Number of Parameters

Newbie
1

Dear experts,
I have a fitting code that works perfectly well but there is something about it that I cannot understand.
In this fitting code, I am using a user defined functions: Polynomial1 with the following definitions:

// Polynomial (Background) //
double Polynomial1(double* x, double* Parameters) {
	double a = Parameters[0];
	double b = Parameters[1];
	return a*x[0]+b;
}

In my fitting code, if I create a TF1 object with Polynomial1 and 2 parameters:

TF1* PolyBL = new TF1("Polynomial Background (Left Side Band)", "Polynomial1", 2.80, 2.95, 2);

the fitting code will fail in the run time with the following message:

/usr/include/c++/11/bits/stl_vector.h:1063: std::vector<_Tp, _Alloc>::const_reference std::vector<_Tp, _Alloc>::operator[](std::vector<_Tp, _Alloc>::size_type) const [with _Tp = double; _Alloc = std::allocator<double>; std::vector<_Tp, _Alloc>::const_reference = const double&; std::vector<_Tp, _Alloc>::size_type = long unsigned int]: Assertion '__n < this->size()' failed.

however, if I create the TF1 object with 3 parameters:

TF1* PolyBL = new TF1("Polynomial Background (Left Side Band)", "Polynomial1", 2.80, 2.95, 3);

everything will work fine.

Can you please explain why do I need to pass 3 parameters when I clearly defined the function with 2 parameters.
Also, do I need to always add 1 to the number of parameters? (e.g. for a second-order polynomial with 3 parameters, should I pass 4)
Thank you in advance

1 Like
2

I guess @moneta can help you.

3

Hi,

What you have written it should work, but you need to make sure the definition of the Polynomial1 function is available when you create or copy the function. I would suggest you to write the function directly using the TFormula code syntax , as:

TF1("Polynomial Background (Left Side Band)", "[a]*x+[b]", 2.80, 2.95, 2);

or using an interpreted lambda:

TF1("Polynomial Background (Left Side Band)", "[](double *x, double * p){ return p[0] * x[0] * p[1]; }", 2.80, 2.95, 2);

Lorenzo

4

Hi Lorenzo,
Thank you very much for the help.