How close is the solution of $xe^x+\ln(x)=c$ to $W_0(c)$ for large $c$?

Question

Denote $f(c)$ to be the solution of the equation $xe^x+\ln(x)=c$

What is the asymptotic behaviour of $f(c)-W_0(c)$ for large $c$ ?

$W_0(c)$ denotes the lambert-w-function.

It seems to be in the order of $\frac{1}{c}$

Answer 1

I tried (with absolutely no sucess) working with asymptotics and, by then end, used numerical evaluations for $10^3 \leq c \leq 10^5$ (using a step size equal to $10^3$).

For this range, using as a model $f(c)-W_0(c)=-\frac a {c^b}$, what I obtained is $(R^2=0.999904)$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.743572 & 0.005750 & \{0.754983,0.732160\} \\ b & 0.908324 & 0.001009 & \{0.906321,0.910326\} \\ \end{array}$$ which is quite close to your assumptions.

Using this model for $c=10^6$, the predicted value is $-2.639 \times 10^{-6}$ while the exact value would be $-2.236\times 10^{-6}$.

Repeating the calculations for $10^4 \leq c \leq 10^6$ (using a step size equal to $10^4$), the same empirical model leads to $(R^2=0.999976)$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.988959 & 0.005177 & \{0.999235,0.978683\} \\ b & 0.938510 & 0.000529 & \{0.937459,0.939560\} \\ \end{array}$$

Edit (after ProfessorVector's comment, LutzL's and user90369's answers)

In the same spirit as LutzL's answer : we are looking for the zero of function $$f(x)=xe^x+\ln(x)-c\qquad \qquad f'(x)=e^x (x+1)+\frac{1}{x}$$ Let us apply one iteration of Newton method, using $x_0=W_0(c)$ to get

$$\Delta x=x_1-x_0=x_1-W(c)=-\frac{W_0(c)\, \log (W_0(c))}{c (W_0(c)+1)+1}\approx \color{red}{-\frac{\log (\log (c))}{c}}$$ Using, for a simple test, $c=10^{50}$, this approximation gives $-4.75\times 10^{-50}$ for an exact value equal to $-4.66\times 10^{-50}$.

Repeating the calculations using $c=10^{100}$, this approximation gives $-5.44\times 10^{-100}$ for an exact value equal to $-5.39\times 10^{-100}$.

Making $c=10^k$, it is quite illustrative to plot of the same graph $$10^k\frac{ W\left(10^k\right) \left(\log \left(10^k\right)-W\left(10^k\right)\right)}{10^k (W\left(10^k\right)+1)+1}\qquad \text{and}\qquad \log \left(\log \left(10^k\right)\right)$$ These two curves almost overlap.

Answer 2

We have $\enspace\displaystyle W_0(x) \approx \ln x - \ln\ln x + \frac{\ln\ln x }{\ln x } \enspace$ for large $\,x\,$ .

It follows:

$\displaystyle f(c)-W_0(c) = x - W_0(xe^x + \ln x) $

$\hspace{2cm}\displaystyle \approx \,\, x -\ln (x e^x + \ln x) + \ln\ln (x e^x + \ln x) - \frac{\ln\ln (x e^x + \ln x) }{\ln (x e^x + \ln x) } $

$\hspace{2cm}\displaystyle \approx \,\, \ln(1+\frac{\ln x}{x}) \,\,\approx \,\, \frac{\ln x}{x} $

$\hspace{2cm}\to \,\, 0\enspace$ for $\enspace x\to\infty$

Note:

$\displaystyle \lim_{x\to\infty}(1 + x - W_0(xe^x + \ln x))^x = 1$

$\displaystyle \lim_{x\to\infty}(1 + x - W_0(xe^x + \ln x))^{x^2} = \infty$

Answer 3

Setting $x=w(1+h)$ with $w=W_0(c)$, $c=we^w$, $\ln(w)=\ln(c)-w$ we get the equation $$ w(1+h)e^{w+wh}+\ln(w)+\ln(1+h)=we^w $$ Inserting linear Taylor polynomials one gets $$ w(1+h)e^w(1+wh)+\ln(w)+h+O(h^2)=we^w\\ h(1+we^w(1+w))+\ln(w)=0\\ h=-\frac{\ln(w)}{1+we^w(1+w)}=-\frac{\ln(w)}{1+c(1+w)} $$ so that in the dominant terms for large $c$ and thus large $w$ $$ f(c)=w(1+h)+O(h^2)\approx w-\frac{\ln(w)}{c}=W_0(c)\left(1+\frac1c\right)-\frac{\ln(c)}{c} $$ so that you also get a logarithmic term in the first correction.