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

01162024, 12:06 PM
(This post was last modified: 01162024 12:08 PM by Gil.)
Post: #1




Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
For my HP50G LambertW program, I was making comparisons with Wolfram Alpha.
In Wolfram Alpha: LambertW(100)=3.2+2.5i LambertW(100)=3.2+2.7i LambertW(1E6)=11.4+2.9i LambertW(1E9)=17. 8+2.98i LambertW(1E15)=31.1+3.04i LambertW(1E30)=64.9+3.09i LambertW(1E30000)=69066.4+3.1415i LambertW(1E3000000000)=6.9E9 +3.141592653i ... Two observations, as real x tends to go to very large negative values : 1) we see that the real part (of LambertW0) seems to —> +infinity; 2) we see that the imaginary part (of LambertW0) seems to —> pi. My questions If 2) is true, it means that lim IM(W0) =pi≠0, when x —> infinity. Then, can we say, considering the above, that lim W0(infinity) = infinity, as does write Wolfram Alpha, leaving out the imaginar part? Or W0(infinity) = infinity+pi*i... = infinity? Or, more generally: if lim "A" = C, when x> B, & C=±infinity +i×(constant K), can we say that the limit C=±infinity, leaving out the imaginary part (the constant K) (the latter, K, being "peanuts" relatively to the ± infinity of the real part)? I thank you in advance for your comments. Gil 

01162024, 02:01 PM
(This post was last modified: 01162024 02:13 PM by Gil.)
Post: #2




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
With Wolfram Alpha
LambertW (k=any value, z=(any real value ±i*infinity) = infinity. Now it seems that lim LambertW, k free, for z=a+ib & with a=±infinity, b=±infinity, gives equally infinity. Could anyone, if possible a "math guy", confirm it? My questions are related of course to HP programs, as I want to implement an output in my Lambert HP50G function for any real or complex "numbers, including the ± infinity input cases, the outputs of which I wish to treat as follows : Wk=0±1...(oo+ib): 'oo' Wk=0±1...(a+ioo): 'oo' Wk=0±1...(±oo±ioo): 'oo' with oo being the infinity sign. Thanks again for your insight. 

01162024, 03:40 PM
Post: #3




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
Wolfram Alpha
LambertW (infinity) = (infinity) But (infinity) * EXP (infinity) can't be equal to infinity Can Wolfram Alpha be wrong? 

01162024, 05:16 PM
(This post was last modified: 01162024 05:26 PM by Nigel (UK).)
Post: #4




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
(01162024 03:40 PM)Gil Wrote: Wolfram Alpha Not in this case (although in general I’m sure it can be!). \(\exp(+\infty+{\rm i}\pi)\) is indeed \(\infty\) because \(\exp({\rm i}\pi)=1\). So far as I can see you’ve already said something like this a couple of posts ago, so maybe I’m not understanding what you are asking? Nigel (UK) 

01162024, 06:11 PM
Post: #5




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
As I said,
LambertW (0,infinity) = (infinity) with Wolfram Alpha. I think it is impossible, as the (given answer = infinity) * EXP (given answer =infinity) can't be equal to infinity Can Wolfram Alpha be wrong? Instead, the answer seems rather to be (infinity+i*pi). Then (infinity+i*pi) × exp(infinity) × cos (pi) = Infinity × 1= infinity. But infinity + i*pi = infinity + number = infinity... In a way, Wolfram may correct. Question, what most "reasonable" output should I give for W0(infinity), W0(±i×infinity), W0(±infinity±i×infinity)? 

01162024, 07:36 PM
Post: #6




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
(01162024 06:11 PM)Gil Wrote: Instead, the answer seems rather to be (infinity+i*pi). (∞ + pi*I) * ∞ = ∞  ∞*I Still does not get back ∞, unless ... (∞ + pi*I) = ∞ ? (∞ + pi*I) * ∞ = ∞ * ∞ = ∞ But then, we are just picking and choosing! We start with saying W(∞) = ∞ + pi*I ≠ ∞ But to make it roundtrip, we had to simplify (∞ + pi*I) = ∞ We may turn it around, and say ∞ = ∞ + (any finite number, real or complex) ∞ * e^∞ = ∞ * e^(∞+θ*i) = ∞ * e^∞ * cis(θ) = ∞ * cis(θ) > W(∞ * cis(θ)) = ∞ // infinity is weird! 

