Tetration for complex bases
William Paulsen
Advances in Computational Mathematics Modelling in Science and Engineering ISSN 1019-7168 Adv Comput Math DOI 10.1007/s10444-018-9615-7
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Author's personal copy Adv Comput Math https://doi.org/10.1007/s10444-018-9615-7
Tetration for complex bases William Paulsen1
Received: 9 January 2018 / Accepted: 22 May 2018 © Springer Science+Business Media, LLC, part of Springer Nature 2018
Abstract In this paper we will consider the tetration, defined by the equation F (z + 1) = bF (z) in the complex plane with F (0) = 1, for the case where b is complex. A previous paper determined conditions for a unique solution the case where b is real and b > e1/e . In this paper we extend these results to find conditions which determine a unique solution for complex bases. We also develop iteration methods for numerically approximating the function F (z), both for bases inside and outside the Shell-Thron region. Keywords Tetration · Abel’s functional equation · Iteration Mathematics Subject Classification (2010) 26A18 · 30D05 · 39B12
1 Background Much research has already been done on the tetration function, whose name comes from tetra- (four) and iteration. Addition by a positive integer n can be thought of as repeated incrementing, multiplication by n is done by repeated addition, and
Communicated by: Aihui Zhou William Paulsen
[email protected] 1
Arkansas State University, Jonesboro, AR, USA
Author's personal copy W. Paulsen
exponentiation by n is repeated multiplication. So a fourth level operation would be repeated exponentiation using the initial base b: b
··
bb n
Although there are many different notations for this operation, we will use the iterated exponential notation bx
expnb (x) = bb
··
with n b’s.
The tetration function for base b can then be expressed as F (n) = expnb (1). Thus we see that F (0) = 1, F (1) = b, F (2) = bb , and so on. When b is complex, we must define a branch cut so that bz is uniquely defined. Once this branch cut is established, we can recursively define F (x) by F (x + 1) = bF (x) . Working backwards from this relation, we see that F (−1) = 0 and F (−2) must be undefined, even for complex b. In [1], Kneser showed that if b is real and b > e1/e , we can extend F (z) to the whole complex plane, minus a branch cut at z ≤ −2. This solution will be referred to as κb (z). Kneser’s solution begins by finding the fixed points of the exponential function bz , that is, points L for which bL = L. To find the principle fixed points, we first define the Wright ω function to be the solution to the initial value problem ω(z) , ω(1) = 1, 1 + ω(z) where the branch cuts are chosen to start at −1 ± π i, run parallel to the real axis in the negative direction [2]. The Wright ω function can in fact be defined in terms of the nth branch of the Lambert’s W function. (z) − π z . ω(z) = Wn (e ), where n = 2π ω (z) =
Note that with this definition, ω(a) = ω(a − ) whenever a is on either of the branch cuts. (In this context, a − means to approach a from below the branch cut rather than from the left. For example, ln(−1− ) = −π i, and ln(−1+ ) = π i.) We prefer a function which is continuous from above the branch cuts, so we define ω2 (z) = ω(z). Then ω2 will be identical to ω except at the branch cuts, where now ω2 (a) = ω2 (a + ) at both branch cuts. The principle fixed point of the exponential function bz can now be expressed by ω2 (ln(ln(b)) − π i) . − ln(b) A second fixed point can be found with a similar formula, L1 =
L2 =
ω2 (ln(ln(b)) + π i) . − ln(b)
In both formulas, we use the principle branch of the logarithm. If b > e1/e , or if b is not real, then we can see that (L1 ) > 0 and (L2 ) < 0.
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Although we have two fixed points for most bases b, there is a question as to whether or not these are attractive fixed points of the function bz . We say that the base b is in the Shell-Thron region if the sequence of values b
bb
{b, bb , bb , bb , . . . }
(1)
converge to a finite fixed point. Note that this can only happen if | ln L| < 1 for one of the fixed points. The Shell-Thron region is depicted in Fig. 1. For b = 1 within this region, the sequence converges to L1 if (b) ≥ 0, and converges to L2 if (b) < 0. We begin by solving the Schr¨oder equation σ (bz ) = sσ (z), where s is the derivative of bz at one of the fixed points. At the fixed point L1 , s1 = ln(L1 ) = L1 ln(b), and at the fixed point L2 , s2 = ln(L2 ) = L2 ln(b). For each fixed point, there is a unique solution to Schr¨oder’s equation which is analytic near the fixed point, and for which the derivative at the fixed point is 1. By equating coefficients of the power series, we find that s12 (1 + 2s1 ) s1 (z − L1 )3 (z − L1 )2 + 2 2L1 (1 − s1 ) 6L1 (1 − s1 )(1 − s12 ) s13 (1+6s1 +5s12 +6s13 ) s 4 (1+14s1+24s12 +45s13 +46s14 +26s15 +24s16 ) (z−L1 )4 + 1 (z−L1 )5 + 3 3 2 24L1 (1 − s1 )(1 − s1 )(1 − s1 ) 120L41 (1 − s1 )(1 − s12 )(1 − s13 )(1 − s14 ) s 5 (1+30s1 +89s12 +214s13 +374s14 +416s15 +511s16 +461s17 +330s18 +154s19 +120s110 ) (z−L1 )6 + 1 720L51 (1 − s1 )(1 − s12 )(1 − s13 )(1 − s14 )(1 − s15 ) +··· .
σ1 (z) = (z − L1 ) +
Fig. 1 The Shell-Thron region
Author's personal copy W. Paulsen s22 (1 + 2s2 ) s2 (z − L2 )3 (z − L2 )2 + 2 2L2 (1 − s2 ) 6L2 (1 − s2 )(1 − s22 ) s23 (1+6s2 +5s22 +6s23 ) s 4 (1+14s2 +24s22 +45s23 +46s24 +26s25 +24s26 ) (z−L2 )4 + 2 (z−L2 )5 + 3 3 2 24L2 (1 − s2 )(1 − s2 )(1 − s2 ) 120L42 (1 − s2 )(1 − s22 )(1 − s23 )(1 − s24 ) s 5 (1+30s2 +89s22 +214s23 +374s24 +416s25 +511s26 + 461s27 +330s28 + 154s29 + 120s210 ) (z−L2 )6 + 2 720L52 (1 − s2 )(1 − s22 )(1 − s23 )(1 − s24 )(1 − s25 ) +··· .
σ2 (z) = (z − L2 ) +
Unfortunately, when b is on the boundary of the Shell-Thron region, one of the series will probably have a zero radius of convergence, requiring us to treat these as asymptotic series. Even for other basis, it is unclear what the radius of convergence will be. However, we can easily extend the region of convergence. In the case where b is within the Shell-Thron region, and the series in Eq. 1 converges to L, then if z is in the basin of attraction for bz , so that z
bz
{z, bz , bb , bb , . . . } also converges to L, then σ (z) = lim s −n (expnb (z) − L), n→∞
where expnb (z) represents applying bz a total of n times. If L is one of the repulsive fixed points of bz , and if z is in the basin of attraction for logb (z), so that repetitively applying the principal value of logb to z will converge to L, then σ (z) = lim s n (lognb (z) − L). n→∞
Next, we let ψ(z) = ln(σ (z))/ ln(s), so that ψ(z) will solve Abel’s equation ψ(bz ) = ψ(z) + 1. We can compute, for j = 1 or 2,
ψj (z) = + + +
sj2 (1 + 5sj ) ln(z − Lj ) sj (z − Lj )2 + (z − Lj ) + ln(sj ) 2Lj ln(sj )(1 − sj ) 24L2j ln(sj )(1 − sj )(1 − sj2 ) sj4 (2 + sj + 3sj2 ) 24L3j ln(sj )(1 − sj )(1 − sj2 )(1 − sj3 )
(z − Lj )3
sj4 (61sj − 1 + 71sj2 + 290sj3 + 299sj4 + 109sj5 + 251sj6 ) 2880L4j ln(sj )(1 − sj )(1 − sj2 )(1 − sj3 )(1 − sj4 )
(z − Lj )4
sj6 (6 + 15sj + 69sj2 + 143sj3 + 115sj4 + 212sj5 + 221sj6 + 155sj7 + 49sj8 + 95sj9 )
+ ··· .
(2)
1440L5j ln(sj )(1 − sj )(1 − sj2 )(1 − sj3 )(1 − sj4 )(1 − sj5 )
(z − Lj )5
Author's personal copy Tetration for complex bases
As long as b is not too close to the boundary of the Shell-Thron region, we can compute ψj (z) very quickly to high levels of precision. If |sj | > 1, and z is in the basin of attraction for logb (z) converging to Lj , then ln(sjn Tm (lognb (z)))
ψj (z) = lim
ln(sj )
n→∞
,
where Tm (z) is the m-th degree Taylor polynomial for σb (z). The larger m is, the faster the convergence. Likewise, if |sj | < 1, and z is in the basin of attraction for bz converging to Lj , then ψj (z) = lim
ln(sj−n Tm (expnb (z))) ln(sj )
n→∞
.
Note that these two are not mutually exclusive—z might be in the basin of attraction for bz converging to L1 and also in the basin of attraction for logb (z) converging to L2 . Next, we find the inverse function ψj−1 (z) = σj−1 (ez ln sj ). This function will satisfy the functional equation −1
ψj−1 (z + 1) = bψj
(z)
.
By reversing the series for σ (z), we find that ψj−1 (z) = Lj + ez ln sj − − −
sj2 (2 + sj ) sj e3z ln sj e2z ln sj + 2 2Lj (1 − sj ) 6Lj (1 − sj )(1 − sj2 )
sj3 (6 + 6sj + 5sj2 + sj3 ) 24L3j (1−sj )(1−sj2 )(1−sj3 )
e4z ln sj +
sj4 (24+36sj +46sj2 +40sj3 +24sj4 + 9sj5 + sj6 ) 120L4j (1−sj )(1 − sj2 )(1 − sj3 )(1 − sj4 )
(3) e5z ln sj
sj5 (120+240sj +390sj2 +480sj3 +514sj4 +416sj5 +301sj6 +160sj7 +64sj8 +14sj9 +sj10 ) 720L5j (1 − sj )(1 − sj2 )(1 − sj3 )(1 − sj4 )(1 − sj5 )
+ ··· .
Again, if z is in the basin of attraction of logb (z) converging to Lj , ψj−1 (z) = lim expnb (Lj + e(z−n) ln sj ), n→∞
and if z is in the basin of attraction of bz converging to Lj , ψj−1 (z) = lim lognb (Lj + e(z+n) ln sj ). n→∞
In fact, we can speed up the convergence by considering the limit expnb (Sm (e(z−n) ln sj )) if |sj | > 1, ψj−1 (z) = lim (z+n) ln sj )) if |s | < 1, n→∞ logn j b (Sm (e
e6z ln sj
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where Sm (z) is the m-th degree Taylor polynomial for σj−1 (z). This function will be entire if |sj | > 1, but will have branch cuts if |sj | < 1. In either case, it should be pointed out that as long as b is not a real number ≤ e1/e , then (ln(s1 )) > 0, since (L1 ) > 0, and likewise (ln(s2 )) < 0. Hence, for any x, lim ψ1−1 (x + iy) = L1
y→∞
and
lim ψ2−1 (x + iy) = L2 .
y→−∞
2 The uniqueness of a complex tetration Even though ψ1−1 (z) and ψ2−1 (z) satisfy the tetration problem F (z + 1) = bF (z) for at least some portion of the complex plane, they are far from unique. We can combine each of these with any periodic function to produce another solution. If we let p(z) be a periodic function of period 1, and let ρj (z) = p(z) + z, then ψj−1 (ρj (z)) will also solve the tetration problem. Since p(z) is periodic, we could express ρj (z) with a complex Fourier series ρj (z) = z +
∞
ck e2πikz .
k=−∞
We would like for the new tetration ψ1−1 (ρ1 (z)) to also have the property that limy→∞ F (x + iy) = L1 . This means that ρ1 (z) must approach ∞i as z approaches ∞i. In order for this to happen in the Fourier series, we see that ck = 0 for all k < 0. Thus, ρ1 (z) = z +
∞
ck e2πikz = z + c0 + c1 e2πiz + c2 e4πiz + c3 e6πiz + · · · .
