# Could real iterates of the Taylor Series expansion of $b^x$ help to find a way to define tetration?

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?

Thanks,

Max

$*$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 $x=1$.

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Punctutation marks go inside the  symbols in displayed equations, and outside the $-symbols for in-line equations. – Arturo Magidin Nov 25 '10 at 19:54 @ Arturo Magidin: oh ok yes that probably looks a lot nicer. – Max Muller Nov 25 '10 at 19:55 You say "we could go on ...until we arrive at$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$is one. – Mariano Suárez-Alvarez Nov 25 '10 at 20:50 @ Mariano Suarez-Alvarez: woops that true... Then the question would be: Is it possible to iterate f(x) a non-integer amount of times? How? – Max Muller Nov 25 '10 at 20:59 add comment ## 1 Answer The process you describe can be better expressed in terms of the Bell-/Carleman-matrix (see Wikipedia ) associated to function$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)) \$

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