lambertw, all branches
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04-07-2023, 01:24 PM
Post: #1
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lambertw, all branches
I am starting a new thread, to discuss issues of solving W(a, k=0), extended to complex numbers.
The thread is based on this post, which supposed guaranteed correct branch, if f=0 (04-02-2023 11:12 PM)Albert Chan Wrote:(03-26-2023 06:43 PM)Albert Chan Wrote: Because solving for y = e^W(a) is easier, W code is not recommended. |
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04-07-2023, 02:47 PM
Post: #2
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RE: lambertw, all branches
f = x + ln(x) - ln(a) - 2*k*pi*I
Solve for f=0 is easy if x is big. For |k| > 1, we can solve directly, since x imaginery part ≈ 2*k*pi (see OP example) 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) In fact, I flip all negative k's. So now, k is non-negative. (04-02-2023 11:12 PM)Albert Chan Wrote: W code is perfect for other branches! (expW code for W branch 0, -1) Again, for now on, we assumed k >= 0 f=0 may still have problem converging if k = 0 or 1, especially for small x It does not matter if we have a really good guess. It is why expW code is needed. It handle this issue nicely. (*) expW code also handled accuracy issue elegantly, when a is close to branch point, -1/e Problem is expW code lost phase information, y = e^(x + 2*pi*k*I) = e^x expW can only get W(a, ±1) with the smaller imaginery part. expW code can get this: >>> lambertw(-1-.1j, 1) (-0.373678737258224 + 1.41169809383182j) But not this: >>> lambertw(-1+.1j, 1) (-2.04424331708213 + 7.48779784363809j) Note that negative a imag part corresponded to smaller W(a, k=1) imag part. It can be deduced from f f = x + ln(x) - ln(a) - 2*k*pi*I = 0 x.imag + phase(x) = 2*k*pi + phase(a) z.imag = abs(z) * sin(phase(z)) a.imag < 0 ⇒ phase(a) < 0 ⇒ RHS smaller ⇒ x.imag smaller (*) Based from tests, solve for y*ln(y)=a is better than x*e^x=a, even with help of Halley's method. mpmath.lambertw make it work only because it throw in extra precisions to compensate. And, it required a lot of code to get a good starting guess. |
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04-07-2023, 06:04 PM
Post: #3
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RE: lambertw, all branches
Side observation. Wouldn't be better to format such posts a bit more and post them in the Article forum? I think they can be more valuable over time if they can get a bit more structured.
Wikis are great, Contribute :) |
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04-07-2023, 07:54 PM
Post: #4
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RE: lambertw, all branches
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. 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 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) |
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04-07-2023, 08:40 PM
(This post was last modified: 04-14-2023 04:40 PM by Albert Chan.)
Post: #5
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RE: lambertw, all branches
I am keeping code simple. Edge cases, example W0(0) = 0, are not implemented.
Code: r, err = 0.36787944117144233, -1.2428753672788363E-17 W01 is solving W branch 0 or an ambiguous ±1 branch (with the small imag part) The code is solving for y*ln(y) = a --> y = e^W(a) --> W(a) = a/y We can "pull out" conjugate: W01(conj(a),k) = conj(W01(a,k)) lua> I.W(3+4*I, 5, true) (0+31.41592653589793*I) (-1.815554248884793+30.714634540472343*I) (-1.8170058914565277+30.71333413897562*I) (-1.8170058918466272+30.713334137004896*I) (-1.8170058918466274+30.713334137004896*I) lua> I.W(3+4*I, -5, true) (-0-31.41592653589793*I) (-1.7565805546696363-28.861921295458465*I) (-1.7547636407605829-28.85710103800315*I) (-1.7547636279545493-28.85710104741841*I) (-1.7547636279545493-28.85710104741841*I) lua> I.W(-0.3,-1,true) (-1.7824743623491122+0*I) W01 (-1.7813374873395937+0*I) W01 (-1.7813370234217047+0*I) W01 (-1.781337023421628+0*I) W01 lua> I.W(-0.03,-1,true) (-3.506557897319982-0*I) (-5.261733759227882-0*I) (-5.144793893666586-0*I) (-5.1444827237861634-0*I) (-5.144482721515682-0*I) (-5.144482721515682-0*I) lua> I.W(-.1+1e-100*I, -1, true) (-2.3025850929940455-1e-099*I) (-3.7769077194622533+1.2592576024726166e-099*I) (-3.579124176402325-8.545840728943592*I) (-3.7689826579869887-0.8337217972782689*I) (-3.5535464061313635-0.025256342802877585*I) (-3.5771468383315437+6.535012002211599e-005*I) (-3.5770902410326704-8.721224618630954*I) (-3.7811926340275566-0.8444302692850991*I) (-3.5536175209072614-0.02647513850865879*I) (-3.5771431199160215+6.824715919089008e-005*I) (-3.5770875001635236+0*I) W01 (-3.5771520641833883-1.3880349425219542e-099*I) W01 (-3.5771520639572962-1.3880252213229777e-099*I) W01 (-3.5771520639572976-1.388025221322979e-099*I) W01 W code f=0 have trouble converging. Switched to W01 help a lot. W01 use W last guess, but dropped the unreliable imaginary part. |
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04-08-2023, 03:21 PM
(This post was last modified: 04-15-2023 01:56 PM by Albert Chan.)