(4)
k=0
Likewise, for the tetration ψ2−1 (ρ2 (z)) to also have the property that limy→−∞ F (x + iy) = L2 , the Fourier series for L2 must be in the form ρ2 (z) = z +
∞
dk e−2πikz = z + d0 + d1 e−2πiz + d2 e−4πiz + d3 e−6πiz + · · · . (5)
k=0
The question is whether we can find ρ1 (z) and ρ2 (z) so that the two iterations ψ1−1 (ρ1 (z)) and ψ2−1 (ρ2 (z)) meet in the middle, that is, one is an analytic continuation of the other. The first step is to show that if this is possible, it is unique. Proposition 1 Suppose that F (z) is an analytic function defined for (z) > −2 for which F (z + 1) = bF (z) , and that for all x > −2, lim F (x + iy) = L1
y→+∞
lim F (x + iy) = L2 .
and
y→−∞
Then there is some M such that for (z) > M, ψ1 (F (z)) can be expressed as ψ1 (F (z)) = z +
∞
k=0
fk e2πikz ,
Author's personal copy Tetration for complex bases
and for (z) < −M, ψ2 (F (z)) can be expressed as ψ2 (F (z)) = z +
∞
gk e−2πikz .
k=0
Proof Since ψ1 (F (z + 1)) = ψ1 (bF (z) ) = ψ1 (F (z)) + 1, we see that ψ1 (F (z)) − z is periodic with period 1. By the limit property, we have that F (z) is in the region where ψ1 (z) is defined whenever 0 ≤ (z) ≤ 1 and (z) > M for sufficiently large M, and by periodicity, we can define ψ1 (F (z)) whenever (z) > M. By the Fourier series, ∞
ψ1 (F (z)) = z +
fk e2πikz .
k=−∞
Thus,
F (z) =
ψb−1
∞
z+
fk e
2πikz
.
k=−∞
If fk = 0 for some negative k, then as (z) → ∞,
2πikz
→ ∞ as (z) → ∞.
fk e If there are many such terms, they will have different exponential growth rates asymptotically, so such terms cannot cancel out asymptotically. Thus
∞
2πikz fk e
→ ∞ as (z) → ∞.
k=−∞
Also note that the argument of this sum will depend on the real part of z. Thus, there is a path C going towards +∞i for which ψ1 (F (z)) is real and positive along this path. That is, for every y > 0 there is an x such that with z = x + iy, ∞
2πikz fk e arg z + = 0. k=−∞
But then along this path, lim F (z) = C
lim ψ1−1 C
z+
∞
k=−∞
fk e
2πikz
= lim ψb−1 (z), z→+∞
and limz→+∞ ψ1−1 (z) does not exist, since the real part is unbounded, and the imaginary part is negative. We can use a similar argument for ψ2 (z), and pick the larger of the two M’s.
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Proposition 2 Suppose that F1 (z) and F2 (z) are analytic functions defined for (z) > −2 for which F (z + 1) = bF (z) , F (0) = 1, and that for all x > −2, lim F1 (x + iy) = lim F2 (x + iy) = L1 ,
y→+∞
y→+∞
and
lim F1 (x + iy) = lim F2 (x + iy) = L2 .
y→−∞
y→−∞
Then F1 (z) = F2 (z) for all (z) > −2. Proof By proposition 1, ψ1 (F1 (z)) = z +
∞
fk e2πikz ,
for (z) > M,
gk e2πikz ,
for (z) > M,
k=0
ψ1 (F2 (z)) = z +
∞
k=0
ψ2 (F1 (z)) = z +
∞
hk e−2πikz ,
for (z) < −M,
jk e−2πikz ,
for (z) < −M.
k=0
ψ2 (F2 (z)) = z +
∞
k=0
Manipulating the series, we can find F1−1 (F2 (z)) = z +
∞
k e2πikz ,
for (z) > M,
k=0
F1−1 (F2 (z))
=z+
∞
mk e−2πikz ,
for (z) < −M.
k=0
We know that F1−1 (F2 (z))−z is periodic, and approaches one constant as (z) → ∞, and another as (z) → −∞. Even though the series may not converge in the same region, the Fourier coefficients are unique, so F1−1 (F2 (z)) = z+C for some constant C. Since F1 (0) = F2 (0) = 1, we have C = 0, and so F1 (z) = F2 (z). Note that this proof does not guarantee existence of the function F (z). In fact, for some bases b, the complex tetration with the required properties is impossible. Nonetheless, if the complex tetration does exist for a base b, then we can define ρ1 (z) = ψ1 (F (z)) and ρ2 (z) = ψ2 (F (z)), for which both ρ1 (z) − z and ρ2 (z) − z will be periodic. In fact, ρ1 (z) and ρ2 (z) must have series expansions of the form of Eqs. 4 and 5. We can even find the domains of ρ1 (z) and ρ2 (z). Lemma 1 If b is not on the boundary of the Shell-Thron region, and there is a F (z) satisfying the conditions of proposition 2, then ρ1 (z) is analytic for (z) > 0, and ρ2 (z) is analytic for (z) < 0.
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Proof If (z) > 0, then F (z−n) is defined for all integer n. If |s1 | > 1 so that L1 is a repulsive fixed point of bz , then limn→∞ F (z − n) = L1 , so F (z) will be in the basin of attraction for logb (z) converging to L1 , hence in the domain of ψ1 (z), so ρ1 (z) is defined. If |s1 | < 1, then L1 is an attractive fixed point of bz , and limn→∞ F (z + n) = L1 , so F (z) is in the basin of attraction for bz converging to L1 , so again, ρ1 (z) = ψ1 (F (z)) will be defined. To show that ρ1 (z) is analytic, note that the series in Eq. 4 can be written as ρ1 (z) = z + g(e2πiz ), where g(z) has the Maclaurin series g(z) =
∞
ck zk .
k=0
Since ρ1 (z) is defined for (z) > 0, g(z) has no singularities within the unit disk, so g(z) will have a radius of convergence of 1, making ρ1 (z) analytic for (z) > 0 whenever b is not on the boundary of the Shell-Thron region. Likewise, ρ2 (z) is analytic for (z) < 0. Even though we have proven uniqueness, we have yet to prove that a tetration exists for complex b. Unfortunately, Kneser’s existence proof for κb (z) in [1] breaks down if b is complex. Nonetheless, we can extend Kneser’s result to the base b in the complex plane. The plan is to form an analytic continuation of Kneser’s solution. Proposition 3 There is an open connected set S in the complex plane containing all real numbers greater than e1/e , such that for b ∈ S, then there will be a κb (z) which satisfies the conditions of Propositions 1 and 2, that is, κb (z +1) = bκb (z) , κb (0) = 1, and that for x > −2, lim κb (x + iy) = L1
and
y→+∞
lim κb (x + iy) = L2 .
