# Upper bound on the Lambert function

Is there an upper bound on the Lambert function $$W(-\frac{k}{e})$$ for $$0 < k < 1$$? Or should the condition be $$0 < k \leq 1$$?

I know that $$W(-\frac{1}{e}) = -1$$; would it be okay to say $$W(-\frac{k}{e}) \leq -1$$?

I just recently learned about Lambert functions and am having a hard time understanding all these. This came from an approximation of an expression given by:

$$\ln \phi_\infty = \mu + \lambda \phi_\infty - \mu e^{-\lambda \phi_\infty},$$

which by letting $$z=\lambda \phi_\infty$$, yields $$z = -\frac{1}{1+\mu}W(-\lambda(1+\mu))$$ which only exists if $$\lambda (1+\mu) \leq \frac{1}{e}.$$ If we let $$\lambda = ke^{-1}\frac{1}{1+\mu}$$ for $$0 < k \leq 1$$, we can have $$z = -\frac{1}{1+\mu}W\bigg(-\frac{k}{e}\bigg).$$

What I need is to represent this last equation in terms of $$\phi_\infty$$ alone (and these parameters $$\lambda, \mu$$) so I was wondering if I could get an upper bound for the Lambert function and do just that.

Thank you so much for your help!

• The red curve on dlmf.nist.gov/4.13 shows it is not OK. Commented May 25, 2023 at 15:17
• Assuming no disingenuousness here, please share your motivation. Commented May 26, 2023 at 4:31
• @Aruralreader have edited the post and added the process for arriving at this question :) Commented May 26, 2023 at 5:29
• Which branch do you denote by $W$? The answer depends on that.
– Gary
Commented May 26, 2023 at 5:33
• @Gary I believe this is for the general Lambert function for either the principal branch $W_0$ or $W_1$. Sorry for the confusion, I am not sure about it yet myself. Commented May 26, 2023 at 5:41

## 1 Answer

$$W$$ without a subscript usually means $$W_0$$.

Let $$f(k)=W\left(-\frac{k}{e}\right)$$ for $$0.

$$\lim_{x\to 0}f(x)=0,\ \ \lim_{x\to 1}f(x)=-1$$

See at Wikipedia: Upper and lower bounds, Bounded function and Bounded set.

$$f$$ is a bounded function: bounded above by $$0$$ and bounded below by $$-1$$.

$$-1