# Shut kind of a partial amount of the power collection of $\exp(x)$

I am seeking a shut kind (preferably shared as primary features) of the function $\exp_n(x) = \sum_{k=0}^n x^k / k!$. I am currently knowledgeable about sharing it in regards to the gamma function.

## History/ Motivation

When counting mixes of things with creating features, it serves to be able to share the partial amount $1 + x + \cdots + x^n$ as $\frac{1-x^{n+1}}{1-x}$. As an example, to count the variety of means to select 5 marbles from a bag of blue, red, and also environment-friendly marbles where we select at the majority of 3 blue marbles and also at the majority of 2 red marbles, we can take into consideration the creating function $f(x) = (1+x+x^2+x^3)(1+x+x^2)(1+x+x^2+\cdots)$.

By utilizing the partial amount identification, we can share it as $f(x) = \left(\frac{1-x^4}{1-x}\right)\left(\frac{1-x^3}{1-x}\right)\left(\frac{1}{1-x}\right)$. Streamline, share as less complex item of collection, and also locate the coefficient of the $x^5$ term.

I intend to have the ability to do the very same for a creating function in the kind $g(x) = \exp_{n_1}(x)^{p_1} \exp_{n_2}(x)^{p_2} \cdots \exp_{n_j}(x)^{p_j}$

The most convenient means to extract the coefficient of an offered term $x^p / p!$ would certainly be to make use of a comparable shut kind expression for $\exp_n(x)$ and also a comparable strategy to $f$.

## Tried Solutions

### Differential formula

Remember that the means to confirm the identification $1+x+x^2+\cdots+x^n = \frac{1-x^{n+1}}{1-x}$ is to specify $S = 1 + x + x^2 + \cdots + x^n$ and also notification that: $S - Sx = 1 - x^{n+1}$. Furthermore, notification that $y(x) = \exp_n(x)$ pleases $y - y' = x^n/n!$. Via SAGE, the remedy is $y(x) = \frac{c+\Gamma(n+1,x)}{n!}e^x$. Our first problem $y(0) = 0$ so $c=0$. By (2), $\Gamma(n+1,x) = n! e^{-x} \exp_n(x)$ so we have $y(x) = \exp_n(x)$.

### Reappearance Relation

Notification that $\exp_n(x) = \exp_{n-1}(x) + x^n/n!$. Making use of the independent Z-Transform and also relevant buildings, we locate that $\mathcal{Z}[\exp_n(x)] = (z e^{x/z})/(z-1)$.

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

$(z^n e^{x/z})/(z-1)$ has 2 selfhoods: $z = 1$ and also $z = 0$. The factor $z = 1$ is a post of order one with deposit $e^x$. To locate the deposit at $z = 0$ take into consideration the item $z^n e^{x/z} (-1/(1-z)) = -z^n \left( \sum_{m=0}^\infty x^m z^{-m} / m! \right) \left( \sum_{j=0}^\infty x^j \right)$. The coefficient of the $z^{-1}$ term is offered when $n - m + j = -1$. The deposit of the factor $z=0$ is after that $-\sum_{m,j} x^m / m! = -\sum_{m=n+1}^\infty x^m / m!$.

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

### Limited Calculus

I've attempted to review the amount making use of limited calculus, yet can not appear to make much progression.

I'm not exactly sure you'll like this, yet in regards to the insufficient $\Gamma$ function, one can get a shut kind as $$\frac{e^{x}\Gamma(n+1,x)}{\Gamma(n+1)}.$$

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

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