Lecture2 Slides 2014

8/19/2019 Lecture2 Slides 2014

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FUNCTIONS OF SEVERAL VARIABLES

FABIZ I, Fall 2014

Luiza Bădin

Department of Applied Mathematics,

Bucharest University of Economic Studies


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Functions of several variables 2

Limits. Continuity

• So far, we have studied functions of one variable, typically written as y = f (x),which represented the variation that occurred in some (dependent) variable y, as

another (independent) variable x changed.

• In the real world, however, it is unusual to deal with functions that depend on asingle variable, and instead of y = f (x), we often work with y = f (x1, x2),

y = f (x1, x2, x3), or even the multivariate case y = f (x1, x2,...,xn).

• Economic models are usually functions of more than one variable, assuming forinstance that output, Q = f (L, K), is a function of two inputs, labor and capital.

• In order to understand the concept of limit and continuity in the general,multivariate case, we have to start with the idea of ”closeness" in the

n-dimensional space.

• In the univariate space, we measure the closeness of two arbitrary points by thelength of the segment joining the points.

• In the n-dimensional space, the distance will be called Euclidian distance.


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Limits. Continuity

Consider the set

Rn = {x = (x1, x2, . . . , xn)| xi ∈ Rn, ∀i = 1, . . . , n} = R× R× . . .× R

n times R

.

Definition 1. An application f : A ⊆ Rn → R, f (x) = f (x1, x2, . . . , xn) is called real function of n variables.Definition 2. An application d : Rn × Rn → [0,∞) is said to be a distance if the

following properties hold:

1. d(x, y) ≥ 0, ∀x, y ∈ Rn and d(x, y) = 0 ⇔ x = y;2. d(x, y) = d(y, x), ∀x, y ∈ Rn;3. d(x, z ) ≤ d(x, y) + d(y, z ), ∀x , y , z ∈ Rn


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Limits. Continuity

Example 1. The function d : Rn × Rn → [0,∞) defined by

d(x, y) = n

i=1(xi − yi)

2

for x = (x1, x2, . . . xn) and y = (y1, y2, . . . , yn) is a distance named Euclidian distance.

For n = 1, we have d(x, y) = |x1 − y1|, where x = x1 and y = y1.For n = 2, we have x = (x1, x2), y = (y1, y2) and d(x, y) =

(x1 − y1)2 + (x2 − y2)2.


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Limits. Continuity

Definition 3. Consider x0 ∈ Rn and r > 0. The set S r(x0) = {x ∈ Rn| d(x0, x) < r}is the open sphere centered at x0 with radius r.

The point x0 ∈

Rn is an interior point of the set A

⊂R

n if and only if ∃

r > 0 such

that S r(x0) ⊆ A.An n-dimensional interval is I 1 × I 2 × . . . I n = {(x1, x2, . . . , xn)|xk ∈ I k, k = 1, . . . n}where I k = (ak, bk), k = 1, . . . n.

Any open sphere centered at x0 contains an n-dimensional interval which includes x0

and conversely.


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Limits. Continuity

For simplicity, all the results are presented for n = 2.

Consider A ⊂ R2, f : A → R and (a, b) an interior point of A.Definition 4. (Limit) lim

(x,y)→(a,b)f (x, y) = l ∈ R if for every (xn, yn)n≥1 ⊂ A with

(xn, yn) → (a, b) and (xn, yn) = (a, b), ∀n ≥ 1 we have limn→∞ f (xn, yn) = l.Equivalently, if l ∈ R, lim

(x,y)→(a,b)f (x, y) = l ∈ R if for every ε > 0 there exists

η = η(ε) > 0 such that for every (x, y) ∈ A with |x − a| < η, |y − b| < η we have|f (x, y) − l| < ε.Definition 5. (Continuity) A function f is continuous at (a, b) if the limit

lim(x,y)→(a,b) f (x, y) exists and it is equal to f (a, b): lim(x,y)→(a,b) f (x, y) = f (a, b).


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Partial Derivatives

Consider a set A ⊂ R2, f : A → R and (a, b) an interior point of A.Definition 6. If lim

x→a

f (x, b)− f (a, b)x− a exists and is finite, we say that f admits a

partial derivative with respect to x at the point (a, b) and we write:

limx→a

f (x, b) − f (a, b)x − a = f

x(a, b) = ∂f ∂x

(a, b).

