aa r X i v : . [ m a t h . C V ] J a n A Tetration Function By Unconventional Means
James David [email protected] 11, 2021
Abstract
The author makes use of infinite compositions and a limiting functionto sketch the construction of a holomorphic tetration function F ( s ) = e ↑↑ s . As a tetration function, F satisfies e F ( s ) = F ( s + 1). Of it, F isholomorphic on the domain C / L where L is a nowhere dense set; and F takes ( − , ∞ ) → R bijectively with strictly monotone growth. Keywords:
Complex Analysis; Infinite Compositions; Tetration.
The study of tetration can be traced back centuries. The first notable momentbeing, when Leonhard Euler proved that when e − e < α < e /e ; the infinitetower converges [2]. Quote unquote, for such α , αα... ( n times) ...α → A as n → ∞ Where α A = A and e − ≤ A ≤ e . This was a monumentous moment inthe study of iterated exponentials; and is probably the first truly publishableresult in the history of tetration–hell, complex dynamics. Though the subject oftetration remains rather dormant; if we do not count numerous forays into thecomplex dynamics of the exponential function; it poses itself as a very interestingproblem. The author would like to think, at the center of the study of iteratedexponentials is the study of tetration functions; and the quest for a good andright solution should be a priority. But as the pulse of the subject is weak, hefeels not many others feel the same way.A tetration function is simple enough to describe, but proves a very difficultfunction to construct (if we want a good and right solution). We’ll restrict ourattention to tetration functions with the base e . For bases b > e /e the resultis similar. If F is a function such that e F ( s ) = F ( s + 1) and F (0) = 1, then wecall F a tetration function. It provides a similar continuation of the sequence1 , e e , e e e , ... as e s provides a continuation of the sequence e, ee, eee, ... . In contrastto the exponential function though, it is a much more volatile construction.One facet of its volatility, tetration functions are highly non-unique. For anytetration function F the function F ( s + sin(2 πs )) is also a tetration function; asis F ( s + θ ( s )) for any 1-periodic function. Alors, a large problem with tetration,is qualifying a tetration function as unique, or as satisfying some property whichcharacterizes it from other tetration functions.There are quite a few trivial ways to construct a tetration function, forinstance letting F ( t ) = 1 + t for t ∈ [ − ,
0] and extending F to R + using thefunctional equation provides such a solution. But necessarily such a solution isnot differentiable on the natural numbers. One can construct many continuousextensions in a parallel manner.If we were to ask for a tetration function, we would at least require that it beanalytic. Even better, that it be holomorphic on some domain in the complexplane including R + .Of that end, Hellmuth Kneser was the first to undoubtedly provide a desir-able solution to the tetration equation [3]. Insofar as it took the real positiveline to itself and was holomorphic. He worked exclusively with the base b = e ,and managed to construct a function H ( s ) = s e holomorphic in C excluding anowhere dense set. His construction, although providing a stable solution, washighly esoteric and deeply expressed the difficulty of this problem. Constructinga holomorphic tetration function is no easy feat. In his defense, his sole goalwas to construct h such that h ( h ( z )) = e z and h was real-valued.More recently, numerous attempts have been made to construct a simplertetration function. There exists quite a few potential candidates for tetration;which exist scattered in the recesses of the internet. Since tetration has notgained widespread recognition as a notable problem–it is difficult to find pub-lished papers on the subject. Kneser’s own paper [3] is in German, and thereexists no English translations–but there are synopses and breakdowns.The problem with most modern approaches to tetration seem to be thatthese candidates can be numerically verified, but never rigorously justified toexist or converge. And in contrast; Kneser’s construction, which is correct, isa rather laborious numerical procedure. Even more troubling with these flurryof candidates to tetration; proofs of analycity become even more difficult–andin most instances do not exist. Alongside a lack of proof for mere convergence,this can be troubling. If that wasn’t bad enough; we