Maths12

Országok listájaHungaryBudapesti Corvinus EgyetemGazdálkodástudományi KarNemzetközi gazdálkodás (angol nyelven)Mathematics1JegyzetekMaths12

Comparative Statics
12.1 Jacobi-matrix and gradient

The multiple variable real functions are maps of form f : Rn R. Further generalization is obtained if not just the domain, but the range of the function is a subset of a vector space. These f : Rn Rm functions can be written in the form
f (x) = (f1 (x), . . . , fm (x))

where fi : Rn R (i = 1, 2, . . . , m) are the coordinate functions of f . Let's assume that the partial derivatives of the coordinate functions are continuous. Then we can form from these partial derivatives the following J matrix:
(f1 )1 (x) (f2 )1 (x) J= . . . (fm )1 (x) (f1 )n (x) (f2 )n (x) . . . . . . (fm )2 (x) . . . (fm )n (x) (f1 )2 (x) (f2 )2 (x) ... ...

This m × n matrix is called the Jacobi-matrix of the vector-vector function f at the point x = (x1 , . . . , xn ). The notation J = f (x) is also often used.
Example 12.1

Let
f2 (x1 , x2 , x3 ) = x2 + x3 + x4 1 2 3

f1 (x1 , x2 , x3 ) = 2x1 x2 x3 x = (x1 , x2 , x3 ) f (x) = 2x2 x3 2x1

Then the Jacobi-matrix (derivative) of the function f = (f1 , f2 ) : R3 R2 at point
2x1 x3 3x2 2 2x1 x2 4x3 3

is a matrix of type 2 × 3.
In particular, if m = 1 then the Jacobi-matrix of the function f : Rn R is a row matrix (vector). This n element vector is called the gradient (derivative) of the function f
f (x) = f (x) = (f1 (x), . . . , fn (x))

The gradient has a geometric interpretation, it determines the direction of the fastest change of f .

Chain rule
Let's take the functions f : Rn Rm and g : Rm Rk and suppose that the range of f is a subset of the domain of g. It is also assumed that the partial derivatives of the functions exist and are continuous. Then the composition of functions can be formed
g f : Rn Rk .

1

12.1.

JACOBI-MATRIX AND GRADIENT

Theorem 12.1 (Chain Rule)

If f is dierentiable at point x and g is dierentiable at point f (x) than g f is also dierentiable at point x and
(g f ) (x) = g (f (x)) · f (x)

The theorem states, that the derivative of the composition of functions, which is a k × n matrix is the product of a k × m and an m × n matrices.
Example 12.2 Let f : R2 R2 be the function converting the polar coordinates of points in the plane to cartesian coordinates, that is

f (r, ) =

r cos r sin

,

and let g : R2 R3 be given as follows:
x2 - xy g(x, y) = y 2 - xy . 2xy

Find the matrix (g f ) (1, /3). Then the derivative of f :
f (r, ) = cos sin -r sin r cos ,

and the derivative of g is
2x - y g (x, y) = -y 2y -x 2y - x . 2x

Therefore, by the dierentiability of the composition of function we have that
(g f ) (1, /3) = 1 - 3/2 -1/2 1/2 = - 3/2 3 - 1/2 · 3/2 3 1 1/2 - 3/2 1/2 - 3/2 = 3/2 - 3/2 1/2 + 3/2 , 3 1 - 3/2 1/2

which is a 3 × 2 matrix. The chain rule often occurs in the following special form. Let f1 , . . . , fn : R R be dierentiable functions and f (t) = (f1 (t), . . . , fn (t)). In addition let g : Rn R be a function, whose partial derivatives exist and continuous. Then n
Proposition 12.2 (Special case)

(g f ) (t) =
k=1

gk (f1 (t), . . . , fn (t))fk (t)

Indeed the Jacobi-matrix (gradient) of g is a row matrix (vector) now:
g(f (t)) = [g1 (f (t)), . . . , gn (f (t))]

while the Jacobi- matrix of f is a column matrix (vector):
f (t) = f1 (t)

. . .

fn (t)

The above formula is obtained as the product of these two vectors. 2

12.1.

