__ Least Squares Parabola__

Solving an over-determined linear system A x = b is generally impossible, but finding an x0 so that A x0 is as close as possible to b leads to the concept of a least squares solution. One can see that such an x0 must satisfy the normal equations

A_tr A x0 = A_tr b

where A_tr is the transpose of the matrix A. Generically this system will have a unique solution. We use this worksheet to study this in the case of fitting a set of points to a parabola.

**Problem**
**: **
Find the best fitting parabola y = ax^2 +bx+c through the following points: (1,2),(1,3),(2,4),(3,5),(4,6)

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**pointlist := [[1,2],[1,3],[2,4],[3,5],[4,6]];**

**Solution: **
The over-determined system here is

1a +1b+c=2

1a +1b+c=3

4a +2b+c=4

9a +3b+c=5

16a+4b +c=6

In matrix form Ax=
B:

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**with (linalg):**

Warning, the protected names norm and trace have been redefined and unprotected

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**A:= array([[1,1,1],[1,1,1],[4,2,1],[9,3,1],[16,4,1]]);**

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**x:= vector ([a, b, c]);**

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**B:= vector([2,3,4,5,6]);**

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**evalm( A &* x = B);**

The normal equations Nx=m.

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**N:= evalm ( transpose(A) &* A );**

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**m:= evalm ( transpose(A) &* B);**

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**x0:= evalm (inverse(N) &* m);**

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**parabola := x -> x0[1] * x^2 + x0[2] * x + x0[3];**

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**plot ({pointlist, parabola});**

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