Definition 7. If limy→b

f (a, y)− f (a, b)y − b exists and is finite, we say that f admits a

partial derivative with respect to y at the point (a, b) and we write:

limy→b

f (a, y)

−f (a, b)

y − b = f

y(a, b) =

∂f

∂y(a, b).


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Multivariate case (n ≥ 2)

If A ⊂ Rn and a = (a1, a2, . . . , an) is an interior point of A, then for every i = 1, . . . , n

limxi→ai

f (a1, . . . , xi, . . . an) − f (a1, . . . , ai, . . . an)xi − ai = f

xi

(a1, . . . , an) = ∂f

∂xi(a1, a2, . . . , an).

If f xi as n-variable function of (x1, x2, . . . , xn) admits first order partial derivatives

with respect to x

j at some point (a1

, a2

, . . . , an) ∈

Rn

, then

(f xi)xj

(a1, a2, . . . , an) = f xixj

(a1, a2, . . . , an) = ∂ 2f

∂x j∂xi(a1, a2, . . . , an)

is the second order partial derivative of function f calculated at (a1, a2, . . . , an).

For a two-variable function f : A → R with A ⊂ R2, if the applications f x, f y : A → Rare defined at any point of A and if they also admit partial derivatives with respect

to x and y, then their partial derivatives are the second order partial derivatives and

the following notations apply: f x2

= (f x)x, f

xy = (f

x)y, f

yx = (f

y)x, f

y2

= (f y)y.


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Examples

Example 2. Find f

x2, f

y2, f

xy, f

yx for the next two-variable functions:1. f (x, y) = x3 + 2xy2 − x

y, y = 0;

2. f (x, y) = ln(1 + x2 + 2y2).


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Differentiability, partial derivatives and continuity

Next results establish the connection between differentiability, partial derivatives andcontinuity.

Theorem 2. If A ⊂ R2 and f : A → R is differentiable at (a, b) ∈ A then f admits first order partial derivatives with respect to x and y at (a, b) and f x(a, b) = λ,

f y(a, b) = µ.

Proof. Consider y = b, x = a such as (x, b) ∈ A. As f is differentiable at (a, b) wehavef (x, y) − f (a, b) = λ(x − a) + µ(y − b) + ω(x, y)ρ(x, y)

⇒ f (x, b) − f (a, b)x − a = λ + ω(x, b)

|x − a|x − a .

Since limx→a,y→b

ω(x, y) = 0 then

limx→a

f (x, b)− f (a, b)x − a = λ + limx→a ω(x, b)

|x − a|x − a = λ ⇒ f

x(a, b) = λ.

Similarly, f y(a, b) = µ.


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Therefore, if the two variable function f is differentiable at (a, b), then we have

f (x, y) − f (a, b) = f x(a, b)(x − a) + f y(a, b)(y − b) + ω(x, y)ρ(x, y).


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Differentiability, partial derivatives and continuity

Theorem 3. If f : A ⊂ R2 → R is differentiable at (a, b) ∈ A then f is continuous at (a, b).

Proof. If f is differentiable at (a, b) then there exists the two-variable function

ω : A → R continuous at (a, b) with ω(a, b) = 0 such thatf (x, y)

−f (a, b) = f x(a, b)(x

−a) + f y(a, b)(y

−b) + ω(x, y)ρ(x, y).

Since ω : A → R is continuous at (a, b) with ω(a, b) = 0, we have limx→a,y→b

ω(x, y) = 0.

Moreover limx→a,y→b

ρ(x, y) = limx→a,y→b

(x − a)2 + (y − b)2 = 0 and so

limx→a,y→b

[f (x, y)−f (a, b)] = limx→a,y→b

f x(a, b)(x − a) + f y(a, b)(y − b) + ω(x, y)ρ(x, y)

= 0

⇒ limx→a,y→b

f (x, y) = f (a, b), which is exactly the continuity of the function f at the

point (a, b).

Next theorem will be presented without proof.


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Theorem 4. If the first order partial derivatives f x, f y of f : A ⊂ R2 → R, are

defined at any point in an open sphere centered at (a, b), S r(a, b) ⊂ A and they are continuous at (a, b), then f is differentiable at (a, b).


