Econ 205 - Slides from Lecture 10

Joel Sobel

September 2, 2010

Econ 205 Sobel

ExampleFind the tangent plane to {x | x1x2 − x2

3 = 6} ⊂ R3 atx̂ = (2, 5, 2).If you let f (x) = x1x2 − x2

3 , then this is a level set of f for value 6.

∇f (x) = (∂f

∂x1,∂f

∂x2,∂f

∂x3)

= (x2, x1,−2x3)

∇f (x̂) = ∇f (x) |x=(2,5,2)

= (5, 2,−4)

Tangent Plane:

{x̂ + y | y · ∇f (x̂) = 0} = {(2, 5, 2) + (y1, y2, y3) | 5y1 + 2y2 − 4y3 = 0}= {x | 5x1 − 10 + 2x2 − 10− 4x3 + 8 = 0}= {5x1 + 2x2 − 4x3 = 12}

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Suppose f (x , y , z) = 3x2 + 2xy − z2.

I ∇f (x , y , z) = (6x + 2y , 2x ,−2z).

I f (2, 1, 3) = 7.

I The level set of f when f (x , y , z) = 7 is{(x , y , z) : f (x , y , z) = 7}.

I This set is a (two dimensional) surface in R3: It can bewritten F (x , y , z) = 0 (for F (x , y , z) = f (x , y , z)− 7).

I The equation of the tangent to the level set of f is a(two-dimensional) hyperplane in R3.

I At the point (2, 1, 3), the hyperplane has normal equal to∇f (2, 1, 3) = (12, 4,−6).

I Hence the equation of the hyperplane to the level set at(2, 1, 3) is equal to:

(12, 4,−6) · (x − 2, y − 1, z − 3) = 0

or12x + 4y − 6z = 10.

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The graph of f is a three-dimensional subset of R4

{(x , y , z ,w) : w = f (x , y , z)}.

A point on this surface is (2, 1, 3, 7) = (x , y , z ,w).The tangent hyperplane at this point can be written:

w − 7 = ∇f (2, 1, 3) · (x − 2, y − 1, z − 3) = 12x + 4y − 6z − 10

or12x + 4y − 6z − w = 3.

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Homogeneous Functions

DefinitionThe function F : Rn −→ R is homogeneous of degree k ifF (λx) = λkF (x) for all λ.

Homogeneity of degree one is weaker than linearity:All linear functions are homogeneous of degree one, but notconversely.For example, f (x , y) =

√xy is homogeneous of degree one but not

linear.

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Theorem (Euler’s Theorem)

If F : Rn −→ R be a differential at x and homogeneous of degreek, then ∇F (x) · x = kF (x).

Proof.Fix x . Consider the function H(λ) = F (λx). This is a compositefunction, H(λ) = F ◦ G (λ), where G : R −→ Rn, such thatG (λ) = λx . By the chain rule, DH(λ) = DF (G (λ))DG (λ). If weevaluate this when λ = 1 we have

DH(1) = ∇F (x) · x . (1)

On the other hand, we know from homogeneity that H(λ) = λkx .Differentiating the right hand side of this equation yieldsDH(λ) = kλk−1F (λx) and evaluating when λ = 1 yields

DH(1) = kF (x). (2)

Combining equations (1) and (2) yields the theorem.

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Comments

1. I A cost function depends on the wages you pay to workers. Ifall of the wages double, then the cost doubles. This ishomogeneity of degree one.

2. A consumer’s demand behavior is homogeneous of degreezero.Demand is a function φ(p,w) that gives the consumer’s utilitymaximizing feasible demand given prices p and wealth w .The demand is the best affordable consumption for theconsumer.The consumptions x that are affordable satisfy p · x ≤ w (andpossibly another constraint like non-negativity).If p and w are multiplied by the same factor, λ, then thebudget constraint remains unchanged.Hence the demand function is homogeneous of degree zero.

