5. Differentiation II Daisuke Oyama Mathematics II April 24, 2020 - PowerPoint PPT Presentation

5. Differentiation II Daisuke Oyama Mathematics II April 24, 2020

Vectors and Matrices ▶ We regard elements in R N as column vectors. ▶ We denote the set of M × N matrices by R M × N ,   a 11 · · · a 1 N . . ...  . .   ∈ R M × N . . .  a M 1 · · · a MN ▶ For A ∈ R M × N , A T ∈ R N × M denotes the transpose of A . ▶ R N and R N × 1 are naturally identified, and we use the natural identification x T y x · y = ���� ���� real number 1 × 1 matrix for x, y ∈ R N or x, y ∈ R N × 1 . 1 / 51

Little o Notation ▶ For functions f, g : U → R , where U ⊂ R N is an open neighborhood of ¯ x ∈ R N , f ( x ) if lim x → ¯ x g ( x ) = 0 and lim x → ¯ g ( x ) = 0 , we write x f ( x ) = o ( g ( x )) as x → ¯ x. ▶ By f ( x ) = h ( x ) + o ( g ( x )) , we mean f ( x ) − h ( x ) = o ( g ( x )) . 2 / 51

Partial Differentiation Let U be a nonempty open subset of R N . ▶ A function f : U → R is partially differentiable with respect to x i at ¯ x ∈ U if the function x i �→ f (¯ x 1 , . . . , ¯ x i − 1 , x i , ¯ x i +1 , . . . , ¯ x N ) is differentiable at ¯ x i . ▶ In this case, the differential coefficient is denoted by x ) , or ∂f f i (¯ x ) , f x i (¯ ∂x i (¯ x ) , and is called the partial differential coefficient of f with respect to x i at ¯ x . We also say that ∂f ∂x i (¯ x ) exists. ▶ f is partially differentiable with respect to x i if it is partially differentiable with respect to x i at all ¯ x ∈ U . ▶ The function x �→ f i ( x ) is called the partial derivative function (or partial derivative ) of f with respect to x i and is denoted by f i , f x i , or ∂f ∂x i . 3 / 51

Gradient Vectors Let U be a nonempty open subset of R N . ▶ For a function f : U → R , if ∂f ∂x i (¯ x ) exists for all i = 1 , . . . , N , we write   ∂f ∂x 1 (¯ x ) .  ∈ R N ,  .  ∇ f (¯ x ) = .  ∂f ∂x N (¯ x ) which is called the gradient vector (or gradient ) of f at ¯ x . 4 / 51

Jacobian Matrices Let U be a nonempty open subset of R N . ▶ For a function f : U → R M , if ∂f j ∂x i (¯ x ) exists for all i = 1 , . . . , N and j = 1 , . . . , M , we write   x ) T ∇ f 1 (¯ .  .  Df (¯ x ) = .   x ) T ∇ f M (¯   ∂f 1 ∂f 1 ∂x 1 (¯ x ) · · · ∂x N (¯ x ) . . ...  . .  ∈ R M × N ,  = . .  ∂f M ∂f M ∂x 1 (¯ x ) · · · ∂x N (¯ x ) which is called the Jacobian matrix (or Jacobian ) of f at ¯ x . x ) T . ▶ For a function f : U → R , Df (¯ x ) = ∇ f (¯ 5 / 51

▶ For a function f ( x, y ) of x ∈ R N and y ∈ R S , we often write   ∂f 1 ∂f 1 ∂x 1 (¯ x, ¯ y ) · · · ∂x N (¯ x, ¯ y ) . . ...  ∈ R M × N ,  . .  D x f (¯ x, ¯ y ) = . .  ∂f M ∂f M ∂x 1 (¯ x, ¯ y ) · · · ∂x N (¯ x, ¯ y ) and   ∂f 1 ∂f 1 ∂y 1 (¯ x, ¯ y ) · · · ∂y S (¯ x, ¯ y ) . . ...  ∈ R M × S ,  . .  D y f (¯ x, ¯ y ) = . .  ∂f M ∂f M ∂y 1 (¯ x, ¯ y ) · · · ∂y S (¯ x, ¯ y ) where ( ) ∈ R M × ( N + S ) . Df (¯ x, ¯ y ) = D x f (¯ x, ¯ y ) D y f (¯ x, ¯ y ) 6 / 51

Differentiation in Several Variables Let U be a nonempty open subset of R N . Definition 5.2 A function f : U → R is differentiable (or totally differentiable ) at p ∈ R N such that x ∈ U if there exists ¯ ¯ f (¯ x + z ) − f (¯ x ) − ¯ p · z lim = 0 , ∥ z ∥ z → 0 or f (¯ x + z ) = f (¯ x ) + ¯ p · z + o ( ∥ z ∥ ) as z → 0 , i.e., for any ε > 0 , there exists δ > 0 such that ⇒ | f (¯ x + z ) − f (¯ x ) − ¯ p · z | 0 < ∥ z ∥ < δ, ¯ x + z ∈ U = < ε. ∥ z ∥ ▶ In this case, ∂f ∂x i (¯ x ) exists for all i = 1 , . . . , N , and p = ∇ f (¯ ¯ x ) . 7 / 51

Differentiability, Continuity, Partial Differentiability Proposition 5.8 If f is differentiable at ¯ x , then it is continuous at ¯ x , and partially differentiable with respect to x i at ¯ x for each i . However, ▶ partial differentiability does not imply differentiability; and ▶ partial differentiability does not even imply continuity. 8 / 51

Continuous Differentiability and Differentiability Let U be a nonempty open subset of R N . ▶ f : U → R is continuously differentiable or of class C 1 if it is partially differentiable with respect to x 1 , . . . , x N and ∂f ∂f its partial derivative functions ∂x 1 , . . . , ∂x N are continuous. Proposition 5.9 If f is continuously differentiable, then it is differentiable. 9 / 51

Vector-Valued Functions Let U be a nonempty open subset of R N .   f 1 ( x ) . ▶ For a function f : U → R M , we write f ( x ) =  .   . .  f M ( x ) f is differentiable if f m is differentiable for all m = 1 , . . . , M . ▶ When f is differentiable, 1 lim ∥ z ∥ ( f (¯ x + z ) − f (¯ x ) − Df (¯ x ) z ) = 0 , z → 0 x ) ∈ R M × N is the Jacobian matrix of f at ¯ where Df (¯ x . ▶ f is of class C 1 if f m is of class C 1 for all m = 1 , . . . , M . 10 / 51

Product Rule Let U ⊂ R N be a nonempty open set. Proposition 5.10 Suppose that f : U → R M and g : U → R M are differentiable. Define the function h : U → R by h ( x ) = f ( x ) T g ( x ) . Then h is differentiable and satisfies = g ( x ) T + f ( x ) T Dh ( x ) Df ( x ) Dg ( x ) � �� � � �� � � �� � � �� � � �� � 1 × N 1 × M M × N 1 × M M × N for all x ∈ U . 11 / 51

Chain Rule Let U ⊂ R N and V ⊂ R S be nonempty open sets. Proposition 5.11 Suppose that g : V → U and f : U → R M are differentiable. Define the function h : V → R M by h ( x ) = f ( g ( x )) . Then h is differentiable and satisfies Dh ( x ) = Df ( g ( x )) Dg ( x ) � �� � � �� � � �� � M × S M × N N × S for all x ∈ V . 12 / 51

Example 1-1 ▶ For a function f : R N → R and y, z ∈ R N , consider the function h : R → R defined by h ( α ) = f ( y + αz ) . ▶ Define the function g : R → R N by g ( α ) = y + αz . Then h ( α ) = f ( g ( α )) . ▶ By the Chain rule, h ′ ( α ) = Dh ( α ) = Df ( g ( α )) Dg ( α ) = Df ( y + αz ) z (matrix product) ���� � �� � N × 1 1 × N = ∇ f ( y + αz ) · z . (inner product) ���� � �� � ∈ R N ∈ R N 13 / 51

Example 1-2 ▶ For a function f : R N → R N and y, z ∈ R N , consider the function k : R → R defined by k ( α ) = z T f ( y + αz ) . ▶ By the Chain rule, [ ] k ′ ( α ) = Dk ( α ) = z T D α f ( y + αz ) = z T Df ( y + αz ) z . ���� ���� � �� � 1 × N N × 1 N × N 14 / 51

Example 2: Slutsky Equation ▶ x : R N ++ × R + → R N + : Walrasian demand function ▶ ▶ h : R N ++ × R → R N + : Hicksian demand function ▶ e : R N ++ × R → R : expenditure function ▶ By duality, we have h ( p ) = x ( p, e ( p )) . (The fixed utility level u is omitted.) I.e., if g : R N ++ → R N ++ × R + is defined by g ( q ) = ( q, e ( q )) , then h ( p ) = x ( g ( p )) .   1 0  , where e n = ∂e ▶ Dg ( q ) = 0 1 (and N = 2 ).  ∂p n e 1 e 2 ▶ We will also write x np k = ∂x n and x nw = ∂x n ∂w . ∂p k 15 / 51

Then by the Chain Rule, Dh ( p ) = Dx ( g ( p )) Dg ( p ) )   1 0 ( x 1 p 1 x 1 p 2 x 1 w = 0 1   x 2 p 1 x 2 p 2 x 2 w e 1 e 2 ( x 1 p 1 ) ( 1 ) ( x 1 w ) ( x 1 p 2 0 ) = + e 1 e 2 x 2 p 1 x 2 p 2 0 1 x 2 w = D p x ( p, e ( p )) + D w x ( p, e ( p )) D p e ( p ) � �� � � �� � � �� � N × 1 1 × N N × N h ( p ) T = D p x ( p, e ( p )) + D w x ( p, e ( p )) , � �� � � �� � � �� � N × N N × 1 1 × N where the last equality follows from ∇ e ( p ) = h ( p ) � �� � ���� N × 1 N × 1 (“Hotelling’s Lemma”) . 16 / 51

Example 3: Homogeneous Functions and Euler’s Formula Definition 5.3 A function f : R N + → R is homogeneous of degree k if f ( tx ) = t k f ( x ) for all t > 0 and all x ∈ R N + . 17 / 51

Proposition 5.12 If f is homogeneous of degree k and differentiable, then for all i , ∂f ∂x i is homogeneous of degree k − 1 . Proof ▶ Since f ( tx ) = t k f ( x ) holds for any value of x i , it holds that ∂ ∂ ∂x i ( LHS ) = ∂x i ( RHS ) . ▶ Since ∂ ( LHS ) = t ∂f ( tx ) , ∂x i ∂x i and ∂ ( RHS ) = t k ∂f ( x ) , ∂x i ∂x i we have ∂f ∂x i ( tx ) = t k − 1 ∂f ∂x i ( x ) . 18 / 51

Proposition 5.13 If f is homogeneous of degree k and differentiable, then ∇ f ( x ) · x = kf ( x ) for all x ∈ R N + . Proof ▶ Since f ( tx ) = t k f ( x ) holds for any value of t , it holds that ∂t ( LHS ) = ∂ ∂ ∂t ( RHS ) . ▶ We have ∂ ∂t ( LHS ) = ∇ f ( tx ) · x, and ∂ ∂t ( RHS ) = kt k − 1 f ( x ) . Since these are equal, evaluating at t = 1 we have ∇ f ( x ) · x = kf ( x ) . 19 / 51

Example 4: A Property of the Hicksian Demand Function ▶ The Hicksian demand function h ( p, u ) is homogeneous of degree 0 in p . ▶ By Proposition 5.13, we have D p h ( p, u ) p = 0 . ���� ���� � �� � N × 1 N × 1 N × N 20 / 51