1
$\begingroup$

Does anyone think that tetration by a non-integer will ever be defined ... really properly?

Great mathematicians struggled with finding an implementation of the non-integer factorial for a long time to no avail ... and then eventually Leonhardt Euler devised a means of doing it, & by a sleight-of-mind that was just so slick & so simple in its essence ... and yet so radical! Is there any scope for another such sleight-of-mind as that, whereby someone might do similarly for tetration, or have they all been used-up? It's impossible to conceive of how there might be any truly new conceptual resource left of that kind. But it's actually tautological that that is so, because the kind I am talking about is precisely the kind that is essentially radically new, & of even how it might be thought hitherto unconceived!

But the attempts I have seen so far at defining tetration by general real number look to me for all the world like mere interpolation - some of them indeed very thorough & cunning & ingenious (insofar as I can follow them atall) - but lacking that spark of essential innovation that is evinced in Euler's definition of the gamma function.

Just incase it seems I have gotten lost in philosophy, I'll repeat the question: will there ever be a definition of tetration by general real number that resolves the matter as thoroughly as Euler's definition of the gamma-function resolved the matter of factorial of general real number?

$\endgroup$
  • $\begingroup$ I am no expert on history, but googling yields the name "Leonhard Euler." I am unsure where you get the "t"... $\endgroup$ – Mohammad Zuhair Khan Nov 21 '18 at 16:25
  • $\begingroup$ I once read the appendix to TE Lawrence's Seven Pillars of Wisdom ... I think it was a bad influence on me! $\endgroup$ – AmbretteOrrisey Nov 21 '18 at 16:35
2
$\begingroup$

We're trying, but it's hard.

A better analogy than the Gamma function would be the way you can now define $x^y$ for real $y$. Why does $x^{3/2}$ make sense? Because I can solve $y^2=x^3$. (Admittedly any solution $y=y_0$ implies $y<y_0$ is a solution too, but we have a convention to get around that for $x>0$.)

So what would $^{3/2}x$ mean? Presumably, a solution of $y^y=x^{x^x}$. Unfortunately, the values of $y^y$ for $y\in (0,\,\frac{1}{e})$ are repeated again for some $y>\frac{1}{e}$ in a... not particularly simple way, so it's already getting confusing. There are similar headaches when trying to define $^{k/(2l)}x$ for $x>0,\,k,\,l\in\mathbb{N},\,2\nmid k$.

I'm not sure whether you can even prove $^y x$ with $y$ irrational can be defined by continuity, i.e. whether we can prove any rational sequence $y_n$ with $\lim_{n\to\infty}y_n=y$ gives the same $n\to\infty$ limit of $^{y_n}x$.

Having said all that, I bet we'll have made a lot of progress within 200 years (even if only in proving what we can't do).

$\endgroup$
  • $\begingroup$ I can't answer this as thoroughly as I would like to at the present moment - but one little item stands out - to me, at least, the most fundamental definition of $x^a$ for general real $a$ is that it is the solution of the differential equation $dy/dx =ay/x$ & $1^a = 1$ forall $a$. But I am very fond of that scheme of mathematics inwhich functions are essentially defined by differential equations - as being solutions of them - that as the axiom of what they are. $\endgroup$ – AmbretteOrrisey Nov 21 '18 at 16:43
0
$\begingroup$

The idea of E.Schroeder in the 19'th century for bases $b$ (allowing two real fixpoints for iterations, for instance $b=\sqrt{2}$) , sometimes called "regular iteration" seems to me a real good one. It allows a meaningful expression for fractional iteration from some starting point $z_0$ towards some endpoint $z_h$ where $h$ means the (possibly fractional, even complex) iteration-(h)eight, such that $z_0$ is the initial value, $z_1 = b^{z_0}$ is the first (integer) iteration and so on.

However, that method needs conjugacy to be able to be applied to all real $z_0$ : one has to choose the appropriate fixpoint for shifting the power-series for the exponentiation with base $b$ and get an evaluatable analytical answer at all.
Unfortunately it has been observed, that in the cases $z_0$ where each fixpoint-conjugacy can be taken, the results of fractional iteration are different - even if only by some 1e-25 or so.

A basic problem for the general $b$ is the multivaluedness of the complex exponentiation/logarithmization, or say, the "clock-arithmetic" with respect to the $2 \pi î$-term in the exponentiation - I have once seen an article (R.Corless & al.) on the "winding number" which tries to make sense of introduction of one more parameter for the complex numbers to overcome that "clock-arithmetic" to a real-arithmetic. But that was no real progress for the problem here.

