• 追加された行はこの色です。
• 削除された行はこの色です。
• 最適化 へ行く。

```*最適化問題 [#k0fe967b]

min　f(x) xはベクトル
gi(x) = 0 i = 1, . . . ,me
gi(x) < 0 i = me + 1, . . . ,m
xl < x < xu
-制約式のない場合は、制約のない最適化問題とよぶ

The general optimization problem has the form:
min　f(x)
subject to:
gi(x) = 0 i = 1, . . . ,me
gi(x) < 0 i = me + 1, . . . ,m
xl < x < xu
In particular, if m = 0, the problem is called an unconstrained optimization
problem.
*解とアルゴリズムの基本的性質：Basic Properties of Solutions and Algorithms [#yedf6bd8]

ミニマム点の定義
-Definition:A point x* is said to be a relative minimum point or a local
minimum point of f if there is an ε > 0 such that f(x*) =< f(x) for all
x such that　||x-x*||<ε. If the inequality is strict for all x ><x* then　x* is said to be a strict relative minimum point.
-Definition: A point x*  is said to be a global minimum point of f if
f(x*) < f(x) for all x . If the inequality is strict for all x without x*
then x* is said to be a strict global minimum point.
*定理１：1次元の最少値の必要条件　First-order necessary conditions [#fc02fdd1]
fがスカラー関数の場合、もしx*がローカルに最少であるならば、任意のベクトルｄに対して内積f'(x*)・dが正または０である。
-Let f in C1. If x* is a relative minimum, then for any vector d which is feasible at x*, we have f'(x)d >= 0. (f'(x) is the gradient, i.e. the vector of partial derivatives of f at x.)
-f'(x)は勾配ベクトル

[証明]
h(t) = f(x* + td)をテーラー展開すると
h(t) = h(0) + t・h'(0) +o(t)
となる。t=０がミニマム値であるから任意のtに対してt・h'(0)>=である。

By a one-sided Taylor expansion of the function h(t) = f(x* + td),
which is defined in the interval [0, α], we obtain that h(t) = h(0) + t・h'(0) +o(t). Since 0 is a relative minimum of h we obtain that th'(0) >= 0, for all　t small enough, which implies that h'(0) >= 0. The claim now follows from
the chain rule.
*補題 [#g84334a9]
もしx*が最少の極値を与えるならば、f'(x*) = 0が成り立つ。

If x* is a local minimum  then f'(x*) = 0.
*勾配：ベクトル [#g9a3fa2e]
The gradient of f is defined to be the vector field whose components are the partial derivatives of f. That is: