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Lagrangian Interpolation - Namir - 12-18-2013 06:04 AM
Langrangian Interpolation of 3 points
Usage
A program prompts you to enter 3 points and the
interpolated x
Enter y1 and the x1 at first prompt
Enter y2 and the x2 at second prompt
Enter y3 and the x3 at third prompt
Enter xint at fourth prompt
Note calculator displays 1, 2, 3 at the first, second,
and third prompt
B program prompts you to enter the interpolated x
Program calculates and display the value of the interpolated Y.
Algorithm
Code:
INPUT N, array X(1..N), Y(1..N), and Xint
Yint = 0
FOR I = 1 TO N
Product = Y(I)
FOR J = 1 to N
IF I <> J THEN
Product = Product * (Xint - X(J)) / (X(I) - X(J))
ENDIF
NEXT J
Yint = Yint + Product
NEXT I
Show YintMemory Map
R00 = x1
R01 = x2
R02 = X3
R03 = y1
R04 = y2
R05 = y3
R06 = x
R07 = y
R08 = Product
R09 = used
A = I for X(I)
B = J for X(J)
C = I for Y(I)
Source Code
Code:
1 LBL A
2 1
3 R/S # enter Y1 and X1
4 STO 0
5 x<>y
6 STO 3
7 2
8 R/S # enter Y2 and X2
9 STO 1
10 x<>y
11 STO 4
12 3
13 R/S # enter Y2 and X3
14 STO 2
15 x<>y
16 STO 5
17 R/S # enter Xint
18 LBL B
19 STO 6
20 2
21 STO B # set J for X(J)
22 5
23 STO C # set I for Y(I)
24 0
25 STO 7 # y = 0
26 LBL 0 # outer loop starts
27 RCL C
28 GSB 9
29 STO 8
30 2
31 STO A # set I fo X(I)
32 LBL 1 # inner loop starts
33 RCL A
34 RCL B
35 x=y? if I = J the skip iteration
36 GTO 2
37 RCL 6
38 RCL A
39 GSB 9
40 STO 9
41 -
42 RCL B
43 GSB 9
44 RCL 9
45 -
46 /
47 STO* 8 # update product
48 LBL 2
49 RCL A
50 1
51 -
52 STO A
53 x>0?
54 GTO 1 # inner loop ends
55 x=0?
56 GTO 1
57 RCL 8
58 STO+ 7
59 RCL B
60 1
61 -
62 STO B
63 RCL C
64 1
65 -
66 STO C
67 RCL B
68 x>0?
69 GTO 0 # outer loop ends
70 x=0?
71 GTO 0
72 RCL 7
73 R/S
74 GTO B
75 LBL 9 # INDIRECT ADDRESS
76 STO I
77 RDN
78 RCL (I)
79 RTNRE: Lagrangian Interpolation - bshoring - 03-05-2015 05:17 AM
Very nice program. Just loaded it and it works very well.
Thank you for posting !
Bob
RE: Lagrangian Interpolation - PedroLeiva - 03-05-2015 09:33 PM
(03-05-2015 05:17 AM)bshoring Wrote: Very nice program. Just loaded it and it works very well.
Thank you for posting !
Bob
Namir, thank you very much for your contribution. I guess it is coded for HP- 67. Could you please include a sample calculation for testing the correct entry of the programming steps.
From already thank you for your time. Pedro
RE: Lagrangian Interpolation - bshoring - 03-07-2015 11:49 PM
A program for the HP-25 gave this example:
X1=-11 Y1=121 X2=3 Y2=9 X3=2 Y3=4
X=2.5 Y=6.25 etc.
You should get the same results on this program.
Bob
RE: Lagrangian Interpolation - Thomas Klemm - 03-08-2015 06:02 AM
You could use the barycentric interpolation formula:
\[
L(x) = \frac{\sum_{j=0}^k \frac{w_j}{x-x_j}y_j}{\sum_{j=0}^k \frac{w_j}{x-x_j}}
\]
This avoids the nested loop at the cost of 3 additional registers for the weights \(w_j\). These weights have to be computed only once for the given data-set.
With only 3 points you could even unroll the loop which would probably speed up the calculation.
Cheers
Thomas
RE: Lagrangian Interpolation - Thomas Klemm - 03-08-2015 07:58 PM
This program uses the method mentioned in my previous post.
