Question

(1 point) Fit a quadratic function of the form f(t) = co + cit + c2t2 to the data points (0,1),(1, -3), (2,5), (3,5), using least squares. f(t) = |

Answer 1

The given points are (0, 1), (1, -3), (2, 5), (3, 5). Squared error is given by

For minima of E

$\begin{align*} &\frac{\partial E}{\partial c_0} = 0\\ &\Rightarrow2(c_0-1)+2(c_0+c_1+c_2+3)+2(c_0+2c_1+4c_2-5)+2(c_0+3c_1+9c_2-5)=0\\ &\Rightarrow4c_0+6c_1+14c_2 = 8 \quad \quad ...(a) \end{align*}$

$\begin{align*} &\frac{\partial E}{\partial c_1} = 0\\ &\Rightarrow 2(c_0+c_1+c_2+3)+4(c_0+2c_1+4c_2-5)+6(c_0+3c_1+9c_2-5)=0\\ &\Rightarrow (c_0+c_1+c_2+3)+2(c_0+2c_1+4c_2-5)+3(c_0+3c_1+9c_2-5)=0\\ &\Rightarrow 6c_0+14c_1+36c_2 = 22 \quad \quad ...(b) \end{align*}$

0 Ос2 2(coc1c23)8(co2c14c2 5)18(co3c19c2- 5) 0 (coC1c2+ 3)4(co2cı 4c2 5)9(co3c1 9c2 5) 0 14co36c1 +98c2 62 ..c)

Solving (a), (b), (c) for

So

$f(t)=-t+t^2.$