Is the General Binomial Coefficient for Any Rational Value Always Defined as 1?

[Galadirith]

Hi everyone, I have been having a problem with the General Binomial Coefficient for any rational value:

$$\left( <br /> \begin{array}{c}<br /> n\\<br /> r\end{array}<br /> \right)<br /> = \frac{1}{r!}\prod_{i=0}^{r-1} (r-i)$$

Now this works fine except when r=0. so 0! is defined to be 1 so the coefficient of the product of the series is 1, but then the cap PI would read:

$$\left( <br /> \begin{array}{c}<br /> n\\<br /> 0\end{array}<br /> \right)<br /> = \frac{1}{0!}\prod_{i=0}^{-1} (r-i)$$

how can that possibly be evaluated, is there a mathematical reason or is it more defined to be 1. I know that this somehow mean the empty product which is defined to be 1, but how is this the empty product. Thanks Guys :-)

 

[HallsofIvy]

Hi everyone, I have been having a problem with the General Binomial Coefficient for any rational value:

$$\left( <br /> \begin{array}{c}<br /> n\\<br /> r\end{array}<br /> \right)<br /> = \frac{1}{r!}\prod_{i=0}^{r-1} (r-i)$$

What is "n" supposed to mean here?

Now this works fine except when r=0. so 0! is defined to be 1 so the coefficient of the product of the series is 1, but then the cap PI would read:

$$\left( <br /> \begin{array}{c}<br /> n\\<br /> 0\end{array}<br /> \right)<br /> = \frac{1}{0!}\prod_{i=0}^{-1} (r-i)$$

how can that possibly be evaluated, is there a mathematical reason or is it more defined to be 1. I know that this somehow mean the empty product which is defined to be 1, but how is this the empty product. Thanks Guys :-)

 

[Galadirith]

Sorry HallsofIvy, i mucked up my latex there a little, equation one should be :

$$<br /> \left( <br /> \begin{array}{c}<br /> n\\<br /> r\end{array}<br /> \right)<br /> = \frac{1}{r!}\prod_{i=0}^{r-1} (n-i)<br />$$

and equation 2 should be :

$$<br /> \left( <br /> \begin{array}{c}<br /> n\\<br /> 0\end{array}<br /> \right)<br /> = \frac{1}{0!}\prod_{i=0}^{-1} (n-i)<br />$$

sorry about that, i typed r instead of n at the very end. my variables n are r are :

n, the order of the coefficient n ∈ $$\mathbb{Q}$$ and r ∈ $$\mathbb{N}_0$$. (in fairness I think this can be expanded so that n is an element of the complex numbers, but I am not worring about that now)

 

[mathman]

A simple way out of this dilemna is to write the comb. expression as n!/[r!(n-r)!]. When r=0, you will simply have 1/0! = 1.

 

[Galadirith]

thank you mathman, however in this situation that actually doesn't work, using the expression you suggested only works with integer values of n that are greater than 0 or n ∈ $$\mathbb{Z}^+$$, I have tried to find a way to adapt the expression you suggested but there is no way, (well there is a way but you end up with the equations from my first post :-)), that I have found at least and I don't think there is one. I could obviously use your expression which is the standard binomial coefficient definition with my original one which is the generalized binomial coefficient and define the coefficient piecewise, but that seem quite un-elegant, and that's not my question, there must be an explanation of how to evaluate my second equation, but Thank you for you suggestion though mathman.

EDIT: Well infact I do know that they are just rearrangements of one another, but still there must be a direct way to evaluate my second equation as apposed to rearrange it, it just seems unusual that the equation is fine to use for every value of r except r=0.

 

[CRGreathouse]

I don't see the problem with the second equation; I think it can simply be evaluated directly. It's a constant: 1/0! times an empty product = 1. prod(i=0, x, ...) = 1 for all x < 0.