Help with a generating function and differential equation

Question

I have a generating function that I'm trying to create. It has a general form to it:

$1 + cx + c(c-1)x^2 + c(c-1)(c-2)x^3 + \dots + c!x^c$

I would like to see a closed form for this function. I'd like something other than a summation. I'm particularly interested if someone could walk through how to derive the form for me. I was hoping that someone could carry on with what I've done and show me how to solve what's left.

My attempt

So I try to create a function $A(x)$ (which will be the closed form) using recurrences. Here I'm using techniques (and especially notation) from Wilf's Generatingfunctionology.

I set $a_0=1$. I then proceed to multiply by $(c-n)$. My guess is that the math should be something like the following:

$a_{n+1}=c \cdot a_n - \frac{d A(x)}{dx}$

I'm trying to say that the next term should be $c$ times the previous term minus $n$ times the previous term. The latter is represented by the derivative, I believe. So I guess that this could be correct. Unfortunately, I don't know for sure, and so I'm kind of lost at this point. I would greatly appreciate if someone could take it from here and explain the rest in detail.

The Differential Equation

Assuming that the last part is correct, the function becomes a differential equation:

$\frac{A(x)-1}{x}=c \cdot A(x) - \frac{d A(x)}{dx}$

Answer 1

I believe that your initial attempt is incorrect. If I solve the suggested differential equation, I get \begin{eqnarray} A(x) = -\frac{1}{cx} \left( 1 - e^{cx} \right) = \sum_{k \geq 1} \frac{(cx)^{k-1}}{k!}, \end{eqnarray} where I have chosen the initial condition to ensure $1$ is the first term of the series expansion. The generating function that you are seeking is the following: \begin{eqnarray} e^{1/x} \left( x^{c} \Gamma(c+1, \tfrac{1}{x}) - c (-x)^{c} \Gamma(0, \tfrac{1}{x}) (1 -c)^{(c)} \right) = 1 + \sum_{k = 1}^{c} c(c-1) \cdots (c-k+1) \ x^{k}, \end{eqnarray} which may be simplified to \begin{eqnarray} e^{1/x} \left( \frac{(-x)^{c} \ \Gamma(0, \frac{1}{x})}{\Gamma(-c)} + x^{c} \ \Gamma(c + 1, \tfrac{1}{x}) \right) = 1 - \sum_{k = 1}^{c} c (1 - c)^{(k-1)} \ (-x)^{k}, \end{eqnarray} where $\Gamma(n,x)$ is the incomplete gamma function defined as \begin{eqnarray} \Gamma(n,x) = \int_{x}^{\infty} t^{n-1} e^{-t} dt = (n-1)! e^{-x} \sum_{k = 0}^{n-1} \frac{x^{k}}{k!} \end{eqnarray} and $(x)^{(n)}$ is the Pochhammer symbol or rising factorial, $x(x+1)\cdots (x+n-1)$. This form makes no assumption on the integrality of $c$. However, if $c$ is a positive integer, then the formula simplifies greatly. For example, if $c = 3$, the left side specializes to \begin{eqnarray} e^{1/x} x^3 \Gamma(4, \tfrac{1}{x}) = 1 + 3 x + 6 x^{2} + 6 x^{3}. \end{eqnarray} NB: There is no extra term. The last exponent of $x$ has coefficient $c!$. In fact, the last two coefficients is $c!$ because $c(c-1) \cdots 2 = c(c-1) \cdots 2 \cdot 1 = c!$.

In general, the generating function assuming integral $c$ is \begin{eqnarray} e^{1/x} x^{c} \ \Gamma(c+1, \tfrac{1}{x}). \end{eqnarray}

Answer 2

The general form of your sum can be expressed succinctly as

$$\sum (-c)_k (-x)^k$$

where $(a)_k$ is a Pochhammer symbol.

We recognize at once that this is a hypergeometric series; specifically, it is a ${}_2 F_0$ :

$${}_2 F_0 \left(-c,1;;-x\right)$$

which can be rewritten as a Tricomi confluent hypergeometric function:

$$\frac1{x}U\left(1,c+2,\frac1{x}\right)$$

which can be re-expressed as an incomplete gamma function:

$$x^c\exp\left(\frac1{x}\right)\Gamma\left(c+1,\frac1{x}\right)$$

As for the differential equation, it can be obtained from this formula to give:

$$x^2\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}-\left(c-2-\frac1{x}\right)x\frac{\mathrm{d}y}{\mathrm{d}x}-cy=0$$

where the solution of interest satisfies the initial conditions $y(0)=1$ and $y^{\prime}(0)=c$.

I'll leave others to flesh out the details.

Answer 3

I found a differential equation which gives the correct result under Mathematica 7.0.1. The differential equation is the same one I derived in the question (see my comment on adding x):

$\frac{A(x)-1}{x}=c \cdot A(x) - \frac{d A(x)}{dx}$

The result it gives as the solution (besides the series) is:

$e^{-1/x}x^{-c}(C_1 - x^{c-1}E_c(-\frac{1}{x}))$

where $E_c(z)$ is the Exponential Integral Function, for which the following equation holds: $-x^{c-1}E_c(b)$ = $b^c x^c \Gamma[1-c,b]$ and $b=-\frac{1}{x}$. $C_1$ is the constant of integration.

Please note my derivation assumed $c \in \mathbb{N}$.

Answer 4

If you know about the incomplete Gamma function then the solution is immediate

since $\rm\quad\quad\quad\quad\quad\ \ \Gamma(s,x)\ =\ (s-1)\ \Gamma(s-1,x) + x^{s-1}\ e^{-x} $

hence $\rm\displaystyle\quad e^x\ \Gamma(s+1,x)\ =\ \sum_{k=0}^s\ \frac{s!}{k!}\: x^k\ \ $ for $\rm\:s\in \mathbb N$

Your method of summing the lower-factorical coefficient recurrence $\rm\ c_{(k+1)}\ = \ (c-k)\ c_{(k)}\ $ will yield a hypergeometric differential equation with the above solution - presuming that you know how to solve such hypergeometric differential equations. But that's overkill compared to the above.

Answer 5

Call your sum $S(x)$. Then you can write: $$ \begin{align*} S(x) &= c! x^c \sum_{0 \le k \le c} \frac{1}{k! x^k} \\ &= c! x^c \left. \exp \right|_c (1 / x) \end{align*} $$ Here $\left. \exp \right|_c (x)$ is the truncated exponential sum (just cut off after the $c$ term).