Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

fauxshaux

  • 5 years ago

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

  • This Question is Closed
  1. druvidfae
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  2. bimol12345
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 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. kh83
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 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: \[A ^{T}e=0\]\[A ^{T}(b-p) = 0\] \[A ^{T}(b-Ax)=0\]Solving the equation for unknow x gives: \[x=(A ^{T}A)^{-1}A ^{T}b\]which contains unknown line coefficients

  4. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy