0
$\begingroup$

Knowing that the Lambert W function is defined as such $$W(xe^{x})=x.$$ Is there any way to simplify and expression of the form $$W(me^{m-2x}),$$ where $m>0$ and $x\ge0$? Clearly $W(me^{m-2x})=m$ when $x=0$ but past that I can't find a neater form for this expression.

Thanks in advance.

  • 0
    $W(me^{m-2x})=y\enspace$ <=> $\enspace W(ye^{y+2x})=m$ ; nothing else to say :-) – 2017-01-24

2 Answers 2

0

$W(m e^{m-2x}) = y$ where $y e^y = m e^{m-2x}$. I doubt that there's much more you can say about it.

0

One way to express $m$ in terms of $y$ and $x$ can be: $m=W(ye^{2x+y})$. Also, $x$ can be expressed as: $x=\frac{m-ln\left(\frac{ye^y}{m}\right)}{2}$