DM15L : Non-linear Curve Fit

** Using DM15L/HP15C built-in L.R function to fit non-linear curves **

The idea is that the 3 different non-linear function forms can be expressed as linear expression y = m.x + c if we manipulate the x or y values before adding them to the statistics accumulator.

The different curves require different treatment of the data to linearize, and the resulting coefficients may require treatment too.

'' Ln(y) = Ln(b) + a.x ''

''y = a.ln(x) + b ''

y = b.eax

This can be expressed as '' Ln(y) = Ln(b) + a.x ''

or

'' Ln(y) = a.x + ln(b) ''

compare with a linear expression y = m.x + c

Using Ln(y) and x as data points and running the L.R. function will give coefficients equal to a and ln(b)

You can obtain the actual value of b finding its antilog with e^x.

Once the L.R. has run you can find estimates of ŷ by running the fŷ.r to get the estimate of Ln(y) and then finding its antilog with e^x

To enter the data points you convert the y data to its Ln first, then accumulate as normal

'' y_i gLn x_i ∑+ ''

Once all the data has been entered you get the Linear Regression coefficients a & b fL.R. and e^x to correct the b LR coefficient.

fClear ∑

''2.8 gLn 0.8 ∑+ ''

''3.6 gLn 1.1 ∑+ ''

''5.8 gLn 1.7 ∑+ ''

''8.0 gLn 2.1 ∑+ ''

All data is in the accumulator, now do the Linear Regression fL.R.

This gives a and Ln b coefficients

Y: 0.805 = a

X: 0.3897 = ln b

so we need to find b so e^x

Y: 0.805 = a

X: 1.476 = b

Equation estimation is therefore y = 1.476 e 0.805x

I made the data table by using the equation y = 1.5 e 0.8x and then used approx. y values for the data table…. so the estimate looks pretty good

To do an estimation of y when x = 0.9

0.9 fŷ.r e^x

Y: 0.999962 = r correlation coefficient

X: 3.047425 = ŷ estimation

''y = a.ln(x) + b ''

Compare to a linear expression '' y = m.x + c '' shows the logarithmic curve is linear if we use Ln(x) as we accumulate data points.

The L.R. will provide the actual a and b coefficients.

To estimate ŷ first find the Ln of x then use the fŷ.r

'' y_i Enter x_i gLn∑+ ''

Calculate LR Coefficients

fL.R.

Calculate estimated ŷ by finding the Ln of your x value first and then using fŷ.r

xgLnfŷ.r

fClear ∑

'' 1.5 Enter 0.8 gLn∑+''

'' 1.7 Enter 1.7 gLn∑+''

'' 1.8 Enter 2.6 gLn∑+''

'' 2.1 Enter 4.9 gLn∑+''

Linear Regression Coefficients : fL.R.

X: 0.324 = a

Y: 1.544 = b

Equation ''y = 0.324 . Ln(x) + 1.544 ''

Estimate ŷ for X = 3

''3 gLn fŷ.r ''

Y: 0.985 = r, correlation coefficient

X: 1.900 = ŷ, estimation

'' y = b.x^a ''

This can be rearranged as '' Ln(y) = Ln(b) + a.Ln(x) '' or '' Ln(y) = a.Ln(x) + Ln(b) ''

Compare to a linear expression '' y = m.x + c ''

Our expression will be a straight line if we accumulate Ln(y) and Ln(x)

The coefficients given by the L.R. function will be a and Ln(b) - so to get the real b coefficient we find its antilog e^x

We get the estimate for ŷ we first find the Ln of our chosen x and then use the ŷ.r to get Ln(ŷ) and then ex to get ŷ

Accumulate using

'' y_i gLn x_i gLn∑+''

Calculate LR Coefficients

fL.R.e^x

Calculate estimated ŷ

xgLnfŷ.rex

fClear ∑

'' 1.3 gLn 0.6 gLn∑+''

'' 2.2 gLn 1.3 gLn∑+''

'' 3.3 gLn 2.4 gLn∑+''

'' 4.4 gLn 3.6 gLn∑+''

Linear Regression Coefficients : fL.R.e^x

Y: 0.678 = a

X: 1.837 = b

Equation : ''y = 1.837x^0.678 ''

Estimate ŷ for X = 3

3 gLnfŷ.rex

Y: 1.000 = r, correlation coefficient

X: 3.869 = ŷ, estimation

— John Pumford-Green 01/02/26 08:42 GMT