HP-15c’s built-in matrix feature can greatly save time and decrease the number of lines when incorporated in a numerical analysis program. I used it extensively during my Numerical Methods subjects in graduate school.
For example, to interpolate (curve fitting) using Newton’s divided difference, we can input the data, x and f(x) using the matrix functionality.
Please see below example and HP15c program:
Program Keystrokes:
000 { }
001 { 42 21 11 } f LBL A
002 { 36 } ENTER
003 { 36 } ENTER
004 { 36 } ENTER
005 { 32 14 } GSB D
006 { 44 3 } STO 3
007 { 34 } x↔y
008 { 1 } 1
009 { 42 16 0 } f MATRIX 0
010 { 42 23 11 } f DIM A
011 { 42 23 12 } f DIM B
012 { 30 } −
013 { 32 14 } GSB D
014 { 44 2 } STO 2
015 { 42 16 1 } f MATRIX 1
016 { 42 21 0 } f LBL 0
017 { 45 0 } RCL 0
018 { 31 } R/S
019 { 44 11 u } STO A
020 { 22 0 } GTO 0
021 { 42 21 1 } f LBL 1
022 { 45 0 } RCL 0
023 { 31 } R/S
024 { 44 12 u } STO B
025 { 22 1 } GTO 1
026 { 42 21 12 } f LBL B
027 { 45 3 } RCL 3
028 { 44 0 } STO 0
029 { 45 2 } RCL 2
030 { 43 44 } g INT
031 { 26 } EEX
032 { 3 } 3
033 { 16 } CHS
034 { 20 } ×
035 { 44 30 0 } STO − 0
036 { 42 21 13 } f LBL C
037 { 1 } 1
038 { 36 } ENTER
039 { 45 40 0 } RCL + 0
040 { 34 } x↔y
041 { 45 43 12 } RCL g B
042 { 45 12 } RCL B
043 { 30 } −
044 { 45 2 } RCL 2
045 { 43 44 } g INT
046 { 45 40 0 } RCL + 0
047 { 1 } 1
048 { 45 43 11 } RCL g A
049 { 45 11 } RCL A
050 { 30 } −
051 { 10 } ÷
052 { 42 31 } f PSE
053 { 42 31 } f PSE
054 { 42 31 } f PSE
055 { 44 12 } STO B
056 { 42 6 0 } f ISG 0
057 { 22 13 } GTO C
058 { 42 6 2 } f ISG 2
059 { 22 12 } GTO B
060 { 43 32 } g RTN
061 { 42 21 14 } f LBL D
062 { 26 } EEX
063 { 3 } 3
064 { 16 } CHS
065 { 20 } ×
066 { 1 } 1
067 { 40 } +
068 { 43 32 } g RTN
To use the HP-15c program:
How to run the program A
N f A
where N= number of data points
example: N=4
X = [5, -1, -3, 2]
Y = f(X) = [120, -6, -32, 3]
4 f A
1.000 (display prompt to enter X1)
R/S (input X1 and press R/S)
2.000 (display prompt to enter X2)
-1 R/S (input X2 and press R/S)
3.000 (display prompt to enter X3)
-3 R/S (input X3 and press R/S)
4.000 (display prompt to enter X4)
2 R/S (input X4 and press R/S)
1.000 (display prompt to enter f(X1))
120 R/S (input f(X1) and press R/S)
2.000 (display prompt to enter f(X2))
-6 R/S (input f(X2) and press R/S)
3.000 (display prompt to enter f(X3))
-32 R/S (input f(X3) and press R/S)
4.000 (display prompt to enter f(X4))
3 R/S (input f(X4) and press R/S)
display the output : (pause 3 seconds after each display)
21 f(X1,X2) coefficient a1
13 f(X2,X3)
7 f(X3,X4)
1 f(X1,X2,X3) coefficient a2
-2 f(X2,X3,X4)
1 f(X1,X2,X3,X4) coefficient a3
and a0 = 120; the first and initial f(X1)
f(x) = a0 + a1(X-5) + a2(X-5)(X- (-1)) + a3(X-5)(X-(-1))(X-(-3))
f(x) = 120 + 21(X-5) + 1(X-5)(X+1) + 1(X-5)(X+1)(X+3)
f(x) = 120 + 21(X-5) + (X-5)(X+1) + (X-5)(X+1)(X+3) --> Perfect Fit!!!