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The total differential

Definition 9. Consider f : A ⊂ R2

→ R and (a, b) an interior point of A such that f is differentiable at (a, b). Then the total differential of the function f at point

(a, b), denoted by df (x, y; a, b) or df (a,b)(x, y) is the two-variable function defined by

df (a,b)(x, y) = f x(a, b)(x − a) + f y(a, b)(y − b).

Remark 1. Consider the two-variable functions on R2, φ, ψ : R2 → R,φ

(x, y

) = x, ψ

(x, y

) = y

that are differentiable on R

2

and φx(x, y) = 1, ψy(x, y) = 1, φ

y(x, y) = 0, ψ

x(x, y) = 0 and so

dφ(a,b)(x, y) = (x− a) notation= dx and dψ(a,b)(x, y) = (y − b) notation= dy.Therefore, if f is an arbitrary function,

df (a,b)(x, y) = f x(a, b)dx + f

y(a, b)dy

or df (a,b) = f

x(a, b)dx + f

y(a, b)dy.

The total differential of function f approximates the variation of f around (a, b).


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The second order total differential of the function f at the point (a, b) is

d2f (a,b)(x, y) = d(df )(a,b)(x, y) = f x2(a, b)dx

2 + f y2(a, b)dy2 + 2f xy(a, b)dxdy.


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Examples

1. Find out the first and the second order partial derivatives of the followingfunctions:

(a) f : A = {(x, y) ∈ R2|y = 0} → R, f (x, y) = xy + xy

(b) f : A = R2 \ {(0, 0)} → R, f (x, y) = x√ x2+y2

(c) f (x, y) = x3 + y3 + 3xy

(d) f (x, y) = x3 + 3xy2 − 12y − 15x(e) f (x, y) = xy + 50

x + 20

y − 3, x = 0, y = 0;

(f) f (x , y , z ) = 2x2 + 2y2 + 2(xy + yz + x + y + 3z );

(g) f (x , y , z ) = x2 + y2 + z 2 − xy + x − 2z .2. Prove that f xy(0, 0) = f yx(0, 0) for the function f : R2 → R,

f (x, y) =xy

x2−y2x2+y2 , (x, y) = (0, 0)

0, (x, y) = (0, 0)(3)

.


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3. Using the definition, prove that the function f : R2 → R, f (x, y) = 3x + y2 isdifferentiable at (2, 1).

4. Is the function f : R2 → R(a) f (x, y) = x3 + xy + y3 differentiable at (1, 1)?

(b) f (x, y) =

x2 + y2 differentiable at (0, 0) ?

(c)

f (x, y) = xyx2−y2

x2+y2 , (x, y) = (0, 0)0, (x, y) = (0, 0)

differentiable at (0, 0)?

5. Consider the function f : R× R→ R, f (x, y) = (x− 2)46(y − 3)44.Then

∂ 82

∂x42

∂y40

f (26, 27)

is equal to: a) 1650; b) 1560; c) 1272; d) 1722; e) 1982; f) 1892; g) 2700; h) 2070;

i) 2256; j) 2526; k) 2450; l) 2540; m) none of the previous.


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Optimization: Finding maxima and minima

Consider f : A ⊂ R2 → R and (a, b) an interior point of A. Assume that f is n-timesdifferentiable at (a, b) and the mixed partial derivatives are equal.

Definition 10. Consider f : A ⊂ R2 → R and (a, b) ∈ A. The point (a, b) is a local maximum (minimum) for f if there exists r > 0 such that S r(a, b)

⊂A and for every

(x, y) ∈ S r(a, b) we have f (a, b) ≥ f (x, y) (respectively f (a, b) ≤ f (x, y)).If (a, b) is a local maximum or a local minimum, then (a, b) is a local extreme point .

In other words, a point is a local maximum if there are no nearby points at which f

takes a larger value. If we want to emphasize that a point (a, b) is a max of f on the

whole domain A, not just a local max, we call (a, b) a global max or an absolute max

of f on A.


