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Math Help - Least square approximation parabola

  1. #1
    Junior Member
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    Least square approximation parabola

    Im working on an assignment which I think I have solved properly but since I dont have the solution Im a little nervous and I wonder if someone could take a look at it.

    Heres the question (translated from another language):

    What is the best adjustment, in the "least square" sense, of the parabola y=c_0+c_1x+c_2x^2 to the data points (x_i,y_i) given by the set {(0,3),(1,2),(2,4),(3,4)}

    Heres my solution:

    I plug the given points in the equation and solve the system of linear equations and find that there is no solution c_0,c_1,c_2. Which means i need to find a parabola which is as close as possible.

    From Ac=b where A= \left[ \begin {array}{ccc} 1&0&0\\ \noalign{\medskip}1&1&1<br />
\\ \noalign{\medskip}1&2&4\\ \noalign{\medskip}1&3&9\end {array}<br />
 \right] , c=  \left[ \begin {array}{c} c_{{0}}\\ \noalign{\medskip}c_{{1}}<br />
\\ \noalign{\medskip}c_{{2}}\end {array} \right] <br />
and b=\left[ \begin {array}{c} 3\\ \noalign{\medskip}2\\ \noalign{\medskip}<br />
4\\ \noalign{\medskip}4\end {array} \right] <br />

    And since the columns in A are linearly independent, A^TA is invertible I get

    c=(A^TA)^{-1}A^Tb=\left[ \begin {array}{ccc} {\frac {19}{20}}&-{\frac {21}{20}}&1/4<br />
\\ \noalign{\medskip}-{\frac {21}{20}}&{\frac {49}{20}}&-3/4<br />
\\ \noalign{\medskip}1/4&-3/4&1/4\end {array} \right] <br />
 \left[ \begin {array}{cccc} 1&1&1&1\\ \noalign{\medskip}0&1&2&3<br />
\\ \noalign{\medskip}0&1&4&9\end {array} \right] <br />
\left[ \begin {array}{c} 3\\ \noalign{\medskip}2\\ \noalign{\medskip}<br />
4\\ \noalign{\medskip}4\end {array} \right] <br />

    Which gives me c= \left[ \begin {array}{c} 11/4\\ \noalign{\medskip}-1/4<br />
\\ \noalign{\medskip}1/4\end {array} \right] <br />

    This means that the parabola that is the best adjustment to the given set of points is

    4y=11-x+x^2

    I really need to get this straight because Im studying math on my own without the help of tutors and peers. So any help is appreciated.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Champaign, Illinois
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    Looks good to me!
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