can’t even prove if twocandidate tetration functions equal or not. But sometimes our floating pointaccuracy can appear to suggest so (or dissuade so).Nonetheless there is still great headway being made in the field. A treasuredexample is in the work of Dmitri Kouznetsov [4]. The author will not go intodetail on Kouznetsov’s method but will simply give a rough heuristic as motiva-tion. The author will blatantly steal this heuristic–but he shall bypass a few ofthe obstacles he believes Kouznetsov built for himself. Though, in truth, muchof the work parallels an extension of his trick.Supposing we had a nice function g ( s ), which has some desirable growthproperties, and we took the limiting function,2 n ( s ) = log log · · · ( n times) · · · log g ( s + n )Then e G n ( s ) = G n − ( s + 1). If the limit were to converge as n → ∞ , thenwe would have our tetration function G n → G . That is, upto a normalizationconstant ω where F ( s ) = G ( s + ω ) to ensure F (0) = 1.Kouznetsov chose a very wonderful function g , such that numerically every-thing worked out and provides us with a calculator’s version of a holomorphictetration function. The function g was constructed through careful fixed-pointanalysis; and looks like an exponential sum of terms e nLs for L a complex fixedpoint of e s . Unfortunately, to rigorously justify convergence proves rather dif-ficult. This becomes a sort of brick-wall. There is no in your hands proof thatKouznetsov’s method actually works.To that end, the goal of this paper is to construct our own g , and showthe convergence of the above limit. Our choice of g will be very manufactured,and requires a familiarity with infinite compositions. To that end, we refer thereader to [6], where sufficient conditions are provided for an infinite compositionto converge–and a familiarity with the subject is created. We shall not needanything from [6], but it exhibits the nuanced detail of the subject a bit moreclearly. We will prove a modified form of the main result of [6]; to keep thispaper self-contained; but we will give little to no motivating intuition in thispaper.The essential trick is to bypass the difficulty of constructing a tetrationfunction using Kouznetsov’s trick by constructing a function which satisfies asimilar functional equation, but exhibits the same growth properties. As youmay guess, tetration grows rapidly and rabidly in the complex plane, so we’llneed something similarly chaotic.To that end, our first goal is to construct an entire function φ ( s ) such that, φ ( s + 1) = e s + φ ( s ) And take φ to be our g from above. The real novelty of the work is held inconstructing φ . But it only really takes us writing out the equation for φ andjustifying convergence. This is more of a taxing process than a difficult one. Itis surprisingly simple to construct φ .* * *We introduce briefly the notation for nested compositions, which allows forour construction of φ . We will restrict from full generality, and only care abouta subset of types of infinite compositions. Therein, if h j ( s, z ) is a sequence ofentire functions in both variables, then, n Ω j =1 h j ( s, z ) • z = h ( s, h ( s, ...h n ( s, z )))Where we are interested in letting n → ∞ . The study of infinite compositionsis very nuanced, and for that reason constructing φ will require care. The type3f convergence we’ll need is one which is a bit simpler than the general case. Towit, we will call, φ n ( s ) = n Ω j =1 e s − j + z • z (cid:12)(cid:12)(cid:12) z =0 Where if h j ( s, z ) = e s − j + z then, φ n ( s ) = h ( s, h ( s, ...h n ( s, e s + φ n ( s ) = φ n +1 ( s + 1), so if this were to converge it wouldequal our desired function. The essential ingredient in our construction is forall compact disks P , K ⊂ C , ∞ X j =1 || h j ( s, z ) || s ∈P ,z ∈K < ∞ Where this is taken to mean the supremum norm. So without further ado,we construct φ . φ The first thing we need to construct φ is a sort of normality condition. For all ǫ >
0, there exists some N , such when m ≥ n > N , | m Ω j = n e s − j + z • z | < ǫ For | z | <
1, and s residing in some compact disk within C . This then impliesas we let m → ∞ , the tail of the infinite composition stays bounded. Forthwith,the infinite composition becomes a normal family, and proving convergence be-comes simpler. We provide a quick proof of this. Lemma 2.1.
For a compact disk
P ⊂ C and | z | ≤ : for all ǫ > , there existssome N , such when m ≥ n > N || m Ω j = n e s − j + z • z || P , | z |≤ < ǫ Proof.
