0

If $$(x_1,y_1), (x_2,y_2),(x_3,y_3)$$ are points in the plane and if $a,b$ are fixed real numbers, how can I visualize $$f=(ax_1+b-y_1)^2+(ax_2+b-y_2)^2+(ax_3+b-y_3)^2$$ as a function from the plane into the reals?

I tried looking at what happens in the case that we only have one point $(x,y):$

$$f=(ax+b-y)^2=ax^2+2abx-2ayx-2by+b^2+y^2$$ which I can think of as the upper half of some ellipse in the plane. I don't see what happens as I include more points in my sum.

The Substitute
  • 3,888
  • Regarding your example with one point: The curve $f=0$ isn't an ellipse, it's just two superimposed copies of the line $ax+b-y = 0$. So $f(x,y)$ is (very roughly) a measure of distance from this line. – bubba Sep 14 '14 at 08:02
  • @bubba do you mean that it is piecewise linear? – The Substitute Sep 14 '14 at 08:46

1 Answers1

1

$|ax_1+b-y_1|$ is the vertical distance of the point $(x_1,y_1)$ from the line $y=ax+b$ Your expression is the sum of the squares of three of these. I would guess it comes from a least squares fit problem, where you are trying to find the $a,b$ that minimize this sum. I don't know an easy to visualize this. It lives in a seven dimensional space, the six coordinates of the points and the function value and since it is one equation it is a six dimensional manifold (except for critical points). I don't know how to visualize those.

Ross Millikan
  • 374,822