Finding polynomials from a set of coordinates
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03-11-2017, 10:25 AM
Post: #9
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RE: Finding polynomials from a set of coordinates
(03-10-2017 07:39 PM)tdh79 Wrote: I need to find polynomial functions from a given set of coordinates, does anyone know how to do this with the 50g? Fitting a polynomial to a set of coordinates can be done by changing your system into a set of linear equations that you then resolve using the Gauss-Jourdain pivot method. One thing you need to know is that you need n+1 points to fit a polynomial of degree n. Eg. you need 2 points to fit a line (a.x + b = 0), 3 points to fit a quadratic (a.x2 + b.x + c = 0) etc. Eg. You have the points (-3,1), (1,0) and (2,1). For the quadratic to fit, all of these points must satisfy the equation, so substituting the coordinates' values of x and y in the equation you end up with something like this: a.(-3)2 + b.(-3) + c = 1 a.12 + b.1 + c = 0 a.22 + b.2 +c = 1 -- or -- 9a -3b +c = 1 a + b + c = 0 4a + 2b + c = 1 That looks suspiciously like a system of 3 linear equations that you can solve directly on the HP 50g. You will then get your values of a, b and c that you can replace in your quadratic equation and hey presto, there's your polynomial that passes through your three points. The solution to this system is a=1/4, b=1/4 and c=-1/2 If you plot f(x)=(x^2)/4 + x/4 - 1/2 you'll see that the curve passes through all three points. |
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