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Is this right?

The Lambert W function, denoted by $W(z)$, is defined as the inverse function of $f(z) = ze^z$. In other words, if $w = W(z)$, then we have $z = w e^w$.

The continuous Newton's method is a technique for finding the roots of a function by following the flow of its gradient vector field. Specifically, given a function $f(x)$ and an initial point $x_0$, the continuous Newton's method generates a solution curve $x(t)$ that satisfies the differential equation:

$$\frac{dx}{dt} = -\nabla f(x)$$

where $\nabla f(x)$ is the gradient vector of $f$ at $x$.

In the case of the function $f(z) = ze^z$, the gradient vector is given by:

$$\nabla f(z) = (1+z)e^z$$

Therefore, the differential equation for the continuous Newton's method is:

$$\frac{dz}{dt} = -(1+z)e^z$$

which is exactly the derivative of the Lambert W function. Therefore, the solution curve generated by the continuous Newton's method starting at a point $z_0$ is given by:

$$z(t) = W(z_0 e^{-t}).$$

In other words, the Lambert W function is the flow of the exponential function under the continuous Newton's method; or equivalently, the W function is the Newton flow of the exponential function.

2 Answers2

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This makes no sense in this form.

The solution of $z'(t) = - \nabla f(z(t))$ can only have critical points of $f$ as limits. If $f(z) = ze^z$, this means that solutions converge to $z = -1$, which is easily seen directly. This is not a value of the $W$-function.

The continuous Newton method, if used to solve the scalar equation $f(z) = y$, considers the differential equation $$ z'(t) = \frac{y - f(z(t))}{\nabla f(z(t))} $$
"in the hope that the limit of $z(t)$ exists", to quote John W. Neuberger, who worked a lot on this method. That limit then should be a zero of $y - f(z)$.

Let $u(t) = f(z(t))$, then $u'(t) = \nabla f(z(t))z'(t)$ and the differential equation may be written as $$ u'(t) = y - u(t) $$ which has the solution $$ u(t) = e^{-t}u(0) + (1-e^{-t})y $$ Therefore we always have $$ z(t) = f^{-1} \left(e^{-t} f(z(0)) + (1-e^{-t}) y\right) $$ for the scalar case of the continuous Newton method.

In the case where $f(z) = ze^z$ and $f^{-1}(u) = W(u)$, the differential equation becomes $$ z' = \frac{y - ze^z}{(1+z)e^z} $$ and the solution is $$ z(t) = W\left(e^{-t}z_0e^{z_0} + (1-e^{-t}) y\right) $$ If $y \ge -e^{-1}$, then the solution is defined for all $t \ge 0$ no matter what $z_0$ is and it converges to $W(y)$. If $y < -e^{-1}$, the solution exists only on a finite $t$-interval $[0,t_0]$ and $z(t_0) = -1$, thus the solution cannot be continued for $t > t_0$.

Gary
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Hans Engler
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    Thank you so much Hans! I didn't get that far .. pinched nerve makes it painful to type and thus think but I knew that something was missing. Neuberger is at UNT which is not far from me, but last time we communicated about 2 years ago he told me that it would take too long to teach me what I need to know without knowing that im a fast learner.. this was before covid, and he's getting up there in age by now but it would be nice to show him my closed-form solution i found for the Newton flow of tanh(ln(1+t^2)) , that I verified in Mathematica and is pretty concise... –  Mar 15 '23 at 00:59
  • there must be a way to use the branches to continue beyond the principal domain... –  Mar 15 '23 at 01:00
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    @crowlogic - Nice to learn that you also know John Neuberger, one of the kindest people I met in my professional career. As to continuing past the singularity at $z = -1$: this would have to go through the complex domain. – Hans Engler Mar 15 '23 at 01:03
  • it was only thru email but i guess that counts, would have loved to chat on the phone or visit in person.. he did invite me to dinner but I was recovering from a major back surgery at the time and realized I had a not to learn before I could talk to him about this stuff without wasting his time. Maybe I'll write something up about my solution for the Newton flow of g(t)=tanh(ln(1+t^2))$ and send it to him, I've found it to be useful in root finding because if t=f(y) then it converts simple roots of f(t) to double-roots of g(t) and thus renders them linearizable –  Mar 15 '23 at 01:11
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    Care to continue this discussion in chat? – Hans Engler Mar 15 '23 at 01:21
  • I'd be delighted.. i dont think I've used chat on here before –  Mar 15 '23 at 01:35
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    Look to the bottom of this page, click on chat, and search for Continuous Newton Method. – Hans Engler Mar 15 '23 at 01:43
  • Hi Hans, it was nice to meet you, I'm leaving this site, it has too many problems with interference from anonymous energy vampires who aren't worth the energy to combat. Happening into meeting you here was a chance encounter that I don't think will recur and hence, hasta la vista, baby. I was sad to learn about professor Neuberger's passing in 2019, I suppose it was shortly after he had replied to me saying he didn't have time to teach me what I needed to know. I thought he was referring to my ability but I guess not :\ In any case, this is the umpteenth time I've been harassed by jealous mods –  Apr 16 '23 at 22:22
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This answer shows a little more clearly the way $z'(t)$ is factored out of the right-hand-side of the equation. It is an attempt at an elaboration of Hans Englers masterful answer given above.

The derivative $z'(t)$ of the solution $z(t)$ to the Newton flow ODE is not supposed to appear on the right-hand side of the solution so it might appear that the formula $u'(t) = \nabla f(z(t))z'(t)$ is incorrect. However, the given formula is right because $\nabla f(z(t))$ drops from the equation when the value of $z'(t)$ is substituted into the differential equation:

$$z'(t) = \frac{y - f(z(t))}{\nabla f(z(t))}$$

Let:

$$u(t) = f(z(t))$$

Differentiate both sides with respect to $t$:

$$u'(t) = \nabla f(z(t))z'(t)$$

Now, using the given differential equation, substitute the value of $z'(t)$:

$$u'(t) = \nabla f(z(t))\frac{y - f(z(t))}{\nabla f(z(t))}$$

This simplifies to:

$$u'(t) = y - u(t)$$

since $\nabla f(z(t))$ gets factored out and $u(t)=f(z(t))$.

Now, the solution to this differential equation is given by:

$$u(t) = e^{-t}u(0) + (1-e^{-t})y$$

And since $u(t) = f(z(t))$, making the substitution we have:

$$z(t) = f^{-1} \left(e^{-t} f(z(0)) + (1-e^{-t}) y\right)$$

This equation represents the scalar case of the continuous Newton method.