[Math] An explicit series representation for the analytic tetration with complex height

analytic-continuationca.classical-analysis-and-odesconjecturesfractional-iterationsequences-and-series

Tetration is the next hyperoperation after more familiar addition, multiplication and exponentiation. It can be seen as a repeated exponentiation, similar to how exponentiation can be seen as a repeated multiplication, and multiplication — as a repeated addition. Tetration is denoted $^n a$, where $a$ is called the base and $n$ is called the height, and is defined for $n\in\mathbb N\cup\{-1,\,0\}$ by the recurrence
$$
{^{-1} a} = 0, \quad {^{n+1} a} = a^{\left({^n a}\right)},\tag1$$
so that
$${^0 a}=1, \quad {^1 a} = a, \quad {^2 a} = a^a, \quad {^3 a} = a^{a^a}, \, \dots \quad {^n a} = \underbrace{a^{a^{{.^{.^{.^a}}}}}}_{n\,\text{levels}},\tag2$$
where power towers are evaluated from top to bottom.
To simplify matters, we restrict our attention to real bases in the interval $1<a<e^{1/e}$. Under this restriction, the sequence $\{^n a\}$ is strictly increasing and converges to a limit:
$$^\infty a=\lim_{n\to\infty} {^n a} = \frac{W(-\ln a)}{-\ln a},\tag3$$
where $W(z)$ denotes the principal branch of the Lambert W-function. To simplify our following exposition it is convenient to introduce a new variable $q$ linked to $a$:
$$q = \ln({^\infty a}) = {^\infty a} \cdot \ln a = -W(-\ln a).\tag4$$
The bounds on $a$ imply $0<q<1$.
The value $q$ is closely related to the asymptotic growth of the tetration. In particular, it appears in remarkable limits:
$$
\lim_{n\to\infty} \, \frac{{^{n+k} a}-{^n a}}{{^{n-k} a}-{^n a}} = -q^k, \quad \lim_{n\to\infty} \, \frac{{^{n+k} a}-{^\infty a}}{{^n a}-{^\infty a}} = q^k
.\tag5$$
Definitions of addition, multiplication and exponentiation for fractional, negative and complex arguments are well-established. A natural extension of the tetration to fractional and complex heights is a long-standing question. It appears to be much less straightforward problem, mainly because exponentiation, unlike addition and multiplication, lacks both commutativity and associativity, and so tetration does not obey simple laws similar to $a^m \cdot a^n = a^{m+n}$ or $(a^m)^n = a^{m\,n}$.

Despite these obstacles, several proposals aiming to extend the definition of the tetration to fractional and complex heights have been put forth. Their common goal is to define a function that agrees with the discrete tetration for all positive integer heights, and satisfies the functional equation
$$^{z+1} a = a^{\left({^z a}\right)}\tag6$$
for all $z$ in its domain. These conditions alone are not sufficient to determine a unique function, so additional natural restrictions (such as continuity, smoothness or analyticity) have been proposed, aiming to narrow down the set of candidate functions to a single and, in some sense, "the most natural" extension. The Wikipedia article on tetration is very cautious and indicates that, apparently, so far there is no consensus about which extension is "the one". Some of the proposals seem to lead to the same function, although apparently no rigorous proof of their equivalence has been published yet. The Citizendum article is written in a much bolder style, and refers to tetration as a unique well-defined function. But, overall, the study of tetration seems to be a quite isolated area of mathematics, with few connections to other topics.

A few months ago I made some interesting conjectures related to tetration that I hoped I would prove soon. It appears that there are some connections between tetration and the theory of $q$-analogs. Unfortunately, I have not made much progress in proving them since then (although I have learnt a lot about $q$-series), so I finally decided to post them here and ask for your help. First, let's recall some definitions.