y→−∞
Proof By the way that Kneser defined κb (z), we see that for each z0 with (z) > −2, κb (z0 ) is a real analytic function of b for b > e1/e . Hence, we can analytically extend κb (z0 ) to the complex b-plane, at least to an open region containing (e1/e , ∞). Since κb (z0 + 1) − bκb (z0 ) = 0 for all real b, this must be true for complex b as well. Letting z0 = 0 shows that κb (0) = 1 for complex b. Let b0 be a point in this open set, and let C be a path from b0 to e in the open set that avoids the point e1/e . Note that for a fixed x0 and b real, limy→∞ κb (x + iy) = L1 (b) uniformly on every closed subset of (e1/e , ∞). Since L1 =
ω2 (ln(ln(b)) − π i) − ln(b)
is obviously an analytic function of b except for a branch cut at e1/e , we can extend this to say that limy→∞ κb (x0 + iy) exists for each b, and converges uniformly on the path C. So limy→∞ κb (x0 + iy) = L1 (b) for all points on C, and in particular b0 . Likewise, lim κb0 (x + iy) = L2 . y→−∞
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So the analytic continuation κb0 (z) will satisfy the conditions of propositions 1 and 2, hence will be the unique tetration which satisfies these conditions.
Of course the main question is determining the open set S for which we can extend the complex tetration. Before we can answer this, we will need a way of computing the κb (z) to a high degree of accuracy.
3 First approximation One of the goals of this paper is to produce an iterative method for calculating the tetration for a given base b. However, this iterative method requires having a first order approximation to begin to process. In [3], the first order approximation was created by forcing a high degree polynomial to satisfy Julia’s equation, which in turn produces a solution to Abel’s equation. This seems to work well if b is real, but when b is complex, it produces a function with a highly periodic component. So we will resort to another method. The goal is to find a function with several key properties: f (−1) = 0,
f (0) = 1,
f (1) = b,
lim f (z) = L1 , and
z→i∞
lim f (z) = L2 .
z→−i∞
We also want the function to be nearly periodic for large imaginary component. From [4], we expect the upper half plane to approach a periodic function with period 2π i/ ln(s1 ) as (z) → ∞, and the lower half plane to approach a periodic function with period 2π i/ ln(s2 ), since these are the periods of ψ1 (z) and ψ2 (z). Thus, we want f (z) ∼ L1 + K1 eln(s1 )z
as x → i∞,
f (z) ∼ L2 + K2 eln(s2 )z
as x → −i∞.
In [4], such a function was found using a piecewise defined function, but we can find a single function having all of these properties by assuming the function is of the form f (z) =
Ae−ir1 z + B + Ceir2 z , De−ir1 z + E + F eir2 z
(6)
with (r1 ) > 0 and (r2 ) > 0. It is not too hard to establish that r1 must be −i ln(s1 ), and r2 must be i ln(s2 ). As long as b is complex, or b > e1/e , then indeed (r1 ) and (r2 ) are both positive. Then we need to find A, B, C, D, E, and F so that As1 + B + Cs2 A+B +C bD bF + bE + s1 s2 A = DL1 ,
= 0, = D + E + F, A C = +B + , s1 s2 and C = F L2 .
Author's personal copy Tetration for complex bases
This gives us 5 equations with 6 unknowns, but we can multiply all 6 variables by a constant to eliminate denominators. We can algebraically solve for the constants to produce A = L21 (L2 ln b − 1)(b − L2 + (b − 1)L22 ln b), B = (L1 −L2 ) ln b(−L1 L2 (L1 +L2 )+b(L21 + L1 L2 + L22 )+bL1 L2 (L1 L2 −L1 −L2 ) ln b), C = −L22 (L1 ln b − 1)(b − L1 + (b − 1)L21 ln b), D = L1 (L2 ln b − 1)(b − L2 + (b − 1)L22 ln b), E = (L1 − L2 )(L1 L2 − b(L1 + L2 − 1) + (−L1 L2 (L1 + L2 ) + b(L21 + L1 L2 + L22 )) ln b + L1 L2 (L1 L2 − L1 − L2 )(ln b)2 ), F = −L2 (L1 ln b − 1)(b − L1 + (b − 1)L21 ln b).
This function may not seem to be that accurate of an initial guess, since the graphs of f (z + 1) and bf (z) are visibly different. However, we do not need an extremely accurate first guess, just one close enough for the double dagger track method to converge. This is akin to the initial guess for Newton’s method, for which even a crude first estimate is good enough for the method to find a root. Coincidentally, when b is close to 1, then f (z + 1) and bf (z) do become very close together, with the differences going to 0 as b approaches 1. Thus, f (z) is a very close approximation to κb (z) for b close to 1. This is interesting because, as we will see later, b = 1 is a singular point for κb (z), since one of the two fixed points goes to ∞ as b → 1.
