# Questions tagged [lambert-w]

For questions related to the Lambert W or product log function, the inverse of $f(z)=ze^z$.

767 questions
Filter by
Sorted by
Tagged with
71 views

### Does $\ln(\frac{2x}{\ln(2x+1)})\sim W(x)$?

The function $f(x)= \ln(\frac{2x}{ln(2x+1)})$ when plotted is similar to the plot of the Lambert W function. The Wolfram Alpha says that the limit $\lim_{x\rightarrow\infty} \frac{W(x)}{f(x)}=1$, but ...
33 views

126 views

### Equations similar to Lambert-$W$ with quadratic exponents

I've seen solutions saying that an equation in the format:$$\ln(x) - \frac{bx}{a} = - \frac{bc}{a}$$ can be solved using the Lambert W function and I am comfortable doing so. My equation however is ...
54 views

### Using Lambert W to solve for time of flight of a projectile with air resistance

picture from Wikipedia page about projectile motion, under air resistance section Wikipedia: Projectile motion - Derivation of the time of flight: $$c_1t+c_2+c_3e^{c_4t}=0$$ I understand how the ...
116 views

### Does the second positive solution (besides $x = 1$) of the equation $e^{x^2-1}=x^3-x\ln x$ have a closed form?

Does the equation $$e^{x^2-1}=x^3-x\ln x$$ have a closed form solution ? The given equation has $2$ positive real roots. Graphically It is not hard to see that $x=1$ is a rational solution. The ...
119 views

### Inverse function for $f(x)=x/(1-e^{-x})$

I'm looking for the inverse function for $f(x)=x/(1-e^{-x})$, over the domain $x>0$. Wolfram says that the answer is $f^{-1}(x)=x+W(-xe^{-x})$, for $x>1$, where $W(x)$ is the Lambert W function. ...
61 views

### Interpreting/understanding the lambertW on Maple software

I decided to use the Maple software to help me solve dynamic optimization problems, and I found this final solution for T. (example in the picture). example What does the LambertW mean for the ...
1 vote
174 views

### If $1 < x < e < y$ and $x^{1/x} = y^{1/y}$ then $y > e^2/x$.
If $1 < x < e < y$ and $x^{1/x} = y^{1/y}$ then $y > e^2/x$. Since $x^{1/x}$ is decreasing for $x > e$, this is equivalent to $x^{1/x} \lt (e^2/x)^{x/e^2}$for $1 < x < e$. Also, ...