# Curve fit on HP-41

Calculator software for curve fit for HP-41. Computes a curve to a set of data points. (linear, exponential, logarithmic and power). The code originates from the PPC ROM Module released by the international group People Programming Computers on December 1981 [1,2]. That in turn was based on Gary Tenzer’s curve fit program published in PPC Journal.

Curve types
name formula restriction
linear $$y=bx + a$$
exponential $$y=ae^{bx}$$ for $$a>0$$
logarithmic $$y=b\ln{x}+1$$
power $$y=ax^{b}$$ for $$a>0$$

The program will compute the coefficients a and b for the curve types as well as compute the coefficient of determination (r2) which is a measure of the goodness of fit. Once a set of data has been fit to a given curve type, a prediction may be made for the y-value given a new x-value, or a prediction may be made for the x-value given a new y-value.

## Instructions

The functions map to the top row on the USER keyboard. These same functions are referenced in the examples by enclosing them in a blue box.

The key mappings are shown in the table below.

Key mappings
USER keyboard description usage
∑+ ∑+ Add a data point x ENTER y ∑+
∑- ∑- Delete a data point x ENTER y ∑+
SOLVEj 1/x Solve type j j SOLVEj
y^ √x Predict y value x y^
x^ LOG Predict x value y x^
SOLVE LN Solve best type
INIT ex Initialize

Operation

1. GTO “CV”
2. SIZE 027
3. USER mode
4. INIT, the display will show 1.
5. key in the data pairs as x ENTER y and push ∑+. The display will stop with the count of the number of the next data pair. This feature makes it possible to enter only the y-values when the x-values are consecutive integers starting at 1. In this case the display provides the x-values which need to be entered. If an improper data pair has just been input with the ∑+ key, then immediately press R/S will delete the pair. Otherwise an improper or undesired data pair can be deleted by re-entering both x and y and pressing ∑-.
6. repeat step until all data pairs are entered.
7. If any x-values are negative or zero, then only types 1 and 2 are feasible curves. If any y-values are negative or zero, then only types 1 and 3 are feasible curves. If in any data pair both and x and y are negative or zero, the type 1 is the only feasible curve. The a coefficient must be positive for curve type 2 and 4.
8. Select the desired curve type. Once the curve type is selected, the HP-41 should not be interrupted.
9. To fit a particular curve type, key in the number 1-4 for that type and press SOLVEj. The stack returns with:
Stack usage
stack register description
Z R10 R2
Y R09 a coefficient
X R08 b coefficient
10. The closer R2 is to one of the extremes (-1 or 1), the better the fit. The sign indicates whether the data is positively or negatively skewed. Note that this step may be repeated any number of times of any of the four curve types.
11. If all data data input is positive, then pressing SOLVE will automatically choose the curve of best fit according to the curve type with largest absolute value of r2. In this case the stack returns with:
Stack usage
stack register description
Z R10 R2
Z R09 a coefficient
Y R08 b coefficient
X R07 j = best curve type
12. Predictions for new x or y values may be made only after the previous step has been completed. Predictions for new values are based on curve type (encoded in flags F08 and F09).
• To predict y given x, key in x and press y^
• To predict x given y, key in y and press x^
13. New data may be added or deleted at any time via the ∑+ and ∑- keys. However the “solve” must be preformed after updating the data before any new predictions can be made. The parameters a and b automatically destroyed after input of new data.

## Examples

We reuse the examples from the PPC ROM User Manual. [hp41.com]

### Straight line

Find the straight line which best fits the data set shown below, and predict y when $$x=20$$ and predict $$x$$ when $$y=25$$.

$$(1.1, 5.2), (4.5, 12.6), (8, 20), (10, 23), (15.6, 34)$$

Minimum size is SIZE 25.

Straight line
key strokes display
GTO “CV”
SIZE 027
USER
INIT 1.0000
1.1 ENTER 5.2 ∑+ 2.0000
4.5 ENTER 12.6 ∑+ 3.0000
8 ENTER 20 ∑+ 4.0000
10 ENTER 23 ∑+ 5.0000
15.6 ENTER 34 6.0000
1 SOLVEj 1.9720

After solving for linear (type 1), the values for $$a$$, $$b$$ and $$R^2$$ are on the stack. We find: $$y=1.972047542x + 3.499147270$$

To determine the points along that line:

Straight line
key strokes display
20 y^ 42.94009811
25 x^ 10.90280649

### Linear or exponential curve

The data shown below fits either a linear or exponential curve. Determine which is more appropriate.

$$(2,12), (-1,2), (3,17), (5,23)$$

Again, to solve this, we use the CV program.