The $q$-Pochhammer symbol:
$$(p;\,q)_\infty = \prod_{k=0}^\infty(1-p\,q^k)\tag7$$
$$(p;\,q)_z = \frac{(p;\,q)_\infty}{(p\,q^z;\,q)_\infty}, \quad z\in\mathbb C\tag8$$
It easily follows that
$$(p;\,q)_n = \prod_{k=0}^{n-1}(1-p\,q^k), \quad n\in\mathbb N\tag9$$

The $q$-binomial coefficients (also known as the Gaussian binomial coefficients):
$${r \brack s}_q = \frac{(q;\,q)_r}{(q;\,q)_s \, (q;\,q)_{r-s}}.\tag{10}$$

From now on we assume that $a$ and, consequently, $q$ are fixed. Let us define an analytic function $t(z)$ by the following series:

$$
t(z) = \sum_{n=0}^\infty \sum_{k=0}^n (-1)^{n-k} \, q^{\binom {n-k} 2} {z \brack n}_q {n \brack k}_q ({^k a}).\tag{11}
$$

This formula can be seen as a combination of the direct and reverse $q$-binomial transforms. It can be proved by induction that for $n\in\mathbb N, \, t(n)={^n a}$, so $t(n)$ satisfies the functional equation for tetration $(6)$ at least for positive integer arguments. Indeed, partial sums of $(11)$ can be seen as Lagrange interpolating polynomials in powers of $q^z$ that exactly reproduce tetration values for progressively larger initial segments of $\mathbb N$. Numeric evidence suggests that adding new points to the interpolating polynomial of this form does not result in erratic oscillations of its arcs between existing points (i.e. Runge's phenomenon does not occur), but rather makes them converge to a monotone function satisfying the functional equation for the tetration $(6)$. So we have:

Conjecture 1.
The series $t(z)$ converges at least in the half-plane $\Re(z)>-2$ to an analytic function that satisfies the functional equation for tetration $t(z+1)=a^{t(z)}$ for all arguments in its domain of analyticity.

Conjecture 2. The function $t(z)$ is periodic with a purely imaginary period $\tau = 2\pi n i/\ln q$, and admits an analytic continuation to the half-plane $\Re(z) \le -2$ by repeated backward application of the functional equation $(6)$. There it has a countable set of isolated singularities at $z = m + \tau\,n, \,m,n\in\mathbb Z, \, m \le -2$, and a countable set of branch cuts that can be made along horizontal rays at $z = x + \tau\,n, \, n\in\mathbb Z, \, x\in\mathbb R, \, x < -2$.

Conjecture 3. For real $x>-2$, the derivative $t'(x)$ is a completely monotone function.

Conjecture 4. There is only one analytic function that vanishes at $z=-1$ and satisfies both the functional equation for the tetration $(6)$, and the complete monotonicity condition from the Conjecture 3.

Best Answer

Not sure if this is an answer but a good suggestion of how to show (1) and (2) (and perhaps an actual solution), and a good start for (3). Consider the Newton series formula for tetration. This series converges for all $\Re(z) > -2$, and is very similar to your formula

$$^za = \sum_{n=0}^\infty \sum_{k=0}^n (-1)^{n-k} \binom{z}{n}\binom{n}{k}(^k a)$$

(The op claims this expression diverges, an alternate expression is

$$\frac{1}{\Gamma(1-z)}\Big{(}\sum_{n=0}^\infty (^{n+1}a)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty \big{(}\sum_{n=0}^\infty (^{n+1}a) \frac{(-x)^n}{n!}\big{)}x^{-z}\,dx\Big{)}$$

of which the rest of this discussion still applies.)

Now, since this solution is bounded in the half plane $\Re(z) > 0$ it does have a uniqueness criterion--quite a good one at that too. It is the only solution to the functional equation to do such. In fact, if a solution $F$ exists where $F(z) = O(e^{\rho|\Re(z)| + \tau|\Im(z)|})$ for $\tau < \pi/2$ then $F$ is still the above solution.

What you have presented is a q-analog of the Newton series I just put above. Sadly I cannot find a rigorous paper detailing these types of series; I don't remember where I have seen them, but I have seen them. This suggests to me, that really you are just interpolating $(^na)$ in the same manner the Newton series does, except you are using q-analogs. It is necessary though that $q$ be fixed to the value you put (explanation below).

Consider the rather beautiful fact that your solution is bounded in the right half plane.