Let | z | ≤ s ∈ P be a compact disk in C . Set h j ( s, z ) = e s − j + z andset || h j ( s, z ) || s ∈P , | z |≤ = ρ j . Pick ǫ >
0, and choose N large enough so when n > N , ρ n < ǫ Denote: φ nm ( s, z ) = Ω mj = n h j ( s, z ) • z = h n ( s, h n +1 ( s, ...h m ( s, z ))). We goby induction on the difference m − n = k . When k = 0 then || φ nn ( s, z ) || | z | < ,s ∈P = || h n ( s, z ) || | z | < ,s ∈P = ρ n < ǫ m − n < k , we show it holds for m − n = k .Observe, || φ nm ( s, z ) || | z | < ,s ∈P = || h n ( s, φ ( n +1) m ( s, z )) || | z | < ,s ∈P ≤ || h n ( s, z ) || | z | < ,s ∈P = ρ n < ǫ Which follows by the induction hypothesis because | φ ( n +1) m ( s, z ) | < ǫ <
1. The next step is to observe that Ω mj =1 h j ( s, z ) is a normal family as m → ∞ ,for | z | < s ∈ P , an arbitrary compact disk. This follows because the tail ofthis composition is bounded. Therefore we can say || Ω mj =1 h j ( s, z ) || | z | < ,s ∈P The expression ∞ Ω j =1 e s − j + z • z (cid:12)(cid:12)(cid:12) z =0 = φ ( s ) is an entire function satisfying the identity e s + φ ( s ) = φ ( s + 1) .Proof. Since φ m ( s, z ) = Ω mj =1 e s − j + z • z are a normal family; there is someconstant M ∈ R + such, || d k dz k φ m ( s, z ) || | z | < ,s ∈P ≤ M · k !Secondly, using Taylor’s theorem, φ m +1 ( s, z ) − φ m ( s, z ) = φ m ( s, e s − m − z ) − φ m ( s, z )= ∞ X k =1 d k dz k φ m ( s, z ) ( e s − m − z − z ) k k != ( e s − m − z − z ) ∞ X k =1 d k dz k φ m ( s, z ) ( e s − m − z − z ) k − k !Setting z = 0, the series on the right can be bounded by some C ∈ R + .Applying the obvious bounds, || φ m +1 ( s, − φ m ( s, || s ∈P ≤ C || e s − m − || s ∈P = Ae − m A ∈ R + . And we can see the telescoping series converges and φ m ( s )must be uniformly convergent for s ∈ P , and therefore defines a holomorphicfunction φ ( s ) as m → ∞ . Naturally e s + φ m ( s ) = φ m +1 ( s + 1), and so thereforethe functional equation is satisfied. τ The main philosophy of our approach to constructing is to add a corrective termto φ such that it becomes a function. The function φ already looks very closeto tetration, satisfying a similar functional equation, φ ( s + 1) = e s + φ ( s ) We will introduce a sequence of correction terms as follows:log log · · · ( n times) · · · log φ ( s + n ) = φ ( s ) + τ n ( s )Where inductively, starting with τ ( s ) = s and τ ( s ) = 0; τ n can be defined, τ n +1 ( s ) = log (cid:0) φ ( s + 1) + τ n ( s + 1) (cid:1) − φ ( s )= log φ ( s + 1) + log (cid:0) τ n ( s + 1) φ ( s + 1) (cid:1) − φ ( s )= s + φ ( s ) + log (cid:0) τ n ( s + 1) φ ( s + 1) (cid:1) − φ ( s )= s + log(1 + τ n ( s + 1) φ ( s + 1) )Our choice of log is defined implicitly by the relation, e φ ( s )+ τ n +1 ( s ) = φ ( s + 1) + τ n ( s + 1)And the restriction that τ n ( s ) is real on the real-line. The first thing tonote, is that for s = t ∈ R + , the sequence of functions τ n converge uniformlyon bounded intervals greater than T . This is because τ n +1 ( t + 1) /φ ( t + 1) < | log(1 + A ) − log(1 + B ) | ≤ | A − B | ,6 τ n +1 ( t ) − τ n ( t ) | ≤ | log(1 + τ n ( t + 1) φ ( t + 1) ) − log(1 + τ n − ( t + 1) φ ( t + 1) ) |≤ φ ( t + 1) | τ n ( t + 1) − τ n − ( t + 1) |≤ φ ( t + 1) φ ( t + 2) | τ n − ( t + 2) − τ n − ( t + 2) | ... ≤ Q nk =1 φ ( t + k ) | τ ( t + n ) − τ ( t + n ) | Recalling that τ ( t ) = t and τ ( t ) = 0 then, | τ n +1 ( t ) − τ n ( t ) | ≤ t + n Q nk =1 φ ( t + k )And here φ ( t ) is monotone increasing and unbounded, so for some T with t > T it is φ ( t ) ≤ λ < 1. Now, | τ m − τ n | ≤ m X j = n λ j | s + j | Of sorts; the telescoping sum converges uniformly on bounded intervals andwe are given a function τ : R + t>T → R + . This function acts such that, e F ( t ) = φ ( t ) + τ ( t )And e F is a function, albeit not yet normalized to e F (0) = 1, but there is anappropriate