So I think, similarly to the full workout of L. Euler about the multivaluedness of the logarithm and then the representation of the gamma-function as an infinite sum of partial products, we need some more idea here - the E. Schroeder-idea seems to me just like a small insular solution, however nice ever...

(just as a remark: you might consider the two common versions of the interpolation of the Fibonacci numbers $fib(n)$ to continuous functions of the index: there is one ansatz providing real numbers for real index, and another ansatz (in analogy perhaps to the Schroeder method) which gives complex numbers for real indexes but seems more smooth when seen overall in the complex plane. See a small essay about this at my homepage)

$\endgroup$
  • $\begingroup$ I presume that by "the first ansatz" you mean the φⁿ±φ⁻ⁿ thing. I have nod idea about the other, but I am very curious about it now, especially as you say it is an instance of the Schroeder method ... which sounds like it could do with 'tapering' by means of a nice familiar & relatively simple example for someone coaching themself in it from the beginning! Anyway - thank-you for the information ... but it dashed my hopes a bit when you began to speak of it being yet another insular effort. But I think (i) it does require fundamentally new thought (ii) it is there - yet to be found! $\endgroup$ – AmbretteOrrisey Dec 7 '18 at 20:36
  • $\begingroup$ @AmbretteOrrisey: thanks for your comment. Unfortunately I'm much busy this and next days and have no space to step in again. Let's see end of next week... $\endgroup$ – Gottfried Helms Dec 8 '18 at 9:12
  • $\begingroup$ Of course I don't expect you to dispense me a course on the Schroeder method! On the contrary, thankyou for taking the trouble to steer my attention in that direction. And I do often just throw ideas out without the expectation that those who catch them process or develop them for me! $\endgroup$ – AmbretteOrrisey Dec 8 '18 at 12:31
  • $\begingroup$ @AmbretteOrrisey: I've found my old discussion about the interpolation of the fibonacci numbers, I've added the link to my remark in my answer. First I thought I'd posted that as Q&A here in MSE but I did not know this place here and communicated via the sci.math-newsgroup in the usenet. Hope the essay is helpful/explanative about my remark. $\endgroup$ – Gottfried Helms Dec 9 '18 at 2:47
  • $\begingroup$ Looks ike you're giving me a course of instruction anyway! That's your work? I think I can discern your style of writing in it. I've only just looked at it - it'll take a while for that to mature. I see the 'other' form of the 'continuous-isation' of the fibonacci numbers - a spiral in the complex plane. Thanks for your ongoing attention ... please don't be tempted to make inroads into your time ... it wasn't meant as a subtle goad or anything when I said I don't expect a course of instruction, and plenty to be fornow.¶ Thanks for that direction - I shall assuredly enjoy delving into that. $\endgroup$ – AmbretteOrrisey Dec 9 '18 at 8:42
0
$\begingroup$

There is a remarkable expression I've found in this connection that might have some bearing on the matter for the iterates the Taylor series of $$\operatorname{f}(x)\to x^{\operatorname{f}(x)}$$ when $x\equiv e^z$, such that $$\operatorname{f_0}(x)\equiv1 ,$$$$\operatorname{f_1}(x)\equiv \exp(z) ,$$$$\operatorname{f_2}(x)\equiv \exp(z\exp(z)) ,$$ etc. The coefficients for $k=0\dots n$ are those of the Lambert W-function; but thereafter, for $k>n$ the coefficients are given by the following recursion. Let $a_{n,0}=1\forall n$, $a_{0,k}=0$ for $k>0$, & thereafter $$a_{n,k}={1\over n}\sum_{j=1}^k ja_{n,k-j}a_{n-1,j-1} .$$ These coefficients are in a sense 'wasted' upthrough $k=n$, inthat they do not actually appear in the Taylor series; and yet they still serve the function of being necessary for the generation of the coefficients that do appear. It is fascinating to my mind, the way there is a kind of discontinuity in the series - the coeffiecients generated by this recursion 'peeling-away' one-at-a -time as $n$ is incremented, 'revealing' the coefficients of the Lambert W-function 'underneath'; and it is a well known result that the limit as the $$\operatorname{f}(x)\to x^{\operatorname{f}(x)}$$ tends to $\infty$ is indeed the Lambert W-function.

