Math Revisited: Gamma and Beta Functions

We begin with the claim that there is the 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) = ∫  ^¥_o    t ^-1/2   exp(-t) dt

 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

 

Problems For The Math Maven:

1. Find:  G (6)

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) )

The Remarkable World of Special Functions

The world of mathematical special functions is as intriguing and interesting as any domain, say in physics, or the natural world. The trouble is that one usually doesn't appreciate these functions enough while taking Calculus or Linear Algebra or Differential Equations, and one must wait until a research focus brings them to prominence.

Let's take the case of the Bessel functions. In solar physics one very key equation (Helmholtz) for which the (axially symmetric- in cylindrical coordinates r, z, t) Bessel function solution is:

B_z (r) = B_o J_o(a r)

B_t(r)) = B_o J_1(ar)

t = theta

J_0(a r) is a Bessel function of the first kind, order zero and J1 (ar) is a Bessel function of the first kind, order unity. The Bessel functions are defined (cf. Menzel, 'Mathematical Physics', 1961, p. 204):

J_m(x) = (1/ 2^m m!) x^m [1 – x^2/ 2^2 1! (m + 1) + x^4/ 2^4 2! (m +1) (m + 2) -…

-(1)^j x^(2j) / 2^(2j) j! (m + 1) (m +2 )……(m + j) + …]

for m = 0 and m = 1 forms one gets:

J_o(x) = 1 - x^2/ 2^2 (1!)^2 + x^4/ 2^4 (2!)^2 - x^6/ 2^6 (3!)^2 + ......

J_1(x) = x/ 2 - x^3/ 2^3*1! 2! + x^5/ 2^5 *2!3! - x^7/ 2^7 *3!4! - .....


The equations in B_z, B_t, with the special Bessel functions at root, are critical in describing the respective magnetic fields for a magnetic tube.

For a cylindrical magnetic flux tube (such as a sunspot represents viewed in cross-section) the “twist” is defined:

T(r) = (L * B_t(r))/ (r * B_z (r))

Where L denotes the length of the sunspot-flux tube dipole. If the twist value exceeds 2(pi) then the magnetic configuration may be approaching instability and a solar flare.

Then there is the Gamma function (call it 'G') for which:

G(a) = (a - 1 )!

where 'a' is a positive integer.

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

Now, there is also the 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)

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

Why all the fuss about Beta and Gamma functions here?

Well, if any readers happened to have caught the PBS series 'The Elegant Universe', they'd have seen string theorist Brian Greene scribbling the Euler equation for string theory on a blackboard:

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

and arriving at the specific string theory form:

Beta([1- alpha(s)][1 - alpha(t)] =

Gamma (1 - alpha(s)) Gamma (1 - alpha(t))/ Gamma(2 - alpha(s) - alpha(t))

Special functions can indeed be loads of fun and have wide applications. In a future foray I'll look at fractional Gamma functions.