Question

2. This problem finds the curve C ++D = b which gives the best least squares fit to the points: t= -2, b=0 t = -1, b=0 t= 0, b=1 t= 1, b=1 t= 2, b=1 (a) (10 points) Write down the 5 equations Ax = b that would be satisfied if the curve went through all 5 points. (b) (10 points) Find the least squares solution = (Ĉ, Ð). (c) (10 points) Find the projection p of b onto the column space of A, and the error e of using p instead of b in solving Ax = b.

Answer 1

solution: Given that C++0=b car. The equation of form ax=b if curve went through given five points are: (1). C-2D =0; c=2D 022 C-D=0 ; c=D (3) c = 1 (42. C+0=1 (52. C+2D =1 -2 o (6) 2

2 .( ط) The least square can be determined by 1 - 2 -2 -1 0 1 2 (6) 2 1 [...] [S] (8)-(3) and lod=3 This give 56=3 35 and c = 1o å = (3/3 Tc )

CC? The project of @) b on colunm space of matrix A is given by P= A CATAJAT - P = -2 -2 1 a 1 2 -1 0 2 2 1 2 - i p - 2 C[5 1 -2 -1 0 1 2 2 p = ( [06])[-e-jói] - 2

4 - 2 PE 50 JL -10 12 2. -10 P = 50 1 20 - 2. s lo 10 29 - 2 - Ps 도 50 2. - 1 | 9 0 - 2. -1 0 12 2. 2. 2. 6 2. 0 Ps.) J | 2 4. 3 2. 0 2 2. 2. 2 2 0 2. 3 5 2 2 0 4 6

porojection of 5 vecton1 b P = Pb 6 P = 4 2 o -2 1o O 4 3 2 0 D 2 2 2 2 2 2 3 4 -2 o 2 4 6 0 p=1 lo 3 6 १ 12 1) x Thank you