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Euler’s Theorem provides a nice decomposition of a function F .Suppose that F describes the profit produced by a team of nagents, when agent i contributes effort xi .How such the team divide the profit it generates?If F is linear, the answer is easy: If F (x) = p · x , then just giveagent i pixi .Here you give each agent a constant “per unit” payment equal tothe marginal contribution of her effort.When F is non-linear, it is harder to figure out the contribution ofeach agent.The theorem states that if you pay each agent her marginalcontribution (Dei f (x)) per unit, then you distribute the surplusfully if F is homogeneous of degree one.

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Higher-Order Derivatives

If f : Rn −→ R is differentiable, then its partial derivatives,Dei f (x) = Di f (x) can also be viewed as functions from Rn to R.You can imagine taking a derivative with respect to one variable,then other then the first again, and so on, creating sequences ofthe form

Dik · · ·Di2Di1f (x).

Provided that all of the derivatives are continuous in aneighborhood of x , the order in which you take partial derivativesdoes not matter. We denote an kth derivative Dk

i1,...,inf (x), where

k is the total number of derivatives and ij is the number of partialderivatives with respect to the jth argument (so each ij is anon-negative integer and

∑nj=1 ij = x . Except for the statement

of Taylor’s Formula, we will have little interest in third or higherderivatives.

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Second derivatives are important

A real-valued function of n variables will have n2 secondderivatives, which we sometimes think of as terms in a squarematrix:

D2f (x)n×n

=

∂2f∂x21

(x) . . . ∂2f∂xn∂x1

(x)

......

∂2f∂x1∂xn

(x) . . . ∂2f∂x2n

(x)

We say f ∈ C k if the kth order partials exist and are continuous.

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Taylor Approximations

I It looks scary, but it really is a one-variable theorem.

I If you know about a function at the point a ∈ Rn and youwant to approximate the function at x .

I Consider the function F (t) = f (xt + a(1− t)). F : R −→ Rand so you can use one-variable calculus.

I F (1) = x and F (0) = a.

I If you want to know about the function f at x usinginformation about the function f at a, you really just want toknow about the one-variable function F at t = 1 knowingsomething about F at t = 0.

I Multivariable version of Taylor’s Theorem: apply the onevariable version of the theorem to F .

I The chain rule describes the derivatives of F (in terms of f )and there are a lot of these derivatives.

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First Order Approximation

Consider f : Rn −→ R such that f is differentiable. At a ∈ Rn the1st degree Taylor Polynomial of f is

P1(x) ≡ f (a) +∇f (a) · (x− a)

The first-order approximation should be familiar. Notice that youhave n “derivatives” (the partials).If we write f (x) = P1(x , a) + E2(x , a) for the first-orderapproximation with error of f at x around the point a, then wehave

limx→a

|E2(x, a)|||x− a||

= limx→a

|f (x)− f (a)− Df (a) · (x− a)|||x− a||

= 0

Thus as before, as x→ a, E2 converges to 0 faster than x to a.

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Second Order Approximation

If f ∈ C 2, the 2nd degree Taylor approximation is

f (x) = f (a) +∇f (a)1×n

(x− a)n×1

+1

2(x− a)′

1×nD2f (a)

n×n(x− a)

n×1︸ ︷︷ ︸P2(x,a)

+E3(x, a)

where

1

2(x− a)′

1×nD2f (a)

n×n(x− a)

n×1︸ ︷︷ ︸1×1

=1

2W

=1

2

n∑i=1

n∑j=1

(xi − ai )∂2f

∂xi∂xj(a)(xj − aj)

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where

W = (x1−a1, . . . , xn−an)·

∂2f∂x21

(a) . . . ∂2f∂xn∂x1

(a)

......

∂2f∂x1∂xn

(a) . . . ∂2f∂x2n

(a)

· x1 − a1

...xn − an

Notice that the error term is a quadratic form.

