The fractional gamma function is an offshoot of how the regular gamma function works. One of the more useful formulas for the latter, introduced in several earlier posts e.g.
is:
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 fractionals we make use of the basic integral that generates: G (½) .
G (½) = ò¥ o [t-1/2 exp(-t)] dt
The resulting integral yields: Öp
The resulting integral yields: Öp
Now let's see how to obtain G (-½):
From the basic Gamma function formula (letting x = -½) :
G (-½) = G (-½ + 1) = -½ G (-½)
Or:
G (-½) = -2G (½) = -2 Öp
Another form of the fractional gamma function appears as decimals - which are merely another form of fraction:
Say you wish to obtain G (-0.30)
In this case, one is assumed to know the basic Gamma function G (1.70) = 0.90864
Then from the gamma function 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)
G ([n – ½] + 1) = (n – ½) G (n – ½)
Þ G (n + ½) = (n – ½) G (n – ½)
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 - ½)(n - 3/2) . .. . .3/2·½·G (½)
But, G (½) = Öp, so:
G (n + ½) = (2n -1)/2 · (2n -3)/2 . . .. 3/2· ½· (p)1/2
Further factoring and additional algebra yields:
G (n + ½) = (2n - 1)! (p)1/2 / 2n n!
This is left as an exercise for the ambitious reader!
Problems for the Math Maven:
1) Find: G (3/2)
2) Find: G (-0.70)
3) Use the form: G (n + ½) = (2n - 1)! (p)1/2 / 2n n!
and show it is equal to the value for G (3/2) you obtained in (1)
4) Hence, or otherwise, compute: G (5/2)
G (-½) = -2G (½) = -2 Öp
Another form of the fractional gamma function appears as decimals - which are merely another form of fraction:
Say you wish to obtain G (-0.30)
In this case, one is assumed to know the basic Gamma function G (1.70) = 0.90864
Then from the gamma function 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)
G ([n – ½] + 1) = (n – ½) G (n – ½)
Þ G (n + ½) = (n – ½) G (n – ½)
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 - ½)(n - 3/2) . .. . .3/2·½·G (½)
But, G (½) = Öp, so:
G (n + ½) = (2n -1)/2 · (2n -3)/2 . . .. 3/2· ½· (p)1/2
Further factoring and additional algebra yields:
G (n + ½) = (2n - 1)! (p)1/2 / 2n n!
This is left as an exercise for the ambitious reader!
Problems for the Math Maven:
1) Find: G (3/2)
2) Find: G (-0.70)
3) Use the form: G (n + ½) = (2n - 1)! (p)1/2 / 2n n!
and show it is equal to the value for G (3/2) you obtained in (1)
4) Hence, or otherwise, compute: G (5/2)
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