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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:

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

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I forgot that I need to multiply the derivative by x. $\frac{A(x)-1}{x}=c \cdot A(x) - (x)\frac{d A(x)}{dx}$ appears to give the correct answer after I solve the differential equation using my math software. Of course, I'd still like to know how to do the differential equation. –  Matt Groff Nov 6 '10 at 6:50
    
I realized that the series should terminate at $c!x^{c-1}$. –  Matt Groff Nov 6 '10 at 13:42
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5 Answers

up vote 4 down vote accepted

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}

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maybe you could add to your answer, and try to show how to make his approach work? also, he was having trouble understanding how to continue with his solution, so you could also show how to solve the differential equation. –  Eric Haengel Nov 6 '10 at 6:28
    
I think I see a mistake I made. I should multiply the derivative by $x$. –  Matt Groff Nov 6 '10 at 6:33
1  
Thanks. I'm not quite finished with my post.... –  user02138 Nov 6 '10 at 6:35
    
Wow! Now I wish I knew ways to arrive at that! –  Matt Groff Nov 6 '10 at 6:37
    
Your function gives an extra $c!x^c$, which is an easy fix. I'm very thankful because yours appearantly works for all integers from what I've seen. However, for the simplification, I get $-(\sum_{k=0}^{c}{c^{\underline{k}}(-x)^k} + c!(-x)^c)$ for your sum, where $c^{\underline{n}}$ is the falling factorial, equaling $c(c-1)(c-2)\dots(c-n+1)$. That result holds for all integers. I think when you introduced $(-1)^k$ it started with the alternating signs, but it's an interesting series as well. –  Matt Groff Nov 6 '10 at 13:09
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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.

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M.: What do you mean by getting the form wrong? Do you perhaps mean that it leads to a tough differential equation? My math software seemed to confirm that the recurrence I got is correct. I just couldn't seem to get the differential equation to work. Maybe the other form helps to solve the differential equation that I got. –  Matt Groff Nov 6 '10 at 9:36
    
M.: By the way, I didn't mean to doubt your commentary on this series. I was simply curious. –  Matt Groff Nov 6 '10 at 9:50
    
Hmm, sorry about that. I misread the first few lines of your question. I'll edit my answer. –  J. M. Nov 6 '10 at 9:57
    
No apology necessary. I'm thankful you helped. But now I wonder about reducing the differential equations. The equation I got has only a first derivative, while yours has a second derivative. Is it common to be able to reduce derivatives like that? It seems odd. Plus, me, you, and user 02138 got different expressions for the same result. –  Matt Groff Nov 6 '10 at 11:17
    
Well, hypergeometrics of this sort tend to have second-order DEs. On the other hand, I had somehow assumed $c$ was a positive integer; if not, then user02138's solution is the correct one. Note that the reciprocal gamma function is zero at the nonpositive integers. –  J. M. Nov 6 '10 at 11:26
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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.

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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 $c \in \mathbb{N}$.

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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).

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