Відео

The problem statement itself limits the scope of answers we want to consider. There is a cubic polynomial that exactly fits the 4 points, which would mean that the sum of the squares would be 0. If you allow any polynomial, you can fit any (finite) number of points (as long as the x-values are distinct). The least squares method allows us to do the "best" we can with a limit on the degree of the desired polynomial. In this case, the problem statement limits us to polynomials of degree at most 2, so higher degree polynomials are not considered. The answer we get yields the lowest possible sum of squares because the result of the least-squares process is a projection. We are projecting the original point in \R^4 (thought of as the vector of 4 y-values) onto the 3-dimensional subspace of \R^4 which contains only those vectors which correspond to polynomials (with degree at most 2). Whenever we project a point onto a subspace, the result is the point in the subspace which is closest to the original point. Closeness (distance) here means the square root of the sum of the squares of the z_i, but this is minimized if and only if the sum of the squares (without the square root) is minimized. For why this minimization happens with projections, check out Khan Academy's video here: ua-cam.com/video/b269qpILOpk/v-deo.htmlsi=SnjOnOt7dy_KCJ78.