Linear or exponential curve
key strokes display description
INIT 1.0000
2 ENTER 12 ∑+ 2.0000
1 CHS ENTER 2 ∑+ 3.0000
3 ENTER 17 ∑+ 4.0000
5 ENTER 23 ∑+ 5.0000
1 SOLVEj 3.5467 linear, b coefficient
R 5.5200 linear, a coefficient
R 0.9976 linear, R2
2 SOLVEj 3.5467 exponential, b coefficient
R 3.8262 exponential, a coefficient
R 0.9586 exponential, R2

Choosing the best $$R^2$$, we find a linear fit is more appropriate. $$y=3.5467x+5.5200$$

Predict $$y$$ when $$x=-10$$. Since we just finished the exponential fit, the exponential parameters are still in the machine, and hence we must go back and solve for linear to predict $$y$$ for $$x=-10$$.

Predict $$y$$ for $$x=-10$$
key strokes display description
1 SOLVEj 3.5467
10 CHS y^ -29.9467

Add the additional data points shown below and solve the same problem.

$$(-4, 0.713), (2.5, 10.93), (6, 47.53), (10, 254.95)$$

Note that the display should show 6 after entering the first new data pair. We try a linear and exponential fit.

key strokes display description
4 CHS ENTER 0.713 ∑+ 6.0000
2.5 ENTER 10.93 ∑+ 7.0000
6 ENTER 47.53 ∑+ 8.0000
10 ENTER 254.95 ∑+ 9.0000
1 SOLVEj 15.3315 linear, b coefficient
R 0.9790 linear, a coefficient
R 0.7657 linear, R2
2 SOLVEj 0.4199 exponential, b coefficient
R 3.8256 exponential, a coefficient
R 0.9936 exponential, R2

Now choosing the best $$R^2$$, we see that the new data reflects a change in the curve type. $$y=3.8256e^{0.4199}$$

Since the exponential parameters are still in the machine, we can predict $$y$$ for $$x=-10$$ as

Predict y for x=-10
key strokes display
10 CHS y^ 0.0574

### Best curve

Fit the best curve to the data points shown below.

$$(1,2), (2,2.828), (3, 3.464), (4, 4), (5, 4.472), (6, 4.899), (7, 5.292), (8, 5.657) and (9, 6)$$

In this example the $$x$$-coordinates start counting from $$1$$ and are consecutive integers. So we need only input the y-coordinates, but they must be in the proper order. The count in the display will serve as the x-coordinates.

Find the best curve
key strokes display
INIT 1.0000
2 ∑+ 2.0000
2.828 ∑+ 3.0000
3.464 ∑+ 4.0000
4 ∑+ 5.0000
4.472 ∑+ 6.0000
4.899 ∑+ 7.0000
5.292 ∑+ 8.0000
5.657 ∑+ 9.0000
6 ∑+ 10.0000
SOLVE 4.0000

We could use the best fit function because all the values were positive. After solving, the stack represents j, b, a and r2. The value j=4 indicates a power curve. Rounded to 2 decimals, this is $$y = 2.00 x^{0.50}$$ using zumzum.com

## Source code

• Requires X-Functions module on the HP-41cv, and a minimum SIZE 25.
• Available as source code, raw code for the V41 emulator and bar code for the HP Wand (HP82153A)
• Size 321 Bytes, 3 magnetic tracks (321/112=2.866)