01162024, 09:34 PM
(This post was last modified: 01162024 09:36 PM by Gil.)
Post: #7




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
Nice summarised presentation, Albert.
For me, the answer given by Wolfram is misleading, as we requested W0(1) and its "answer is infinity", but "answer is infinity" ×EXP("answer is infinity") clearly cannot give back 1, unless we assume that, in one step, infinity+iPI≠ infinity to assume at the end that infinity+iPI=infinity, which is contradictory. What would you think if, for such cases, Wolfram Alpha would reply here "undefined", for x—> oo, but curiously also for x—>+oo? Indeed, x—> +oo = +oo + "Somewhat", in particular, oo+ipI, and, as we saw, (x+ipI) × EXP(x+iPI) might give, with Euler and cos PI =  1, a result of oo (by special "picking up"), though clearly the graphs shows the "right" answer having rather to be +oo. 

01162024, 09:53 PM
Post: #8




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
I read your last post again, Albert, and think that is a similar reasoning that might have incited the Wolfram team to give a possible math answer for W(oo)=W(+oo) = +oo.


01162024, 11:45 PM
(This post was last modified: 01172024 12:13 AM by Albert Chan.)
Post: #9




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
Not all math software follow Wolfram Alpha's logic.
p2> from mpmath import * p2> mp.pretty = 1 p2> z = inf p2> lambertw(z), log(z) ((+inf + 3.14159265358979j), (+inf + 3.14159265358979j)) p2> z = mpc(0,inf) p2> lambertw(z), log(z) ((+inf + 1.5707963267949j), (+inf + 1.5707963267949j)) p2> z = inf p2> lambertw(z), log(z) (+inf, +inf) I am OK with ∞ + 1 = ∞, but not so much for ∞ + i = ∞ I like mpmath style better. p2> inf + 1 +inf p2> inf + j (+inf + 1.0j) 

01172024, 12:36 AM
(This post was last modified: 01172024 01:02 AM by Gil.)
Post: #10




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
Yes, W0(inf) = inf+i×pi is the limit that appears when x tends to very large negative real numbers.
But, again, (inf+i×pi)*EXP(inf+i×pi) (1) =(inf+i×pi)*EXP(inf) *EXP(i×pi) =(inf+i×pi)*inf*(cos pi +i× sin pi) =(inf+i×pi)*inf*(1) =(inf+i×inf)*(1) =infi*inf ≠ inf if I am not mistaken, unless you force and say that first member in (1), inf+i×PI, = inf, with no more imaginary part (but that sounds strange : infinity of bananas [real part of number] + 1 orange×[imaginary part] =? infinity of bananas; on the other hand, infinity of bananas means that there are everywhere bananas in the universe, and consequently no place for a single orange). And then annoying again, mixing up the definitions. By the way, what does give your software for W0(infinity +10i). By what I understood, the output should be infinity + 10i, vs inf with Wolfram. 

01172024, 12:57 AM
Post: #11




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
Yes, W0(inf) = inf+i×pi is the limit that appears when x tends to very large negative real numbers.
But, again, (inf+i×pi)*EXP(inf+i×pi) (1) =(inf+i×pi)*EXP(inf) *EXP(i×pi) =(inf+i×pi)*inf*(cos pi +i× sin pi) =(inf+i×pi)*inf*(1) =(inf+i×inf)*(1) =infi*inf ≠ inf if I am not mistaken, unless you force and say that first member in (1), inf+i×PI, = inf, with no more imaginary part (but that sounds strange : infinity of bananas [real part of number] + 1 orange×[imaginary part] =? infinity of bananas; on the other hand, infinity of bananas means that there are everywhere bananas in the universe, and consequently no place for a single orange). And then annoying again, mixing up the definitions. 

01172024, 02:06 AM
(This post was last modified: 01172024 10:43 PM by Albert Chan.)
Post: #12




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
Update with generalized W formula. Set θ=pi, we have W0(∞) = ∞ + pi*I
Let z = ∞*cis(θ), r = ln(z) We want to show w = W_{0}(z) = ln(z) = r + θ*I By checking for roundtrip, i.e. w * e^w = z w = r + θ*I ≈ r * cis(θ/r) // polar form e^w = e^(r + θ*I) = e^r * cis(θ) w * e^w ≈ r*e^r * cis(θ/r + θ) z → ∞ > r → ∞ too w * e^w = ∞ * cis(0 + θ) = z From definition of kth branch of W k = (w + ln(w)  ln(z)) / (2*pi*i) 2*pi*k = im(w + ln(w)  ln(z)) = θ + atan(θ/r)  θ = 0 > k = 0 > W_{0}(∞*cis(θ)) = ln(∞*cis(θ)) = ∞ + θ*I Again, from definition of kth branch: > W_{k}(∞*cis(θ)) = ln(∞*cis(θ)) + 2*k*pi*I 

