I would like to know what is known about Taylor series for tetration (and other hyper-exponentiations).

Surprisingly, such information is rare on internet. Numerical values for expansion in hyperexponent for tetration with basis $e$ can be found here:


However I am interested in expansion in the basis of tetration. I look for

$$ x \uparrow \uparrow m = \sum_{n=0}^{\infty} c_n(m)x^n $$

Wolfram mathematica gives me wired result (obviously a bug) at $x=0$ for $x \uparrow 3$:


where in the output $log(x)$ appears (not a polynomial).

It is better at $x=1$


I imagine the point $x=0$ may be "peculiar", but at least expansion at $x=1$ should be possible.

List of my questions:

1) Is tetration analytic at $x=0$? At $x=1$?

2) If yes, is an explicit closed-form formula for $c_n(m)$ known? (at any of these points)

3) If not, is an explicit closed-form formula for $c_n(m)$ know for some specific values of $m$?

4) Same questions for extension to higher hyper-exponentiations...

Thank you.

  • 1
    $\begingroup$ I little advice, don't loose too much time on these themes : either on the point of view of practical applications or on the theoretical side, hyper-exponentiations of different kinds are a little a deadend. $\endgroup$
    – Jean Marie
    Dec 8, 2017 at 15:13
  • $\begingroup$ I know it's a long time now, but I happened to be looking up the very same question. I couldn't help being amused by your little admonishon - yes it is a very frustrating subject all this hyperoperations business; and one gets the distinct impression that the series of mathematical operations defined recursively as the previous one applied to n instances of the variable does not (except for natural numbers alone) extend beyond the third - exponentiation. However, I do think this little 'island' of tractibility is truly a gem; and I would say to OP "have this amongst your crown jewels"! $\endgroup$ Nov 19, 2018 at 7:14
  • 1
    $\begingroup$ There should be a method for solving finding coefficients at $x=1$ of height $m$ and at $x=0$ for $e^x\uparrow\uparrow m$ as multiple finite sums like here. $\endgroup$ Mar 10 at 22:40
  • $\begingroup$ Late hint: after Tyme Gaidash's answer I remembered I had done such a discussion in 2008, but in terms of a "Bell-"/"Carleman-" matrix and had found some formalism in terms of powers of such a matrix and a final binomial transformation appended. Anyone might like to read through go.helms-net.de/math/tetdocs/TTetrationExactEntries_short.htm $\endgroup$ Apr 21 at 8:43

5 Answers 5


$x \uparrow \uparrow m$ is never analytic at $x = 0$ for $m > 1$. For example, consider $m = 2$, then you have the function $x \uparrow \uparrow 2 = x^x = e^{x \ln x}$ which clearly has a branch point at $x = 0$ and hence is not analytic. You can see this if we substitute $\ln x = 2\pi i n + \log x$ where $\log$ represents the principal branch of the logarithm, in which case we have $x^x = e^{x (2\pi i n + \log x)} = e^{x 2 \pi i n} e^{x\log x}$, which certainly depends on $n$ as long as $x$ is not an integer. You have similar problems for all $m > 1$.

The function $x \uparrow \uparrow m$ is always analytic at $1$, which can be seen inductively: Clearly for $m=1$ it is analytic at $1$. If $x \uparrow \uparrow m$ is analytic at 1, then $x \uparrow \uparrow{(m+1)} = e^{(\ln x)(x\uparrow \uparrow m)}$ is analytic at $1$ because $\exp$ is analytic everwhere and $\ln$ is analytic at 1. The Taylor series of $x \uparrow \uparrow m$ at $1$ is not very nice, but the terms can be computed explicitly. Wolfram Mathworld tells us that $$ (e^x) \uparrow\uparrow m = \sum_{n=0}^m \frac{(n+1)^n}{(n+1)!} x^n + \sum_{n = m+1}^\infty a_{m n} x^n$$ where $a_{m n}$ have a rather complicated recursive formula. To compute the Taylor series of $x \uparrow\uparrow m$ at $1$, you could compose the above series with the series for $\ln(x)$ at $1$ using Faa di Bruno's formula. Not very simple, but it probably as close to a closed form as you will get.

  • 1
    $\begingroup$ Surprisingly, discussing the tetration of the pascalmatrix $P$, I arrived at the same powerseries solution for the tetration as your's here. If interested, see go.helms-net.de/math/tetdocs/PascalMatrixTetrated.pdf at my webspace. $\endgroup$ Oct 14, 2020 at 11:20
  • 1
    $\begingroup$ Interesting (and probably a computationally friendlier method) but it shouldn't be surprising. Power series are unique, so whatever other tetration techniques you have they must agree with this for integer heights. $\endgroup$ Oct 14, 2020 at 16:50

Consider the tetration of the function $e^x$ $$^n(e^x)=(e^x)_1^{(e^x)_2^{(e^x)_3^{_..^{.e^x_n}}}}$$

For a natural number n, the taylor series of that function is


where $T(n,k)$ is the OEIS A210725 (if k<n, $T(n,k)=T(k,k)$)

then, for $n\in\mathbb{N}$:


On the limit for $n\to\infty$, the T(n,k)=T(k,k) are the coefficients of the Lambert w function.

