Curve Fitting: Fitting to the Curve y = a*e^(-b*x^2)
Introduction
This blog is to fit data to the equation y =
a*e^(-b*x^2). A famous example is the
equation:
y = 2/√(π) * e^(-x^2)
which the indefinite integral of 0 to real number t is the
basis of the error function.
Turning y = a*e^(-b*x^2) into a Linear Form
y = a*e^(-b*x^2)
ln y = ln (a*e^(-b*x^2))
ln y = ln a + (-b)*x^2
Y’ = A’ + B’ * x^2
Where: Y’ = ln
y, A’ = ln a, B’ = -b
Results: e^A, B
Inputs: square each
of the data points of X, take the natural logarithm for Y.
Example
I use the HP Prime for this example.
Fit the data below to the curve y = a*e^(-b*x^2).
|
x
|
y
|
|
-0.75
|
0.642931
|
|
-0.35
|
0.998284
|
|
0
|
1.128379
|
|
0.35
|
0.998284
|
|
0.62
|
0.768267
|
|
0.94
|
0.466951
|
Adjust data:
|
x
|
C1 = x^2
|
y
|
C2 = ln y
(to 6 decimal places)
|
|
-0.75
|
0.5625
|
0.642931
|
-0.441718
|
|
-0.35
|
0.1225
|
0.998284
|
-0.001717
|
|
0
|
0
|
1.128379
|
0.120782
|
|
0.35
|
0.1225
|
0.998284
|
-0.001717
|
|
0.62
|
0.3844
|
0.768267
|
-0.263618
|
|
0.94
|
0.8836
|
0.466951
|
-0.761531
|
Running the linear_regression(C1^2, LN(C2)) returns the
matrix [B’, A’]:
[-0.998758, 0.120567]
The results must be adjusted: e^(.120567) = 1.128136
Final adjusted curve:
y ≈ 1.128136 * e^(-0.998758*x^2)
Using the Exponential Regression Model
We can use exponential regression function as an alternative
to the linear regression.
For the input, square all of the x data.
|
x
|
C1 = x^2
|
C2 = y
|
|
-0.75
|
0.5625
|
0.642931
|
|
-0.35
|
0.1225
|
0.998284
|
|
0
|
0
|
1.128379
|
|
0.35
|
0.1225
|
0.998284
|
|
0.62
|
0.3844
|
0.768267
|
|
0.94
|
0.8836
|
0.466951
|
The function exponential_regression from the HP Prime that
fits data to the model y = B * A^x.
To fit this model asked, y = a*e^(-b*x^2), we will need to
take the natural logarithm of B.
Hence:
exponential_regression(C1^2, C2) returns
[0.368337, 1.128136]
With ln(0.368337) ≈ -0.998757, the required equation is
y = 1.128136*e^(-0.998757*x^2)
Summary
To fit data to the curve y = a*e^(-b*x^2):
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Linear Regression, Y’ = A’ + B’*X’
|
Exponential Regression, Y’ = B’*A’^X’
|
|
Input: X’ = x^2, Y’
= ln y
Output: a = e^A’, b
= B’
|
Input: X’ = x^2, Y’ = y
Output: a = A’, b = ln(B’)
|
Eddie
This blog is property of Edward Shore, 2018.
