Question

Fit a quadratic function of the form ????(????)=c0+c1????+c2????2f(t)=c0+c1t+c2t2 to the data points (0,10)(0,10), (1,6)(1,6), (2,26)(2,26), (3,30)(3,30), using least squares.

Answer 1

Therefore the quadratic function is

`f(x=(19)/(2)+(17)/(2)t -\frac12 t^2`

Explanation:

We want find a quadratic function

`f(t)=c_0+c_1t+c_2t^2`

such that, f(0)=10, f(1)=16, f(2)=26, f(3)=30

These conditions gives the following system of linear equations

`c_0+0.c_1+0.c_2=10`

`c_0+1.c_1+1.c_2=16`

`c_0+2.c_1+4.c_2=26`

`c_0+3.c_1+9.c_2=30`

Let

`A=\left[\begin{array}{ccc}1&0&0\\1&1&1\\1&2&4\\1&3&9\end{array}\right]`,
`\vec x= \left[\begin{array}{c}c_0\\c_1\\c_2\end{array}\right]` and
`\vec b=\left[\begin{array}{ccc}10\\16\\26\\30\end{array}\right]`

The unique least square solution is
`A\vec x=\vec b`

`\vec x= (A^TA)^(-1)A^T \vec b`

`\left[\begin{array}{c}c_0\\c_1\\c_2\end{array}\right]=\left(\left[\begin{array}{ccc}4&6&14\\6&14&36\\14&36&98\end{array}\right] \right)^(-1)\left[\begin{array}{cccc}1&1&1&1\\0&1&2&3\\0&1&4&9\end{array}\right] \left[\begin{array}{ccc}10\\16\\26\\30\end{array}\right]`

`\left[\begin{array}{c}c_0\\c_1\\c_2\end{array}\right]=\left[\begin{array}{ccc}(19)/(20)&-(21)/(20)&\frac14 \\ \\-(21)/(20)&(49)/(20)&-(3)/(4)\\ \\ \frac14&-\frac34& \frac14 \end{array}\right] \left[\begin{array}{cccc}1&1&1&1\\0&1&2&3\\0&1&4&9\end{array}\right] \left[\begin{array}{ccc}10\\16\\26\\30\end{array}\right]`

`\left[\begin{array}{c}c_0\\c_1\\c_2\end{array}\right]= \left[\begin{array}{c}(19)/(2)\\ \\(17)/(2)\\ \\-\frac12\end{array}\right]`

Therefore
`c_0 = (19)/(2)`,
`c_1=(17)/(2)` and
`c_2=-\frac12`.

Therefore the quadratic function is

`f(x=(19)/(2)+(17)/(2)t -\frac12 t^2`