lambertw, all branches

[Albert Chan]

Hi, pier4r

I am still learning, and math is not up to par into article section, where no feedback allowed.
Hopefully, someone reading this may suggest a better way.

(04-07-2023 02:47 PM)Albert Chan Wrote:  For k=-1, solve for f=0 is especially difficult.
Fortunately, we can flip it, to solve for k=1 instead. W(z, k) == conj(W(conj(z), -k)

Example, as I am writing code, I realized k "flipping" is unnecessary.
With equivalent starting guess, convergence is the same.

It is just here, "equivalent" may mean a difference of +0*I vs -0*I

Below is work in progress code, without flipping k to non-negative.
It simply solve f=0 by Newton's method, with a supplied guess x

lua> a, k = -0.1, -1
lua> x = I.log(-a)
lua> I.W(a, k, true, x)
(-2.3025850929940455+0*I)
(-3.7769077194622533-11.106812831380271*I)
(-3.949948702089913-0.9503184433887455*I)
(-3.558355491208183-0.04104686782028688*I)
(-3.577077533853747+8.297706605468053e-005*I)
(-3.5770735614781484-8.721290183674583*I)
...

lua> I.W(a, k, true, I.conj(x)) -- +0*I to -0*I fixed it!
(-2.3025850929940455-0*I)
(-3.7769077194622533-0*I)
(-3.579124176402325-0*I)
(-3.5771522746958055-0*I)
(-3.5771520639573-0*I)
(-3.577152063957297-0*I)

(04-07-2023 02:47 PM)Albert Chan Wrote:  f = x + ln(x) - ln(a) - 2*k*pi*I = 0
x.imag + phase(x) = 2*k*pi + phase(a)

Guess x with +0*I is a lousy one, with phase(x) = +pi
LHS = 0 + (+pi) = +pi
RHS = 2*(-1)*pi + (pi) = -pi

guess of conj(x) with -0*I is much better, LHS = 0 + (-pi) = -pi = RHS

---

The fix is simple, argument for log, instead of wrong implicit I.new(-a), use explicit -I.new(a)

lua> I.log(-I.new(a)) -- good W_-1 guess
(-2.3025850929940455-0*I)