# Closed form of a partial sum of the power series of $\exp(x)$

I am looking for a closed form (ideally expressed as elementary functions) of the function $\exp_n(x) = \sum_{k=0}^n x^k / k!$. I am already aware of expressing it in terms of the gamma function.

## Background / Motivation

When counting combinations of objects with generating functions, it is useful to be able to express the partial sum $1 + x + \cdots + x^n$ as $\frac{1-x^{n+1}}{1-x}$. For example, to count the number of ways to pick 5 marbles from a bag of blue, red, and green marbles where we pick at most 3 blue marbles and at most 2 red marbles, we can consider the generating function $f(x) = (1+x+x^2+x^3)(1+x+x^2)(1+x+x^2+\cdots)$.

By using the partial sum identity, we can express it as $f(x) = \left(\frac{1-x^4}{1-x}\right)\left(\frac{1-x^3}{1-x}\right)\left(\frac{1}{1-x}\right)$. Simplify, express as simpler product of series, and find the coefficient of the $x^5$ term.

I want to be able to do the same for a generating function in the form $g(x) = \exp_{n_1}(x)^{p_1} \exp_{n_2}(x)^{p_2} \cdots \exp_{n_j}(x)^{p_j}$

The easiest way to extract the coefficient of a given term $x^p / p!$ would be to use a similar closed form expression for $\exp_n(x)$ and a similar technique to $f$.

## Attempted Solutions

### Differential equation

Recall that the way to prove the identity $1+x+x^2+\cdots+x^n = \frac{1-x^{n+1}}{1-x}$ is to define $S = 1 + x + x^2 + \cdots + x^n$ and notice that: $S - Sx = 1 - x^{n+1}$. Likewise, notice that $y(x) = \exp_n(x)$ satisfies $y - y' = x^n/n!$. Via SAGE, the solution is $y(x) = \frac{c+\Gamma(n+1,x)}{n!}e^x$. Our initial condition $y(0) = 0$ so $c=0$. By (2), $\Gamma(n+1,x) = n! e^{-x} \exp_n(x)$ so we have $y(x) = \exp_n(x)$.

### Recurrence Relation

Notice that $\exp_n(x) = \exp_{n-1}(x) + x^n/n!$. Using the unilateral Z-Transform and related properties, we find that $\mathcal{Z}[\exp_n(x)] = (z e^{x/z})/(z-1)$.

Therefore, $\exp_n(x) = \mathcal{Z}^{-1}\left[(z e^{x/z})/(z-1)\right] = \frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz$.

$(z^n e^{x/z})/(z-1)$ has two singularities: $z = 1$ and $z = 0$. The point $z = 1$ is a pole of order one with residue $e^x$. To find the residue at $z = 0$ consider the product $z^n e^{x/z} (-1/(1-z)) = -z^n \left( \sum_{m=0}^\infty x^m z^{-m} / m! \right) \left( \sum_{j=0}^\infty x^j \right)$. The coefficient of the $z^{-1}$ term is given when $n - m + j = -1$. The residue of the point $z=0$ is then $-\sum_{m,j} x^m / m! = -\sum_{m=n+1}^\infty x^m / m!$.

Let $C$ by the positively oriented unit circle centered at the origin. By Cauchy's Residue Theorem, $\frac{1}{2 \pi i} \oint_C z^n e^{x/z}/(z-1)\;dz = \frac{1}{2 \pi i} 2 \pi i \left(e^x - \sum_{m=n+1}^\infty x^m / m!\right) = \exp_n(x)$.

### Finite Calculus

I've tried to evaluate the sum using finite calculus, but can't seem to make much progress.

• It's already an elementary function: you can't get any more elementary than a polynomial. – Qiaochu Yuan Jul 31 '10 at 4:13
• Well that's true. I'm looking to something analogous to 1 + x + ... + x^n = (1-x^(n+1))/(1-x). Whatever that would be called. – Alex Jeffries Jul 31 '10 at 4:38
• I don't think that's likely. For one thing, the roots of exp_n are much more complicated. – Qiaochu Yuan Jul 31 '10 at 4:48
• @Qiaochu: would not "closed form" imply an absence of summation symbols? While each instance is a polynomial, he is asking for an expression for the family of such polynomials, which is closed form as a function of n. – Niel de Beaudrap Jul 31 '10 at 10:39
• Fair enough. Let me make the following suggestion: the factorization you give basically reflects the fact that the products you're interested in can be evaluated by pretending that all of the factors are 1/(1 - x) and then using inclusion-exclusion. The analogous construction in the egf case is probably the best you're going to get. – Qiaochu Yuan Jul 31 '10 at 18:36

## 1 Answer

I'm not sure you'll like this, but in terms of the incomplete $\Gamma$ function, one can get a closed form as $$\frac{e^{x}\Gamma(n+1,x)}{\Gamma(n+1)}.$$

The incomplete $\Gamma$ function is defined as $$\Gamma(s,x) = \int_x^{\infty} t^{s-1}e^{-t}dt$$.

• His second sentence says "I am already aware of expressing it in terms of the gamma function," though I am no longer sure what sort of answer(s) is he expecting since an incomplete gamma is not satisfactory for him as a closed form... – J. M. is a poor mathematician Aug 4 '10 at 6:30
• @J. Mangaldan: I would prefer the closed form to be just as BlueRaja describes it in the comment to the original question. That is, no summations, recursive definitions, integrals, derivatives, limits, etc. The presence of the integral makes extracting the coefficient of a certain power of x very difficult, which is really the root of this question. Mathworld gives an equivalence for Γ(s,x) but it unfortunately is in terms of \exp_n(x). – Alex Jeffries Aug 5 '10 at 6:55
• "That is, no summations, recursive definitions, integrals, derivatives, limits, etc." My point being, this very restrictive condition of yours just took out a great number of the transcendental functions used in applications. To use an elementary example, the properties of the integral of the reciprocal function can all be derived ab initio without knowing that it is in fact what others call the "natural logarithm". – J. M. is a poor mathematician Aug 6 '10 at 3:54
• Right, that is one way to do it. Likewise, I'm aware of the formulation of a closed form of exp_n(x) via the incomplete Gamma function, but I would really prefer a closed form expression with the requirements that I provide. If there was a closed form expression that made taking powers of exp_n(x) and extracting coefficients of an arbitrary x^n / n! then it would be okay if it's outside my restrictions. However, I've asked for those restrictions because it's analogous to the geometric series. – Alex Jeffries Aug 6 '10 at 21:00