4 The double dagger track method The first step in finding the analytic continuation of κb (z) is to consider b to be outside the Shell-Thron region. Technically b = 0 is outside this region, and will be a singular point of κb (z), so to avoid this we will only consider b to be in the connected portion of the exterior of the Shell-Thron region with (b) ≥ 0. As long as b is not too close to the boundary, we can modify the cross-track method of [3] to apply to complex bases. The key difference is that we cannot use the symmetry Fb (z) = Fb (z), so we will have to use both ψ1 (z) and ψ2 (z). We also will modify the contour to one which optimizes computations for (b) ≥ 0, shown in Fig. 2, involving the parameter A, which will typically have a value of 1. A: B: C: D: E:
integrate along the line x = 1 from t = 1 − Ai to t = 1 + Ai. integrate along the upper half of the circle x 2 + (y − A)2 = 1 counterclockwise. integrate along the line x = −1 from t = −1 + Ai to t = −1 − Ai. integrate along the line x+2y = −1−2A from t = −1−Ai to t = −(A+1/2)i. integrate along the line x − 2y = 1 + 2A from t = −(A + 1/2)i to t = −1 + Ai. Then by the Cauchy contour integral, we have for z within , κb (t) 1 κb (z) = dt. 2π i t − z
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Fig. 2 The contour along with the interior sample points that will be used to numerically compute the contour integral
Note that along the integral A, we can let t = 1 + iy, and use the fact that κb (z + 1) = bκb (z) to simplify. Thus, we see that 1 2π i
A
κb (t) 1 dt = t −z 2π
A −A
bκb (iy) dy. 1 + iy − z
Likewise, integrating along C can be done by letting t = −1 + iy, along with the identity κb (z − 1) = logb (κb (z)). 1 2π i
C
κb (t) 1 dt = − t −z 2π
A −A
logb (κb (iy)) dy. −1 + iy − z
We can put these two pieces together to form the function A logb (κb (iy)) bκb (iy) 1 dy. − H (z) = 2π −A 1 + iy − z −1 + iy − z
(7)
Integrating along the curve B is a bit trickier. We can use the substitution t = Ai +eiθ to produce G(z) =
1 2π i
B
κb (t) 1 dt = t −z 2π
π 0
κb (Ai + eiθ )eiθ dθ. Ai + eiθ − z
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To integrate along the curve D, we can use the substitution t = −1 − Ai + (2 − i)u for 0 ≤ u ≤ 1/2, to produce 1 κb (t) 1 + 2i 1/2 κb (−1 − Ai + 2u − ui) dt = du. 2π i D t − z 2π Ai + 1 − 2u + ui + z 0 We can maintain symmetry to integrate along the curve E, using the substitution t = 1 − Ai − (2 + i)u. This travels the contour E backwards, so we need to add a minus sign to get 1 κb (t) 2i − 1 1/2 κb (1 − Ai − 2u − ui) dt = du. 2π i E t − z 2π Ai − 1 + 2u + ui + z 0 These last two integrals can be combined to form J (z) =
1 2π
0
1/2 (1 + 2i)κ (−1 − Ai b
+ 2u − ui) (2i − 1)κb (1 − Ai − 2u − ui) du. + Ai + 1 − 2u + ui + z Ai − 1 + 2u + ui + z
Putting the pieces together, we see that for z within the racetrack , κb (z) = H (z) + G(z) + J (z).
(8)
To find the approximation for κb (z), we will use the following steps: 1) After picking a value for A (usually A = 1), pick a value of n. This will determine the 4n nodes creating the double dagger formation (‡) in Fig. 2. Note that if we have one approximation Fj (z) for κb (z) known at the 4n nodes, we can use the Gaussian-Legendre quadrature on equation 8 to find a better approximation Fj +1 (z) evaluated at the same 4n nodes. 2) We will use F0 (z) =
Ae−ir1 z + B + Ceir2 z De−ir1 z + E + F eir2 z
from Section 3, and evaluate this at the 4n nodes. 3) Since we have one approximation Fj (z) known at the 4n nodes, we can find a better approximation. We start with finding a better approximation for ρ1 (z) and ρ2 (z). The Fourier series on ρ1 (z) − z = ψ1 (κb (z)) − z gives us 1/2 fk = ekAπ (ψ1 (Fj (x + Ai/2)) − (x + Ai/2))e−2kπix dx. (9) −1/2
which can be evaluated using the n equally spaced nodes on the top crossbar of the double dagger. Likewise, we can find 1/2 gk = ekAπ (ψ2 (Fj (x − Ai/2)) − (x − Ai/2))e2kπix dx. (10) −1/2
using the lower crossbar of the double dagger. By lemma 1, both of the series produced will converge in a half plane, so by computing n/3 terms of the series, we will get almost n/A digits of accuracy for |(z)| ≥ A.
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4) We now have a way of estimating G(z), using the Gaussian quadrature with 2n nodes on the integral π −1 ψ1 (ρ1 (Ai + eiθ ))eiθ 1 Gj (z) = dθ. 2π 0 Ai + eiθ − z where we use the Fourier series to approximate ρ1 (z). We can also use the Gaussian quadrature with n nodes on the integral Jj (z) =
1 2π
1/2 0
(1 + 2i)ψ2−1 (ρ2 (−1−Ai +2u−ui)) (2i −1)ψ2−1 (ρ2 (1−Ai −2u−ui)) du. + Ai + 1 − 2u + ui + z Ai − 1 + 2u + ui + z
We can also use the Gaussian quadrature with the 2n nodes along the vertical portion of the double dagger to estimate H (z), which we can call Hj (z). Finally, we define Fj +1 (z) = Hj (z) + Gj (z) + Jj (z). 5) We want to incorporate the initial condition F (0) = 1 at each iteration. If we let aj =
Fj +1 (0) − 1 , Fj +1 (0)
then Fj +1 (− aj ) will be much closer to 1 than Fj +1 (0). We can estimate Fj +1 (0) by computing 1 1 Fj +1 (0) ≈ 50 Fj +1 − Fj +1 − . 100 100 Then we evaluate Fj +1 (z − aj ) for each of the 4n nodes of the double dagger, and we are ready for the next iteration. 6) Repeat steps 3 through 5 for a fixed n until the digits stabilize to m digits of accuracy for all of the values of Fj (xk ) and Fj (yk ). For most values of b, we gain a digit of accuracy for every two iterations. We can, in fact, increase n dynamically (between steps 3 and 4) which speeds up the process. For values of b outside the Shell-Thron region and (b) ≥ 0, this method works very similar to the cross-track method of [3]. Figure 3 shows a typical example with b = 3 + i. Provided that the double-dagger track method converges to a function, it is clear that this function satisfies the conditions of Propositions 1 and 2. Since the analytic continuation of κb (z) also satisfies these properties, we have a way of computing the analytic continuation of κb (z).
