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.

2019-05-05 23:36:12
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Answers: 1

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 $$\Gamma(s,x) = \int_x^{\infty} t^{s-1}e^{-t}dt$$.

2019-05-08 08:00:35