Observe if

$[z]_q = \frac{1-q^z}{1-q}$, $[z]_q$ is periodic in $z$.

$${z \brack n}_q = \frac{[z]_q!}{[n]_q![z-n]_q!} = \frac{[z]_q[z-1]_q...[z-n+1]_q}{[1]_q[2]_q...[n]_q}$$

And each of these functions have a period, namely $2\pi i/\log(q)$..

Therefore your final expansion has an imaginary period, it tends to $-W(\log(a))/\log(a)$ as $\Re(z) \to \infty$; thus, it is bounded in the right half plane.

From this it follows your function is the standard tetration function (If it converges).

So $t$ is bounded. The first obvious fact is $t(n) = (^n a)$. Next consider the wondrous power of a little known identity theorem for functions bounded in the right half plane. Namely, if $F(z)$ and $G(z)$ are bounded in the right half plane and $F\Big{|}_{\mathbb{N}} = G\Big{|}_{\mathbb{N}}$ then $F = G$. This follows because of Ramanujan's master theorem.

I'll explain the identity theorem for a second, and then explain how this applies to your case. If $F$ and $G$ are bounded, then surely

$$f(x) = \sum_{n=0}^\infty F(n+1)\frac{(-x)^n}{n!}$$ $$g(x) = \sum_{n=0}^\infty G(n+1)\frac{(-x)^n}{n!}$$

are both entire functions, and by Ramanujan both satisfy

$$\int_0^\infty f(x)x^{z-1}\,dx = \Gamma(z)F(1-z)$$ $$\int_0^\infty g(x)x^{z-1}\,dx = \Gamma(z)G(1-z)$$

But $f = g$ and therefore $F = G$.

Look at $t(z+1)$ and $a^{t(z)}$, both of these functions are bounded and agree on the naturals ($t(n+1) = a^{t(n)}$), therefore they are equal everywhere.

Now as to the fact that this function is completely monotone: On the tetration forum Sheldon made a great point. I'm not sure of this fact, but I'm betting it's true. If $0 < \lambda < 1$ then $f(z) = \lambda^z$ for $\Re(z) > 0$ is the only completely monotone function on $\mathbb{R}^+$ such that $\lambda f(z) = f(z+1)$. Another function would have to be $f(z+\theta(z))$ for $\theta$ a 1-periodic function, but this will most assuredly destroy the completely monotone structure.

What's great about the standard solution (the one that is periodic, the one that is the Schroder iteration, the one that is bounded), is that it can be expanded in an exponential series for $\Re(z) > 0$. This is just Fourier series, essentially.

$$^za = F(z) = \sum_{k=0}^\infty a_k \lambda^{kz}$$

$$F'(x) \sim C\lambda^x$$

Since another solution is just $F(z+\theta(z))$ for $\theta$ periodic, we can most likely form a contradiction by using our knowledge of the uniqueness of the exponential function under these conditions.

All in all, I mostly just had to weigh in my two cents, but I think all the conjectures you put are true and that your solution is in fact the 'standard' solution, which is: the periodic one, the bounded one, the Schroder one, and hopefully, the completely monotone one. Hopefully the rough proof I gave of (1) and (2) is good enough for you.

Related Solutions

[Math] Curious $q$-analogues

This is not a definitive answer, since it doesn't contain any more examples, but more of an observation.
If you have a sequence $P_n(x)$ for polynomials defined recursively by the formula $$P_n(x)=x^{a_1}P_{n-1}(x)+x^{a_2}P_{n-2}(x)+\cdots+x^{a_k}P_{n-k}(x),$$ and an operator $B(q)$ with $\lim_{q \to 1} B(q)=X$, we get: $$P_n(B)1=B^{a_1}P_{n-1}(B)1+B^{a_2}P_{n-2}(B)1+\cdots+B^{a_k}P_{n-k}(B)1.$$

Denoting $P_n(B)1$ as $\hat P_n(x)$ and supposing that $B$ is linear (which it is, if it depends only on $X$ and $D_q$, like all your examples suggest), we get: $$\lim_{q\to 1}\hat P_n(x)=\lim_{q\to 1}x^{a_1}\hat P_{n-1}(x)+x^{a_2}\hat P_{n-2}(x)+\cdots+x^{a_k}\hat P_{n-k}(x),$$

Thus, $\lim_{q\to 1}\hat P_n(x)$ satisfies the same recursion as $P_n(x)$ does and if the first $k$ terms are equal, then $\lim_{q\to 1}\hat P_n(x) = P_n(x)$ for all $n$. This is precisely the definition of a $q$-analog. Whether it will be interesting is another matter, but I think this must be decided case-by-case.