5 Within the Shell-Thron region We have seen that κb (z) can be analytically extended for b outside the Shell-Thron region, but can we analytically extend this to inside the region? There is a simple way of showing that we can. In [4], another way of finding κb (z) is given, which we will refer to as the Kouznetsov method. A large value of A is used in the contour of Fig. 2. We can approximate the function F (z) around the contour B simply with the constant
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Fig. 3 Level curves for (κb (z)) and (κb (z)) = 0, ±1, ±2, ±3, ±4 for b = 3 + i
L1 , and the function around the contours D and E with L2 . The integral for Hj (z) will require many more nodes to accurately integrate using the Gaussian quadrature. Although this method converges more slowly than the double dagger track method, it has the distinct advantage of not requiring either ψ1 (z) or ψ2 (z), only the values of L1 and L2 . Thus, the Kouznetsov method is totally blind as to whether the point b is inside the Shell-Thron region or outside. By proposition 2, the Kouznetsov method will converge to the same function as the double dagger track method as A → ∞. Each approximation uses the value of the function at a finite number of points, and the value at each of these points would vary continuously as the base b varies. Thus, the tetration κb (z) will vary continuously as b varies, even across the boundary of the Shell-Thron region. The only exception to this is the point b = e1/e . At this point L1 = L2 = e, and we are evaluating ω2 (z) at its singular points. Thus, there is a potential branch cut to κb (z) at b = e1/e . Even though the double dagger track method fails for this base, the tetration for b = e1/e was throughly investigated in [5]. For other points on the boundary of the Shell-Thron region, we see that κb (z) is holomorphic, since both κb (z) and the function produced by Kouznetsov’s method satisfies the conditions of Propositions 1 and 2. We will begin with discussing how to compute κb (z) when b = 1 is within the Shell-Thron region, and later consider the case where b is on the boundary. Inside the Shell-Thron region, one of the two fixed points will be an attractive fixed point of bz . Without loss of generality, we can assume that (b) ≥ 0, so that L1 is the attractive fixed point. We will still be able to use the double dagger track method, but there will be some new issues. To begin, even though ψ1 (z) still has a branch cut that may have to be adjusted, it will only be defined on the basin of attraction of L1 . However, if we have a tetration κb (z), then ρ1 (z) − z = ψ1 (κb (z)) − z will be periodic for (z) sufficiently large. By Lemma 1, ρ1 (z) can be analytically continued to the upper half plane. So κb (z)
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will be in the basin of attraction of L1 , whenever (z) > 0. Thus, even though ψ1 (z) is not defined for the whole plane, we can safely evaluate ψ1 (Fj (z)) along the top crossbar from (−1 + i)/2 to (1 + i)/2. Also, ψ1−1 (z) is no longer an entire function, but has a branch cut. Since we use −1 ψ1 (z) to evaluate the tetration along part B of the contour, we must be careful that the branch cut is positioned so that the evaluations along B do not cross the branch cut. Thus, with a few minor changes, the double dagger track method works for most points within the Shell-Thron region. Figure 4 shows the tetration for the base b = i. This graph explains the unusual shape of the contour . The main “pineapple top” is fairly close to the origin, so that the tetration is rapidly changing in the region of 1−i. By using A = 1/2, we avoid the part of the pineapple where there is rapid change, so that integral around the contour can be accurately computed. This specialized contour allows us to answer the conundrum asked in [6]: κi (i) ≈ 0.50134608238017127458221942256989499070737981445660 +0.33020066879235626265604272037746894978172757389433 i. There is another difference for bases within the Shell-Thron region. The branch cut at z = −2 is clear, but there are other “spontaneous” branch cuts at z ≈ −4.8101919687+0.1456703034 i and z ≈ −7.5906468833+0.2732864288 i. These branch cuts are unavoidable, because κi (−3.8101919687+0.1456703034 i) ≈ 0, and κi (−6.5906468833+0.2732864288 i) ≈ 0.
Actually, since the psuedo-period for the upper half plane is 2π i/ ln(s1 ), which has a negative imaginary component, there will actually be an infinite number of spontaneous branch cuts, going up and to the left. Spontaneous branch cuts also exist for bases outside the Shell-Thron region, but one must cross the branch cut at z = −2 to another sheet of the Riemann surface to see them.
Fig. 4 Level curves for (κb (z)) and (κb (z)) = 0, ±1, ±2, ±3, ±4 for b = i
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There is another interesting question we can ask when b is within the Shell-Thron region. As b approaches the real axis from the upper complex plane, does κb (z) approach a function that√is real whenever z is real? One real base that has stud√ been z ied extensively is√ b = 2. In fact, there are 4 different tetrations for 2 given in [7]. If we let b = 2 + i, does κb (z) approach one of the four functions given in [7] as → 0? √ Figure 5 shows κb (z) for b = 2+i/100. The familiar “pineapple top” has moved down to around the position 1.5−17.5i. It is still true that limy→−∞ κb (x +iy) = L2 , which is close to 4, but only after getting past this “pineapple.” Only one of the four functions in [7] satisfies the condition F (0) = 1, namely F2,1 (z). This function can be expressed as ψ1−1 (z + k), where ψ1 is the function at the attractive fixed point L1 = 2, and k ≈ 1.2515514788221865095737713539516428646086989358005409587170658 574886949888276516169343 +8.571574089677423552089688951087337604341737977394035019241728 0618645702439744277372740 i In spite of appearances, this function is real whenever z is real, for example, ψ1−1 (1/2 + k) ≈ 1.2436216276685218042950989836094029316881983566155237864263842667 39854041905083929661122493. In order to see whether κ√2+ i (z) approaches this function as → 0+ , we can √ actually use the double dagger track method with b = 2, so that L1 = 2 and
Fig. 5 Level curves for b = shown by the inset
√ 2 + i/100. The main “pineapple” has moved down the complex plane, as
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L2 = 4. It is natural to call this limit κ√2+ (z). Using n = 60, we can find κ√2+ (z) to 49 places of accuracy, κ√2+ (1/2) ≈ 1.2436216276685218042950989836094029316881983566155, which agrees with ψ1−1 (1/2 + k) to all 49 places. Proposition 4 For real 1 < b < e1/e , let σ1 (z) be the Schr¨oder function for the attractive fixed point L1 , and ψ1 (z) = ln(σ1 (z))/ ln(s1 ). Let kb = ψ1 (1), so that fb (z) = ψ1−1 (z + kb ) will be real whenever z is real, and fb (z + 1) = bfb (z) and fb (0) = 1. Then lim κb+ i (z) →0+
does not converge to fb (z) for all 1 < b < e1/e . Proof Suppose that lim κb+ i (z) = fb (z)
→0+
for all b in an open subinterval of 1 < b < e1/e . It should be noted that since fb (z) is defined using Taylor series, fb (z) extends to a holomorphic function of both z and b. This holomorphic function would extend to all b within the Shell-Thron region. However, fb (z) cannot extend outside of the Shell-Thron region, because the Taylor series diverges on the boundary. Even if we could extend fb (z) to the outside of the Shell-Thron region, it would not be a real function at b = e, since fb (z) only uses one of the complex fixed points. But we have seen that κb (z) is a holomorphic function of b across the boundary of the Shell-Thron region, and of course κe (z) is real when z is real. If κb (z) extended to the real axis for 1 < b < e1/e and agreed with fb (z) on an open subinterval, then the two functions must be identical everywhere, which we see cannot be the case. Hence, it cannot be true that κb (z) = fb (z) for all b in an open interval. A maxim of Sherlock Holmes was that “when you have eliminated the impossible, whatever remains, however improbable, must be the truth” [8]. Even though κ√2+ (z)
and ψb−1 (z + kb ) agree to 49 places of accuracy, they cannot be the same function. By increasing n to 120, we can obtain more digits for κ√2+ (1/2): 1.24362162766852180429509898360940293168819835661554061856203329 176722789043816675442745 − 1.18899697185401045226976795872715599664 · 10−48 i. Thus, we see that κ√2+ (z) is not a real function after all. This value was computed using both A = 1 and A = 2 with identical results. The difference is more clear if we look at the graph, shown in Fig. 6. The function approaches ψ1 (z + c1 ) as z → ∞i for some c1 , and approaches ψ2 (z + c2 ) as z → −∞i. With a moment’s reflection, we see that this is essential to maintain analytic continuity, since these limits hold for all other bases.
