Lambert $W$-functions

Branko Ćurgus

Definition

Consider the following function \[ g: [-1,+\infty) \to [-1/e,+\infty), \qquad g(x) = x \exp(x). \] This function is represented by its navy blue graph in the picture below. The First Monotonicity Test and the Mean Value Theorem imply that this function is an increasing bijection. The inverse of this function is called the Lambert $W_0$ function. It is represented by its maroon graph in the picture below.

$ X = [-1,+\infty), \ Y = [-1/e,+\infty), \ W_0 = \bigl\{ \bigl(x e^x,x \bigr) \ \bigl| \bigr. \ x \in X \bigr\} $

The Lambert $W_0$ function
Consider the following function \[ h: (-\infty, -1] \to [-1/e,0), \qquad h(x) = x \exp(x). \] This function is represented by its navy blue graph in the picture below. The Second Monotonicity Test and the Mean Value Theorem imply that this is a decreasing bijection. The inverse of this function is called the Lambert $W_{-1}$ function. It is represented by its maroon graph in the picture below.

$X = (-\infty,-1), \ Y = (-1/e,0), \ W_{-1} = \bigl\{ \bigl(x e^x,x \bigr) \ \bigl| \bigr. \ x \in X \bigr\}$

The Lambert $W_{-1}$ function
Mathematica's notation for $W_0(x)$ function is ProductLog[x], or, equivalently ProductLog[0,x]. Mathematica's notation for $W_{-1}(x)$ function is ProductLog[-1,x].
The Wikipidia's page on Lambert W function has a lot of information about this function. However, this page deals with the complex $W$ function so it might be difficult for you to extract useful information.

Examples

Solve $2^x = 5 x$ for $x \in \mathbb R$. This is how this equation is being solved. Rewrite the equation using the exponential function $e^x$: \[ e^{x \ln 2} = 5 x. \] Next bring the equation to the form $Y = X e^X$ where $Y$ will be a specific number. Then we can use the Lambert functions to solve this equation. Here we will bring the exponential part together with $x$. \[ \frac{1}{5} = x\,e^{-x \ln 2} . \] Now we recognize that our $X$ must be $-x \ln 2$, so we have to multiply the equation by $-\ln 2$: \[ -\frac{\ln 2}{5} = - (x \ln 2) \,e^{-(x \ln 2)}. \] Since $-1/e \lt -(\ln 2)/5 \lt 0$ the above equation has two solutions. The solutions are \[ - (x_1 \ln 2) = W_{0}\bigl(-(\ln 2)/5\bigr) \quad \text{and} \quad - (x_2 \ln 2) = W_{-1}\bigl(-(\ln 2)/5\bigr). \] Hence, \[ x_1 = - \frac{1}{\ln 2} W_{0}\bigl(-(\ln 2)/5\bigr) \approx 0.235456 \quad \text{and} \quad x_2 = - \frac{1}{\ln 2} W_{-1}\bigl(-(\ln 2)/5\bigr) \approx 4.488001. \]
Solve $x^x = y$ for $x \gt 0$ and appropriate $y \gt 0$. This is a little trickier. Apply $\ln$ to both sides to get \[ x \ln x = \ln y. \] Rewrite the equation using the exponential function $X \, e^X$: \[ (\ln x) \, e^{\ln x} = \ln y. \] It just happens that this equation is exactly in the form $X\,e^X = Y$. So this equation has a solution if and only if $\ln y \geq -1/e$. This is the case if and only if $y \geq e^{-1/e}$. The equation has the unique solution if $y = e^{-1/e}$ and that solution is $x = 1/e$. If $e^{-1/e}\lt y \lt 1$, the equation has two solutions given by \[ \ln (x_1) = W_{-1}\bigl( \ln y \bigr) \quad \text{and} \quad \ln (x_2) = W_{0}\bigl( \ln y \bigr) \] Hence \[ x_1 = \exp\Bigl( W_{-1}\bigl( \ln y \bigr) \Bigr) = \frac{\ln y}{W_{-1}\bigl( \ln y \bigr)} \quad \text{and} \quad x_2 = \exp\Bigl( W_{0}\bigl( \ln y \bigr) \Bigr) = \frac{\ln y}{W_{0}\bigl( \ln y \bigr)}. \] The last identity might look obscure at first. It follows from the fact that $X = W(Y)$ if and only if $X\,e^X = Y$, with $X$ and $Y$ in the appropriate domains. Therefore, $W(Y)\,e^{W(Y)} = Y$, with $Y$ in the appropriate domain. Hence, \[ e^{W(Y)} = \frac{Y}{W(Y)}. \] Here $W$ stands for either $W_0$ or $W_{-1}$. In each case we have to adjust the domains for $X$ and $Y$ appropriately.
If $y \geq 1$ the equation has a unique solution which is given by \[ x = \exp\Bigl( W_{0}\bigl( \ln y \bigr) \Bigr) = \frac{\ln y}{W_{0}\bigl( \ln y \bigr)}. \]
Prove that the function $f:\mathbb R \to \mathbb R$ defined by $f(x) = x \exp(x^2)$ is a bijection. Find a formula for its inverse.

The claim that $f$ is a bijection follows from Monotonicity tests and the fact that $f$ is continuous and $f$ is not bounded. Let $y \gt 0$. Here we will demonstrate how to solve \[ x \exp(x^2) = y. \] Since $y \gt 0$ a solution $x$ must be positive. Therefore the given equation is equivalent to \[ x^2 \exp(2\,x^2) = y^2. \] The last equation is equivalent to \[ 2 x^2 \exp(2\,x^2) = 2 y^2. \] Since $2 y^2 \gt 0$ the last equation is equivalent to \[ 2\, x^2 = W_0(2\, y^2). \] Hence the unique solution of the original equation is \[ x = \sqrt{ \frac{W_0(2\, y^2)}{2} }. \] Since the given function is odd, its inverse will also be odd. Therefore the inverse function of $f$ is \[ f^{-1}(x) = (\operatorname{sgn} x) \sqrt{ \frac{W_0(2\, x^2)}{2} }. \]

$f(x)$ in navy blue and its inverse $f^{-1}(x)$ in maroon