Given the real or complex-valued function \(F(x) = xe^x\), the Lambert W function\(W(y)\) or product logarithm is the inverse of \(F(x)\).
Definition of Lambert W function
The Lambert W function can be defined as:
\[\begin{array}{rrl} & F(x) &= xe^x\\ \Leftrightarrow & F^{-1}(F(x)) &= F^{-1}(xe^x)\\ \Leftrightarrow & W(F(x)) &= W(xe^x)\\ \Leftrightarrow & x &= W(xe^x) \end{array} \]
or equivalently
\[W(y)e^{W(y)} = y\]
From the definion the following properties can be derived
\[W(0) = 0\]
\[W(e) = 1\]
\[W'(0) = 0\]
\[a e^{-W(a)} = W(a)\]
Example \(x=W(1)=\Omega\)
Using Newton’s method, we are looking for the root of
\[f(x) = xe^x - 1\]
which stems from the definition with \(y=1\):
\[xe^x=1 \Leftrightarrow xe^x-1=0\]
The root of \(f(x)\) should be on \((0, 1)\), which we conclude by the Intermediate value theorem, since probing the function with the interval boundaries gives
\[\begin{array}{l} f(0) = 0\cdot e^0 - 1 < 0\\ f(1) = 1\cdot e^1 - 1 > 1 \end{array}\]
Therefore the desired \(\Omega\) should be between 0 and 1. Applying Newtons method using the derivative \(f'(x)= (x+1)e^x = xe^x+e^x\):
\[\begin{array}{rl} x_{n+1} &= x_n - \frac{f(x_n)}{f'(x_n)}\\ &= x_n - \frac{x_ne^{x_n}-1}{x_ne^{x_n}+e^{x_n}}\\ &= \frac{x_n^2 + e^{-x_n} y}{x_n + 1} \end{array}\]
And get the following approximation:
\[ x_{1}=0.50000\dots\\x_{2}=0.57102\dots\\x_{3}=0.56716\dots\\x_{4}=0.56714\dots\\x_{5}=0.56714\dots\\ \]
Therefore the Omega Constant that satisfies \(\Omega e^\Omega = 1\) is approximately \(\Omega \approx 0.567143\)
Example \(x^x=2\)
\[\begin{array}{rrl} &x^x &= 2\\ \Leftrightarrow&\ln(x^x) &= \ln(2)\\ \Leftrightarrow&x\ln(x) &= \ln(2)\\ \Leftrightarrow&\ln(x) e^{\ln(x)} &= \ln(2)\\ \Leftrightarrow&W(\ln(x) e^{\ln(x)}) &= W(\ln(2))\\ \Leftrightarrow&\ln(x) &= W(\ln(2))\\ \Leftrightarrow&x&= e^{W(\ln(2))}\\ \Leftrightarrow& x & \approx 1.55961047 \end{array}\]
Example \(x = \ln(4x)\)
\[\begin{array}{rrl} & x &= \ln(4x)\\ \Leftrightarrow& e^x &= 4x\\ \Leftrightarrow& \frac{1}{x}e^x &= 4\\ \Leftrightarrow& -xe^{-x} &= -\frac{1}{4}\\ \Leftrightarrow& W(-xe^{-x}) &= W(-\frac{1}{4})\\ \Leftrightarrow& -x &= W(-\frac{1}{4})\\ \Leftrightarrow& x &= -W(-\frac{1}{4})\\ \Leftrightarrow& x & \approx 0.357402956 \end{array}\]
Example \(x^2e^x = 2\)
\[\begin{array}{rrl} & x^2e^x &= 2\\ \Leftrightarrow& \sqrt{x^2e^x} &= \sqrt{2}\\ \Leftrightarrow& xe^{\frac{x}{2}} &= \sqrt{2}\\ \Leftrightarrow& \frac{x}{2}e^{\frac{x}{2}} &= \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\\ \Leftrightarrow& W\left(\frac{x}{2}e^{\frac{x}{2}}\right) &= W\left(\frac{1}{\sqrt{2}}\right)\\ \Leftrightarrow& \frac{x}{2} &= W\left(\frac{1}{\sqrt{2}}\right)\\ \Leftrightarrow& x &= 2W\left(\frac{1}{\sqrt{2}}\right)\\ \Leftrightarrow& x & \approx 0.901201032 \end{array}\]
Example \(x+e^x = 2\)
\[\begin{array}{rrl} &x +e^x&= 2\\ \Leftrightarrow&e^x&= 2 -x \\ \Leftrightarrow&1 &= (2 -x)^{e^{-x}}\\ \Leftrightarrow&e^2 &= (2 -x)^{e^{2-x}}\\ \Leftrightarrow&W(e^2) &= W((2 -x)^{e^{2-x}})\\ \Leftrightarrow&W(e^2) &= 2 -x\\ \Leftrightarrow&x &= 2 - W(e^2)\\ \Leftrightarrow&x &\approx 0.442854401 \end{array}\]
Lambert W function Calculator
\(W(\)\() = W( x\cdot e^x) = x \approx\)0.5671432904097838