Solutions to Math Problems

We look at the solutions to the problems from the blog before last, dealing with Gamma and Beta functions.

1. G (6) = (6 - 1)! = 5! = 5·4 · 3 · 2 · 1  = 120

Or:  G (6) = G (5 + 1) = 5 G (5) where:

G (5) = (5 - 1)! = 4! = 4 · 3 · 2 · 1  = 24,  and

5 G (5)  = 5 x 24 = 120

2. b(3, 6) = G (3) G (6)/ G (3 + 6)

Whence: G (3) = (3 - 1)! = 2! = 2•1 = 2

And: G (6) = 120 (from 1), so:

b(3, 6) =  (2) (120)/ G (9)

And:   G (9) = (9 - 1)! = 8! = 40320, then:

b(3, 6) =  (2) (120)/ 40320 = 240/ 40320 = 5.95 x 10^-3

3.  We have (from the link):

G (½)  =   ò^¥ _o  [ exp(-t)/  Öt ]  dt 

=  2  ò^¥ _o   exp(-u² ) du    = 2 Öp/  2 =  Öp