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Analysis question

  1. Apr 11, 2007 #1

    kreil

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    1. The problem statement, all variables and given/known data

    Show that the calculation of an arithmetic mean for a given set of values y1,y2....yn is equivalent to fitting the line y=y_av to the data under the least square criterion

    2. Relevant equations

    distance between the data points and the best fit curve:
    [tex]d(y_n,f(x))= \Sigma_{i=1}^{n}(y_i-f(x_i,a_1,a_2,...,a_n))^2[/tex]

    arithmetic mean of a set of values y_n:
    [tex]y_av= \frac{1}{n} \Sigma_{i=1}^{n}y_i[/tex]

    3. The attempt at a solution

    I need help getting started please!
     
  2. jcsd
  3. Apr 11, 2007 #2

    Dick

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    In your distance formula, just put f=y_av. Now how would you minimize d with respect to y_av?
     
  4. Apr 11, 2007 #3

    kreil

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    set the partial derivative of d equal to zero?
     
    Last edited: Apr 11, 2007
  5. Apr 11, 2007 #4

    Dick

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    Absolutely. What do you get?
     
  6. Apr 11, 2007 #5

    kreil

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    [tex] \frac{d(R^2)}{dy_{av}}= -2 \Sigma_{i=1}^n(y_i-y_{av})=0[/tex]
     
  7. Apr 11, 2007 #6

    Dick

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    What's your next move?
     
  8. Apr 11, 2007 #7

    kreil

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    i was thinking maybe distribute the sum, but i dont know how the -2 helps, since im trying to get the formula in the form of the arithmetic mean
     
  9. Apr 11, 2007 #8

    Dick

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    -2*x=0 means x=0, doesn't it?
     
  10. Apr 11, 2007 #9

    kreil

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    how is this:

    [tex] \frac{d(R^2)}{dy_{av}}= -2 \Sigma_{i=1}^n(y_i-y_{av})=0 \implies \Sigma_{i=1}^n(y_i-y_{av})=0 \implies \Sigma_{i=1}^ny_i=\Sigma_{i=1}^ny_{av}=y_{av}[/tex]
     
  11. Apr 11, 2007 #10

    Dick

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    The sum from 1 to n of y_av is NOT = y_av.
     
  12. Apr 11, 2007 #11

    kreil

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    oh ok that sum is n y_av, so the sum of 1/ny_i is y_av correct?
     
  13. Apr 11, 2007 #12

    Dick

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    That's what you are trying to prove, right?
     
  14. Apr 11, 2007 #13

    kreil

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    right! thanks a lot man
     
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