lambertw, all branches

[Albert Chan]

Hi, Gil

This is more lua language question than math question.
see http://lua-users.org/wiki/ExpressionsTutorial

x = A and B or C      → x = ((A and B) or C)      // parenthesis added to show precedence order

Code:

if A then
    local T = B
    if T then x=T else x=C end
else
    x = C  
end

Test expression, false or nil are false, everything else (e.g. 0.0, "", {}) are true.

Quote:< if k==0 then x = I.abs(a+1)<.6 and T/2 or I.log1p(a) end

Code:

if k==0 then
    if I.abs(a+1)<.6 then
        x = T/2
    else
        x = I.log1p(a)
    end
end

(01-21-2024 01:53 PM)Gil Wrote:  C) About T/2: it is an interval from [pi/4 pi/2]×pi, yes?

For k=0, I.W code ensured imaginery part is positive (x>0, y≥+0):

W₀(-x-y*I) = conj(W_-0(conj(-x-y*I))) = conj(W₀(-x+y*I))

T = (arg(-x+y*I) + 2*0*pi*I) = arg(-x+y*I)*I = (pi - atan(y/x))*I = (pi/2 .. pi)*I
T/2 = (pi/4 .. pi/2)*I

Quote:> if k==0 then x = I.log1p(2*a)/2 end

This is my 2nd iteration of W0 guess:
log1p(-1) = -∞ occurs when a = -0.5, inside lyuka branch |a+1/e|<.25, thus are safe.

if |a|→0, then x = log1p(2a)/2 ≈ a-a*a, matching W0(a) taylor series
if |a|→∞, then x = log1p(2a)/2 ≈ ln(2a)/2 ≈ ln(a)/2

Not quite match asymptote, but it will, for next iteration

f = x + ln(x) - ln(a) = ln(a)/2 - ln(ln(a)/2) - ln(a) ≈ -ln(a)/2
f' = 1 + 1/x ≈ 1

x - f/f' ≈ ln(a)/2 + ln(a)/2 = ln(a)