Question

What is the purpose of the Gamma function?

Answer 1

Start with the basic idea of interpolation , that is finding a best fit continuous smooth curve to a set of discrete data points. The interpolation function is constructed in such a way that it passes through all of the discrete data pints (so that we have exact results for these data points), but also allows the values outside of the data points to be estimated (i.e. it provides a continuous extension to discrete data).

If we had two discrete points we could perform linear interpolation and find a best fit straight line. This is how the trapezium rule works for numerical integration

If we had three discrete points we could perform quadratic interpolation and find a best fit quadratic curve. This is how Simpson's rule works for numerical integration

Next let us understand the factorial function `n!` which is the product of the first `n` integers, thus

`1! = 1`
`2! = 1 xx 2 = 2`
`3! = 1 xx 2 xx 3 = 6`
`vdots`
`n! = n xx (n-1)!`

etc, and in order for this recurrence relation to be extended to `n=0`, we define :

`0! = 1`

So the question posed by Euler & Legendre was can we find a smooth function that interpolates all the factorial values, and the solution is the Gamma Function :

`Gamma(z) = int_0^oo \ x^(z-1) \ e^(-x) \ dx`

The Gamma Function has the properties that:

`Gamma(n)=(n-1)!` where `n in NN`

And direct calculation yields some interesting results for non-positive integers, e.g.

`Gamma(1/2) = sqrt(pi)`

Which is sometimes used as an "extension" to the factorial function, and so one might see such notation as:

`(1/2)! = sqrt(pi)`

We can graph the Gamma Function, as follows:
graph{x! [-11.29, 8.71, -2.84, 7.16]}

Answer 2

The answer above is great but I'd just like to add that if your students are familiar with reduction formulae you can explain to them why `Gamma(n) = (n-1)!`.

Define `I_n = int_0^oo \ x^(n) \ e^(-x) \ "d"x`, and integrate by parts, differentiating `x^n` and integrating `e^(-x)`.

`I_n = [-e^(-x)x^n]_0^oo - int_0^oo nx^(n-1)(-e^(-x))"d"x`,

For any `n>=0`, `lim_{x \to \pm \infty} e^(-x)x^n = 0`.

Then,

`I_n = n int_0^oo x^(n-1)e^(-x)"d"x`
`I_n = n * I_(n-1)`.

Then,
`I_0 = int_0^oo e^(-x) "d"x`,
`I_0 = [-e^(-x)]__0^oo`,
`I_0 = 1`.

The factorial function is defined by `0! = 1` and `n! = n*(n-1)!`. So, our integral `I_n` is equivalent to `n!` when evaluated at integer `n`.

The gamma function is defined,

`Gamma(z) = int_0^oo \ x^(z-1) \ e^(-x) \ "d"x`, so,

`Gamma(z) = (z-1)!` for integer `z`.

A high school student with knowledge of reduction formulae could follow this argument relatively easily.