Mathematical Economics: Lecture 10econren.weebly.com/uploads/9/0/1/5/9015734/lecture10.pdfmath...

Mathematical Economics:Lecture 10

Yu Ren

WISE, Xiamen University

October 22, 2012

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Chapter 15: Implicit Functions and Their Derivatives

Outline

1 Chapter 15: Implicit Functions and TheirDerivatives

Yu Ren Mathematical Economics: Lecture 10

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New Section

Chapter 15: ImplicitFunctions and Their

Derivatives


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Implicit Function

Explicit function: y = F (x1, x2, · · · , xn)

Implicit function G(x1, x2, · · · , xn, y) = 0


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Examples

Example 15.1 the equation 4x + 2y = 5 or4x + 2y − 5 = 0 express y as an implicitfunction of x .write y as an explicit function of x : y = 2.5− 2x


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Examples

Example 15.2 Consider the equation:

y2 − 5xy + 4x2 = 0

convert it into an explicit function:

y =5x ±

√25x2 − 16x2

2=

12

(5x ± 3x) =

{4x

x


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Questions

The fact that we can write down an implicitfunction G(x , y) = 0 does not mean that thisequation automatically defines y as a function ofx . example: x2 + y2 = 1


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Questions

two questions:(a) Given the implicit equation G(x , y) = cand a point (x0, y0) such that G(x0, y0) = c,does there exist a continuous functiony = y(x) defined on the interval I s.t.G(x , y(x)) = c for all x in I and y(x0) = y0

(b) if y(x) exists and differentiable, what isy ′(x0)?


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Implicit Function Theorem

Theorem 15.1 Let G(x , y) be a C1 function on aball about (x0, y0) in R2. Suppose thatG(x0, y0) = c and consider the expressionG(x , y) = c. If (∂G/∂y)(x0, y0) 6= 0, then thereexists a C1 function y = y(x) defined on aninterval I about the point x0 s.t.(a) G(x , y(x)) ≡ c for all x in I(b) y(x0) = y0

(c) y ′(x0) = −∂G∂x (x0,y0)∂G∂y (x0,y0)


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ExampleExample 15.7 Consider the equation

G(x , y) ≡ x2 − 3xy + y3 − 7 = 0

one computes that∂G∂x

= 2x − 3y = −1 at(4,3)

∂G∂y

= −3x + 3y2 = 15 at(4,3)

y ′(x0) = −∂G∂x (x0, y0)∂G∂y (x0, y0)

=115.

y1 ≈ y0 + y ′(x0)∆x = 3 + (115

) · 3 = 3.02

with the actual y1 = 3.01475 . . .Yu Ren Mathematical Economics: Lecture 10

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ExampleExample 15.8 the equation

x2 + y2 = 1

First note that

y ′(x)|x=0 = −∂G/∂x∂G/∂y

= −2x2y

= −02

= 0

an explicit formula

y(x) =√

1− x2

y ′(x) =−x√

1− x2


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Higher Order derivatives & Hessian

Theorem 15.2 Let G(x1, · · · , xk , y) be a C1

function on a ball about (x∗1 , · · · , x∗k , y∗).Suppose

G(x∗1 , · · · , x∗k , y∗) = c∂G∂y

(x∗1 , · · · , x∗k , y∗) 6= 0


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Higher Order derivatives & Hessian

Theorem 15.2 Then, there is a C1 functiony = y(x1, · · · , xn) defined on an open ball Babout (x∗1 , · · · , x∗k )(a) G(x∗1 , · · · , x∗k , y(x∗1 , · · · , x∗k )) ≡ c for all(x1, · · · , xk) in B(b) y∗ = y(x∗1 , · · · , x∗k )

(c) ∂y∂xi

(x∗1 , · · · , x∗k ) = −∂G∂x (x

∗1 ,··· ,x∗

k ,y∗)

∂G∂y (x

∗1 ,··· ,x∗

k ,y∗)


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Level curves and their tangents

Definition: A point (x0, y0) is called a regularpoint of the C1 function G(x , y) if ∂G

∂x (x0, y0) 6= 0or ∂G

∂y (x0, y0) 6= 0. If every point (x , y) on thelocus G(x , y) = c is a regular point of G, thenwe call the level set a regular curve


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Theorem 15.3 Let (x0, y0) be a point on thelocus of points G(x , y) = c in the plane, whereG is a C1 function of two variables. If(∂G/∂y)(x0, y0) 6= 0, then G(x , y) = c defines asmooth curve around (x0, y0) which can bethought of as the graph of a C1 functiony = f (x). Furthermore, the slope of this curve is:

−∂G∂x (x0, y0)∂G∂y (x0, y0)

.


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Theorem 15.3 If ∂G/∂y(x0, y0) = 0, but∂G/∂x(x0, y0) 6= 0, then the Implicit FunctionTheorem tells us that the locus of pointG(x , y) = c is a smooth curve about (x0, y0),which we can consider as defining x as afunction of y . It also tells us that the tangent lineto the curve at (x0, y0) is parallel to the y− axis.


