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Consider the equation

$$y=x e^x.$$ Its real solution is given by $$x=W(y),$$ where $W$ is the Lambert W function (or product log).

Can the function

$$f(x) = W\left(-\frac{1}{r}xe^x\right)$$

be written in simple terms of $x$?

What I've tries so for is writing this as a power series expansion

$$W(x)=\sum_{n=1}^{\infty}\frac{(-1)^n n^{n-2}}{(n-1)!}x^n$$

therefore with the constant I've introduced $$ \begin{split} W\left(-\frac{1}{r}x\right) & =\sum_{n=1}^{\infty}\frac{(-1)^{n-1} n^{n-2}}{(n-1)!}\left(-\frac{x}{r}\right)^n\\ & =\sum_{n=1}^{\infty}-\frac{(-1)^{n-1}(-1)^{n-1} n^{n-2}}{(n-1)!}\left(\frac{x}{r}\right)^n \\ & =\sum_{n=1}^{\infty}-\frac{n^{n-2}}{(n-1)!}\left(\frac{x}{r}\right)^n \end{split} $$ but got stuck here.

I've also tried guessing many functions in mathematica that yeilded no result.

As I see it now the problem is not solvable

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    $\begingroup$ One thing to be aware of is that the domain of the Lambert function isn't closed under $y \mapsto -\frac1 r y$, so you'll have to be careful about where your $f$ is defined. $\endgroup$
    – LSpice
    Commented Jul 1, 2019 at 20:25
  • $\begingroup$ @LSpice, $x$ is positive and real, and $r>0$, does that help? $\endgroup$
    – jarhead
    Commented Jul 1, 2019 at 20:27
  • $\begingroup$ Probably not much; still the function is defined only on $(0, W(r/e))$. $\endgroup$
    – LSpice
    Commented Jul 1, 2019 at 20:28
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    $\begingroup$ What have you tried? Did you graph the function? Did you calculate a few terms of its Taylor series? If so, what did you see? $\endgroup$
    – Daniel McLaury
    Commented Jul 1, 2019 at 21:34
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    $\begingroup$ I did not downvote the post. $\endgroup$
    – Daniel McLaury
    Commented Jul 2, 2019 at 4:46

3 Answers 3

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If what you want is a series expansion in powers of $x$ and $t:=1/r$, we can write $$f(x,t)=W\big(- {txe^x}\big)=-\sum_{n=1}^\infty {q_n(t)} \,x^n=-\bigg[tx+\frac{2t+2t^2}{2!}\, {x^2} +\frac{3t+12t^2+9t^3}{3!}\,x^3+\dots\bigg]$$ The coefficient $q_n$ is a polynomial of degree $n$, precisely $$q_n(t)=\frac{1}{n!}(tD)^{n-1}(1+t)^n$$ where $D$ denotes differentiation in $t$.

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  • $\begingroup$ I did not understand what you have done, $r$ is a parameter. I'm also interested in a closed form, or the closest thing I can get to it. Getting to a closed form through a power series is one option out of many I was hoping people would lay out here. $\endgroup$
    – jarhead
    Commented Jul 9, 2019 at 15:10
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    $\begingroup$ I just wrote the power series expansion of $f(x)=W(-x e^x/r)=\sum_{n=1}^\infty q_nx^n$. Of course he coefficients $q_n$ depend on the parameter $r$; in fact they are polynomials in $t:=1/r$. $\endgroup$
    – Pietro Majer
    Commented Jul 9, 2019 at 15:15
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    $\begingroup$ @jarhead -- the closed form exists, it is $y = W(-r^{-1}xe^x)$, resumming the power series, either this one or the one in my answer, will just give you back that closed-form expression. $\endgroup$
    – Carlo Beenakker
    Commented Jul 9, 2019 at 15:16
  • $\begingroup$ Indeed it seems difficult to have a better closed form. Instead, I would look at the dependence from $r$; for instance, if I'm not wrong, $f_s\circ f_r=f_r\circ f_s=W\circ f_{-sr}$ $\endgroup$
    – Pietro Majer
    Commented Jul 9, 2019 at 15:39
  • $\begingroup$ Can you please elaborate on what is $D$, I apologize as it seems basic, but if I take a derivative of $r$ with respect to $x$ then I will get $0$ as it is a parameter. $\endgroup$
    – jarhead
    Commented Jul 10, 2019 at 7:52
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Maple did the following. (Of course once you see it you can do it yourself...)

$$ y = W(-\frac{1}{r}xe^x) $$ solve for $x$; result $$ x = W(-rye^y) $$

Probably even your original assertion needs adjustment. We may think that $$ W(xe^x) = x $$ but perhaps not. Here is the graph for $W_0(xe^x)$ W0graph

Only part of that is $y=x$.

For more explanation, see this, where $W_0(xe^x)$ is in red and $W_{-1}(xe^x)$ is in blue:

two Ws

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    $\begingroup$ thanks for your effort but this not address the question. $\endgroup$
    – jarhead
    Commented Jul 2, 2019 at 12:23
  • $\begingroup$ also I'm only worried about $x>0$ so the $-1$ real branch is not an issue. Still no solution. $\endgroup$
    – jarhead
    Commented Jul 2, 2019 at 12:33
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    $\begingroup$ I expect the best solution is $x = W(-rye^y)$ which has no simpler form for general $r$. $\endgroup$
    – Gerald Edgar
    Commented Jul 2, 2019 at 13:15
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As mentioned by Gerald Edgar, you seek to simplify $$y = W(-r^{-1}xe^x)\Rightarrow x=W(-rye^y),$$ for $r>0$. Here is a series expansion in powers of $r$,

$$x=-\sum_{n=1}^\infty n^{n-1}\frac{1}{n!}(rye^y)^n.$$

The plot compares the exact result for $x$ (blue curve) with the series expansion up to $n=8$ (gold) at $y=-1$, as a function of $r$. (I take $y<0$ because the OP says the interest is in $x>0$.) The series expansion remains quite accurate even for $r$ approaching unity.

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  • $\begingroup$ and can you transform this series to obtain a continuous function of $f(x)$ when $y=f(x)$? $\endgroup$
    – jarhead
    Commented Jul 4, 2019 at 16:54
  • $\begingroup$ I'm not quite sure what you mean; the series $\sum_{n=1}^\infty n^{n-1}\frac{1}{n!}(rye^y)^n$ is a contiuous function of $y$; if you are asking whether the series can be summed in closed form, the answer is "yes", but that would bring you back to the Lambert W-function, which you wish to avoid. $\endgroup$
    – Carlo Beenakker
    Commented Jul 4, 2019 at 17:46
  • $\begingroup$ please refer again to my question $\endgroup$
    – jarhead
    Commented Jul 5, 2019 at 9:09

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