simplification of $W(x\cdot e^{a+x})$

Question

Is it possible to simplify $W(x\cdot e^{a+x})$?

Because $W(x\cdot e^{x})=x$

So I was wondering if it was possible to simplify this expression.

Assuming $a$ is real, this would be the same as simplifying $W(cxe^x)$ for $c>0$ (I cannot think of anything helpful for this case) — parsiad, Aug 19 '15 at 18:20
Really? Is it impossible to find a method to remove the lambert w term? In any case? — Idrees Samim, Aug 19 '15 at 18:25
I dont think that there will be a simplification. Because there is a linear scaling factor $k=e^a$ which shifts the value of the function, again as Lambert-$W$. This is basically true for any shifting. Especially, $W$ is non-linear and one cannot separate it. But, one can approximate it and find some approximate simplifications. See the answer. — Seyhmus Güngören, Aug 02 '17 at 23:52

Simply Beautiful Art · Answer 1 · 2017-08-02T23:22:53.340

1

I'd have to say no, this is not simplify-able.

You can however attempt to obtain an expansion via logarithms:

$$W(x)=\ln(x)-\ln(\ln(x))+\mathcal O(\ln(\ln(\ln(x))))$$

So,

$$W(xe^{a+x})=a+x+\ln(x)-\ln(a+x+\ln(x))+\mathcal O(\ln(\ln(x)))$$

as $x\to\infty$.

edited Aug 02 '17 at 23:22

answered Dec 19 '15 at 18:32

there is no $n$ defined, anyways. – Seyhmus Güngören Aug 03 '17 at 00:35
@SeyhmusGüngören I've searched my screen for the letter $n$, but you and the related links are the only ones mentioning the variable $n$. Could you clarify on what you meant? – Simply Beautiful Art Aug 03 '17 at 00:37
I do use glasses but... $n\rightarrow\infty$.. and now it is $x\rightarrow\infty$ magic!) – Seyhmus Güngören Aug 03 '17 at 00:40
xD Funny thing is this happened to me just this morning!. – Simply Beautiful Art Aug 03 '17 at 00:41

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