I noticed that all of your examples have the form $B(q)=X+a(q-1)D_q$. This looks somewhat like a linear term of a series in $D_q$ of the form $$B(q)=X+a_1(q-1)D_q+a_2(q-1)^2D_q^2+\cdots,$$ so maybe if you could find examples with $B$ not linear in $D_q$, it would be very interesting.

Classical Analysis – Finding Closed Forms/Related Constants to a Limit Involving Tetration

There is an implementation around (Pari/GP; in the tetration-forum) which claims to have a Kneser-implementation. It is a bit difficult to handle, so I'll show here a simpler version (essentially polynomial) of a tetration-function which seems to approximate that Kneser function when the polynomial's order is increased.
With a polynomial of order 32 I got for your problem a value of about $a=1.2329216$

Added: I got the mentioned Kneser-implementation work; it gives a taylorseries in $h$ (where the spurious imaginary parts of the coefficients are deleted) as

 f(h)=  2.0 
    + 1.2329216 21568706952691849*h
    + 0.3920521 821375754199047770*h^2 
    + 0.2175339 103974922699983933*h^3 
    + 0.087772 79346256013703882103*h^4
    + 0.0409164 3821897543029382908*h^5 + O(h^6)

where the coefficient at $h^1$ is the first derivative with respect to $h$ which I've approximated by my method.
To compare, the method described below gives (when the same parameters are inserted)

q&d(h) = 2 
    + 1.2329216 07*h
    + 0.3920521 614*h^2
    + 0.2175339 098*h^3
    + 0.087772 80375*h^4
    + 0.0409164 4709*h^5 + O(h^6)

[end addition]

This is how I got my approximate solution:

get pc as the vector of first 32 coefficients of the power series for $f(x)=2^x$
From pc generate the (truncated!) Carlemanmatrix F for the function $f(x)$.
The vector pc and thus F should have infinite size, but of course we must cut it to something we can handle and which gives not too much crap, at least in a certain range of data.
By the principle of (infinite) Carlemanmatrixes, powers of F give iterates of $f(x)$. Fractional powers give fractional iterates- of course, practically this is always only approximated.
Fractional powers of F can be achieved, when we diagonalize F = M * D * W(where I write Wfor M^-1 for simplicity) and compute $F^h = M*D^h*W$

The approximation runs now

 z_0=1;
   m_z0 = vector(32,j, sum(k=0,31, z_0^k * M[1+k,j] )   )     
 h=1.1
   z_h=sum(k=0,31, mz_0[1+k]  *  D[1+k]^h  *  W[1+k,2])

$z_h$ is now (approximately) the $h$'th iterate from $z_0$ to $z_h$.

Letting $h$ go downwards to $h=1$ it seems that we approximate the value I've given above. And because I've tested earlier this method against the (claimed) Kneser's method and have empirically found improving convergence to that results when increasing the size of the matrix (which means increased order of the underlying polynomial) I think, that we are near the true Kneser's solution here.

Remark: the matrices M and W look ugly in the view of giving coefficients for a polynomials (ideally we should have power series with infinitely many coefficients) and should surely be evaluated only for $z_0$ and $h$ near zero. But tests with $z_0=1$ and various heights $h$ gave meaningful results such that $z_{h_1+h_2} = (z_{h1})_{h_2}$ .