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Theorem 15.4: Let G be a C1 function on aneighborhood of (x0, y0). Suppose that (x0, y0) isa regular point of G. Then the gradient vectorOG(x0, y0) is perpendicular to the level set of Gat (x0, y0).


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Definition: A point (x0, y0) is called a regularpoint of the C1 function F (x1, · · · , xn) ifOF (x∗) 6= 0, that is, if some (∂F/∂xi)(x∗) is notzero. If every point (x , y) on the level setFc = {(x1, · · · , xn) : F (x1, · · · , xn) = c} is aregular point of F , then we call the level set Fc aregular surface.


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Theorem 15.6 If F : Rn → R1 is a C1 function, ifx∗ is a point in Rn, and if some (∂F/∂xi)(x∗) 6= 0then: (a) the level set of F through x∗Fc = {(x1, · · · , xn) : F (x1, · · · , xn) = c} can beviewed as the graph of a real valued C1 functionof (n-1) variables in a neighborhood of x∗ (b) thegradient vector OF (x∗), considered as a vectorat x∗, is perpendicular to the tangent hyperplaneof FF (x∗) at x∗


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Nonlinear Systems

F1(y1, y2, · · · , ym, x1, · · · , xn) = c1

F2(y1, y2, · · · , ym, x1, · · · , xn) = c2... =

...Fm(y1, y2, · · · , ym, x1, · · · , xn) = cm


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Nonlinear Systems

Question:

What is ∂yi∂xj

(x∗, y∗)?


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Nonlinear Systems

∂F1

∂y1dy1 + · · · ∂F1

∂ymdym +

∂F1

∂x1dx1 + · · · ∂F1

∂xndxn = 0

... =...

∂Fm

∂y1dy1 + · · · ∂Fm

∂ymdym +

∂Fm

∂x1dx1 + · · · ∂Fm

∂xndxn = 0


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Nonlinear Systems

∂F1∂y1· · · ∂F1

∂ym... · · · ...∂Fm∂y1· · · ∂Fm

∂ym

dy1

...dym

=

−

Σni=1

∂F1∂xi

dxi...

Σni=1

∂Fm∂xi

dxi


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Nonlinear Systems

dy1...

dym

=

−

∂F1∂y1· · · ∂F1

∂ym... · · · ...∂Fm∂y1· · · ∂Fm

∂ym

−1 Σn

i=1∂F1∂xi

dxi...

Σni=1

∂Fm∂xi

dxi


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Example

Example 15.15 Consider the system ofequations

F1(x , y ,a) ≡ x2 + axy + y2 − 1 = 0F2(x , y ,a) ≡ x2 + y2 − a2 + 3 = 0

the Jacobian of (F1,F2) with respect to theendogenous variable x and y at the point x = 0,y = 1, a = 2:


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Example

det(∂F1

∂x∂F1∂y

∂F2∂x

∂F2∂y

)(0,1,2) = det

(2 20 2

)= 4 6= 0

we can solve the system for x and y as functionsof a near (0, 1, 2)


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Example

dyda

= −det ∂(F1,F2)

∂(x ,a)

det ∂(F1,F2)∂(x ,y)

= −det(2x+ay xy

2x -2a)

det(2x+ay ax+2y

2x 2y)

dyda

= −det( 2 0

0 -4)

det(2 20 2

) =84

= 2 > 0


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Example

if a increases to 2.1, the corresponding y willincrease to about 1.2.Let’s use another method to compute the effecton x:

(2x + ay)dx + (ax + 2y)dy + xyda = 02xdx + 2ydy − 2ada = 0


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Example

plug in x = 0, y = 1, a = 2:

2xdx + 2ydy = 0da0dx + 2ydy = 4da

so if a increases to 2.1, the corresponding x willdecrease roughly to -.2.


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Comparative Statics

Economic Environment: pure exchangeeconomy, two consumers 1 and 2, twogoods x and y, initial endowments: (e1,0),(0,e2), utility functions: U1,U2:Ui(xi , yi) = αui(xi) + (1− α)ui(yi), pricelevels: p,q.


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Comparative Statics

Maximizing the utility functions, we have inequilibrium

α

1− αu′1(x1)− pu′1(y1) = 0

px1 + y1 − pe1 = 0α

1− αu′2(x2)− pu′2(y2) = 0

x1 + x2 − e1 = 0y1 + y2 − e2 = 0


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Comparative Statics

Question:how a change in the initial endowmente2 affects the equilibrium consumption bundleswhile keeping e1 fixed


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Comparative Statics

Differentiate the equations

α

1− αu′′1(x1)dx1 − pu′′1(y1)dy1 − u′1(y1)dp = 0

pdx1 + dy1 − (x1 − 1)dp = 0α

1− αu′′2(x2)dx2 − pu′′2(y2)dy2 − u′2(y2)dp = 0

dx1 + dx2 = 0dy1 + dy2 = de2


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Comparative Statics

Solve the above equations, we can get equation(50) and (52) in page 363


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