There are other series, that you can find here.

For code able to calculate tetrations with high precision, (for a limited set of bases, and $n\in\mathbb{R}$) check Sheldonison's fatou.gp code

In Wolframalpha, you can get the series of $^n(x+1)$, and it will produce the Stirling transform of $^n(e^x)$. The series has power of x, and no logarithms.


Looking online, no explicit formula was found either. You can find an exact form for the coefficients at $z=a$, but the coefficients are simpler at $z=1$ or at $e^z=1$ respectively as multiple finite series. One can probably also use the Ramanujan Master theorem to integrate to find coefficients. However, the method for finding coefficients of the inverse of $^nz$ also works; one uses repeated general Leibniz rule, $e^y$’s Maclaurin series, and Kronecker delta properties for a very similar derivation. $$^kz=\sum_{n=0}^\infty\frac{(z-1)^n}{n!}\left.\frac{d^n}{dz^n}\,^kz\right|_1=\sum_{n=0}^\infty\frac{(z-1)^n}{n!}\sum_{m=1}^nS_n^{(m)} \left.\frac{d^m}{dz^m}\,^k(e^z)\right|_0 $$ We can use $\frac{\ln^k(x)}{k!} =\sum_\limits{n=k}^{\infty } \frac{(x-1)^n S_n^{(k)}}{n!},|x|<1$ or $^k z=\sum_\limits{n=0}^\infty \frac{\ln^n(x)}{n!}\left.\frac{d^n}{dz^n}\,^k(e^z)\right|_0$ both of which change the radius of convergence. $\frac{(-\ln(x))^k}{k!}=\sum_\limits{n=k}^{\infty } \frac{S_n^{(k)}}{n!} (\frac1x-1)^n,|x|>1$ may not apply here. Finally, we get: $$\bbox[2px,border: 3px solid red]{^k z=\sum_{n=0}^\infty\sum_{m=0}^n\sum_{n_1=0}^m\dots\sum_{n_{k-1}=0}^{m-\sum\limits_{j=1}^{k-2}n_j}\frac{(z-1)^nS_n^{(m)}m!n_{k-1}^{m-\sum\limits_{j=1}^{k-1}n_j}}{\left(m-\sum\limits_{j=1}^{k-1}n_j\right)!n_{k-1}!n!}\prod_{j=1}^{k-2}\frac{n_j^{n_{j+1}}}{n_j!}= \sum_{n=0}^\infty\sum_{n_1=0}^n\dots\sum_{n_{k-1}=0}^{n-\sum\limits_{j=1}^{k-2}n_j}\frac{\ln^n(z)n_{k-1}^{n-\sum\limits_{j=1}^{k-1}n_j}}{\left(n-\sum\limits_{j=1}^{k-1}n_j\right)!n_{k-1}!}\prod_{j=1}^{k-2}\frac{n_j^{n_{j+1}}}{n_j!}}$$

where the ratio of factorials are multinomial coefficients. We find the following special cases:

$$^1z=z\\^2z=\sum_{n=0}^\infty\sum_{m=0}^n\sum_{k=0}^m\frac{m!(z-1)^nS_n^{(m)}k^{m-k}}{(m-k)!k!n!}=\sum_{n=0}^\infty\sum_{k=0}^n\frac{\ln^n(z)k^{n-k}}{(n-k)!k!}\\^3z=\sum_{n=0}\sum_{m=0}^n\sum_{k=0}^m\sum_{j=0}^{m-k}\frac{m!(z-1)^nS_n^{(m)}k^j j^{m-k-j}}{(m-k-j)!k!j!n!}=\sum_{n=0}^\infty\sum_{k=0}^n\sum_{j=0}^{n-k}\frac{\ln^n(z)k^jj^{n-k-j}}{(n-k-j)!k!j!}\\^4z=\sum_{n=0}^\infty\sum_{m=0}^n\sum_{k=0}^m\sum_{j=0}^{m-k}\sum_{a=0}^{m-k-j}\frac{m!(z-1)^nS_n^{(m)}k^jj^aa^{m-k-j}}{(m-k-j-a)!k!j!a!n!}=\sum_{n=0}^\infty\sum_{k=0}^n\sum_{j=0}^{n-k}\sum_{a=0}^{n-k-j}\frac{\ln^n(z)k^jj^aa^{n-k-j-a}}{(n-k-j-a)!k!j!a!}\\^5z=\sum_{n=0}^\infty\sum_{m=0}^n\sum_{k=0}^m\sum_{j=0}^{m-k}\sum_{a=0}^{m-k-j}\sum_{b=0}^{m-k-j-a}\frac{m!(z-1)^nS_n^{(m)}k^jj^aa^bb^{m-k-j-a-b}}{(m-k-j-a-b)!k!j!a!b!n!}=\sum_{n=0}^\infty\sum_{k=0}^n\sum_{j=0}^{n-k}\sum_{a=0}^{n-k-j}\sum_{b=0}^{n-k-j-a}\frac{\ln^n(z)k^jj^aa^bb^{n-k-j-a-b}}{(n-k-j-a-b)!k!j!a!b!}\\\vdots$$

shown here. $\displaystyle 0^0\mathop=^\text{conv}1$ when $n=0$. However, extracting it seems to give the series expansion for $^{k-2}z$.

  • 1
    $\begingroup$ Tyma Gaidash - perhaps I've another way which leads to the same results but with less "literal overflow". This uses the notation in terms of (Carleman)-matrices and their powers. You might look at go.helms-net.de/math/tetdocs/TTetrationExactEntries_short.htm , an older work-out, which I think meets your results. (Unfortunately I seem to be strongly in the ageing-process and mind becomes too weak to engage much and so I don't attempt to prove (or disprove) the identity of my results with yours here.) $\endgroup$ Apr 21 at 8:07

I think really the extension of the progression of operations by introducing a new one that is the $n$-fold iteration of the thus-far highest one is a process that on the face of it kindof looks like it ought to be susceptible of being plied indefinitely ... but at the end of the day isn't! (Except in this case to the natural numbers only.) You get it with complex-quaternion-octonion-etc also - that progression seems also to fizzle-out rather disappointingly; and a similar principle seems to be abroad in very many areas of mathematics. Sometimes I so want this or that progression to be extensible without limit ... but then I find eventually that I need to resign myself to the first two or three items of the progression being beautiful special cases. I must admit, though ... sometimes it's actually quite a relief!

However - I do think that business of relating the tetration by $n$ of the exponential function to the Lambert w function as $n\rightarrow\infty$, and the way there is a taylor series for the function at finite $n$ with coefficients that can be found by a recursion algorithm - and the way they 'peel-away' term-by-term to leave the coefficients of the lambert w function: I do find that to be an extraordinarily sweet little oasis in the midst of a parched desert ... and it is certainly an item that I treasure amongst my crown jewels!


Just for your curiosity.

Expanding as Taylor series aound $x=1$ as $$f_n=\sum_{k=0}^\infty c_k\,(x-1)^k$$ see how fast converge the coefficients $$f_1=\left\{1,1,1,\frac{1}{2},\frac{1}{3},\frac{1}{12},\frac{3}{40},-\frac{1}{120},\frac{59} {2520},-\frac{71}{5040},\frac{131}{10080}\right\}$$ $$f_2=\left\{1,1,1,\frac{3}{2},\frac{4}{3},\frac{3}{2},\frac{53}{40},\frac{233}{180},\frac{56 27}{5040},\frac{2501}{2520},\frac{8399}{10080}\right\}$$ $$f_3=\left\{1,1,1,\frac{3}{2},\frac{7}{3},3,\frac{163}{40},\frac{1861}{360},\frac{33641}{504 0},\frac{8363}{1008},\frac{22391}{2160}\right\}$$ $$f_4=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{243}{40},\frac{3421}{360},\frac{71861}{504 0},\frac{54371}{2520},\frac{69281}{2160}\right\}$$ $$f_5=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4321}{360},\frac{102941}{50 40},\frac{85871}{2520},\frac{61333}{1080}\right\}$$ $$f_6=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4681}{360},\frac{118061}{50 40},\frac{106661}{2520},\frac{81583}{1080}\right\}$$ $$f_7=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4681}{360},\frac{123101}{50 40},\frac{115481}{2520},\frac{93013}{1080}\right\}$$ $$f_8=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4681}{360},\frac{123101}{50 40},\frac{118001}{2520},\frac{97333}{1080}\right\}$$ $$f_9=\left\{1,1,1,\frac{3}{2},\frac{7}{3},4,\frac{283}{40},\frac{4681}{360},\frac{123101}{50 40},\frac{118001}{2520},\frac{97333}{1080}\right\}$$

  • $\begingroup$ Are these continued fractions? $\endgroup$ Apr 3, 2021 at 10:56
  • $\begingroup$ @RanjitKumarSarkar. No, they are the coefficients $c_k$ $\endgroup$ Apr 3, 2021 at 13:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .