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+--- Thread: Bernoulli Num,ber using Series Approximation (/thread-147.html)
Bernoulli Num,ber using Series Approximation - Namir - 12-18-2013 05:57 AM
Bernoulli Numbers using Series Approximations
Algorithm
Given integer n and tolerance toler
Code:
IF n=1 THEN Return -0.5
IF n Mod 2 = 1 THEN Return 0
IF n=1 THEN
RETURN -0.5
ELSE
sum=1
i=2
term=2*toler
WHILE term>toler DO
term=1/i^n
sum=sum+term
i=i+1
END
IF n MOD 4 == 0 THEN chs=-1 ELSE chs=1
t=chs*(n!)/pi^n/2^(n-1)
Return t*sum
ENDHP-41C Implementation
Memory Map
R0 = n
R1 = toler
R2 = i
R3 = Sum
R4 = term
R6 = used in LBL 0
Implementation
Code:
1 LBL A # toler Enter n and press [A]
2 STO 0
3 X<>Y
4 STO 1
5 1
6 RCL 0
7 X=Y?
8 GTO 8
9 2
10 GSB 0 # MOD
11 1
12 X=Y?
13 GTO 9
14 1
15 STO 3 # sum = 1
16 STO 2 # i = 1
17 LBL 1 # Start summation loop
18 1
19 STO+ 2 # i=i+1
20 RCL 2
21 RCL 0
22 Y^X
23 1/X # Calculate 1/i^n
24 STO+ 3 # sum = sum + 1/i^n
25 RCL 0
26 X<=Y? # toler <= term?
27 GTO 1 # resume to the next iteration
28 CF 0 # start calculating other factor
29 RCL 0
30 4
31 GSB 0
32 X=0?
33 SF 0
34 RCL 0
35 N! # n!
36 PI
37 RCL 0
38 Y^X # pi^n
39 /
40 2
41 RCL 0
42 1
43 -
44 Y^X # 2^(n-1)
45 /
46 RCL 03
47 *
48 F? 0 # change sign?
49 CHS
50 CF 0
51 RTN
52 LBL 0 # Implement MOD operator
53 STO 6
54 X<>Y
55 STO 7
56 X<>Y
57 /
58 INT
59 RCL 6
60 *
61 RCL 7
62 X<>Y
63 -
64 RTN
65 LBL 9
66 0
67 RTN
68 LBL 8
69 0
70 .
71 5
72 CHS
73 RTN