Thursday, July 17, 2008

Fractional Gamma Functions

Earlier we explored how the Gamma function works. 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 Gamnma 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) = INT (0 to oo) t^-1/2 exp(-t) dt

where 'INT' denotes an integral, taken in this case from 0 to infinity. The resulting integral yields:

G(1/2) = pi

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 pi

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

Use same sort of procedure:

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

(Readers are invited here to find G(5/2) on their own!)

Another application, decimals - which are merely another form of fraction:

Say you wish to obtain G(-0.30)

(In this case, one already is assumed to know the basic Gamma function G(1.70) = 0.90864)

In this case, from the Gamma formula given earlier:

G(-0.30) = G (1 - 0.30) = -0.30 G (-0.30)

G(0.70)/ (-0.30) = G(-0.30)

But: G(0.70) = G(1.70)/ 0.70 = (0.90864)/ 0.70 = 1.29805

So: G(-0.30) = G(0.70)/ -0.30 = 1.29805/ -0.30 = -4.32683

Fractional sequences can also come into play, e.g.


Find: G(n + ½):

G(n + ½) = (n- ½) G(n – ½)

Since: we use x = (n – ½) in: G(x + 1) = xG(x)

Thus:

G([n – ½] + 1) = (n – ½) G(n – ½)

-> G(n + ½) = (n – ½) G(n – ½)

And we can go further, focusing on treating the right hand side:

(n – ½) G(n – ½) = (n – ½) (n – 3/2) G(n –3/2)

= (n - 1/2)(n - 3/2) . .. . .3/2* 1/2 * G(1/2)

But, G(1/2) = pi, so:

G(n + 1/2) = (2n -1)/2 * (2n -3)/2 . . .. 3/2* 1/2* (pi)^1/2

Further factoring and additional algebra will yield:

G(n + 1/2) = (2n)! (pi)^1/2 / 2^n n!

This is left as an exercise for the ambitious reader!

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