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√ Fig. 6 The limiting function as the base b approaches 2 from above. There are no branch cuts below the reals axis, even though there are branch cuts above it
Thus, we see that the point b = e1/e begins a “slight” branch cut in the b-plane. For 1 < b < e1/e , the limits lim κb+ i (z)
→0+
and
lim κb− i (z)
→0+
will be very slightly different for real z. This is not the first time in which two tetrations to the same base are extremely close, yet not the same. In [7], two of the four functions agreed to 24 places of accuracy on the real axis, yet√ were different holomorphic functions. We now have two more tetrations for b = 2, namely κ√2+ (z) and κ√2− (z) = κ√2+ (z). The branch cut is much more pronounced if e−e < b < 1. A typical example is b = (1/4)+ which has the attractive fixed point L1 = 1/2. Figure 7 shows that the tetration is far from symmetrical. If we compare this with the tetration of b = (1/4)− , the L1 and L2 will switch places, producing the mirror image of Fig. 7. There is a simple explanation for this radical change of behavior from b > 1 to b < 1. When b < 1, the derivative at the fixed points are negative, so there is not a tetration which is real whenever z is real. This also points out that b = 1 will be a singular point for κb (z). There is one more issue to consider when using the double dagger track method for computing the tetration. Notice that the “pineapple” in Fig. 7 is even further to the left than Fig. 4. Even using a small value of A, the integral for part E will involve a fairly rapidly changing function with an amplitude of over 100. This would make numerical integration extremely difficult, and the double-dagger method will fail to converge.
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Fig. 7 Level curves for b = (1/4)+ , using L1 = 1/2
So how was Fig. 7 created? Rather than using the initial condition F (0) = 1, we used initial condition F (3/4) = 1. This shifts the graph of the solution 3/4 units to the right, so that the contour is no longer going through the heart of the pineapple top. The initial guess is adjusted accordingly. Once the iteration method has converged to a solution, it is easy to shift it back to the standard tetration. Proposition 2 assures us that this operation will not affect the final solution. This trick of shifting the initial condition is limited, though. The branch cut at z = −2 must obviously stay outside of the contour . There will be some base b for which the “pineapple” will be even further left than in Fig. 7, and the doubledagger track method will fail. Nonetheless, the double-dagger track works whenever (b) ≥ 0, except if we are too close to either b = 0, b = 1, or b = e1/e .
6 The Shell-Thron boundary There still remains the problem of evaluating κb (z) to high precision when b is on or very near the boundary of the Shell-Thron region. We again can assume without loss of generality that (b) > 0. There will not be any issues with ψ2 (z), but the series for ψ1 (z) and ψ1−1 (z) in equations 2 and 3 will probably not converge, since |s1 | = 1, and the denominators will get arbitrarily close to 0. We still can improve the Kouznetsov method by using the dagger track method for the lower half-plane, as in Fig. 8. However, we can view equation 3 as an asymptotic series, which means if we only consider a finite number of terms from the series, this will be accurate as z → ∞i. If we express the series for ψ1−1 (z) as ψ1−1 (z) ∼ L1 +
∞
n=1
an enz ln s1 ,
Author's personal copy Tetration for complex bases Fig. 8 The “sword-track” contour combines the cross-track method with the Kouznetsov method
then since ρ1 (z) also has an asymptotic series ρ1 (z) ∼ z + c0 + c1 e2πiz + · · · , we can find the asymptotic behavior of κb (z) = ψ1−1 (ρ1 (z)) as z → ∞i. κb (z) ∼ L1 +
∞
nc0
an s1
enz ln s1 + nc1 ln s1 e(2πi+n ln s1 )z
n=1
+
(nc1 ln s1 )2 + nc2 ln s1 e(4πi+n ln s1 )z + · · · . 2
In [4], the integral along the top part of the integral from 1 + Ai to −1 + Ai, κb (t) dt B t −z
(11)
was estimated by replacing κb (t) with its asymptotic approximation, L1 . But we now have a better asymptotic series for κb (t), which we can integrate term by term. We must be very careful with branch cuts as we evaluate this integral. So instead of using the familiar exponential integral Ei(z), we will use the entire exponential integral z 1 − e−t Ein(z) = dt = E1 (z) + ln(z) + γ . t 0
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Then we can compute B
−1+Ai Kt−Kz 1 e −1 dt + eKz dt t −z 1+Ai B t −z 1 − z + Ai Kz = e π i − ln + Ein(Kz−K −AKi)−Ein(Kz+K −AKi) . 1 + z − Ai
eKt dt = eKz t −z
This can be used to integrate equation 11 asymptotically term by term. If we let 2π 1 N= , + (ln s1 ) 2 then using the first N terms of the series gives us accuracy up to e2πiz . Also, if s1 is close to a principle root of unity e2πi/n , then N = n, so that we will include the first large term of the series with (1 − s n ) in the denominator. We will only need to know the c0 term for the series of ρ1 (z). We can use one point z0 at the tip of the sword to estimate the value of c0 , using the first N terms of the series for ψ1 (z). c0 = ψ1 (Fj (z0 )) − z0 . At which point we can compute Gj (z) =
L1 1 − z + Ai iL1 + ln 2 2π 1 + z − Ai N
an nc 1 − z + Ai + s1 0 enz ln s1 π i − ln 2π i 1 + z − Ai n=1
+Ein(n ln s1 (z − 1 − Ai)) − Ein(n ln s1 (z + 1 − Ai)) .
