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Someone 3 days ago

A power of 1 doesn’t guarantee a unique solution. A simple example has 3 points:

  (0,0), (1,0), (1,1)
Any y = a × x with a between zero and one gives you a sum of errors of 1.

Powers less than 1 have the undesired property that they will prefer making one large error over multiple small ones. With the same 3-point example and a power of ½, you get:

- for y = 0, a cumulative error of 1

- for y = x/2, a cumulative error of 2 times √½. That’s √2 or about 1.4

- for y = x, a cumulative error of 1

(Underlying reason is that √(|x|+|y|) < √|x| + √|y|. Conversely for powers p larger than 1, we have *(|x|+|y|)^p > |x|^p + |y|^p)

Odd powers would require you to take absolute differences to avoid getting, for example, an error of -2 giving a contribution of (-2)³ = -8 to the cumulative error. Otherwise they would work fine.

The math for squares (https://en.wikipedia.org/wiki/Simple_linear_regression) is easy, even when done by hand, and has some desirable properties (https://en.wikipedia.org/wiki/Simple_linear_regression#Numer...)