When does this least squares analytical solution based on zeros of partial derivatives start providing more than one solution?If I want to fit a quadratic function of two variables to some data, I can use f ( x , y ) = c 1 x 2 + c 2 x y + c 3 y 2 + c 4 x + c 5 y + c 6 ∂ ∂ c i ∑ j ( z j − f ( x j , y j ) ) 2 = 0to obtain six equations, and then endeavor to solve them.I've done it for one variable not two, but I'm guessing the process is straightforward.If I extend this to more variables, and to higher order than 2, when will the analytical expressions be at risk for having multiple solutions?

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When does this least squares analytical solution based on zeros of partial derivatives start providing more than one solution?If I want to fit a quadratic function of two variables to some data, I can use  f  (  x  ,  y  )  =      c    1        x    2    +      c    2    x  y  +      c    3        y    2    +      c    4    x  +      c    5    y  +      c    6        ∂          ∂              c        i                  ∑    j              (              z        j            −      f      (              x        j            ,              y        j            )      )        2    =  0to obtain six equations, and then endeavor to solve them.I&#039;ve done it for one variable not two, but I&#039;m guessing the process is straightforward.If I extend this to more variables, and to higher order than 2, when will the analytical expressions be at risk for having multiple solutions?

Zoey Benitez · Accepted Answer

Step 1For example:      ∂          ∂              c        1                  ∑    j              (              z        j            −      f      (              x        j            ,              y        j            )      )        2    =  2      ∑    j        (          z      j        −    f    (          x      j        ,          y      j        )    )              ∂      f              ∂              c        1            and since             ∂      f              ∂              c        1             is now constant and       (          z      j        −    f    (          x      j        ,          y      j        )    )   is linear in       c    1  , there will always be one solution.Step 2The number of variables and the power to which they are raised is unimportant. It&#039;s only the power of 2 in &quot;least squares&quot; that we need to notice here.

When does this least squares analytical solution based on zeros of partial derivatives start providing more than one solution? If I want to fit a quadratic function of two variables to some data, I can use f(x, y)=c_1x^2+c_2xy+c_3y^2+c_4x+c_5y+c_6

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