Whether this is susceptible of treatment by Schroeder's method I would not venture definitely to say at the present time, as, though I see in outline how that method can be applied to something like the recursion that gives the Fibonacci numbers, I am rather daunted by that discontinuity in the generation of the coefficients in this case; and I cannot see at a glance how it would be encoded.

$\endgroup$
0
$\begingroup$

I just found some discussion about exactly your ansatz in your own answer, but without successful route to an end at, for instance, a half-iterate of the function. I'd posted this in the "tetration-forum" in about 2009

The final conclusion of that study was (see bottom of this answer): So for that approach: it looks as if we cannot express a half-iterate based on this type of powerseries. Pity.... Maybe we can find a workaround - change order of summation or something else, don't have an idea.

Now to my message itself (date-of-saving:16 Mar 2009):


Here I present three postings in sci.math. It seems, that the method is not well suited for the interpolation to fractional heights (as I hoped it would be). But - perhaps we can find a workaround. On the other hand: it is not needed that many different methods exist, so...

Also Ioannis (Galidakis) reminded me of the entry in mathworld,"powertower", where he already characterized this type of series. (http://mathworld.wolfram.com/PowerTower.html)

Here the current msgs aus sci.math: ( some edits in double-brackets [< >])

*subject: tetration: another family of powerseries for fractional iteration*         

Maybe this is all known; I didn't see it so far. The idea was triggered by
the comments of V Jovovic in the OEIS concerning the below generating functions.

Consider the sequence of functions

T0(x) = 1, T1(x) = exp(x*1), T2(x) = exp(x*exp(x)), T_h(x) = exp(x*T_{h-1}(x)),...

They are also the generation-functions for the following sequence of powerseries:
T0:  1 + 0 +   0   + ....
T1:  1 + x + 1/2*x^2 +  1/6*x^3 +   1/24*x^4 +    1/120*x^5 +     1/720*x^6 +      1/5040*x^7 ...
T2:  1 + x + 3/2*x^2 + 10/6*x^3 +  41/24*x^4 +  196/120*x^5 +  1057/720*x^6 +   6322/5040*x^7 +...
T3:  1 + x + 3/2*x^2 + 16/6*x^3 + 101/24*x^4 +  756/120*x^5 +  6607/720*x^6 + 160504/5040*x^7 + ...
T4:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1176/120*x^5 + 12847/720*x^6 + 229384/5040*x^7 + ...
T5:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16087/720*x^6 + 257104/5040*x^7 + ...
T6:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ...
T7:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ...
T8:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ...
T9:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ...
...
Too: 1 + x + 3/2*x^2 + 4^2/3!*x^3 + 5^3/4!*x^4 + 6^4/5!*x^5 +    7^5/6!*x^6 +      8^6/7!*x^7 + ... //limit h->inf

That means, if x = log(b), we have by this

 T0(x) = 1
 T1(x) = b      = b^^1
 T2(x) = b^b    = b^^2
 T3(x) = b^b^b  = b^^3
 ...
 Too(x) = ...^b^b   = b^^oo

and for the limit h->inf we have with Too(x) the series for the h-function of b: Too(x) = h(b)
which is convergent for |x|<exp(-1)

[<...>]

The 2.nd msg:

> > (Galidakis replies) : 
> >                          However, the recursive expression for the coefficients
> > given in (6) [<in mathworld, G.H.>] does not seem to allow that.
> > 
> > If you can find a way to interpolate between those coefficients for non-natural
> > heights using your matrix method AND at the same time you manage to preserve the
> > functional equation F(x + 1) = e^{x*F(x)}, then, by Jove, you've got a nice
> > analytic solution to tetration :-)

Ok, let's give a start. Recall:

T0:  1 + 0 +   0   + ....
T1:  1 + x + 1/2*x^2 +  1/6*x^3 +   1/24*x^4 +    1/120*x^5 +     1/720*x^6 +      1/5040*x^7 ...
T2:  1 + x + 3/2*x^2 + 10/6*x^3 +  41/24*x^4 +  196/120*x^5 +  1057/720*x^6 +   6322/5040*x^7 +...
T3:  1 + x + 3/2*x^2 + 16/6*x^3 + 101/24*x^4 +  756/120*x^5 +  6607/720*x^6 + 160504/5040*x^7 + ...
T4:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1176/120*x^5 + 12847/720*x^6 + 229384/5040*x^7 + ...
T5:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16087/720*x^6 + 257104/5040*x^7 + ...
T6:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ...
T7:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ...
T8:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ...
T9:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ...
...
Too: 1 + x + 3/2*x^2 + 4^2/3!*x^3 + 5^3/4!*x^4 + 6^4/5!*x^5 +    7^5/6!*x^6 +      8^6/7!*x^7 + ... //limit h->inf

We want to interpolate for coefficients of T0.5, means between rows T0 and T1.

