This question have been driving me crazy for months now. This comes from work on multiple integrals and convolutions but is phrased in terms of formal power series.

We start with a formal power series

$P(C) = \sum_{n=0}^\infty a_n C^{n+1}$

where $a_n = (-1)^n n!$

With these coefficients the formal power series can be expressed as a hypergeometric function

$P(C) = C \, _2F_0(1,1;;-C)$

I'm then interested in the formal power series $P_T(C)=\frac{P}{1-P}$ as well as, if possible, the series $P^n$ for arbitrary positive integer n (where this is the power series P raised to the nth power).

Specifically if

$P_T(C) = \sum_{n=0}^\infty b_n C^{n}$

then want to construct the function

$f(x) = \sum_{n=0}^\infty \frac{b_{n+1}}{(n!)^2} x^{n}$

which, from other results, should converge for all x. We can think of this as the doubly-exponential generating function for the $b_n$ sequence.

There are rules for multiplying and dividing formal power series (see here) and I've used these to get a recurrence relation for the coefficients in $P_T(C)$ (as well as P^n(C)) but I've been unable to solve these recurrence relations explicitly. They're in a form where each $b_n$ depends on all the previous $b_n$'s and I've not been able to make progress with them.

Explicitly the recurrence relation for the $b_n$ is $b_0 = 1$, $b_n = \sum_{k=1}^n b_{n-k} a_k$ (for n > 1). This looks simple enough but I don't think has a nice closed-form expression.

Nevertheless I do know what the $b_n$ are. They are the sequence A052186 (up to plus and minus signs). So

$P_T(C) = C+C^3-3 C^4+14 C^5-77 C^6+497 C^7-3676 C^8+\ldots$


$f(x) = 1 + \frac{1}{4}x^2 - \frac{1}{12}x^3 + \frac{7}{288}x^4 - \frac{77}{14400} x^5 + \frac{497}{518400}x^6 +\ldots$

The question is, is it possible to figure out the function $f(x)$!? Does it have a nice closed form? Perhaps in the form of a hypergeometric function? If it does, great, if it doesn't then at least I can stop searching for it!

  • $\begingroup$ There is a formula that describes the coefficients of the multiplicative inverse of a power series in closed-form (i.e. not as recurrence relation). Have you tried to apply it ? $\endgroup$ – Ralph Jul 26 '11 at 9:46
  • $\begingroup$ That sounds helpful. Do you know what the formula is? Do you think it could be applied to the multiplicative inverse of 1-P? $\endgroup$ – StevenJ Jul 26 '11 at 13:45

Of course your series $P(C)$ diverges. But it is a transseries. Or an asymptotic series. In fact, one of the best known. The series $$ \sum_{n=0}^\infty (-1)^n n! C^{n+1} $$ is the asymptotic series (as $C \downarrow 0$) for the function $$ p(C) = -e^{1/C} \mathrm{Ei}(-1/C) . $$ So, of course, your series $P_T(C)$ is the asymptotic series for $$ p_T(C) = -1 + \frac{1}{1+e^{1/C} \mathrm{Ei}(-1/C)} . $$ Now all we need is to remember the formal relation between an ordinary generating function and an exponential generating function, and apply it twice to $p_T$. Maybe that is not so easy?

  • $\begingroup$ I suppose the question then is, what's the nth coefficient in the expansion of the function P_T(C) you have above? Is there a general relationship between the ordinary and exponential generating functions? It seems like there should/might be but I don't know what it is. $\endgroup$ – StevenJ Jul 27 '11 at 22:06
  • $\begingroup$ Oh, and also, if $P_T(C)$ was a hypergeometric function then it's easy to add factors of 1/n! into the denominator of the coefficients. I don't know if that suggests that $P_T$ is not a hypergeometric function because if it was then it would be simple to work backwards and express the $b_n$ series in terms of factorials. It seems like if the nth term in that series could be expressed simply someone would have found the expression by now... $\endgroup$ – StevenJ Jul 27 '11 at 22:51

I post this as an answer since this way it's more easy to read. But it's a successive comment to my comment above.

The formula I mean is the following: Let $$f(x) = 1 + \sum_{n >0}b_nx^n$$ be a formal power series over a commutative ring with unit. Then $$1/f = 1 + \sum_{n>0}c_nx^n,$$ $$c_n = \sum_{k=1}^n(-1)^k \sum b_{i_1}\cdots b_{i_k}$$ where the inner sum is taken over all tuples $(i_1, \dots, i_k) \in \lbrace 1, \dots, n \rbrace ^k$ such that $i_1 + \dots + i_k = n$.

Applied to $1-P$ this yields $1/(1-P) = 1 + \sum_{n>0}c_nx^n$ with $$c_n = (-1)^n\sum_{k=1}^n(-1)^k \sum (i_1-1)! \cdots (i_k-1)!$$ But actually I don't believe that this is very helpful at all.

