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Abstract and Figures

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.
Content may be subject to copyright.
Tetration for Complex Bases
William Paulsen
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 bis complex. A previous paper determined conditions for a unique
solution the case where bis 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.
2010 Mathematics Subject Classification: 26A18, 30D05, 39B12
Keywords: Tetration, Abel’s functional equation, iteration.
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 ncan be thought of as repeated incrementing, multiplication by n
is done by repeated addition, and exponentiation by nis repeated multiplication. So a fourth level operation
would be repeated exponentiation using the initial base b:
bb·
·b

n
Although there are many different notations for this operation, we will use the iterated exponential notation
expn
b(x) = bb·
·bx
with n b’s.
The tetration function for base bcan then be expressed as F(n) = expn
b(1). Thus we see that F(0) = 1,
F(1) = b,F(2) = bb, and so on. When bis complex, we must define a branch cut so that bzis 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 bis 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 Lfor
which bL=L. To find the principle fixed points, we first define the Wright ωfunction to be the solution to
the initial value problem
ω(z) = ω(z)
1 + ω(z), ω(1) = 1,
1
where the branch cuts are chosen to start at 1±πi, run parallel to the real axis in the negative direction
[2]. The Wright ωfunction can in fact be defined in terms of the nth branch of the Lambert’s Wfunction.
ω(z) = Wn(ez),where n=(z)π
2π.
Note that with this definition, ω(a) = ω(a) whenever ais on either of the branch cuts. (In this context,
ameans to approach afrom below the branch cut rather than from the left. For example, ln(1) = πi,
and ln(1+) = πi.) We prefer a function which is continuous from above the branch cuts, so we define
ω2(z) = ω(z).
Then ω2will be identical to ωexcept at the branch cuts, where now ω2(a) = ω2(a+) at both branch cuts.
The principle fixed point of the exponential function bzcan now be expressed by
L1=ω2(ln(ln(b)) πi)
ln(b).
A second fixed point can be found with a similar formula,
L2=ω2(ln(ln(b)) + πi)
ln(b).
In both formulas, we use the principle branch of the logarithm. If b > e1/e, or if bis not real, then we can
see that (L1)>0 and (L2)<0.
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 bis in the Shell-Thron region if the sequence
of values
{b, bb, bbb, bbbb
, . . .}(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 figure 1. For b̸= 1 within this region, the sequence converges to L1if
(b)0, and converges to L2if (b)<0.
We begin by solving the Schr¨oder equation σ(bz) = (z), where sis the derivative of bzat one of the
fixed points. At the fixed point L1,s1= ln(L1) = L1ln(b), and at the fixed point L2,s2= ln(L2) = L2ln(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
σ1(z) = (zL1) + s1
2L1(1 s1)(zL1)2+s2
1(1 + 2s1)
6L2
1(1 s1)(1 s2
1)(zL1)3
+s3
1(1 + 6s1+ 5s2
1+ 6s3
1)
24L3
1(1 s1)(1 s2
1)(1 s3
1)(zL1)4+s4
1(1 + 14s1+ 24s2
1+ 45s3
1+ 46s4
1+ 26s5
1+ 24s6
1)
120L4
1(1 s1)(1 s2
1)(1 s3
1)(1 s4
1)(zL1)5
+s5
1(1 + 30s1+ 89s2
1+ 214s3
1+ 374s4
1+ 416s5
1+ 511s6
1+ 461s7
1+ 330s8
1+ 154s9
1+ 120s10
1)
720L5
1(1 s1)(1 s2
1)(1 s3
1)(1 s4
1)(1 s5
1)(zL1)6
+···.
σ2(z) = (zL2) + s2
2L2(1 s2)(zL2)2+s2
2(1 + 2s2)
6L2
2(1 s2)(1 s2
2)(zL2)3
2
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........
........
........
........
e1/e
ee
1 1 2 3
2
1
1
2
Figure 1: The Shell-Thron region.