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Extreme points and partial derivatives

Proposition 1. If (a, b) ∈ A ⊂ R2 is a local extreme point for the function f : A → R and if ∃r > 0 such that the partial derivatives f x, f y exist on S r(a, b) ⊂ Aand are defined at any (x, y)

∈S r(a, b), then f

x(a, b) = 0, f

y(a, b) = 0.

Proof. Let (x, b) ∈ S r(a, b) and consider the function φ(x) = f (x, b). As (a, b) is alocal extreme point of f , it comes that x = a is a local extreme point for φ. Because

φ(a) exists, by Fermat theorem we have φ(a) = 0 and so

f x(a, b) = limx→a

f (x, b) − f (a, b)x − a = φ

(a) = 0. Similarly, f y(a, b) = 0.


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Stationary points and saddle points

Definition 11. An interior point (a, b) of A, is called a stationary point of f , if

f x(a, b) = 0 and f

y(a, b) = 0.

Any local extreme point (a, b), interior of A, is a stationary point of f (x, y). The

reciprocal is not true: there are stationary points that are not extremes.

Definition 12. Stationary points that are not extreme points are called saddle points .


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Finding Extremes

Theorem 5. Consider a subset A ⊂ R2, f : A → R and (a, b) a stationary point for the function f . Assume that ∃r > 0 such that the second order derivatives f x2

, f y2

, f xy, f yx are continuous on S r(a, b). Let H (a, b) = (f

xixj

(a, b))i,j=1,2 be the

hessian matrix and let ∆1(a, b) = f x2(a, b), ∆2(a, b) = det H (a, b). Then:

• if ∆2(a, b) > 0, (a, b) is a local extreme point:– if ∆1(a, b) > 0 then (a, b) is a local minimum;

– if ∆1(a, b) < 0 then (a, b) is a local maximum.

• if ∆2(a, b) < 0, then (a, b) is not an extreme point, is a saddle point.


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Finding extremes

1. If ∆2(a, b) = 0 we can conclude nothing and the investigation has to be

continued some other way. For instance we might check the sign of

f (x, y) − f (a, b) on S r(a, b).2. We note that when ∆2(a, b) > 0, the second order partial derivatives

f x2

(a, b), f y2

(a, b) have the same sign, since f x2

(a, b)f y2

(a, b) > 0, so we could as

well check whether f y2

is positive or negative if that were easier.


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Multivariate case n ≥ 2

Consider A ⊂ Rn, f : A → R, a = (a1, a2, . . . , an) ∈ A a stationary point for thefunction f such as its second order partial derivatives are continuous on an open

sphere S r(a). Then the Hessian matrix associated to f at a ∈ A isH (a) = (f xixj(a))i,j=1,...,n.

Consider the following determinants:

∆1(a) = f x21

(a),

∆2(a) =

f x21

(a) f x1x2(a)

f x2x1(a) f x22

(a)

= f x21(a)f x22(a) − f x1x2(a)f x2x1(a),. . . . . . . . .

∆n(a) = det H (a)and assume ∆i(a) = 0, ∀i = 1, . . . , n.


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Multivariate case n ≥ 2

The matrix H (a) is called positive definite if ∆1(a) > 0, ∆2(a) > 0, . . . , ∆n(a) > 0

and negative definite if ∆1(a) < 0, ∆2(a) > 0, . . . , (−1)n∆n(a) > 0.Then if H (a) is

• positive definite, then x = a is a local minimum.• negative definite, then x = a is a local maximum;• indefinite (neither positive nor negative definite), then x = a is a saddle point.


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Examples

Find the local extreme points of the following functions:Example 3. f : R2 → R, f (x, y) = x2 + y2

−2

−1

0

1

2

−2

−1

0

1

20

2

4

6

8

x valuesy values

z

=

f ( x , y

)

Figure 1: f (x, y) = x2 + y2


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Examples

Example 4. f : R2 → R, f (x, y) = x2 − y2

−2−1

01

2

−2

−1

0

1

2

−4

−2

0

2

4

x values

y values

z

=

f ( x , y )

Figure 2: f (x, y) = x2 − y2


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Examples

Example 5. f : R2

→ R, f (x, y) = xe−(x2+y2)

−2

−10

12

−2

−1

0

1

2−0.5

0

0.5

x valuesy values

z

=

f ( x , y

)