Memory Map
\(
\begin{matrix}
I & : & x \\
R_0 & : & x_1 \\
R_1 & : & x_2 \\
R_2 & : & x_3 \\
R_3 & : & y_1 \\
R_4 & : & y_2 \\
R_5 & : & y_3 \\
R_6 & : & w_1 \\
R_7 & : & w_2 \\
R_8 & : & w_3 \\
R_{S4} & : & \Sigma x \\
R_{S8} & : & \Sigma xy \\
\end{matrix}
\)
Code:
001 31 25 11 : LBL A
002 01 : 1
003 84 : R/S
004 33 00 : STO 0
005 35 52 : x<>y
006 33 03 : STO 3
007 02 : 2
008 84 : R/S
009 33 01 : STO 1
010 35 52 : x<>y
011 33 04 : STO 4
012 03 : 3
013 84 : R/S
014 33 02 : STO 2
015 35 52 : x<>Y
016 33 05 : STO 5
017 01 : 1
018 42 : CHS
019 33 06 : STO 6
020 33 07 : STO 7
021 33 08 : STO 8
022 34 00 : RCL 0
023 34 01 : RCL 1
024 51 : -
025 33 81 06 : STO/ 6
026 33 81 07 : STO/ 7
027 34 01 : RCL 1
028 34 02 : RCL 2
029 51 : -
030 33 81 07 : STO/ 7
031 33 81 08 : STO/ 8
032 34 02 : RCL 2
033 35 00 : RCL 0
034 51 : -
035 33 81 08 : STO/ 8
036 33 81 06 : STO/ 6
037 31 42 : P<>S
038 84 : R/S
039 31 25 12 : LBL B
040 31 43 : CL REG
041 35 33 : ST I
042 31 42 : P<>S
043 34 03 : RCL 3
044 34 06 : RCL 6
045 35 34 : RC I
046 34 00 : RCL 0
047 51 : -
048 81 : /
049 21 : Σ+
050 34 04 : RCL 4
051 34 07 : RCL 7
052 35 34 : RC I
053 34 01 : RCL 1
054 51 : -
055 81 : /
056 21 : Σ+
057 34 05 : RCL 5
058 34 08 : RCL 8
059 35 34 : RC I
060 34 02 : RCL 2
061 51 : -
062 81 : /
063 21 : Σ+
064 31 42 : P<>S
065 34 08 : RCL 8
066 34 04 : RCL 4
067 81 : /
068 35 22 : RTNExplanation
Lines 001-016 are the same as in Namir's orginal program.
Lines 017-036 calculate the weights
\(w_j = \frac{1}{\prod_{i=0,i \neq j}^k(x_j-x_i)}\).
However this formula is slightly changed so that we only need the differences \(x_1-x_2\), \(x_2-x_3\) and \(x_3-x_1\):
\(w_1=\frac{-1}{(x_1-x_2)(x_3-x_1)}\)
\(w_2=\frac{-1}{(x_2-x_3)(x_1-x_2)}\)
\(w_3=\frac{-1}{(x_3-x_1)(x_2-x_3)}\)
As you can see the same factor can be used in two of the weights. Nothing fancy it's just to get the sign correct. That's why we start with \(-1\) in lines 017-021.
Lines 043-063 recall \(y_j\) and calculate \(\frac{w_j}{x-x_j}\) and then use \(\Sigma +\) to calculate both \(\sum_{j=0}^k \frac{w_j}{x-x_j}\) and \(\sum_{j=0}^k \frac{w_j}{x-x_j}y_j\) in one single step.
Lines 065-067 finally recall these values and calculate \(\frac{\Sigma xy}{\Sigma x}\).
