00:01
Hi, let's start the solution.
00:02
In this question, we have given data points are 1 ,1 .9 ,2 ,2 .6 ,3 ,3 .2 ,4 ,3 .6 ,5 ,3 .8.
00:25
Then part a, the objective is to find the design matrix and observation vector for unknown parameter vector that is beta is beta 1, beta 2.
00:41
So the required function is in the form of y equal to beta 1 x plus beta 2 x square.
00:50
Say this one is equation 1.
00:53
Now substitute x equal to 1 and y equal to 1 .9 in equation 1, we get beta 1 plus beta 2 equal to 1 .9.
01:10
Now similarly put x equal to 2, y equal to 2 .6 in equation 1, then we get 2 beta 1 plus 4 beta 2 equal to 2 .6.
01:31
Now put x equal to 3, y equal to 3 .2 in equation 1, then we get 3 beta 1 plus 9 beta 2 equal to 3 .2 and put x equal to 4, y equal to 3 .6 in equation 1, we get 4 beta 1 plus 16 beta 2 equal 3 .6 and last put x equal to 5 and y equal to 3 .8 in equation 1, then we get 5 beta 1 plus 25 beta 2 equal to 3 .8.
02:23
So the values of x and y from each data point are substituted in the least square function to obtain the equation.
02:33
So the equation obtained are equations are that is beta 1 plus beta 2 equal to 1 .9, 2 beta 1 plus 4 beta 2 equal to 2 .6 and 3 beta 1 plus 9 beta 2 equal to 3 .2 and 4 beta 1 plus 16 beta 2 equal to 3 .6 and 5 beta 1 plus 25 beta 2 equal to 3 .8.
03:14
This can be written as metric equation a beta y equal to 0.
03:21
So design matrix a is 1, 2, 3, 4, 5, 1, 4, 9, 16, 25.
03:37
Now observation vector, observation vector say y that is equal to 1 .9, 2 .6, 3 .2, 3 .6, 3 .8.
04:00
So the solution of the matrix equation a beta equal to y is given by beta equal to at into a whole inverse y.
04:15
Now part b, the objective is to find the least square curve for the given data...