Question

Fit a quadratic function of the form f(t)=c0+c1t+c2t2f(t)=c0+c1t+c2t2 to the data points (0,−5)(0,−5), (1,5)(1,5), (2,−3)(2,−3), (3,11)(3,11), using least squares.

Answer 1

Answer:

Therefore the required quadratic polynomial is

$f(t)=\frac{5}{3} -13t +\frac{17}{3} t^2$

Step-by-step explanation:

Given quadratic function is

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

Given data (0,-5), (1,5) , (2,-3) and (3,11)

Its means f(0)= -5, f(1)= 5, f(2) = -3 and f(3)=11

$c_0+0.c_1+0.c_2= -5$

$c_0+1.c_1+1^2.c_2= 5$

$c_0+2.c_1+2^2.c_2= -3$

$c_0+3.c_1+3^2.c_2= 11$

Let

$A=\left[\begin{array}{ccc}1&0&0\\1&1&1\\1&2&4\\1&3&9\end{array}\right]$   ,   $\vec B=\left[\begin{array}{c}-5\\5\\-3\\11\end{array}\right]$      and $\vec x=\left[\begin{array}{ccc}c_0\\c_1\\c_2\end{array}\right]$

We can write this system as $A\vec x=\vec B$

Since

  $rref A=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\\0&0&0\end{array}\right]$

we have ,ker A= {0}

$A^T=\left[\begin{array}{cccc}1&1&1&1\\0&1&2&3\\0&1&4&2\end{array}\right]$

$\vec x = (A^TA)^{-1}(A^T\vec B)$

   $=\left( \left[\begin{array}{cccc}1&1&1&1\\0&1&2&3\\0&1&4&2\end{array}\right]\left[\begin{array}{ccc}1&0&0\\1&1&1\\1&2&4\\1&3&9\end{array}\right]\right)^{-1} \left(\left[\begin{array}{cccc}1&1&1&1\\0&1&2&3\\0&1&4&2\end{array}\right]\left[\begin{array}{c}-5\\5\\-3\\11\end{array}\right] \right)$

  $=\left(\left[\begin{array}{ccc}4&6&14\\6&14&36\\7&15&35\end{array}\right] \right)^{-1}\left[\begin{array}{c}8\\32\\15\end{array}\right]$

$=\left[\begin{array}{ccc}\frac{5}{6}&0&-\frac{1}{3}\ \\-\frac{7}{10} &-\frac{7}{10}&1\\\frac{2}{15} &\frac{3}{10} &-\frac{1}{3}\end{array}\right] \left[\begin{array}{c}8\\32\\15\end{array}\right]$

$=\left[\begin{array}{c}\frac{5}{3} \\-13\\\frac{17}{3} \end{array}\right]$

Therefore$c_0=\frac{5}{3} , c_1=-13 \ and \ c_2=\frac{17}{3}$

Putting the values of $c_0, c_1 \ and \ c_2$ in the given quadratic equation

$f(t)=\frac{5}{3} -13t +\frac{17}{3} t^2$

Therefore the required quadratic polynomial is

$f(t)=\frac{5}{3} -13t +\frac{17}{3} t^2$