Post: #6
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RE: lambertw, all branches
(04-07-2023 07:54 PM)Albert Chan Wrote: It is just here, "equivalent" may mean a difference of +0*I vs -0*I Just to show how sensitive this ±0*I business is ... lua> I.W(-1e-9,0,true) (-1.0000000005000001e-009+0*I) (-1.000000001e-009+0*I) (-1.0000000010000002e-009+0*I) lua> I.W(I.conj(-1e-9),0,true) (-1.0000000005000001e-009+0*I) (-1.000000001e-009+6.283185316604365e-009*I) (3.1450880362038825e-008-4.729032266472273e-010*I) (-7.848729544901978e-008-9.717506161329017e-008*I) ... 2nd should be complex conjugate of first. W(-1E-9 ± 0*I) ≈ -1.000000001E-9 ± 0*I Imaginary part sign of (a, W0(a)) should match. At first, I thought it is also a case of bad W guess. To keep x imag part intact, x+1 should be written as x-(-1) I added +0 to correction too, just to be sure. Technically, it is not needed. Signed zero: (±0) - 0 = (±0) , (±0) + 0 = 0 Code: < function complex.log1p(x) local y = x+1; return I.log(y) - (y-1-x)/y end This get W0 guess phase correct, however iterations still diverges. It is interesting Newton's 1st iteration essentially undo log1p patches. lua> I.W(I.conj(-1e-9),0,true) (-1.0000000005000001e-009-0*I) (-1.000000001e-009+0*I) (-1.0000000010000002e-009+6.283185319745956e-009*I) (3.145088037733907e-008-4.729032324603788e-010*I) ... The problem is Newton correction: 1/slope = x/(x+1). We fix it the same way: Code: < h = x/(x+1) * s(x) lua> I.W(I.conj(-1e-9),0,true) (-1.0000000005000001e-009-0*I) (-1.000000001e-009-0*I) (-1.0000000010000002e-009-0*I) I had applied above 2 patches to previous post. |
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04-08-2023, 05:54 PM
(This post was last modified: 04-09-2023 08:31 PM by Albert Chan.)
Post: #7
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RE: lambertw, all branches
(04-08-2023 03:21 PM)Albert Chan Wrote: The problem is Newton correction: 1/slope = x/(x+1). We fix it the same way: Patch is wrong. It get the correct result by accident. The problem is (±0) + 0 = 0, losing sign information. If h imag part = -0, x = x - h may lose x sign information. We could do x = x - (h + 0), but that is ugly. I am sick of this ±0*I, and simply flip k to avoid it. I.W(I.conj(-1e-9), 0) --> I.conj(I.W(-1e-9, -0)) We flip k to ensure x imag part always non-negative. Losing sign information now becomes an advantage. x = x - h will not generate x with imag part of -0. Only update is complex.W code, the rest remains unaffected. I am leaving 1/slope = x/(x-(-1)), but x/(x+1) will work as well. Code: function complex.W(a, k, verbal, x) |
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04-09-2023, 03:59 AM
(This post was last modified: 04-22-2023 08:29 PM by Albert Chan.)