ω such that F ( t ) = φ ( t + ω ) + τ ( t + ω ) is a true tetration function.Also by taking logarithms, the domain can be extended to its maximal ( − , ∞ ).Call this function F .Going through the same motions as above one can derive that F ′ is a contin-uous function and that τ ′ n → τ ′ uniformly on bounded intervals. This is reallyno different then what we’ve already written. The function, τ ′ n +1 ( t ) = 1 + (cid:16) 11 + τ n ( t +1) φ ( t +1) (cid:17)(cid:16) τ ′ n ( t + 1) φ ( t + 1) − τ n ( t + 1) φ ( t + 1) φ ′ ( t + 1) (cid:17) Grinding the gears of this expression we get | τ ′ n +1 − τ ′ n | is a convergent seriesof the same form as above. This is more of a task than a problem. It is left tothe reader.The next brief fact we need is that F ′ ( t ) > t ∈ ( − , ∞ ). From theexpression above, τ ′ ( t ) − → t → ∞ . So, eventually τ ′ ( t ) > t ≥ T . It is no hard fact to notice φ ′ ( t ) > t ∈ R . Therefore, since,7 ′ ( t − 1) = ddt log F ( t )= F ′ ( t ) F ( t )By infinite descent we must have F ′ ( t ) > 0. Therefore of this nature wehave a differentiable (equally as bijective) inverse A = F − ( t ) : R → ( − , ∞ ).In laymen’s terms, amongst the jargon of people who study tetration; one callsthis the super-logarithm. It is a continuously differentiable Abel function of e t .En drame, A ( e t ) = A ( t ) + 1These facts will be reinforced throughout this paper. Nonetheless it helps tointroduce them when they can be conveniently introduced. We state this lessthan drastic theorem below. Theorem 3.1. For some ω ∈ R there exists a continuously differentiable tetra-tion function F ( t ) : ( − , ∞ ) → R such that F ′ ( t ) > and F is a bijection. Thisfunction F ( t ) can be expressed as, F ( t ) = lim n →∞ log log · · · ( n times ) · · · log φ ( t + ω + n ) Where, φ ( t ) = ∞ Ω j =1 e t − j + z • z (cid:12)(cid:12)(cid:12) z =0 This provides us with a continuously differentiable function F defined for( − , ∞ ), but it sadly says nothing of the case for complex numbers. This provesto be a much more exhausting challenge. But the challenge is perfectly man-ageable.To set the stage we’ll work on an easier case. The function φ is periodicwith period 2 πi , and therefore a very similar argument as that of above allowsus to construct F ( s ) for s ∈ R + 2 πik for k ∈ Z when k = 0. F will not bereal valued on these lines, but φ will be, which allows for the same bounds. Thesame initial conditions are given as τ ( s ) = s and τ ( s ) = 0, so the convergencefollows for t > T .As that, | τ n +1 ( t + 2 πik ) − τ n ( t + 2 πik ) | ≤ | τ n ( t + 1 + 2 πik ) − τ n − ( t + 1 + 2 πik ) || φ ( t + 1) |≤ | t + 2 πik + n | Q nj =1 | φ ( t + j ) | F ( t + 2 πik ) = φ ( t + ω ) + τ ( t + ω + 2 πik )Is a continuously differentiable function. Going further, if F ( s ) = 0 then s = − F to R + 2 πik for all k ∈ Z with k = 0 bytaking repeated logarithms.An important note, as s → −∞ , the functions τ ( s + 2 πik ) → L k where L k is a fixed point of the exponential map. The derivation of this comes from thelimit lim s →−∞ φ ( s ) = 0 so,lim s →−∞ e φ ( s )+ τ ( s +2 πik ) = lim s →−∞ φ ( s + 1) + τ ( s + 1 + 2 πik )lim s →−∞ e τ ( s +2 πik ) = lim s →−∞ τ ( s + 1 + 2 πik ) e τ ( −∞ +2 πik ) = τ ( −∞ + 2 πik ) e L k = L k The fixed points satisfy the conjugate identity L k = L − k . Once we’ve shown F is analytic; this can be derived because F ( s ) = F ( s ). Which follows because F : R + → R + . The idea then, depending on how we limit to negative infinity, weget different fixed points. This relationship will be mirrored for lim t →−∞ F ( t + iy ) = L for arbitrary y ∈ R . In that, as ℜ ( s ) → −∞ the function F ( s ) → L (orwe hit a branch-cut along the way) where L is a fixed point of the exponentialmap. This can be better codified if one thinks: iterated logarithms