### Listing

Available through the repository

; /———————————————————————\ ; | C u r v e F i t | ; | | ; | for the HP-41 | ; \———————————————————————/ ; ; 1.00 ; PPC ROM ; ; https://coertvonk.com/technology/hp41/curve-fit-4581 ; BLOCK CLEAR ROUTINE, stores zeroes in a block of registers ; ; uses the complete form of the general block control word bbb.eeeii an ; can thus be used to clear blocks of consecutive registers or can be used ; to sip over registers within a block. 01 LBL “BC” 02 LBL 16 03 SIGN 04 CLX 05 LBL 20 06 STO IND L 07 ISG L 08 GTO 20 09 RTN 10 LBL 17 11 LBL “CV” 12 GTO IND 06 ; provides access to all numeric labels within CV ; INPUT A DATA POINT ; ; All summations are updated when this routine is called. These summations ; include sums of x, x^2, y, y^2, xy, ln(x), ln(x)^2, ln(y)^2, ln(x).ln(y), ; x.ln(y), y.ln(x). 13 LBL A ; [sigma+] 14 LBL 01 15 CF 10 16 LBL 06 17 STO 09 18 X<>Y 19 STO 08 20 sREG 13 21 FC? 10 22 s+ 23 FS? 10 24 s- 25 RDN 26 RCL 08 27 ENTER^ 28 X>0? 29 LN 30 ST* Z 31 RCL 09 32 X>0? 33 LN 34 ST* Z 35 X<>Y 36 sREG 19 37 FC? 10 38 s+ 39 FS? 10 40 s- 41 R^ 42 FS? 10 43 CHS 44 ST+ 12 45 R^ 46 FS? 10 47 CHS 48 ST+ 11 49 X<> Z 50 SIGN 51 ST+ L 52 RCL 08 53 RCL 09 54 X<> L 55 TONE 9 56 RTN 57 RCL 08 58 RCL 09 ; REMOVE A DATA POINT ; 59 LBL a 60 SF 10 ; signal “remove data point” 61 GTO 06 ; jump to “add data point” ; SOLVE SPECIFIED CURVE TYPE J ; ; b is stored in R08, a is stored in R09, r2 is stored in R10 62 LBL B ; [SOLVEj] 63 LBL 02 64 CF 08 65 CF 09 66 STO 07 67 2 68 XY 99 RCL 16 100 RCL 15 101 X^2 102 RCL 18 103 ST/ 09 104 / 105 – 106 * 107 SQRT 108 ST/ 10 109 XEQ IND 07 110 8 111 ST- 07 112 RCL 10 113 RCL 09 114 FS? 08 115 E^X 116 STO 09 117 RCL 08 118 TONE 5 119 RTN ; A SERIES OF INTERTWINED SUBROUTINES, called in the curve fitting process. ; ; These routine simply perform a series of register exchanges which place ; the proper sums in the sigma registers for the calculation of the ; parameters a, b and r2 depending on the curve type selected. Since the ; exchange is performed twice (line 076, line 109) all registers are ; returned to their original state. 120 LBL 10 121 RCL 11 122 X<> 17 123 STO 11 124 LBL 13 125 RCL 21 126 X<> 15 127 STO 21 128 RCL 22 129 X<> 16 130 STO 22 131 LBL 09 132 RTN 133 LBL 11 134 RCL 12 135 X<> 17 136 STO 12 137 LBL 14 138 RCL 19 139 X<> 13 140 STO 19 141 RCL 20 142 X<> 14 143 STO 20 144 RTN 145 LBL 12 146 RCL 23 147 X<> 17 148 STO 23 149 XEQ 14 150 GTO 13 ; PREDICT THE Y VALUE 151 LBL C ; [Y^] 152 LBL 03 153 FS? 09 154 LN 155 RCL 08 156 * 157 RCL 09 158 FS? 08 159 LN 160 + 161 FS? 08 162 E^X 163 RTN ; PREDICT THE X VALUE 164 LBL D ; [X^] 165 LBL 04 166 FS? 08 167 LN 168 RCL 09 169 FS? 08 170 LN 171 – 172 RCL 08 173 / 174 FS? 09 175 E^X 176 RTN ; INITIALIZE, clear the data registers used to accumulate the sums 177 LBL e ; [INIT] 178 LBL 00 179 11.024 180 XEQ 16 181 E 182 RTN ; SOLVE BEST CURVE TYPE 183 LBL E ; [SOLVE] 184 LBL 05 185 . 186 STO 25 187 4 188 STO 07 189 LBL 07 190 RCL 07 191 XEQ B 192 RCL 25 193 RCL 10 194 ABS 195 X<=Y? 196 GTO 15 197 STO 25 198 RCL 07 199 STO 26 200 LBL 15 201 DSE 07 202 GTO 07 203 RCL 26 204 XEQ 02 205 RCL 26 206 TONE 5 207 END[/code]

## References

  CV – Curve FitGary Tenzer, Keith Jarret and John Kennedy, August 1981PPC User Manual, page 110-111  BC – Block ClearJohn Kennedy, August 1981PPC User Manual, page 50-51

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