Post: #8
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RE: lambertw, all branches
complex.W code version 2, with many improvements.
Code: r, err = 0.36787944117144233, -1.2428753672788363E-17 lua> I.W(3+4*I, 5, true) (0+32.34322175389954*I) (-1.8166634488635778+30.71625713741827*I) (-1.8170058961963056+30.713334138441088*I) (-1.8170058918466274+30.713334137004896*I) true lua> I.W(3+4*I, -5, true) (0-30.48863131789632*I) (-1.754507767775116-28.860288696238875*I) (-1.7547636338945982-28.857101048916533*I) (-1.7547636279545493-28.85710104741841*I) true lua> I.W(-0.3678, 0, true) (-0.9793607152032429+0*I) (-0.9793607149578304+0*I) (-0.9793607149578306+0*I) true lua> I.W(-.1+1e-100*I,-1,true) (-2.3025850929940455-1e-099*I) (-3.7769077194622533-5.084465616380436e-100*I) (-3.579124176402325-5.180347740277737e-100*I) (-3.5771522746958055-5.181454351722726e-100*I) (-3.5771520639573-5.181454470168089e-100*I) (-3.577152063957297-5.18145447016809e-100*I) (-3.577152063957297-1.3880252213229779e-099*I) true lua> I.W(-.1+1e-100*I,0,true) (-0.10536051565782631+1.1111111111111111e-100*I) (-0.11161906744205384+1.1848854610650868e-100*I) (-0.11183232962808998+1.1874337731206376e-100*I) (-0.11183255915869776+1.1874365171431562e-100*I) (-0.11183255915896297+1.1874365171463266e-100*I) (-0.11183255915896298+1.259138243719729e-100*I) true lua> I.W(-.1+1e-100*I,1,true) (0+9.42477796076938*I) (-4.330508133087425+7.394500450320804*I) (-4.448949088733658+7.307107936743056*I) (-4.449098178536345+7.307060789152774*I) (-4.44909817870089+7.3070607892176085*I) true lua> x = 5*I lua> a = x * I.exp(x) lua> x + x:log() - a:log() (0+6.283185307179586*I) lua> I.W(a, 1) (3.1321400622130957e-068+5*I) false |
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04-09-2023, 04:36 PM
(This post was last modified: 04-10-2023 10:31 PM by Albert Chan.)
Post: #9
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RE: lambertw, all branches
(04-09-2023 03:59 AM)Albert Chan Wrote: [*] if |Im(x)| < pi, we are on W branch 0, or branch ±1 with small imag part. (04-07-2023 02:47 PM)Albert Chan Wrote: f = x + ln(x) - ln(a) - 2*k*pi*I = 0 W code forced k non-negative, to guarantee x.imag ≥ 0 Since x.imag and phase(x) have same sign, we have: k=0: (x.imag < phase(a) < pi) // for k=0, we forced (0 ≤ phase(a) ≤ pi) k=1: (x.imag < pi) ⇒ (LHS < pi + pi) ⇒ (phase(a) < 0), thus W branch ±1 with small imag part. Quote:To be safe, I lower the limit, from pi to 3. Experiments suggested |Im(x)| < pi is a hard limit. (why?) lua> a = -100-92*I lua> x = I.W(a, 1) lua> x -- abs imag part < pi (3.382739150014567+3.1375382604161515*I) lua> x/(x+1) * (I.log(a/x) + 1) -- y = e^x Newton step (3.382739150014567+3.1375382604161515*I) If |Im(x)| > pi, y = e^W Newton step will not improve, but overshoot toward W branch 0. lua> a = -100-93*I lua> x = I.W(a, 1) lua> x -- abs imag part > pi (3.386390672646167+3.1426530733361844*I) lua> x/(x+1) * (I.log(a/x) + 1) -- y = e^x Newton step (4.064553974086096-2.1939787611722497*I) Hard pi limit puzzle solved! ln(a/x) ≈ ln(e^x) = Re(x) + i * smod(Im(x), 2*pi) // signed mod to limit arg(z) within ±pi If |Im(x)| < pi, ln(a/x) ≈ ln(e^x) = x → x/(x+1) * (ln(a/x) + 1) ≈ x/(x+1) * (x+1) = x // Newton's step work as expected If pi < |Im(x)| < 3*pi, ln(a/x) ≈ x ± 2*pi*i // sign opposite of Im(x) → x/(x+1) * (ln(a/x) + 1) ≈ x + x/(x+1)*(±2*pi*i) // Newton's step overshoot toward W branch 0 |
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04-10-2023, 04:44 PM
Post: #10
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RE: lambertw, all branches