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Notation

Define Dkh f to be a kth derivative:

Dkh f =

∑j1+···+jn=k

(k

j1 · · · jn

)hj11 · · · h

jnn D j1

1 · · ·Djnn f ,

where the summation is taken over all n-tuples of j1, . . . , jn ofnon-negative integers that sum to k and the symbol(

k

j1 · · · jn

)=

k!

j1! · · · jn!.

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General Form of Taylor’s Theorem

Theorem (Taylor’s Theorem)

If f is a real-valued function in C k+1 defined on an open setcontaining the line segment connecting a to x, then there exists apoint η on the segment such that

f (x) = Pk(x , a) + Ek(x , a)

where Pk(x , a) is a kth order Taylor’s Approximation:

Pk(x , a) =k∑

r=0

Drx−a(a)

r !

and Ek(x , a) is the error term:

Ek(x , a) =Dk+1x−a (η)

(k + 1)!.

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Coda

Moveover, the error term satisfies:

limh→0

Ek(a + h, a)

||h||k= 0

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Inverse and Implicit Functions: Introduction

Theme:Whatever you know about linear functions is true locally aboutdifferentiable functions.

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Inverse Functions: Review

1. One variable, linear case: f (x) = ax . Invertible if and only ifa 6= 0.

2. General:

DefinitionWe say the function f : X −→ Y is one-to-one if

f (x) = f (x ′) =⇒ x = x ′

Recall f −1(S) = {x ∈ X | f (x) ∈ S} for S ⊂ Y .Now let’s consider f −1(y) for y ∈ Y . Is f −1 a function?If f is one-to-one, then f −1 : f (X ) −→ X is a function.

3. f is generally not invertible as the inverse is not one-to-one.But in the neighborhood (circle around a point x0), it may bestrictly increasing so it is one-to-one locally and thereforelocally invertible.

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TheoremIf f : R −→ R is C 1 and

f ′(x0) 6= 0,

then ∃ ε > 0 such that f is strictly monotone on the open interval(x0 − ε, x0 + ε).

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Locally

DefinitionThe function f : Rn −→ Rn is locally invertible at x0 if there is aδ > 0 and a function g : Bδ(f (x0)) −→ Rn such that f ◦ g(y) ≡ yfor y ∈ Bδ(f (x0)) and g ◦ f (x) ≡ x for x ∈ Bδ(x0).

So f is locally invertible at x0, then we can define g on(x0 − ε, x0 + ε) such that

g(f (x)) = x

In the one-variable case, a linear function has (constant) derivative.When derivative is non zero and not equal to zero, the function isinvertible (globally).Differentiable functions with derivative not equal to zero at a pointare invertible locally.For one variable functions, if the derivative is always non zero andcontinuous, then the inverse can be defined on the entire range ofthe function.

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Higher Dimensions

I When is f : Rn −→ Rn invertible?

I Linear functions can be represented as multiplication by asquare matrix.

I Invertibility of the function is equivalent to inverting thematrix.

I A linear function is invertible (globally) if its matrixrepresentation is invertible.

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Inverse Function Theorem

TheoremIf f : Rn −→ Rn is differentiable at x0 and Df (x0) is invertible,then f is locally invertible at x0. Moreover, the inverse function, gis differentiable at f (x0) and Dg(f (x0)) = (Df (x0))−1.

The theorem asserts that if the linear approximation of a functionis invertible, then the function is invertible locally.Unlike the one variable case, the assumption that Df is globallyinvertible does not imply the existence of a global inverse.

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One Variable Again

g ′(y0) =1

f ′(x0)

so the formula for the derivative of the inverse generalizes theone-variable formula.

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More General Problem

Given G : Rm+n −→ Rn. Let x ∈ Rm and y ∈ Rn.Can we solve the system of equations: G (x , y) = 0?n equations in n + m variables.Search for a solution to the equation that gives x as a function ofy .The problem of finding an inverse is really a special case wheren = m and G (x , y) = f (x)− y .

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Who Cares?

1. Equations characterize an economic equilibrium (marketclearing price; first-order condition)

2. y variables are parameters.

