Transcript
Derivatives
We first consider a univariate function g : R→ R.
Definition (Derivatives)
The derivative of g at x , g ′(x), if it exists, is
limy→x
g(y)− g(x)
y − x= lim
h→0
g(x + h)− g(x)
h,
and g is differentiable at x . g is differentiable if it is differentiableon every point in its domain.
If g ′(x) exists, the limit of the quotient g(x+h)−g(x)h from the left
(h→ 0−) must equal to the limit from the right (h→ 0+).
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The derivative at x can be seen as the limit of the secants between(x , g(x)) and (y , g(y)) as y approaches x .g ′(x) describes the slope at x , or the rate of change of g(x) at x .Higher order derivatives can be defined similarly. g ′′ describes thecurvature of g .
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Derivatives
TheoremIf g is differentiable at x, it is also continuous at x.
Proof.Use the result that if {xn} → x and {yn} → y , then{xnyn} → xy .
If the n-th order derivative, gn exists and is also continuous, g iscalled n-th order continuously differentiable.
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Derivatives
Let f and g be functions R→ R and assume that they aredifferentiable at x0. The following are some rules fordifferentiations.
• (f + g)′(x0) = f ′(x0) + g ′(x0)
• (product rule) Define p(x) = f (x)g(x).p′(x0) = f (x0)g ′(x0) + f ′(x0)g(x0).
• Define p(x) = f (x)/g(x) and g(x0) 6= 0.
p′(x0) = g(x0)f ′(x0)−g ′(x0)f (x0)g2(x0)
.
• (chain rule) Define p(x) = f (g(x)). p′(x0) = f ′(g(x0))g ′(x0).
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Definition (Local Maximizer)
x0 is a local maximizer of f if for some δ > 0, f (x0) ≥ f (x) for allx ∈ Bδ(x0). Local minimizer can be defined similarly.
The following theorem provides a necessary condition for x0 as alocal maximizer for an interior point.
Theorem (first-order condition)
f : (a, b)→ R is differentiable. If x0 is a local maximizer, thenf ′(x0) = 0.
Proof.Consider the limit as h→ 0− and h→ 0+. The derivative at x0exists only if the left and right limits are equal.
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Exercise (sufficient condition for a local maxmizer)
f : (a, b)→ R is twice continuously differentiable and x0 ∈ (a, b).Show that if f ′(x0) = 0 and f ′′(x0) < 0, then x0 is a localmaximizer of f .
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Theorem (Rolle’s Theorem)
f : [a, b]→ R is differentiable. If f (a) = f (b) = 0, then thereexists some c ∈ (a, b) such that f ′(c) = 0.
Proof.Use the Weierstrass Theorem.
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Theorem (Mean-Value Theorem)
f : R→ R is differentiable. There exists a point c ∈ (a, b) suchthat f (b)− f (a) = (b − a)f ′(c).
Proof.Subtract the secant line between (a, f (a)) and (b, f (b)) from f (x),and apply Rolle’s theorem.
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We state the following results without proof.
Theorem (Taylor’s Formula)
f : R→ R is n-th order differentiable on (a, b). Forx , x + h ∈ (a, b), there exists a point c ∈ (x , x + h),
f (x + h) = f (x) +n−1∑k=1
f (k)(x)
k!hk +
f (n)(c)
n!hn.
Theorem (L’Hopital’s Rule)
f , g : R→ R are differentiable on an open interval containingpoint a, f (a) = g(a) = 0, then
limx→a
f (x)
g(x)= lim
x→a
f ′(x)
g ′(x).
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Derivatives of a Multivariate FunctionConsider a multivariate function f : Rn → R. Let u be a vector inRn. For some point x0 ∈ Rn, x0 + αu, α ∈ R describes a straightline in Rn through x0 in the direction of u. The derivative of f atx0 if we approach it along this line is the directional derivative of fin the direction of u at x0.
Definition (Directional Derivative)
Df (x0, u) = limα→0
f (x0 + αu)− f (x0)
α,
where α ∈ R and ||u|| = 1.
Definition (Partial Derivative)
Let e i ∈ Rn where all components are zero except for the i-thelement which is 1. The partial derivative of f with respect to itsi-th argument at x0 is Df (x0, e i ), which can be denoted as
Dxi f (x0), fxi (x0), fi (x0), or ∂f (x0)∂xi
.
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Definition (Gradient Vector)
∇f (x) = (f1(x), · · · , fn(x)) ,
an 1× n vector.
ExerciseLet f (x1, x2) = x1x2, u = (35 ,
45) and x0 = (1, 2). Calculate
Df (x0, u) according to the definition on the previous slide, andshow that it equals ∇f (x0)u.
ExerciseLet
f (x1, x2) =
{xy2
x2+y4 if x 6= 0
0 if x = 0
and x0 = (0, 0). Calculate Df (x0, u), and show that it does notequal ∇f (x0)u.
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The previous exercise shows that the existence of directionalderivatives is not sufficient for continuity. Another example is
f (x , y) =
{1 if y = x2 and x 6= 0
0 otherwise.
The directional derivative of f at (0, 0) along any direction existsand is 0, but f is not continuous at (0, 0).
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Differentiability of a Multivariate FunctionFor f : R→ R, from the Taylor formula, for some c ∈ (x , x + h),
f (x + h) = f (x) + f ′(x)h +f ′′(c)
2h2.
The linear terms f (x) + f ′(x)h approximates f (x + h), as the error
term E (h) = f ′′(c)2 h2 quickly vanishes as h→ 0, limh→0
E(h)h = 0.