(04-09-2023 03:59 AM)Albert Chan Wrote: [*] if |Im(x)| < pi, we are on W branch 0, or branch ±1 with small imag part. Another way to look at this, if above is true, we can simplify f: < f = x - (ln(a) - ln(x)) - 2*k*pi*i > f = x - ln(a/x) f general form has issues with catastrophic cancellation, simplifed f is much better. The problem is we don't know |Im(x)| < pi, until we solve it. Also, f simplifed form have multiple roots. A good guess is needed to get correct branch. For f general form, with 1 root, guess good enough for convergence is all that is needed. We use f general form for *good* guess, correct with using simplified f (if applicable) lua> a = 1e-100-1e-10*I lua> I.W(a, 0, true) (1e-100-1e-010*I) (1.000000082740371e-020-1e-010*I) (1.000000082740371e-020-1e-010*I) -- general f=0 solution (1.0000000000000002e-020-1e-010*I) true Here, initial guess log1p(a) ≈ a is not good (I.log code need work ...) But, it doesn't matter. Guess is good enough for general f to converge. f simplified form, with 1 Newton step, we get W(a) ≈ a - a*a |
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04-10-2023, 06:47 PM
Post: #11
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RE: lambertw, all branches
(04-10-2023 04:44 PM)Albert Chan Wrote: Here, initial guess log1p(a) ≈ a is not good (I.log code need work ...) As expected, better complex.log make little difference. I considered Kahan's clog(z), but Python's cmath.log code is simpler. Code: function complex.log(z) lua> a = 1e-100-1e-10*I lua> I.W(a, 0, true) (5.0000000000000005e-021-1e-010*I) -- log1p(a) ≈ a - a^2/2 (1.0000000413701856e-020-1e-010*I) (1.0000000413701856e-020-1e-010*I) (1.0000000000000001e-020-1e-010*I) true |
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04-13-2023, 03:03 PM
(This post was last modified: 04-17-2023 10:16 PM by Albert Chan.)
Post: #12
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RE: lambertw, all branches
Here is equivalent complex.W version 2, adapted to machine without signed zero.
Without signed zero, this trick is out: W(a, k) = conj(W(conj(a), -k) Instead, let W(a, k) = x = re(z) + s*im(z)*I , such that s = ±1, arg(z) ≥ 0 s = sign(arg(x)) = ±1 |x| = |z| arg(x) = s * arg(z) lua> I.W(-1e-100, -1, true) (-230.25850929940458-0*I) (-235.72143712473638-0*I) (-235.72115887568603-0*I) (-235.72115887568532-0*I) (-235.72115887568535-0*I) true Above example, s=-1, all x iterations, arg(x) = s * pi Almost identical, for efficiency: arg(x) ≠ bad = s * -pi In other words, iteration overshoot allowed, where arg(z) ≥ 0 does not hold. But, not to the extreme opposite direction, i.e. arg(z) ≠ -pi We need to add arg(x) edge case, to make W (without signed zero) work the same as I.W Code: h = arg(x) I use Python mpmath, which does not have signed zero. Code: from mpmath import * # no signed zero support >>> W(-1e-100, -1, True) -230.258509299405 (-235.721437124736 + 0.0j) (-235.721158875686 + 0.0j) (-235.721158875685 + 0.0j) (mpc(real='-235.72115887568535', imag='0.0'), True) |
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04-13-2023, 04:14 PM
Post: #13
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RE: lambertw, all branches
Nice. Since I see the lambert as usefull for infinite iteration, do you have any idea how to use the lambert (with "jumping" into branches?) for following type of problem
a) z = x + i *y b) f(z) = sqrt(x*y) + i* (x+y)/2 c) where finally f(f(f..(z)))) = a + i*a (a=AGM(x,y). AGM arithemic geometric mean) HP71B 4TH/ASM/Multimod, HP41CV/X/Y & Nov64d, PILBOX, HP-IL 821.62A & 64A & 66A, Deb11 64b-PC & PI2 3 4 w/ ILPER, VIDEO80, V41 & EMU71, DM41X |
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04-19-2023, 01:30 AM
(This post was last modified: 02-01-2024 03:33 PM by Albert Chan.)
