By using the method of least squares, find the best parabola through the points: (-1,-3). (1,3). (-2,-3). (0,-3) Step 1. The general equation of a parabola is c₂+₁ +₂2²=3. Plugging the data points into this formula gives a matrix equation Ac = y. ATA- Step 2. The matrix equation Ac=y has no solution, so instead we use the normal equation ATA è = A¹y ATy Step 3. Solving the normal equation gives the answer CO 0 0 which corresponds to the formula 866
By using the method of least squares, find the best parabola through the points: (-1,-3). (1,3). (-2,-3). (0,-3) Step 1. The general equation of a parabola is c₂+₁ +₂2²=3. Plugging the data points into this formula gives a matrix equation Ac = y. ATA- Step 2. The matrix equation Ac=y has no solution, so instead we use the normal equation ATA è = A¹y ATy Step 3. Solving the normal equation gives the answer CO 0 0 which corresponds to the formula 866
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:By using the method of least squares, find the best parabola through the points:
(-1,-3). (1,3). (-2,-3). (0,-3)
Step 1. The general equation of a parabola is c₂+₁+c₂2²-3. Plugging the data points into this formula gives a matrix equation Ac=y.
1
ATA-
ATy=
1
Step 3. Solving the normal equation gives the answer
-8
which corresponds to the formula
a
Step 2. The matrix equation Ac=y has no solution, so instead we use the normal equation A¹A = A¹y
O
0
3
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