Base zero

The exponential ${\displaystyle 0^{0}}$ is not consistently defined. Thus, the tetrations ${\displaystyle \,{^{n}0}}$ are not clearly defined by the formula given earlier. However, ${\displaystyle \lim _{x\rightarrow 0}{}^{n}x}$ is well defined, and exists:^[14]

${\displaystyle \lim _{x\rightarrow 0}{}^{n}x={\begin{cases}1,&n{\text{ even}}\\0,&n{\text{ odd}}\end{cases}}}$

Thus we could consistently define ${\displaystyle {}^{n}0=\lim _{x\rightarrow 0}{}^{n}x}$. This is analogous to defining ${\displaystyle 0^{0}=1}$.

Under this extension, ${\displaystyle {}^{0}0=1}$, so the rule ${\displaystyle {^{0}a}=1}$ from the original definition still holds.

Complex bases

Tetration by period

Tetration by escape

Since complex numbers can be raised to powers, tetration can be applied to bases of the form z = a + bi (where a and b are real). For example, in ⁿz with z = i, tetration is achieved by using the principal branch of the natural logarithm; using Euler's formula we get the relation:

${\displaystyle i^{a+bi}=e^{{\frac {1}{2}}{\pi i}(a+bi)}=e^{-{\frac {1}{2}}{\pi b}}\left(\cos {\frac {\pi a}{2}}+i\sin {\frac {\pi a}{2}}\right)}$

This suggests a recursive definition for ⁿ⁺¹i = a′ + b′i given any ⁿi = a + bi:

${\displaystyle {\begin{aligned}a'&=e^{-{\frac {1}{2}}{\pi b}}\cos {\frac {\pi a}{2}}\\[2pt]b'&=e^{-{\frac {1}{2}}{\pi b}}\sin {\frac {\pi a}{2}}\end{aligned}}}$

The following approximate values can be derived:

Values of tetration of complex bases

${\textstyle {}^{n}i}$	Approximate value
${\textstyle {}^{1}i=i}$	i
${\textstyle {}^{2}i=i^{\left({}^{1}i\right)}}$	0.2079
${\textstyle {}^{3}i=i^{\left({}^{2}i\right)}}$	0.9472 + 0.3208i
${\textstyle {}^{4}i=i^{\left({}^{3}i\right)}}$	0.0501 + 0.6021i
${\textstyle {}^{5}i=i^{\left({}^{4}i\right)}}$	0.3872 + 0.0305i
${\textstyle {}^{6}i=i^{\left({}^{5}i\right)}}$	0.7823 + 0.5446i
${\textstyle {}^{7}i=i^{\left({}^{6}i\right)}}$	0.1426 + 0.4005i
${\textstyle {}^{8}i=i^{\left({}^{7}i\right)}}$	0.5198 + 0.1184i
${\textstyle {}^{9}i=i^{\left({}^{8}i\right)}}$	0.5686 + 0.6051i

Solving the inverse relation, as in the previous section, yields the expected ⁰i = 1 and ⁻¹i = 0, with negative values of n giving infinite results on the imaginary axis. Plotted in the complex plane, the entire sequence spirals to the limit 0.4383 + 0.3606i, which could be interpreted as the value where n is infinite.

Such tetration sequences have been studied since the time of Euler, but are poorly understood due to their chaotic behavior. Most published research historically has focused on the convergence of the infinitely iterated exponential function. Current research has greatly benefited by the advent of powerful computers with fractal and symbolic mathematics software. Much of what is known about tetration comes from general knowledge of complex dynamics and specific research of the exponential map.

