Maths11

Országok listájaHungaryBudapesti Corvinus EgyetemGazdálkodástudományi KarNemzetközi gazdálkodás (angol nyelven)Mathematics1JegyzetekMaths11

Multiple variable functions
11.1 Partial derivatives

Let us consider a two variable f : R2 R function. Set the variable y = b and study the obtained one variable function
x f (x, b).

Assuming that this function is dierentable at x = a nd its derivative.
(a, b) with respect to the variable x. It is denoted by: f (a, b) = f1 (a, b) x

Denition 11.1 The above derivative is called the partial derivative of f at point

Sometimes the notation fx (a, b) is also used.

Example 11.1 Find the partial derivative of f (x, y) = (x + 2y)ex+3y-1 with respect to the variable x at point (1, 1). Now f (x, 1) = (x + 2)ex+2 , and its derivative function is
f1 (x, 1) = ex+2 + (x + 2)ex+2 = (x + 3)ex+2 .

If x = 1 then t takes the value f1 (1, 1) = 4e3 .

Example 11.2 We could determine the partial derivative of f with respect to the variable x at an arbitrarily xed second variable y and then we could substitute a for x and b for y , but sometimes it is not suggested, as it is reasoned by the following example: Let
f (x, y) = x2 + y 2 + 5 · e-2x+y · cos(y + /2)

and nd its partial derivative with respect to x at the point (1, 0). To nd the partial derivatives with respect to x at an arbitrary point takes a long time, but if we use the denition and rst determine the function in one variable f (x, 0) then it turns out that
f (x, 0) = 0

for any x, therefore f1 (1, 0) = 0.

The partial derivative of a function f : Rn R in n variables with respect to its variable xk similarly can be dened.

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PARTIAL DERIVATIVES

Tangent Planes

The partial derivatives of multiple variable functions has a geometric interpretation (similarly to the geometric interpretation of the derivative of one variable functions). Let's take the two variable function f : R2 R. Its graph is a surface in the 3dimensional space. Chose a point
P (a, b, f (a, b))

on this surface. Intersecting the surface and the plane with equation y = b passing through the point P a curve is obtained. The slope of the tangent line to the curve at point P is the partial derivative f1 (a, b). Similarly, the slope of the tangent line to the curve in the plane with equation x = a is f2 (a, b). The normal vector of the plane containing the two tangent lines is
n = (f1 (a, b), f2 (a, b), -1)

thus the equation of the plane is
f1 (a, b)(x - a) + f2 (a, b)(y - b) - (z - c) = 0 ,

where c = f (a, b). This is the tangent plane to the surface at point P .
Second order partial derivatives

Let's consider the function f : Rn R of n variables and suppose that its partial derivative with respect to the variable xi exists and dierentiable with respect to the variable xk . Then the second order partial derivatives of f can be formed. Notation:
2f (x1 , . . . , xn ) = fik (x1 , . . . , xn ) xi xk

If i = k then the second order partial derivative of f :
2f (x1 , . . . , xn ) = fii (x1 , . . . , xn ) x2 i

Example 11.3 If the the function is f (x, y) = 2x3 + 5x2 y 3 - ln(xy 2 ) then
f1 (x, y) f2 (x, y) f11 (x, y) f22 (x, y) f12 (x, y) = = = = 6x2 + 10xy 3 - 1/x 15x2 y 2 - 1/y 2 12x + 10y 3 - 1/x2 30x2 y - 2/y 3

= f21 (x, y) = 30xy 2

The second order partial derivatives of a function f with n variables is arranged in an n × n matrix that is called Hesse-matrix of f at point x = (x1 , . . . , xn ):
f11 (x) f12 (x) . . . f1n (x) f21 (x) f22 (x) . . . f2n (x) H= . . . . . . . . . fn1 (x) fn2 (x) . . . fnn (x)

Theorem 11.1 (Young) If the second order partial derivatives of n variable function f exist and are continuous then the Hesse-matrix of f is symmetric, that is,
fik (x) = fki (x)

for any indices i, k = 1, 2, . . . , n.

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PARTIAL DERIVATIVES

11.1.1

Quadratic forms

Quadratic forms are n variable functions often used in applications.

Denition 11.2 Let A = [aij ] be a symmetric matrix of order n and x Rn . The map Q(x) = Ax, x dened on Rn is called quadratic form. In details
n

Q(x) =
i,j=1

aij xi xj = a11 x2 + a12 x1 x2 + . . . + ann x2 , 1 n

where x1 . . . , xn are the coordinates of vector x. Using matrix arithmetical notation the quadratic form can be written in the form Q(x) = x Ax.

