Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to solve an equation for B where

$2 B e^{(1-B)}=1$

I plugged it into wolfram alpha and got:

enter image description here

What is that $W_n$ mean? How do I intrepret this answer?

(Feel free to retag it, I wasn't entirely sure what it falls under!)

share|improve this question
Isn't there a tiny box that says "W_k is the analytic continuation of the product log function" with a link to some documentation there? ( reference.wolfram.com/mathematica/ref/ProductLog.html ) –  Myself Mar 1 '11 at 0:06
Lambert W function. mathworld.wolfram.com/LambertW-Function.html. –  user17762 Mar 1 '11 at 0:07
@Myself nope, it just mentions Z being integers. –  corsiKa Mar 1 '11 at 0:10
@Sivaram are you saying that B is a set of values, not a particular solvable value? –  corsiKa Mar 1 '11 at 0:10
There are two real solutions and infinite complex solutions. –  user17762 Mar 1 '11 at 0:19
show 6 more comments

1 Answer 1

up vote 8 down vote accepted

$W(z)$ is the Lambert $W$ function.

$W_k(-\frac{1}{2e})$ where $k \in \mathbb{Z}$ denotes the $k^{th}$ root of the equation $xe^x = -\frac{1}{2e}$

enter image description here

share|improve this answer
The text in that image is hardly readable on my browser... Why not just include a short writeup? +1 Anyway :-) –  Aryabhata Mar 1 '11 at 0:13
@Moron it was great for me when I clicked on the image to snap it to a better resolution. @Sivaram that's so weird, I can't get that output to come up. It would appear my result is .23 and 2.67 - Thank you so much! :) –  corsiKa Mar 1 '11 at 0:15
@Moron: Added. I think the image will be better if you try opening the image on a new tab. –  user17762 Mar 1 '11 at 0:30
What is a new tab? Just kidding :-) In any case, you need some supporting write up. I am pretty sure glowcoder could see the same thing before even creating this question :-) –  Aryabhata Mar 1 '11 at 0:37
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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