Infinite heights

${\displaystyle \textstyle \lim _{n\rightarrow \infty }{}^{n}x}$ of the infinitely iterated exponential converges for the bases ${\displaystyle \textstyle \left(e^{-1}\right)^{e}\leq x\leq e^{\left(e^{-1}\right)}}$

The function ${\displaystyle \left|{\frac {\mathrm {W} (-\ln {z})}{-\ln {z}}}\right|}$ on the complex plane, showing the real-valued infinitely iterated exponential function (black curve)

Tetration can be extended to infinite heights; i.e., for certain a and n values in ${\displaystyle {}^{n}a}$, there exists a well defined result for an infinite n. This is because for bases within a certain interval, tetration converges to a finite value as the height tends to infinity. For example, ${\displaystyle {\sqrt {2}}^{{\sqrt {2}}^{{\sqrt {2}}^{\cdot ^{\cdot ^{\cdot }}}}}}$ converges to 2, and can therefore be said to be equal to 2. The trend towards 2 can be seen by evaluating a small finite tower:

${\displaystyle {\begin{aligned}{\sqrt {2}}^{{\sqrt {2}}^{{\sqrt {2}}^{{\sqrt {2}}^{{\sqrt {2}}^{1.414}}}}}&\approx {\sqrt {2}}^{{\sqrt {2}}^{{\sqrt {2}}^{{\sqrt {2}}^{1.63}}}}\\&\approx {\sqrt {2}}^{{\sqrt {2}}^{{\sqrt {2}}^{1.76}}}\\&\approx {\sqrt {2}}^{{\sqrt {2}}^{1.84}}\\&\approx {\sqrt {2}}^{1.89}\\&\approx 1.93\end{aligned}}}$

In general, the infinitely iterated exponential ${\displaystyle x^{x^{\cdot ^{\cdot ^{\cdot }}}}\!\!}$, defined as the limit of ${\displaystyle {}^{n}x}$ as n goes to infinity, converges for e^−e ≤ x ≤ e^1/e, roughly the interval from 0.066 to 1.44, a result shown by Leonhard Euler.^[15] The limit, should it exist, is a positive real solution of the equation y = x^y. Thus, x = y^1/y. The limit defining the infinite exponential of x does not exist when x > e^1/e because the maximum of y^1/y is e^1/e. The limit also fails to exist when 0 < x < e^−e.

This may be extended to complex numbers z with the definition:

${\displaystyle {}^{\infty }z=z^{z^{\cdot ^{\cdot ^{\cdot }}}}={\frac {\mathrm {W} (-\ln {z})}{-\ln {z}}}~,}$

where W represents Lambert's W function.

As the limit y = ^∞x (if existent on the positive real line, i.e. for e^−e ≤ x ≤ e^1/e) must satisfy x^y = y we see that x ↦ y = ^∞x is (the lower branch of) the inverse function of y ↦ x = y^1/y.

Negative heights

We can use the recursive rule for tetration,

${\displaystyle {^{k+1}a}=a^{\left({^{k}a}\right)},}$

to prove ${\displaystyle {}^{-1}a}$:

${\displaystyle ^{k}a=\log _{a}\left(^{k+1}a\right);}$

Substituting −1 for k gives

${\displaystyle {}^{-1}a=\log _{a}\left({}^{0}a\right)=\log _{a}1=0}$.^[13]

Smaller negative values cannot be well defined in this way. Substituting −2 for k in the same equation gives

${\displaystyle {}^{-2}a=\log _{a}\left({}^{-1}a\right)=\log _{a}0=-\infty }$

which is not well defined. They can, however, sometimes be considered sets.^[13]

For ${\displaystyle n=1}$, any definition of ${\displaystyle \,\!{^{-1}1}}$ is consistent with the rule because

${\displaystyle {^{0}1}=1=1^{n}}$ for any ${\displaystyle \,\!n={^{-1}1}}$.

Real heights

At this time there is no commonly accepted solution to the general problem of extending tetration to the real or complex values of n. There have, however, been multiple approaches towards the issue, and different approaches are outlined below.

In general, the problem is finding — for any real a > 0 — a super-exponential function ${\displaystyle \,f(x)={}^{x}a}$ over real x > −2 that satisfies

• ${\displaystyle \,{}^{-1}a=0}$
• ${\displaystyle \,{}^{0}a=1}$
• ${\displaystyle \,{}^{x}a=a^{\left({}^{x-1}a\right)}}$for all real ${\displaystyle x>-1.}$^[16]

To find a more natural extension, one or more extra requirements are usually required. This is usually some collection of the following:

• A continuity requirement (usually just that ${\displaystyle {}^{x}a}$ is continuous in both variables for ${\displaystyle x>0}$).
• A differentiability requirement (can be once, twice, k times, or infinitely differentiable in x).
• A regularity requirement (implying twice differentiable in x) that:

  ${\displaystyle \left({\frac {d^{2}}{dx^{2}}}f(x)>0\right)}$ for all ${\displaystyle x>0}$

The fourth requirement differs from author to author, and between approaches. There are two main approaches to extending tetration to real heights; one is based on the regularity requirement, and one is based on the differentiability requirement. These two approaches seem to be so different that they may not be reconciled, as they produce results inconsistent with each other.

