1

So today I came across an answer on MathStackExchange. I read it and found it well explained. But I stuck on a step where the user write "Solve using the properties of $W$ function". I don't know what property the user used.

What I know about Lambert $W$ function is that, $W(x) = f^{-1} (x)$, where $f(x) = x\cdot e^x$.

Can anyone solve the question or just tell me what the property the answerer used in the question which I have linked to you?

DMcMor
  • 9,407
  • 1
    I'm not sure that really helps, since values of $W$ aren't really computable in any sort of closed form anyway. It's sort of a cheat to say "oh just use $W$ to get your answers." – Randall Oct 22 '21 at 18:20
  • @amWhy that's the question OP is asking clarification for. – Umberto P. Oct 22 '21 at 18:23
  • 1
    @amWhy did you even read my question? Or just you read the title and voted to close it? –  Oct 22 '21 at 18:23
  • The property of the $W$ function is that $a = W(b)$ if and only if $ae^a = b$ (with $a \ge -1$ and $b \ge -\dfrac 1e$). The linked question uses $a = -\dfrac{y\log(x)}{x}$ and $b = -\dfrac{\log(x)}{x}$. – Umberto P. Oct 22 '21 at 18:24

2 Answers2

0

The step in question is how do the properties of the Lambert $W$ function, give us the following $$-\dfrac{y\log(x)}{x}\exp\left(-\dfrac{y\log(x)}{x}\right) = -\dfrac{\log(x)}{x}\implies-xy\log(x) = \operatorname{W}\left(-\dfrac{\log(x)}{x}\right).$$

There is actually a typo in the linked answer, (which I corrected), the desired result should be $$-\dfrac{y\log(x)}{x}\exp\left(-\dfrac{y\log(x)}{x}\right)= -\frac{\log(x)}{x}\implies= -\frac{y\log(x)}{x} = \operatorname{W}\left(-\dfrac{\log(x)}{x}\right).$$

Here we will use the fact that $$z = we^w \iff w = W(z).$$

We first make the substitutions $$s = -\frac{y\log(x)}{x},\qquad z = -\frac{\log(x)}{x},$$

and rewrite the first equation as $se^{s} =z.$

Using the noted property of $W$, then, we can rewrite this as $s = W(z)$, which reversing our substitutions gives:

$$-\frac{y\log(x)}{x} = W\left(-\frac{\log(x)}{x}\right).$$

amWhy
  • 209,954
DMcMor
  • 9,407
-1

as @Randall mentioned, using the $W$ function is really just putting a name on it, only more complex maths programs are actually going to return values for you. That said: $$y^x=x^y\\ y^{1/y}=x^{1/x}\\ \frac1y\ln y=\frac1x\ln x$$ now say $y=e^{-u}$ so: $$ue^u=-\frac1x\ln x\\ u=W_k\left(-\frac1x\ln x\right)\\ y=\exp\left[-W_k\left(-\frac1x\ln x\right)\right]$$ where $k$ denotes the branch of the $W$ function

Henry Lee
  • 12,215