When we consider the Taylor Series expansion of $f(x)=b^x$ for some $b \in \mathbb{R}$, we see that $$b^x = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}x^n.$$ We can substitute $x$ for $b^x$ to find that $$ b^{b^{x}} = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{xn}.$$

Now, let's say that we want to find a function that accurately describes how we should raise $b$ to $b$'th power $x$ times. We say that $b^{(b)}_x$ says that $b$ should be raised to itself $x$ times, for some $x\in\mathbb{R}$. We can generalize the previous examples, and write $$b^{(b)}_x = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{(b)n}_{x-1}.\qquad \text{(1)}$$ (Please note that $b^{(b)n}_{x}$ is not equal to $b^{(bn)}_{x}$.) Using this equation we can also state that $$b^{(b)}_{x-1} = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{(b)n}_{x-2}.\qquad\text{(2)}$$ When we substitute (2) in (1), we find that $$ b^{(b)}_x = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}\Big( 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}b^{(b)n}_{x-2} \Big).$$ From now on, we write $ \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!} = A$, because then the next equations look more clear. We could go on substituting in this manner $*$, until we arive at $b^{(b)}_{(x-(x-1))}=b$. We then see that the equality $$ b^{(b)}_x = 1 + \sum A \Big( 1 + \sum A \Big(\cdots \Big(1 + \sum_{n=1}^{\infty} \frac{(\log(b^b))^n}{n!}\Big)\Big)\Big) $$ holds when we iterate $f(x) = 1 + \sum_{n=1}^{\infty}\frac{(\log(b))^n}{n!}$ an $x-2$ amount of times, for real $x$.

Question 1: Is this possible?

Question 2: If so, how is this done? How do you describe the formula that precisely defines how $f(x)$ looks like after being iterated $x-2$ times?

Question 3: If such a formula is found, would it imply that a nice way of finding the fourth hyper operator (or "tetration") is found for real numbers?



$*$EDIT: When we proceed in this manner when $x$ is not an integer, we will not find $x=1$. We should be able to iterate the function a real amount of times to find $x=1$.

  • 2
    $\begingroup$ You say "we could go on ...until we arrive at $b^{(b)}_{(x-(x-1))}=b$", but since you are iterating, the subindices you find as you go on are all of the form $x-n$ with $n$ an integer. So you will not find $x-1$ unless $x-1$ is one. $\endgroup$ – Mariano Suárez-Álvarez Nov 25 '10 at 20:50
  • $\begingroup$ @ Mariano Suarez-Alvarez: woops that true... Then the question would be: Is it possible to iterate f(x) a non-integer amount of times? How? $\endgroup$ – Max Muller Nov 25 '10 at 20:59
  • $\begingroup$ I've done a somehow similar discussion (some years ago, sloppy written: I had not much experience with writing at that time) but where I focused/arrived at a slightly different representation of the iteration on the power series. Perhaps you find this interesting: go.helms-net.de/math/tetdocs/TTetrationExactEntries_short.htm I've done a nice implementation in Pari/GP of this - however in the same way as @mariano remarked: this leads only to integer iterates and not to iterations to fractional heights. $\endgroup$ – Gottfried Helms Nov 19 '17 at 3:14

The process you describe can be better expressed in terms of the Bell-/Carleman-matrix (see Wikipedia ) associated to function $f(x) = b^x$ . I got myself used to the following notation:
let $V(x)$ denote a rowvector of infinite dimension of consecutive powers of an argument x $V(x)=[1,x,x^2,x^3, \dots]$ then let B denote the infinite matrix which performs $ V(x) * B = V(b^x) $ . The columns of B contain the coefficients for the powerseries for the consecutive powers of $(b^x)^0, (b^x)^1 , (b^x)^2, \ldots $. B is then the (transposed) Carleman-matrix.

What your formula and iteration constructs can simply be expressed by the notion of powers of B; the h-fold nested infinite sums (your A) are captured by the formal matrixproducts $ V(x)*B*B*\ldots*B = V(x)*B^h = V(b^(b^(b^\ldots (b^x))) $ . Unfortunately the fractional part of iteration must then be expressed by a fractional power of B --- which is not trivial.

But that is only one problem. We have already the problem of powers of B (or nested summation of your example): the convergence using matrices of finite size gets lost after few iterations and it is a special art to find closed forms of arbitrarily precise computable expressions for the powerseries only for third or fourth iteration. I've tried this different ways and could not get significant improvement of the convergence behaviour as long as I used that nesting which you describe above for more than some exotic bases b near 1.

But there are ways to express the coefficients of the powerseries in finite expressions/sums including exp/log which we can assume to be available always in arbitrary precision. This can be found by triangular decomposition of B and a modified way to arrive at the required powers. (See for a not very explanative technical description exact entries, remark: there might be better descriptions...)

