0

Sure I can just get an answer from wolfram alpha, but I want to know the steps involved.

I noticed the title equation while reading this: https://cs.uwaterloo.ca/research/tr/1993/03/W.pdf

The paper shows you how to get to the equation, and how its solvability for $α = 1$ implies solvability for all $α \neq 0$, but not how to actually solve it. So, can someone please guide me through the general solution method?

Stumbleine75
  • 209

1 Answers1

1

Introducing $y=-\nu\alpha x^\alpha$ one must solve $y\mathrm e^y=-\nu\alpha$ hence indeed Lambert W function is involved, since $$x=\left(\frac{W(-\nu\alpha)}{-\nu\alpha}\right)^{1/\alpha}.$$ Note: whether this formula can be useful at all is somewhat doubtful.

Did
  • 279,727
  • How do you get to $ye^y = -v\alpha$? Does it have to do with $W(z)e^{W(z)} = z$? – Stumbleine75 Sep 19 '14 at 06:23
  • Surely, reading this, you tried to compute $ye^y$ when $y$ is as defined in the answer and $x$ solves the equation in your question. What did it lead you to? – Did Sep 19 '14 at 06:25
  • Yeah I see it now, and feel bad for missing it. $y$ as you defined it completes the identity and then it is trivial to manipulate it into the answer you gave. Now, can I ask how you knew to define y the way you did? – Stumbleine75 Sep 19 '14 at 07:57
  • The natural way: I considered the exponential of the identity in your post and tried to massage it into an identity involving $ye^y$ for some well chosen $y$. – Did Sep 19 '14 at 08:46
  • Massage? Is that a mathematical term? – JacksonFitzsimmons Dec 21 '15 at 02:25