Cheers
Thomas
For those with a HP-15C here's the corresponding program:
Code:
001 - 42,21,11 LBL A
002 - 1 1
003 - 31 R/S
004 - 44 8 STO 8
005 - 34 x<>y
006 - 44 .1 STO .1
007 - 2 2
008 - 31 R/S
009 - 44 9 STO 9
010 - 34 x<>y
011 - 44 .2 STO .2
012 - 3 3
013 - 31 R/S
014 - 44 .0 STO .0
015 - 34 x<>y
016 - 44 .3 STO .3
017 - 1 1
018 - 16 CHS
019 - 44 .4 STO .4
020 - 44 .5 STO .5
021 - 44 .6 STO .6
022 - 45 8 RCL 8
023 - 45,30, 9 RCL- 9
024 - 44,10,.4 STO/ .4
025 - 44,10,.5 STO/ .5
026 - 45 9 RCL 9
027 - 45,30,.0 RCL- .0
028 - 44,10,.5 STO/ .5
029 - 44,10,.6 STO/ .6
030 - 45 .0 RCL .0
031 - 45,30, 8 RCL- 8
032 - 44,10,.6 STO/ .6
033 - 44,10,.4 STO/ .4
034 - 31 R/S
035 - 42,21,12 LBL B
036 - 44 0 STO 0
037 - 42 32 CLEAR Σ
038 - 45 .1 RCL .1
039 - 45 .4 RCL .4
040 - 45 0 RCL 0
041 - 45,30, 8 RCL- 8
042 - 10 /
043 - 49 Σ+
044 - 45 .2 RCL .2
045 - 45 .5 RCL .5
046 - 45 0 RCL 0
047 - 45,30, 9 RCL- 9
048 - 10 /
049 - 49 Σ+
050 - 45 .3 RCL .3
051 - 45 .6 RCL .6
052 - 45 0 RCL 0
053 - 45,30,.0 RCL- .0
054 - 10 /
055 - 49 Σ+
056 - 45 7 RCL 7
057 - 45,10, 3 RCL/ 3
058 - 43 32 RTNAdded a card of the program that can be used with Jacques Laporte's HP-67 Microcode Simulator.
RE: Lagrangian Interpolation - bshoring - 03-09-2015 03:30 AM
Thomas,
I assume pressing Keys A & B are the same as in Namir's, am I right?
However after entering three data pairs, when I enter an X value and press B I am getting an Error message.
For example, if I enter these X & Y values:
100 Enter 1
105 Enter 2
110.25 Enter 3
and then press 1 B, I either get an error message or get a number like 0.49 that doesn't correspond. With Namir's program, pressing 1 B would give me 100.
I have checked and rechecked the program listing and can't find any errors. Am I entering the numbers wrong ?
Thanks,
Bob
RE: Lagrangian Interpolation - PedroLeiva - 03-09-2015 03:37 AM
(03-07-2015 11:49 PM)bshoring Wrote: A program for the HP-25 gave this example:
X1=-11 Y1=121 X2=3 Y2=9 X3=2 Y3=4
X=2.5 Y=6.25 etc.
You should get the same results on this program.
Bob
Thank you very much Bob, I'll try the values in my HP 35s. Pedro
RE: Lagrangian Interpolation - Thomas Klemm - 03-09-2015 09:36 AM
(03-09-2015 03:30 AM)bshoring Wrote: I have checked and rechecked the program listing and can't find any errors. Am I entering the numbers wrong ?
No. That's the problem with this formula:
\(\ell_j(x) = \ell(x)\frac{w_j}{x-x_j}\)
This leads to a division by 0 if \(x=x_j\). The reasoning is that you usually don't want to evaluate the interpolation at the known points.
You could try a value very close to 1 instead.