+s3
2(1 + 6s2+ 5s2
2+ 6s3
2)
24L3
2(1 s2)(1 s2
2)(1 s3
2)(zL2)4+s4
2(1 + 14s2+ 24s2
2+ 45s3
2+ 46s4
2+ 26s5
2+ 24s6
2)
120L4
2(1 s2)(1 s2
2)(1 s3
2)(1 s4
2)(zL2)5
+s5
2(1 + 30s2+ 89s2
2+ 214s3
2+ 374s4
2+ 416s5
2+ 511s6
2+ 461s7
2+ 330s8
2+ 154s9
2+ 120s10
2)
720L5
2(1 s2)(1 s2
2)(1 s3
2)(1 s4
2)(1 s5
2)(zL2)6
+···.
Unfortunately, when bis 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 bis within the Shell-Thron region, and the series in Eq. 1 converges to L, then if zis in the
basin of attraction for bz, so that
{z, bz, bbz, bbbz
, . . .}
also converges to L, then
σ(z) = lim
n→∞ sn(expn
b(z)L),
where expn
b(z) represents applying bza total of ntimes. If Lis one of the repulsive fixed points of bz, and
if zis in the basin of attraction for logb(z), so that repetitively applying the principal value of logbto zwill
converge to L, then
σ(z) = lim
n→∞ sn(logn
b(z)L).
3
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) = ln(zLj)
ln(sj)+sj
2Ljln(sj)(1 sj)(zLj) + s2
j(1 + 5sj)
24L2
jln(sj)(1 sj)(1 s2
j)(zLj)2(2)
+s4
j(2 + sj+ 3s2
j)
24L3
jln(sj)(1 sj)(1 s2
j)(1 s3
j)(zLj)3
+s4
j(61sj1 + 71s2
j+ 290s3
j+ 299s4
j+ 109s5
j+ 251s6
j)
2880L4
jln(sj)(1 sj)(1 s2
j)(1 s3
j)(1 s4
j)(zLj)4
+s6
j(6 + 15sj+ 69s2
j+ 143s3
j+ 115s4
j+ 212s5
j+ 221s6
j+ 155s7
j+ 49s8
j+ 95s9
j)
1440L5
jln(sj)(1 sj)(1 s2
j)(1 s3
j)(1 s4
j)(1 s5
j)(zLj)5
+···.
As long as bis 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 zis in the basin of attraction for logb(z) converging to Lj, then
ψj(z) = lim
n→∞
ln(sn
jTm(logn
b(z)))
ln(sj),
where Tm(z) is the m-th degree Taylor polynomial for σb(z). The larger mis, the faster the convergence.
Likewise, if |sj|<1, and zis in the basin of attraction for bzconverging to Lj, then
ψj(z) = lim
n→∞
ln(sn
jTm(expn
b(z)))
ln(sj).
Note that these two are not mutually exclusive—zmight be in the basin of attraction for bzconverging to
L1and also in the basin of attraction for logb(z) converging to L2.
Next, we find the inverse function ψ1
j(z) = σ1
j(ezln sj). This function will satisfy the functional equation
ψ1
j(z+ 1) = bψ1
j(z).
By reversing the series for σ(z), we find that
ψ1
j(z) = Lj+ezln sjsj
2Lj(1 sj)e2zln sj+s2
j(2 + sj)
6L2
j(1 sj)(1 s2
j)e3zln sj(3)
s3
j(6 + 6sj+ 5s2
j+s3
j)
24L3
j(1 sj)(1 s2
j)(1 s3
j)e4zln sj+s4
j(24 + 36sj+ 46s2
j+ 40s3
j+ 24s4
j+ 9s5
j+s6
j)
120L4
j(1 sj)(1 s2
j)(1 s3
j)(1 s4
j)e5zln sj
s5
j(120 + 240sj+ 390s2
j+ 480s3
j+ 514s4
j+ 416s5
j+ 301s6
j+ 160s7
j+ 64s8
j+ 14s9
j+s10
j)
720L5
j(1 sj)(1 s2
j)(1 s3
j)(1 s4
j)(1 s5
j)e6zln sj
+···.
