Given,
X-data = [1 2 3 4 5 6]
Y-data = [0.1, 0.3, 0.7, 1, 2.5, 2.7]
Model equation y = a X e^(b X)
where,
X = dataset from X-data
Y = dataset from Y-data
a and b are the coefficients
e = exponential/exp symbol
How can I calculate the value of the coefficient of a and b?
Hi, to determine the values of a and b you need to solve a least square problem:
Min Sum (axi.exp(bxi) - yi)^2 , i=1, ...,n
a,b
You can use MATLAB for example to solve this problem.
Thanks, Richard. For your valuable reply. Is there any way to perform the calculations by hand (without solvers)?
Many thanks,
Pankaj
Hi, a way to address the problem "by hand" is to write down the necessary optimality conditions of the problem above:
dF/da = 0
dF/db=0
where function F is the least square objective function and "d" denotes partial derivative.
By doing so, you will obtain two nonlinear equations to solve.
Nevertheless, it will be not easy to obtain a closed-form expression of a and b. A numerical root-finding algorithm (Newton-like) should be used. Best
consult
Article Fitting Curves to Data Using Nonlinear Regression: A Practic...
Model equation is: y = aX·exp(bX).
The correlation can first be linearized as:
ln(|y|) = ln(|a|) + ln(|X|) + bX (*); or ln(|y/X|) = ln(|a|) + bX
or with Y ≡ ln(|y/X|):
Y = ln(|a|) + bX
Even considering that Y and X are not read from orthogonal axis; the parameters can, possibly, still be reasonably well-fitted by least-squares regression. These are, however, biased estimates, even for the linearized correlation.The corresponding linearized plot can be used to display scattering around the trendline.
Previous estimates of the parameters can possibly be found sufficiently approximate to be refined by iterative nonlinear least-squares regression, to find unbiased least-squares estimates for the original correlation.
This can be done iteratively using this easy software:
www.lerenisplezant.be/fitting.htm
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