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Caclulus III problem involving mean.

  1. Apr 27, 2014 #1
    http://i.imgur.com/DkYMylb.png, probably easier to understand than me trying to type it out.

    so far I tried exapanding the (y_i - y(x_i))^2 and and then subbing in y_i = ax + b. This gives me part of the answer but i'm not sure if it's legal. I'm generally struggling to find a workable relationship between yi, y and y(bar), same for x.

    If nothing else, I could use some clarification as to what the y is (y_i - y(x_i))^2. I thought maybe it was like f(x) but if that's not the case, what would it be if not y_i?

    Thanks for your help.
    Last edited: Apr 27, 2014
  2. jcsd
  3. Apr 27, 2014 #2


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    The problem is to find the equation of the straight line, y= ax+ b, that is "closest" (in the "least squares" sense) to the given point. That is what "[itex]y(x_i)[/itex]" is: the value of that function at [itex]x_i[/itex]. [itex]y(x_i)= ax_i+ b[/itex].

    No, you do not substitute [itex]ax+ b[/itex] for [itex]y_i[/itex]. "[itex]y_i[/itex]" is the given y value of the ith point.
  4. Apr 27, 2014 #3

    Ray Vickson

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    It is saying that ##E(a,b) = \sum_{i=1}^n ( y_i - a - b x_i)^2.##
  5. Apr 27, 2014 #4


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    It is not asking you to find the line of best fit. Just expand ##E(a,b)## out, use the definition of ##\bar z## (for ##z## whatever variable you need) and see if you don't get the answer. (Also note that Ray inadvertently left off the ##\frac 1 n## in his formula above in post #3 ##E(a,b) =\color{red}{\frac 1 n} \sum_{i=1}^n ( y_i - a - b x_i)^2##).
    Last edited: Apr 27, 2014
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