convergeto a repelling fixed point of e s or we get a branching problem (I.e: F hits zerosomewhere).Of the following discussion we can see that F ( s ) is continuously differentiableon the lines R + 2 πik for k = 0 and continuously differentiable on ( − , ∞ ). Wewere able to obtain this result because τ n +1 − τ n was bounded by a product ofthe form | t + n + 2 πik | Q nk =1 φ ( t + k ) . Since φ ( t ) grows super-exponentially, this product isvery well behaved. We had yet to mention this fact but 1 φ ( t ) is smaller than anyiterate of the exponential as t grows. Which is φ ( t ) ≤ ◦ n ( t ) for large enough t ≥ T .In order to get τ to be holomorphic it would be required that we can controlthis product well in the complex plane. Doing this is a bit of a hassle. Itnecessitates us understanding how φ grows. The main hurdle of this papercomes in what follows. 9 Obtaining lower bounds on φ This section will focus on the function φ ( s ) in the strip 0 ≤ ℑ ( s ) = y ≤ π forlarge positive values of t = ℜ ( s ). The idea is to show for some large value T the function | φ ( t + iy ) | > ǫ for some ǫ > t > T . Despite: thesimplicity of this statement; the fact φ grows super-exponentially on the realline; this statement proves to be the most difficult part of this construction todigest.Commence by assuming that | φ ( t + iy ) | < M ∈ R + for all t ≥ T and y fixed.Then by the functional equation, | φ ( t + 1 + iy ) | = e t + ℜ φ ( t + iy ) ≥ e t − M But this equation tends to infinity as t → ∞ . Therefore, we know instantlythat φ ( t + iy ) is unbounded as t → ∞ . The more important question is whetherit grows towards infinity or oscillates back and forth.As some preliminary remarks; this result is highly non-trivial. It mesheswell with intuition, but requires a good amount of heavy lifting. We’ll need toreduce the case for all 0 ≤ y ≤ π to the case y = π ; which in essence is theworst our function can grow like. The author had a more ambitious idea for thissection, but kept on failing; and upon failure gathered what he could to provethat eventually φ ( t + iπ ) is greater than 1 + ǫ and stays greater. The ambitiousproof the author, at first, thought was self-apparent, was that φ ( t + πi ) growsasymptotic to log ◦ ǫ ( t ) for ǫ → t → ∞ . And that which he cannot prove forthe life of him. But luckily, we can at least prove it stays above a point. Andthat’s really all that’s needed–fortunately.In short; φ does not grow super-exponentially in the complex plane as itdoes on the real positive line. In fact, if | φ ( t + πi ) | > t then | φ ( t + 1 + πi ) | = e t −| φ ( t + πi ) | ≤ 1. As such φ ( t + πi ) must be o ( t ).The only fact the author can think of to explain why this is so relies on aheavy study of complex dynamics. But, if N is a neighborhood in C then theorbits exp ◦ k ( N ) are dense in C [5]. Therefore, the neighborhoods F ( N + k )must be dense in C . And to do this, it must grow relaxed in some parts. Inaccordance, φ must grow slow in some parts; or otherwise the entire constructionwill collapse.The author couldn’t derive an exact asymptotic for φ , but he guesses themain term is something just less than t ; he hopes something like log ◦ /t ( t );which can be expressed as F ( A ( t ) − /t ).