But all this covers only the part of integer height (integer number of iterations). And for this we do not really need that formal powerseries/matrixpowers approximated by finitely truncated matrices: we have precise exponentiation for each base at hand. The crux is the part of the fractional height/iteration count. Here there are some different approaches available; one of it is the approach to compute fractional powers of B by diagonalization to get the coefficients for a formal powerseries which represents the fractional iteration of $b^x$ However - to have series with real coefficients we must restrict ourselves to bases b between $exp(-exp(1))\ldots exp(exp(-1)) $

  • $\begingroup$ Thanks Gottfried. I guess a lot of research still has to be done on tetration. $\endgroup$ – Max Muller Nov 26 '10 at 17:31

On Wikipedia, it is suggested that you use a piecewise to represent the xth tetration of b. How you do this depends on what level of differentiability you want for your representation. For example, if you just want a simple representation of $^xb$, where it’s continuous but not differentiable, you could represent it as

$^xb$ = $x+1, -1<x<0$

$b^{x}, 0<x<1$

$b^{b^{x-1}}, 1<x<2$

$b^{b^{b^{x-2}}}, 2<x<3$

And so on and so forth. As I’ve said, this function is continuous, though it’s not differentiable (except in the special case where b=e, as in Euler’s number). It is also worth noting that this function gives exact values for all integer values of x.

The base function of this graph is linear (x+1). However, if you want, you could use a quadratic form instead of a linear.

$^xb$ = $1+\frac{2ln(b)}{1+ln(b)}x-\frac{1-ln(b)}{1+ln(b)}x², -1<x<0$

$b^{1+\frac{2ln(b)}{1+ln(b)}(x-1)-\frac{1-ln(b)}{1+ln(b)}(x-1)²}, 0<x<1$

$b^{b^{1+\frac{2ln(b)}{1+ln(b)}(x-2)-\frac{1-ln(b)}{1+ln(b)}(x-2)²}}, 1<x<2$


In this form, the function is both continuous and differentiable, as well as a more accurate approximation.

If we use a cubic approximation, we can have it be twice differentiable for any real value x. For this, you have to define two functions: f(x) and g(x). We’ll assume $g(x)=b^{f(x-1)}$ and that f(x) is a cubic defined as $Ax^3+Bx²+Cx+1$

f(x) will be the approximation for $^xb$ for -1

In order to solve for the values of A, B, and C, you have to assume that when x=0, f and g have identical y values, derivatives, and second derivatives. From here, well end up using 3 variable substitution. You’ll end up solving for two of the three variables as a function of the other one and then solve for that one using the quadratic equation. (When I did it, I solved for B and C as functions of A and solved for A using the quadratic equation). Even though I solved this already, I tried typing it out and it doesn’t really fit when using the scripting that this website runs off of (in other words it’s too long).

Using this method, you can also solve for a quartic or quintic representation, and piecewise it out. If you wanted to go any further, you’d probably have to use numerical approximation methods to get any more accurate. That being said, you also might find yourself running into an issue where you find yourself with multiple possible solutions to the polynomial representation of f, so as a helpful hint, the value of y at x=0 should always be 1 and the 1st derivative at that point should be close to 1, depending on how big the value of b (your tetration base) is. Hope this helps, sorry if it’s a little hard to understand.

  • $\begingroup$ This can be for instance (and I think: with advantage) be expressed by truncated "Carleman-matrices" - but of increasing size. The larger the size the higher is the approximating polynomial which is generated when fractional powers for the involved functions/the basic Carlemanmatrix are made by diagonalization. For exponentiation with base $e$ it seems, that this method approximates the solution which was proposed by H. Kneser in the 50ties (?). See for some more ideas my small "comparision-of-methods" essay: go.helms-net.de/math/tetdocs/ComparisionOfInterpolations.pdf $\endgroup$ – Gottfried Helms Nov 19 '17 at 2:51
  • $\begingroup$ There is also the linear approximation for base $a>exp(1/e)$, where $a2=\frac{1}{\ln(a)};\;\;\;a1=\ln_a(a2)$, then if sexp_a(x1)=a1, then sexp_a(x1+1)=a2, and sexp'(x1)=sexp'(x1+1). Which leads to the the "linear" piece-wise once differentiable approximation approximation for any bases $a > \exp(1/e)$. This answer is also similar to Andrew Robbin's slog (inverse of tetration) piece-wise approximation which has a cleaner matrix solution. see allergrootste.com/big/Source/articles/… Kneser's solution is preferred to piece-wise equations. $\endgroup$ – Sheldon L Nov 19 '17 at 18:18

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