2011. 1. 16. 17:40 Mathematics
Lagrange Multipliers - 라그랑주 승수법
\text{To find the maximum of }G=G(x_1,\cdots,x_n)\\\text{Find }(\chi_1,\cdots,\chi_n) \text{ where } \nabla G=0
\text{But }R=R(x_1,\cdots,x_n)\text{must satisfy the relation }R=0
\text{To handle the problem, let }n=2\\\text{The exact differential of }R \text{ becomes}\\dR=\frac{\partial R}{\partial x_1}dx_1+\frac{\partial R}{\partial x_2}dx_2\\\text{when infinitesimal movement does not violate the requirement;}\\dR=0
\text{When the function }G\text{ takes the extremum at the point}\\\text{The exact differential of }G \text{ also satisfies}\\dG=\frac{\partial G}{\partial x_1}dx_1+\frac{\partial G}{\partial x_2}dx_2=0\\\text{under the condition that }dR=0
\frac{\partial R}{\partial x_1}dx_1+\frac{\partial R}{\partial x_2}dx_2=0\\\therefore dx_1=-\frac{\frac{\partial R}{\partial x_2}}{\frac{\partial R}{\partial x_1}}dx_2\\\therefore dG=\frac{\partial G}{\partial x_1}dx_1+\frac{\partial G}{\partial x_2}dx_2\\=\left[-\frac{\partial G}{\partial x_1}\frac{\frac{\partial R}{\partial x_2}}{\frac{\partial R}{\partial x_1}}+\frac{\partial G}{\partial x_2}\right]dx_2\\=0\\\text{However, we are free to choose } dx_2 \text{, which implies}\\-\frac{\partial G}{\partial x_1}\frac{\frac{\partial R}{\partial x_2}}{\frac{\partial R}{\partial x_1}}+\frac{\partial G}{\partial x_2}=0
dR=0, dG=0\\\therefore dR-\alpha dG\\=\left[\frac{\partial R}{\partial x_1}-\alpha\frac{\partial G}{\partial x_1}\right]dx_1\\+\left[\frac{\partial R}{\partial x_2}-\alpha\frac{\partial G}{\partial x_2}\right]dx_2\\=0
\text{However, as the restriction is still not removed,}\\\frac{\partial R}{\partial x_1}dx_1+\frac{\partial R}{\partial x_2}dx_2=0\\\therefore dx_1=-\frac{\frac{\partial R}{\partial x_2}}{\frac{\partial R}{\partial x_1}}dx_2
\text{Therefore under this restriction, we can freely choose }dx_2\\\frac{\partial R}{\partial x_1}dx_1+\frac{\partial R}{\partial x_2}dx_2=0
\text{Assume we choose }\alpha\text{ so that}\\\frac{\partial R}{\partial x_1}-\alpha\frac{\partial G}{\partial x_1}=0\\\text{Then }dR-\alpha dG =0 \text{ reduces to}\\\left[\frac{\partial R}{\partial x_2}-\alpha\frac{\partial G}{\partial x_2}\right]dx_2=0\\\text{As we are free to choose }dx_2 \text{, we must conclude that}\\\frac{\partial R}{\partial x_2}-\alpha\frac{\partial G}{\partial x_2}\text{ must be zero as well}
'Mathematics' 카테고리의 다른 글
개드립의 마지막 정리 (0) | 2013.10.29 |
---|---|
델타 분포 만들기 (6) | 2012.08.23 |
경계조건의 중요성 - Boundary condition (2) | 2010.08.21 |
Involute 곡선 (10) | 2010.05.01 |
수학의 아름다움 (2) | 2010.04.24 |