Post: #14
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RE: lambertw, all branches
(04-07-2023 02:47 PM)Albert Chan Wrote: (*) Based from tests, solve for y*ln(y)=a is better than x*e^x=a, even with help of Halley's method. Not when W guess is extremely close to true root. We can use Newton's method, f = x*e^x - a, for finishing touches Since we are now using W loops for guess, we might as well make it slightly worse. Instead of log(abs(x)/A), make it log(abs(x)) - log(A), literally, f = x + ln(x) - lnk(a) Code is easier to read, and cheaper to compute. (last term is a constant) Instead of comparing *both* complex parts converged, we test for relative errors. This solved the problem if any complex parts is bouncing around zero. For symmetry, I used relative error test for Y loop as well. With relative error test, it should not get stuck in infinte loops → loop limits removed complex.W code, version 3 Code: r, err = 0.36787944117144233, -1.2428753672788363E-17 (04-09-2023 03:59 AM)Albert Chan Wrote: lua> x = 5*I complex.W version 2 was not able to converge after 40 loops. Redo with version 3: lua> a (4.794621373315692+1.4183109273161312*I) lua> I.W(a, 1, true) (0+6.570796326794897*I) (-0.033367069043767406+4.994921913968373*I) (1.97521636071743e-005+5.000010560789135*I) (3.753082249819371e-012+5.000000000009095*I) (1.0141772755203484e-016+5*I) (-6.474631941441244e-017+5*I) We need more precisions to get real part right. For reference, with hexfloat of a: W1(a) ≈ 4.51974e-018 + 5.00000 I Udpate 1/19/24: added |a| = ∞ or 0 edge cases lua> I.W(inf), I.W(-inf) (inf+0*I) (inf+3.141592653589793*I) lua> z = 0 lua> I.W(-z,-1), I.W(z,-1) (-inf-0*I) (-inf-3.141592653589793*I) Update 1/29/24 < h = (a + r + err) > h = I(a:real() + r + err, a:imag()) -- keep -0*I sign Update 2/01/24 lyuka branch relative limit reduced, from 1e-12 to 1e-9 Although convergence near -1/e is hard, lyuka (improved) formula is very accurate there. Further away, guess formula is not as good, but convergence O(rel_err^2) constant < 1 I also removed lua implementation of log1p(x). it had moved to C code |
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04-23-2023, 02:49 PM
(This post was last modified: 04-25-2023 10:32 AM by Albert Chan.)
Post: #15
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RE: lambertw, all branches
Previous post, complex.W version 3 old guess code (OK, but now replaced)
Note: k flipped to non-negative, to force Im(x) non-negative too, to avoid -0*I issue. Code: local A, T = I.abs(a), (2*k*pi+I.arg(a))*I k=0 and Im(a)≥+0, W0(a) guess of log1p(a) is good, except around log1p singularity, a ≈ -1 If a is real, inside (-1,-1/e), iterations may get stuck on real line, or take a while to get out. That's why I had added a "hole" around -1, with purely complex guess. x + ln(x) = ln(|a|) + T Im(x + ln(x)) = T/i = [pi/2, pi] a ≈ -1 → guess x = T/2 = [pi/4, pi/2]*i We could hide log1p singularity inside branch region, for simpler and better W0 guess Code: < if k==0 then x = I.abs(a+1)<.6 and T/2 or I.log1p(a) end New guess is better for small a. W0(ε) = ε - ε² + 3/2*ε³ - ... log(1+2ε)/2 = ε - ε² + 4/3*ε³ - ... New guess is worse for huge a. (factor of 2) But, it doesn't matter. x ≫ ln(x), curve is very flat. Cost is at most 1 extra iteration. Update: perhaps we can adjust constant 2, to have best of both worlds. Constant is 2 when a is small, 1 when a is huge (getting back log1p(a)) Code: < if k==0 then x = I.log1p(2*a)/2 end k=1 and Im(a)≤-0, W1(a) guess of log(-a) is good, except a ≈ -1 (log(1)=0, terrible W guess) If a is real < -1/e, iterations may get stuck on real line, or take a while to get out. My solution was only use log(-a) for a ≈ 0, and purely imaginary T as default W guess. T/i = 2*pi + arg(a) re(a)≤0: T/i = 2*pi .