* * *To better accustom ourselves to the behaviour of φ in the complex plane wecan rewrite our function. It’ll look a bit stranger in this form but, ψ y ( t ) = e − iy φ ( t + iy ) = ∞ Ω j =1 et − j + e iy z • z (cid:12)(cid:12)(cid:12) z =0 ψ y ( t ) = et − e iy et − e iy e ... By analyzing the partial compositions, ψ n ( t ) = n Ω j =1 et − j + e iy z • z (cid:12)(cid:12)(cid:12) z =0 They satisfy the identity, ψ n +1 ( t + 1) = et + e iy ψ n ( t )Therefore, ψ y ( t + 1) = et + e iy ψ y ( t )And we can obtain the crude estimate, | ψ y ( t + 1) | ≥ et − | ψ y ( t ) | Now this estimate gets its worse when y = π where, ψ π ( t + 1) = et − ψ π ( t )So, the heuristic is, if we could construct a lower bound for this case we couldbound from below the other cases. So the manner of proof will involve workingfirstly on the case where y = π . As a preliminary remark, if the reader hasn’tnoticed: When y = π the function ψ π ( t ) is real valued and strictly positive.To this end, ψ ( t ) = ψ π ( t ) = ∞ Ω j =1 e t − j − z • z (cid:12)(cid:12) z =0 Where we notice that all we’ve really done is swap the sign in the z termfrom our definition of φ ( t ). A subtle difference, which as Dorothy would putit–I don’t think we’re in Kansas anymore, Toto. It is hereupon we have to fiddlewith our notation a bit. We will write, ψ m ( t, z ) = m +1 Ω j =1 e t − j − z • z The reader should note the 2 m + 1 in the upper index of our composition.And they should note that ψ m ( t, 0) converges uniformly to a continuously dif-ferentiable function as m → ∞ . Where now our recursion is a tad differentbut, ψ m +1 ( t, z ) = ψ m ( t, et − m − − e t − m − − z )We’ve doubled up on the exponentials here because we want to use that,11 t +1 − e t − z Starts to look like e − δe t for some δ > 0; and this has very fast convergenceto zero t → ∞ . But additionally, et − m − − e t − m − − z Tends to zero like O ( e − m ) as m → ∞ as well. We’ve chosen the upper index2 m + 1 because each of these ψ m ( t ) grow exponentially. This leads us to a moreregular idea of what our infinite compositions behave like; which is, ψ m ( t, z ) → ψ ( t ) = lim m →∞ ψ m ( t, z -value we choose the result still convergesto the same limit as when we set z = 0. We can think of this as iterationsconverging to a fixed point locally in z . This is perhaps the best way to thinkof it. So, as per the above, Theorem 4.1. The function φ ( s ) can be represented as, φ ( s ) = ∞ Ω j =1 e s − j + z • z for all z ∈ C .Proof. The limit function, φ ( s, z ) = ∞ Ω j =1 e s − j + z • z = m Ω j =1 e s − j + z • ∞ Ω j = m +1 e s − j + z • z = m Ω j =1 e s − j + z • ǫ • z per Lemma 2.1 → ∞ Ω j =1 e s − j + z • z (cid:12)(cid:12)(cid:12) z =0 as m → ∞ Because as m → ∞ we get ǫ → m > M we can take | z | < t > T such that inf | z | < | ψ m ( t, z ) | > ǫ . From this, | ψ m +1 ( t, z ) | = | ψ m ( t, et − m − − e t − m − − z ) | > ǫ Because, || et − m − − e t − m − − z || | z | < ≤ δ ≤ m > M and t > T . Therefore, by induction the result follows.12 heorem 4.2. Let, ψ ( t ) = ∞ Ω j =1 e t − j − z • z Then for some ǫ > there exists some T > such that, for all t > T : ψ ( t ) ≥ ǫ To derive the result for the general case, we need only run a cranking mech-anism. The function, ddy | ψ y ( t ) | = 0 iff y = kπ for k ∈ Z Where when k is even corresponds to maxima; and when k is odd corre-sponds to minima. Since this function is periodic in y with period 2 π we knowinstantaneously that | ψ y ( t ) | ≥ ψ π ( t ). We leave the cranking to the reader; it’sfairly laborious to say the least; but doesn’t take much more than punching inthe values.From henceforth, we’ll write λ = ǫ < 1. And we can see that,1 | φ ( t + iy ) | ≤ λ < τ is holomorphic To show τ is holomorphic only requires a few careful observations. Firstly, thereexists some T ∈ R + such that for all t = ℜ ( s ) ≥ T the value | φ ( s ) | ≥ ǫ .Therefore, of this form, 1 Q nk =1 | φ ( s + k ) | ≤ λ n The nature of which is that this product geometrically converges to 0. It isnot quite super-exponential–like what happens on the real-line, but nonethelesswe’ve done fairly well for ourselves. This is a very convenient bound. When welook at our τ functions, | τ n +1 ( s ) − τ n ( s ) | ≤ | φ ( s + 1) | | τ n ( s + 1) − τ n − ( s + 1) |≤ λ | τ n ( s + 1) − τ n − ( s + 1) | Because | log(1 + z ) − log(1 + z ) | ≤ | z − z | so long as | z | , | z | ≥ 