*Gamma function*(G) for which:

G (a) = (a - 1)!

where 'a' is a positive integer. (Note: for a detailed derivation of the Gamma function, see: http://www.frm.utn.edu.ar/analisisdsys/material/funcion_gamma.pdf )

Thus, for a = 3:

G (3) = (3 - 1)! = 2! = 2·1 = 2

One can also make use of a recursion formula:

G(a + 1) = a G(a)

For example: G (4) = G (3 + 1) = 3 G (3) = 3 (2) = 6

Check this from the earlier formula: G (a) = (a - 1)!

G (4) = (4 - 1)! = 3! = 3·2·1 = 6

There is also the composite Beta function, call it b(u,v) which can be expressed in terms of the Gamma functions

G(u), G (v).

Thus:

b(u,v) = G (u) G (v)/ G (u + v)

Some alert readers may recall this was the form used by cosmologist Brian Greene, when he sought to explicate string theory on the PBS special, The Elegant Universe. The specific form Greene used in that chalk board segment was:

b(p, q) = G (p) G (q)/ G (p + q)

and arriving at the unique string theory form:

b([1- a(s)][1 - a(t)] =

G (1 - a(s)) G (1 - a(t))/ G (2 - a(s) - a(t))

Now, to fix ideas, consider the Beta function b(3, 4):

b(3, 4) = G (3) G (4)/ G (3 + 4) = (2) (6)/ G (7) = 12 / G (7)

where: G (7) = (7 - 1)! = 6! = 6·5·4·3·2·1 = 720

so b(3,4) = 12/ 720 = 1/60

__Fractional Gamma Functions__:

One of the more useful formulas for generalizing integral forms was found to be:

G (x + 1) = x G (x)

This will also be found very useful in working with fractional Gamma functions, as I will show in this article. Most solutions of fractional G (x) entail already knowing at least one basic form, usually obtained from a special integral.

For example, working with most fractional halves we make use of the basic integral that generates:

G (1/2)

This is defined:

G(1/2) = ∫

The resulting integral yields:

G (1/2) = p

Now let's see how it works, say to obtain G(-1/2):

From the basic Gamma function formula (letting x = -1/2) :

G(-1/2) = G(-1/2 + 1) = -1/2 G (-1/2)

Or:

G(-1/2) = -2 G (1/2) = -2 p

That was easy enough. Now what about G (3/2)?

Use the same sort of procedure:

G (3/2) = G(1 + 1/2) = 1/2 G(1/2) = p /2

G (x + 1) = x G (x)

This will also be found very useful in working with fractional Gamma functions, as I will show in this article. Most solutions of fractional G (x) entail already knowing at least one basic form, usually obtained from a special integral.

For example, working with most fractional halves we make use of the basic integral that generates:

G (1/2)

This is defined:

G(1/2) = ∫

^{¥}_{o}^{ }t ^-1/2 exp(-t) dtThe resulting integral yields:

G (1/2) = p

Now let's see how it works, say to obtain G(-1/2):

From the basic Gamma function formula (letting x = -1/2) :

G(-1/2) = G(-1/2 + 1) = -1/2 G (-1/2)

Or:

G(-1/2) = -2 G (1/2) = -2 p

That was easy enough. Now what about G (3/2)?

Use the same sort of procedure:

G (3/2) = G(1 + 1/2) = 1/2 G(1/2) = p /2

__Problems For The Math Maven__:

1. Find: G (6)

3. Using the link given earlier, for the detailed Gamma function. derivation, show how:

G (½) = Öp

2. Find: b(3,6)

3. Using the link given earlier, for the detailed Gamma function. derivation, show how:

G (½) = Öp

_{}

^{}

4) Find G(5/2)

5) Find G(-0.30)

If the basic Gamma function G(1.70) = 0.90864

6) Find: G(n + ½):

(You may use x = (n – ½) in: G(x + 1) = xG(x) )

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