When ${\displaystyle \,{}^{x}a}$ is defined for an interval of length one, the whole function easily follows for all x > −2.

Linear approximation for real heights

${\displaystyle \,{}^{x}e}$ using linear approximation

A linear approximation (solution to the continuity requirement, approximation to the differentiability requirement) is given by:

${\displaystyle {}^{x}a\approx {\begin{cases}\log _{a}\left(^{x+1}a\right)&x\leq -1\\1+x&-1<x\leq 0\\a^{\left(^{x-1}a\right)}&0<x\end{cases}}}$

hence:

Linear approximation values

Approximation	Domain
${\textstyle {}^{x}a\approx x+1}$	for −1 < x < 0
${\textstyle {}^{x}a\approx a^{x}}$	for 0 < x < 1
${\textstyle {}^{x}a\approx a^{a^{(x-1)}}}$	for 1 < x < 2

and so on. However, it is only piecewise differentiable; at integer values of x the derivative is multiplied by ${\displaystyle \ln {a}}$. It is continuously differentiable for ${\displaystyle x>-2}$ if and only if ${\displaystyle a=e}$. For example, using these methods ${\displaystyle {}^{\frac {\pi }{2}}e\approx 5.868...}$ and ${\displaystyle {}^{-4.3}0.5\approx 4.03335...}$

A main theorem in Hooshmand's paper^[6] states: Let ${\displaystyle 0<a\neq 1}$. If ${\displaystyle f:(-2,+\infty )\rightarrow \mathbb {R} }$ is continuous and satisfies the conditions:

• ${\displaystyle f(x)=a^{f(x-1)}\;\;{\text{for all}}\;\;x>-1,\;f(0)=1,}$
• ${\displaystyle f}$ is differentiable on (−1, 0),
• ${\displaystyle f^{\prime }}$ is a nondecreasing or nonincreasing function on (−1, 0),
• ${\displaystyle f^{\prime }\left(0^{+}\right)=(\ln a)f^{\prime }\left(0^{-}\right){\text{ or }}f^{\prime }\left(-1^{+}\right)=f^{\prime }\left(0^{-}\right).}$

then ${\displaystyle f}$ is uniquely determined through the equation

${\displaystyle f(x)=\exp _{a}^{[x]}\left(a^{(x)}\right)=\exp _{a}^{[x+1]}((x))\quad {\text{for all}}\;\;x>-2,}$

where ${\displaystyle (x)=x-[x]}$ denotes the fractional part of x and ${\displaystyle \exp _{a}^{[x]}}$ is the ${\displaystyle [x]}$-iterated function of the function ${\displaystyle \exp _{a}}$.

The proof is that the second through fourth conditions trivially imply that f is a linear function on [−1, 0].

The linear approximation to natural tetration function ${\displaystyle {}^{x}e}$ is continuously differentiable, but its second derivative does not exist at integer values of its argument. Hooshmand derived another uniqueness theorem for it which states:

If ${\displaystyle f:(-2,+\infty )\rightarrow \mathbb {R} }$ is a continuous function that satisfies:

• ${\displaystyle f(x)=e^{f(x-1)}\;\;{\text{for all}}\;\;x>-1,\;f(0)=1,}$
• ${\displaystyle f}$ is convex on (−1, 0),
• ${\displaystyle f^{\prime }\left(0^{-}\right)\leq f^{\prime }\left(0^{+}\right).}$

then ${\displaystyle f={\text{uxp}}}$. [Here ${\displaystyle f={\text{uxp}}}$ is Hooshmand's name for the linear approximation to the natural tetration function.]

