For all positive integers $n$, $^n(-1) = -1$, thus I thought ${^{\infty}}(-1)$ could be $\displaystyle \lim_{n \to \infty} {^n}(-1) = -1$.

But $\dfrac{W(-\ln z)}{-\ln z}$, analytic continuation of infinite power tower, gives some imaginary number.

Which one is correct for this tetration?

  • $\begingroup$ Maybe the limit does not exist. $\endgroup$
    – NoChance
    Feb 16, 2023 at 22:02

2 Answers 2


Lambert-W gives the real value $-1$ when we use the branch at $k=1$ (in some software at $k=-1$).


  1. W|A gives for exp(-productlog(1,-log(-1))) the result: -1
  2. An implementation in Pari/GP gives for exp(-LambertW(-log(-1),-1)) the result -1.000000000000000 + 3.300 E-213*I
    (the spurious imaginary part is due to internal float precision calculations and no default rounding/truncation is performed)

Your value is given when the default branch of the Lambert-W is chosen.

Using a Pari/GP implementation of the Lambert-W I find the following exemplaric list (from an infinite number of solutions) :

$$\small \begin{array} {rrl} k & \exp(-\operatorname{LambertW}(-\log(-1),k)) \\ \hline -9 & 17.98372743-0.9201621357*I \\ -8 & 15.98243823-0.8826775326*I \\ -7 & 13.98089425-0.8401770770*I \\ -6 & 11.97900898-0.7911056923*I \\ -5 & 9.976653691-0.7330485254*I \\ -4 & 7.973635250-0.6619491593*I \\ -3 & 5.969690012-0.5701679520*I \\ -2 & 3.964786505-0.4404083501*I \\ -1 & 1.964999038-0.2169439255*I \\ 0 & 0.2660365993+0.2942900219*I & *** \\ 1 & -1.000000000+3.308E-213*I \\ 2 & -2.962792822-0.3479062027*I \\ 3 & -4.967312786-0.5118935159*I \\ 4 & -6.971795660-0.6193684989*I \\ 5 & -8.975242573-0.6994852445*I \\ 6 & -10.97790036-0.7634014098*I \\ 7 & -12.98000110-0.8165873435*I \\ 8 & -14.98170272-0.8621368113*I \\ 9 & -16.98311045-0.9019717088*I \end{array} $$ The marked entry (with positive or negative imaginary component) should be the one that you've got yourself, correct?

  • 1
    $\begingroup$ Furthermore, it seems $\dfrac{W_k(-(\operatorname{Log}(-1)+2k\pi i))}{-(\operatorname{Log}(-1)+2k\pi i)} = \dfrac{W_{-k}(2k\pi i + \operatorname{Log}(-1))}{2k\pi i + \operatorname{Log}(-1)} = -1$, for all positive integers k, according to WolframAlpha. If this is right, it depends branch of both: logarithm and Lambert-W. $\endgroup$
    – user404273
    Feb 19, 2023 at 23:58
  • $\begingroup$ Yes, it's that value I got. P.S ) You used $\log(-1)$ instead of $-\log(-1)$ $\endgroup$
    – user404273
    Feb 21, 2023 at 4:57
  • $\begingroup$ @cplex - upps, corrected (note the sign-change in the imaginary component; don't know though whether this implementation in Pari/GP matches that in W|A) $\endgroup$ Feb 21, 2023 at 9:38
  • $\begingroup$ According to this, its implementation is experimental stage. So We may have to implement it manually. pari.math.u-bordeaux.fr/dochtml/html-stable/… $\endgroup$
    – user404273
    Feb 21, 2023 at 9:47
  • $\begingroup$ @cplex - hmm, in this case it is a user-implementation (years before the official one in Pari/GP - surprised that they write: "experimental") The implementation has been done by a user of the tetration-forum. So far I seem never to have encountered an error (however, didn't run any serious testsuite myself). If you like I can send you the source or you look at it in the forum (math.eretrandre.org/tetrationforum/index.php) look for user "mike3" (no more active) and "lambertw" and "branch". $\endgroup$ Feb 21, 2023 at 11:11

If we think of the infinite power tower as a limit over a real height, I think the -1 case is comparable to the limit of $\cos 2 \pi n$. For integer $n$, it is always 1, but the limit over real $n$ doesn't exist.

If you define a crude version of fractional tetrarion -- eg $^{2.5}(x)=x^{x^\sqrt x}$, then for $0<x\le e^{1/e}$, the limit of tetrarion over the reals equals the one over the integers.

On the other hand, this breaks down for $-1$. To summarize, I think the limit existing is tied to some definition of fractional tetrarion. Even though I don't think that's a well established thing, it's hard to imagine it being "nice" for negative numbers.


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