Multivariate Calculus - Bruce E. Shapiro

112 LECTURE 14. UNCONSTRAINED OPTIMIZATION
Dividing by 2 and separating the three sums as before
0 =
=
n∑
x i (mx i + b − y i )
i=1
n∑
mx 2 i + b
i=1
= m
n∑ n∑
x i − x i y i
i=1
n∑
x 2 i + bX −
i=1
i=1
i=1
n∑
x i y i
where X is defined in equation 14.8. Next we define,
A =
C =
n∑
x 2 i (14.11)
i=1
n∑
x i y i (14.12)
i=1
so that
0 = mA + bX − C (14.13)
Equations ?? and ?? give us a a system of two linear equations in two variables
m and b. Multiplying equation 14.10 by A and equation 14.13 by X gives
0 = A (mX + nb − Y ) = AXm + Anb − AY (14.14)
0 = X (mA + bX − C) = AXm + X 2 b − CX (14.15)
Subtracting these two equations gives
and therefore
0 = Anb − AY − X 2 b + CX = b(An − X 2 ) + CX − AY
b =
n∑
x 2 i
AY − CX
An − X 2 = i=1
n∑ ∑
y i − n x i y i
i=1
i=1
n∑
∑
n n ( n∑
) 2
x 2 i − x i
i=1 i=1
x i
i=1
If we instead multiply equation 14.10 by X and equation 14.13 by n we obtain
Subtracting these two equations,
0 = X (mX + nb − Y ) = mX 2 + nXb − Y X
0 = n (mA + bX − C) = nAm + nXb − nC
0 = m ( X 2 − nA ) − (Y X − nC)
Solving for m and substituting the definitions of A, C, X and Y , gives
Revised December 6, 2006. Math 250, Fall 2006