I'll rewrite the powerseries without the powers of x. And since we do
the binomial composition of coefficients at like powers of x, we compose
the coefficients down a column; so the common denominator(the factorial) of a
column can be omitted for the scheme.
Thus I get for the original coefficients, only rescaled

  T0:   1  0  0   0    0     0      0       0   ...
  T1:   1  1  1   1    1     1      1       1   ...
  T2:   1  1  3  10   41   196   1057    6322
  T3:   1  1  3  16  101   756   6607   65794
  T4:   1  1  3  16  125  1176  12847  160504
  T5:   1  1  3  16  125  1296  16087  229384
  T6:   1  1  3  16  125  1296  16807  257104
  T7:   1  1  3  16  125  1296  16807  262144
  T8:   1  1  3  16  125  1296  16807  262144
  T9:   1  1  3  16  125  1296  16807  262144
    ...
The first binomial-composition along the columns gives
  X0:   1   0    0    0     0      0       0        0 ...
  X1:   0   1    1    1     1      1       1        1 ...
  X2:   0  -1    1    8    39    194    1055     6320
  X3:   0   1   -3  -11   -19    171    3439    46831
  X4:   0  -1    5    8   -37   -676   -7243   -64744
  X5:   0   1   -7    1   105   1021    7357    21589
  X6:   0  -1    9  -16  -161  -1026   -3301    67304
  X7:   0   1  -11   37   181    631   -3605  -168125
  X8:   0  -1   13  -64  -141    104   10961   246224
  X9:   0   1  -15   97    17   -999  -16007  -278711
  ...       ...
The second binomial-composition (using h=0.5)
[< Table 5: this will be the reference-table for the composition of coefficients of T05 >]
  Y0:   1          0             0            0            0              0                0  ...
  Y1:   0        1/2           1/2          1/2          1/2            1/2              1/2  ...
  Y2:   0        1/8          -1/8           -1        -39/8          -97/4          -1055/8
  Y3:   0       1/16         -3/16       -11/16       -19/16         171/16          3439/16
  Y4:   0      5/128       -25/128        -5/16      185/128         845/32        36215/128
  Y5:   0      7/256       -49/256        7/256      735/256       7147/256        51499/256
  Y6:   0    21/1024     -189/1024        21/64    3381/1024      10773/512       69321/1024
  Y7:   0    33/2048     -363/2048    1221/2048    5973/2048     20823/2048     -118965/2048
  Y8:   0  429/32768   -5577/32768      429/512  60489/32768     -5577/4096   -4702269/32768
  Y9:   0  715/65536  -10725/65536  69355/65536  12155/65536  -714285/65536  -11445005/65536  ...
     ...      ...
----------------------------------------------------------------------------------------------------
sum.    s0     s1             s2           s3     ...
====================================================================================================
T0.5:   c0     c1             c2           c3     ...

and                T0.5(x) = c0 + c1*x + c2*x^2/2! + c3*x^/3! + ...

the interpolated coefficients c0,c1,c2,... for h=0.5 should then be computed by the
column-sums (and finally the rescaling by the omitted factorials).

The partial sums in the columns converge only badly if at all, so let's look,
whether we can find some analytic solution.
The denominators in the rows can be majorized by powers of 4, and all can then be divided by
2, so let's rewrite this
                                                                                                       common scaling
  Y0:   1/2     0       0       0      0         0          0           0            0             0   *2 /4^0
  Y1:     0     1       1       1      1         1          1           1            1             1   *2 /4^1
  Y2:     0     1      -1      -8    -39      -194      -1055       -6320       -41391       -293606   *2 /4^2
  Y3:     0     2      -6     -22    -38       342       6878       93662      1219314      16331654   *2 /4^3
  Y4:     0     5     -25     -40    185      3380      36215      323720      2128445      -5199340   *2 /4^4
  Y5:     0    14     -98      14   1470     14294     102998      302246     -9722034    -332756410   *2 /4^5
  Y6:     0    42    -378     672   6762     43092     138642    -2826768    -93176118   -1954258068   *2 /4^6
  Y7:     0   132   -1452    4884  23892     83292    -475860   -22192500   -463551132   -7659247332   *2 /4^7
  Y8:     0   429   -5577   27456  60489    -44616   -4702269  -105630096  -1778712507  -23047084632   *2 /4^8
  Y9:     0  1430  -21450  138710  24310  -1428570  -22890010  -398556730  -5760084330  -51266562490   *2 /4^9
     ...      ...
----------------------------------------------------------------------------------------------------
sum.    s0     s1     s2    s3     ...
====================================================================================================
T0.5:   c0     c1     c2    c3     ...

and                T0.5(x) = c0 + c1*x + c2*x^2/2! + c3*x^/3! + ...