For a differentiable function, linear approximation can approximatethe function value well locally.
Definition (Differentiability)
Let f : Rn → Rm and X an open set in Rn. f is differentiable atx ∈ X if there exists a m × n matrix Ax , such that
lim||h||→0
||f (x + h)− f (x)− Axh||||h||
= 0,
h is a vector in Rn and || · || is the Euclidean norm, ||h|| =√
h · h.f is differentiable if it is differentiable at every point in its domain.
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TheoremIf f : Rn → Rm is differentiable at x ∈ X ⊂ Rn, the derivative isthe m × n matrix of first partial derivatives of f ,
Ax = Df (x) =
Df 1(x)· · ·
Df m(x)
=
∇f 1(x)· · ·
∇f m(x)
=
f 1x1(x) · · · f 1
xn(x)· · · · · · · · ·
f mx1 (x) · · · f m
xn (x)
.
The matrix Df (x) is called the Jacobian of f at x .
Proof.Consider the ith component of f , f i : Rn → R and show that f isdifferentiable if and only if each f i is differentiable. To show thatDf m(x) = ∇f m(x), use e i .
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The previous theorem shows that if f is differentiable at x , itsdirectional derivative Df (x , h) = Df (x)h.
TheoremIf f : Rn → Rm is differentiable at x ∈ X ⊂ Rn, it is continuous atx.
Proof.||f (x + h)− f (x)|| ≤ ||Df (x)h||+ ||f (x + h)− f (x)−Df (x)h||.
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The proof of the following results can be found in the textbook.
TheoremLet f : Rn → R. If all the partial derivatives of f exist and arecontinuous at an interior point x, then f is differentiable.
Theorem (Chain Rule for Multivariate Composite Functions)
Let f : Rn → Rm, g : Rm → Rp and F (x) = g(f (x)),DF (x) = Dg(y)Df (x) with y = f (x).
Theorem (Mean Value Theorem for Multivariate Functions)
Let f : Rn → Rm be differentiable on an open set X ⊂ Rn.Suppose that x , y ∈ X and the line segment between x and y isalso in X . For each a ∈ Rm, there exists λ ∈ [0, 1], such that
a′ (f (y)− f (x)) = a′ (Df (λx + (1− λ)y)(y − x)) .
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Taylor’s Formula for MultivariateFunctions
Theoremf : Rn → R is second order differentiable. For some λ ∈ (0, 1),
f (x + h) = f (x) + Df (x)h +1
2h′D2f (x + λh)h.
Proof.Define a univariate function g(α) = f (x + αh) and apply theunivariate Taylor’s formula.
The formula provides a first order approximation for a multivariatefunction. The remainder term is quadratic in h.
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Continuous Differentiability
Definition (Continuously Differentiable Function)
If function f : Rn → Rm has continuous partial derivatives on someopen set X ⊂ Rn, it is continuously differentiable in X . All suchfunctions are denoted C 1 functions.
Similarly, a C 0 function is a continuous function. A C k functionfor some k ≥ 1 has continuous k-th order partial derivatives.
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Implicit Function
The graph of x and y that satisfy u(x , y) = c describes anindifference curve. We may be interested in knowing how much theconsumption of y needs to change with some change in x in orderfor one to be indifferent. If we can write y = f (x), u(x , f (x)) = cand take derivative with respect to x ,
ux(x , y) + uy (x , y)f ′(x) = 0
The marginal rate of substitution, f ′(x) is
f ′(x) = −ux(x , y)
uy (x , y).
For a general function f (x , y), we would like to know when y canbe expressed as a function of x .
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Implicit Function Theorem
Theorem (Implicit Function Theorem)
f : R2 → R is continuously differentiable. Suppose f (x0, y0) = c.If fy (x0, y0) 6= 0, in some neighborhood of (x0, y0), y = g(x) and gis continuously differentiable, that satisfies f (x , g(x)) = c.
Example
Let f (x , y) = y − g(x). Consider f (x , y) = 0. The inverse functionof g , x = g−1(y) satisfies f (g(y), y). From the implicit functiontheorem, g−1 exists if fx(x , y) 6= 0.
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Homogeneous Functions
A set X ⊂ Rn is a cone if for any x ∈ X , λx ∈ X for any λ > 0.
Definition (Homogeneous Function)
f is homogenous of degree k in a cone X if for all λ > 0 andx ∈ X , f (λx) = λk f (x).
Examples of homogeneous functions
• demand function x(p, y)
• f (x1, x2) = x1x2
• Cobb-Douglas function f (x) = A∏n
i=1 xαii is homogeneous of
degree∑n
i=1 αi .
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Euler’s Theorem
Theorem (Euler’s Theorem)
f : Rn → R has continuous partial derivatives on an open cone X .It if homogeneous of degree k in X if and only if ∀x ∈ X ,
n∑i=1
fi (x)xi = kf (x).
Example
If the production function f (x1, x2) is homogeneous of degree 1,and the price of each input relative to the output equals itsmarginal product, Euler’s theorem implies that the profit is zero.
ExerciseUse Euler’s theorem to show that f (x) = A
∏ni=1 xαi
i ishomogeneous of degree
∑ni=1 αi .
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ExerciseShow that if f is homogeneous of degree k, then fi is homogeneousof degree k − 1.
• If utility function u(x1, x2) is homogenous, the indifferencecurves are radial expansions of each other. The marginal rateof substitution between x1 and x2, or the slope of indifferencecurves, is constant along each straight line from the origin.
• The same geometric properties also hold for a homotheticfunction which is an increasing transformation of ahomogeneous function.
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