Again, if zis in the basin of attraction of logb(z) converging to Lj,
ψ1
j(z) = lim
n→∞ expn
b(Lj+e(zn) ln sj),
4
and if zis in the basin of attraction of bzconverging to Lj,
ψ1
j(z) = lim
n→∞ logn
b(Lj+e(z+n) ln sj).
In fact, we can speed up the convergence by considering the limit
ψ1
j(z) = lim
n→∞ expn
b(Sm(e(zn) ln sj)) if |sj|>1,
logn
b(Sm(e(z+n) ln sj)) if |sj|<1,
where Sm(z) is the m-th degree Taylor polynomial for σ1
j(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 bis not a real number e1/e, then (ln(s1)) >0,
since (L1)>0, and likewise (ln(s2)) <0. Hence, for any x,
lim
y→∞ ψ1
1(x+iy) = L1and lim
y→−∞ ψ1
2(x+iy) = L2.
2 The Uniqueness of a complex tetration
Even though ψ1
1(z) and ψ1
2(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 ψ1
j(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+
k=−∞
cke2πikz .
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 ias zapproaches i. In order for this to happen in the Fourier
series, we see that ck= 0 for all k < 0. Thus,
ρ1(z) = z+
k=0
cke2πikz =z+c0+c1e2πiz +c2e4π iz +c3e6πiz +···.(4)
Likewise, for the tetration ψ1
2(ρ2(z)) to also have the property that limy→−∞ F(x+iy) = L2, the Fourier
series for L2must be in the form
ρ2(z) = z+
k=0
dke2πikz =z+d0+d1e2πiz +d2e4π iz +d3e6πiz +···.(5)
The question is whether we can find ρ1(z) and ρ2(z) so that the two iterations ψ1
1(ρ1(z)) and ψ1
2(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
y+F(x+iy) = L1and lim
y→−∞ F(x+iy) = L2.
5
Then there is some Msuch that for (z)> M,ψ1(F(z)) can be expressed as
ψ1(F(z)) = z+
k=0
fke2πikz ,
and for (z)<M,ψ2(F(z)) can be expressed as
ψ2(F(z)) = z+
k=0
gke2πikz .
Proof:
Since ψ1(F(z+ 1)) = ψ1(bF(z)) = ψ1(F(z)) + 1, we see that ψ1(F(z)) zis periodic with period 1. By
the limit property, we have that F(z) is in the region where ψ1(z) is defined whenever 0 ≤ ℜ(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+
k=−∞
fke2πikz .
Thus,
F(z) = ψ1
bz+
k=−∞
fke2πikz .
If fk̸= 0 for some negative k, then as (z)→ ∞,
fke2πikz → ∞ as (z)→ ∞.
If there are many such terms, they will have different exponential growth rates asymptotically, so such terms
cannot cancel out asymptotically. Thus
k=−∞
fke2πikz → ∞ as (z)→ ∞.
Also note that the argument of this sum will depend on the real part of z. Thus, there is a path Cgoing
towards +ifor 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,
arg z+
k=−∞
fke2πikz = 0.
But then along this path,
lim
CF(z) = lim
Cψ1
1z+
k=−∞
fke2πikz = lim
z+ψ1
b(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.
6
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
y+F1(x+iy) = lim
y+F2(x+iy) = L1,and
lim
y→−∞ F1(x+iy) = lim
y→−∞ F2(x+iy) = L2.
Then F1(z) = F2(z) for all (z)>2.
Proof:
By proposition 1,
ψ1(F1(z)) = z+
k=0
fke2πikz ,for (z)> M,
ψ1(F2(z)) = z+
k=0
gke2πikz ,for (z)> M,
ψ2(F1(z)) = z+
k=0
hke2πikz ,for (z)<M,
ψ2(F2(z)) = z+
k=0
jke2πikz ,for (z)<M.
Manipulating the series, we can find
F1
1(F2(z)) = z+
k=0
ke2πikz ,for (z)> M,
F1
1(F2(z)) = z+
k=0
mke2πikz ,for (z)<M.
We know that F1
1(F2(z)) zis 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+Cfor 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)zand
ρ2(z)zwill 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 bis 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.