2; whichbecause τ n looks like s , and s is very large; we are all good. Remembering that τ ( s ) = 0 and τ ( s ) = s , 13 τ n +1 ( s ) − τ n ( s ) | ≤ λ n | s + n | This converges uniformly in the strip T ≤ t = ℜ ( s ) ≤ T ′ and y = |ℑ ( s ) | ≤ Y for T, T ′ , Y > τ ( s ) is holomorphic. The next stepis to extend τ ’s domain of holomorphy, from these boxes to a maximal domain.Using the functional equation, since, φ ( s + 1) + τ ( s + 1) = eφ ( s ) + τ ( s )By taking logarithms τ can be extended. The only difficulty we could haveis if τ ( s + 1) = − φ ( s + 1); wherein no logarithm can be taken and τ must havea singularity. We will defer a proof that this is impossible; the author doesn’tknow. But nonetheless, if we continue this implicit definition, the worst thathappens when extending F is that we hit a branch cut at some point. So thenice way to say this is that, F ( s ) is holomorphic on C upto a nowhere dense setWhich is really just a nice way of saying–I have no idea where the branchcuts are, or if there are even branch cuts; I’m just playing it safe. Don’t sue me.With this we state our titular theorem: Theorem 5.1 (The Tetration Existence Theorem) . The function F from The-orem 3.1 can be analytically continued to a function which satisfies the followingproperties:1. F is holomorphic for s ∈ C / L where L is a nowhere dense set.2. F ( s + 1) = e F ( s ) F (0) = 1 F : R + → R + , and F ′ : R + → R + It is best described; for some ω ∈ R ; as, lim n →∞ log log ... ( n times ) .... log φ ( s + ω + n ) = F ( s ) = e ↑↑ s In Conclusion We have constructed a function, but there lies a more daunting problem. It isnecessary to describe this function more acutely. The author is unsure of thebehaviour e ↑↑ t + iy as y → ∞ , nor of which fixed points of e s the function e ↑↑ t + iy tends to as t → −∞ depending on y . Just as well, he knowsno method of constructing a uniqueness criterion for this function dependingentirely on its behaviour on the real positive line. Though, he suspects the eventual monotonicity of each derivative may suffice–perhaps hinging on some14xtraneous growth condition. We haven’t proved this here; the author considersit an insufficient theorem, and would much rather have monotonicity on all of R + for all F ( n ) –not just eventually for large enough t > T . Can the readerprove this: For all n there exists T such for t > T the function F ( n ) ( t ) > ?If they can, can they prove the stronger result that F ( n ) ( t ) > for all t ∈ R + ?The author would be eternally grateful. We do not know much about this solution qualitatively. Does this solutionagree with Kneser’s solution? Does this solution agree with any other solution?Where are the zeroes of e ↑↑ s ? And consequently, where are the branch cuts?The author doesn’t; but he hopes to find out; and hopes others do as well. References [1] Amdeberhan, Tewodros. Espinosa, Olivier. Gonzalez, Ivan. Harrison, Mar-shall. Moll, Victor. Straub, Armin. (2012). Ramanujan’s Master Theorem .The Ramanujan Journal.[2] Euler, Leonhard. (1921). De serie Lambertina Plurimisque eius insignibusproprietatibus. Acta Acad. Scient. Petropol. 2, 29-51, 1783. Reprinted inEuler, L. Opera Omnia, Series Prima, Vol. 6: Commentationes Algebraicae.Leipzig, Germany: Teubner, pp. 350-369.[3] Kneser, Hellmuth. (1950). Reelle analytische Losungen der Gleichung ϕ ( ϕ ( x )) = e x und verwandter Funktionalgleichungen. Journal fur die reineund angewandte Mathematik. 187: 56-67[4] Kouznetsov, Dmitrii. (2009). Solution of F ( z + 1) = exp F ( z ) in complex z -plane. Mathematics of Computation. 78. 10.1090/S0025-5718-09-02188-7.[5] Milnor, John. (2006). Dynamics In One Complex Variable. Princeton Uni-versity Press.[6] Nixon, James D., (2019). ∆ y = e sy , Or: How I Learned To Stop WorryingAnd Love The Γ -Function.-Function.