In this video I start building some intuition for the Lambert W function by plotting it in comparison to its inverse. For more videos in this series, visit: • Lambert W Function
Hello, A day has passed since I watched this video having commented yesterday what's below. Today is May 12 2021. Say I have a value for x some number. Then I can write: (some number) = yexpy I was thinking that the whole problem with the Lambert W Function is that this equation cannot be solved for y and yet there is some other number for y such that this equation is true. So glancing at what I wrote below I can calculate a value and get some number for in my case: Number 2 = Exp(1 -4at/c) by selecting a value for t and entering values for the constants a and c. I then can write: Number 2 = yexpy such that the value of y that satisfies this equation is: y = W(Number 2) the value of the Lambert W Function. This then, in my case, will yield the speed of the particle. I think I might understand it all a bit better now, All the best and many thanks, Peter Nolan. Ph.D.(physics). Dublin. Ireland. Hello, Your account of The Lambert W Function is very good indeed. The graph is most excellent. I'm interested in this differential equation: dv/dt = a((1 -2v/c)/(1 - v/c)) such that v is zero when t is zero and a and c are constants. According to Wolfram's Mathematica the solution is: v(t) = 1/2(c -cProductLog(E^(1 - 4at/c))) or: v(t) = 1/2(c - cW(E^(1 - 4at/c))) where W is The Lambert W Function and where E is "e" the base of natural logarithms. I'm struggling a bit however. I'm hoping that since t varies from zero to infinity I can select any t and calculate the value of: E^(1 - 4at/c) but how then do I calculate the value of y = W(x) where x = E^(1 - 4at/c)? I'm sorry to be confused. All the best and many thanks, Peter Nolan. Ph.D.(physics). Dublin. Ireland. I'm 67yo.
The inverse of a function is just the reflection with respect to the line y = x . let's try it with polynomials and trigonometric functions or any other functions other than exponential
This is a good trick and sound manipulation. What is lacking is how to transform your scalar W(64ln2)/ln2 into a numerical value. What is the function of W?
Lets see,if i take x and aply into the function :X . e^ X so i obtain Y. If i take now this Y,under certain conditions supposed acomplished;i want to know how to operate with the operation to be back to Y. Thanks.
