Question

10. This data set is super small. It's really kinda cute. t 1.4 2.7 3.8 y 1.0 -2.3 1.6 I think a function of the form f(t) = c Int+2 cost could fit the data well. Let's find the best fit of this form in the least squares sense. Set your calculator in radian mode. (a) (4 points) Write the square error to be minimized to find the best choices for and cu (b) (8 points) Let c be a column vector containing the unknowns. Write the normal equations in Me= form by filling in the blanks below with numbers (round to the nearest hundredth): M = (e) (2 points) Now that we've defined M and 7, write a single line MATLAB com- mand to solve the system from part (b).

Answer 1

a) The square error would be the following:

b) Now the normal equations can be found by differentiating with respect to the coefficients and then take the RHS to be 0. Doing this, we obtain the following:

$\\ 2(1-0.34c_1-0.17c_2)(-0.34)+2(-2.3-0.99c_1+0.9c_2)(-0.99)+2(1.6-1.34c_1+0.79c_2)(-1.34) = 0 \\ \Rightarrow 5.7826c_1 - 3.7836c_2 - 0.414= 0 \\ \\ 2(1-0.34c_1-0.17c_2)(-0.17)+2(-2.3-0.99c_1+0.9c_2)(0.9)+2(1.6-1.34c_1+0.79c_2)(0.79) = 0 \\ \Rightarrow -3.7836c_1 + 2.926c_2 - 1.952= 0$

Trasforming it into a matrix equation, we get that

$M{\vec c} = \vec v$

where

$M = \begin{bmatrix} 5.8726 & 3.7836\\ -3.7836 & 2.926 \end{bmatrix} \\ \vec v = \begin{bmatrix} 0.414\\ 1.952 \end{bmatrix}$

c) The following command in MATLAB solves the system from part b)

>linsolve([5.8726 3.7836;-3.7836 2.926],[0.414;1.952])
ans =

  -0.19601
   0.41366