3. “Solve” a model for a fixed value of the parameters.

4. What happens when parameters change?

5. The implicit function theorem says (under a certain condition),if you can solve the system at a give y0, then you can solvethe system in a neighborhood of y0. Furthermore, it gives youexpressions for the derivatives of the solution function.

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Why Call it Implicit Function Theorem?

I If you could write down the system of equations and solvethem to get an explicit representation of the solution function,great.

I . . . explicit solution and a formula for solution for derivatives.

I Life is not always so easy.

I IFT assumes existence of solution and describes “sensitivity”properties even when explicit formula is not available.

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Examples

f : R2 −→ R

Suppose f (x , z) = 0 is an identity relating x and z .How does z depends on x?

f (x , z) = x3 − z = 0

Explicit solution possible.

x2z − z2 + sin x log z + cos x = 0,

then there is no explicit formula for z in terms of x .

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Lower-order case

If solving f (x , z) = 0 gives us a function

g : (x0 − ε, x0 + ε) −→ (z0 − ε, z0 + ε)

such thatf (x , g(x)) = 0,

∀ x ∈ (x0 − ε, x0 + ε)then we can define

h : (x0 − ε, x0 + ε) −→ R

byh(x) = f (x , g(x))

h(x) = 0 for an interval, then h′(x) = 0 for any x in the interval.

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Also . . .

h′(x) = D1f (x , g(x)) + D2f (x , g(x))Dg(x)

Since this expression is zero, it yields a formula for Dg(x) providedthat D2f (x , g(x)) 6= 0.

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Explicit Calculation of Implicit Function Theorem FormulaIf f and g are differentiable, we calculate

y =

(xz

), G (x) =

(x

g(x)

)h(x) = [f ◦ G ](x)

h′(x) = Df (x0, z0)DG (x0)

=

(∂f

∂x(x0, z0),

∂f

∂z(x0, z0)

)·(

1g ′(x0)

)=∂f

∂x(x0, z0) +

∂f

∂z(x0, z0)g ′(x0)

= 0

This gives us

g ′(x0) = −∂f∂x (x0, z0)∂f∂z (x0, z0)

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More Generally

f : R× Rm −→ Rm

f (x , z) =

f1(x , z)f2(x , z)

...fm(x , z)

=

00...0

= 0

where x ∈ R, and z ∈ Rm.And we have

g : R −→ Rm

z = g(x)Econ 205 Sobel

IFT

TheoremSuppose

f : R× Rm −→ Rm

is C 1 and write F (x , z) where x ∈ R and z ∈ Rm. Assume

|Dz f (x0, z0)| =

∣∣∣∣∣∣∣∂f1∂z1

. . . ∂f1∂zm

......

∂fm∂z1

. . . ∂fm∂zm

∣∣∣∣∣∣∣6= 0

andf (x0, z0) = 0.

There exists a neighborhood of (x0, z0) and a functiong : R −→ Rm defined on the neighborhood of x0, such thatz = g(x) uniquely solves f (x , z) = 0 on this neighborhood.

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This didn’t fit

Furthermore the derivatives of g are given by

Dg(x0)m×1

= −[Dz f (x0, z0)]−1

m×mDx f (x0, z0)

m×1

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Comments

1. This is hard to prove.

2. A standard proof uses a technique (contraction mappings)that you’ll see in macro.

3. Hard part: Existence of the function g that gives z in terms ofx .

4. Computing the derivatives of g is a simple application of thechain rule.

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The Easy Part

f (x , g(x)) = 0

And we defineH(x) ≡ f (x , g(x))

And thus

DxH(x) = Dx f (x0, z0) + Dz f (x0, z0)Dxg(x0)

= 0

=⇒ Dz f (x0, z0)Dxg(x0) = −Dx f (x0, z0)

Multiply both sides by the inverse:

[Dz f (x0, z0)]−1 · [Dz f (x0, z0)]︸ ︷︷ ︸=Im

·Dxg(x0) = −[Dz f (x0, z0)]−1DxF (x0, z0)

=⇒ Dxg(x0) = −[Dz f (x0, z0)]−1DxF (x0, z0)

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Reminders

1. The implicit function theorem thus gives you a guarantee thatyou can (locally) solve a system of equations in terms ofparameters.

