To find the least squares polynomial of degree 2 to(adsbygoogle = window.adsbygoogle || []).push({});

approximate points (X,Y) given in the table

X_____________Y

1_____________36

1.9_____________-49

2.9_____________23

4_____________-38

5.1_____________-14

the normal equations to be solved have the form

A c = v, where A is the matrix (NOT NECESSARILY DISPLAYED TO THE

PRECISION USED FOR THE CALCULATIONS)

_ _

| |

| _____ 1______ _____ 2______ _____ 3______ |

| _____ 4______ _____ 5______ _____ 6______ |

| _____ 7______ _____ 8______ _____ 9______ |

|_ _|

and c is the coefficient vector for the polynomial;

and v is the transpose of the vector:

_ _

| |

| _____10______ _____11______ _____12______ |

|_ _|

The solution vector c (FOUND USING DOUBLE PRECISION

CALCULATIONS with the ORIGINAL DATA above) is

the transpose of vector:

_ _

| |

| _____13______ _____14______ _____15______ |

|_ _|

2. Relevant equations

Ac=v

3. The attempt at a solution

I believe I've got the first part; i.e. solving the matrix:

5.1________14.9_________55.03

14.9_______55.03________228.899

55.03______228.899______1017.2803

Now, I'm clueless as to how to proceed from here on.

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# Homework Help: Matrix Vector Transpose

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