Let's look at the columnsums of the table; that sums, divided by the factorial, give the coefficients
c_k for the T0.5(x)-powerseries.

First, s0 = 1, (remember the scaling extracted to the rhs) ,
    so c0 = 1

Next, s1. Here we recognize, that the numbers are the catalan-numbers, and, with the
current scaling have the generation-function  1- sqrt(1-z). Since we want to know
simply the sum, we set z=1 and get for the sum
        s1 = 1- sqrt(1-1) = 1
 so     c1 =1

Next, s2. It becomes more difficult. We can add columns s2 and s1 to get a sequence,
which can formally be expressed as the derivative of the sqrt(1 - z)-function, where
possibly we need also a scaling at z, so likely something like

  1 - sqrt(1 - a z)'

It looks, as if the series is divergent, too, so we'll have to see, whether this
operation (and the following, which surely are similar) can be justified/make sense
at all.

-----------------              

I proceeded for the first few terms s2,s3,s4,s5... Things seem to come out uneasy... :( Now follows msg 3:

(...)

Formally composed by derivatives of sqrt(1-z) I get for the series s1,s2,s3,... the following 
generating functions

s0:    1
s1:    1 -    1*sqrt(1-z)
s2:    3 -    3*sqrt(1-z) +    2*z*(sqrt(1-z)')
s3:   16 -   16*sqrt(1-z) +   15*z*(sqrt(1-z)') -   3*z^2*(sqrt(1-z)'')
s4:  125 -  125*sqrt(1-z) +  124*z*(sqrt(1-z)') -  42*z^2*(sqrt(1-z)'') +  4*z^3*(sqrt(1-z)''')
s5: 1296 - 1296*sqrt(1-z) + 1295*z*(sqrt(1-z)') - 550*z^2*(sqrt(1-z)'') + 90*z^3*(sqrt(1-z)''')  - 5*z^4*(sqrt(1-z)'''')
 ...
which have to be evaluated at z=1 to give the value for the sums. Now the derivatives have 
a vertical asymptote at z=1, so here are infinities everywhere...

Even more obvious, if I expand the derivatives into terms of sqrt(1-z) I get the following
explicite generating functions for the series of s0,s1,s2,...:

s0: 1
s1: 1     - sqrt(1-z)        * (   1 )
s2: 3     - sqrt(1-z)/(1-z)^1* (   3 -    4/2*z)
s3: 16    - sqrt(1-z)/(1-z)^2* (  16 -   49/2*z +    31/4*z^2 )
s4: 125   - sqrt(1-z)/(1-z)^3* ( 125 -  626/2*z +   962/4*z^2 -   408/8*z^3 )
s5: 1296  - sqrt(1-z)/(1-z)^4* (1296 - 9073/2*z + 22784/4*z^2 - 23462/8*z^3 +  7561/16*z^4)

where all except the first two grow unboundedly, if z->1

So for that approach: it looks as if we cannot express a half-iterate based on
this type of powerseries. Pity.... Maybe we can find a workaround - change order
of summation or something else, don't have an idea.

Another idea around?


(end of that msg to the tetration-forum)

$\endgroup$
  • $\begingroup$ Been a tad absent from here lately ... and I see you've been rather busy at my post in the meantime. I've actually been marshaling some thoughts on the inverse Ackermann function, particularly in connection with Davenport Schinzel sequences, and have a post about it nearly ripe. I think it will chime with what you have contributed here! $\endgroup$ – AmbretteOrrisey Jan 2 at 6:14
  • $\begingroup$ @AmbretteOrrisey: you're welcome. And happy new year! Unfortunately I likely shall have no new ideas in all this, but of course would like it much if there comes out some connections with my own older stuff. With the inverse of the Ackermann it is perhaps useful to contact Mr. Daniel Geisler who is founding member of the tetration-forum and has also an account here in MSE or MO and is spuriously active on questions on tetration. Perhaps via email you might be able to install some helpful connection. $\endgroup$ – Gottfried Helms Jan 2 at 9:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.