### Pascal’s rule for Beta functions

Recently I became acquainted something called the Beta function. It is defined in terms of the Gamma function (a continuous generalization of the factorial function):

The first thing I’m tempted to do here is try to write this in terms of a binomial coefficient. If we replace each by the (equivalent for integers) , then we get

Now I’m more happy since I see an expression in terms of things I understand. But I would also like to know: is there a simple recurrence relation between the Beta function values, similar to Pascal’s rule ? It seems a little hopeless, since these guys are not only reciprocals, but they have the additional term complicating things.

Surprisingly, these two complications work with each other in some way and permit a really beautiful recurrence relation

which is even symmetric! You can prove this using the same simple algebraic approach that you can use to prove Pascal’s rule:

which is just the definition of when you group the terms.

When you learn about Pascal’s rule in a combinatorics class, you learn not only the simple algebraic proof in the above vein, but also the combinatorial “proof from the book:” the number of ways to pick k+1 things from n+1 involves the initial choice of including the initial item (leaving k things to pick from n things) or excluding the initial item (leaving k+1 things to pick from n things). Is there a similarly slick proof of this Beta function identity?

Filed under: math | 2 Comments

It seems to me that if you wanted a “similarly slick” proof for the beta function identity then you’d need a slick definition for the beta function first. Is there one?

One try: for integer values, the Beta function B(x, y) is the inverse of the multinomial coefficient for (x-1, y-1, 1). I.e. if we arrange the numbers from -x+1 to y-1 in a random order, it is the probability that the negatives come before 0 and the positives come after 0. Could this help?

For non-integer values, I wouldn’t know where to start, but doing the integer case would be awesome enough.