Example 11.4 Let

Q(x1 , x2 ) = 3x2 + 4x1 x2 + 4x2 . 1 2

Then using matrix multiplication it can be written in the form
Q(x1 , x2 ) = x1 x2 3 2 2 4 x1 x2 .

Denition 11.3 The quadratic form Q is said to be positive denite, if Q(x) > 0 for all x = 0 and positive semidenite, if Q(x) 0 for all x Rn . Q is negative (semi)denite, if -Q is positive (semi)denite. If Q takes both positive and negative values then it is called indenite.
If the matrix of the quadratic form is diagonal, then it is easy to decide if it is positive -, negative denite or indenite, because then
Q(x) = 1 x2 + . . . + n x2 1 n

that is, Q(x) is linear combination of squares. This is the canonical form of Q. The quadratic form Q given in canonical form is positive denite i i > 0 for all i = 1, . . . , n.

Theorem 11.2 Let A = [aij ] be a symmetric matrix of order n and Q(x) = x Ax be the corresponding quadratic form. Then
(a) Q (b) Q (c) Q (d) Q (e) Q

is positive denite each eigenvalue of A is positive, is positive semidenite each eigenvalue of A is greater than or equal to 0, is negative denite each eigenvalue of A is negative, is negative semidenite each eigenvalue of A is less than or equal to 0, is indenite A has both positive and negative eigenvalues.

Example 11.5 Determine if the given quadratic forms are positive denite/semidenite, negative denite/semidenite or indenite.
(a) Q(x1 , x2 ) = x2 - 6x1 x2 + x2 ; 1 2 (b) Q(x1 , x2 , x3 ) = 2x2 + 4x1 x3 + 4x2 + 5x2 . 1 2 3

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PARTIAL DERIVATIVES

Solution: (a) First nd the symmetric matrix A of the quadratic form: Q(x) = x Ax
x R2 . A= 1 -3 -3 1- -3 1

The characteristic equation of the matrix A is
|A - E| = 1- -3 = ( + 2) ( - 4) = 0

Thus the eigenvalues of the matrix are: 1 = -2 and 2 = 4 . Since A has a positive and a negative eigenvalue, the quadratic form Q is indenite. (b)
Q(x1 , x2 , x3 ) = x1 x2 x3 2 0 2 0 4 0
A

x1 2 0 x2 5 x3

The characteristic equation of A:
|A - E| = 2- 0 4 0 2 4- 0 0 5- =

= - ( - 1) ( - 4) ( - 6) = 0.

Thus the eigenvalues: 1 = 1, 2 = 4 és 3 = 6 are positive, therefore the quadratic form Q is positive denite. Let A = [aij ] be an n × n matrix. The principal minors in the top left corner of the matrix A are the determinants:
Dk = a11 a21

. . .

a12 a22

. . .

··· ··· ···

a1k a2k

. . .

,

(k = 1, . . . , n) .

ak1

ak2

akk

Theorem 11.3 Let A = [aij ] be a symmetric matrix of order n and Dk (k = 1, . . . , n) be the principal minors in the top left corner of A. Then the quadratic form Q(x) = x Ax is
(a) (b)

positive denite Dk > 0, for all k = 1, . . . , n, negative denite (-1)k Dk > 0, for all k = 1, . . . , n.
3 A= 1 2 1 4 1 2 1 3

Example 11.6 Let

and Q(x) = x Ax the corresponding quadratic form. The principle minors in the top left corner of A:
D1 D3 = 3 > 0, 3 1 1 4 2 1 D2 = 2 1 3 3 1 1 4 = 11 > 0,

=

= 18 > 0.

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PARTIAL DERIVATIVES

Thus the quadratic form is positive denite. (remark: the eigenvalues of A are 1, 3 and 6)

Example 11.7 Let

-2 A= 1 0



1 -2 0

0 0 . -3

Now the principle minors in the top left corner of A :
D1 D3 = -2 < 0, -2 1 0 1 -2 0 D2 = 0 0 -3 -2 1 1 -2 = 3 > 0,

=

= -9 < 0.

Therefore the quadratic form Q(x) = x Ax is negative denite.
Partial derivatives of quadratic forms

Let A be a symmetric matrix and Q(x) = xAx, be the induced quadratic form. In details n n
Q(x) =
i=1 k=1

aik xi xk

Taking into account that A is a symmetric matrix, the partial derivative with respect to the variable xi is
Qi (x) = 2(Ax)i ,

where the right side is the ith coordinate of the vector Ax. Similarly can be obtained that the Hesse-matrix of the quadratic form Q is
H = 2A.

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