The proof is much the same as before; the recursion equation ensures that ${\displaystyle f^{\prime }(-1^{+})=f^{\prime }(0^{+}),}$ and then the convexity condition implies that ${\displaystyle f}$ is linear on (−1, 0).

Therefore, the linear approximation to natural tetration is the only solution of the equation ${\displaystyle f(x)=e^{f(x-1)}\;\;(x>-1)}$ and ${\displaystyle f(0)=1}$ which is convex on (−1, +∞). All other sufficiently-differentiable solutions must have an inflection point on the interval (−1, 0).

Higher order approximations for real heights

A comparison of the linear and quadratic approximations (in red and blue respectively) of the function ${\displaystyle ^{x}0.5}$, from x = −2 to x = 2

Beyond linear approximations, a quadratic approximation (to the differentiability requirement) is given by:

${\displaystyle {}^{x}a\approx {\begin{cases}\log _{a}\left({}^{x+1}a\right)&x\leq -1\\1+{\frac {2\ln(a)}{1\;+\;\ln(a)}}x-{\frac {1\;-\;\ln(a)}{1\;+\;\ln(a)}}x^{2}&-1<x\leq 0\\a^{\left({}^{x-1}a\right)}&x>0\end{cases}}}$

which is differentiable for all ${\displaystyle x>0}$, but not twice differentiable. For example, ${\displaystyle {}^{\frac {1}{2}}2\approx 1.45933...}$ If ${\displaystyle a=e}$ this is the same as the linear approximation.^[1]

Because of the way it is calculated, this function does not "cancel out", contrary to exponents, where ${\displaystyle \left(a^{\frac {1}{n}}\right)^{n}=a}$. Namely,

${\displaystyle {}^{n}\left({}^{\frac {1}{n}}a\right)=\underbrace {\left({}^{\frac {1}{n}}a\right)^{\left({}^{\frac {1}{n}}a\right)^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{\left({}^{\frac {1}{n}}a\right)}}}}}}} _{n}\neq a}$.

Just as there is a quadratic approximation, cubic approximations and methods for generalizing to approximations of degree n also exist, although they are much more unwieldy.^[1][17]

Complex heights

Drawing of the analytic extension ${\displaystyle f=F(x+{\rm {i}}y)}$ of tetration to the complex plane. Levels ${\displaystyle |f|=1,e^{\pm 1},e^{\pm 2},\ldots }$ and levels ${\displaystyle \arg(f)=0,\pm 1,\pm 2,\ldots }$ are shown with thick curves.

It has now been proven^[18] that there exists a unique function F which is a solution of the equation F(z + 1) = exp(F(z)) and satisfies the additional conditions that F(0) = 1 and F(z) approaches the fixed points of the logarithm (roughly 0.318 ± 1.337i) as z approaches ±i∞ and that F is holomorphic in the whole complex z-plane, except the part of the real axis at z ≤ −2. This proof confirms a previous conjecture.^[19] The construction of such a function was originally demonstrated by Kneser in 1950.^[20] The complex map of this function is shown in the figure at right. The proof also works for other bases besides e, as long as the base is bigger than ${\displaystyle e^{\frac {1}{e}}\approx 1.445}$. Subsequent work extended the construction to all complex bases.^[21]

The requirement of the tetration being holomorphic is important for its uniqueness. Many functions S can be constructed as

${\displaystyle S(z)=F\!\left(~z~+\sum _{n=1}^{\infty }\sin(2\pi nz)~\alpha _{n}+\sum _{n=1}^{\infty }{\Big (}1-\cos(2\pi nz){\Big )}~\beta _{n}\right)}$

where α and β are real sequences which decay fast enough to provide the convergence of the series, at least at moderate values of Im z.

The function S satisfies the tetration equations S(z + 1) = exp(S(z)), S(0) = 1, and if α_n and β_n approach 0 fast enough it will be analytic on a neighborhood of the positive real axis. However, if some elements of {α} or {β} are not zero, then function S has multitudes of additional singularities and cutlines in the complex plane, due to the exponential growth of sin and cos along the imaginary axis; the smaller the coefficients {α} and {β} are, the further away these singularities are from the real axis.

The extension of tetration into the complex plane is thus essential for the uniqueness; the real-analytic tetration is not unique.