2. Theorem is a local version of a result about linear systems.

3. Above, only one one parameter.

4. In general, parameters x ∈ Rn rather than x ∈ R.

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IFT - the real thing

TheoremSuppose

f : Rn × Rm −→ Rm

is C 1 and write F (x, z) where x ∈ Rn and z ∈ Rm.Assume

|Dz f (x0, z0)| =

∣∣∣∣∣∣∣∂f1∂z1

. . . ∂f1∂zm

......

∂fm∂z1

. . . ∂fm∂zm

∣∣∣∣∣∣∣6= 0

andf (x0, z0) = 0.

There exists a neighborhood of (x0, z0) and a functiong : Rn −→ Rm defined on the neighborhood of x0, such thatz = g(x) uniquely solves f (x, z) = 0 on this neighborhood.

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Still Doesn’t Fit

Furthermore the derivatives of g are given by implicitdifferentiation (use chain rule)

Dg(x0)m×n

= −[Dz f (x0, z0)]−1

m×mDx f (x0, z0)

m×n

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More Comments1. Verifying the formula is just the chain rule.2. Keep track of the dimensions of the various matrices.3. Keep in mind the intuitive idea that you usually need exactly

m variables to solve m equations is helpful.4. This means that if the domain has n extra dimensions,

typically you will have n parameters – the solution functionwill go from Rn into Rm.

5. The implicit function theorem proves that a system ofequations has a solution if you already know that a solutionexists at a point.

6. Repeat: Theorem says: If you can solve the system once,then you can solve it locally.

7. Theorem does not guarantee existence of a solution.8. In this respect linear case is special.9. The theorem provides an explicit formula for the derivatives of

the implicit function. Don’t memorize it. Compute thederivatives of the implicit function by “implicitlydifferentiating” the system of equations.Econ 205 Sobel

Example

A monopolist produces a single output to be sold in a singlemarket. The cost to produce q units is C (q) = q + .5q2 dollars

and the monopolist can sell q units for the price of P(q) = 4− q5

6dollars per unit. The monopolist must pay a tax of one dollar perunit sold.

1. Show that the output q∗ = 1 that maximizes profit (revenueminus tax payments minus production cost).

2. How does the monopolist’s output change when the tax ratechanges by a small amount?

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The monopolist picks q to maximize:

q(4− q5

6)− tq − q − .5q2.

The first-order condition is

q5 + q − 3 + t = 0

and the second derivative of the objective function is −5q4−1 < 0.Conclude: at most one solution to this equation; solution must bea (global) maximum.

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Details

I Plug in q = 1 to see that this value does satisfy the first-ordercondition when t = 1.

I How the solution q(t) to:

q5 + q − 3 + t = 0

varies as a function of t when t is close to one?

I We know that q(1) = 1 satisfies the equation.

I LHS of the equation is increasing in q, so IFT holds.

I Differentiation yields:

q′(t) = − 1

5q4 + 1. (3)

In particular, q′(1) = −16 .

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I To obtain the equation for q′(t) you could use the generalformula or differentiate the identity:

q(t)5 + q(t)− 3 + t ≡ 0

with respect to one variable (t) to obtain

5q(t)q′(t) + q′(t) + 1 = 0,

and solve for q′(t).

I Equation is linear in q′.

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Implicit Differentiation

I This technique of “implicit differentiation” is fully general.

I In the example you have n = m = 1 so there is just oneequation and one derivative to find.

I In general, you will have an identity in n variables and mequations.

I If you differentiate each of the equations with respect to afixed parameter, you will get m linear equations for thederivatives of the m implicit functions with respect to thatvariable.

I The system will have a solution if the invertibility condition inthe theorem is true.

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