(11C) Summation of infinite, alternating series - Thomas Klemm - 03-16-2014 05:16 AM
Code:
001 - 42,21,11 LBL A 029 - 43, 4, 0 SF 0 057 - 44,30,24 STO- (i)
002 - 43, 5, 0 CF 0 030 - 43 35 CLx 058 - 45 .1 RCL .1
003 - 44 .1 STO .1 031 - 44 25 STO I 059 - 45 25 RCL I
004 - 34 x<>y 032 - 42,21, 2 LBL 2 060 - 42 30 x!=y
005 - 44 .0 STO .0 033 - 45 8 RCL 8 061 - 22 0 GTO 0
006 - 1 1 034 - 32 12 GSB B 062 - 45 .2 RCL .2
007 - 44 8 STO 8 035 - 44 24 STO (i) 063 - 44 25 STO I
008 - 0 0 036 - 42 6 ISG 064 - 43 40 x=0
009 - 44 9 STO 9 037 - 42, 7, 4 FIX 4 065 - 22 4 GTO 4
010 - 44 25 STO I 038 - 1 1 066 - 1 1
011 - 42,21, 1 LBL 1 039 - 44,40, 8 STO+ 8 067 - 22 6 GTO 6
012 - 32 12 GSB B 040 - 45 .1 RCL .1 068 - 42,21, 4 LBL 4
013 - 45 8 RCL 8 041 - 45 25 RCL I 069 - 45 24 RCL (i)
014 - 44,30, 8 STO- 8 042 - 42 10 x<=y 070 - 45 8 RCL 8
015 - 44,30, 8 STO- 8 043 - 22 2 GTO 2 071 - 10 /
016 - 20 x 044 - 2 2 072 - 44,40, 9 STO+ 9
017 - 44,40, 9 STO+ 9 045 - 43, 6, 0 F? 0 073 - 2 2
018 - 42 6 ISG 046 - 16 CHS 074 - 16 CHS
019 - 42, 7, 4 FIX 4 047 - 44 8 STO 8 075 - 44,20, 8 STOx 8
020 - 45 .0 RCL .0 048 - 43 16 ABS 076 - 42 6 ISG
021 - 45 25 RCL I 049 - 42,21, 6 LBL 6 077 - 42, 7, 4 FIX 4
022 - 42 10 x<=y 050 - 30 - 078 - 45 .1 RCL .1
023 - 22 1 GTO 1 051 - 44 25 STO I 079 - 45 25 RCL I
024 - 44 8 STO 8 052 - 44 .2 STO .2 080 - 42 10 x<=y
025 - 2 2 053 - 42,21, 0 LBL 0 081 - 22 4 GTO 4
026 - 10 / 054 - 45 24 RCL (i) 082 - 45 9 RCL 9
027 - 42 44 FRAC 055 - 42 6 ISG 083 - 43 32 RTN
028 - 43 30 x!=0 056 - 42, 7, 4 FIX 4 084 - 42,21,12 LBL B
Example:
Find the sum of S = 1 -1/2 + 1/3 - 1/4 + 1/5 - ...
- we define the i-th term = 1/(1+i):
GTO B, P/R, 1, +, 1/x, RTN, P/R
- we'll use PSum = 10, NDif = 7 for maximum accuracy:
10, ENTER, 7, A
- after just one minute, we get the sum = 0.693147182 in the display. The
theoretically correct value is Ln(2) = 0.693147181, so we've got 9 decimals accuracy.
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