Figure 3: f (x, y) = xe−(x2+y2)


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Examples

Example 6. f : R2

→ R, f (x, y) = xye−(x2+y2)

−2

−1

0

1

2

−2

−1

0

1

2−0.2

−0.1

0

0.1

0.2

x valuesy values

z

=

f ( x , y

)

Figure 4: f (x, y) = xye−(x2+y2)


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Examples

Find the local extreme points of the following functions:

1. f : R2 → R, f (x, y) = x3 + y3 + 3xy;

2. f :R

2

→ R, f (x, y) = x3

− y2

− 4x;3. f : R2 → R, f (x, y) = x3 + 3xy2 − 12y − 15x;4. f (x, y) = xy + 50

x + 20

y − 3, x = 0, y = 0;

5. f : R3 → R, f (x,y,z ) = 2x2 + 2y2 + 2(xy + yz + x + y + 3z );6. f : R3

→R, f (x,y,z ) = x2 + y2 + z 2

−xy + x

−2z .


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Least Squares Method

• The Least Squares Method (LSM) was first described by Gauss around 1794 andthe most important application of the LSM is in data fitting. The best fit in the

least-squares sense minimizes the sum of squared residuals, a residual being the

difference between an observed value and the fitted value according to a given

model.

• Least squares problems fall into two categories: linear or ordinary least squaresand non-linear least squares, depending on whether or not the residuals are

linear in all unknowns.

• Researchers studying the data from experiments are often interested indiscovering whether the variables under study are linearly related, or in finding

the linear approximation which best fits the data points according to some

specific criterion. This may help detecting any possible underlying patterns and

also predict future values.


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Suppose the data points are (x1, y1), . . . , (xn, yn), n ≥ 3.For any given line y = ax + b we can measure the distance from any of these points

to the line yi = axi + b, by

d2i = (yi − yi)2 = (yi − (axi + b))2.

The line which minimizes the sum of squared residuals:

S (a, b) =n

i=1

[yi − (axi + b)]2

is called the least squares line.

The corresponding method is called least squares method or ordinary least squares

(OLS) and occurs in linear regression analysis.

The values a∗ and b∗ that minimize S (a, b) are usually called least squares

approximations (estimators) and under specific assumptions on the data generating

process, they have important statistical properties.


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S a(a, b) = −2n

i=1

[yi − (axi + b)]xi

S b(a, b) = −2n

i=1

[yi − (axi + b)].(4)

a

ni=1

x2i + bn

i=1

xi =n

i=1

xiyi

a

ni=1

xi + nb =n

i=1

yi.

(5)

The equation system (5) is called Gauss normal equations system and it can be

proved that it has a unique solution, which is the global minimum point for the sum

of squared residuals, S (a, b).


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∆ =

ni=1 x

2i

ni=1 xin

i=1 xi n

= nn

i=1

x2i −

ni=1

xi

2= 0

∆a =n

i=1 xiyin

i=1 xini=1 yi n

= nn

i=1

xiyi −n

i=1

xi

ni=1

yi

∆b =

n

i=1 x2i

ni=1 xiyi

n

i=1 xi n

i=1 yi

=n

i=1

x2i

ni=1

yi −n

i=1

xi

ni=1

xiyi


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a∗

= ∆a

∆ = nni=1 xiyi −

n

i=1 xini=1 yinn

i=1 x2i − (ni=1 xi)2 (6)

b∗ = ∆b

∆ =

ni=1 x

2i

ni=1 yi −

ni=1 xi

ni=1 xiyi

nn

i=1 x2i − (

ni=1 xi)

2 (7)


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Example 7. Find the line which best fits the data points: (0, 4), (3, 3), (4, 2), (3, 1),

(5, 0).

Answer: 5x + 7y = 29.

Example 8. Consider the following time series corresponding to the monthly sales of

some company:

t 1 2 3 4 5 6 7 8 9 10

y(t) 10 12 12 12 14 15 15 15 17 18

i) Using the Least Squares Method, find parameters a and b such that equation

y(t) = a + bt provides the best linear fit for the given data.

ii) Using the result of (i), predict the sales for November (t=11) and December

(t=12).Answer: a = 9, 6 and b = 0, 8.

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Lecture2 Slides 2014