01172024, 11:34 AM
(This post was last modified: 01172024 11:42 AM by Gil.)
Post: #13




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
"w = r + pi*I ≈ r * cis(pi/r) // polar form
e^w = e^(r*cos(pi/r)) * cis(r*sin(pi/r)) w * e^w = r*e^(r*cos(pi/r)) * cis(pi/r + r*sin(pi/r))" How to you go from line 2 to 3 w = r * cis(pi/r) w= r * cos (pi/r) + I*r*sin(pi/r) e^w = e^(r*cos(pi/r)) * e^(I*r*sin(pi/r)) =e^(r*cos(pi/r)) * cis(r*sin(pi/r)) With w≈ r * cis(pi/r) w*e^w=r * cis(pi/r) × e^(r*cos(pi/r)) * cis(r*sin(pi/r)) =? r*e^(r*cos(pi/r)) * cis(pi/r + r*sin(pi/r)) cis (A)* cis (B) =? +cis (A+B) 

01172024, 12:04 PM
(This post was last modified: 01172024 12:05 PM by Gil.)
Post: #14




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
I am not so familiar with this notation.
I discovered it thanks you, Albert. And checked that cis (A+B) = cis A × cis B. By the way, with your software, may I kindly ask you if it were possible to check, for Wk=0, Wk=±1, Wk=±2, W±3 and arguments x = ± infinity +i*20, 20±I*inf and ±inf±I×inf, the given outputs of Wk(x)? Thanks again for your painstaking. Your insights are always most previous, for me and, certainly, beyond the HP community. 

01172024, 01:51 PM
Post: #15




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
Some "guessed" results for input z =a+ib, with a = INF:
W3(INF +i20) = infi(5pi) W3(INF  i20) = infi(7pi) W2(INF +i20) = infi(3pi) W2(INF  i20) = infi(5pi) W1(INF +i20) = infi(pi) W1(INF  i20) = infi(3pi) W0(INF +i20) = inf+i(pi) W0(INF  i20) = infi(pi) W1(INF + i20) = inf+i(5pi) W1(INF  i20) = inf+i(3pi) W2(INF + i20) = inf+i(5pi) W2(INF  i20) = inf+i(pi) W3(INF + i20) = inf+i(7pi) W3(INF  i20) = inf+i(5pi) Wk(INF, +ib, b>0) = inf+i(2k+1)(pi) Wk(INF, +ib, b<0) = inf+i(2k+12)(pi) Wk(INF, +ib, b>0) = inf+i(2k+1)(pi) Wk(INF, +ib, b<0) = inf+i(2k1)(pi) Conclusion Wk(INF, +ib, b≠0) = inf+i(2k+sign(b))(pi) & we may check Wk(INF, +ib, b=0) = inf+i(2k+1)(pi) Summary Wk(INF, +ib) =inf+i(pi) *(2k+1, if b>=0; 2k1, if b<0) 

01172024, 02:01 PM
Post: #16




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
Some "guessed" results for input z =a+ib, with a = INF:
W3(INF +i20) = infi(5pi) W3(INF  i20) = infi(7pi) W2(INF +i20) = infi(3pi) W2(INF  i20) = infi(5pi) W1(INF +i20) = infi(pi) W1(INF  i20) = infi(3pi) W0(INF +i20) = inf+i(pi) W0(INF  i20) = infi(pi) W1(INF + i20) = inf+i(5pi) W1(INF  i20) = inf+i(3pi) W2(INF + i20) = inf+i(5pi) W2(INF  i20) = inf+i(pi) W3(INF + i20) = inf+i(7pi) W3(INF  i20) = inf+i(5pi) Wk(INF, +ib, b>0) = inf+i(2k+1)(pi) Wk(INF, +ib, b<0) = inf+i(2k+12)(pi) Wk(INF, +ib, b>0) = inf+i(2k+1)(pi) Wk(INF, +ib, b<0) = inf+i(2k1)(pi) Conclusion Wk(INF, +ib, b≠0) = inf+i(2k+sign(b))(pi) & we may check Wk(INF, +ib, b=0) = inf+i(2k+1)(pi) Summary Wk(INF, +ib) =inf+i(pi) *(2k+1, if b>=0; 2k1, if b<0) 

