最优化计算方法-第2章（基本理论） - 图文(3)

例1 z?2x1?3x2?6x3?2????z?3????6???000?2???z?000????000??例2 例2 z?2x2?3x212?4x1x2?z???4x1?4x2??4x?1?6x2??2z???44?常量矩阵

?46??2z?2x3?3x212?4x1x2?z???6x1?4x2??4x?1?6x2?2z???12x14?函数矩阵

?46??11

?u,x?S?R，p?R,t?R二阶可导，f(u)例6 已知

nn1?(t)?f(x?tp)，求??(t),???(t)解设u?(u1,u2,?,un),x?(x1,x2,?,xn),TTp?(p1,p2,?,pn)T?(t)?f(x?tp)?f(x1?tp1,x2?tp2,?,xn?tpn)?f(x?tp)d(xn?tpn)?f(x?tp)d(x1?tp1)?????(t)?dtdt?un?u1?f(x?tp)?f(x?tp)?f(x?tp)p1p???np2???un?u2?u112

?p1???f(x?tp)?f(x?tp)?f(x?tp)??p?2??,,?,????u2?un??u1????T?p???f(x?tp)p?n?ndd?f(x?tp)???(t)?(??(t))??pidtdti?1?uinn??d?f(x?tp)d??f(x?tp)??pi????pi??dt??ui?1i??dt?ui?1i??n22nnn?f(x?tp)?f(x?tp)??(?pj)pi???pipj?ui?uj?ui?uji?1j?1i?1j?1?p?f(x?tp)p13

T26、泰勒公式

定理1（一阶泰勒公式）设n元函数f(x)(x?R)在U(x)n0内连续可微,则?x?U(x)有

0f(x)?f(x)??f(x)(x?x)?o(x?x)或

00T00f(x)?f(x)??f(ξ)(x?x)000T0其中ξ?x??(x?x).定理2（二阶泰勒公式）设

．

(0???1)n元函数f(x)(x?R)在U(x)0n0有内二阶连续可微，则?x?U(x)有

10T20002f(x)?f(x)??f(x)(x?x)?(x?x)?f(x)(x?x)?o(||x?x||)2或

100T00T20f(x)?f(x)??f(x)(x?x)?(x?x)?f(ξ)(x?x)200T014

f(x)若为非二次函数

f(x)?f(x0)??f(x0)T(x?x0)?10T2002(x?x)?f(x)(x?x)f(x)若为二次函数

f(x)?f(x0)??f(x0)T(x?x0)?10T2002(x?x)?f(x)(x?x)15

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