can find regression and curve fit for given equation
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can this equation use for reggresion and curve fitting
where dy is partial differensiation of y and dx is partial differentiation of x
and x= linspace(0,600,1400)
y = linspace(0,1,1400)
m = a+ b*exp(c*(dy/dx))
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RANGA BHARATH
il 13 Giu 2023
Hi @Arvind. Here is the solution for your question. And the question can be framed in a better way as follows:
Question:
Can the below equation can be used for Regression and Curve fitting?
m = a + b*exp(c*(dy/dx)), where x = linspace(0,600,1400), y = linspace(0,1,1400) and dy/dx is the differentiation of y with respect to x.
Solution:
It is esential to remember that Regression and Curve Fitting are related concepts, and frequently involved in fitting a model to data.
But the difference lies in the dependency of variables i.e Regression focuses on fitting a model (line or a curve) that best represents the relationship between a dependent variable and one or more independent variables, whereas Curve Fitting does not necessarily involve a dependent variable or pre-specified independent variable.
Additionally, the equation you presented doesn't seem suited to a standard regression problem since it has an explicit function for the independent variable x. If the data includes error, it would be difficult to estimate the parameters for multiple y values at various x values. Thus, curve fitting might be the more appropriate technique here.
However, the expression dy/dx in the equation suggests that a derivative is being taken, which means that the data you have might be some sort of time-series or function analyzed using differential calculus. In this case, you may need to ensure that the data is differentiable.
In conclusion, the equation you provided can be used for curve fitting cautiously, making sure that the data allows for accurate differentiation.
Assumption:
Data is differentiable and taking the derivarives is meaningful i.e the function to be differentiated is continuous and differentiable at the point in the question.
Torsten
il 17 Giu 2023
Modificato: Torsten
il 17 Giu 2023
Since your equation can be rearranged as
dy/dx = 1/c * log((m-a)/b)
your model assumes that dy/dx is equal to a constant that you try to estimate.
That means that your model for y is linear in x:
y = c1 + c2*x
The constants c1 and c2 can be estimated by standard methods:
You won't be able to estimate m separately in your original model, only 1/c * log((m-a)/b) which is equal to c2.
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