01172024, 02:39 PM
(This post was last modified: 01182024 12:11 AM by Gil.)
Post: #17




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
Now Wk(inf +ib)
W3(INF +i20) = infi(6pi) W3(INF  i20) = infi(6pi) W3(INF  i20) = infi(5pi) W3(INF + i20) = infi(7pi) W3(INF  i0) = infi(5pi) W2(INF +i20) = infi(4pi) W2(INF  i20) = infi(4pi) W1(INF +i20) = infi(2pi) W1(INF  i20) = infi(2pi) W0(INF +i20) = infi(0pi) W0(INF  i20) = infi(0pi) W1(INF +i20) = inf+i(2pi) W1(INF  i20) = inf+i(2pi) W2(INF +i20) = inf+i(4pi) W2(INF  i20) = inf+i(4pi) W2(INF +i0) = inf+i(4pi) Conclusion & summary for Wk(inf+ib) Wk(INF +ib) = inf+2ik z=a+ib, a=±infinity, b≠±infinity Wk(±infinity+ib) =+inf + i × (2k, if a>0; (else 2k+1, if b>=0; 2k1, else)) 

01172024, 02:44 PM
Post: #18




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
That seems to make sense and should be acceptable for my HP50G result.


01172024, 03:07 PM
(This post was last modified: 01172024 03:40 PM by Gil.)
Post: #19




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
From previous post:
A) z=a+ib, a=±infinity, b≠±infinity Wk(±infinity+ib) =+inf + i × (2k, if a>0; (else 2k+1, if b>=0; 2k1, else)) And to complete Cases z=a≠±inf + i*INF, i*(INF) W3(20, i*INF) = inf  i(5.5pi) W3(20, i*INF) = inf  i(6.5pi) W2(20, i*INF) = inf  i(3.5pi) W2(20, i*INF) = inf  i(4.5pi) W1(20, i*INF) = inf  i(1.5pi) W1(20, i*INF) = inf  i(2.5pi) W0(20, i*INF) = inf +i(0.5pi) W0(20, i*INF) = inf  i(0.5pi) W1(20, i*INF) = inf +i(2.5pi) W1(20, i*INF) = inf+i(1.5pi) B) Wk(a+ib,a≠±infinity, b=±in) = inf+i×(2k+0.5, if b>0 ; 2k0.5, else) Cases z=a+ib, a=±inf, b=±inf W3(inf, i×inf) = inf  i(6pi) W3(inf, i×inf) =inf  i(6pi) W3(inf, i×inf) = inf  i(5pi) W3(inf, i×inf) =inf  i(7pi) C) Conclusion Wk(a=+inf, b=±inf) = inf + i(2k) Wk(a=inf) = inf +i(2k+1, if b=inf; 2k1, if b=inf) Summary of A & C Wk(a=+inf, b=free) = inf + i(2k) Wk(a=inf) = inf +i(2k+1, if b>≠0; 2k1, else And case B repeated B) Wk(a+ib,a≠±infinity, b=±in) = inf+i×(2k+0.5, if b>0 ; 2k0.5, else) 

01172024, 06:34 PM
(This post was last modified: 01192024 12:10 AM by Albert Chan.)
Post: #20




RE: Lambert function and Wolfram or "±infinity+i×K=±infinity" ?
I have extended proof for W0(∞) = ∞ + pi*I, for generalized case
(01172024 02:06 AM)Albert Chan Wrote: ... (01172024 03:07 PM)Gil Wrote: Cases z=a+ib, a=±inf, b=±inf Unless z is in polar form, we don't know θ. (∞/∞ is indeterminate form) mpmath, with slope=nan, θ = atan2(b,a) = nan p2> log(mpc(inf,inf)) (+inf + nanj) Comment: Technically we do know θ. Above example, 0 ≤ θ ≤ pi/2. IEEE 7542008 standard pick the midpoint for θ, instead of nan. lua> I.log(I.new(inf,inf)) (inf+0.7853981633974483*I) p3> from cmath import * p3> log(complex(inf,inf)) (inf+0.7853981633974483j) 

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