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math_proof

  • one year ago

method of least squares fit, y=Bx to the following n=6 points (3,4),(1,2),(5,4),(6,8),(3,6),(4,5) and it has to go through the origin can someone help me get started?

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  1. tkhunny
    • one year ago
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    Did you answer the question last night? \(\sum{xy} = B\cdot\sum{x^{2}}\) Have you seen it? Can you create it?

  2. math_proof
    • one year ago
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    you have to minimize the B right?

  3. math_proof
    • one year ago
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    how did you already get the summation etc without minimazing and taking derivatives?

  4. tkhunny
    • one year ago
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    No, you are minimizing the error. You must select the B that accomplishes this task.

  5. tkhunny
    • one year ago
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    You are asking good questions. This suggests that you have seen the development of Normal Equations.

  6. tkhunny
    • one year ago
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    \(I(B) = \left(y - Bx\right)^{2}\) Find \(\dfrac{dI}{dB}\).

  7. math_proof
    • one year ago
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    2x(y-Bx)?

  8. math_proof
    • one year ago
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    I don't get how did you get the summation equation that you posted, i know how to solve it from there, but how did you figured it out so fast

  9. tkhunny
    • one year ago
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    Negative x?

  10. tkhunny
    • one year ago
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    It doesn't matter, in this case. I just wanted to check to see that you had it right and were moving on.

  11. math_proof
    • one year ago
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    the question is if there is other way to coming up with the equation to the problem you came up with, instead of going through the minimalization formula then partial derivative =0

  12. tkhunny
    • one year ago
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    Well, there are some Matrix methods, but I'm guessing we don't want to talk about those, today. You have it just fine, excepting that this is a one-parameter derivation so it's just a Derivative, not a Partial Derivative. -2x(y-Bx) = 0 leads to what?

  13. math_proof
    • one year ago
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    -2xy-2bx^2 =0

  14. math_proof
    • one year ago
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    -2xy+Bx^2=0 sorry

  15. math_proof
    • one year ago
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    2Bx^2=+2xy B=2xy/2x^2 B=xy/x^2

  16. math_proof
    • one year ago
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    ii see it now

  17. tkhunny
    • one year ago
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    More importantly, \(xy = Bx^{2}\). Resist the temptation to solve for B. If you've more than one parameter, you WILL become confused. Reach for the standard Normal Equation form.

  18. math_proof
    • one year ago
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    i am not really familiar with Normal Equation yet

  19. tkhunny
    • one year ago
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    Pretty much just: 1) Everything without a parameter on one side, and 2) Everything with a parameter on the other side - in ascending Order of the terms.

  20. tkhunny
    • one year ago
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    Solving for y = a + bx comes out: y = a + bx yx = ax + bx^2 For example.

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