American Mathematical Monthly Problem 11426 by M. L. Glasser
Let Γ denote the usual gamma function, defined by Γ (z) =R∞
0 tz−1e−tdt. Simplify Γ
1 14
Γ
9 14
Γ
11 14
Γ 3
14
Γ 5
14
Γ 13
14 .
Solution (by Darij Grinberg).
The duplication formula states that Γ (z) Γ
z+1
2
= 21/2−2z√
2πΓ (2z). Applying this duplication formula to z = 1
14, z= 2
14, and toz = 4
14, we obtain
Γ 1
14
Γ
1 14+ 1
2
| {z }
=
8 14
= 21/2−2·1/14√ 2πΓ
2· 1
14
| {z }
=
2 14
;
Γ 2
14
Γ
2 14+ 1
2
| {z }
= 9 14
= 21/2−2·2/14√ 2πΓ
2· 2
14
| {z }
= 4 14
;
Γ 4
14
Γ
4 14+ 1
2
| {z }
=11 14
= 21/2−2·4/14√ 2πΓ
2· 4
14
| {z }
= 8 14
.
Multiplying these three equations and dividing by Γ 2
14
Γ 4
14
Γ 8
14
(which is 6= 0 since Γ (x)6= 0 for allx >0), we get
Γ 1
14
Γ 9
14
Γ 11
14
= 21/2−2·1/14·21/2−2·2/14·21/2−2·4/14·√ 2π3
. (1)
1
On the other hand, the duplication formula, applied to z = 3
14, z = 6
14, and to z = 12
14,yields
Γ 3
14
Γ
3 14 +1
2
| {z }
=
10 14
= 21/2−2·3/14√ 2πΓ
2· 3
14
| {z }
=
6 14
;
Γ 6
14
Γ
6 14 +1
2
| {z }
=13 14
= 21/2−2·6/14√ 2πΓ
2· 6
14
| {z }
=12 14
;
Γ 12
14
Γ
12 14 +1
2
| {z }
= 5 14+1
| {z }
= 5 14Γ
0
@
5 14
1 A
= 21/2−2·12/14√ 2πΓ
2· 12
14
| {z }
=10 14+1
| {z }
=10 14Γ
0
@
10 14
1 A
.
Multiplying these three equations and dividing by 5 14Γ
10 14
Γ
6 14
Γ
12 14
(which is 6= 0 since Γ (x)6= 0 for allx >0) results in
Γ 3
14
Γ 5
14
Γ 13
14
= 2·21/2−2·3/14·21/2−2·6/14·21/2−2·12/14·√ 2π3
. (2)
From (1) and (2), we conclude that Γ
1 14
Γ
9 14
Γ
11 14
Γ 3
14
Γ 5
14
Γ 13
14
= 21/2−2·1/14·21/2−2·2/14·21/2−2·4/14· √ 2π3
2·21/2−2·3/14·21/2−2·6/14·21/2−2·12/14· √ 2π3
= 2(1/2−2·1/14)+(1/2−2·2/14)+(1/2−2·4/14)−1−(1/2−2·3/14)−(1/2−2·6/14)−(1/2−2·12/14)
= 21 = 2.
2
