A Nonstandard Equation | Two Methods? | Problem 365

[@scottleung9587]

Yes, they're equal. If you e (or exponentiate) both sides, you'll get W(1)=1/e^(W(1)). Then after multiplying both sides by the exponential, you'll have W(1)*e^(W(1))=1. Finally, after applying Lambert's W function once more, you'll have W(1) on both sides and thus they'll agree.

[@XJWill1]

No need to "apply Lambert's W function once more". Since W(x) and f(x) = x*exp(x) are inverse functions, f( W(x) ) = x

[@XJWill1]

There are an infinite number of solutions. As usual for this type of equation, you can visualize the solutions by splitting the complex-valued equation into two real-valued equations and making an implicit plot of the real-valued equations to see the point(s) of intersection. y^2 = exp(2*x) - x^2 2*atan( tan(y/2) ) = atan2( -y , -x ) As usual, if you are using desmos then atan2( -y , -x ) is entered as arctan( -y , -x ) Looking at the graph, there is only one solution with negative real part. And there are an infinite number of solutions with positive real-part and those come in complex-conjugate pairs.

[@mikecaetano]

The Corless et al. paper, "On the Lambert W Function", asserts the equation logW(z) = logz - W(z) is valid for the principal branch when z > 0. (See expression 3.8)

[@XJWill1]

What do you mean by z > 0 ???

[@XJWill1]

Anyway, there is nothing profound about that result. If f(x) = x*exp(x) and W(x) is the inverse function of f(x) then by definition of inverse functions f( W(x) ) = x W(x) * exp( W(x) ) = x and then the result follows by simple manipulation and applying the log() function.

[@trojanleo123]

z = -W(1)

[@trojanleo123]

Noooo. Please don't leave us in suspense. Please I really want to know why Method 1 and Method 2 produced results that look different. I'm hoping you'll explain to us in a comment or maybe make a companion video explaining this. Thank you.

[@trojanleo123]

Actually, never mind. I figured out the answer. The result from Method 1 is indeed equal to the result in Method 2. Although they look different, they have the same value.

[@aplusbi]

Thanks! 😁

[@dorkmania]

For some k = t•e^t, W(k) = W(t•e^t) = t Substituting W(k) for t, k = W(k)•e^W(k). When k = 1, W(1)•e^W(1) = 1 => W(1) = e^-W(1) => ln(W(1)) = -W(1)