Solutions 281
f(x*+h)^f(x*)+
1
-h
T
V
2
f(x*)h.
Suppose that V^2 /(x*) is not positive semi-definite. Then,
3«eln 9 vTV^2 f(x*)v < 0;
even for the remainder term, vTV^2 f(£)v < 0 if \\x* — £|| is small enough. If
we take h as being along v, we should have f(x*) > f(x* + h), Contradiction
to the local minimality of f(x*). Thus, V^2 /(x*) is positive semi-definite.
If we combine the first order necessary condition and the second order
necessary condition after deleting the term -semi-, we will arrive at the suffi-
ciency condition for x* being the strict local minimizer.
c) At every iteration, we will approximate f(x) by a quadratic function Q(p)
using the first three terms of its Taylor series about the point xk-\:
V/(xfc_i) + V^2 /(a*-i)rP * V/(xfc_! + p).
f(xk-i + p) « /(xfe-i) + V/(xfc_ 1 )rp + -pTV^2 f(xk^)p = Q(p);
and we will minimize Q as a function of p, then we will finally set xk =
Xk-l +Pk-
Let us take the derivative of Q:
dQ
dp
Since we expect 6 = V/(xfc_i + pk) « V/(xfc_i) + V^2 /(xfc_i)rpfc,
V^2 /(fc-i)rPfc = -V/(xfc_x) &pk = -[V^2 /(xft_ 1 )]-^1 V/(xfc_ 1 ).
This method of finding a root of a function is known as Newton's method,
which has a quadratic rate of convergence except in some degenerate cases.
Newton's method for finding V/(x) = 8 is simply to iterate as xk = xk-i —
[vvock-or'v/fo-i).
d) See Figure S.24 for the plot of the bivariate function, /(xi,x 2 ) = x\ +
2x\ + 24x^2 + x\ + 12x2, m tne question.
4xf + ®xi + 48xi
4x1 + 24x 2
v/(
v
2
/(
Let X(o) =^1
1
Xi
X 2
Xi
X2
)"
=^v/^1
1
)"[
12xf
)"
12xf + 12xi + 48 0
0 12x2 + 24^1 1™
58
28 , V
(^2) / 72 0
0 36
. Then,