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1、經濟數學方法壹、 矩陣與行列式 定義: 階矩陣為一包括列和行的數字的方形排列，若以A代表此矩陣，則 例: 分別為4和2矩陣 定義: 若 則 =C 例: 則 定義:若A=(為矩陣，B=(為矩陣，則A和B的 乘積AB為矩陣C 例: 求AB及BA = = BA無法計算 行列式: Cramers Rule 已知 例:解下列聯立方程式: 貳、微分 微分公式: 若 設與皆存在: j k l 鏈鎖律(chain rule): 設函數f與g皆可微分 反函數 (inverse function): 設函數f與g滿足 f(g(Y)=Y 函數g為f之反函數 g(f(X)=X 且g=f 偏微分: 例: 全微分: 例:
2、 TE=PQ 自然對數(e)與自然指數(ln): xyexlnx11 性質: (1) 、 、 (2) (3)設存在 (4) (5) (6) (7) (8) (9) (10) (11) (12) 且(x0,y0)y=f(x)yx (13) 切線與射線: j給定切線上任一點(X, Y) k 射線角度值 函數的高階導數: 、 函數的臨界點及反曲點: (一) 若f/(x)0X1X2f(x2)f(x1)XYab則為函數f之臨界點 (二) 函數f在為嚴格遞增 f/(x)0f/(x)0XY上凹f/(x)0XY上凹concave downwardf/(x)0f/(x)0f/(x)0XY (三) 上凹反曲點(i
3、nflection point)上凹下凹xyC0 故函數遞增遞減性，函數凹性 (四)第一導數檢驗定理:或 XC 切記 - + f(C)為局部極小值 + - f(C)為局部極大值 - - + + f(C)為非局部極值xyf(C1)f(C2)C2局部最小值C1局部最大值 第二導數檢驗定理: j k l 本定理失敗參、積分 (一) 不定積分(Indefinite integral) 而F之導函數、F為f之 反導數故F為f之反導數 性質: j k l m (二) 定積分 (definite integral) xyf(x)ab 性質: j k l m n oX=a被定義 p q 肆、齊次函數與尤拉定理
4、 (一) 階齊次函數 (homogeneous function of degree n) 定義: 若則稱階齊次函數 (二) 尤拉定理 (Euler Theorem) 定義:若 則 証明: 對入微分: 令 (三) 齊序函數 (同位函數) (homothetic function) 定義: (一階齊次函數的正單調上升轉換稱之)若 為H.O.D 1 且 稱之。I.C.C.I=1500I=1000CDbook40606090 例: 若有齊次偏好，所得1000元，買40本書，60張CD, 當所得為1500時，而書，CD價格不變，會買60本書，90張CD 伍、古典規劃分析:最適化(Optimizatio
5、n) (一) 未受限制下的極大與極小 單變數函數(X) 1. 極大: 2. 極小: (二) 多變數函數( 1. j k 有限制條件下之極值分析: ( 正負相間(Max) 全為正 (Min) 陸、古典規劃分析應用: Optimization(1) max Q (2) min C=W 3個主要問題類型 (3) max f(x) max U(x, y) x or s.t The Structure of an Optimization Problem Max f(x) f(X)=objective function X: choice variables S: feasible set soluti
6、ons: Important general problems about the solutions to any optimization problem: (1) Existence of Solutions Propositions: An optimization problem always has a solution if (1) the objective function is “ continuous” (2) the feasible set is “nonempty, close and bounded” (2) Local and Global Optima Pre
7、positions: A local maximum is always a global maximum if (1) the objective function is quasiconcave.(2) the feasible set is convex. (3) Uniqueness of SolutionPropositions: Given an optimization problems in which the feasible set is convex and the objective function is nonconstant and quasiconcave, a
8、 solution is unique if: (1) the feasible set is strictly convex, or (2) the objective function is strictly quasiconcave, or (3) both (4) Interior and Boundary Optima (5) Location of the Optimum min max f(x) F.O.C XR S.O.C (多變數) Multivarial Case F.O.C Gradient vector of f S.O.C Hessian of f now, max
9、f( S.O.C (負定) ( Quadratic Forms and their Signs symmetric: X A X=( = (1) Negative Semidefinite (2) Negative definite (3) Positive Semidefinite (4) Positive definite ex n=2 = = = -Negative definite and - Positive definite: and 續 Hessian; H is negative definite if H is positive definite if General Cas
10、e A =( Negative definite: MAX . Positive definite: MIN Optimizations: The unconstrained case I. may f( Min F.O.C Gradient Veotor S.O.C Hessian Matrix Necessary conditions Sufficient conditions Df=0 H is definite f is concave (dx)H(dx)0 ex 1. F.O.C S.O.C 2. F.O.C S.O.C H is negative definite f is con
11、cave. II. The Constrained Case s.t g(=b Lagrangian Function: L ( constraint gualification: F.O.C D S.O.C s.t. Dg( 全微分 Bordered Hessian S.O.C. for The naturally ordered principled mincrs of the bordered (all be negative)guaslconcave Hessianmatrix alternate in sign, the sign of the first being positiv
12、e i.e ex min Lagrangian funotion: F.O.C. Nonlinear Programming Max f( inequality constraint F.O.C Max f(x) Langrangian Function: Max Ex = F.O.C 因有ineguediy, . 所以要多考慮這些可能 ex “” min s.t F.O.C 檢查這些條件是否都符合 限制式中 共有四種組合四種可能情況 Case 1 ( 代入 (2) 式) Case 2 step2 ( step2 用第2種生產要素 Case 3 用第1種生產要素 Case 4 Ex Kuhn-
13、Tucker Formulation Kuhn-Tucker Conditions s.t. (K-T conditions): Utility Maximization Problem max u(x, y) x, y s.t Comparative Statics F.O.C Implicit Functions Implicit Functions Theorem If D= = -totally differentiating the system D (無限制式) ex max F.O.C Totally differentiate F.O.C with respect to By Cramers Rule 同理 (有限制式) max s.t. F.O.C S.O.C = = (p 0 0 From totally differentiating with respect to w1: By the Cramers Rule: 把算出，代入利潤函數中，即可得:*profit Function But 此題中可直接代入為one decision的問題，不需如此麻煩。24
