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(20S) Gamma Function Approximation (Stirling) - Eddie W. Shore - 10-19-2022 05:43 AM
The gamma function uses the approximation sequence:
Let t = x + 9
Then calculate:
Let G = Γ(t) ≈
exp( ln √(2 × π ÷ t) + t × ln t - t + (12 × t)^-1 - (360 × t^3)^-1 + (1260 × t^5)^-1 )
Note:
(360 × t^3)^-1 = (12 × t)^-1 × (30 × t^2)^-1
(1260 × t^5)^-1 = (360 × t^3)^-1 × (3.5 × t^2)^-1
While x > t:
G = G ÷ x
x = x + 1
End Loop
Display G as the final answer
The approximation polynomial is used for higher values because for the approximation is more accurate for higher values.
HP 20S: Gamma Approximation
(63 steps)
Key Code: { Key }
Code:
61, 41, b : { LBL B }
21, 1 : { STO 1 }
75 : { + }
9 : { 9 }
74 : { = }
21, 2 : { STO 2 }
32 : { +/- }
21, 0 : { STO 0 }
32 : { +/- }
55 : { × }
13 : { LN }
74 : { = }
21, 75, 0 : { STO+ 0 }
2 : { 2 }
55 : { × }
61, 22 : { π }
45 : { ÷ }
22, 2 : { RCL 2 }
74 : { = }
11 : { √ }
13 : { LN }
21, 75, 0 : { STO+ 0 }
1 : { 1 }
2 : { 2 }
55 : { × }
22, 2 : { RCL 2 }
74 : { = }
15 : { 1/x }
21, 75, 0 : { STO+ 0 }
45 : { ÷ }
3 : { 3 }
0 : { 0 }
45 : { ÷ }
22, 2 : { RCL 2 }
51, 11 : { x^2 }
74 : { = }
21, 65, 0 : { STO- 0 }
45 : { ÷ }
3 : { 3 }
73 : { . }
5 : { 5 }
45 : { ÷ }
22, 2 : { RCL 2 }
51, 11 : { x^2 }
74 : { = }
21, 75, 0 : { STO+ 0 }
22, 0 : { RCL 0 }
12 : { e^x }
21, 0 : { STO 0 }
61, 41, 0 : { LBL 0 }
22, 1 : { RCL 1 }
21, 45, 0 : { STO÷ 0 }
1 : { 1 }
21, 75, 1 : { STO+ 1 }
22, 2 : { RCL 2 }
31 : { INPUT }
22, 1 : { RCL 1 }
61, 42 : { x≤y? }
51, 41, 1 : { GTO 1 }
51, 41, 0 : { GTO 0 }
61, 41, 1 : { LBL 1 }
22, 0 : { RCL 0 }
61, 26 : { RTN }Examples
Γ(0.5) returns 1.77245385109
Γ(4.4) returns 10.1361018514
Calculate the gamma function, press [ XEQ ] B.
This program is based on the approximation code of the HP 25.
Source:
Davidson, Jim, and John Vlissides. "HP-25 Program-Gamma Function" ENTER: 65 NOTES Vol. 3 No. 10 December 1976
RE: (20S) Gamma Function Approximation (Stirling) - Gil - 10-19-2022 10:25 AM
Very nice indeed.
Thanks.