Appendix for minf

With y = cx, our functional becomes

sqrt(F) = Sigma min(d(data_i, f(x)), which we write as

sqrt(f(x_i)) = d(x_i, f(x)).

We have

f =~ d_i², so we begin

f(x) = (x - x_i)² + (cx - y_i)²,

f'(x) = 2(x - x_i) + 2(cx - y_i), so the critical number is

x = (x_i + cy_i)/(1 + c²).

Plugging this into F = Sigma d_i²

(omitting sigma for now)

= ([x_i + cy_i - x_i(1+c²)]² + [c(x_i + cy_i) - y_i(1+c²)]²) / (1 + c²)²

= ([cy_i - y_ic²]² + [cx_i - y_i]²) / (1 + c²)²

= [c²(y_i - x_ic)² + (-1)²(y_i - cx_i)²] / (1 + c²)²

= (c² + 1)(y_i - cx_i)²/(1 + c²)

= (y_i - cx_i)²/(1 + c²).

So F(c) = Sigma d_i² = 1/(1 + c²) Sigma (y_i - cx_i)².

At this point, our task stopped looking daunting. Rather than pursuing it for curiosity, we expect to be able to finish it.

We still need to find min F(c).

So we have F'(c) = 1/(1 + c²)² Sigma [2(1+c²)(y_i - cx_i)(x_i) + 2c(y_i - cx_i)²]

= -2/(1+c²)² Sigma [(y_i - cx_i)[1*],

[where 1* = (1 + c²)x_i + c(y_i - cx_i)

= x_i + c²x_i - cy_i - c²x_i

= x_i - cy_i]

= -2/(1 + c²)² Sigma (y_i - cx_i)(x_i - cy_i).

Find zeros of F'(c), i.e., find zeros of

Sigma (y_i - cx_i)(x_i - cy_i)

= Sigma y_ix_i - c(y_i² + x_i²) + c²x_iy_i as a function of c.

This is a quadratic equation, with

c = a = Sigma x_iy_i, and b = -Sigma y_i² + x_i².