8/19/2019 Lecture2 Slides 2014
1/37
FUNCTIONS OF SEVERAL VARIABLES
FABIZ I, Fall 2014
Luiza Bădin
Department of Applied Mathematics,
Bucharest University of Economic Studies
8/19/2019 Lecture2 Slides 2014
2/37
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.
8/19/2019 Lecture2 Slides 2014
3/37
Functions of several variables 3
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
8/19/2019 Lecture2 Slides 2014
4/37
Functions of several variables 4
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.
8/19/2019 Lecture2 Slides 2014
5/37
Functions of several variables 5
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.
8/19/2019 Lecture2 Slides 2014
6/37
Functions of several variables 6
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).
8/19/2019 Lecture2 Slides 2014
7/37
8/19/2019 Lecture2 Slides 2014
8/37
Functions of several variables 8
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).
8/19/2019 Lecture2 Slides 2014
9/37
Functions of several variables 9
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.
8/19/2019 Lecture2 Slides 2014
10/37
Functions of several variables 10
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).
8/19/2019 Lecture2 Slides 2014
11/37
8/19/2019 Lecture2 Slides 2014
12/37
Functions of several variables 12
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.
8/19/2019 Lecture2 Slides 2014
13/37
Functions of several variables 13
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).
8/19/2019 Lecture2 Slides 2014
14/37
Functions of several variables 14
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.
8/19/2019 Lecture2 Slides 2014
15/37
Functions of several variables 15
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).
8/19/2019 Lecture2 Slides 2014
16/37
Functions of several variables 16
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).
8/19/2019 Lecture2 Slides 2014
17/37
Functions of several variables 17
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.
8/19/2019 Lecture2 Slides 2014
18/37
Functions of several variables 18
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)
.
8/19/2019 Lecture2 Slides 2014
19/37
Functions of several variables 19
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.
8/19/2019 Lecture2 Slides 2014
20/37
Functions of several variables 20
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.
8/19/2019 Lecture2 Slides 2014
21/37
Functions of several variables 21
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.
8/19/2019 Lecture2 Slides 2014
22/37
Functions of several variables 22
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 .
8/19/2019 Lecture2 Slides 2014
23/37
Functions of several variables 23
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.
8/19/2019 Lecture2 Slides 2014
24/37
Functions of several variables 24
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.
8/19/2019 Lecture2 Slides 2014
25/37
Functions of several variables 25
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.
8/19/2019 Lecture2 Slides 2014
26/37
Functions of several variables 26
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.
8/19/2019 Lecture2 Slides 2014
27/37
Functions of several variables 27
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
8/19/2019 Lecture2 Slides 2014
28/37
Functions of several variables 28
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
8/19/2019 Lecture2 Slides 2014
29/37
Functions of several variables 29
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)
8/19/2019 Lecture2 Slides 2014
30/37
Functions of several variables 30
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)
8/19/2019 Lecture2 Slides 2014
31/37
Functions of several variables 31
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 .
8/19/2019 Lecture2 Slides 2014
32/37
Functions of several variables 32
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.
8/19/2019 Lecture2 Slides 2014
33/37
Functions of several variables 33
Least Squares Method
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.
8/19/2019 Lecture2 Slides 2014
34/37
Functions of several variables 34
Least Squares Method
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).
8/19/2019 Lecture2 Slides 2014
35/37
Functions of several variables 35
Least Squares Method
∆ =
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
8/19/2019 Lecture2 Slides 2014
36/37
Functions of several variables 36
Least Squares Method
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)
8/19/2019 Lecture2 Slides 2014
37/37
Functions of several variables 37
Least Squares Method
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.