Problem with fitting simple quadratic function to 3 data points

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The discussion revolves around fitting a quadratic function to three data points representing chi^2 values for a fit. The user is seeking a simple method to derive the coefficients a and b, given that c is constrained to the minimum value. Suggestions include using linear algebra to create three equations from the data points, which can then be solved for the unknowns. The Lagrange Polynomial is mentioned as a potential method for interpolation, though the user is hesitant to implement complex changes to their Fortran-77 program. Overall, the focus is on finding a straightforward solution to the quadratic fitting problem without extensive modifications to existing code.
johnismint
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Hello PF,

Ok so here is my problem. I have an array of data points which are the chi^2 values for a fit I have performed. I wish to calculate the 1-sigma error for this fit, so I figure I could find the lowest chi^2 value, and the two points either side that value, and fit a quadratic function to this. Then 'fill' this such that deltachi^2=1 and calculate the deviations either side. But this is somewhat irrelevant for now...

I have a complete mental block, and I'm really struggling to figure out how to fit a quadratic to these 3 data points. I'm editing a fortran-77 program to do this (which I'm not that familiar with) so would like to keep it as simple as possible. If my function is F(x)= a*x**2 + b*x + c, I already have c constrained to be my minimum value, however, I can't seem to figure out how to get these other co-efficients. I know there are methods such as Cramer's rule, however this would require significant changes to the program, which I would like to avoid. Can anyone think of an easier way to implement this?

Many thanks
 
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If you plug in each of your data points, you'll get three linear equations in three unknowns (a, b, c). Then it's just a bit of linear algebra.
 
rochfor1 said:
If you plug in each of your data points, you'll get three linear equations in three unknowns (a, b, c). Then it's just a bit of linear algebra.

I think that's what he was referring to by Cramer's Rule in his original post, he doesn't want to do that.
 

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