Square super-root

The graph ${\displaystyle y={\sqrt {x}}_{s}}$

The 2nd-order super-root, square super-root, or super square root has two equivalent notations, ${\displaystyle \mathrm {ssrt} (x)}$ and ${\displaystyle {\sqrt {x}}_{s}}$. It is the inverse of ${\displaystyle ^{2}x=x^{x}}$ and can be represented with the Lambert W function:^[22]

${\displaystyle \mathrm {ssrt} (x)=e^{W(\ln x)}={\frac {\ln x}{W(\ln x)}}}$

The function also illustrates the reflective nature of the root and logarithm functions as the equation below only holds true when ${\displaystyle y=\mathrm {ssrt} (x)}$:

${\displaystyle {\sqrt[{y}]{x}}=\log _{y}x}$

Like square roots, the square super-root of x may not have a single solution. Unlike square roots, determining the number of square super-roots of x may be difficult. In general, if ${\displaystyle e^{-1/e}<x<1}$, then x has two positive square super-roots between 0 and 1; and if ${\displaystyle x>1}$, then x has one positive square super-root greater than 1. If x is positive and less than ${\displaystyle e^{-1/e}}$ it does not have any real square super-roots, but the formula given above yields countably infinitely many complex ones for any finite x not equal to 1.^[22] The function has been used to determine the size of data clusters.^[23]

At ${\displaystyle x=1}$:

$${\displaystyle \mathrm {ssqrt} (x)=1+(x-1)-(x-1)^{2}+{\frac {3}{2}}(x-1)^{3}-{\frac {17}{6}}(x-1)^{4}+{\frac {37}{6}}(x-1)^{5}-{\frac {1759}{120}}(x-1)^{6}+{\frac {13279}{360}}(x-1)^{7}+{\mathcal {O}}{\left((x-1)^{8}\right)}}$$

Other super-roots

The graph ${\displaystyle y={\sqrt[{3}]{x}}_{s}}$

For each integer n > 2, the function ⁿx is defined and increasing for x ≥ 1, and ⁿ1 = 1, so that the nth super-root of x, ${\displaystyle {\sqrt[{n}]{x}}_{s}}$, exists for x ≥ 1.

One of the simpler and faster formulas for a third-degree super-root is the recursive formula, if: x^xx = a, and next x (n + 1) = exp (W (W (x (n) ln (a)))), for example x (0) = 1.

However, if the linear approximation above is used, then ${\displaystyle ^{y}x=y+1}$ if −1 < y ≤ 0, so ${\displaystyle ^{y}{\sqrt {y+1}}_{s}}$ cannot exist.

In the same way as the square super-root, terminology for other super roots can be based on the normal roots: "cube super-roots" can be expressed as ${\displaystyle {\sqrt[{3}]{x}}_{s}}$; the "4th super-root" can be expressed as ${\displaystyle {\sqrt[{4}]{x}}_{s}}$; and the "nth super-root" is ${\displaystyle {\sqrt[{n}]{x}}_{s}}$. Note that ${\displaystyle {\sqrt[{n}]{x}}_{s}}$ may not be uniquely defined, because there may be more than one n^th root. For example, x has a single (real) super-root if n is odd, and up to two if n is even.

Just as with the extension of tetration to infinite heights, the super-root can be extended to n = ∞, being well-defined if 1/e ≤ x ≤ e. Note that ${\displaystyle x={^{\infty }y}=y^{\left[^{\infty }y\right]}=y^{x},}$ and thus that ${\displaystyle y=x^{1/x}}$. Therefore, when it is well defined, ${\displaystyle {\sqrt[{\infty }]{x}}_{s}=x^{1/x}}$ and, unlike normal tetration, is an elementary function. For example, ${\displaystyle {\sqrt[{\infty }]{2}}_{s}=2^{1/2}={\sqrt {2}}}$.

It follows from the Gelfond–Schneider theorem that super-root ${\displaystyle {\sqrt {n}}_{s}}$ for any positive integer n is either integer or transcendental, and ${\displaystyle {\sqrt[{3}]{n}}_{s}}$ is either integer or irrational.^[24] It is still an open question whether irrational super-roots are transcendental in the latter case.