph of $$g(x) = W(x) - 1.006 \log _{3.96} (x+1), x=5 \: \text{to} \: x=105$$ on WolframAlpha, $$f(x)$$ seems to be within $$\approx\pm 0.02$$ of $$W(x)$$. Is this a good approximation? If not, how can I make it better? Furthermore, is there a way to find a good approximation algebraically rather than graphically?

• To make my last formula looking nicer, making the numers rational, use $$\frac{79}{102} \log \left(x+\frac{119}{52}\right)-\frac{33}{164}$$ Commented Mar 29, 2021 at 11:14
• Sorry if this is a stupid question but does $\log(x)$ here refer to $\log _{10} (x)$ or $\log _{e} (x)$ ?
– user905694
Commented Mar 29, 2021 at 11:19
• There is no stupid question ! To tell the truth, I only know one logarithm : $\log_e(.)$. Commented Mar 29, 2021 at 11:21
• Okay, thank you!
– user905694
Commented Mar 29, 2021 at 11:24
• @Claude Here you go Commented Mar 29, 2021 at 14:23

Considering the bounds

$$\log (x)-\log (\log (x))+\frac12\frac{\log (\log (x))}{ \log (x)} < W(x)$$ $$W(x)< \log (x)-\log (\log (x))+\frac e{e-1 }\frac{ \log (\log (x))}{ \log (x)}$$ for a specific range, we can numerically minimize $$\Phi(k)=\int_a^b \Big[\log (x)-\log (\log (x))+k\frac{\log (\log (x))}{ \log (x)}-W(x)\Big]^2\,dx$$

For $$a=5$$ and $$b=105$$, this gives $$k\sim 0.881076$$ but the formula is a bit more complex than your.

Trying something of the same shape as your $$W(x) \sim a+b \log_e(x+c)$$ a nonlinear regression gives $$(R^2>0.999999)$$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -0.201141 & 0.004453 & \{-0.209981,-0.192302\} \\ b & +0.774451 & 0.000974 & \{+0.772518,+0.776385\} \\ c & +2.288360 & 0.034073 & \{+2.220730,+2.356000\} \\ \end{array}$$ which leads to a maximum absolute error of $$0.005$$.

Congratulations for your idea !

Edit

Since we are using totally empirical models, let us try

$$W(x) \sim a+b \Big[\log_e(x^d+c)\Big]^f$$

$$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.256813 & 0.001340 & \{0.254152,0.259474\} \\ b & 0.815983 & 0.001525 & \{0.812955,0.819012\} \\ c & 0.547202 & 0.001079 & \{0.545059,0.549345\} \\ d & 0.678026 & 0.000656 & \{0.676723,0.679329\} \\ f & 1.172580 & 0.000318 & \{1.171940,1.173210\} \\ \end{array}$$ which reduces the previous sum of squares by a factor close to $$800,000$$ (!!) and leads to a maximum absolute error equal to $$5\times 10^{-6}$$. Better, isn't it ?