By using the method of least squares, find the best line through the points: (−2,−1), (0,−2), (−1,3). Step 1. The general equation of a line is co + c₁x = y. Plugging the data points into this formula gives a matrix equation Ac = y. 1 0 0 0 1 [co] 0 0 0 0 Step 2. The matrix equation Ac = y has no solution, so instead we use the normal equation ATAC = ATy 5 -1 ATA= -1 14 0 ATy= 0 Step 3. Solving the normal equation gives the answer 0 E 0 which corresponds to the formula y = || = =
By using the method of least squares, find the best line through the points: (−2,−1), (0,−2), (−1,3). Step 1. The general equation of a line is co + c₁x = y. Plugging the data points into this formula gives a matrix equation Ac = y. 1 0 0 0 1 [co] 0 0 0 0 Step 2. The matrix equation Ac = y has no solution, so instead we use the normal equation ATAC = ATy 5 -1 ATA= -1 14 0 ATy= 0 Step 3. Solving the normal equation gives the answer 0 E 0 which corresponds to the formula y = || = =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:||
=
which corresponds to the formula
y =
Analysis. Compute the predicted y values: y = Ac.
ŷ
Compute the error vector: e = y - y.
e =
Compute the total error: SSE = e² + e² + e²3.
SSE
=
||
Transcribed Image Text:By using the method of least squares, find the best line through the points:
(−2,−1), (0,−2), (−1,3).
Step 1. The general equation of a line is co + C₁x = y. Plugging the data points into this formula
gives a matrix equation Ac = y.
1
0
0
0
1
0
=
0
0
0
Step 2. The matrix equation Ac = y has no solution, so instead we use the normal equation
ATAC = A¹y
5
-1
ATA =
14
0
ATy
0
Step 3. Solving the normal equation gives the answer
0
18
which corresponds to the formula
y =
=
=
-1
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