Method for Nonlinear Least Squares Problems

所需积分/C币:10 2016-01-04 14:33:32 402KB PDF

详细的讲解了求解各类最小二乘问题的方法:高斯牛顿法 梯度下降算法
F(x+h)=F(x)+hg+ih Hh+O(hll F(x +h)=F(x)+ih Hh+O(hl g H=FX OF X F'(X G>0 O hHh>‖hl2 a2F H≡F"(x) 0:2O g≡F(x)=0 g=F(x)=0 F h +h2 o(le e e 0 h- h 尸 F(xk+1)<F(Xk) 0 k:=0 0 h h (x +ah;k:=k+1 ek+1|<|ek‖k>K e F F(x+ah)=F(x)+chF(x)+O( Xk F(x)+ahF( h ek+1≤alek‖l ek‖l ;0<a<1 F x h'F(x)<0 F(x+ahh F(x*)=0 F(x+h)=F(x)+F(xh+O(h F(x)+F"(x)h‖h Fx )-F(x+ah H=F"(x), a→0 hF(x=-F(x) h F(x) H Fx 1Hu>0 h=h 0<h'hh=-h F(x))))) F"(x) (C)=F(x+ah >0 F F∥(x h: =h h: = h + ah h (0)=hF(x)<0 (a)<y(0) F"( H*=FX* (x)≥ Q1. 0 C∈ B (a)≤r|y(0), L(h h (a) ( ak)=F(x+akh P(ak)=h F(x+anh) a-1 h h= h 1|≤△L(h)} (a)≤g(0)+:g(0) 0<~1<1 h h LL(h)+ouhh 1 号hl (a, p(a)) y(a)≥72·(0) n1<2<1 F(x+h) <F(x) x+h △ 0 L h xtcl F(x+h)aL(h)=F(x+hc+h Bh <0.25 (h) F(x+h) h Q>0.75 4/3 0.25 0.75 p 0.250.75 F(x-F(x+h) L(0)-J(h) 0 1:三1*max{3,1-(2g-1)3];b:=2 2 0.25 Q>0. max{△,3*‖b‖} 0> 0.250.75 2 L(h) F(x+h ( h=L(h)+ouh h Bh C 4(h)=L'(h)+h=0 L(h) (B+uIh B+I L B L(h B=F"( (F"(x)+Ih=-F(x) Bh +c=-Ah F'(x) lh L(h)y L(h) B OF dfi (x)-∑f(x)mn( F O2F' )=∑ ) dfi (x)21(x)+f(x) f ar; dCk arar F"(x)=J(x)J(x)+>fi(r)f, C F(x)=∑((x)2=f(x)2=r(x)f(x) r(X b-Ax A∈ mx n J(x) A a(b-Ax f(x+h)=f(x)+J(x)h+o(h) (AAX=A'b nx n A d(x))i]: (x)

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