Lambert function and Wolfram or "±infinity+i×K=±infinity" ?

[Gil]

Are my previous post results pure non-sense?

Could you just try, at random, one of my cases?

I entered each time in Wolfram the values k and z=a+ib, with |a| or |b| or both very large (>1E100000) to find the above outputs formulae j, equal to "+inf±i × f,j(k)".

Thanks in advance.

[Albert Chan]

(01-17-2024 08:30 PM)Gil Wrote:  Could you just try, at random, one of my cases?

I did, but mpmath (version 1.1.0) had not implemented W_k(∞*cis(θ)), for nonzero k

p2> from mpmath import *
p2> mp.pretty = 1
p2> lambertw(mpc(20,inf), k=0)
(+inf + 1.5707963267949j)
p2> lambertw(mpc(20,inf), k=-3)
(+inf + 1.5707963267949j)

Instead, I posted formula (my previous post), when z = ∞*cis(θ)

W_k(z) = ln_k(z) = ln(z) + 2*k*pi*I

[Gil]

Thanks again, Albert, for the time you always take to answer our posts or questions, always trying to develop clean and detailed arguments or solutions.

Sometimes even too technical posts with your program Lua, sometimes only you answering your own commentaries!

But great, and always nice written.

Congratulations for partaking your knowledge and valuable time.

Really amazing — and very much appreciated.

[Gil]

For k=0 and z=20+i×1E490,
I find the approached result you mentioned.

However, for k=-3, my HP50G program gives quite different multiple values of pi for the imaginary part.

See my results:
W-3((20.,1.E400)): (914.215792373,-17.2598823971) —>-i5.5pi
W-3((20.,-1.E400)): (914.215721783,-20.3980438811)—>-i6.5pi
W-3((20.,9.E490)): (1123.43965171,-17.2633942517) —>-i5. 5pi
W-3((20.,-1.E490)): (1121.24433569,-20.4021582594) —>-i6.5pi
W0((20.,1.E490)):(1121.24450008,1.56939663599) —>i1/2pi

[Gil]

For k=0 and z=20+i×1E490,
I find the approached result you mentioned.

However, for k=-3, my HP50G program gives quite different multiple values of pi for the imaginary part.

See my results:
W-3((20.,1.E400)): (914.215792373,-17.2598823971)
-17.26i = about -i5.5pi or iPI(2k+0.5),
according to the formula in a previous post

W-3((20.,-1.E400)): (914.215721783,-20.3980438811)
-20.40i = about -i6.5pi or iPI(2k-0.5),
according to the formula in a previous post

W-3((20.,9.E490)): (1123.43965171,-17.2633942517)
-17.26 pi = about -i5.5pi

W-3((20.,-1.E490)): (1121.24433569,-20.4021582594)
-20.40i = about -i6.5pi

W0((20.,1.E490)):(1121.24450008,1.56939663599)
1.57pi = about 1/2pi

My result for your case with k=-3 correspond to the snapshot included here taken from Wolfram Alpha.

Have you a way to check again your result?

[Albert Chan]

(01-17-2024 09:28 PM)Gil Wrote:  However, for k=-3, my HP50G program gives quite different multiple values of pi for the imaginary part.

See my results:
W-3((20.,1.E400)): (914.215792373,-17.2598823971) —>-i5.5pi
W-3((20.,-1.E400)): (914.215721783,-20.3980438811)—>-i6.5pi
...

If |z| → ∞, W_k(z) = ln_k(z) = ln(z) + 2*k*pi*I

Imag part with different multiples of pi is due to ln(z) = ln(|z|*cis(arg(z))) = ln(|z|) + arg(z)*I

arg((20,+1e400)) + 2*-3*pi ≈ +pi/2 - 6*pi = -5.5*pi
arg((20,−1e400)) + 2*-3*pi ≈ −pi/2 - 6*pi = -6.5*pi
...

[Gil]

Yes, but you wrote in a previous post
"p2> lambertw(mpc(20,inf), k=-3)
(+inf + 1.5707963267949j)"

instead of — according to my results, of course... — (+inf + (2k+.5)j)".

[Gil]

Input {k X},
k for the branch
& X, real or a complex number,

with real part of X = ± inf
and/or imaginary part of X = ±inf.

Here is a program or rather a subroutine summarising the above discussed cases, with output in form of a string like "oo ± iPI#",
with oo standing for "limit tending to +inf"
& with iPI together, to remember the Euler's formula when EXP(iPI#):

Taking the real part of the given result, we get
"something—> inf" × EXP( "something—>inf")
And that latter expression in bold is "something—> inf".

This last limit is still to be multiplied by a "fixed number", ie, EXP(iPI#), here always a real number of the form (±1+i0) or a complex number of the form (0±i).

Consequently, the final result of the above mentioned multiplication will always be either
+inf, or -inf, or 0+i×inf, or 0-i×inf.

In other words...
Having for instance x=a+i×inf, a≠0 & a≠±inf
W0(x) = (inf + iPI/2)
But (that last result=inf + iPI/2) × EXP(that last result = inf + iPI/2) = "something tending to inf" × (0+i×inf) =(0+i×inf),
and never the initial value of x=a+i×inf.

Code:

\<< OBJ\-> DROP \-> k x
  \<< x RE ABS \->NUM \oo \->NUM == x IM ABS \->NUM \oo \->NUM == OR
    IF
    THEN "\oo "
      CASE x RE \->NUM \oo \->NUM ==
        THEN 2 k * DUP 0 \>=
          IF
          THEN "+ i\pi\183"
          ELSE "- i\pi\183"
          END SWAP ABS R\->I
        END x RE \->NUM \oo \->NUM NEG ==
        THEN 2 k * 1 x IM \->NUM 0 >
          IF
          THEN +
          ELSE -
          END DUP 0 \>=
          IF
          THEN "+ i\pi\183"
          ELSE "- i\pi\183"
          END SWAP ABS R\->I
        END 2 k * .5 x IM \->NUM 0 >
        IF
        THEN +
        ELSE -
        END DUP 0 \>=
        IF
        THEN "+ i\pi\183"
        ELSE "- i\pi\183"
        END SWAP ABS \->Q \->STR "'" "" SREPL DROP
      END + + "\1831/" "/" SREPL DROP "<U>" + "i" "<U>i" SREPL DROP "<U>" 19 CHR 3 CHR OVER + + SREPL DROP
    ELSE
    END
  \>>
\>>