JACOBI-MATRIX AND GRADIENT

Take a function g : Rn R whose partial derivatives exist and continuous, and let v Rn be a given vector. Then the derivative of the one variable function
Example 12.3

F (t) = g(a + tv)

is by the above formula:
F (t) = g(a + tv), v

in particular F (0) =

g(a), v

.

Implicit functions
The following problem often occurs in economics. An equation
F (x, y) = 0

is given. Wether the variable y can be expressed unambiguously as a function of the variable x. With other words: is there a function y = f (x), such that the identity
F (x, f (x)) = 0

holds for all x. Such a function not necessarily exists. For instance, for the equation
F (x, y) = x2 + y 2 - 1 = 0

of the unit circle around the origin, the variable y can not be expressed unambiguously as a function of x. The geometric meaning is that the points of the plane satisfying the equation F (x, y) = 0 do not points of the graph of a function. The reason is that there are lines parallel to the y axis, which intersects the above curve at two points. It may also happen that the variable y can not be expressed from the implicit equation using algebraic conversions. The equation
F (x, y) = ex+y - 2 cos y + 1 = 0

is such an example. It can be seen that the point (x, y) = (0, 0) satises the equation, but the variable y can not be expressed. The following question is also raised: if the function F is dierentiable, then under what conditions can be expressed the variable y as a dierentiable function of x. The answer is given by the following theorem.
Theorem 12.3 (Implicit Function Theorem)

Assume that at a point (x0 , y0 )

F (x0 , y0 ) = 0

and the partial derivatives of F are continuous in a neighborhood of this point and
F2 (x0 , y0 ) = 0.

Then there exists a unique continuously dierentiable f function in a neighborhood of the point x, such that
· f (x0 ) = y0 , · F (x, f (x)) = 0

for all x in the neighborhood,

· f (x) = -F1 (x, f (x))/F2 (x, f (x)).

3

12.1.

JACOBI-MATRIX AND GRADIENT

We have to attract attention to the fact, that the continuity of the partial derivatives imply that F2 (x, f (x)) = 0 in a neighborhood of x0 . The geometric interpretation of the theorem is, that if the tangent line to the curve with equation F (x, y) = 0 at the point (x0 , y0 ) is not parallel to the y axis, then in a neighborhood of this point the variable y can be expressed as a dierentiable function of x.
Example 12.4

Take the implicit equation
F (x, y) = ex+y + x + y - 1 = 0.

The point (0, 0) satises this equation. Furthermore
F2 (0, 0) = 2.

Thus F meets the conditions of the Implicit Function Theorem, therefore the equation uniquely determines a dierentiable y = f (x) function. It follows from the theorem, that
f (x) = -F1 (x, f (x))/F2 (x, f (x)) 1 = - x+f (x) · (ex+f (x) + 1) = -1 e +1

at all point x. Since f (0) = 0, this means that
f (x) = -x

is the only solution.
Example 12.5

Consider the above mentioned implicit equation
F (x, y) = ex+y - 2 cos y + 1 = 0.

The point (0, 0) satises this equation. Furthermore
F2 (0, 0) = 1,

that is, the conditions of the Implicit Function Theorem hold. Therefore the equation determines a unique dierentiable f function, though this function can not be obtained using algebraic conversions from the equation.

4



Hasonló témájú dokumentumok

Hozzászólások

11.05-2 18 2.előadás 2011 3 eloadas 4.óra ábra balázs jános élete etnicitás eupol európai unió fogaskerék fogyasztás gazdaságtörténelem hatalom hónolás információelmélet internet intézményi gyakorlat irodalomesztétika jogviszony kik könyv 2 környezetgazdaságtan környezeti katalízis kronológia lengyel ferenc leon festinger lévi-strauss magyarország geográfiája marketing mechanika3 zh mérleg mikrobiológai montázs műemlékvédelem növényélettan polgár political science pszichológia reakció rézsűállékonyság segédlet tb nemzetközi tektonika termelési tényezők tőkeelmélet vám vizsgatételek