By using the method of least squares, find the best parabola through the points: (−2,1), (1,–3), (−1,3), (0,0) Step 1. The general equation of a parabola is co+c₁x + ₂x² = y. Plugging the data points into this formula gives a matrix equation Ac = y. tep 2. The matrix equation Ac = y has no solution, so instead we use the normal equation A¹A = ATy ATA= ATy = Step 3. Solving the normal equation gives the answer Ĉ= which corresponds to the formula y = 8

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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By using the method of least squares, find the best parabola through the points:
(−2,1), (1,–3), (−1,3), (0,0)
Step 1. The general equation of a parabola is c + c₁x + c₂x² = y. Plugging the data points into this formula gives a matrix equation Ac = y.
Step 2. The matrix equation Ac = y has no solution, so instead we use the normal equation A¹A c = A¹y
ATA =
ATy=
Step 3. Solving the normal equation gives the answer
Ĉ
which corresponds to the formula
y =
Co
5 S
=
Transcribed Image Text:By using the method of least squares, find the best parabola through the points: (−2,1), (1,–3), (−1,3), (0,0) Step 1. The general equation of a parabola is c + c₁x + c₂x² = y. Plugging the data points into this formula gives a matrix equation Ac = y. Step 2. The matrix equation Ac = y has no solution, so instead we use the normal equation A¹A c = A¹y ATA = ATy= Step 3. Solving the normal equation gives the answer Ĉ which corresponds to the formula y = Co 5 S =
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