  • $\begingroup$ Well it's good to know a formula exists but I agree it seems pretty formidable in this scenario. $\endgroup$ – StevenJ Jul 27 '11 at 22:51

I stumbled across a similar question on math overflow here. That had lot of insights into P including a simple continued fraction expression for P. Details in a paper here.

They give $P(x) = [1, x, x, 2x, 2x, 3x, 3x, . . . , nx, nx, . . .]$

in continued fraction notation.

This surely means that

$P_T(x)+1 = \frac{1}{1-P(x)}= [1, 1, x, x, 2x, 2x, 3x, 3x, . . . , nx, nx, . . .]$

Perhaps written like this maybe it is simple enough to have a nice solution. Maybe?


[update] remark: I shifted the original answer to the bottom although it was missing the question to keep it as reference.
I've done a nice table, which shows the coefficients of a powerseries with constant term =1; for any p 'th integer power of the function you insert p into the formula.
Unfortunately the Latex-notation of the table was too messy, so I insert it as an image here:

(source: helms-net.de)

If you have Pari/GP you can easily manipulate power series; the formula $ \small P \cdot (1 - P)^{-1}$, where P is defined to be a powerseries can directly be evaluated :

P = Ser( sum(k=0,20, (-1)^k*k! * x * x^k ))
print ( P*(1 - P)^-1 + O(x^16) )

The matrix-operator is simply the coefficients of each power of the taylor series of a function in a column of an (ideally infinite sized) matrix.

[original answer] Assumed that I got your function correct ( $ \small P(x) = \sum_{k=1}^{\infty} (-1)^{k-1} \cdot (k-1)! \cdot x^k $ ) I have polynomials for the coefficients of $ \small P^{\text{ o } h} (x) $ Here I assume, that your notation $ \small P^n$ means iteration and not power. (If that was wrong I can delete this answer) I constructed the matrix-logarithm of the matrix-operator $ \small M$ for the function $ \small P(x)$ and got by $ \small \exp(h \cdot \log(M)) $ the following powerseries, where the coefficients at x are polynomials in the iteration-parameter h :

$ \qquad \small \begin{array} {l} P^{\text{ o } h} (x) = & 1 \cdot x \\\ & - h \cdot x^2 \\\ & + (h^2 + h) \cdot x^3 \\\ & + (-h^3 - 5/2 \cdot h^2 - 5/2 \cdot h) \cdot x^4 \\\ & + ( h^4 + 13/3 \cdot h^3 + 9 \cdot h^2 + 29/3 \cdot h) \cdot x^5 \\\ & + (-h^5 - 77/12 \cdot h^4 - 125/6 \cdot h^3 - 511/12 \cdot h^2 - 295/6 \cdot h) \cdot x^6 \\\ & + O(x^8) \end{array} $

I was not yet able to construct the other function $ \small P_T(C)$ although I've tried with a certain vague idea; please show some of the coefficents $ \small b_n $ so that I can compare.

  • $\begingroup$ Thanks a lot for responding. Unfortunately $P^n$ does refer to power, not iteration. I've edited the question to make that clearer as well as to explicitly write the $b_n$ coefficients and the start of $P_T(C)$ and f(x). Although I'm interested to see what you were able to construct. What is the matrix-operator M for the function P(x)? $\endgroup$ – StevenJ Jul 25 '11 at 21:48
  • $\begingroup$ Thanks for taking the time to make the table! Certainly using a computer I can calculate the coefficients of $P_T$ or $P^n$ to any order. The key difficulty is to calculate the coefficients analytically so that I can determine if f(x) has a closed-form solution. Carleman matrices are new to me but seem to do the job pretty well. Do you think it it possible to calculate the Carleman matrix for P(C) in a simple way? From my experiments the off-diagonal terms have a fairly complicated structure which doesn't obviously allow you to write down the whole matrix if you know the coefficients $a_n$. $\endgroup$ – StevenJ Jul 26 '11 at 14:02
  • $\begingroup$ @STevenj: but to construct the Carlemanmatrix is easy! Just assume exponent p=0 in the above table and compute the coefficients - these are just the coefficients of the column c=0 of the Carleman-matrix (actually it is everywhere zero except at x^0). Then use p=1 - the new coefficients are that of column c=1; use p=2 in the above table compute the coefficients at each x and have the coefficients of column c=2 of the Carleman-matrix. And so on. No recursion needed... $\endgroup$ – Gottfried Helms Jul 26 '11 at 16:27
  • $\begingroup$ Got it. Do you think it might help figure out f(x)? $\endgroup$ – StevenJ Jul 27 '11 at 22:43

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