+ (-pi, -pi/2] = (pi, 3/2*pi] re(a)≥0: T/i = 2*pi .+ [-pi/2, -0.] = [3/2*pi, 2*pi] Another approach is to add reasonable complex part to log(-a). if Re(h = a+1/e) < 0, W is complex. Code: < if small_imag and A<.6 then x = I.log(-a) end New guess, Re(a) < -1/e, which covered a ≈ -1: arg(a) = (-pi, -pi/2] im(log(-a) + T/2) = (arg(a)+pi) + (pi+arg(a)/2) = (0, pi/2] + (pi/2, 3/4*pi] = (pi/2, 5/4*pi] I've patched new W guess to complex.W, version 3. Better W guess welcome ... Update: it seems to fit better if we again treat a ≈ 0 as the hole. Code: < if small_imag then x = I.log(-a) + (I.real(h)<0 and T/2 or 0) end Outside the hole, with Re(a) > 0, we have same formula, but different domain. arg(a) = [-pi/2, -0] im(log(-a) + T/2) = (arg(a)+pi) + (pi+arg(a)/2) = [pi/2, pi] + [3/4*pi, pi] = [5/4*pi, 2*pi] Added T/2 match asymptotes imaginary part better. lua> I.W(1e300-I, 1, true) (690.7755278982137+6.283185307179586*I) -- 2*pi (684.2471217875482+6.274017030794511*I) (684.2471666555265+6.274016338969102*I) (684.2471666555265+6.274016338969102*I) -- ≈ 2*pi lua> I.W(1-1e300*I, 1, true) (690.7755278982137+3.9269908169872414*I) -- 5/4*pi (684.2471409153251+4.705523410032926*I) (684.2471850188219+4.705512170223622*I) (684.2471850188219+4.705512170223621*I) -- ≈ 3/2*pi |
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04-23-2023, 04:40 PM
(This post was last modified: 02-01-2024 03:35 PM by Albert Chan.)
Post: #16
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RE: lambertw, all branches
complex.W version 3, adapted for system without signed zero.
This time I made a customized LN(x) to honor signed zero. Code: from mpmath import * # no signed zero support Trivia: W(a < -1/e) is complex, W0(a) = conj(W-1(a)) x + ln(x) = lnk(a) ln0(a) = ln(|a|) - pi*j + 2*pi*j = ln1(conj(a)) W0(a) = W1(conj(a)) = conj(W-1(a)) >>> a = -pi/2 >>> W(a, 0, True) (0.375191803678016 + 1.59508875325509j) (0.0128176527641888 + 1.58759064221342j) (-5.58573510823744e-5 + 1.57084664181356j) (6.00809277877961e-10 + 1.57079632603836j) (5.72587285769658e-17 + 1.5707963267949j) (1.30059124689582e-17 + 1.5707963267949j) >>> W(a, -1, True) (0.451582705289455 - 1.5707963267949j) (0.0200654830289762 - 1.59320451813763j) (-9.57300550001525e-5 - 1.57091376376939j) (1.39642133562001e-9 - 1.5707963231279j) (-3.14444301571107e-17 - 1.5707963267949j) (-5.01152163329281e-17 - 1.5707963267949j) Confirm x * exp(x) = a, both at the same time. (±pi/2*j) * exp(±pi/2*j) = (±pi/2*j) * (±j) = pi/2 * j² = -pi/2 Udpate 1/19/24: added |a| = ∞ or 0 edge cases p2> W(inf), W(-inf) ((+inf + 0.0j), (+inf + 3.14159265358979j)) p2> W(0, -1) (-inf - 3.14159265358979j) Default setup is 0 == +0, that is why we get W-1(+0) = -∞ - pi*I To get W-1(0) = -∞, we need special case for k=-1, 0 == −0 + 0*I Real part -0, Imag part +0 ... it is just too confusing. I am keeping 0 == +0, for consistency. ln(0) = ln(+0) = -∞ --> Wk(+0) = 0 if k==0 else -∞ + (2*k-sign(k))*pi*I Of course, we are free to patch it to make it pick what we wanted. But then, what is Wk(0) in general ? Is 0 still meant +0 + 0*I ? This is a weakness of not having signed zero. Update 2/01/24 lyuka branch relative limit reduced, from 1e-12 to 1e-9 (see post #14) |
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01-19-2024, 04:14 PM
Post: #17
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RE: lambertw, all branches
I have added w = Wk(a), for edge cases, |a| = ∞ or 0 (updated to post 14 and 16)
For |a| → ∞, we have Wk(a) = lnk(a) see https://www.hpmuseum.org/forum/thread-21...#pid182967 For |a| → 0, k=0 XCas> series(LambertW(x)) → x - x^2 + 3/2*x^3 - 8/3*x^4 + 125/24*x^5 For |a| → 0, k≠0, we use W branch k definition w + ln(w) = lnk(a) = ln(|a|) + (arg(a) + 2*k*pi)*I Let w = r + θ*I, matching complex parts --> (r + re(ln(r)) ≈ r) = ln(|a|) = -∞ --> (θ + sign(θ)*pi) = (arg(a) + 2*k*pi) --> sign(θ + sign(θ)*pi) = sign(arg(a) + 2*k*pi) --> Since k≠0, sign(θ) = sign(k) = ±1 Wk(a → 0) = a if k==0 else lnk(a) - sign(k)*pi*I This depends on meaning of arg(a) = atan2(im(a), re(a)) = atan2(±0, ±0) IEEE starndard assumed real part is (ε → 0), but imaginery part is truly 0. IEEE: arg(±0 ± 0*I) → arg(±ε ± 0*I) → ±0 or ±pi Technically, we don't really know arg(0), except its full range (-pi, pi] WA avoid all these, ∞ + (finite number, real or complex) = ∞ WA also don't have signed zero, ±0 ±0*I collapsed to plain 0 WA: Wk(0) = 0 if k==0 else -∞ |
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01-20-2024, 04:48 PM
Post: #18
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RE: lambertw, all branches
Newton's method convergence to prove previous post.
x * e^x = a // x = Wk(a) x + ln(x) = lnk(a) = c If c = ±∞ + θ*I, then (|x| → ∞) ⇒ (slope = 1+1/x ≈ 1) Iteration formula is practically equivalent to Newton's method. x = c - ln(x) x1 = c // guess x2 = c - ln(x1) x3 = c - ln(x2) // ... until x's converged if a = ∞*cis(θ), r = re(c) = ∞ c = r + θ*I ≈ |r| * cis(θ/r) ln(c) ≈ ln(|r|) + (θ/r)*I x2 ≈ c - ln(c) ≈ (r-ln(|r|)) + (θ-(θ/r))*I ≈ (r + θ*I) = x1 --> Wk(∞*cis(θ)) = lnk(∞*cis(θ)) if a = 0*cis(θ) and k≠0, r = -∞, sign(θ) = sign(k) c = r + θ*I ≈ |r| * cis(sign(k)*(pi - |θ/r|)) ln(c) ≈ ln(|r|) + (sign(k)*(pi - |θ/r|))*I ≈ ln(|r|) + sign(k)*pi*I x2 = c - ln(c) ≈ (r-ln(|r|)) + (θ-sign(k)*pi)*I ≈ r + θ2*I sign(θ2) = sign(θ) = sign(k) // 1-sided convergence (★) ln(x2) ≈ ln(|r|) + (sign(k)*(pi - |θ2/r|))*I ≈ ln(|r|) + sign(k)*pi*I x3 = c - ln(x2) ≈ c - ln(c) = x2 --> Wk≠0(0*cis(θ)) = lnk(0*cis(θ)) - sign(k)*pi*I (★) Without slope correction, convergence is 1-sided, not crossing discontinuity. lua> c = I(-99999*log(10), -pi) -- = ln-1(-10^-99999) lua> c (-230256.2067143116-3.141592653589793*I) lua> c - I.log(_) (-230268.55366222185-1.3643899976489848e-05*I) lua> c - I.log(_) (-230268.55371584298-5.925215873503475e-11*I) lua> c - I.log(_) (-230268.5537158432+0*I) -- W-1(-10^-99999) conjugate Note that final imaginery part should be -0 This is why I.W(a, k) flip to positive k, iterate, then flip back for negative k lua> c = c:conj() lua> c (-230256.2067143116+3.141592653589793*I) lua> c - I.log(_) (-230268.55366222185+1.3643899976489848e-05*I) lua> c - I.log(_) (-230268.55371584298+5.925215873503475e-11*I) lua> c - I.log(_) (-230268.5537158432+0*I) lua> _:conj() (-230268.5537158432-0*I) -- W-1(-10^-99999) |
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01-20-2024, 10:52 PM
(This post was last modified: 01-20-2024 11:49 PM by Gil.)