But I think your point is valid and it made me wonder if we could just use the Lagrange polynomials as Namir did but still unroll the loops. Thus the weights are now calculated in lines 017-031:
\(w_1=\frac{y_1}{(x_1-x_2)(x_3-x_1)}\)
\(w_2=\frac{y_2}{(x_2-x_3)(x_1-x_2)}\)
\(w_3=\frac{y_3}{(x_3-x_1)(x_2-x_3)}\)
Memory Map
\(
\begin{matrix}
I & : & x \\
R_0 & : & x_1 \\
R_1 & : & x_2 \\
R_2 & : & x_3 \\
R_3 & : & w_1 \\
R_4 & : & w_2 \\
R_5 & : & w_3 \\
\end{matrix}
\)
By factoring the sum I was able to calculate it using only the stack in lines 035-057:
\((x-x_2)(x-x_3)w_1+(x-x_1)(x-x_3)w_2+(x-x_1)(x-x_2)w_3=((x-x_2)w_1+(x-x_1)w_2)(x-x_3)+(x-x_1)(x-x_2)w_3\)
Code:
001 31 25 11 : LBL A
002 01 : 1
003 84 : R/S
004 33 00 : STO 0
005 35 52 : x<>y
006 33 03 : STO 3
007 02 : 2
008 84 : R/S
009 33 01 : STO 1
010 35 52 : x<>y
011 33 04 : STO 4
012 03 : 3
013 84 : R/S
014 33 02 : STO 2
015 35 52 : x<>Y
016 33 05 : STO 5
017 34 00 : RCL 0
018 34 01 : RCL 1
019 51 : -
020 33 81 03 : STO/ 3
021 33 81 04 : STO/ 4
022 34 01 : RCL 1
023 34 02 : RCL 2
024 51 : -
025 33 81 04 : STO/ 4
026 33 81 05 : STO/ 5
027 34 02 : RCL 2
028 34 00 : RCL 0
029 51 : -
030 33 81 05 : STO/ 5
031 33 81 03 : STO/ 3
032 84 : R/S
033 31 25 12 : LBL B
034 35 33 : ST I
035 34 04 : RCL 4
036 35 34 : RC I
037 34 00 : RCL 0
038 51 : -
039 71 : *
040 34 05 : RCL 5
041 35 82 : LST x
042 71 : *
043 35 34 : RC I
044 34 01 : RCL 1
045 51 : -
046 71 : *
047 35 52 : x<>y
048 34 03 : RCL 3
049 35 82 : LST x
050 71 : *
051 61 : +
052 35 34 : RC I
053 34 02 : RCL 2
054 51 : -
055 71 : *
056 61 : +
057 42 : CHS
058 35 22 : RTNIt made the program even a little shorter.
Cheers
Thomas
RE: Lagrangian Interpolation - bshoring - 03-09-2015 09:50 PM
Thomas,
Thanks for the updated code. This works just great ! And it's fast, too.
Regards,
Bob
RE: Lagrangian Interpolation - Thomas Klemm - 03-11-2015 08:04 AM
For the sake of completeness I wrote a program using the Newton Polynomial
\(N(x) = [y_0] + [y_0,y_1](x-x_0) + \cdots + [y_0,\ldots,y_k](x-x_0)(x-x_1)\cdots(x-x_{k-1})\)
where
\([y_0,\ldots,y_j]\)
is the notation for divided differences.
Code:
001 31 25 11 : LBL A
002 01 : 1
003 84 : R/S
004 33 00 : STO 0
005 35 52 : x<>y
006 33 03 : STO 3
007 02 : 2
008 84 : R/S
009 33 01 : STO 1
010 35 52 : x<>y
011 33 04 : STO 4
012 03 : 3
013 84 : R/S
014 33 02 : STO 2
015 35 52 : x<>y
016 34 04 : RCL 4
017 51 : -
018 34 02 : RCL 2
019 34 01 : RCL 1
020 51 : -
021 81 : /
022 34 04 : RCL 4
023 34 03 : RCL 3
024 51 : -
025 34 01 : RCL 1
026 34 00 : RCL 0
027 51 : -
028 81 : /
029 33 04 : STO 4
030 51 : -
031 34 02 : RCL 2
032 34 00 : RCL 0
033 51 : -
034 81 : /
035 34 05 : STO 5
036 84 : R/S
037 31 25 12 : LBL B
038 35 33 : ST I
039 34 01 : RCL 1
040 51 : -
041 34 05 : RCL 5
042 71 : *
043 34 04 : RCL 4
044 61 : +
045 35 34 : RC I
046 34 00 : RCL 0
047 51 : -
048 71 : *
049 34 03 : RCL 3
050 61 : +
051 35 22 : RTNIt is a little shorter and maybe a little faster as well.