{update
A much better way to compute the derivative of d/dh F^h is to use the derivative of the diagonalmatrix $D^h$ with repect to h. The code

 z_0=1;
    m_z0=vector(32,j,   sum(k=0,31, z_0^k * M[1+k,j] )  )     
 h=1 
    z_h=sum(k=0,31, m_z0[1+k] * log(D[1+k]) * D[1+k]^h * W[1+k,2])

$\hspace{120 px}$ represents the derivative of F with respect to $h$ and gives immediately the first derivative with respect to $h$ at $z_0=1$ and $h=1$ }

This Q&D-method can easily be reconstructed using Pari/GP. Note that the diagonalization needs large value for the internal decimal precision, I just used prec=800 (dec digits) for the matrixsize 32x32.

Short protocol ("a" means your sought value $a={z_h-2 \over h-1}$):

  h                   z_h          a
1.0001,          2.0001233, 1.2329608
1.000001,        2.0000012, 1.2329220
1.000000001,     2.0000000, 1.2329216
1.0000000000001, 2.0000000, 1.2329216

I did not find some closed form for this via google, W|A and inverse symbolic calculator, so possibly the same coefficient, just for the base $e$ instead, might be better known:

The computation for the base $b=e$ instead of $b=2$ gives the following approximation:

1.1000000, 3.0407711, 3.2248931
1.0100000, 2.7481969, 2.9915037
1.0010000, 2.7212519, 2.9700850
1.0001000, 2.7185786, 2.9679609
1.0000100, 2.7183115, 2.9677487
1.0000010, 2.7182848, 2.9677274
1.0000001, 2.7182821, 2.9677253
1.0000000, 2.7182819, 2.9677251
1.0000000, 2.7182818, 2.9677251
1.0000000, 2.7182818, 2.9677251

But no success for finding a closed form representation so far.

Added: With that idea of a derivative with respect to the height-parameter the idea of constructing a powerseries from the sequence of higher derivatives comes to mind. With the adapting of my derivative-formula (see par. "update") I construct a powerseries in $h$ which gives the $h$'th iterates starting from $z_0=0$. This are the first 32 coefficients: $$ g(h)= \small \begin{array} {rl} 1 & \\ +0.8893649182531790 & *h \\ +0.008676688896772925 & *h^{2} \\ +0.09523868310780171 & *h^{3} \\ -0.005752348511600163 & *h^{4} \\ +0.01296662374986303 & *h^{5} \\ -0.002196108907293045 & *h^{6} \\ +0.001996803031014021 & *h^{7} \\ -0.0005633994435825211 & *h^{8} \\ +0.0003482750168520301 & *h^{9} \\ -0.0001285546189885988 & *h^{10} \\ +0.00006709638266032520 & *h^{11} \\ -0.00002830779288402342 & *h^{12} \\ +0.00001380563705669089 & *h^{13} \\ -0.000006205175034368587 & *h^{14} \\ +0.000002957461339961192 & *h^{15} \\ -0.000001369764406300663 & *h^{16} \\ +0.0000006496675894767698 & *h^{17} \\ -0.0000003055056195438239 & *h^{18} \\ +0.0000001451337771274319 & *h^{19} \\ -0.00000006884715898369447 & *h^{20} \\ +0.00000003282131706893498 & *h^{21} \\ -0.00000001565967274219413 & *h^{22} \\ +0.000000007493152321891950 & *h^{23} \\ -0.000000003590500482888162 & *h^{24} \\ +0.000000001723905030908289 & *h^{25} \\ -0.0000000008288804738313791 & *h^{26} \\ +0.0000000003991564395933366 & *h^{27} \\ -1.924678487785590E-10 & *h^{28} \\ +9.292387572960278E-11 & *h^{29} \\ -4.491503965074198E-11 & *h^{30} \\ +2.173332915755159E-11 & *h^{31} \end{array} $$

This is then good at least for $h= 0 \ldots 1$ and should reproduce the Kneser-solution by $z_h=\exp_2^{\circ h}(0)=g(h-1)$ .
(This part is just current experimenting and might have errors)

Best Answer

Related Solutions

[Math] Curious $q$-analogues

Classical Analysis – Finding Closed Forms/Related Constants to a Limit Involving Tetration

Related Question