Post: #19
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RE: lambertw, all branches
I looked your case k=-1 & k=1, x=-.1 + i ×1E-99.
With my initial guess= 'LN(((-.1,1.E-100)+1./e)*(e-\v/2.-1.)+\v/(((-.1,1.E-100)+1./e)*2./e)+1./e)' = (-.11284787341,1.26802807724E-100), the built-in multisolver from HP50 runs endless. I looked in your programs, but could not understand which initial Xo formulae I should use for tiny imaginary part. I tried, from what I guessed from your code, as a guess xo=ln(-a), and it worked. When should I used that guess formula for my initial guess? For example, xo=ln(-a) failed to produce any answer for W0((1.E-100,-.0000000001)): a bad guess. I had to use instead the first formulae to get an answer for W0((1.E-100,-.0000000001)): (9.99905166733E-21,-1.00000000005E-10). And what do you mean by I.log(-a)? Is I.log(-x) a special function or is it i×ln(-x). Could you help me on that matter? Thanks in advance. |
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01-21-2024, 01:14 AM
(This post was last modified: 01-24-2024 07:27 PM by Albert Chan.)
Post: #20
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RE: lambertw, all branches
(01-20-2024 10:52 PM)Gil Wrote: I looked your case k=-1 & k=1, x=-.1 + i ×1E-99. x = Wk(a) Newton's method required a good guess to start. (iterations cannot cross discontinuity!) lyuka's e^W formula is only used when a ≈ -1/e, and (k=0 or k=±1 and x with small imag part) Above example probably too far away from -1/e (my code use lyuka if |a+1/e| < 0.25) Also, your formula is only good for k=0. For k=-1, you need to flip sqrt sign. lua> a, e = I(-0.1, 1e-100), exp(1) lua> I.log((a+1/e)*(e-sqrt(2)-1) + I.sqrt((2/e)*(a+1/e)) + 1/e) -- x guess for k=0 (-0.1128478734106899+1.268028077236754e-99*I) lua> I.log((a+1/e)*(e-sqrt(2)-1) − I.sqrt((2/e)*(a+1/e)) + 1/e) -- x guess for k=-1 (-5.225138594746179-9.75118016335884e-98*I) If x imag part is big, it just use that. -- local A, T = I.abs(a), I.new(0,2*k*pi+I.arg(a)) -- local x = T -- W rough guess lua> I.W(a, 1, true) (0+9.42477796076938*I) (-4.330508133087424+7.394500450320804*I) (-4.448949088733658+7.307107936743056*I) (-4.449098178536345+7.307060789152774*I) (-4.44909817870089+7.3070607892176085*I) (-4.44909817870089+7.3070607892176085*I) If k=-1 and x imag part is small, we wanted real guess when a is negative real. Also, x→-∞ if a→0, log(-a) fit the bill. But if a is too negative (left of -1/e), code added some imaginery part. Imag part is needed because a=-1 --> log(-a) = log(1) = 0, very bad x guess! -- if small_imag then x = I.log(-a) + (A<.5 and 0 or T/2) end lua> I.W(a, -1, true) (-2.3025850929940455-1e-99*I) (-3.776907719462253-5.084465616380437e-100*I) (-3.5791241764023245-5.180347740277738e-100*I) (-3.577152274695805-5.181454351722727e-100*I) (-3.5771520639572993-5.18145447016809e-100*I) (-3.577152063957297-1.388025221322979e-99*I) |
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