# Elegant Notation

## Main Question or Discussion Point

Hi guys, just wondering if you can give me some advice on how to write certain co-efficients with in a compact elegant way.

The co-efficients are given by the following rule:

a0=1
a1=z
a2=z(z-1)
a3=z(z-1)(z-2)
.
.
.
an=z(z-1)(z-2)...(z-(n-1))

where a = any real number

I was thinking of using the following definition, an = z!/(z-n)! which seems to give me the correct results but im worried about whether or not the ! can be defined on real numbers. Like is there a problem with this definition?

## Answers and Replies

quasar987
Homework Helper
Gold Member
The ! can't really be defined on reals analogously to on N, but the expression "z!/(z-n)!" as a whole makes a lot of sense. Analogously to N, it means multiply z by z-1 by z-2, etc up to z-n+1.

morphism
Homework Helper
How about (for n>1)

$$a_n = \prod_{i=0}^{n-1} (z-i)$$

Written differently, you can define it recursively by

$$a_0 = 1$$

and

$$a_n = a_{n-1} \cdot (z - (n-1)), \ \text{for } \, n \geq 1$$

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Pythagorean
Gold Member
How about (for n>1)

$$a_n = \prod_{i=0}^{n-1} (z-i)$$

Written differently, you can define it recursively by

$$a_0 = 1$$

and

$$a_n = a_{n-1} \cdot (z - (n-1)), \ \text{for } \, n \geq 1$$

What's that thing called anyway? I know that it is to multiplication what the sigma is to addition, but is it just a capital pi or what?

What's that thing called anyway? I know that it is to multiplication what the sigma is to addition, but is it just a capital pi or what?
I am pretty sure it is just a capital pi.

jim mcnamara
Mentor
I learned it as "product". But that was fifty years ago....

Gib Z
Homework Helper
Lol maybe you misunderstood, yes they mean it as product as well jim, but they're just saying, the actual letter used to show that is a capital pi from the greek alphabet.

And just incase everyones forgotten, these can be very easily written with binomial notation?

EDIT: Shoot my last statement, though it could still help a tiny bit.

$${r \choose k} &{}= {1 \over k!}\prod_{n=0}^{k-1}(r-n)=\frac{r(r-1)(r-2)\cdots(r-(k-1))}{k!}$$

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Try $$\frac{\Gamma(z+1)}{\Gamma(z-n+1)}$$
The gamma function is the unique extension of the factorial function to the real numbers. $$\Gamma(z+1) = z!$$

By the way, if you are only doing this with integers, the accepted notation is $$(z)_n$$ It is known as the falling factorial, or Pochhammer symbol

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Hurkyl
Staff Emeritus
Gold Member
The gamma function is the unique extension of the factorial function to the real numbers.
It's not unique. For example,

$$f(z) = \Gamma(z + 1) + \sin (\pi z)$$

also agrees with the factorial function on the natural numbers. And so does

$$f(z) = \begin{cases} 0 & z < 0\\ \lfloor z \rfloor ! & z \geq 0$$

and

$$f(z) = \begin{cases} z! & z \in \mathbb{N} \\ -14 & z \notin \mathbb{N}$$

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It's not unique. For example,

$$f(z) = \Gamma(z + 1) + \sin (\pi z)$$

also agrees with the factorial function on the natural numbers. And so does

$$f(z) = \begin{cases} 0 & z < 0\\ \lfloor z \rfloor ! & z \geq 0$$

and

$$f(z) = \begin{cases} z! & z \in \mathbb{N} \\ -14 & z \notin \mathbb{N}$$
What I meant is that it's unique in that it extends all of the properties of the factorial function to the real (and complex) numbers so that for all complex numbers (except 0) $$\Gamma(z+1) = z*\Gamma(z)$$ and so that it is complex differentiable at all points except 0,-1,-2,...

However, I haven't studied the Gamma function, so I don't know much else about it.

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