Cheers
Thomas
Thanks to its extended register arithmetic the program for the HP-15C is even 10 lines shorter:
Code:
001 - 42,21,11 LBL A
002 - 1 1
003 - 31 R/S
004 - 44 0 STO 0
005 - 34 x<>y
006 - 44 3 STO 3
007 - 2 2
008 - 31 R/S
009 - 44 1 STO 1
010 - 34 x<>y
011 - 44 4 STO 4
012 - 3 3
013 - 31 R/S
014 - 44 2 STO 2
015 - 34 x<>y
016 - 45,30, 4 RCL- 4
017 - 45 2 RCL 2
018 - 45,30, 1 RCL- 1
019 - 10 /
020 - 45 4 RCL 4
021 - 45,30, 3 RCL- 3
022 - 45 1 RCL 1
023 - 45,30, 0 RCL- 0
024 - 10 /
025 - 44 4 STO 4
026 - 30 -
027 - 45 2 RCL 2
028 - 45,30, 0 RCL- 0
029 - 10 /
030 - 44 5 STO 5
031 - 31 R/S
032 - 42,21,12 LBL B
033 - 44 25 STO I
034 - 45,30, 1 RCL- 1
035 - 45,20, 5 RCLx 5
036 - 45,40, 4 RCL+ 4
037 - 45 25 RCL I
038 - 45,30, 0 RCL- 0
039 - 20 x
040 - 45,40, 3 RCL+ 3
041 - 43 32 RTNRE: Lagrangian Interpolation - bshoring - 03-13-2015 05:33 AM
Thanks, Thomas for posting the listing for the Newton polynomial.
Looks nice. Haven't had time to try it yet. It also looks like it will work on an HP-25 as well as the HP-67.
Regards,
Bob
RE: Lagrangian Interpolation - Newton Polynomial - bshoring - 04-04-2015 07:40 PM
Newton Polynomial Version works just great. Thanks Thomas for posting !
Bob
RE: Lagrangian Interpolation - Thomas Klemm - 03-09-2019 07:08 PM
(03-13-2015 05:33 AM)bshoring Wrote: It also looks like it will work on an HP-25 as well as the HP-67.
This program just fits in the 49 steps available for the HP-25:
Code:
01: 01 : 1
02: 74 : R/S
03: 23 00 : STO 0
04: 21 : x<>y
05: 23 03 : STO 3
06: 02 : 2
07: 74 : R/S
08: 23 01 : STO 1
09: 21 : x<>y
10: 23 04 : STO 4
11: 03 : 3
12: 74 : R/S
13: 23 02 : STO 2
14: 21 : x<>y
15: 24 04 : RCL 4
16: 41 : -
17: 24 02 : RCL 2
18: 24 01 : RCL 1
19: 41 : -
20: 71 : ÷
21: 24 04 : RCL 4
22: 24 03 : RCL 3
23: 41 : -
24: 24 01 : RCL 1
25: 24 00 : RCL 0
26: 41 : -
27: 71 : ÷
28: 23 04 : STO 4
29: 41 : -
30: 24 02 : RCL 2
31: 24 00 : RCL 0
32: 41 : -
33: 71 : ÷
34: 23 05 : STO 5
35: 74 : R/S
36: 23 06 : STO 6
37: 24 01 : RCL 1
38: 41 : -
39: 24 05 : RCL 5
40: 61 : ×
41: 24 04 : RCL 4
42: 51 : +
43: 24 06 : RCL 6
44: 24 00 : RCL 0
45: 41 : -
46: 61 : ×
47: 24 03 : RCL 3
48: 51 : +
49: 13 35 : GTO 35Definition of the Polynomial
Example:
Find a quadratic polynomial given these 3 points: \(P_1(-5, 12)\), \(P_2(1, 13)\) and \(P_3(2, 11)\)
CLEAR PRGM
R/S
1.0000
12 ENTER -5
R/S
2.0000
13 ENTER 1
R/S
3.0000
11 ENTER 2
R/S
These are the coefficients of the Newton polynomial:
R3: 12.000000
R4: 0.166667
R5: -0.309524
This leads to the formula:
\(f(x) = 12 + (x+5)(\frac{1}{6} - (x-1)\frac{13}{42})\)
Interpolation of the Polynomial
Example:
Evaluate the polynomial at \(x=0.5\).
0.5
R/S
13.7679
Hint: You can just enter new values and hit R/S to evaluate the polynomial.
Cheers
Thomas
RE: Lagrangian Interpolation - PedroLeiva - 03-14-2019 03:55 PM
Here is my compilation for HP 25, I have not modified anything, just for your file
To make things clear, in the HP 25 Application Programs manual, Chap 5 Numerical Methods: Linear interpolation, page 85. cites a program that works with two points instead of three. This means that when the function is a straight line I have to use this and when it is a curve the Lagrangian method?