7
Proof:
If (z)>0, then F(zn) is defined for all integer n. If |s1|>1 so that L1is a repulsive fixed point
of bz, then limn→∞ F(zn) = 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 L1is an attractive fixed point of
bz, and limn→∞ F(z+n) = L1, so F(z) is in the basin of attraction for bzconverging 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) =
k=0
ckzk.
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 bis 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 bis complex. Nonetheless, we can
extend Kneser’s result to the base bin the complex plane. The plan is to form an analytic continuation of
Kneser’s solution.
Proposition 3:
There is an open connected set Sin the complex plane containing all real numbers greater than e1/e,
such that for bS, 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
y+κb(x+iy) = L1and lim
y→−∞ κb(x+iy) = L2.
Proof:
By the way that Kneser defined κb(z), we see that for each z0with (z)>2, κb(z0) is a real analytic
function of bfor 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 bas well. Letting z0= 0 shows that κb(0) = 1 for complex b.
Let b0be a point in this open set, and let Cbe a path from b0to ein the open set that avoids the point
e1/e. Note that for a fixed x0and breal, limy→∞ κb(x+iy) = L1(b) uniformly on every closed subset of
(e1/e,). Since
L1=ω2(ln(ln(b)) πi)
ln(b)
is obviously an analytic function of bexcept 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
y→−∞ κb0(x+iy) = L2.
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 Sfor 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.
8
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 bis real, but when bis 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
zif(z) = L1,and lim
z→−if(z) = L2.
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 2πi/ ln(s1) as (z)→ ∞, and the lower half
plane to approach a periodic function with period 2πi/ ln(s2), since these are the periods of ψ1(z) and ψ2(z).
Thus, we want
f(z)L1+K1eln(s1)zas xi, f(z)L2+K2eln(s2)zas 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) = Aeir1z+B+C eir2z
Deir1z+E+F eir2z,(6)
with (r1)>0 and (r2)>0. It is not too hard to establish that r1must be iln(s1), and r2must be
iln(s2). As long as bis complex, or b > e1/e , then indeed (r1) and (r2) are both positive. Then we need
to find A,B,C,D,E, and Fso that
As1+B+Cs2= 0,
A+B+C=D+E+F,
bD
s1
+bE +bF
s2
=A
s1
+B+C
s2
,
A=DL1,and C=F L2.
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=L2
1(L2ln b1)(bL2+ (b1)L2
2ln b),
B= (L1L2) ln b(L1L2(L1+L2) + b(L2
1+L1L2+L2
2) + bL1L2(L1L2L1L2) ln b),
C=L2
2(L1ln b1)(bL1+ (b1)L2
1ln b),
D=L1(L2ln b1)(bL2+ (b1)L2
2ln b),
E= (L1L2)(L1L2b(L1+L21) + (L1L2(L1+L2) + b(L2
1+L1L2+L2
2)) ln b
+L1L2(L1L2L1L2)(ln b)2),
F=L2(L1ln b1)(bL1+ (b1)L2
1ln 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
9
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 bis close to 1, then f(z+ 1) and bf(z)do become very close together, with the
differences going to 0 as bapproaches 1. Thus, f(z) is a very close approximation to κb(z) for bclose 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 b1.
4 The Double dagger track Method
The first step in finding the analytic continuation of κb(z) is to consider bto 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 bto be in the connected portion of the exterior of the Shell-Thron region with (b)0. As
long as bis 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 Ω to one which optimizes computations for (b)0,
shown in figure 2, involving the parameter A, which will typically have a value of 1.
A: integrate along the line x= 1 from t= 1 Ai to t= 1 + Ai.
B: integrate along the upper half of the circle x2+ (yA)2= 1 counterclockwise.
C: integrate along the line x=1 from t=1 + Ai to t=1Ai.
D: integrate along the line x+ 2y=12Afrom t=1Ai to t=(A+ 1/2)i.
E: integrate along the line x2y= 1 + 2Afrom t=(A+ 1/2)ito t=1 + Ai.
Then by the Cauchy contour integral, we have for zwithin Ω,
κb(z) = 1
2πi
κb(t)
tzdt.
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
2πi