By using the method of least squares, find the best line through the points: (2,-3), (-2,0), (1,-1). Step 1. The general equation of a line is co + C₁ = 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 A™A ĉ=A¹y ATA = A¹y = Step 3. Solving the normal equation gives the answer Ĉ= 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 + e +e3. SSE =

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 line through the points:
(2,-3), (-2,0), (1,-1).
Step 1. The general equation of a line is co + C₁ = y. Plugging the data points into this formula gives a matrix equation Ac = y.
6
Step 2. The matrix equation Ac = y has no solution, so instead we use the normal equation ATA ĉ = A¹y
ATA=
ATY
Step 3. Solving the normal equation gives the answer
Ĉ=
which corresponds to the formula
y =
Analysis. Compute the predicted y values: y = Ac.
Compute the error vector: e = y - y.
e =
Compute the total error: SSE = e³ + e² + e3.
SSE =
Transcribed Image Text:By using the method of least squares, find the best line through the points: (2,-3), (-2,0), (1,-1). Step 1. The general equation of a line is co + C₁ = y. Plugging the data points into this formula gives a matrix equation Ac = y. 6 Step 2. The matrix equation Ac = y has no solution, so instead we use the normal equation ATA ĉ = A¹y ATA= ATY Step 3. Solving the normal equation gives the answer Ĉ= which corresponds to the formula y = Analysis. Compute the predicted y values: y = Ac. Compute the error vector: e = y - y. e = Compute the total error: SSE = e³ + e² + e3. SSE =
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