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I'm wondering if there is a way to obtain the inverse of the function $y=xe^x-x$. I am aware of the use of Lambert's W function in the inverse of $xe^x$ but as can be seen this is a different animal altogether. I have plotted the function and the inverse looks close to the W function.

This is related to a problem I'm working on in which I need to find the density of the random variable $Y=Xe^X-X$, where I know the density of X. I am aware that this function is not one-to-one. For my purposes, we can restrict to $x>0$. Any help would be greatly appreciated!!

giorgiomugnaini
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royalT
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    this function has no inverse function since $f(x)$ is not monoton – Dr. Sonnhard Graubner Dec 02 '14 at 19:26
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    Caleb inverses only exist for 1-1 functions. Perhaps you want to restrict to $x>0$? – Cheerful Parsnip Dec 02 '14 at 19:53
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    I think there are two problems involved here: is there an inverse? and can we find a "formula" for the inverse? As Grumpy Parsnip has written, if x > 0, there is an inverse. But it's impossible to find a simple "formula" for it, though you may find as many points (x,y) on its graph by solving numerically the equation x = f ( y ) for every x you want. – MasB Dec 02 '14 at 20:15
  • @BernardMassé What if we allow $W$ (Labert's W function) to be part of the formula for the inverse? – Akiva Weinberger Dec 02 '14 at 20:54
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    @columbus8myhw It all depends what we are ready to call "simple functions". This category usually includes by definition (and quite arbitrarily) polynomial, rational, trigonometric, logarithmic and exponential functions. The Lambert, gamma and other more exotic creations are left out in the cold. One could even decide that the inverse of f ( x ) when x > 0 in this case to be an important function and give it a name, for example the "december function" because it was first deemed to be important in the month of december. But usually such named functions answer an important problem. – MasB Dec 02 '14 at 21:53
  • possible duplicate of Is it possible to solve this equation using lambert function – leonbloy Dec 02 '14 at 22:15
  • @BernardMassé So, the question becomes, "Can the 'December function' be expressed as a combination of elementary functions and Lambert Ws?" – Akiva Weinberger Dec 02 '14 at 22:21
  • @leonbloy Disagree with this being a duplicate. The OP does not insist on expressing the inverse in terms of $W$. –  Dec 03 '14 at 00:17
  • Thank you all for your answers so far. As I mentioned in the initial problem statement, I am aware that this functions is not 1-1 across the entire real line. I edited it to put a restriction to the positive real line. I would also be interested in answers for when $x<0$. I saw the question in the link provided by @leonbloy and as I also mentioned in the initial problem statement Lambert's W function is not the solution. I feel that any function of Lambert's W would also not be a solution. – royalT Dec 18 '14 at 18:09

2 Answers2

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This equation can be placed in the form:

$$e^x=\frac{x+f}{x}$$

We know that mixed exponential/bilinear equations can be solved by the extended Lambert function $W_r(x)$, which can be represented by the following Lagrange inverting series [note: $L_n$ are the Laguerre polynomials]:

$$z(A,t,s)=t- (t-s) \sum_{n=1} \frac{L_n' (n(t-s))}{n} e^{-nt} A^n$$

which is the solution of:

$$e^z=A\frac{z-t}{z-s}$$

Related questions

Related question concerning equations of the form: $$e^z=A\frac{z-t}{z-s}$$

Lambert W function with rational polynomial

Solve $-B \ln y -A y \ln y + A y- A =0$ for $y$

Transcendental equations involving more than 2 terms

How to solve this equation for x? $0 = (x+k)e^{-(x+k)^2}+(x-k)e^{-(x-k)^2}$

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References

On the generalization of the Lambert $W$ function with applications in theoretical physics, http://arxiv.org/abs/1408.3999

[68] C. E. Siewert and E. E. Burniston, "Solutions of the Equation $ze^z=a(z+b)$," Journal of Mathematical Analysis and Applications, 46 (1974) 329-337. http://www4.ncsu.edu/~ces/pdfversions/68.pdf

About Laguerre Polynomials:

https://en.wikipedia.org/wiki/Laguerre_polynomials

giorgiomugnaini
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Beside the formal answer in terms of the generalized Lambert function, we can obtain a good approximation of the inverse at the price of a single iteration of Halley method using $x_0=W(y)$. This gives $$x_1=W(y)+\frac{2 ((y-1) W(y)+y)}{2 \left(2 (y-1) e^{W(y)}+e^{2 W(y)}+y (y-1)+1\right)+y W(y)}$$ We could even do better changing from Halley to Householder (the formula is too long to be typed here). $$\left( \begin{array}{cccc} y & x_ 1\text{ (Halley)}& x_ 1\text{ (Householder)} & \text{solution} \\ 1 & 0.79480270709 & 0.80794504316 & 0.80646599424 \\ 2 & 1.05568629122 & 1.06039805820 & 1.06009031989 \\ 3 & 1.23095181927 & 1.23340361714 & 1.23329258687 \\ 4 & 1.36596225621 & 1.36747274984 & 1.36742103137 \\ 5 & 1.47681961058 & 1.47784536126 & 1.47781737575 \\ 6 & 1.57134651217 & 1.57208888765 & 1.57207216458 \\ 7 & 1.65400279510 & 1.65456484098 & 1.65455411438 \\ 8 & 1.72759676052 & 1.72803683495 & 1.72802957840 \\ 9 & 1.79402191858 & 1.79437560652 & 1.79437048920 \\ 10 & 1.85462068870 & 1.85491094750 & 1.85490721630 \end{array} \right)$$

Claude Leibovici
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