lambertw, all branches
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01-21-2024, 08:39 PM
(This post was last modified: 01-21-2024 09:14 PM by Gil.)
Post: #28
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RE: lambertw, all branches
One more question, if I may:
You wrote "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" a) Suppose that k=0, s=(+1.E-100-i.0000000001). —> abs(a) < 0.6 Then initial xo = T/2. Could you give the different step to find the value of T/2? I found, for xo =T/2, pi/4× PI, but that value leads to endless loops on my program, whereas xo =ln(1+a) gives the expected result. But here I was lucky. As I was not sure of the value of T/2, b) for the other case k=0, a=(-.1,+i×1.E-100), though abs(a) <0.6, I tried as the above mentioned case, xo=ln(1+a), but then I got endless loops. What would be the correct step to find here also T/2. Meanwhile,Takayuki HOSODA's formulae for initial xo gives good, approximate results first digits exact) in both cases a) & b) Thanks again for your comprehension and possible help. |
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