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If I have an equation of the form $a = b \cdot e^b$ then the Lambert W function tells me that $W(a) = b$.

But what if I instead have $a \geq b \cdot e^b$ ? Do I then have $W(a) \leq b$ or $W(a) \geq b$, or does it depend on the values of $a$ and $b$?

The nature of my problem makes me believe that $W(a) \leq b$ is the right answer, but I need someone to convince me that this is the case and if possible explain why. I can't seem to find anything online. If it matters, I'm working with the -1 branch of the Lambert W function.

(The actual problem can be found in this question that I posted earlier, but I didn't find it to be of importance to this question)

AstridNeu
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  • Note that the real branch $W_0$ is strictly increasing. – gammatester Sep 27 '18 at 13:39
  • @gammatester, I do not see what you are hinting at. – AstridNeu Sep 28 '18 at 09:03
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    If $a=be^b$ then $W_0(a)=b$. For positive $\epsilon$ you get $W_0(a+\epsilon) > W_0(a) = b$, i.e. if $a > be^b$ you have $W(a) > b$. Example: With $b=2$ you have $a= 14.77811220$, check $W_0(a)=2$. Now compute $W_0(a+0.001)=2.000045111 > 2$. If this is not you want please edit your question. – gammatester Sep 28 '18 at 09:58
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    Thank you. I understand now. And as I am working with the -1 branch (I do mention this in the question, but should have made it more clear, sorry), which is strictly decreasing, the same reasoning will give me $W(a) \leq b$, which was exactly what I needed. Thanks. – AstridNeu Sep 28 '18 at 10:17
  • Correct. Sorry for not using $W_{-1}$ in my comments. – gammatester Sep 28 '18 at 10:25

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