Lambert W Function Intuition

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  • Опубликовано: 3 ноя 2024

Комментарии • 12

  • @jamesmccamish3901
    @jamesmccamish3901 Год назад +1

    You do a great job of simplifying things as much as they can be. Much appreciated.

  • @marpin6162
    @marpin6162 4 года назад +8

    You deserve more subscribers.

  • @alijoueizadeh2896
    @alijoueizadeh2896 7 месяцев назад

    Thank you for your precious time.

  • @peternolan814
    @peternolan814 3 года назад +1

    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.

  • @jamesmorton5017
    @jamesmorton5017 Год назад

    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?

  • @NXT_LVL_DVL
    @NXT_LVL_DVL 11 месяцев назад

    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

  • @hectortroncoso322
    @hectortroncoso322 4 года назад

    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.

  • @CaptainBlackthorn
    @CaptainBlackthorn 4 года назад +1

    Great video

  • @АлексейЕринский

    Решено измерение скорости имея координаты X1,X2,Y W-функция Ламберта вместо Евклидова пространства