1
$\begingroup$

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

Improve this question
$\endgroup$
9
  • 2
    $\begingroup$ 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
    Jul 1, 2019 at 20:25
  • $\begingroup$ @LSpice, $x$ is positive and real, and $r>0$, does that help? $\endgroup$
    – jarhead
    Jul 1, 2019 at 20:27
  • $\begingroup$ Probably not much; still the function is defined only on $(0, W(r/e))$. $\endgroup$
    – LSpice
    Jul 1, 2019 at 20:28
  • 2
    $\begingroup$ 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
    Jul 1, 2019 at 21:34
  • 1
    $\begingroup$ I did not downvote the post. $\endgroup$
    – Daniel McLaury
    Jul 2, 2019 at 4:46

3 Answers 3

Reset to default
2
$\begingroup$

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 $t$.

Improve this answer
$\endgroup$
7
  • $\begingroup$ 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
    Jul 9, 2019 at 15:10
  • 1
    $\begingroup$ 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
    Jul 9, 2019 at 15:15
  • 2
    $\begingroup$ @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
    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
    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
    Jul 10, 2019 at 7:52
1
$\begingroup$

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

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ thanks for your effort but this not address the question. $\endgroup$
    – jarhead
    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
    Jul 2, 2019 at 12:33
  • 3
    $\begingroup$ I expect the best solution is $x = W(-rye^y)$ which has no simpler form for general $r$. $\endgroup$
    – Gerald Edgar
    Jul 2, 2019 at 13:15
0
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ and can you transform this series to obtain a continuous function of $f(x)$ when $y=f(x)$? $\endgroup$
    – jarhead
    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
    Jul 4, 2019 at 17:46
  • $\begingroup$ please refer again to my question $\endgroup$
    – jarhead
    Jul 5, 2019 at 9:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.