The Omega Constant and the Lambert W Function

I first encountered the Omega constant in the following video by blackpenredpen:

In this video, he looks at the function $y=xe^x-1$ and finds where the $y$ value is zero. He does this using Newton's Method which relies on the iteration: $$x_{n+1}=x_n-\frac{f(x)}{f'(x)} \text{ for any function } f(x)$$ Figure 1 shows the SageMath code and permalink that I used to generate the digits of this constant, beginning with a value of $x=0.5$ since we know the zero lies somewhere between 0 and 1 because: $$y=-1<0\ \text{ when } x=0 \text{ and } y=e-1>0 \text{ when } x=1$$ :

Figure 1: permalink

As can be seen, the constant works out to be $0.5671432904 \text{ ... }$ and it is this constant that is given the name Omega constant and the symbol $\Omega$. It is a transcendental number and has the property that $\Omega e^{\Omega}=1$. Thus $\Omega$ is the value of $x$ in $x \, e^x$ that makes it equal to 1. Figure 2 shows the graph of $y=x \, e^x$ with key points on the graph noted.

Figure 2

The minimum turning point occurs when $y'=(1+x) \, e^x=0\ $ and this occurs when $x=-1$. The point of inflexion occurs when $y''=(2+x) \, e^x=0$ and this occurs when $x=-2$. $\Omega$ is shown with its corresponding $y$ value of 1 noted.

The Lambert W function is simply the inverse function for $y=x \, e^x$ and is thus $x=y \, e^y$ but needs to be broken into two parts to conform to the requirement that a single $x$ value cannot be associated with more than one $y$ value. It is the same graph as in Figure 2 but reflected about the line $y=x$. See Figure 3 (source). The Lambert W function is also called the omega function and the product logarithm function.

Figure 3: The graph of y = W(x) for real x < 6 and y > −4. 

The upper branch (blue) with y ≥ −1 is the graph of the function W₀ (principal branch).

The lower branch (red) with y ≤ −1 is the graph of the function W₋₁. 

The minimum value of x is at {-1/e,-1}

So the solution to the equation $x \, e^x=2$ is W(2) $\approx 0.852605502013726$ and so on. We can also solve $x^x=2$ because $$\log(x^x)=x \, \log(x)=\log(2)\\ \text{ let } u=\log(x) \text{ and so }u \, e^u= \log(2)\\ u=W(\log(2)) \text{ and so } \log(x)=W(\log(2))\\ \therefore x=e^{W(\log(2))} \approx 1.55961046946...$$ Here are links to blackpenredpen's videos on:

• the graph of $y=x \, e^x$
• the Lambert W function. and 
• solving equations by using the Lambert W function

The last mentioned video I'll cover in detail here because the solutions to the two equations are very satisfying. The first equation is $x^2 \, e^x=2$ and the second is $x+e^x=2$. Let's solve each in turn: $$x^2 \, e^x=2\\ \text{Take the square root of both sides}\\x \,e^{\frac{x}{2}}=\pm \sqrt{2}\\ x \text{ must be } \geq -1/e \text{ (see Figure 3)}\\ \therefore x \,e^{\frac{x}{2}}=\sqrt{2}\\ \frac{x}{2} \,e^{\frac{x}{2}}= \frac{\sqrt{2}}{2}\\W \bigg ( \frac{x}{2} \,e^{\frac{x}{2}} \bigg )= W \bigg (  \frac{\sqrt{2}}{2} \bigg )\\ \frac{x}{2}=W \bigg (  \frac{\sqrt{2}}{2} \bigg )\\x=2 \times W \bigg (  \frac{\sqrt{2}}{2} \bigg ) \approx 0.90120$$ Figure 4 shows the graphical situation:

Figure 4

Now for the second equation: $$x+e^x=2\\ e^x=2-x\\ \frac{e^x}{e^x}=\frac{2-x}{e^x}\\1=(2-x) \, e^{-x}\\e^2=(2-x) \, e^{2-x}\\W  (e^2 )=W  ( (2-x) \, e^{2-x}  )\\W  (e^2  )=2-x\\x=2-W  (e^2  ) \approx 0.44285$$ Figure 5 shows the graphical situation:

Figure 5