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.\(\)
| 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.
| 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
- GTO “CV”
- SIZE 027
- USER mode
- INIT, the display will show 1.
- 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 ∑-.
- repeat step until all data pairs are entered.
- 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.
- Select the desired curve type. Once the curve type is selected, the HP-41 should not be interrupted.
-
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 - 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.
-
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 -
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^
- 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 \(y=25\).
Minimum size is SIZE 25.
| 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:
To determine the points along that 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.
Again, to solve this, we use the CV program.
| 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.
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 \(x=-10\).
| 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.
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.
Since the exponential parameters are still in the machine, we can predict \(y\) for \(x=-10\) as
| key strokes | display |
|---|---|
| 10 CHS y^ | 0.0574 |
Best curve
Fit the best curve to the data points shown below.
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.
| 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
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
References
| [1] | CV – Curve Fit Gary Tenzer, Keith Jarret and John Kennedy, August 1981 PPC User Manual, page 110-111 |
|
| [2] | BC – Block Clear John Kennedy, August 1981 PPC User Manual, page 50-51 |