from scipy.special import lambertw
import numpy as npFrom [1]:
The Lambert W Function is denoted \(W(x)\) and is the inverse of the function \(f(x) = xe^{x}\).
Thus, \(W(x) = f^{-1}(x)\).
For \(f(x)\) we have that:
- The derivative \(f'(x) = e^{x}(x+1)\)
- \(f''(x) = 2e^{x} + xe^{x}\) and \(f''(-1) = 1\)
- \(f(x)\) is minimized at \(x = -1\)
- Domain of \(f(x)\) is \([-1,\infty)\)
- Range of \(f(x)\) is \([-\frac{1}{e},\infty)\)
The last two bullet points mean that the domain and range of \(W(x)\) are \([-\frac{1}{e},\infty)\) and \([-1,\infty)\) respectively.
For \(W(x)\) we also have that:
- \(W(f(x)) = W(xe^{x}) = x\)
- \(f(W(x)) = W(x)e^{W(x)} = x\). In other words, \(e^{W(x)} = \frac{x}{W(x)}\)
Solve \(x^{x} = 2\)
Take the log of both sides to get \(x\ln(x) = \ln(2)\) or that \(e^{\ln(x)}\ln(x) = \ln(2)\). Rearranging we get \(\ln(x)e^{\ln(x)} = \ln(2)\).
Applying the Lambert W function on both sides produces \(W(\ln(x)e^{\ln(x)}) = W(\ln(2))\) or that \(\ln(x) = W(\ln(2))\).
Finally, we get that \(x = e^{W(\ln(2))}\).
x = np.exp( lambertw( np.log(2) ) )
x, np.allclose( np.power(x,x) , 2 )((1.5596104694623694+0j), True)Thus, the solution of \(x^{x} = 2\) is \(x = 1.55961\)
Solve \(x^{2}e^{x} = 2\)
Take the square root in both sides to get \(xe^{x/2} = \sqrt 2\). Divide both sides by \(2\) to get \(0.5xe^{0.5x} = \frac{1}{\sqrt 2}\).
Applying the Lambert W function on both sides produces \(W(0.5xe^{0.5x}) = W(\frac{1}{\sqrt 2})\) or \(0.5x = W(\frac{1}{\sqrt 2})\).
Finally, we get \(x = 2W(\frac{1}{\sqrt 2})\).
x = 2 * lambertw(1./np.sqrt(2))
x, np.allclose( x*x*np.exp(x), 2 )((0.9012010317296661+0j), True)Solve \(x + e^{x} = 2\)
Exponentiating both sides we get \(e^{x}e^{e^{x}} = e^{2}\). Applying the Lambert W function on both sides produces \(W( e^{x}e^{e^{x}} ) = W(e^{2})\) or \(e^{x} = W(e^{2})\)
Thus, \(x = \ln(W(e^{2}))\)
x = np.log( lambertw( np.exp(2) ) )
x, np.allclose( x + np.exp(x) , 2 )((0.4428544010023887+0j), True)Find the minimizer of \(x \ln(\frac{x}{u}) + \frac{1}{2\lambda}(x-v)^{2}\) (suppose \(\lambda >0\)).
This is a sum of convex functions so it is convex and a minimizer exists.
Taking the derivative and setting equal to zero we get: \[1 + \ln x - \ln u + \frac{x}{\lambda} - \frac{v}{\lambda} = 0\]
Hence, \(\ln x + \frac{x}{\lambda} = \ln u + \frac{v}{\lambda} - 1\).
Exponentiating both sides we get:
\(e^{\ln x + \frac{x}{\lambda}} = e^{\ln u + \frac{v}{\lambda} - 1}\) or \(xe^{\frac{x}{\lambda}} = ue^{\frac{v}{\lambda} - 1}\).
Dividing both sides by \(\lambda\) we have \(\frac{x}{\lambda}e^{\frac{x}{\lambda}} = \frac{u}{\lambda}e^{\frac{v}{\lambda} - 1}\).
Applying the Lambert W function on both sides gives: \[ W(\frac{x}{\lambda}e^{\frac{x}{\lambda}}) = W(\frac{u}{\lambda}e^{\frac{v}{\lambda} - 1})\] or
\(\frac{x}{\lambda} = W(\frac{u}{\lambda}e^{\frac{v}{\lambda} - 1})\).
Thus, \[x = \lambda W(\frac{u}{\lambda}e^{\frac{v}{\lambda} - 1})\]
References
[1]
blackpenredpen, “<A href="https://www.youtube.com/watch?v=sWgNCra93D8">"lambert w function intro and x^x=2"</a>.” https://www.youtube.com/watch?v=sWgNCra93D8, 2018.