Fig. 9 Level curves for (κb (z)) and (κb (z)) = 0, ±1, ±2, ±3, ±4 for b = 2 + i
κs (z)
1.00000000000000000000000000000000000000000000000000 + 0.00000000000000000000000000000000000000000000000000 i
1.06461506378360449145504943689911966810171696543874 + 0.00118689963189625015607374362277463470792862745352 i
1.12573323536699582285076577499990051018374881693187 + 0.00252015363521040593531272992048888734956332799993 i
1.18374058213588977132134067251971035508701187517556 + 0.00399230200758573597178146554856089363715593275967 i
1.23897035166988908773194787918100031932378550804792 + 0.00559746613418730947715879388224567602638073040156 i
1.29171208219992887866969870767002501172173866594472 + 0.00733111006196133463799523266873194228451770379388 i
1.34221887359500688681719380864654816349727422382336 + 0.00918985309423200652762750432456272241318913679580 i
1.39071324395805633394586031758365499346658855224331 + 0.01117132207855846280239114544921770278918103114513 i
1.43739188725488151278412307544709508929637380535724+ 0.01327403479122457366760953881527486074969215533831 i
1.48242956881836215417913821165867916323906522348329 + 0.01549730799555231588528307547056479505182849686716 i
1.52598233851700000000000000000000000000000000000000 + 0.01784118533210000000000000000000000000000000000000 i
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Table 1 Table of values for κs (z), where s is the Sheldon base, for various real z
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Tetration for complex bases
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Fig. 10 Level curves for (κb (z)) and (κb (z)) = 0, ±1, ±2, ±3, ±4 for the Sheldon base
The estimation of the integral for parts D and E can be done as before. Finally, we can let Fj +1 (z) = Hj (z) + Gj (z) + Jj (z). Iterating this procedure, we can achieve κb (z) to a large degree of accuracy. Increasing the value of A will improve this accuracy. For example, the point b = 2 + i is very near to the boundary of the Shell-Thron region. In fact, s1 is very close to i. If we compute 1 2π , + N= (ln s1 ) 2 we obtain N = 5, so we will use 5 terms of the series for ψ1 (z), which includes the term involving (1 − s14 ) in the denominator. The resulting function is depicted in Fig. 9. Another example of a point close near to the boundary of the Shell-Thron region is the Sheldon base. This value, whose exact value is given as s = 1.525982338517 + 0.0178411853321i, was given by Sheldon Levenstein as an example of a base for which the tetration would be very hard to compute [9]. As such, it has become the challenge problem to accurately compute the tetration to this base. The sword-track method is able to obtain tetration to over 50 places of accuracy. Because s is so close to e1/e , N turns out to be 12, so we compute 12 terms of the ψ1−1 (z) series. Since we anticipate the amount of error to be e−2πA , we let A = 19. Finally, we use 50 nodes per unit, so the main part of the sword uses 1000 nodes. Table 1 gives a table of values that produces 50 places of accuracy. Even though the convergence was slower than normal for this base (requiring 10 iterations to gain 1 place of accuracy) we are able to compute the tetration within a reasonable amount of time. The contour plot is given in Fig. 10.
7 Conclusion We have been able to analytically extend Kneser’s tetration κb (z) to complex values of b. That is, for a fixed value z0 with (z0 ) > −2, κb (z0 ) will be a holomorphic function, at least in the right half-plane, with a branch cut running to the left of b = e1/e . Furthermore, we have established a method for computing κb (z) for b inside, outside, and on the boundary of the Shell-Thron region. The Mathematica notebook
Author's personal copy Tetration for complex bases
that created these figures is available at http://myweb.astate.edu/wpaulsen/tetration. html. This site also links to a tetration calculator that uses the sword-track method to compute the tetration to 20 places of accuracy for various bases. By extending the definition of the tetration to complex bases, we establish the tetration as a special function. There is still an issue of notation. We intentionally avoided the established notation tetb (z), because the analytic continuation of Kneser’s tetration actually disagrees with tetrations defined in other papers. Since holomorphism in the variable b is of highest priority, we propose defining tetb (z) = κb (z) when b is not a real number ≤ e1/e . Then for real 1 < b ≤ e1/e , want tetb (z) to be a real function, so we can still define tetb (z) = ψ1−1 (z+ψ1 (1)) as in [5] and [7]. The tetration will not be continuous across the branch cut, and in fact tet√2+ (z), tet√2− , and tet√2 (z) will be three different functions. This convention should create the least confusion.
References 1. Kneser, H.: Reelle analytishe L¨osungen der Gleichung ϕ(ϕ(x)) = ex und verwandter Funktionalgleichungen. J. reine angew. Math. 187, 56–67 (1950) 2. Lawrence, P.W., Corless, R.M., Jeffrey, D.J.: Algorithm 917: Complex double-precision evaluation of the wright omega function. ACM Trans. Math. Softw. 38, 3, 20 (2012). 1–17 3. Paulsen, W., Cowgill, S.: Solving F (z + 1) = bF (z) in the complex plane. Adv. Comput. Math. 43(6), 1261–1282 (2017). https://doi.org/10.1007/s10444-017-9524-1 4. Kouznetsov, D.: Solution of F (z + 1) = exp(F (z)) in the complex z-plane. Math. Comput. 78(267), 1647–1670 (2009) 5. Trappmann, H., Kouznetsov, D.: Computation of the two regular super-exponentials to base e1/e . Math. Comput. 81, 2207–2227 (2012). https://doi.org/10.1090/S0025-5718-2012-02590-7 6. Mathematics Stack Exchange, What is i exponentiated to itself i times? (2013), https://math. stackexchange.com/questions/280251 7. Kouznetsov, D., Trappmann, H.: Portrait of the four regular super-exponentials to base sqrt(2). Math. Comput. 79, 1727–1756 (2010). https://doi.org/10.1090/S0025-5718-10-02342-2 8. Doyle, A.C.: The Sign of Four. The Complete Sherlock Holmes. Garden City Publishing Company, New York (1938) 9. Levenson, S.: Complex base tetration program. Tetration and Related Topics. Tetration Forum. http:// math.eretrandre.org/tetrationforum/showthread.php?tid=729 (2012)
