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fauxshaux
Group Title
How can I use the method of least squares to fit a line to the points { (0,0), (1,2), (2,1), (3,4) }?
 3 years ago
 3 years ago
fauxshaux Group Title
How can I use the method of least squares to fit a line to the points { (0,0), (1,2), (2,1), (3,4) }?
 3 years ago
 3 years ago

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druvidfae Group TitleBest ResponseYou've already chosen the best response.0
the method of least squares gives rise to the entire concept of karl pearsons coefficient.Try doing some research on that subject.And then retry the question.
 3 years ago

bimol12345 Group TitleBest ResponseYou've already chosen the best response.0
let the best fit line be y = ax + b then u using least square method "a" and "b" satisfies \[(\sum_{1}^{4} x _{i} ^{2}).a + (\sum_{1}^{4} x _{i}).b = \sum_{1}^{4} x _{i} y _{i}\] \[(\sum_{1}^{4} x _{i}).a + 4b = \sum_{1}^{4} y _{i}\] 14a + 7b = 16 7a + 4b = 7 on solving a= 15/7 , b = 2 => best fit y=(15/7)x 2
 3 years ago

kh83 Group TitleBest ResponseYou've already chosen the best response.0
Write down the equations of the line (for all the points) in a form \[Mx_{i}+N=y_{i}\] Then assemble the matrix equation: Ax = b, Where x is the vector of unknown line coeffiients x= [M,N] In general this equation cannot be solved unless vector b is in the column space of matrix A. Solve equation Ax = p instead, where p is projection of b on the column space of A. The error you make by taking the projection and not the real vector is e = b  p e is perpendicular to p and all vectors in column space of A so: \[A ^{T}e=0\]\[A ^{T}(bp) = 0\] \[A ^{T}(bAx)=0\]Solving the equation for unknow x gives: \[x=(A ^{T}A)^{1}A ^{T}b\]which contains unknown line coefficients
 3 years ago
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