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 $y=x{e}^x-1$ and finds where the $y$ value is zero. He does this using Newton's Method which relies on the iteration:$$x_{n+1}=x_n-\frac{f(x)}{f^{\prime }(x)}\text{ for any function }f(x)$$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 $\mathrm{\Omega }$. It is a transcendental number and has the property that $\mathrm{\Omega }{e}^{\mathrm{\Omega }}=1$. Thus $\mathrm{\Omega }$ 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 $y^{\prime }=(1+x)e^x=0$ and this occurs when $x=-1$. The point of inflexion occurs when $y^{″}=(2+x)e^x=0$ and this occurs when $x=-2$. $\mathrm{\Omega }$ 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$$\begin{gathered} \log (x^x)=x\log (x)=\log (2) \\ \text{ let }u=\log (x)\text{ and so }u{e}^u=\log (2) \\ u=W(\log (2))\text{ and so }\log (x)=W(\log (2)) \\ \therefore x=e^{W(\log (2))}\approx 1.55961046946... \end{gathered}$$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:$$\begin{gathered} x^2{e}^x=2 \\ \text{Take the square root of both sides} \\ x{e}^{\frac{x}{2}}=\pm \sqrt{2} \\ x\text{ must be }\ge -1/e\text{ (see Figure 3)} \\ \therefore x{e}^{\frac{x}{2}}=\sqrt{2} \\ \frac{x}{2}{e}^{\frac{x}{2}}=\frac{\sqrt{2}}{2} \\ W(\frac{x}{2}{e}^{\frac{x}{2}})=W(\frac{\sqrt{2}}{2}) \\ \frac{x}{2}=W(\frac{\sqrt{2}}{2}) \\ x=2\times W(\frac{\sqrt{2}}{2})\approx 0.90120 \end{gathered}$$Figure 4 shows the graphical situation:

Figure 4

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

Figure 5