I apologize for my little knowledge in mathematics
Pedro
RE: Lagrangian Interpolation - Thomas Klemm - 03-14-2019 05:56 PM
(03-14-2019 03:55 PM)PedroLeiva Wrote: This means that when the function is a straight line I have to use this and when it is a curve the Lagrangian method?
You could enter a dummy point (e.g. 0 ENTER) as third point.
Just make sure that the x-value is different from the other 2 points.
And then clear register 5:
CLx
STO 5
If you evaluate now the polynomial of the previous example at \(x=2\) you get:
2
R/S
13.1667
Or then you can just shorten the existing program to:
Code:
01: 01 : 1
02: 74 : R/S
03: 23 00 : STO 0
04: 21 : x<>y
05: 23 01 : STO 1
06: 02 : 2
07: 74 : R/S
08: 21 : x<>y
09: 24 01 : RCL 1
10: 41 : -
11: 21 : x<>y
12: 24 00 : RCL 0
13: 41 : -
14: 71 : ÷
15: 23 02 : STO 2
16: 74 : R/S
17: 24 00 : RCL 0
18: 41 : -
19: 24 02 : RCL 2
20: 61 : ×
21: 24 01 : RCL 1
22: 51 : +
23: 13 16 : GTO 16HTH
Thomas
PS: You could also use the mid-point of the first two points as a third point.
This would lead to a "natural" way to make the entry in register 5 equal to 0.
RE: Lagrangian Interpolation - PedroLeiva - 03-14-2019 07:22 PM
Both altrnatives (0 ENTER Xn) plus Clx STO 5, or (12,5 ENTER -2) gives me the same results as yours: 13.1667. So it works as you suggest. So this is a way to use Lagrangian for linear interpolation (1st. order polynomials), only two points.
For 2nd order polynomials we can use your program, just loading three points; so we cannot use it for 3rd. or 4th. order, considering we will need 4 or 5 points. Am I right?
Pedro
RE: Lagrangian Interpolation - Thomas Klemm - 03-14-2019 07:58 PM
(03-14-2019 07:22 PM)PedroLeiva Wrote: so we cannot use it for 3rd. or 4th. order, considering we will need 4 or 5 points. Am I right?
Correct.
In this case you could use (42S) Newton Polynomial Interpolation.
It allows to enter as many points as you like and memory allows.
Given the restrictions of the HP-25 I'm afraid we can't go further than 3 points.
Cheers
Thomas
RE: Lagrangian Interpolation - PedroLeiva - 03-14-2019 08:12 PM
I think I was asking HP25 too much. Anyway, I found that the example of the manual for linear interpolation works very well in both cases………….
Example from HP 25 Application Programs, Chapter 5 Numerical Methods/Linear Interpolation: p86
Given
f(7.3)= 1.9879
f(7.4)= 2.0015
find by linear interpolation f(7.37)
A. CLEAR PRGM
R/S
1.0000
1.9879 ENTER 7.3
R/S
2.0000
2.0015 ENTER 7.4
R/S
3.0000
0 ENTER 7
[R/S]
CLx STO 5
7.37 [R/S] 1.9974
B. CLEAR PRGM
R/S
1.0000
1.9879 ENTER 7.3
R/S
2.0000
2.0015 ENTER 7.4
R/S
3.0000
1.9947 [ENTER] 7.35
[R/S]
7.37 [R/S] 1.9974
I appreciate your attention, Pedro
RE: Lagrangian Interpolation - Thomas Klemm - 07-23-2022 10:20 AM
(03-14-2019 08:12 PM)PedroLeiva Wrote: Example from HP 25 Application Programs, Chapter 5 Numerical Methods/Linear Interpolation: p86
Given
f(7.3)= 1.9879
f(7.4)= 2.0015
find by linear interpolation f(7.37)
For those who want to give it a try:
Code:
01: 23 04 : STO 4
02: 24 00 : RCL 0
03: 41 : -
04: 24 03 : RCL 3
05: 61 : *
06: 24 02 : RCL 2
07: 24 04 : RCL 4
08: 41 : -
09: 24 01 : RCL 1
10: 61 : *
11: 51 : +
12: 24 02 : RCL 2
13: 24 00 : RCL 0
14: 41 : -
15: 71 : /
16: 13 00 : GTO 00References
- HP-25C emulator
- HP-25 Applications Programs
- (25) 55 Applications Programs for Teenix MultiCalculator Emulator and Hardware