Econ 205 - Slides from Lecture 10Joel SobelSeptember 2, 2010Econ 205 SobelExampleFind the tangent plane to {x | x1x2− x23= 6} ⊂ R3atbx = (2, 5, 2).If you let f (x) = x1x2− x23, then this is a level set of f for value 6.∇f (x) = (∂f∂x1,∂f∂x2,∂f∂x3)= (x2, x1, −2x3)∇f (bx) = ∇f (x) |x=(2,5,2)= (5, 2, −4)Tangent Plane:{bx + y | y · ∇f (bx) = 0} = {(2, 5, 2) + (y1, y2, y3) | 5y1+ 2y2− 4y3= 0}= {x | 5x1− 10 + 2x2− 10 −4x3+ 8 = 0}= {5x1+ 2x2− 4x3= 12}Econ 205 SobelSuppose f (x, y , z) = 3x2+ 2xy −z2.I∇f (x, y , z) = (6x + 2y, 2x, −2z).If (2, 1, 3) = 7.IThe level set of f when f (x, y , z) = 7 is{(x, y, z) : f (x, y , z) = 7}.IThis 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).IThe equation of the tangent to the level set of f is a(two-dimensional) hyperplane in R3.IAt the point (2, 1, 3), the hyperplane has normal equal to∇f (2, 1, 3) = (12, 4, −6).IHence 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) = 0or12x + 4y − 6z = 10.Econ 205 SobelThe 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 − 10or12x + 4y − 6z −w = 3.Econ 205 SobelHomogeneous FunctionsDefinitionThe 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 notlinear.Econ 205 SobelTheorem (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 haveDH(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 yieldsDH(1) = kF (x). (2)Combining equations (1) and (2) yields the theorem.Econ 205 SobelComments1. 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.Econ 205 SobelEuler’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 (Deif (x)) per unit, then you distribute the surplusfully if F is homogeneous of degree one.Econ 205 SobelHigher-Order DerivativesIf f : Rn−→ R is differentiable, then its partial derivatives,Deif (x) = Dif (x) can also be viewed as functions from Rnto R.You can imagine taking a derivative with respect to one variable,then other then the first again, and so on, creating sequences ofthe formDik···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 Dki1,...,inf (x), wherek is the total number of derivatives and ijis the number of partialderivatives with respect to the jth argument (so each ijis anon-negative integer andPnj=1ij= x. Except for the statementof Taylor’s Formula, we will have little interest in third or higherderivatives.Econ 205 SobelSecond derivatives are importantA real-valued function of n variables will have n2secondderivatives, 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 ∈ Ckif the kth order partials exist and are continuous.Econ 205 SobelTaylor ApproximationsIIt looks scary, but it really is a one-variable theorem.IIf you know about a function at the point a ∈ Rnand youwant to approximate the function at x.IConsider the function F (t) = f (xt + a(1 − t)). F : R −→ Rand so you can use one-variable calculus.IF (1) = x and F (0) = a.IIf 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.IMultivariable version of Taylor’s Theorem: apply the onevariable version of the theorem to F .IThe chain rule describes the derivatives of F (in terms of f )and there are a lot of these derivatives.Econ 205 SobelFirst Order ApproximationConsider f : Rn−→ R such that f is differentiable. At a ∈ Rnthe1st degree Taylor Polynomial of f isP1(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 wehavelimx→a|E2(x, a)|||x − a||= limx→a|f (x) − f (a) − Df (a) · (x − a)|||x − a||= 0Thus as before, as x → a, E2converges to 0 faster than x to a.Econ 205 SobelSecond Order ApproximationIf f ∈ C2, the 2nd degree Taylor approximation isf (x) = f (a) + ∇f (a)1×n(x − a)n×1+12(x − a)01×nD2f (a)n×n(x − a)n×1| {z }P2(x,a)+E3(x, a)where12(x − a)01×nD2f (a)n×n(x − a)n×1| {z }1×1=12W=12nXi=1nXj=1(xi− ai)∂2f∂xi∂xj(a)(xj− aj)Econ 205 SobelwhereW =