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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