3
$\begingroup$

I have asked a similar question about this one particular limit:

\begin{equation} A=\lim_{c\to 1}\exp\left[ -\left(\frac{1}{1-c}\right)\left(W_{0}\left[ B\left( 1+\frac{x}{rc}\right) \right]-W_{0}[B]\right)\right] \end{equation} where $r>0$, $x \in \mathbb{R}$, $W_0$ is the $k=0$ branch of the Lambert-$W$ function defined as: \begin{equation} x=W(x)e^{W(x)} \end{equation} and B is defined as: \begin{equation} B=\frac{(1-c)r}{1-(1-c)r}\exp\left[ \frac{(1-c)r}{1-(1-c)r} \right] \end{equation} Now, I tried with Mathematica and it is shown that $A \to e^{-x}$ when $c\to 1$ but I have trouble proving that analytically.

If anyone could assist me with this one I would be grateful. Thank you!

$\endgroup$
3
$\begingroup$

First note that as $c\rightarrow 1$, $(1-c)\rightarrow 0$, and so $$ B \sim (1-c)r e^{(1-c)r} \sim (1-c) r. $$ Moreover, for small $x$, $W_0(x) \sim x$. So $$ W_0\left[B\left(1+\frac{x}{rc}\right)\right]-W_0[B]\sim\frac{B x}{rc}\sim(1-c)x. $$ Hence $$ A=\lim_{c\rightarrow 1}\exp\left[-\left(\frac{1}{1-c}\right)\left((1-c)x\right)\right]=e^{-x}. $$

$\endgroup$
  • $\begingroup$ Thank you for your answer. I have two questions: 1) Why would you consider so freely that $B \approx (1-c)e^{(1-c)r}$. Is there any way you could prove this claim? 2) I am not familiar with this $W_0(x)$ approximation, therefore is there any source I could see this? Anything would do, even the wikipedia page. Thank you again. $\endgroup$ – Mitscaype Jun 28 '16 at 19:12
3
$\begingroup$

This is not an answer but it is too long for a comment.

mjqxxxx's answer contains all the required steps.

Concerning the approximation made for $B$, consider the definition $$B=\frac{(1-c)r}{1-(1-c)r}\exp\left[ \frac{(1-c)r}{1-(1-c)r} \right]$$ and let us define $a=(1-c)r$ which makes $$B=\frac a{1-a}\exp\left[ \frac{a}{1-a} \right]$$ and develop as a Taylor series built at $a=0$; this gives $$B=a+2 a^2+O\left(a^3\right)$$

Similarly, for small values of $y$, using Taylor again $$W(B(1+y))-W(B)=\frac{ W(B)}{1+W(B)}y+O\left(y^2\right)$$ and, for small $B$ $$\frac{ W(B)}{1+W(B)}=B-2 B^2+O\left(B^3\right)$$ Combining all the above leads to

$$W_{0}\left[ B\left( 1+\frac{x}{rc}\right) \right]-W_{0}[B]=(1-c) x+(1-c)^2 \left(x-x^2\right)+O\left((c-1)^3\right)$$ $$ -\left(\frac{1}{1-c}\right)\left(W_{0}\left[ B\left( 1+\frac{x}{rc}\right) \right]-W_{0}[B]\right)=-x-(1-c) \left(x-x^2\right)+O\left((c-1)^2\right) $$

$\endgroup$
  • $\begingroup$ Googling "limit Lambert function" led me to another answer of you... $\endgroup$ – jlandercy Dec 4 '18 at 9:35
  • $\begingroup$ @jlandercy. This is a fantastic function with so many applications ! Cheers $\endgroup$ – Claude Leibovici Dec 4 '18 at 10:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.